Work, Power & Energy- JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Simulator

Q. A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 kg and 0.72 kg. Taking g = 10 $\mathrm{m} / \mathrm{s}^{2}$, find the work done (in joules) by the string on the block of mass 0.36 kg during the first second after the system is released from rest.

[IIT-JEE-2009]

Sol. 8

$\mathrm{a}=\frac{(0.73-0.36) \times 10}{(0.72+0.36)}$

$=\frac{10}{3} \mathrm{m} / \mathrm{s}^{2}$

$\mathrm{T}=0.36\left(10+\frac{10}{3}\right)=4.8 \mathrm{N}$

$\Delta y=\frac{1}{2} \times \frac{10}{3} \times 1^{2}$

$\mathrm{W}_{\mathrm{T}}=4.8 \times \frac{5}{3}=8 \mathrm{J}$

Q. A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 kg and 0.72 kg. Taking g = 10 m/s2, find the work done (in joules) by the string on the block of mass 0.36 kg during the first second after the system is released from rest.

[IIT-JEE-2009]

Sol.

$-\frac{1}{2} \times 2 \times(0.06)^{2}-0.18 \times 0.06=-\frac{1}{2} \times 0.18 \times v^{2}$

$\Rightarrow 0.0036+0.0108=0.09 \mathrm{v}^{2}$

$\Rightarrow \mathrm{v}=0.4=\frac{4}{10} \mathrm{m} / \mathrm{sec}$

Q. A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is un-stretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in m/s is V= N/10. Then N is

[IIT-JEE-2011]

Sol. 0

$\overrightarrow{\mathrm{F}}=\mathrm{k}\left[\frac{\overrightarrow{\mathrm{r}}}{\mathrm{r}^{3}}\right] \Rightarrow$ force is radial.

So W = 0

Q. A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle. If the initial speed $\left(\text { in } \mathrm{ms}^{-1}\right)$ of the particle is zero, the speed $\left(\text { in } \mathrm{ms}^{-1}\right)$ after 5 s is.

Sol. 5t

$\mathrm{P}=\mathrm{F} \cdot \mathrm{v} \Rightarrow 0.5=0.2 \mathrm{v} \frac{\mathrm{dv}}{\mathrm{dt}} \Rightarrow \mathrm{v}^{2}=5 \mathrm{t}$

Paragraph for Questions 5 and 6

A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 $\mathrm{m} \mathrm{s}^{-2}$)

Q. The magnitude of the normal reaction that acts on the block at the point Q is

(A) 7.5 N (B) 8.6 N (C) 11.5 N (D) 22.5 N

Sol. (A)

Q. The speed of the block when it reaches the point Q is

(A) $5 \mathrm{ms}^{-1}$

(B) $10 \mathrm{ms}^{-1}$

(C) $10 \sqrt{3} \mathrm{ms}^{-1}$

(D) $20 \mathrm{ms}^{-1}$

Sol. (B)

Q. Consider an elliptically shaped rail PQ in the vertical plane with OP = 3 m and OQ = 4m. A block of mass 1 kg is pulled along the rail from P to Q with a force of 18 N, which is always parallel to line PQ (see the figure given). Assuming no frictional losses, the kinetic energy of the block when it reaches Q is (n × 10) Joules. The value of n is (take acceleration due to gravity = 10 $\mathrm{ms}^{-2}$ )

Sol. 5

$\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}=\mathrm{W}_{\mathrm{all}}$

$\mathrm{K}_{\mathrm{f}}=\mathrm{W}_{\mathrm{ext}}+\mathrm{W}_{\mathrm{gr}}$

= 18 × 5 J – 1 × 10 × 4 J = 50 J = 5 × 10 J

Q. A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is :-

Sol. (D)

Q. A particle of mass m is initially at rest at the origin. It is subjected to a force and starts moving along the x-axis. Its kinetic energy K changes with time as dK/dt = $\gamma \mathrm{t}$, where  is a positive constant of appropriate dimensions. Which of the following statements is (are) true ?

(A) The force applied on the particle is constant

(B) The speed of the particle is proportional to time

(C) The distance of the particle from the origin increses linerarly with time

(D) The force is conservative

Sol. (A,B,D)

$\frac{\mathrm{d} \mathrm{k}}{\mathrm{dt}}=\gamma \mathrm{t}$ as $\mathrm{k}=\frac{1}{2} \mathrm{mv}^{2}$

$\therefore \frac{\mathrm{dk}}{\mathrm{dt}}=\mathrm{mv} \frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=\gamma \mathrm{t}$

$\therefore \mathrm{m} \int_{0}^{\mathrm{v}} \mathrm{vdv}=\gamma \int_{0}^{\mathrm{t}} \mathrm{tdt}$

$\frac{\mathrm{mv}^{2}}{2}=\frac{\gamma \mathrm{t}^{2}}{2}$

$\mathrm{v}=\sqrt{\frac{\gamma}{\mathrm{m}} \mathrm{t}} \textrm{ } \quad \ldots \ldots$ (i)

$\mathrm{a}=\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=\sqrt{\frac{\gamma}{\mathrm{m}}}=\mathrm{constant}$

since F = ma

$\therefore \mathrm{F}=\mathrm{m} \sqrt{\frac{\gamma}{\mathrm{m}}}=\sqrt{\gamma \mathrm{m}}=\mathrm{constant}$

Wave on String – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Simulator

Q. A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/hr towards a tall building which reflects the sound waves. The speed of sound in air is 320 m/s. The frequency of the siren heard by the car driver is ?

(A) 8.50 kHz         (B) 8.25 kHz        (C) 7.75 kHz         (D) 7.50 kHz

[JEE 2011]

Sol. (A)

$f^{\prime}=\left(\frac{v}{v-v_{s}}\right)\left(\frac{v+v_{0}}{v}\right) f \Rightarrow f^{\prime}=\left(\frac{320}{320-10}\right)\left(\frac{320+10}{320}\right) \times 8 \Rightarrow f^{\prime} \approx 8.50 \mathrm{kHz}$

Q. Column-I shows four systems, each of the same length L, for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as $\lambda_{\mathrm{f}}$. Match each system with statements given in Column-II describing the nature and wavelength of the standing waves.

[JEE 2011]

Sol. ((A) $\mathrm{P}, \mathrm{T}(\mathrm{B}) \mathrm{P}, \mathrm{S}(\mathrm{C}) \mathrm{Q}, \mathrm{S}(\mathrm{D}) \mathrm{Q}, \mathrm{R}$)

Q. A person blows into open-end of a long pipe. As a result, a high-pressure pulse of air travels down the pipe. When this pulse reaches the other end of the pipe.

(A) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is open

(B) a low -pressure pulse starts travelling up the pipe, if the other end of the pipe is open

(C) a low pressure pulse starts travelling up the pipe, if the other end of the pipe is closed

(D) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed

[JEE 2012]

Sol. ($\mathrm{B}, \mathrm{D}$)

Information Based (Refer NCERT)

Q. A student is performing the experiment of Resonance Column. The diameter of the column tube is 4 cm. The frequency of the tuning fork is 512 Hz. The air temperature is $38^{\circ} \mathrm{C}$in which the speed of sound is 336 m/s. The zero of the meter scale coincides with the top end of the Resonance column tube. When the first resonance occurs, the reading of the water level in the column is :-

(A) 14.0 cm        (B) 15.2 cm         (C) 16.4 cm         (D) 17.6 cm

[JEE 2012]

Sol. (B)

In resonance column experiment $\frac{\lambda}{4}=\ell_{1}+e$ so, $\ell_{1}=\frac{\lambda}{4}-e=\frac{v}{4 f}-0.3 d$

$=\frac{336 \times 100 \mathrm{cm}}{4 \times 512}-(0.3)(4 \mathrm{cm})=16.4-1.2=15.2 \mathrm{cm}$

Q. A student is performing an experiment using a resonance column and a tuning fork of frequency

$244 \mathrm{s}^{-1}$ . He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is (0.350 $\pm 0.005)$ m, the gas in the tube is

(Useful information $: \sqrt{167 \mathrm{RT}}=640 \mathrm{J}^{1 / 2}$ mole $^{-1 / 2} ; \sqrt{140 \mathrm{RT}}=590 \mathrm{J}^{1 / 2}$ mole $^{-1 / 2} .$ The molar masses M in grams are given in the options. Take the values of $\sqrt{\frac{10}{\mathrm{M}}}$ for each gas as given there. )

(A) Neon $\left(\mathrm{M}=20, \sqrt{\frac{10}{20}}=\frac{7}{10}\right)$

(B) Nitrogen $\left(\mathrm{M}=28, \sqrt{\frac{10}{28}}=\frac{3}{5}\right)$

(C) Oxygen $\left(\mathrm{M}=32, \sqrt{\frac{10}{32}}=\frac{9}{16}\right)$

(D) Argon $\left(\mathrm{M}=36, \sqrt{\frac{10}{36}}=\frac{17}{32}\right)$

Sol. (D)

Q. Two loudspeakers M and N are located 20m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car is initially at a point P, 1800 m away from the midpoint Q of the line MN and moves towards Q constantly at 60 km/hr along the perpendicular bisector of MN. It crosses Q and eventually reaches a point R, 1800 m away from Q. Let n(t) represent the beat frequency measured by a person sitting in the car at time t. Let $v_{\mathrm{p}}, v_{\mathrm{Q}}$ and $v_{\mathrm{R}}$ be the beat frequencies measured at locations P, Q and R, respectively. The speed of sound in air is 330 $\mathrm{ms}^{-1}$. Which of the following statement(s) is(are) true regarding the sound heard by the person ?

(A) The plot below represents schematically the variation of beat frequency with time

(B) The plot below represents schematically the variations of beat frequency with time

(C) The rate of change in beat frequency is maximum when the car passes through Q

(D) $v_{\mathrm{p}}+v_{\mathrm{R}}=2 v_{\mathrm{Q}}$

Sol. (A,C,D)

Q. A stationary source emits sound of frequency $\mathrm{f}_{0}=492 \mathrm{Hz}$ . The sound is reflected by a large car approaching the source with a speed of $2 \mathrm{ms}^{-1}$. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz ? (Given that the speed of sound in air is 330 $\mathrm{ms}^{-1}$ and the car reflects the sound at the frequency it has received).

Sol. 6

Frequency of sound as received by large car approaching the source.

Q. Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed 1.0 $\mathrm{ms}^{-1}$ and the man behind walks at a speed 2.0 $\mathrm{ms}^{-1}$. A third man is standing at a height 12 m above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency 1430 Hz. The speed of sound in air is 330 $\mathrm{ms}^{-1}$. At the instant, when the moving men are 10 m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is _____

Sol. 5.00 Hz

Q. In an experment to measure the speed of sound by a resonating air column, a tuning fork of frequency 500 Hz is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonances are heard at air columns of length 50.7 cm and 83.9 cm. Which of the following statements is (are) true ?

(A) The speed of sound determined from this experiment is $332 \mathrm{ms}^{-1}$

(B) The end correction in this experiment is 0.9 cm

(C) The wavelength of the sound wave is 66.4 cm

(D) The resonance at 50.7 cm corresponds to the fundamental harmonic

Sol. (A,C)

Let $\mathrm{n}_{1}$ harmonic is corresponding to 50.7 cm & $\mathrm{n}_{2}$ harmonic is corresponding 83.9 cm. since both one consecutive harmonics.

$\therefore$ their difference $=\frac{\lambda}{2}$

$\therefore \frac{\lambda}{2}=(83.9-50.7) \mathrm{cm}$

$\frac{\lambda}{2}=33.2 \mathrm{cm}$

$\lambda=66.4 \mathrm{cm}$

$\therefore \frac{\lambda}{4}=16.6 \mathrm{cm}$

length corresponding to fundamental mode must be close to \frac{\lambda}{4} \& 50.7 cm must be an odd multiple of this length 16.6 × 3 = 49.8 cm. therefore 50.7 is 3^{| \mathrm{rd}}harmonic

If end correction is e, then

\mathrm{e}+50.7=\frac{3 \lambda}{4}

\mathrm{e}=49.8-50.7=-0.9 \mathrm{cm}

\text { speed of sound, } \mathrm{v}=\mathrm{f} \lambda

\therefore \mathrm{v}=500 \times 66.4 \mathrm{cm} / \mathrm{sec}=332.000 \mathrm{m} / \mathrm{s}

Thermodynamics – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Simulator

Q. Among the following, the state function(s) is (are)

(A) Internal energy

(B) Irreversible expansion work

(C) Reversible expansion work

(D) Molar enthalpy

[JEE 2009]

Sol. (A,D)

Q. One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is ws and that along the dotted line path is $\mathrm{w}_{\mathrm{d}},$ then the integer close to the ratio $\mathrm{w}_{\mathrm{d}} / \mathrm{w}_{\mathrm{s}}$ is-

[JEE 2010]

Sol. 2

$\mathrm{W}_{\mathrm{d}}=-(4 \times 1.5)+(0)-(1 \times 1)-\left(\frac{2}{3} \times 2.5\right)$

$\mathrm{W}_{\mathrm{S}}=-\mathrm{P}_{1} \mathrm{V}_{1} \ln \frac{\mathrm{V}_{2}}{\mathrm{V}_{1}}$

$=-4 \times 05 \times \ln \frac{5.5}{0.5}$

Q. Match the transformations in Column-I with appropriate option in Column-II

[JEE 2011]

Sol. $A \rightarrow(P, R, S) B \rightarrow(R, S) C \rightarrow T, D \rightarrow(P, Q, T)$

Q. The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement(s) is (are) correct ?

(A) $\mathrm{T}_{1}=\mathrm{T}_{2}$

(B) $\mathrm{T}_{3}>\mathrm{T}_{1}$

(C) $\mathrm{W}_{\text {isothermal }}>\mathrm{W}_{\text {adiabatic }}$

(D) $\Delta \mathrm{U}_{\text {isothemal }}>\Delta \mathrm{U}_{\text {adibaic }}$

[JEE 2012]

Sol. (A,D)

Q. For an ideal gas, consider only P-V work in going from an initial state X to the final state Z. The final state Z can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct ? [take S as change in entropy and w as work done]

(A) $\Delta \mathrm{S}_{\mathrm{x} \rightarrow \mathrm{z}}=\Delta \mathrm{S}_{\mathrm{x} \rightarrow \mathrm{y}}+\Delta \mathrm{S}_{\mathrm{y} \rightarrow \mathrm{z}}$

(B) $\mathrm{W}_{\mathrm{x} \rightarrow \mathrm{z}}=\mathrm{W}_{\mathrm{x} \rightarrow \mathrm{y}}+\mathrm{W}_{\mathrm{y} \rightarrow \mathrm{z}}$

(C) $\mathrm{W}_{\mathrm{x} \rightarrow \mathrm{y} \rightarrow \mathrm{z}}=\mathrm{W}_{\mathrm{x} \rightarrow \mathrm{y}}$

(D) $\Delta \mathrm{S}_{\mathrm{x} \rightarrow \mathrm{y} \rightarrow \mathrm{z}}=\Delta \mathrm{S}_{\mathrm{x} \rightarrow \mathrm{y}}$

[JEE 2012]

Sol. (A,C)

Paragraph for Question 6 and 7

A fixed mass ‘m’ of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure.

Q. The pair of isochoric processes among the transformation of states is

(A) K to L and L to M

(B) L to M and N to K

(C) L to M and M to N

(D) M to N nd N to K

[JEE 2013]

Sol. (B)

Isochoric $\Rightarrow \mathrm{V}-$ constant

Q. The succeeding operations that enable this transformation of states are

(A) Heating, cooling, heating, cooling

(B) cooling, heating, cooling, heating

(C) Heating, cooling, cooling, heating

(D) Cooling, heating, heating, cooling

[JEE 2013]

Sol. (C) Isochoric $\Rightarrow \mathrm{V}-$ constant

Q. An ideal gas in thermally insulated vessel at internal pressure = $\mathrm{P}_{1}$, volume = $V_{1}$ and absolute temperature = $\mathrm{T}_{1}$ expands irrversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are $P_{2}$, $\mathrm{V}_{2}$ and $\mathrm{T}_{2}$, respectively. For this expansion,

(A) q = 0

(B) $\mathrm{T}_{2}=\mathrm{T}_{1}$

(C) $\mathrm{P}_{2} \mathrm{V}_{2}=\mathrm{P}_{1} \mathrm{V}_{1}$

$(\mathrm{D}) \mathrm{P}_{2} \mathrm{V}_{2}^{\gamma}=\mathrm{P}_{1} \mathrm{V}_{1}^{\gamma}$

[JEE 2014]

Sol. 1,2,3

$\mathrm{q}=0 ; \mathrm{w}=0 ; \Delta \mathrm{U}=0 ; \mathrm{T}_{1}=\mathrm{T}_{2} ; \mathrm{PV}=\mathrm{constant}$

Q. One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings $\left(\Delta \mathrm{S}_{\text {surr }}\right)$ in $\mathrm{J} \mathrm{K}^{-1}$ is –

(1 L atm = 101.3 J)

(A) 5.763 (B) 1.013 (C) –1.013 (D) –5.763

Sol. (C)

Q. For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by

(A) With increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive

(B) With increase in temperature, the value of K for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases

(C) With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases

(D) With increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system negative

Sol. (B,C)

Q. An ideal gas is expanded from $\left(p_{1}, V_{1}, T_{1}\right)$ to $\left(p_{2}, V_{2}, T_{2}\right)$ under different conditions. The correct statement(s) among the following is(are)

(A) The work done on the gas is maximum when it is compressed irreversibly from ($\mathrm{p}_{2}$ , $\mathrm{V}_{2}$) to ($\mathrm{p}_{1}, \mathrm{V}_{1}$) against constant pressure $\mathrm{p}_{1}$

(B) The work done on the gas is less when it is expanded reversibly from $\mathrm{V}_{1}$ to $\mathrm{V}_{2}$ under adiabatic conditions as compared to that when expanded reversibly from $\mathrm{V}_{1}$ to $\mathrm{V}_{2}$ under isothermal conditions.

(C) The change in internal energy of the gas (i) zero, if it is expanded reversibly with $\mathrm{T}_{1} \mathrm{T}_{2}$, and (ii) positive, if it is expanded reversibly under adiabatic conditions with $\mathrm{T}_{1} \neq \mathrm{T}_{2}$

(D) If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic

Sol. 1,2,4

(3) (i) $\left.\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}=0 \text { (isothermal hence } \Delta \mathrm{T}=0\right)$

(ii) $\Delta \mathrm{U}=\mathrm{q}+\mathrm{w}=-\mathrm{ve}(\mathrm{q}=0, \mathrm{w}<0)$

$\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T} \Rightarrow \Delta \mathrm{T}<0$

\begin{aligned}(\mathbf{4}) \mathrm{q} &=0(\text { adiabatic }), \mathrm{w}=0(\text { free expansion }) \\ \Delta \mathrm{U} &=0 \Rightarrow \Delta \mathrm{T}=0 \text { (isothermal) } \end{aligned}

Q. A reversible cyclic process for an ideal gas is shown below. Here, P , V and T are pressure , volume and temperature , respectively. The thermodynamic parameters q, w, H and U are heat, work, enthalpy and internal energy, respectively.

The correct option(s) is (are)

(A) $\mathrm{q}_{\mathrm{AC}}=\Delta \mathrm{U}_{\mathrm{BC}}$ and $\mathrm{w}_{\mathrm{AB}}=\mathrm{P}_{2}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$

(B) $\mathrm{w}_{\mathrm{BC}}=\mathrm{P}_{2}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$ and $\mathrm{q}_{\mathrm{BC}}=\Delta \mathrm{H}_{\mathrm{AC}}$

(C) $\Delta \mathrm{H}_{\mathrm{CA}}<\Delta \mathrm{U}_{\mathrm{CA}}$ and $\mathrm{q}_{\mathrm{AC}}=\Delta \mathrm{U}_{\mathrm{BC}}$

(D) $\mathrm{q}_{\mathrm{BC}}=\Delta \mathrm{H}_{\mathrm{AC}}$ and $\Delta \mathrm{H}_{\mathrm{CA}}>\Delta \mathrm{U}_{\mathrm{CA}}$

Sol. (B,C)

Q. For a reaction, $\mathbf{A} \square \mathbf{P}$, the plots of [A] and [P] with time at temperatures $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ are given below.

If $\mathrm{T}_{2}>\mathrm{T}_{1},$ the correct statement(s) is (are)

(Assume $\Delta \mathrm{H}^{\theta}$ and $\Delta \mathrm{S}^{\theta}$ are independent of temperature and ratio of $\ln \mathrm{K}$ at $\mathrm{T}_{1}$ to $\ln \mathrm{K}$ at $\mathrm{T}_{2}$ is greater than $\mathrm{T}_{2 / \mathrm{T}_{\mathrm{T}}}$. Here H, S, G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

(A) $\Delta \mathrm{H}^{\theta}<0, \Delta \mathrm{S}^{\theta}<0$

(B) $\Delta \mathrm{G}^{\theta}<0, \Delta \mathrm{H}^{\theta}>0$

(C) $\Delta \mathrm{G}^{\theta}<0, \Delta \mathrm{S}^{\theta}<0$

(D) $\Delta \mathrm{G}^{\theta}<0, \Delta \mathrm{S}^{\theta}>0$

Sol. (A,C)

Thermal Expansion – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Simulator

Q. Steel wire of length ‘L’ at $40^{\circ}$ C is suspended from the ceiling and then a mass ‘m’ is hung from its free end. The wire is cooled down from $40^{\circ}$ to $30^{\circ}$ C to regain its original length ‘L’. The coefficient of linear thermal expansion of the steel is $10^{-5} /^{\circ} \mathrm{C}$, Young’s modulus of steel is 1011 N/m2 and radius of the wire is 1 mm. Assume that L>> diameter of the wire. Then the value of ‘m’ in kg is nearly –

[JEE 2011]

Sol. $\Delta L_{1}=\frac{F L}{A Y}=\frac{m g L}{\pi r^{2} Y}=$Increase in length

$\Delta L_{2}=L \alpha \Delta T=$ Decrease in length

$\Delta L_{1}=\Delta L_{2}$

Sound Wave – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Simulator

Q. A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/hr towards a tall building which reflects the sound waves. The speed of sound in air is 320 m/s. The frequency of the siren heard by the car driver is ?

(A) 8.50 kHz (B) 8.25 kHz (C) 7.75 kHz (D) 7.50 kHz

[JEE 2011]

Sol. (A)

$f^{\prime}=\left(\frac{v}{v-v_{s}}\right)\left(\frac{v+v_{0}}{v}\right) f \Rightarrow f^{\prime}=\left(\frac{320}{320-10}\right)\left(\frac{320+10}{320}\right) \times 8 \Rightarrow f^{\prime} \approx 8.50 \mathrm{kHz}$

Q. Column-I shows four systems, each of the same length L, for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as $\lambda_{\mathrm{f}}$. Match each system with statements given in Column-II describing the nature and wavelength of the standing waves.

[JEE 2011]

Sol. ((A) $\mathrm{P}, \mathrm{T}(\mathrm{B}) \mathrm{P}, \mathrm{S}(\mathrm{C}) \mathrm{Q}, \mathrm{S}(\mathrm{D}) \mathrm{Q}, \mathrm{R}$)

Q. A person blows into open-end of a long pipe. As a result, a high-pressure pulse of air travels down the pipe. When this pulse reaches the other end of the pipe.

(A) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is open

(B) a low -pressure pulse starts travelling up the pipe, if the other end of the pipe is open

(C) a low pressure pulse starts travelling up the pipe, if the other end of the pipe is closed

(D) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed

[JEE 2012]

Sol. ($\mathrm{B}, \mathrm{D}$)

Information Based (Refer NCERT)

Q. A student is performing the experiment of Resonance Column. The diameter of the column tube is 4 cm. The frequency of the tuning fork is 512 Hz. The air temperature is $38^{\circ} \mathrm{C}$in which the speed of sound is 336 m/s. The zero of the meter scale coincides with the top end of the Resonance column tube. When the first resonance occurs, the reading of the water level in the column is :-

(A) 14.0 cm (B) 15.2 cm (C) 16.4 cm (D) 17.6 cm

[JEE 2012]

Sol. (B)

In resonance column experiment $\frac{\lambda}{4}=\ell_{1}+e$ so, $\ell_{1}=\frac{\lambda}{4}-e=\frac{v}{4 f}-0.3 d$

$=\frac{336 \times 100 \mathrm{cm}}{4 \times 512}-(0.3)(4 \mathrm{cm})=16.4-1.2=15.2 \mathrm{cm}$

Q. A student is performing an experiment using a resonance column and a tuning fork of frequency

$244 \mathrm{s}^{-1}$ . He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is (0.350 $\pm 0.005)$ m, the gas in the tube is

(Useful information $: \sqrt{167 \mathrm{RT}}=640 \mathrm{J}^{1 / 2}$ mole $^{-1 / 2} ; \sqrt{140 \mathrm{RT}}=590 \mathrm{J}^{1 / 2}$ mole $^{-1 / 2} .$ The molar masses M in grams are given in the options. Take the values of $\sqrt{\frac{10}{\mathrm{M}}}$ for each gas as given there. )

(A) Neon $\left(\mathrm{M}=20, \sqrt{\frac{10}{20}}=\frac{7}{10}\right)$

(B) Nitrogen $\left(\mathrm{M}=28, \sqrt{\frac{10}{28}}=\frac{3}{5}\right)$

(C) Oxygen $\left(\mathrm{M}=32, \sqrt{\frac{10}{32}}=\frac{9}{16}\right)$

(D) Argon $\left(\mathrm{M}=36, \sqrt{\frac{10}{36}}=\frac{17}{32}\right)$

Sol. (D)

Q. Two loudspeakers M and N are located 20m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car is initially at a point P, 1800 m away from the midpoint Q of the line MN and moves towards Q constantly at 60 km/hr along the perpendicular bisector of MN. It crosses Q and eventually reaches a point R, 1800 m away from Q. Let n(t) represent the beat frequency measured by a person sitting in the car at time t. Let $v_{\mathrm{p}}, v_{\mathrm{Q}}$ and $v_{\mathrm{R}}$ be the beat frequencies measured at locations P, Q and R, respectively. The speed of sound in air is 330 $\mathrm{ms}^{-1}$. Which of the following statement(s) is(are) true regarding the sound heard by the person ?

(A) The plot below represents schematically the variation of beat frequency with time

(B) The plot below represents schematically the variations of beat frequency with time

(C) The rate of change in beat frequency is maximum when the car passes through Q

(D) $v_{\mathrm{p}}+v_{\mathrm{R}}=2 v_{\mathrm{Q}}$

Sol. (A,C,D)

Q. A stationary source emits sound of frequency $\mathrm{f}_{0}=492 \mathrm{Hz}$ . The sound is reflected by a large car approaching the source with a speed of $2 \mathrm{ms}^{-1}$. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz ? (Given that the speed of sound in air is 330 $\mathrm{ms}^{-1}$ and the car reflects the sound at the frequency it has received).

Sol. 6

Frequency of sound as received by large car approaching the source.

Q. Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed 1.0 $\mathrm{ms}^{-1}$ and the man behind walks at a speed 2.0 $\mathrm{ms}^{-1}$. A third man is standing at a height 12 m above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency 1430 Hz. The speed of sound in air is 330 $\mathrm{ms}^{-1}$. At the instant, when the moving men are 10 m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is _____

Sol. 5.00 Hz

Q. In an experment to measure the speed of sound by a resonating air column, a tuning fork of frequency 500 Hz is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonances are heard at air columns of length 50.7 cm and 83.9 cm. Which of the following statements is (are) true ?

(A) The speed of sound determined from this experiment is $332 \mathrm{ms}^{-1}$

(B) The end correction in this experiment is 0.9 cm

(C) The wavelength of the sound wave is 66.4 cm

(D) The resonance at 50.7 cm corresponds to the fundamental harmonic

Sol. (A,C)

Let $\mathrm{n}_{1}$ harmonic is corresponding to 50.7 cm & $\mathrm{n}_{2}$ harmonic is corresponding 83.9 cm. since both one consecutive harmonics.

$\therefore$ their difference $=\frac{\lambda}{2}$

$\therefore \frac{\lambda}{2}=(83.9-50.7) \mathrm{cm}$

$\frac{\lambda}{2}=33.2 \mathrm{cm}$

$\lambda=66.4 \mathrm{cm}$

$\therefore \frac{\lambda}{4}=16.6 \mathrm{cm}$

length corresponding to fundamental mode must be close to \frac{\lambda}{4} \& 50.7 cm must be an odd multiple of this length 16.6 × 3 = 49.8 cm. therefore 50.7 is 3^{| \mathrm{rd}}harmonic

If end correction is e, then

\mathrm{e}+50.7=\frac{3 \lambda}{4}

\mathrm{e}=49.8-50.7=-0.9 \mathrm{cm}

\text { speed of sound, } \mathrm{v}=\mathrm{f} \lambda

\therefore \mathrm{v}=500 \times 66.4 \mathrm{cm} / \mathrm{sec}=332.000 \mathrm{m} / \mathrm{s}

Simple Harmonic Motion – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

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Simulator

Q. The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t = 4/3 s is

[IIT JEE 2009]

Sol. (D)

$\tau=8$

$\therefore \omega=\frac{\pi}{4}$

$\mathrm{x}=\sin \left(\frac{\pi}{4} \mathrm{t}\right)$

$\therefore a=-\frac{\pi^{2}}{16} \sin \left(\frac{\pi}{4} t\right)$

$-\frac{\pi^{2} \sqrt{3}}{32} \mathrm{cm} / \mathrm{s}^{2}$

Q. The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is

[IIT JEE 2009]

Sol. (D)

$\frac{\mathrm{k}_{1} \mathrm{k}_{2}}{\mathrm{k}_{1}+\mathrm{k}_{2}} \mathrm{A}=\mathrm{k}_{1} \mathrm{x}=\mathrm{F}$

$\therefore \mathrm{x}=\frac{\mathrm{k}_{2} \mathrm{A}}{\mathrm{k}_{1}+\mathrm{k}_{2}}$

Q. A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring constants k. The springs are fixed to rigid supports as shown in the figure, and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle $\theta$ in one direction and released. The frequency of oscillation is :-

[IIT JEE 2009]

Sol. (C)

$\tau=2 \mathrm{k} \cdot \frac{\ell \theta}{2} \cdot \frac{\ell}{2}=\frac{\mathrm{m} \ell^{2}}{12} \cdot \omega^{2} \theta$

$\therefore \omega=\sqrt{\frac{6 \mathrm{k}}{\mathrm{m}}}$

$\therefore \mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{6 \mathrm{k}}{\mathrm{m}}}$

Paragraph for Question No. 4 to 6

When a particle of mass m moves on the x–axis in a potential of the form $\mathrm{V}(\mathrm{x})=\mathrm{k} \mathrm{x}^{2}$, it performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{\frac{m}{k}}$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from $\mathrm{kx}^{2}$ and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x–axis. Its potential energy is $\mathrm{V}(\mathrm{x})=\alpha \mathrm{x}^{4}(\alpha>0)$ for $|\mathrm{x}|$ near the origin and becomes a constant equal to $\mathrm{V}_{0}$ for $|\mathrm{x}| \geq \mathrm{X}_{0}$ (see figure)

Q. If the total energy of the particle is E, it will perform periodic motion only if :-

(A) $\mathrm{E}<0$

(B) E > 0

(C) $\mathrm{V}_{0}>\mathrm{E}>0$

(D) $\mathrm{E}>\mathrm{V}_{0}$

[IIT-JEE 2010]

Sol. (C)

Particle must be trapped.

$\therefore \mathrm{V}_{0}>\mathrm{E}>0$

Q. For periodic motion of small amplitude A, the time period T of this particle is proportional to :-

(A) $A \sqrt{\frac{m}{\alpha}}$

(B) $\frac{1}{A} \sqrt{\frac{m}{\alpha}}$

(C) $A \sqrt{\frac{\alpha}{m}}$

(D) $\frac{1}{A} \sqrt{\frac{\alpha}{m}}$

[IIT-JEE 2010]

Sol. (B)

Dimentions are

$\propto=\frac{\left[\mathrm{ML}^{2} \mathrm{T}^{-2}\right]}{\left[\mathrm{L}^{4}\right]}=\left[\mathrm{ML}^{-2} \mathrm{T}^{-2}\right]$

$\therefore \mathrm{T} \propto \frac{1}{\mathrm{A}} \sqrt{\frac{\mathrm{m}}{\alpha}}$

Q. The acceleration of this particle for $|x|>X_{0}$ is :-

(A) proportional to $V_{0}$

(B) proportional to $\frac{V_{0}}{m X_{0}}$

(C) proportional to $\sqrt{\frac{V_{0}}{m X_{0}}}$

(D) Zero

[IIT-JEE 2010]

Sol. (D)

As $\mathrm{U}=\mathrm{V}_{0}=$ constant

$\therefore \mathrm{a}=0$

Q. A point mass is subjected to two simultaneous sinusoidal displacements in x-direction,

$x_{1}(t)=A \sin \omega t$ and $x_{2}(t)=A \sin \left(\omega t+\frac{2 \pi}{3}\right)$. Adding a third sinusoidal displacement

$x_{3}(t)=B \sin (\omega t+\phi)$ brings the mass to a complete rest. The values of $\mathrm{B}$ and $\phi$ are :-

[IIT-JEE 2011]

Sol. (B)

$\mathrm{x}^{\prime}=\mathrm{A} \sin (\omega \mathrm{t})+\mathrm{A} \sin \left(\omega \mathrm{t}+\frac{2 \pi}{3}\right)$

$=2 \mathrm{A} \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}\right)$

$=\mathrm{A} \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right)$

$\therefore \mathrm{B}=\mathrm{A} \quad \phi=\frac{4 \pi}{3}$

Q. A metal rod of length ‘L’ and mass ‘m’ is pivoted at one end. A thin disk of mass ‘M’ and radius ‘R’ (< L) is attached at its center to the free end of the rod. Consider two ways the disc is attached: (case A). The disc is not free to rotate about its center and (case B) the disc is free to rotated about its center. The rod-disc system performs SHM in vertical plane after being released from the same displaced position. Which of the following statement(s) is(are) true?

(A) Restoring torque in case A = Restoring torque in case B

(B) Restoring torque in case A < Restoring torque in case B

(C) Angular frequency for case A> Angular frequency for case B

(D) Angular frequency for case A< Angular frequency for case B

[IIT-JEE 2011]

Sol. (A,D)

$\tau=\frac{\mathrm{mL}}{2} \mathrm{g} \cos \theta+\mathrm{mL} \mathrm{g} \cos \theta \mathrm{same}$

$\frac{\omega}{2 \pi}=\mathrm{f}$ for $\mathrm{A} \mathrm{mg} \mathrm{H}=\frac{1}{2} \mathrm{mv}^{2} \&$ for $\mathrm{B}=\operatorname{mg} \mathrm{x}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$

$\therefore \omega$

Q. A small block is connected to one end of a massless spring of un-stretched length 4.9 m. The other end of the spring (see the figure) is fixed. They system lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t = 0. It then executes simple harmonic motion with angular frequency $\omega=\frac{\pi}{3}$ rad/s. Simultaneously at t = 0, a small pebble is projected with speed v from point P at an angle of 45° as shown in the figure. Point P is at a horizontal distance of 10m from O. If the pebble hits the block at t = 1s, the value of v is:- (take $\left.\mathrm{g}=10 \mathrm{m} / \mathrm{s}^{2}\right)$

[IIT-JEE 2012]

Sol. (A)

Time of

$\therefore \mathrm{v}=\frac{\mathrm{g}}{\sqrt{2}}=\sqrt{50} \mathrm{m} / \mathrm{s}$

Q. A particle of mass m is attached to one end of a mass-less spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t = 0 with an initial velocity $\mathbf{u}_{0}$. When the speed of the particle is 0.5 $\mathbf{u}_{0}$, it collides elastically with a rigid wall. After this collision :-

(A) the speed of the particle when it returns to its equilibrium position is $\mathrm{u}_{0}$

(B) the time at which the particle passes through the equilibrium position for the first

time is $t=\pi \sqrt{\frac{m}{k}}$

(C) the time at which the maximum compression of the spring occurs is $\mathrm{t}=\frac{4 \pi}{3} \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

(D) the time at which the particle passes through the equilibrium position for the second

time is $t=\frac{5 \pi}{3} \sqrt{\frac{m}{k}}$

Sol. (A,D)

Collision is elastic so energy is conserve in given question so when it return to equilibrium, its speed is $\mathrm{u}_{0}$

Q. Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies  and have total energies $\mathrm{E}_{1}$ and $\mathrm{E}_{2}$, respectively. The variations of their momenta p with positions x are shown in the figures. If $\frac{\mathrm{a}}{\mathrm{b}}=\mathrm{n}^{2}$ and $\frac{\mathrm{a}}{\mathrm{R}}=\mathrm{n}$, then the correct equation (s) is (are)

Sol. (B,D)

$\mathrm{P}_{1 \max }=\operatorname{mac}_{1}=\mathrm{b}$

$\mathrm{P}_{2 \max }=\mathrm{mR} \omega_{2}=\mathrm{R}$

$\frac{\omega_{1}}{\omega_{2}}=\frac{1}{n^{2}}$

$\frac{\omega_{2}}{\omega_{1}}=\mathrm{n}^{2}$

$\mathrm{E}_{1}=\frac{1}{2} \mathrm{m} \omega_{1}^{2} \mathrm{a}^{2}$

Q. A particle of unit mass is moving along the x-axis under the influence of a force and its total enegy is conserved. Four possible forms of the potential energy of the particle are given in column I (a and $\mathrm{U}_{0}$ are constants). Match the potential energies in column I to the corresponding statement(s) in column-II

Sol. ($(\mathrm{A})-\mathrm{P}, \mathrm{Q}, \mathrm{R}, \mathrm{T} ;(\mathrm{B})-\mathrm{Q}, \mathrm{S} ;(\mathrm{C})-\mathrm{P}, \mathrm{Q}, \mathrm{R}, \mathrm{S} ;(\mathrm{D})-\mathrm{P}, \mathrm{R}, \mathrm{T}$)

Q. A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position x0. Consider two cases : (i) when the block is at x0; and (ii) when the block is at x = x0 + A. In both the cases, a particle with mass m (<M) is softly placed on the block after which they stick to each other. Which of the following statement(s) is(are) true about the motion after the mass m is placed on the mass M?

(A) The amplitude of oscillation in the first case changes by a factor of , whereas $\sqrt{\frac{\mathrm{M}}{\mathrm{m}+\mathrm{M}}}$ in the second case it remains unchanged

(B) The final time period of oscillation in both the cases is same

(C) The total energy decreases in both the cases

(D) The instantaneous speed at x0 of the combined masses decreases in both the cases.

Sol. (A,B,D)

$\mathrm{T}_{\mathrm{i}}=2 \pi \sqrt{\frac{\mathrm{M}}{\mathrm{K}}}, \mathrm{T}_{\mathrm{f}}=2 \pi \sqrt{\frac{\mathrm{M}+\mathrm{m}}{\mathrm{K}}}$

case (i) :

M $(\mathrm{A} \omega)=(\mathrm{M}+\mathrm{m}) \mathrm{V}$

$\therefore$ Velocity decreases at equilibrium position.

By energy conservation

$\mathrm{A}_{\mathrm{f}}=\mathrm{A}_{\mathrm{i}} \sqrt{\frac{\mathrm{M}}{\mathrm{M}+\mathrm{m}}}$

case ( ii ) :

No energy loss, amplitude remains same

At equilibrium $\left(\mathrm{x}_{0}\right)$ velocity $=\mathrm{A} \omega .$

In both cases $\omega$ decreases so velocity decreases in both cases

Q. Two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ are defined as $\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{A}}=$ ai and $\overrightarrow{\mathrm{B}}=\mathrm{a}(\cos \omega \mathrm{ti}+\sin \omega \mathrm{t} \hat{\mathrm{j}}),$ whre a is a constant and

$\omega=\pi / 6 \mathrm{rad} \mathrm{s}^{-1} .$ If $|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{3}|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|$ at time $\mathrm{t}=\tau$ for the first time, the value of $\tau,$ in seconds, is

Sol. 2.00 Sec

Rotational Dynamics – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Simulator

Q. A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is $\sqrt{3}$. The inclination $\theta$ of this inclined plane from the horizontal plane is gradually increased from $0^{\circ}$. Then :-

(A) at $\theta=30^{\circ}$, the block will start sliding down the plane

(B) the block will remain at rest on the plane up to certain $\theta$ and then it will topple

(C) at $\theta=60^{\circ}$, the block will start sliding down the plane and continue to do so at higher angles

(D) at $\theta=60^{\circ}$, the block will start sliding down the plane and on further increasing q, it will topple at certain q

[IIT-JEE 2009]

Sol. (B)

Q. If the resultant of the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that

(A) linear momentum of the system does not change in time

(B) kinetic energy of the system does not change in time

(C) angular momentum of the system does not change in time

(D) potential energy of the system does not change in time

[IIT-JEE 2009]

Sol. (A)

Q. A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then

(A) $\vec{v}_{C}-\vec{v}_{A}=2\left(\vec{v}_{B}-\vec{v}_{C}\right)$

(B) $\vec{v}_{C}-\vec{v}_{B}=\vec{v}_{B}-\vec{v}_{A}$

(C) $\left|\vec{v}_{C}-\vec{v}_{A}\right|=2\left|\vec{v}_{B}-\vec{v}_{C}\right|$

(D) $\left|\vec{v}_{C}-\vec{v}_{A}\right|=4\left|\vec{v}_{B}\right|$

[IIT-JEE 2009]

Sol. (B,C)

Q. A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick applies a force of 2N on the ring and rolls it without slipping with an acceleration of 0.3 $\mathrm{m} / \mathrm{s}^{2}$. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is (P/10). The value of P is

[IIT-JEE 2011]

Sol. 4

Q. Four solid spheres each of diameter $\sqrt{5}$ cm and mass 0.5 kg are placed with their centers at the
corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is $\mathrm{N} \times 10^{-4} \mathrm{kg}-\mathrm{m}^{2}$, then N is

[IIT-JEE 2011]

Sol. 9

Q. A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1 m/s. A small ball of mass 0.1 kg, moving with velocity 20 m/s in the opposite direction, hits the ring at a height of 0.75 m and goes vertically up with velocity 10 m/s. Immediately after the collision

(A) the ring has pure rotation about its stationary CM

(B) the ring comes to a complete stop

(C) friction between the ring and the ground is to the left

(D) there is no friction between the ring and the ground

[IIT-JEE 2011]

Sol. (A,C or C)

Q. A thin uniform rod, pivoted at OP, is rotating in the horizontal plane with constant angular speed $\omega$. as shown in the figure. At time t = 0, a small insect starts from O and moves with constant speed v with respect to the rod towards the other end. If reaches the end of the rod at t = T and stops. The angular speed of the system remains $\omega$ throughout. The magnitude of the torque $(|\tau|)$ on the system about O, as a function of time is best represented by which plot?

[IIT-JEE 2012]

Sol. (B)

Q. A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in the x-y plane with centre at O and constant angular speed $\omega$. If the angular momentum of the system, calculated about O and P are denoted by $\vec{L}_{o}$ and $\vec{L}_{p}$ respectively, then

(A) $\vec{L}_{o}$ and $\vec{L}_{p}$ do not vary with time

(B) $\vec{L}_{o}$ varies with time while $\vec{L}_{p}$ remains constant

(C) $\vec{L}_{o}$ remains constant while $\vec{L}_{P}$ varies with time

(D) $\vec{L}_{o}$ and $\vec{L}_{P}$ both vary with time

[IIT-JEE 2012]

Sol. (C)

Q. A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in the figure. The moment of inertia of this lamina about axes assing through O and P is $\mathrm{I}_{0}$ and $\mathrm{I}_{\mathrm{P}}$ respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $\frac{I_{P}}{I_{O}}$ to the nearest integer is

[IIT-JEE 2012]

Sol. 3

$\mathrm{I}_{0}=\mathrm{I}_{\text {biggle }}-\mathrm{I}_{\text {mall }}$

$=\frac{4 \mathrm{m}(2 \mathrm{R})^{2}}{2}-\left(\frac{\mathrm{mR}^{2}}{2}+\mathrm{mR}^{2}\right)$

$=8 \mathrm{mR}^{2}-\frac{3}{2} \mathrm{mR}^{2}=\frac{13}{2} \mathrm{mR}^{2}$

$\mathrm{I}_{\mathrm{P}}=\left(\frac{4 \mathrm{m}(2 \mathrm{R})^{2}}{2}+4 \mathrm{m}(2 \mathrm{R})^{2}\right)-\left(\frac{\mathrm{mR}^{2}}{2}+\mathrm{m} 4 \mathrm{R}^{2}\right)=20-\frac{1}{2}=\frac{39}{2}$

$\therefore$ ratio $=\frac{39}{13}=3$

Paragraph for Questions 10 and 11

The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass. These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed , the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the z-axis, and (ii) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both these motions have the same angular speed $\omega$ in this case.

Now consider two similar systems as shown in the figure : case (A) the disc with its face vertical and parallel to x-z plane; Case (B) the disc with its face making an angle of $45^{\circ}$ with x-y plane and its horizontal diameter parallel to x-axis. In both the cases, the disc is welded at point P, and the systems are rotated with constant angular speed  about the z-axis.

Q. Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct?

(A) It is $\sqrt{2 \omega}$ for both the cases.

(B) It is $\omega$ for case $(a) ;$ and $\frac{\omega}{\sqrt{2}}$ for case $(b)$

(C) It is $\omega$ for case $(a) ;$ and $\sqrt{2 \omega}$ for case (b).

(D) It is $\omega$ for both the cases.

[IIT-JEE 2012]

Sol. (D)

$\omega$ for both the cases.

$\therefore$ in same t $2 \pi$ angle.

Q. Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct?

(A) It is vertical for both the cases (a) and (b).

(B) It is vertical for case (a); and is at $45^{\circ}$ to the x-z plane and lies in the plane of the disc for case (b).

(C) It is horizontal for case (a); and is at $45^{\circ}$ to the x-z plane and is normal to the plane of the disc for case (b).

(D) It is vertical for case (a); and is at $45^{\circ}$ to the x-z plane and is normal to the plane of the disc for case (b)

[IIT-JEE 2012]

Sol. (A)

vertical for both.

Q. The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed  and (ii) an inner disc of radius 2R rotating anti-clockwise with angular speed $\omega /$2. The ring and disc are separated by frictionless ball bearing. The system is in the x-z plane. The point P on the inner disc is at a distance R from the origin, where OP makes an angle of $30^{\circ}$ with the horizontal. Then with respect to the horizontal surface,

(A) the point O has a linear velocity $3 R \omega \hat{i}$

(B) the point $P$ has a linear velocity $\frac{11}{4} R \omega \hat{i}+\frac{\sqrt{3}}{4} R \omega \hat{k}$

(C) the point P has a linear velocity $\frac{13}{4} R \omega \hat{i}-\frac{\sqrt{3}}{4} R \omega \hat{k}$

(D) the point P has a linear velocity $\left(3-\frac{\sqrt{3}}{4}\right) R \omega \hat{i}+\frac{1}{4} R \omega \hat{k}$

[IIT-JEE 2012]

Sol. (A,B)

Q. Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is(are) correct?

(A) Both cylinders P and Q reach the ground at the same time.

(B) Cylinder P has larger acceleration than cylinder Q.

(C) Both cylinders reach the ground with same translational kinetic energy.

(D) Cylinder Q reaches the ground with larger angular speed.

[IIT-JEE 2012]

Sol. (D)

Q. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of
10 rad $\mathrm{s}^{-1}$ about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s–1) of the system is

[IIT-JEE 2013]

Sol. 8

by angular momentum conservation

$\mathrm{I}_{\text {disc }} \omega_{0}=\left(\mathrm{I}_{\text {disc }}+2 \mathrm{I}_{\text {ring }}\right) \omega$

Q. A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about its axis. Two massless spring toy-guns, each carrying a steel ball of mass 0.05 kg are attached to the platform at a distance 0.25 m from the centre on its either sides along its diameter (see figure). Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the balls have horizontal speed of 9 ms$-1$with respect to the ground. The rotational speed of the platform in rad s$-1$after the balls leave the platform is

Sol. 4

Since no external torque acts therefore angular momentum remains conserved.

Angular momentum of ball = Angular momentum of plateform

$0.05 \times 9 \times 0.25 \times 2=\frac{1}{2} \times 0.45 \times 0.5 \times 0.5 \times \omega$

$\omega=4 \mathrm{rad} / \mathrm{s}$

Q. A uniform circular disc of mass 1.5 kg and radius 0.5 m is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude F = 0.5 N are applied simultaneously along the three sides of an equilateral triangle XYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc in rad $\mathrm{s}^{-1}$ is

Sol. 2

Angular impulse = change in angular momentum

$\tau \Delta \mathrm{t}=\mathrm{I} \omega$

$3 \times \mathrm{F} \times \mathrm{R} \sin 30 \times \Delta \mathrm{t}=\mathrm{I} \omega$

$3 \times 0.5 \times 0.5 \times \frac{1}{2} \times 1=\frac{1}{2} \times 1.5 \times 0.5 \times 0.5 \times \omega$

$\omega=2 \mathrm{rad} / \mathrm{s}$

Q. Two identical uniform discs roll without slipping on two different, surfaces AB and CD (see figure) starting at A and C with linear speeds $\mathrm{v}_{1}$ and $\mathrm{v}_{2}$ respectively, and always remain in contact with the surfaces. If they reach B and D with the same linear speed and $\mathrm{v}_{1}=3 \mathrm{m} / \mathrm{s},$ then $\mathrm{v}_{2}$ in m/s is (g = 10 $\mathrm{m} / \mathrm{s}^{2}$)

Sol. 7

Q. A ring of mass M and radius R is rotating with angular speed $\omega$about a fixed vertical axis passing through its centre O with two point masses each of mass $\frac{\mathrm{M}}{8}$ at rest at O. These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is $\frac{8}{9} \omega$ and one of the masses is at a distance of $\frac{3}{5}$ R from O. At this instant the distance of the other mass from O is :

(A) $\frac{2}{3} \mathrm{R}$

(B) $\frac{1}{3} \mathrm{R}$

(C) $\frac{3}{5} \mathrm{R}$

(D) $\frac{4}{5} \mathrm{R}$

Sol. (C,D)

Q. The densities of two solid spheres A and B of the same radii R vary with radial distance r as $\rho_{\mathrm{A}}(\mathrm{r})=\mathrm{k}\left(\frac{\mathrm{r}}{\mathrm{R}}\right)$ and $\rho_{\mathrm{B}}(\mathrm{r})=\mathrm{k}\left(\frac{\mathrm{r}}{\mathrm{R}}\right)^{5}$ , respectively, where k is a constant. The moments of inertia of the indivisual spheres about axes passing through their centres are $\mathrm{I}_{\mathrm{A}}$ and $\mathrm{I}_{\mathrm{B}}$, respectively. If $\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}}=\frac{\mathrm{n}}{10}$, the value of n is.

Sol. 6

Q. A uniform wooden stick of mass 1.6 kg and length  rests in an inclined manner on a smooth, vertical wall of height h (<) such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of $30^{\circ}$ with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio h/ and the frictional force f at the bottom of the stick are: $\left(\mathrm{g}=10 \mathrm{ms}^{-2}\right)$

(A) $\frac{\mathrm{h}}{\ell}=\frac{\sqrt{3}}{16}, \mathrm{f}=\frac{16 \sqrt{3}}{3} \mathrm{N}$

(B) $\frac{\mathrm{h}}{\ell}=\frac{3}{16}, \mathrm{f}=\frac{16 \sqrt{3}}{3} \mathrm{N}$

(C) $\frac{\mathrm{h}}{\ell}=\frac{3 \sqrt{3}}{16}, \mathrm{f}=\frac{8 \sqrt{3}}{3} \mathrm{N}$

(D) $\frac{\mathrm{h}}{\ell}=\frac{3 \sqrt{3}}{16}, \mathrm{f}=\frac{16 \sqrt{3}}{3} \mathrm{N}$

Sol. (D)

Q. The position vector $\overrightarrow{\mathrm{r}}$ of a particle of mass $\mathrm{m}$ is given by the following equation $\overrightarrow{\mathrm{r}}(\mathrm{t})=\alpha \mathrm{t}^{2} \hat{\mathrm{i}}+\beta t \hat{\mathrm{i}}+\beta t \hat{\mathrm{j}}$

where $\alpha=\frac{10}{3} \mathrm{ms}^{-3}, \beta=5 \mathrm{ms}^{-2}$ and $\mathrm{m}=0.1 \mathrm{kg} .$ At $\mathrm{t}=1 \mathrm{s},$ which of the following statement(s)

(A) The velocity $\overrightarrow{\mathrm{v}}$ is given by $\overrightarrow{\mathrm{v}}=(10 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}) \mathrm{ms}^{-1}$

(B) The angular momentum $\overrightarrow{\mathrm{L}}$ with respect to the origin is given by $\overrightarrow{\mathrm{L}}=-\left(\frac{5}{3}\right) \hat{\mathrm{k}} \mathrm{Nms}$

(C) The force $\overrightarrow{\mathrm{F}}$ is given by $\overrightarrow{\mathrm{F}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}) \mathrm{N}$

(D) The torque $\vec{\tau}$ with respect to the origin is given by $\vec{\tau}=-\left(\frac{20}{3}\right) \hat{\mathrm{k}} \mathrm{Nm}$

Sol. (A,B,D)

$\overrightarrow{\mathrm{r}}=\alpha \mathrm{t}^{3} \hat{\mathrm{i}}+\beta \mathrm{t}^{2} \hat{\mathrm{j}}$

$\overrightarrow{\mathrm{v}}=\frac{\mathrm{d} \overrightarrow{\mathrm{r}}}{\mathrm{dt}}=3 \alpha \mathrm{t}^{2} \hat{\mathrm{i}}+2 \beta \hat{\mathrm{t} \hat{\mathrm{j}}}$

$\overrightarrow{\mathrm{a}}=\frac{\mathrm{d}^{2} \overrightarrow{\mathrm{r}}}{\mathrm{dt}^{2}}=6 \alpha t \hat{\mathrm{i}}+2 \beta \hat{\mathrm{j}}$

At t = 1

(A) $\overrightarrow{\mathrm{v}}=3 \times \frac{10}{3} \times 1 \hat{\mathrm{i}}+2 \times \mathrm{J} \times 1 \hat{\mathrm{j}}$

$=10 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}$

$(\mathrm{B}) \quad \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}$

$=\left(\frac{10}{3} \times 1 \hat{i}+5 \times 1 \hat{j}\right) \times 0.1(10 \hat{i}+10 \hat{j})$

$=-\frac{5}{3} \hat{\mathrm{k}}$

(C) $\overrightarrow{\mathrm{F}}=\mathrm{m} \times\left(6 \times \frac{10}{3} \times 1 \hat{\mathrm{i}}+2 \times 5 \hat{\mathrm{j}}\right)=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}$

$(\mathrm{D}) \vec{\tau}=\mathrm{r} \times \overrightarrow{\mathrm{F}}$

$=\left(\frac{10}{3} \hat{\mathrm{i}}+5 \hat{\mathrm{j}}\right) \times(2 \hat{\mathrm{i}}+\hat{\mathrm{j}})$

$=+\frac{10}{3} \hat{\mathrm{k}}+10(-\hat{\mathrm{k}})$

$=-\frac{20}{3} \hat{\mathrm{k}}$

Q. Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a massless, right rod of length $\ell=\sqrt{24} \mathrm{a}$ throught their center. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is $\omega$. The angular momentum of the entire assembly about the point ‘O’ is (see the figure). Which of the following statement(s) is(are) true ?

(A) The magnitude of angular momentum of the assembly about its center of mass is $17 \mathrm{ma}^{2} \omega / 2$

(B) The magnitude of the z-component $\overrightarrow{\mathrm{L}}$ is $55 \mathrm{ma}^{2} \omega$

(C) The magnitude of angular momentum of center of mass of the assembly about the point O is 81 $\mathrm{ma}^{2} \mathrm{\omega}$

(D) The center of mass of the assembly rotates about the z-axis with an angular speed of $\omega$/ 5

Sol. (A,D)

By no slip condition, here $\omega^{\prime}$ is angular velocity about z axis

$\omega^{\prime} x=\omega r$

$\omega^{\prime}=\frac{\omega r}{x}=\omega \sin \theta=\frac{\omega}{5}$

Paragraph 1

A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity  is an example of a non-inertial frame of reference. The relationship between the force $\overrightarrow{\mathrm{F}}_{\mathrm{rot}}$ experienced by a particle of mass m moving on the rotating disc and the force $\overrightarrow{\mathrm{F}}_{\mathrm{in}}$ experienced by the particle in an inertial frame of reference is

$\overrightarrow{\mathrm{F}}_{\mathrm{rot}}=\overrightarrow{\mathrm{F}}_{\mathrm{in}}+2 \mathrm{m}\left(\overrightarrow{\mathrm{v}}_{\mathrm{rot}} \times \vec{\omega}\right)+\mathrm{m}(\vec{\omega} \times \overrightarrow{\mathrm{r}}) \times \vec{\omega}$

where $\vec{v}_{\text {rot }}$ is the velocity of the particle in the rotating frame of reference and $\overrightarrow{\mathrm{r}}$ is the position

vector of the particle with respect to the centre of the disc.

Now consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed  about its vertical axis through its center. We assign a coordinate system with the origin at the centre of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along the rotation axis $(\vec{\omega}=\omega \hat{k})$ A small block of mass m is gently placed in the slot at $\overrightarrow{\mathrm{r}}=(\mathrm{R} / 2) \hat{\mathrm{i}}$ at $\mathrm{t}=0$ and is constrained to move only along the slot.

Q. The distance r of the block at time t is :

(A) $\frac{\mathrm{R}}{4}\left(\mathrm{e}^{2 \omega \mathrm{t}}+\mathrm{e}^{-2 \omega \mathrm{t}}\right)$

(B) $\frac{\mathrm{R}}{2} \cos 2 \omega \mathrm{t}$

(C) $\frac{\mathrm{R}}{2} \mathrm{cos} \omega \mathrm{t}$

(D) $\frac{\mathrm{R}}{4}\left(\mathrm{e}^{\mathrm{ot}}+\mathrm{e}^{-\mathrm{ot}}\right)$

Sol. (D)

Force on block along slot $=m \omega^{2} \mathrm{r}=\mathrm{ma}=\mathrm{m}\left(\frac{\mathrm{vdv}}{\mathrm{dr}}\right)$

$\int_{0}^{\mathrm{v}} \mathrm{vd} \mathrm{v}=\int_{\mathrm{R} / 2}^{\mathrm{r}} \omega^{2} \mathrm{rdr}$

$\frac{\mathrm{v}^{2}}{2}=\frac{\omega^{2}}{2}\left(\mathrm{r}^{2}-\frac{\mathrm{R}^{2}}{4}\right) \Rightarrow \mathrm{v}=\omega \sqrt{\mathrm{r}^{2}-\frac{\mathrm{R}^{2}}{4}}=\frac{\mathrm{dr}}{\mathrm{dt}}$

$\Rightarrow \int_{\mathrm{R} / 4}^{\mathrm{r}} \frac{\mathrm{dr}}{\sqrt{\mathrm{r}^{2}-\frac{\mathrm{R}^{2}}{4}}}=\int_{0}^{\mathrm{t}} \omega \mathrm{dt}$

$\ell_{\mathrm{n}}\left(\frac{\mathrm{r}+\sqrt{\mathrm{r}^{2}-\frac{\mathrm{R}^{2}}{4}}}{\frac{\mathrm{R}}{2}}\right)-\ell \mathrm{n}\left(\frac{\mathrm{R} / 2+\sqrt{\frac{\mathrm{R}^{2}}{4}-\frac{\mathrm{R}^{2}}{4}}}{\frac{\mathrm{R}}{2}}\right)=\omega \mathrm{t}$

$\Rightarrow \mathrm{r}+\sqrt{\mathrm{r}^{2}-\frac{\mathrm{R}^{2}}{4}}=\frac{\mathrm{R}}{2} \mathrm{e}^{\omega \mathrm{t}}$

$\Rightarrow \mathrm{r}^{2}-\frac{\mathrm{R}^{2}}{4}=\frac{\mathrm{R}^{2}}{4} \mathrm{e}^{2 \omega \mathrm{t}}+\mathrm{r}^{2}-2 \mathrm{r} \frac{\mathrm{R}}{2} \mathrm{e}^{\omega \mathrm{t}}$

$\Rightarrow \mathrm{r}=\frac{\frac{\mathrm{R}^{2}}{4} \mathrm{e}^{2 \omega \mathrm{t}}+\frac{\mathrm{R}^{2}}{4}}{\mathrm{Re}^{\mathrm{ot}}}=\frac{\mathrm{R}}{4}\left(\mathrm{e}^{\omega \mathrm{t}}+\mathrm{e}^{-\omega \mathrm{t}}\right)$

Q. The net reaction of the disc on the block is :

(A) $-\mathrm{m\omega}^{2} \mathrm{R}$ cos $\omega$ tì $\hat{\mathrm{j}}-\mathrm{mg} \hat{\mathrm{k}}$

(B) $\mathrm{m\omega}^{2} \mathrm{R} \sin \omega \hat{\mathrm{j}}-\mathrm{mg} \hat{\mathrm{k}}$

(C) $\frac{1}{2} \mathrm{m\omega}^{2} \mathrm{R}\left(\mathrm{e}^{\omega \mathrm{t}}-\mathrm{e}^{-\omega \mathrm{t}}\right) \hat{\mathrm{j}}+\mathrm{mg} \hat{\mathrm{k}}$

(D) $\frac{1}{2} \mathrm{m\omega}^{2} \mathrm{R}\left(\mathrm{e}^{2 \omega \mathrm{t}}-\mathrm{e}^{-2 \omega \mathrm{t}}\right) \hat{\mathrm{j}}+\mathrm{mg} \hat{\mathrm{k}}$

Sol. (C)

Q. A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time, the angle made by the bar with the vertical is $\theta$. Which of the following statements about its motion is/are correct ?

(A) When the bar makes an angle $\theta$ with the vertical, the displacement of its midpoint from the initial position is proportional to $(1-\cos \theta)$

(B) The midpoint of the bar will fall vertically downward

(C) Instantaneous torque about the point in contact with the floor is proportional to $\sin \theta$

(D) The trajectory of the point $A$ is a parabola

Sol. (A,B,C)

When the bar makes an angle $\theta ;$ the height of its COM (mid point) is $\frac{\mathrm{L}}{2} \cos \theta$

$\therefore$ displacement $=\mathrm{L}-\frac{\mathrm{L}}{2} \cos \theta=\frac{\mathrm{L}}{2}(1-\cos \theta)$

Since force on COM is only along the vertical direction, hence COM is falling vertically downward.

Instantaneous torque about point of contact is

$\mathrm{Mg} \times \frac{\mathrm{L}}{2} \sin \theta$

Q. A wheel of radius R and mass M is placed at the bottom of a fixed step of height R as shown in the figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque about an axis normal to the plane of the paper passing through the point Q. Which of the following options is/are correct ?

(A) If the force is applied normal to the circumference at point X then $\tau$ is constant

(B) If the force is applied tangentially at point S then $\tau \neq 0$ but the wheel never climbs the step

(C) If the force is applied normal to the circumference at point P then $\tau$ is zero

(D) If the force is applied at point P tangentially then t decreases continuously as the wheel climbs

Sol. (C)

(A) is incorrect.

if force is applied at P tangentially the

$\tau=\mathrm{F} \times 2 \mathrm{R}=\mathrm{constant}$

Paragraph for Questions 27 and 28

One twirls a circular ring (of mass M and radius R) near the tip of one’s finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity $\omega_{0}$. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is µ and the acceleration due to gravity is g.

Q. The total kinetic energy of the ring is :-

(A) $\mathrm{M} \omega_{0}^{2} \mathrm{R}^{2}$

(B) $\operatorname{Mo}_{0}^{2}(\mathrm{R}-\mathrm{r})^{2}$

(C) $\frac{1}{2} \mathrm{M} \omega_{0}^{2}(\mathrm{R}-\mathrm{r})^{2}$

(D) $\frac{3}{2} \mathrm{M} \omega_{0}^{2}(\mathrm{R}-\mathrm{r})^{2}$

Sol. Bonus

Q. The minimum value of $\omega_{0}$ below which the ring will drop down is :-

(A) $\sqrt{\frac{3 \mathrm{g}}{2 \mu(\mathrm{R}-\mathrm{r})}}$

(B) $\sqrt{\frac{\mathrm{g}}{\mu(\mathrm{R}-\mathrm{r})}}$

(C) $\sqrt{\frac{2 \mathrm{g}}{\mu(\mathrm{R}-\mathrm{r})}}$

(D) $\sqrt{\frac{\mathrm{g}}{2 \mu(\mathrm{R}-\mathrm{r})}}$

Sol. (B)

$\mu \mathrm{M} \omega_{0}^{2}(\mathrm{R}-\mathrm{r})=\mathrm{Mg}$

$\omega_{0}=\sqrt{\frac{\mathrm{g}}{\mu(\mathrm{R}-\mathrm{r})}}$

Q. The potential energy of a particle of mass m at a distance r from a fixed point O is given by $\mathrm{V}(\mathrm{r})=\mathrm{kr}^{2} / 2$, where k is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius R about the point O. If v is the speed of the particle and L is the magnitude of its angular momentum about O, which of the following statements is (are) true ?

(A) $v=\sqrt{\frac{\mathrm{k}}{2 \mathrm{m}}} \mathrm{R}$

(B) $v=\sqrt{\frac{k}{m}} R$

(C) $\mathrm{L}=\sqrt{\mathrm{mk}} \mathrm{R}^{2}$

Sol. (B,C)

Q. Consider a body of mass $1.0 \mathrm{kg}$ at rest at the origin at time $\mathrm{t}=0 .$ A force $\overrightarrow{\mathrm{F}}=(\alpha \mathrm{ti}+\beta \hat{\mathrm{j}})$ is

applied on the body, where $\alpha=1.0 \mathrm{Ns}^{-1}$ and $\beta=1.0 \mathrm{N}$. The torque acting on the body about

the origin at time $\mathrm{t}=1.0 \mathrm{s}$ is $\vec{\tau} .$ Which of the following statements is (are) true?

(A) $|\vec{\tau}|=\frac{1}{3} \mathrm{Nm}$

(B) The torque $\vec{\tau}$ is in the direction of the unit vector $+\hat{\mathrm{k}}$

(C) The velocity of the body at $\mathrm{t}=1 \mathrm{s}$ is $\overrightarrow{\mathrm{v}}=\frac{1}{2}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}) \mathrm{ms}^{-1}$

(D) The magnitude of displacement of the body at $t=1$ s is $\frac{1}{6} m$

Sol. (A,C)

Q. A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle 60° with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $(2-\sqrt{3}) / \sqrt{10} \mathrm{s}$, then the height of the top of the inclined plane, in meters, is _____. Take g = 10 $\mathrm{ms}^{-2}$.

Sol. 0.75

Q. In the List-I below, four different paths of a particle are given as functions of time. In these functions,  and  are positive constants of appropriate dimensions and . In each case, the force acting on the particle is either zero or conservative. In List–II, five physical quantities of the particle are mentioned; $\overrightarrow{\mathrm{p}}$ is the linear momentum $\overrightarrow{\mathrm{L}}$ is the angular momentum about the origin, K is the kinetic energy, U is the potential energy and E is the total energy. Match each path in List–I with those quantities in List–II, which are conserved for that path

Sol. (A)

Sound Waves – JEE Advanced Previous Year Questions with Solutions

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Photoelectric Effect – JEE Advanced Previous Year Questions with Solutions

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Simulator

Paragraph for Question Nos. 1 to 3

When a particle is restricted to move along x-axis between x = 0 and x = a, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as \mathrm{E}=\frac{\mathrm{p}^{2}}{2 \mathrm{m}}. Thus, the energy of the particle can be denoted by a quantum number ‘n’ taking values 1,2,3, … (n = l, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving in the line x = 0 to x = a.

Take h = 6.6 \times 10^{-34} \mathrm{Js} \text { and } \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}.

Q. The allowed energy for the particle for a particular value of n is proportional to

(A) $\mathrm{a}^{-2}$

(B) $\mathrm{a}^{-3 / 2}$

(C) $\mathrm{a}^{-1}$

$(\mathrm{D}) \mathrm{a}^{2}$

[JEE-2009]

Sol. (A)

Q. If the mass of the particle is m = $1.0 \times 10^{-30}$ kg and a = 6.6 nm, the energy of the particle in its ground state is closest to :

(A) 0.8 meV (B) 8 meV (C) 80 meV (D) 800 meV

[JEE-2009]

Sol. (B)

$\mathrm{E}=\frac{\mathrm{h}^{2}}{2 \mathrm{m} 4 \mathrm{a}^{2}} \Rightarrow \frac{\left(6.6 \times 10^{-34}\right)^{2}}{2\left(1.0 \times 10^{-30}\right) \times 4 \times\left(6.6 \times 10^{-9}\right)^{2}} \frac{1}{\mathrm{e}}$

E = 8 meV

Q. The speed of the particle, that can take discrete values, is proportional to

(A) $\mathrm{n}^{-3 / 2}$

$(\mathrm{B}) \mathrm{n}^{-1}$

(C) $\mathrm{n}^{1 / 2}$

(D) n

[JEE-2009]

Sol. (D)

$\mathrm{mv}=\frac{\mathrm{h}}{\lambda} \Rightarrow \mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \lambda} \Rightarrow \frac{\mathrm{hn}}{\mathrm{m}(2 \mathrm{a})} \Rightarrow \mathrm{v} \propto \mathrm{n}$

Q. An –particle and a proton are accelerated from rest by a potential difference of 100 V. After this, their de Broglie wavelengths are $\lambda_{\alpha}$ and $\lambda_{p}$ respectively. The ratio $\frac{\lambda_{\mathrm{p}}}{\lambda_{\alpha}}$, to the nearest integer, is ?

[JEE 2010]

Sol. 3

$\frac{\mathrm{h}}{\lambda}=\mathrm{p}=\sqrt{2 \mathrm{mqV}}$

$\Rightarrow \frac{\lambda_{\mathrm{p}}}{\lambda_{\alpha}}=\sqrt{\frac{\mathrm{m}_{\alpha} \mathrm{q}_{\alpha}}{\mathrm{m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}}}}=\sqrt{4 \times 2}=\sqrt{8} \approx 3$

Q. A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free-space. It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is $\mathrm{A} \times 10^{2}$ (where 1 < A < 10). The value of ‘Z’ is ?

[JEE 2011]

Sol. 7

$\frac{h c}{\lambda}-\phi=e V=e \frac{(N e) K}{R}$

$\left(\frac{1240}{200}-4.7\right) 1.6 \times 10^{-19}=\frac{N\left(1.6 \times 10^{-19}\right)^{2} 9 \times 10^{9}}{1 / 100}$

$\frac{15}{1.6} \times 10^{7}=N$

Q. A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is ?

(Take : The proton mass, $\mathrm{m}_{\mathrm{P}}=(5 / 3) \times 10^{-27} \mathrm{kg} ; \mathrm{h} / \mathrm{e}=4.2 \times 10^{-15} \mathrm{J} . \mathrm{s} / \mathrm{C} ; \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{m} /$

$\left.\mathrm{F} ; 1 \mathrm{fm}=10^{-15} \mathrm{m}\right)$

[JEE 2012]

Sol. 7

Q. The work functions of Silver and sodium are 4.6 and 2.3 eV, repetitively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is.

Sol. 1

$\mathrm{eV}_{\mathrm{s}}=\mathrm{hv}-\phi \Rightarrow \mathrm{V}_{\mathrm{s}}=\left(\frac{\mathrm{h}}{\mathrm{e}}\right) \mathrm{v}-\frac{\phi}{\mathrm{e}}$

Q. A metal surface is illuminated by light of two different wavelength 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are $\mathrm{u}_{1}$ and $\mathrm{u}_{2}$, respectively. If the ratio $\mathrm{u}_{1}: \mathrm{u}_{2}$ = 2 : 1 and hc = 1240 eV nm, the work function of the metal is nearly :-

(A) 3.7 eV (B) 3.2 eV (C) 2.8 eV (D) 2.5 eV

Sol. (A)

Q. Planck’s constant h, speed of light c and gravitational constant G are used to form a unit of length L and a unit of mass M. Then the correct option(s) is(are) :-

(A) $\mathrm{M} \propto \sqrt{\mathrm{c}}$

(B) $\mathrm{M} \propto \sqrt{\mathrm{G}}$

(C) $\mathrm{L} \propto \sqrt{\mathrm{h}}$

$(\mathrm{D}) \mathrm{L} \propto \sqrt{\mathrm{G}}$

Sol. (A,C,D)

Q. For photo-electric effect with incident photon wavelength $\lambda$, the stopping potential is $\mathrm{V}_{0}$. Identify the correct variation(s) of $\mathrm{V}_{0}$ with $\lambda$ and 1/$\lambda$.

Sol. (A,C)

Stopping potential

$\mathrm{eV}_{0}=\frac{\mathrm{hC}}{\lambda}-\phi$

hence $\mathrm{V}_{0}$ v/s $\lambda$ curve will be hyperbola and $\mathrm{V}_{0}$ v/s $\frac{1}{\lambda}$ curve will be straight line with slope

positive.

hence (A,C)

Q. An electron in an excited state of $\mathrm{Li}^{2+}$ ion has angular momentum 3h/2\pi. The de Broglie

wavelength of the electron in this state is p\pia, (where a is the Bohr radius). The value of $\mathrm{p}$ is?

Sol. 2

From Bohr’s law

$\operatorname{mvr}=\frac{\mathrm{nh}}{2 \pi}=\frac{3 \mathrm{h}}{2 \pi}(\text { from ques. })$

$\Rightarrow \mathrm{n}=3$

and momentum $=\mathrm{mv}=\frac{3 \mathrm{h}}{2 \pi \mathrm{r}}$

Now, radius of $\mathrm{n}^{\mathrm{th}}$ shell, $\mathrm{r}=\left(\frac{\mathrm{n}^{2}}{\mathrm{z}}\right) \mathrm{a}_{0}$

$\Rightarrow \mathrm{r}=\frac{(3)^{2}}{3} \cdot \mathrm{a}_{0} \quad\left[\because \mathrm{Z}_{\mathrm{Li}}=3\right]$

$\Rightarrow \mathrm{r}=3 \mathrm{a}_{0}$

wavelength $=\frac{\mathrm{h}}{\text { momentum }}$

$\Rightarrow \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\frac{3 \mathrm{h}}{2 \pi} \mathrm{r}}$

$\Rightarrow \lambda=\frac{2 \pi \mathrm{r}}{3}=\frac{2 \pi}{3} \times 3 \mathrm{a}_{0}$

$\lambda=2 \pi \mathrm{a}_{0}=\mathrm{p} \pi \mathrm{a}_{0}$

$\Rightarrow \mathrm{P}=2$

Q. In a historical experiment to determine Planck’s constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength ($\lambda$) of incident light and the corresponding stopping potential ($\mathrm{V}_{0}$) are given below:

Given that $c=3 \times 10^{8} \mathrm{ms}^{-1}$ and $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C},$ Planck’s constant (in units of $\mathrm{J}$ s) found

from such an experiment is :

(A) $6.0 \times 10^{-34}$

(B) $6.4 \times 10^{-34}$

(C) $6.6 \times 10^{-34}$

(D) $6.8 \times 10^{-34}$

Sol. (B)

$\mathrm{KE}_{\mathrm{max}}=\frac{\mathrm{hC}}{\lambda}-\phi$

$\mathrm{eV}_{\mathrm{s}}=\frac{\mathrm{hC}}{\lambda}-\phi$

Q. Light of wavelength $\lambda_{\mathrm{ph}}$ falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is $\phi$ and the anode is a wire mesh of conducting material kept at a distance d from the cathode. A potential difference V is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is $\lambda_{\mathrm{e}}$, which of the following statement(s) is(are) true ?

(A) For large potential difference $(\mathrm{V}>>\phi / \mathrm{e}), \lambda_{\mathrm{e}}$ is approximately halved if $\mathrm{V}$ is made four

times

(B) $\lambda_{\mathrm{e}}$ increases at the same rate as $\lambda_{\mathrm{ph}}$ for $\lambda_{\mathrm{ph}}<\mathrm{hc} / \phi$

(C) $\lambda_{\mathrm{e}}$ is approximately halved, if d is doubled

(D) $\lambda_{\mathrm{e}}$ decreases with increase in $\phi$ and $\lambda_{\mathrm{ph}}$

Sol. (A)

$\mathrm{K}_{\max }=\frac{\mathrm{hc}}{\lambda_{\mathrm{Ph}}}-\phi$

kinetic energy of $\mathrm{e}^{-}$ reaching the anode will be

$\mathrm{K}=\frac{\mathrm{hc}}{\lambda_{\mathrm{Ph}}}-\phi+\mathrm{eV}$

Now

$\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{m}\left(\frac{\mathrm{hc}}{\lambda_{\mathrm{Ph}}}-\phi+\mathrm{eV}\right)}}$

If $\mathrm{eV}>>\phi$

$\lambda_{e}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{m}\left(\frac{\mathrm{hc}}{\lambda_{\mathrm{ph}}}+\mathrm{eV}\right)}}$

If $\mathrm{V}_{\mathrm{f}}=4 \mathrm{V}_{\mathrm{i}}$

Q. A photoelectric material having work-function $\phi_{0}$ is illuminated with light of wavelength $\lambda$

$\left(\lambda<\frac{\mathrm{hc}}{\phi_{0}}\right) .$ The fastest photoelectron has a de-Broglie wavelength $\lambda_{\mathrm{d}}$. A change in wavelength

of the incident light by $\Delta \lambda$ results in a change $\Delta \lambda_{\mathrm{d}}$ in $\lambda_{\mathrm{d}} .$ Then the ratio $\Delta \lambda_{\mathrm{d}} / \Delta \lambda$ is proportional to :-

$(A) \lambda_{d}^{3} / \lambda^{2}$

(B) $\lambda_{\mathrm{d}}^{3} / \lambda$

(C) $\lambda_{\mathrm{d}}^{2} / \lambda^{2}$

(D) $\lambda_{d} / \lambda$

(A)

According to photo electric effect equation :

$\mathrm{KE}_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi_{0}$

$\frac{\mathrm{p}^{2}}{2 \mathrm{m}}=\frac{\mathrm{hc}}{\lambda}-\phi_{0}$

$\frac{\left(\mathrm{h} / \lambda_{\mathrm{d}}\right)^{2}}{2 \mathrm{m}}=\frac{\mathrm{hc}}{\lambda}-\phi_{0}$

Assuming small changes, differentiating both sides,

$\frac{\mathrm{h}^{2}}{2 \mathrm{m}}\left(-\frac{2 \mathrm{d} \lambda_{\mathrm{d}}}{\lambda_{\mathrm{d}}^{3}}\right)=-\frac{\mathrm{hc}}{\lambda^{2}} \mathrm{d} \lambda$

$\frac{\mathrm{d} \lambda_{\mathrm{d}}}{\mathrm{d} \lambda} \propto \frac{\lambda_{\mathrm{d}}^{3}}{\lambda^{2}}$

Sol.

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

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Simulator

Q. To determine the half life of a radioactive element, a student plots a graph of $\ell n\left|\frac{d N(t)}{d t}\right|$ versus t. Here $\frac{d N(t)}{d t}$ is the rate of radioactive decay at time t. If the number of radioactive nuclei of this element decreases by a factor of p after 4.16 years, the value of p is ?

[JEE- 2010]

Sol. 8

$\left|\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}\right|=\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda t} \therefore \operatorname{en}\left|\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}\right|=\ln \left(\mathrm{N}_{0} \lambda\right)-\lambda \mathrm{t}$

$\therefore \ln \left|\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}\right|=\ln \left(\mathrm{N}_{0} \lambda\right)-\lambda \mathrm{t} \Rightarrow-\lambda=\mathrm{slope}=-\frac{1}{2} \mathrm{year}^{-1}$

$\Rightarrow \lambda=\frac{1}{2} y r^{-1}$

$\therefore \mathrm{t}_{1 / 2}=\frac{0.693}{\lambda}=1.386$ given time is 3 times of $\mathrm{t}_{1 / 2}$

$\therefore$ value of $\mathrm{p}$ is 8

Q. The activity of a freshly prepared radioactive sample is $10^{10}$ disintegrations per second, whose mean life is $10^{9}$ s. The mass of an atom of this radioisotope is $10^{-25}$kg. The mass (in mg) of the radioactive sample is

[JEE-2011]

Sol. 1

$\mathrm{A}=\lambda \mathrm{N} \Rightarrow 10^{10}=\lambda \mathrm{N} \Rightarrow \mathrm{N}=\frac{10^{10}}{\lambda}=\left(10^{10}\right) \tau=10^{10} \times 10^{9}=10^{19}$

$\mathrm{M}=\mathrm{Nm}=\left(10^{19}\right)\left(10^{-25}\right)=10^{-6} \mathrm{kg}=1 \mathrm{mg}$

Paragraph for Questions 3 and 4

The -decay process, discovered around 1900, is basically the decay of a neutron (n). In the laboratory, a proton (p) and an electron $\left(\mathrm{e}^{-}\right)$ are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, $\mathrm{n} \rightarrow \mathrm{p}+\mathrm{e}^{-}+\vec{v}_{e}$ , around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $\left(\vec{v}_{e}\right)$ to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is $0.8 \times 10^{6} \mathrm{eV}$ . The kinetic energy carried by the proton is only the recoil energy.

Q. If the anti-neutrino had a mass of $3 \mathrm{eV} / \mathrm{c}^{2}$ (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K, of the electron?

(A) $0 \leq K \leq 0.8 \times 10^{6} e V$

(B) $3.0 e V \leq K \leq 0.8 \times 10^{6} e V$

(C) $3.0 e V \leq K<0.8 \times 10^{6} e V$

(D) $0 \leq K<0.8 \times 10^{6} e V$

[JEE 2012]

Sol. (D)

Q. A freshly prepared sample of a radioisotope of half-life 1386 s has activity $10^{3}$ distintegrations per second. Given that n 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after preparation of the sample is.

Sol. 4

Given $\mathrm{T}_{1 / 2}=1386 \mathrm{sec}: ;\left|\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}\right|=10^{3}$

fraction $=\frac{\mathrm{N}_{0}-\mathrm{N}}{\mathrm{N}_{0}}=\frac{\mathrm{N}_{0}-\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}}{\mathrm{N}_{0}}=1-\mathrm{e}^{-\lambda \mathrm{t}}$

Q. A nuclear power plant supplying electrical power to a village uses a radioactive material of half life T years as the fuel. The amount of fuel at the beginning is such that the total power requirement of the village is 12.5% of the electrical power available from the plant at that time. If the plant is able to meet the total power needs of the village for a maximum period of nT years, then the value of n is.

Sol. 3

$12.5 \% \rightarrow\left(\frac{1}{8}\right)^{\mathrm{th}}$ of initial power

$\because$ After each $\mathrm{T},$ P will be half

$\mathrm{P} \stackrel{\mathrm{T}}{\longrightarrow} \frac{\mathrm{P}}{2} \stackrel{\mathrm{P}}{\longrightarrow} \frac{\mathrm{P}}{\longrightarrow} \stackrel{\mathrm{T}}{\longrightarrow} \frac{\mathrm{P}}{8}$

$\therefore$ Total time $=3 \mathrm{T}$

n = 3

Q. For a radioactive material, its activity A and rate of change of its activity R are defined as $A=-\frac{d N}{d t}$ and $\mathrm{R}=-\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}$, where N(t) is the number of nuclei at time t. Two radioactive sources P (mean life $\tau$) and Q (mean life $2 \tau$) have the same activity at t = 0. Their rates of change of activities at t = $2 \tau$ are $\mathrm{R}_{\mathrm{p}}$ and $\mathrm{R}_{\mathrm{Q}}$, respectively. If $\frac{\mathrm{R}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{Q}}}=\frac{\mathrm{n}}{\mathrm{e}}$, then the value of n is :

Sol.

$\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$

$\mathrm{A}=\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$

$\mathrm{R}=\lambda^{2} \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$

$A_{P}=A_{Q}$ at $t=0$

$\lambda_{\mathrm{p}} \mathrm{N}_{\mathrm{p}} \mathrm{e}^{-\lambda \mathrm{Pt}}=\lambda_{0} \mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{Qt}}$ at $\mathrm{t}=0$

$\lambda_{\mathrm{p}} \mathrm{N}_{\mathrm{p}}=\lambda_{\mathrm{o}} \mathrm{N}_{\mathrm{O}} \quad \ldots$ (i)

$\frac{\mathrm{R}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{Q}}}=\left(\frac{\lambda_{\mathrm{P}}}{\lambda_{\mathrm{Q}}}\right)^{2}\left(\frac{\mathrm{N}_{\mathrm{P}}}{\mathrm{N}_{\mathrm{Q}}}\right) \frac{\mathrm{e}^{-\lambda_{\mathrm{P}} 2 \tau}}{\mathrm{e}^{-\lambda_{\mathrm{Q}} 2 \tau}}$

$\frac{\mathrm{R}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{Q}}}=\left(\frac{\lambda_{\mathrm{P}}}{\lambda_{\mathrm{Q}}}\right)^{2}\left(\frac{\mathrm{N}_{\mathrm{P}}}{\mathrm{N}_{\mathrm{Q}}}\right) \frac{1}{\mathrm{e}} \ldots$ (ii)

$\left[\lambda_{\mathrm{P}}=\frac{1}{\text { mean life }}\right]$

from equation (i) and equation (ii)

$\frac{\mathrm{R}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{Q}}}=\frac{2}{\mathrm{e}}$

Ans. 2

Q. An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half-life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use?

(A) 64 (B) 90 (C) 108 (D) 120

Sol.

Nuclear Physics – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

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Simulator

Paragraph for Question Nos. 1 to 3

Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, $_{1}^{2} \mathrm{H}$, known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The

D-D reaction is $_{1}^{2} \mathrm{H}+_{1}^{2} \mathrm{H} \rightarrow_{2}^{3} \mathrm{He}+\mathrm{n}+$ energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of $_{1}^{2} \mathrm{H}$ nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time $\mathrm{t}_{0}$ before the particles fly away from the core. If n is the density (number/volume) of deuterons, the product $\mathrm{nt}_{0}$ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than $5 \times 10^{14} \mathrm{s} / \mathrm{cm}^{3}$. It may be helpful to use the following :

Boltzmann constant $\mathrm{k}=8.6 \times 10^{-5} \mathrm{eV} / \mathrm{K} ; \frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}=1.44 \times 10^{-9} \mathrm{eVm}$

Q. In the core of nuclear fusion reactor, the gas becomes plasma because of –

(A) strong nuclear force acting between the deuterons

(B) Coulomb force acting between the deuterons

(C) Coulomb force acting between deuteron-electron pairs

(D) the high temperature maintained inside the reactor core

[JEE-2009]

Sol. (D)

Due to the high temperature developed as a result of collision & fusion causes the core of fusion reactor to plasma.

Q. Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy 1.5 kT, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature T required for them to reach a separation of 4 × $10^{-15}$ m is in the range

(A) $1.0 \times 10^{9} \mathrm{K}<\mathrm{T}<2.0 \times 10^{9} \mathrm{K}$

(B) $2.0 \times 10^{9} \mathrm{K}<\mathrm{T}<3.0 \times 10^{9} \mathrm{K}$

(C) $3.0 \times 10^{9} \mathrm{K}<\mathrm{T}<4.0 \times 10^{9} \mathrm{K}$

(D) $4.0 \times 10^{9} \mathrm{K}<\mathrm{T}<5.0 \times 10^{9} \mathrm{K}$

[JEE-2009]

Sol. (A) $\left(\frac{3}{2} \mathrm{KT}\right) 2=\frac{\mathrm{Kq}_{1} \mathrm{q}_{2}}{\mathrm{r}} 3 \mathrm{KT}=\frac{\mathrm{Kq}_{1} \mathrm{q}_{2}}{\mathrm{r}}$

$\mathrm{T}=\frac{1.44 \times 10^{-9}}{4 \times 10^{-15} \times 3 \times 8.6 \times 10^{-5}}$

Q. Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which of these is most promising based on Lawson criterion?

(A) deuteron density $=2.0 \times 10^{12} \mathrm{cm}^{-3},$ confinement

time $=5.0 \times 10^{-3} \mathrm{s}$

(B) deuteron density $=8.0 \times 10^{14} \mathrm{cm}^{-3},$ confinement time $=9.0 \times 10^{-1} \mathrm{s}$

(C) deuteron density $=4.0 \times 10^{23} \mathrm{cm}^{-3},$ confinement time $=1.0 \times 10^{-11} \mathrm{s}$

(D) deuteron density $=1.0 \times 10^{24} \mathrm{cm}^{-3},$ confinement time $=4.0 \mathrm{x} 10^{-12} \mathrm{s}$

[JEE-2009]

Sol. (B)

Q. What is the maximum energy of the anti-neutrino?

(A) zero

(B) much less than $0.8 \times 10^{6} \mathrm{eV}$

(C) Nearly $0.8 \times 10^{6} \mathrm{eV}$

(D) Much larger than $0.8 \times 10^{6} \mathrm{eV}$

[JEE 2012]

Sol. (C)

Paragraph for Questions 5 and 6

The mass of a nucleus $_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}$ is less than the sum of the masses of (A – Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ only if $\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)$ < M. Also two light nuclei of masses $\mathrm{m}_{3}$ and $\mathrm{m}_{4}$ can undergo complete fusion and form a heavy nucleus of mass M’ only if $\left(\mathrm{m}_{3}+\mathrm{m}_{4}\right)$ > M’. The masses of some neutral atoms are given in the table below :-

Q. The kinetic energy (in keV) of the alpha particle, when the nucleus 210

84 $P O$ at rest undergoes alpha decay, is :-

(A) 5319 (B) 5422 (C) 5707 (D) 5818

Sol. (A)

Q. The correct statement is :-

(A) The nucleus $_{3}^{6}$ Li can emit an alpha particle

(B) The nucleus $_{84}^{210}$ Po can emit a proton

(C) Deuteron and alpha particle can undergo complete fusion

(D) The nuclei $_{30}^{70} \mathrm{Zn}$ and $_{34}^{82} \mathrm{Se}$ can undergo complete fusion

Sol. (C)

Q. Match List I of the nuclear processes with List II containing parent nucleus and one of the end products of each process and then select the correct answer using the codes given below the lists:

Sol. (C)

Q. A fission reaction is given by $_{92}^{236} \mathrm{U} \rightarrow_{54}^{140} \mathrm{Xe}+_{38}^{94} \mathrm{Sr}+\mathrm{x}+\mathrm{y},$ where $\mathrm{x}$ and $\mathrm{y}$ are two particles.

Considering $_{92}^{236} \mathrm{U}$ to be at rest, the kinetic energies of the products are denoted by

$\mathrm{K}_{\mathrm{Xe}}, \mathrm{K}_{\mathrm{sr}}, \mathrm{K}_{\mathrm{x}}(2 \mathrm{MeV})$ and $\mathrm{K}_{\mathrm{v}}(2 \mathrm{MeV}),$ respectively. Let the binding energies per nucleon of

Sol. (A)

Q. The isotope $_{5}^{12} \mathrm{B}$ having a mass 12.014 u undergoes $\beta$ -decay to $_{6}^{12} \mathrm{C} ._{6}^{12} \mathrm{C}$ has an excited state of

the nucleus $\left(_{6}^{12} \mathrm{C}^{*}\right)$ at $4.041 \mathrm{MeV}$ above its ground state. If 12

5 $\mathrm{decays}$ to $_{6}^{12} \mathrm{C}^{*},$ the maximum

kinetic energy of the $\beta$ -particle in units of MeV is $\left(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2}\right.$, where c is the speed of light in vacuum).

Sol. 9

Q. The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is given by $\mathrm{E}=\frac{3}{5} \frac{\mathrm{Z}(\mathrm{Z}-1) \mathrm{e}^{2}}{4 \pi \varepsilon_{0} \mathrm{R}}$

(A) 2.85 fm (B) 3.03 fm (C) 3.42 fm (D) 3.80 fm

Sol. (C)

Magnetic Effect of Current – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Simulator

Q. Six point charges, each of the same magnitude q, are arranged in different manners as shown in Column II. In each case, a point M and a line PQ passing through M are shown. Let E be the electric field and V be the electric potential at.M (poteptial at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let B be the magnetic field at M and µ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current.

(t) Charges are placed on two coplanar, identical insulating rings at equal intervals. M is the mid-point btween the centres of the rings.

PQ is perpendicular to the line joining the centres and coplanar to the rings.

[JEE-2009]

Sol. ($(\mathrm{A}) \rightarrow \mathrm{p}, \mathrm{r}, \mathrm{s}(\mathrm{B}) \rightarrow \mathrm{r}, \mathrm{s}(\mathrm{C}) \rightarrow \mathrm{p}, \mathrm{q}, \mathrm{t}(\mathrm{D}) \rightarrow \mathrm{r}, \mathrm{s}$)

Q. Column II shows five systems in which two objects are labelled as X and Y. Also in each case a point P is shown. Column I gives some statements about X and/or Y. Match these statements to the appropriate system(s) from Column II

[JEE-2009]

Sol. ($(A) \rightarrow p, t(B) \rightarrow q, s, t(C) \rightarrow p, r, t(D) \rightarrow q$)

(p)

(A) Total contact force = $\sqrt{\mathrm{N}^{2}+\mathrm{f}^{2}}$ = mg

(B) X is fixed therefore P.E. remain same

(C) Potential energy of ‘Y’ is decreasing as it slide down the plane and K.E. is not increasing

(D) Toque of weight of y is

$\tau=\operatorname{mg} \ell \cos \theta \Rightarrow \tau \neq 0$

(q)

(A) Force due to both ‘X’ and ‘Z’ on ‘Y’ is equal to mg

(B) As the system is moving up P.E. of system is increasing

(C) Mechanical energy of total system is increasing

(D) As the force passes through ‘p’ torque about p is zero

(r)

(A) Force on ‘X’ due to Y

$\mathrm{F}_{\mathrm{net}}=\sqrt{2} \mathrm{T} \Rightarrow \sqrt{2} \mathrm{mg}$

(B) System is moving down therefore P.E. is decreasing

(C) System is moving down with constant speed with constant speed therefore machanical energy is decreasing

(D) $\tau=\operatorname{mg} \ell \Rightarrow \tau \neq 0$

(s)

(A) Force exerted by (X) on ‘Y’ is less than mg and its magnitude $\mathrm{f}_{\mathrm{B}}=\mathrm{Vd}_{\mathrm{t}} \mathrm{g}$

(B) As the ball moves down it displaces the water to its higher position hence P.E. of water increases

(C) As no desssipitative force acts on it mechanical energy is conserved

(D) Torque is non-zero about p

(t) Similar to ‘s’ but it is viscous fluid therefore total mechanical energy decreases with time.

Q. A steady current I goes through a wire loop PQR having shape of a right angle triangle with PQ = 3x, PR = 4x and QR = 5x. If the magnitude of the magnetic field at P due to this loop is $\left(\frac{\mu_{0} I}{48 \pi x}\right)$, find the value of k.

[2009, 6M]

Sol. 7

Q. A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is :

(A) IBL (B) $\frac{I B L}{\pi}$ (C) $\frac{I B L}{2 \pi}$ $(\mathrm{D}) \frac{I B L}{4 \pi}$

[2010, 3M]

Sol. (C)

Comprehension Type

Electrical resistance of certain materials, known as superconductors, changes abruptly from a nonzero value to zero as their temperature is lowered below a critical temperature $\mathrm{T}_{\mathrm{C}}$ (0). An interesting property of superconductors is that their critical temperature becomes smaller than $\mathrm{T}_{\mathrm{C}}$ (0) if they are placed in a magnetic field, i.e., the critical temperature $\mathrm{T}_{\mathrm{C}}$ (B) is a function of the magnetic field strength B. The dependence of $\mathrm{T}_{\mathrm{C}}$ (B) on B is shown in the figure.

Q. In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields $\mathrm{B}_{1}$ (sold line) and $\mathrm{B}_{2}$ (dashed line). If $\mathrm{B}_{2}$ is larger than $\mathrm{B}_{1}$, which of the following graphs shows the correct variation of R with T in these fields?

[JEE2010]

Sol. (A)

If $\mathrm{B}_{2}>\mathrm{B}_{1}$, critical temperature, (at which resistance of semiconductors abrupty becomes zero) in case 2 will be less than compared to case 1.

Q. A superconductor has $\mathrm{T}_{\mathrm{C}}$ (0) = 100 K. When a magnetic field of 7.5 Tesla is applied, its $\mathrm{T}_{\mathrm{C}}$ decreases to 75 K. For this material one can definitely say that when

(A) $\mathrm{B}=5$ Tesla, $\mathrm{T}_{\mathrm{C}}(\mathrm{B})=80 \mathrm{K}$

(B) $\mathrm{B}=5$ Tesla, $75 \mathrm{K}<\mathrm{T}_{\mathrm{C}}(\mathrm{B})<100 \mathrm{K}$

(C) $\mathrm{B}=10$ Tesla, $75 \mathrm{K}<\mathrm{T}_{\mathrm{c}}(\mathrm{B})<100 \mathrm{K}$

(D) $\mathrm{B}=10$ Tesla, $\mathrm{T}_{\mathrm{c}}(\mathrm{B})=70 \mathrm{K}$

[JEE2010]

Sol. (B)

With increase in temperature, TC is decreasing.

$\mathrm{T}_{\mathrm{c}}$ (0) = 100 K

$\mathrm{T}_{\mathrm{c}}$ = 75 K at B = 7.5 T

Hence, at B = 5T, $\mathrm{T}_{\mathrm{c}}$ should line between 75 K and 100 K

Q. An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi-infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement(s) is/are true?

(A) they will never come out of the magnetic field region

(B) they will come out travelling along parallel paths

(C) they will come out of the same time

(D) they will come out at different times

[IIT-JEE 2011]

Sol. (B,D)

By diagram B is true.

Q. A long insulated copper wire is closely wound as a spiral of ‘N’ turns. The spiral has inner radius ‘a’ and outer radius ‘b’. The spiral lies in the X-Y plane and a steady current ‘I’ flows through the wire. The Z-component of the magnetic field at the center of the spiral is

[IIT-JEE 2011]

Sol. (A)

Taking an elemental strip of radius x and width dx.

Area of strip = $2 \pi \mathrm{x}$ dx

Number of turns through area $=\frac{N}{b-a} d x$

$\int d B=\int_{a}^{b} \frac{\mu_{0} \frac{N}{(b-a)} I d x}{2 x}=\frac{\mu_{0} N I \ell n\left(\frac{b}{a}\right)}{2(b-a)} |$

Q. Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields $\vec{E}=E_{0} \hat{j}$ and $\vec{B}=B_{0} \hat{j}$ and . At time t =0, this charge has velocity in the x-y plane, making an angle $\theta$ with the x-axis. Which of the following option (s) is (are) correct for time t > 0?

(A) If $\theta=0^{\circ},$ the charge moves in a circular path in the $\mathrm{x}-\mathrm{z}$ plane.

(B) If $\theta=0^{\circ},$ the charge undergoes helical motion with constant pitch along the y-axis.

(C) If $\theta=10^{\circ},$ the charge undergoes helical motion with its pitch increasing with time, along the

$y$ -axis

(D) If $\theta=90^{\circ},$ the charge undergoes linear but accelerated motion along the $\mathrm{y}$ -axis

[IIT-JEE 2012]

Sol. (C,D)

Q. A cylindrical cavity of diameter a exists inside a cylinder of diameter

2a as shown in the figure. Both the cylinder and the cavity are infinitely long. A uniform current density J flows along the length. If the magnitude of the magnetic field at the point P is given by $\frac{N}{12} \mu_{0} a J$, then the value of N is

[IIT-JEE 2012]

Sol. (5)

Q. An infinitely long hollow conducting cylinder with inner radius R/2 and outer radius R carries a uniform current density along its length. The magnitude of the magnetic field, $|\vec{B}|$ as a function of the radial distance r from the axis is best represented by

[IIT-JEE 2012]

Sol. (D)

Q. A particle of mass M and positive charge Q, moving with a constant velocity $\overrightarrow{\mathrm{u}}_{1}=4 \hat{\mathrm{i}} \mathrm{ms}^{-1}$, enters a region of uniform static magnetic field normal to the x-y plane. The region of the magnetic field extends from x = 0 to x = L from all values of y. After passing through this region, the particle emerges on the other side after 10 milliseconds with a velocity $\overrightarrow{\mathrm{u}}_{2}=2(\sqrt{3 \hat{\mathrm{i}}}+\hat{\mathrm{j}}) \mathrm{ms}^{-1}$. The correct statements(s) is (are) :-

(A) The direction of the magnetic field is –z direction.

(B) The direction of the magnetic field is +z direction.

(C) The magnitude of the magnetic field $\frac{50 \pi \mathrm{M}}{3 \mathrm{Q}}$ units.

(D) The magnitude of the magnetic field is $\frac{100 \pi \mathrm{M}}{3 \mathrm{Q}}$ units.

[IIT-JEE 2013]

Sol. (A,C)

by direction of $\overrightarrow{\mathrm{F}} ;$ magnetic field is in $-\mathrm{z}$ direction

Time $=\frac{\theta}{\omega}=\frac{\pi / 6}{\mathrm{QB} / \mathrm{M}}=\frac{\mathrm{M} \pi}{6 \mathrm{QB}}$

$\Rightarrow \mathrm{B}=\frac{\mathrm{M} \pi}{6 \mathrm{Q}\left(10 \times 10^{-3}\right)}=\frac{50 \pi \mathrm{M}}{3 \mathrm{Q}}$

Q. A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is placed coaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carries a steady current I. Consider a point P at a distance r from the common axis. The correct statement(s) is (are) :-

(A) In the region 0 < r < R, the magnetic field is non-zero

(B) In the region R < r < 2R, the magnetic field is along the common axis

(C) In the region R < r < 2R, the magnetic field is tangential to the circle of radius r, centredon the axis

(D) In the region r > 2R, the magnetic field is non-zero

Sol. (A,D)

Q. Two parallel wires in the plane of the paper are distance $\mathrm{X}_{0}$ apart. A point charge is moving with speed u between the wires in the same plane at a distance $\mathbf{X}_{1}$ from one of the wires. When the wires carry current of magnitude I in the same direction, the radius of curvature of the path of the point charge is $\mathrm{R}_{1}$. In contrast, if the currents I in the two wires have directions opposite to each other, the radius of curvature of the path is $\mathrm{R}_{2}$. If $\frac{X_{0}}{X_{1}}=3$ , and value of $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}$ is ?

Sol. 3

Paragraph for Questions 15 & 16

The figure shows a circular loop of radius a with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the loop is d. The loop and the wires are carrying the same current I. The current in the loop is in the counterclockwise direction if seen from above.

Q. When d $\approx$ a but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height h above the loop. In that case

(A) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h $\approx$ a

(B) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h $\approx$ a

(C) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h $\approx$ 1.2 a

(D) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h $\approx$ 1.2 a

Sol. (C)

$\mathrm{B}_{\text {wire }}=\frac{\mu_{0} \mathrm{i}}{2 \pi \sqrt{\mathrm{a}^{2}+\mathrm{h}^{2}}} 2 \sin \theta$

$=\frac{\mu_{0} \mathrm{ia}}{\pi\left(\mathrm{a}^{2}+\mathrm{h}^{2}\right)}$

$\mathrm{B}_{\text {loop }}=\frac{\mu_{0} \mathrm{ia}^{2}}{2\left(\mathrm{a}^{2}+\mathrm{h}^{2}\right)^{3 / 2}}$

$\mathrm{B}_{\text {wire }}=\mathrm{B}_{\text {loop }}$

$\frac{\mu_{0} \mathrm{ia}}{\pi\left(\mathrm{a}^{2}+\mathrm{h}^{2}\right)}=\frac{\mu_{0} \mathrm{i} \mathrm{a}^{2}}{2\left(\mathrm{a}^{2}+\mathrm{h}^{2}\right)^{3 / 2}}$

$\sqrt{a^{2}+h^{2}}=\frac{\pi a}{2}$

$a^{2}+h^{2}=2.5 \mathrm{a}^{2}$

$\mathrm{h}^{2}=1.5 \mathrm{a}^{2}$

$\mathrm{h} \approx 1.2 \mathrm{a}$

Q. Consider d >> a, and the loop is rotated about its diameter parallel to the wires by $30^{\circ}$ from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop)

(A) $\frac{\mu_{0} \mathrm{I}^{2} \mathrm{a}^{2}}{\mathrm{d}}$

(B) $\frac{\mu_{0} I^{2} a^{2}}{2 d}$

(C) $\frac{\sqrt{3} \mu_{0} \mathrm{I}^{2} \mathrm{a}^{2}}{\mathrm{d}}$

(D) $\frac{\sqrt{3} \mu_{0} \mathrm{I}^{2} \mathrm{a}^{2}}{2 \mathrm{d}}$

Sol. (B)

$|\vec{\tau}|=|\vec{M} \times \vec{B}|=M B \sin 30^{\circ}$

where $\mathrm{M}=\left(\pi \mathrm{a}^{2}\right) \mathrm{I}$ and $\mathrm{B}=2 \times \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{d}}=\frac{\mu_{0} \mathrm{I}}{\pi \mathrm{d}}$ Therefore

$\tau=\left(\frac{\mu_{0} \mathrm{I}}{\pi \mathrm{d}}\right)\left(\pi \mathrm{a}^{2} \mathrm{I}\right)\left(\frac{1}{2}\right)=\frac{\mu_{0} \mathrm{I}^{2} \mathrm{a}^{2}}{2 \mathrm{d}}$

Q. A conductor (shown in the figure) carrying constant current I is kept in the x-y plane in a unfirom magnetic field $\overrightarrow{\mathrm{B}}$. If F is the magnitude of the total magnetic force acting on the conductor,

(A) If $\overrightarrow{\mathrm{B}}$ is along $\hat{\mathrm{z}}, \mathrm{F} \propto(\mathrm{L}+\mathrm{R})$

(B) If $\overrightarrow{\mathrm{B}}$ is along $\hat{\mathrm{x}}, \mathrm{F}=0$

(C) If $\overrightarrow{\mathrm{B}}$ is along $\hat{\mathrm{y}}, \mathrm{F} \propto(\mathrm{L}+\mathrm{R})$

(D) If $\overrightarrow{\mathrm{B}}$ is along $\hat{\mathrm{z}}, \mathrm{F}=0$

Sol. (A,B,C)

A charged particle (electron or proton) is introduced at the origin (x = 0, y = 0, z = 0) with a given initial velocity $\overrightarrow{\mathbf{v}}$. A uniform electric field $\overrightarrow{\mathrm{E}}$ and a uniform magnetic field $\overrightarrow{\mathrm{B}}$ exist everywhere. The velocity $\overrightarrow{\mathbf{V}}$, electric field $\overrightarrow{\mathrm{E}}$ and magnetic field $\overrightarrow{\mathrm{B}}$ are given in column 1, 2 and 3, respectively. The quantities $E_{0}, B_{0}$ are positive in magnitude.

Q. In which case will the particle move in a straight line with constant velocity ?

(A) (II) (iii) (S) (B) (IV) (i) (S) (C) (III) (ii) (R) (D) (III) (iii) (P)

Sol. (A)

Q. In which case will the particle describe a helical path with axis along the positive z-direction ?

(A) (II) (ii) (R) (B) (IV) (ii) (R) (C) (IV) (i) (S) (D) (III) (iii) (P)

Sol. (C)

For path to be helix with axis along +ve z-direction, particle should experience a centripetal acceleration in x-y plane.

For the given set of options only option (C) satisfy the condition. Path is helical with increasing pitch.

Q. In which case would the particle move in a straight line along the negative direction of y-axis (i.e., move along $-\hat{y}$) ?

(A) (II) (iii) (S) (B) (IV) (i) (S) (C) (III) (ii) (R) (D) (III) (iii) (P)

Sol. (D)

For particle to move in -ve y-direction, either its velocity must be in –ve y-direction (if initial velocity $\neq$ 0) & force should be parallel to velocity or it must experience a net force in –ve y-direction only (if initial velocity = 0)

Q. A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The distance between the diametrically opposite vertices of the star is 4a. The magnitude of the magnetic field at the center of the loop is :

Sol. (B)

The given points (1, 2, 3, 4, 5, 6) makes $360^{\circ}$ angle at ‘O’. Hence angle made by vertices 1 & 2 with ‘O’ is $60^{\circ}$.

Direction of magnetic field at ‘O’ due to each segment is same. Since it is symmetric star shape, magnitude will also be same.

Magnetic field due to section BC.

$\left(\mathrm{B}_{1}\right)=\frac{\mathrm{k} \mathrm{i}}{\mathrm{a}}(\sin (+60)-\sin 30)=\frac{\mathrm{ki}}{2 \mathrm{a}}(\sqrt{3}-1)$

$\mathrm{B}_{\mathrm{net}}=12 \times \mathrm{B}_{1}=\frac{6 \mathrm{ki}}{\mathrm{a}}(\sqrt{3}-1) \&\left(\mathrm{k}=\frac{\mu_{0}}{4 \pi}\right)$

Q. A uniform magnetic field B exists in the region between x = 0 and x = $\frac{3 \mathrm{R}}{2}$ (region 2 in the figure) pointing normally into the plane of the paper. A particle with charge +Q and momentum p directed along x-axis enters region 2 from region 1 at point $P_{1}$ (y = –R). Which of the following options(s) is/are correct ?

(A) For $\mathrm{B}=\frac{8}{13} \frac{\mathrm{p}}{\mathrm{QR}}$, the particle will enter region 3 through the point $P_{2}$ on x-axis

(B) For $\mathrm{B}>\frac{2}{3} \frac{\mathrm{p}}{\mathrm{QR}}$, the particle will re-enter region 1

(C) For a fixed B, particle of same charge Q and same velocity v, the distance between the point $P_{1}$ and the point of re-entry into region 1 is inversely proportional to the mass of the particle.

(D) When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the chage in its linear momentum between point $P_{1}$ and the farthest point from y-axis is $\frac{\mathrm{p}}{\sqrt{2}}$.

Sol. (A,B)

For $\mathrm{B}=\fra c{8}{13} \frac{\mathrm{p}}{\mathrm{QR}},$ radius of path

$\mathrm{R}^{\prime}=\frac{\mathrm{p}}{\mathrm{Q.B}}=\frac{\mathrm{p} \times 13 \mathrm{QR}}{\mathrm{Q} 8 \mathrm{p}}=\frac{13}{8} \mathrm{R}$

using pythagorous theorem, $\mathrm{y}=\frac{5 \mathrm{R}}{8}$

$\therefore$ particle will enter region 3 through point $\mathrm{P}_{2}$

for $\mathrm{B}>\frac{2}{3} \frac{\mathrm{p}}{\mathrm{QR}}$

Radius of path $<\frac{3 \mathrm{PQR}}{\mathrm{Q} 2 \mathrm{p}}=\frac{3}{2} \mathrm{R}$

$\therefore$ Particle will not enter in region $3 \&$ will re-enter region 1

charge in momentum = $\sqrt{2} \mathrm{p}$. When particle enters region 1 between entry point & farthest point from y-axis.

Q. Two infinitely long straight wires lie in the xy-plane along the lines x = $\pm R$. The wire located at x = +R carries a constant current $\mathbf{I}_{1}$ and the wire located at x = –R carries a constant current $\mathrm{I}_{2}$. A circular loop of radius R is suspended with its centre at $(0,0, \sqrt{3} \mathrm{R})$ and in a plane parallel to the xy-plane. This loop carries a constant current I in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the $+\hat{j}$ direction. Which of the following statements regarding the magnetic field $\overrightarrow{\mathrm{B}}$ is (are) true ?

(A) If $\mathbf{I}_{1}$ = $\mathbf{I}_{2}$, then cannot be equal to zero at the origin (0, 0, 0)

(B) If $\mathbf{I}_{1}$ > 0 and $\mathbf{I}_{2}$ < 0, then can be equal to zero at the origin (0, 0, 0)

(C) If $\mathbf{I}_{1}$ < 0 and $\mathbf{I}_{2}$ > 0, then can be equal to zero at the origin (0, 0, 0)

(D) If $\mathbf{I}_{1}$ = $\mathbf{I}_{2}$, then the z-component of the magnetic field at the centre of the loop is $\left(-\frac{\mu_{0} I}{2 R}\right)$

Sol. (A,B,D)

Q. In the x-y-plane, the region y > 0 has a uniform magnetic field $B_{1} \hat{k}$ and the region y < 0 has a another uniform magnetic field $B_{2} \hat{k}$. A positively charged particle is projected from the origin along the positive y-axis with speed $\mathrm{v}_{0}=\pi \mathrm{ms}^{-1}$ at $\mathrm{t}=0$ as shown in the figure. Neglect gravity in this problem. Let t = T be the time when the particle crosses the x-axis from below for the first time. If $\mathrm{B}_{2}=4 \mathrm{B}_{1}$, the average speed of the particle, in $\mathrm{ms}^{-1}$, along the x-axis in the time interval T is _______.

Sol. 2

(1) Average speed along x-axis

Q. A moving coil galvanometer has 50 turns and each turn has an area $2 \times 10^{-4} \mathrm{m}^{2}$. The magnetic field porduced by the magnet inside the galvanometer is 0.02 T. The torsional constant of the suspension wire is $10^{-4} \mathrm{N} \mathrm{m} \mathrm{rad}^{-1}$. When a current flows through the galvanometer, a full scale deflection occurs if the coil rotates by 0.2 rad. The resistance of the coil of the galvanometer is 50 $\Omega$. This galvanometer is to be converted into an ammeter capable of measuring current in the range 0 – 1.0 A. For this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance, in ohms, is………..

Sol. 5.55

Kinetic Theory of Gases – JEE Advanced Previous Year Questions with Solutions

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Simulator

Q. A real gas behaves like an ideal gas if its ?

(A) pressure and temperature are both high

(B) pressure and temperature are both low

(C) pressure is high and temperature is low

(D) pressure is low and temperature is high

Sol. (D)

$\operatorname{High} \vec{P},$ Low $P$

Q. A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic

mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds $\left(\frac{v_{r m s}(\text {helium})}{v_{r m s}(\text {arg on})}\right)$ is ?

(A) 0.32         (B) 0.45            (C) 2.24              (D) 3.16

[JEE-2012]

Sol. (D)

$v_{m s}=\sqrt{\frac{3 R T}{M}} ; \frac{\left(v_{m s}\right)_{H e}}{\left(v_{m s}\right)_{A r}}=\sqrt{\frac{M_{A r}}{M_{H e}}}=\sqrt{\frac{40}{4}}=\sqrt{10}=3.16$

Q. Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is :-

(A) 1 : 4          (B) 1 : 2           (C) 6 : 9            (D) 8 : 9

[JEE-2013]

Sol. (D)

$\mathrm{P}=\frac{\rho}{\mathrm{M}} \mathrm{RT}$

$\Rightarrow \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\rho_{1} \mathrm{M}_{2}}{\rho_{2} \mathrm{M}_{1}}=\frac{4}{3} \Rightarrow \frac{\rho_{1}}{\rho_{2}}=\frac{4}{3} \times \frac{2}{3}=\frac{8}{9}$

Q. A container of fixed volume has a mixutre of one mole of hydrogen and one mole of helium in equilibrium at temperature T. Assuming the gases are ideal, the correct statement(s) is (are) :-

(A) The average energy per mole of the gas mixture is 2RT.

(B) The ratio of speed of sound in the gas mixture to that in helium gas is $\sqrt{6 / 5}$.

(C) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/2.

(D) The ratio of the rms speed of helium atoms to that of hydrogen molecules is $1 / \sqrt{2}$.

Sol. (A,B,D)

$C_{V(\operatorname{mix})}=\frac{(1)\left(\frac{3}{2} R\right)+(1)\left(\frac{5}{2} R\right)}{2}=2 R$

$\mathrm{C}_{\mathrm{P}(\mathrm{mix})}=3 \mathrm{R} \quad \gamma_{\operatorname{mix}}=\frac{3}{2} \Rightarrow \mathrm{f}=4$

Average energy/mole $=\mathrm{f} \frac{1}{2} \mathrm{RT}=2 \mathrm{RT}$

$\frac{\left(\mathrm{V}_{\mathrm{sound}}\right)_{\mathrm{mixture}}}{\left(\mathrm{V}_{\mathrm{sound}}\right)_{\mathrm{He}}}=\frac{\sqrt{\frac{\mathrm{RT}}{2}}}{\sqrt{\frac{5 \mathrm{RT}}{12}}}=\sqrt{\frac{6}{5}}$

$\frac{\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{He}}}{\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}}=\frac{\sqrt{\frac{3 \mathrm{RT}}{4}}}{\sqrt{\frac{3 \mathrm{RT}}{2}}}=\frac{1}{\sqrt{2}}$

$\therefore(\mathrm{A}, \mathrm{B}, \mathrm{D})$

Alternating Current – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Q. You are given many resistances, capacitors and inductors. These are connected to a variable DC voltage source (the first two circuits) or an AC voltage source of 50 Hz frequency (the next three circuits) in different ways as shown in Column II. When a current I (steady state for DC or rms for AC) flows through the circuit, the corresponding voltage $\mathrm{V}_{1}$ and $\mathrm{V}_{2}$ (indicated in circuits) are related as shown in Column I. Match the two

[JEE 2010]

Sol. (A) RST (B) QRST (C) PQ (D) QRST

Q. An AC voltage source of variable angular frequency  and fixed amplitude $\mathrm{V}_{0}$ is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When $\omega$ is increased

(A) the bulb glows dimmer

(B) the bulb glows brighter

(C) total impedance of the circuit is unchanged

(D) total impedance of the circuit increases

[JEE 2010]

Sol. (B)

Q. A series R-C circuit is connected to AC voltage source. Consider two cases ; (A) when C is without a dielectric medium and (B) when C is filled with dielectric of constant 4. The current $\mathrm{I}_{\mathrm{R}}$ through the resistor and voltage $\mathrm{V}_{\mathrm{c}}$ across the capacitor are compared in the two cases. Which of the following is/are true?

(A) $I_{R}^{A}>I_{R}^{B}$

$(\mathrm{B}) \quad I_{R}^{A}<I_{R}^{B}$

(C) $V_{c}^{A}>V_{c}^{B}$

(D) $V_{C}^{A}<V_{C}^{B}$

[JEE 2011]

Sol. (B,C)

$\mathrm{x}_{\mathrm{c}}$ decreases therefore impedence decreases and current increases. $\mathrm{I}_{\mathrm{B}}>\mathrm{I}_{\mathrm{A}}$

As $\mathrm{I}_{\mathrm{B}}$ increases the voltage across ‘R’ increases therefore $\mathrm{V}_{\mathrm{c}}$ decreases.

Q. A series R-C combination is connected to an AC voltage of angular frequency $\omega=500$radian/s. If the impedance of the R-C circuit is $R \sqrt{1.25}$, the time constant (in millisecond) of the circuit is ?

[JEE 2011]

Sol. 4

$1.25 \mathrm{R}^{2}=\mathrm{R}^{2}+\left(\frac{1}{\omega c}\right)^{2}$

$0.25 \mathrm{R}^{2}=\left(\frac{1}{\omega c}\right)^{2} ; 0.5 \mathrm{R}=\frac{1}{500 \times C} ; \mathrm{C}=\frac{1}{250 R} ; \mathrm{RC}=\frac{1}{250} \mathrm{sec}$

$\tau=4$ millisecond; $\tau=4$

Q. In the given circuit, the AC source has $\omega=100 \mathrm{rad} / \mathrm{s}$. Considering the inductor and capacitor to be ideal, the correct choice (s) is(are)

(A) The current through the circuit, I is $0.3 \mathrm{A}$.

(B) The current through the circuit, is $0.3 \sqrt{2 \mathrm{A}}$

(C) The voltage across $100 \Omega$ resistor $=10 \sqrt{2} \mathrm{V}$

(D) The voltage across $50 \Omega$ resistor $=10 \mathrm{V}$

[JEE 2012]

Sol. (C or A,C)

Paragraph for Questions 6 and 7

A thermal power plant produces electric power of 600 kW and 4000 V, which is to be transported to a place 20 km away from the power plant for consumers’ usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers’ end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with a power factor unity. All the currents and voltages mentioned are rms values.

Q. In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1 : 10. If the power to the consumers has to be supplied at 200 V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is

(A) 200 : 1

(B) 150 : 1

(C) 100 : 1

(D) 50 : 1

Sol. (A)

Q. If the direct transmission method with a cable of resistance 0.4 W km–1 is used, the power dissipation (in %) during transmission is

(A) 20

(B) 30

(C) 40

(D) 50

Sol. (B)

Current in transmission line $=\frac{\text { Power }}{\text { Voltage }}$

$=\frac{600 \times 10^{3}}{40,000}=150 \mathrm{A}$

Resistance of line $=0.4 \times 20=8 \Omega$

Power loss in line $=\mathrm{i}^{2} \mathrm{R}=(150)^{2} 8$

$=180 \mathrm{KW}$

percentage of power dissipation in during transmission $=\frac{1800 \times 10^{3}}{600 \times 10^{3}} \times 100=30 \%$

Q. At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current $\mathrm{I}(\mathrm{t})=\mathrm{I}_{0} \cos (\omega \mathrm{t})$,with I0 = 1A and $\omega=500 \mathrm{rad} \mathrm{s}^{-1}$ starts flowing in it with the initial direction shown in the figure. At $=\frac{7 \pi}{6 \omega}$, the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If $\mathrm{C}=20 \mu \mathrm{F}, \mathrm{R}=10 \Omega$ and the battery is ideal with emf of 50 V, identify the correct statement (s).

(A) Magnitude of the maximum charge on the capacitor before $\mathrm{t}=\frac{7 \pi}{6 \omega}$ is $1 \times 10^{-3} \mathrm{C}$

(B) The current in the left part of the circuit just before $\mathrm{t}=\frac{7 \pi}{6 \omega}$ is clockwise.

(C) Immediately after A is connected to D, the current in $\mathrm{R}$ is $10 \mathrm{A}$

(D) $\mathrm{Q}=2 \times 10^{-3} \mathrm{C}$

Sol. (C,D)

Current $\mathrm{I}=\mathrm{I}_{0} \cos (\omega \mathrm{t})$

$\frac{\mathrm{d} q}{\mathrm{dt}}=\mathrm{I}_{0} \cos (\omega \mathrm{t})$

$\Rightarrow \mathrm{q}=\frac{\mathrm{I}_{0}}{\omega} \sin (\omega \mathrm{t})$

$\Rightarrow \mathrm{q}=\frac{1}{500} \sin (\omega \mathrm{t})$

$\Rightarrow \mathrm{q}=\left(2 \times 10^{-3}\right) \sin (\omega \mathrm{t})$

So, maximum charge $=2 \times 10^{-3} \mathrm{C}$

immediately before $\mathrm{t}=\frac{7 \pi}{6 \omega}$

Current in left part just before $\mathrm{t}=\frac{7 \pi}{6 \omega}$

$\mathrm{I}=\mathrm{I}_{0} \cos \left(\omega \times \frac{7 \pi}{6 \omega}\right)=-\frac{\mathrm{I}_{0} \sqrt{3}}{2}$

Since current is negative hence current will be anticlockwise.

immediately after $\mathrm{t}=\frac{7 \pi}{6 \omega}$

$\mathrm{q}=\left(2 \times 10^{-3}\right) \sin \left(\omega \times \frac{7 \pi}{6 \omega}\right)$

$\mathrm{q}=\left(2 \times 10^{-3}\right) \sin \left(\omega \times \frac{7 \pi}{6 \omega}\right)$

$=-1 \times 10^{-3} \mathrm{C}$

Current in $10 \Omega$ resistance,

$\mathrm{I}=\frac{100}{10}=10 \mathrm{A}$

At steady state, potential difference of capaictor is same as of battery,

So, final charge is

$\mathrm{Q}_{\mathrm{f}}=\mathrm{C} \varepsilon=(20 \mu \mathrm{F})(50 \mathrm{V})=+1 \times 10^{-3} \mathrm{C}$

change in charge $=+10^{-3}-\left(-10^{-3}\right)=2 \times 10^{-3} \mathrm{C}$

Q. In the circuit shown, $\mathrm{L}=1 \mu \mathrm{H}, \mathrm{C}=1 \mu \mathrm{F}$ and $\mathrm{R}=1 \mathrm{k} \Omega$.They are connected in series with an a.c. source $\mathrm{V}=\mathrm{V}_{0}$ $\sin \omega t$ as shown. Which of the following options is/are correct ?

(A) The frequency at which the current will be in phase with the voltage is independent of R.

(B) At $\omega \sim 0$ the current flowing through the circuit becomes nearly zero

(C) At $\omega>>10^{6}$ rad.s $^{-1},$ the circuit behaves like a capacitor.

(D) The current will be in phase with the voltage if $\omega=10^{4} \mathrm{rad.s}^{-1}$

Sol. (A,B)

Q. The instantaneous voltages at three terminals marked X, Y and Z are given by $\mathrm{V}_{\mathrm{x}}=\mathrm{V}_{0} \sin$

$\omega \mathrm{t} \mathrm{V}_{\mathrm{Y}}=\mathrm{V}_{0} \sin \left(\omega \mathrm{t}+\frac{2 \pi}{3}\right)$ and $\mathrm{V}_{\mathrm{Z}}=\mathrm{V}_{0} \sin \left(\omega \mathrm{t}+\frac{4 \pi}{3}\right)$

An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points X and Y and then between Y and Z. The reading(s) of the voltmeter will be:-

(A) $\mathrm{V}_{\mathrm{XY}}^{\mathrm{rms}}=\mathrm{V}_{0}$

(B) $\quad \mathrm{V}_{\mathrm{YZ}}^{\mathrm{ms}}=\mathrm{V}_{0} \sqrt{\frac{1}{2}}$

(C) Independent of the choice of the two terminals

(D) $\quad V_{\mathrm{XY}}^{\mathrm{ms}}=\mathrm{V}_{0} \sqrt{\frac{3}{2}}$

Sol. (C,D)

Potential difference between $X \& Y=V_{X}-V_{Y}$

Potential difference between $\mathrm{Y} \& \mathrm{Z}=\mathrm{V}_{\mathrm{Y}}-\mathrm{V}_{7}$

Phasor of the voltages :

$\therefore \mathrm{V}_{\mathrm{X}}-\mathrm{V}_{\mathrm{Y}}=\sqrt{3} \mathrm{V}_{0}$

$\mathrm{V}_{\mathrm{XY}}^{\mathrm{ms}}=\frac{\sqrt{3} \mathrm{V}_{0}}{\sqrt{2}}$

similarly $\mathrm{V}_{\mathrm{YZ}}^{\mathrm{ms}}=\frac{\sqrt{3} \mathrm{V}_{0}}{\sqrt{2}}$

Also difference is independent of choice of two terminals.

Calorimetry – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Simulator

Q. A piece of ice (heat capacity = 2100 J $\mathrm{kg}^{-1}$ $^{\circ} \mathrm{C}^{-1}$ and latent heat = 3.36 × $10^{5} \mathrm{J} \mathrm{kg}^{-1}$) of mass m grams is at $-5^{\circ}$C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice–water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m is ?

[JEE 2010]

Sol. 8

$420=\left(m \times 10^{-3}\right)(2100)(5)+\left(1 \times 10^{-3}\right) \times\left(3.36 \times 10^{5}\right)$

$m=8 \mathrm{gm}$

Q. A water cooler of storage capacity 120 litres can cool water at constant rate of P watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3 kW of heat (thermal load). The temperature of water fed into the device cannot exceed $30^{\circ}$C and the entire stored 120 litres of water is initially cooled to 10°C. The entire system is thermally insulated. The minimum value of P (in watts) for which the device can be operated for 3 hours is :

(Specific heat of water is $4.2 \mathrm{kJ} \mathrm{kg}^{-1} \mathrm{K}^{-1}$ and the density of water is $1000 \mathrm{kg} \mathrm{m}^{-3}$ )

(A) 1600

(B) 2067

(C) 2533

(D) 3933

Sol. (B)

$3000-\mathrm{P}=(120 \times 1)\left(4.2 \times 10^{3}\right) \frac{\mathrm{dT}}{\mathrm{dt}}$

$\frac{\mathrm{dT}}{\mathrm{dt}}=\frac{20}{60 \times 60 \times 3}$

$\mathrm{P}=2067 \mathrm{W}$

Atomic Structure – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Simulator

Paragraph for Question Nos. 1 to 3

The key feature of Bohr’s theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr’s quantization condition.

Simulator

Paragraph for Question Nos. 1 to 3

The key feature of Bohr’s theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr’s quantization condition.

Q. A diatomic molecule has moment of inertia I. By Bohr’s quantization condition its rotational energy in the nth level (n =0 is not allowed) is

$(\mathrm{A}) \frac{1}{\mathrm{n}^{2}}\left(\frac{\mathrm{h}^{2}}{8 \pi^{2} \mathrm{I}}\right)$

$(\mathrm{B}) \frac{1}{\mathrm{n}}\left(\frac{\mathrm{h}^{2}}{8 \pi^{2} \mathrm{I}}\right)$

$(\mathrm{C}) \mathrm{n}\left(\frac{\mathrm{h}^{2}}{8 \pi^{2} \mathrm{I}}\right)$

$(\mathrm{D}) \mathrm{n}^{2}\left(\frac{\mathrm{h}^{2}}{8 \pi^{2} \mathrm{I}}\right)$

[JEE 2010]

Sol. (D)

$\mathrm{E}_{\mathrm{n}}=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{(\mathrm{l} \omega)^{2}}{2 \mathrm{l}}=\frac{(\mathrm{nh} / 2 \pi)^{2}}{2 \mathrm{l}}=\frac{\mathrm{n}^{2} \mathrm{h}^{2}}{8 \pi^{2} \mathrm{I}}$

Q. It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close to $\frac{4}{\pi} \times 10^{11} \mathrm{Hz}$ Hz. Then the moment of inertia of CO molecule about its center of mass is close to $\left[\text { Take } \mathrm{h}=2 \pi \times 10^{-34} \mathrm{Js}\right)$

(A) $2.76 \times 10^{-46} \mathrm{kg} \mathrm{m}^{2}$

(B) $1.87 \times 10^{-46} \mathrm{kg} \mathrm{m}^{2}$

(C) $4.67 \times 10^{-47} \mathrm{kg} \mathrm{m}^{2}$

(D) $1.17 \times 10^{-47} \mathrm{kg} \mathrm{m}^{2}$

[JEE 2010]

Sol. (B)

Q. In a CO molecule, the distance between C (mass = 12 a.m.u.) and O (mass = 16 a.m.u.), where 1 a.m.u. $=\frac{5}{3} \times 10^{-27} \mathrm{kg},$ is close to

(A) $2.4 \times 10^{-10} \mathrm{m}$

(B) $1.9 \times 10^{-10} \mathrm{m}$

(C) $1.3 \times 10^{-10} \mathrm{m}$

(D) $4.4 \times 10^{-11} \mathrm{m}$

[JEE 2010]

Sol. (C)

Moment of inertia of CO molecule about centre of mass : $\mathrm{I}=\mu \mathrm{r}^{2}$ where $\mu=\frac{\mathrm{m}_{1} \mathrm{m}_{2}}{\mathrm{m}_{1}+\mathrm{m}_{2}}$

Q. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is ?

[JEE 2011]

Sol. (A)

Q. The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a0, where a0 is the Bohr radius. Its orbital angular momentum is $\frac{3 \mathrm{h}}{2 \pi}$. It is given that h is Planck constant and R is Rydberg constant. The possible wavelength (s), when the atom de-excites, is (are) :-

(A) $\frac{9}{32 \mathrm{R}}$

(B) $\frac{9}{16 \mathrm{R}}$

(C) $\frac{9}{\text { sR }}$

(D) $\frac{4}{3 \mathrm{R}}$

Sol. (A,C)

for $\mathrm{n}=2$ to $\mathrm{n}=1 \quad \frac{1}{\lambda}=4 \mathrm{R}\left[\frac{1}{1}-\frac{1}{4}\right] \Rightarrow \lambda=\frac{3}{\mathrm{R}}$

Q. Consider a hydrogen atom with its electron in the nth orbital. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of n is (hc = 1242 eV nm).

Sol. 2

$\mathrm{E}_{\mathrm{Ph}}=\frac{\mathrm{hc}}{\lambda}=\frac{1242}{90}=13.8 \mathrm{eV}$

$\mathrm{E}_{\mathrm{Ph}}=\Delta \mathrm{E}+(\mathrm{K.E.})$

$13.8=\Delta \mathrm{E}+10.4$

$\Delta E=3.4 \mathrm{eV}$

so electron initially was in n = 2

Q. Highly excited states for hydrogen like atoms (also called Rydberg states) with nuclear charge Ze are defined by their principal quantum number n, where n >> 1. Which of the following statement(s) is (are) true?

(A) Relative change in the radii of two consecutive orbitals does not depend on Z

(B) Relative change in the radii of two consecutive oribitals varies as 1/n

(C) Relative change in the energy of two consecutive orbitals varies as $1 / \mathrm{n}^{3}$

(D) Relative change in the angular momenta of two consecutive orbitals varies as 1/n

Sol. (A,B,D)

As radius $\mathrm{r} \propto \frac{\mathrm{n}^{2}}{\mathrm{z}}$

as energy $\mathrm{E} \propto \frac{\mathrm{z}^{2}}{\mathrm{n}^{2}}$

Q. A hydrogen atom in its ground state is irradiated by light of wavelength 970 $\hat{\mathbf{A}}$. Taking

hc/e = $1.237 \times 10^{-6}$ eV m and the ground state energy of hydrogen atom as –13.6 eV, the number of lines present in the emission spectrum is ?

Sol. 6

Q. An electron in a hydrogen atom undergoes a transition from an orbit with quantum number $\mathrm{n}_{\mathrm{i}}$ to another with quantum number $\mathrm{n}_{\mathrm{f}} \cdot \mathrm{V}_{\mathrm{i}}$ and $\mathrm{V}_{\mathrm{f}}$ are respectively the initial and final potential energies of the electron. If $\frac{\mathrm{V}_{\mathrm{i}}}{\mathrm{V}_{\mathrm{f}}}=6.25$, then the smallest possible $\mathbf{n}_{\mathrm{f}}$ is.

Sol. 5

Math Topic-wise JEE Advanced Previous Year Question with Solutions

Practicing JEE Advanced Previous Year Question Papers will help you in many ways in your Exam preparation. It will help you to boost your confidence level. Students can check where they are lagging through practicing these previous year question papers. Here you will get the  last 10 years JEE Advanced previous year question paper questions with solutions.
These JEE Advanced Previous Year Questions for Mathematics plays an important role in IIT-JEE preparation. We are providing IIT-JEE Advanced Previous Year Questions with detailed Solution.

We have tried our best to provide you last 10 years question with solutions.
This set of question papers serves as a very important source to revise the important topics and gain an understanding into the pattern of questions asked in previous years. Practicing these papers will positively help students to gain confidence over their learning.
The benefits of solving IIT-JEE previous years papers is that aspirants get to know the type of questions asked in the JEE exam. JEE aspirants can evaluate their preparation after finishing the entire syllabus, topics and chapters. They can get the experience of giving real exam while solving a past year JEE question paper to become confident for the upcoming JEE exam.

While preparing for the IIT-JEE exam, aspirants should be aware about the question paper structure and the format of questions to be asked in this exam. This will help to make an effective preparation strategy for the exam. JEE Advanced Previous Year Question Papers are the best resources to prepare for exam. This will help an individual to understand the exam pattern of JEE. This will also enhance your level of preparation. JEE aspirants must solve multiple sample papers and analyse their performances in order to recognize their strengths and weaknesses.

Here are the Math Topic-wise Previous year question for JEE Advanced:

We have tried our best to provide you last 10 years question with solutions.
This set of question papers serves as a very important source to revise the important topics and gain an understanding into the pattern of questions asked in previous years. Practicing these papers will positively help students to gain confidence over their learning.
The benefits of solving IIT-JEE previous years papers is that aspirants get to know the type of questions asked in the JEE exam. JEE aspirants can evaluate their preparation after finishing the entire syllabus, topics and chapters. They can get the experience of giving real exam while solving a past year JEE question paper to become confident for the upcoming JEE exam.

Matrices – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Q. Let A be a 2 × 2 matrix

Statement- $1: \operatorname{adj}(\operatorname{adj} A)=A$

Statement-2: $|$ adj $A|=| A |$

(1) Statement–1 is true, Statement–2 is false.

(2) Statement–1 is false, Statement–2 is true.

(3) Statement–1 is true, Statement–2 is true;Statement–2 is a correct explanation for Statement–1.

(4) Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for statement–1.

[AIEEE- 2009]

Sol. (4)

Let $A=\left(\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right)$

$\operatorname{adj}(\mathrm{A})=\left(\begin{array}{cc}{\mathrm{d}} & {-\mathrm{b}} \\ {-\mathrm{c}} & {\mathrm{a}}\end{array}\right)$

$\operatorname{adj}(\operatorname{adj} A)=\left(\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right)=A$ statement is right

Statement 2 We know $\mathrm{A}$ adj $(\mathrm{A})=|\mathrm{A}| \mathrm{I}_{\mathrm{n}}$ taking determinant $|\mathrm{A} \cdot \operatorname{adj}(\mathrm{A})|=\| \mathrm{A}\left|\mathrm{I}_{\mathrm{n}}\right|$

$\Rightarrow|\operatorname{adj}(\mathrm{A})|=|\mathrm{A}|^{\mathrm{n}-1}$

Here $\mathrm{n}=2$ (order)

so $|\operatorname{adj} \mathrm{A}||\mathrm{A}|^{2-1}=|\mathrm{A}|$

so statement 2 is also true and 2 is not explanation for statement 1

Q. The number of 3× 3 non-singular matrices, with four entries as 1 and all other entries as 0, is :-

(1) Less than 4            (2) 5             (3) 6              (4) At least 7

[AIEEE-2010]

Sol. (4)

Q. Let A be a $2 \times 2$ matrix with non-zero entries and let $\mathrm{A}^{2}=\mathrm{I},$ where I is $2 \times 2$ identity matrix. Define $\operatorname{Tr}(\mathrm{A})=$ sum of diagonal elements of $\mathrm{A}$ and $|\mathrm{A}|=$ determinant of matrix $\mathrm{A}$. Statement- $1: \operatorname{Tr}(\mathrm{A})=0$

Statement-2: $|\mathrm{A}|=1$

(1) Statement–1 is true, Statement–2 is true; Statement–2 is a correct explanation for

Statement–1.

(2) Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for

statement–1.

(3) Statement–1 is true, Statement–2 is false.

(4) Statement–1 is false, Statement–2 is true.

[AIEEE-2010]

Sol. (3)

Statement 1:

Let $\mathrm{A}=\left(\begin{array}{ll}{\mathrm{a}} & {\mathrm{b}} \\ {\mathrm{c}} & {\mathrm{d}}\end{array}\right) \mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d},$ are non zero

$A^{2}=\left(\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right)\left(\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right)=\left(\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right)$

$\Rightarrow a^{2}+b c=1$

$\Rightarrow \mathrm{ab}+\mathrm{bd}=0 \Rightarrow \mathrm{b}(\mathrm{a}+\mathrm{d})=0$

so $\mathrm{b} \neq 0,(\mathrm{a}+\mathrm{d})=0$

$a+d=0 \Rightarrow \operatorname{tr}(A)=0$

Statement 2:

$|A|=a d-b c=-a^{2}-b c=-\left(a^{2}+b c\right)=-1$

So Statement 2 is false.

Q. Let A and B be two symmetric matrices of order 3.

Statement-1 : A(BA) and (AB)A are symmetric matrices.

Statement-2 : AB is symmetric matrix if matrix multiplication of A with B is commutative.

(1) Statement-1 is true, Statement-2 is false.

(2) Statement-1 is false, Statement-2 is true

(3) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

(4) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

[AIEEE-2011]

Sol. (4)

$A^{T}=A$

$\mathrm{B}^{\mathrm{T}}=\mathrm{B}$

Statement- 1 $(\mathrm{A}(\mathrm{BA}))^{\mathrm{T}}=(\mathrm{BA})^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}$ $=\mathrm{A}^{\mathrm{T}} \mathrm{B}^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}=\mathrm{A}(\mathrm{BA}) \rightarrow \text { symmetric }$

$((\mathrm{AB}) \mathrm{A}))^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}=(\mathrm{AB}) \mathrm{A} \rightarrow$ symmetric

Statement – 1 is true

Statement- 2:

$(\mathrm{AB})^{\mathrm{T}}=\mathrm{B}^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}=\mathrm{B} \mathrm{A}$

if $\mathrm{AB}=\mathrm{B} \mathrm{A}$ then

$(\mathrm{AB})^{\mathrm{T}}=\mathrm{B} \mathrm{A}=\mathrm{AB}$

Statement- 2 is true

but Not a correct expalnation.

Q. Statement-1 : Determinant of a skew-symmetric matrix of order 3 is zero.

Statement-1 : For any matrix A, det(AT) = det(A) and det(–A) = –det(A).

Where det(B) denotes the determinant of matrix B. Then :

(1) Statement-1 is true and statement-2 is false

(2) Both statements are true

(3) Both statements are false

(4) Statement-1 is false and statement-2 is true.

[AIEEE-2011]

Sol. (1)

Statement- 1: The value of determinant of skew symmetric matrix of odd order is always zero. So Statement-I. is true. Statement-II : This Statement is not always true depends on the order of matrix. $|-A|=-|A|$ if order is odd, so Statement–II is wrong. Statement-I is true and Statement-II is false.

Q. Let $\mathrm{A}=\left(\begin{array}{lll}{1} & {0} & {0} \\ {2} & {1} & {0} \\ {3} & {2} & {1}\end{array}\right) .$ If $\mathrm{u}_{1}$ and $\mathrm{u}_{2}$ are column matrices such that $\mathrm{Au}_{1}=\left(\begin{array}{l}{1} \\ {0} \\ {0}\end{array}\right)$ and $\mathrm{Au}_{2}=\left(\begin{array}{l}{0} \\ {1} \\ {0}\end{array}\right),$ then $\mathrm{u}_{1}+\mathrm{u}_{2}$ is equal to :

[AIEEE-2012]

Sol. (1)

Q. If $P=\left[\begin{array}{lll}{1} & {\alpha} & {3} \\ {1} & {3} & {3} \\ {2} & {4} & {4}\end{array}\right]$ is the adjoint of a $3 \times 3$ matrix $A$ and $|A|=4,$ then $\alpha$ is equal to

(1) 4                  (2) 11                      (3) 5                     (4) 0

[JEE(Main) – 2013]

Sol. (2)

taking determinant

Q. If $\mathrm{A}$ is an $3 \times 3 \times 3$ non-singular matrix such that $\mathrm{AA}^{\prime}=\mathrm{A}^{\prime} \mathrm{A}$ and $\mathrm{B}=\mathrm{A}^{-1} \mathrm{A}^{\prime},$ the BB’ equals :

(1) I + B (2) I (3) $\mathrm{B}^{-1}$ (4) $\left(B^{-1}\right)^{\prime}$

[JEE(Main) – 2014]

Sol. (2)

Q. If $A=\left[\begin{array}{ccc}{1} & {2} & {2} \\ {2} & {1} & {-2} \\ {a} & {2} & {b}\end{array}\right]$ is a matrix satisfying the equation $A A^{T}=9$, where I is $3 \times 3$ identity matrix, then the ordered pair $(\mathrm{a}, \mathrm{b})$ is equal to :

(1) (2, 1)               (2) (–2, –1)                (3) (2, –1)                (4) (–2, 1)

[JEE(Main)-2015]

Sol. (2)

Q. If $\mathrm{A}=\left[\begin{array}{cc}{5 \mathrm{a}} & {-\mathrm{b}} \\ {3} & {2}\end{array}\right]$ and $\mathrm{A}$ adj $\mathrm{A}=\mathrm{A} \mathrm{A}^{\mathrm{T}},$ then $5 \mathrm{a}+\mathrm{b}$ is equal to :

(1) 13                (2) –1                (3) 5                       (4) 4

[JEE(Main)-2016]

Sol. (3)

Q. If $A=\left[\begin{array}{cc}{2} & {-3} \\ {-4} & {1}\end{array}\right],$ then adj $\left(3 A^{2}+12 A\right)$ is equal to :-

[JEE(Main)-2017]

Sol. (3)

Q. Let $\mathrm{A}=\left[\begin{array}{lll}{1} & {0} & {0} \\ {1} & {1} & {0} \\ {1} & {1} & {1}\end{array}\right]$ and $\mathrm{B}=\mathrm{A}^{20} .$ Then the sum of the elements of the first column of $\mathrm{B}$ is :

(1) 211               (2) 251               (3) 231                 (4) 210

[JEE(Main)-2018]

Sol. (3)

Q. Let A be a matrix such that A. $\left[\begin{array}{ll}{1} & {2} \\ {0} & {3}\end{array}\right]$ is a scalar matrix and $|3 \mathrm{A}|=108 .$ Then $\mathrm{A}^{2}$ equals :

[JEE(Main)-2018]

Sol. (1)

Q. Suppose $\mathrm{A}$ is any $3 \times 3$ non-singular matrix and $(\mathrm{A}-3 \mathrm{I})(\mathrm{A}-5 \mathrm{I})=0,$ where $\mathrm{I}=\mathrm{I}_{3}$ and $\mathrm{O}$ $=\mathrm{O}_{3} .$ If $\alpha \mathrm{A}+\beta \mathrm{A}^{-1}=4 \mathrm{I},$ then $\alpha+\beta$ is equal to :

(1) 13                 (2) 7                  (3) 12                    (4) 8

[JEE(Main)-2018]

Sol. (4)

Logarithm – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.