Mind Map for Modern Physics: Radioactivity Revision – Class XII, JEE, NEET

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Radioactivity in Modern Physics comprises variety of cases with important formulae and key points. So here is the mind map to help you in remembering all the formulas and important key concepts on finger tips.

Wave Optics – JEE Advanced Previous Year Questions with Solutions

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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.

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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.

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Q. Column I shows four situations of standard Young’s double slit arrangement with the screen placed far away from the slits $S_{1}$ and $S_{2}$. In each of these cases $S_{1} P_{0}$ = $S_{2} P_{0}$, $S_{1} P_{1}$$S_{2} P_{1} = \lambda / 4 and S_{1} P_{2}$$S_{2} P_{2}=\lambda / 3$, where $\lambda$ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index $\mu$ and thickness t is pasted on slit S2. The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by $\delta$ (P) and the intensity by I(P). Match each situation given in Column I with the statement(s) in Column II valid for that situation.

[IIT-JEE-2009]

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

(A) $\Delta \mathrm{x}=\mathrm{S}_{2} \mathrm{P}-\mathrm{S}_{1} \mathrm{P}=0$

$\delta\left(\mathrm{P}_{0}\right)=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=0$

$\Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}_{1}-\mathrm{S}_{2} \mathrm{P}_{1}=\frac{\lambda}{4}$

$\delta\left(\mathrm{P}_{1}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2}$

$\mathrm{I}=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\Delta \phi}{2}\right)$

$\mathrm{I}\left(\mathrm{P}_{1}\right)=\mathrm{I}_{1}=\mathrm{I}_{\max } \cos ^{2} \frac{\delta}{2}=\frac{\mathrm{I}_{\max }}{2}$

$\delta\left(\mathrm{P}_{2}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}=\frac{2 \pi}{3}$

$\mathrm{I}\left(\mathrm{P}_{2}\right)=\mathrm{I}_{2}=\mathrm{I}_{\max } \cos ^{2} \frac{\pi}{3}=\frac{\mathrm{I}_{\max }}{4}$

$\mathrm{I}\left(\mathrm{P}_{0}\right)>\mathrm{I}\left(\mathrm{P}_{1}\right)$

$(\mathrm{B}) \Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}-\left[\mathrm{S}_{2} \mathrm{P}+(\mu-1) \mathrm{t}\right]$

$\Delta \mathrm{x}_{1}=\mathrm{S}_{1} \mathrm{P}_{1}-\mathrm{S}_{2} \mathrm{P}_{1}-(\mu-1) \mathrm{t}$

$\Delta \mathrm{x}_{1}=\frac{\lambda}{4}-\frac{\lambda}{4}=0$

$8\left(\mathrm{P}_{1}\right)=0 ; \mathrm{I}\left(\mathrm{P}_{1}\right)=\mathrm{I}_{\max }$

$8\left(\mathrm{P}_{0}\right)=\frac{\pi}{2} \delta\left(\mathrm{P}_{0}\right) \neq 0$

$\mathrm{I}\left(\mathrm{P}_{0}\right)=\mathrm{I}_{\max } / 2$

$\Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}_{2}-\mathrm{S}_{1} \mathrm{P}_{2}-(\mu-1) \mathrm{t}$

$=\frac{\lambda}{3}-\frac{\lambda}{4}=\frac{\lambda}{12}$

$8\left(\mathrm{P}_{2}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{12}=\frac{\pi}{6}$

$\mathrm{I}\left(\mathrm{P}_{2}\right)=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\pi}{12}\right)$

Q. Young’s double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are $\beta_{G}, \beta_{R}$ and $\beta_{B},$ respectively. Then

(A) $\beta_{G}>\beta_{B}>\beta_{R}$

(B) $\beta_{B}>\beta_{G}>\beta_{R}$

(C) $\beta_{R}>\beta_{B}>\beta_{G}$

(D) $\beta_{R}>\beta_{G}>\beta_{B}$

[IIT-JEE-2012]

Sol. (D)

$\beta=\frac{\mathrm{D} \lambda}{\mathrm{d}}$

$\lambda_{\mathrm{R}}>\lambda_{\mathrm{a}}>\lambda_{\mathrm{B}}$

Q. In the Young’s double slit experiment using a monochromatic light of wavelength $\lambda$, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is :-

(A) $(2 n+1) \frac{\lambda}{2}$

(B) $(2 n+1) \frac{\lambda}{4}$

(C) $(2 n+1) \frac{\lambda}{8}$

$(D)(2 n+1) \frac{\lambda}{16}$

Sol. (B)

$\frac{\mathrm{I}_{\max }}{2}=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)$

$\cos ^{2}\left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)=\frac{1}{2}$

$\cos \left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)=\pm \frac{1}{\sqrt{2}}$

$\frac{\pi}{\lambda} \Delta \mathrm{x}=\mathrm{n} \pi \pm \frac{\pi}{4}$

$\Delta \mathrm{x}=\left(\mathrm{n} \pm \frac{1}{4}\right) \lambda$

Q. A light source, which emits two wavelengths $\lambda_{1}=400 \mathrm{nm}$ and $\lambda_{2}=600 \mathrm{nm},$ is used in a Young’s double slit experiment. If recorded fringe widths for $\lambda_{1}$ and $\lambda_{2}$ are $\beta_{1}$ and $\beta_{2}$ and the number of fringes for them within a distance y on one side of the central maximum are $\mathrm{m}_{1}$ and $\mathrm{m}_{2},$ respectively, then :-

(A) $\beta_{2}>\beta_{1}$

(B) $\mathrm{m}_{1}>\mathrm{m}_{2}$

(C) From the central maximum, $3^{\mathrm{rd}}$ maximum of $\lambda_{2}$ overlaps with $5^{\text {th }}$ minimum of $\lambda_{1}$

(D) The angular separation of fringes of $\lambda_{1}$ is greater than $\lambda_{2}$

Sol. (A,B,C)

$\beta=\frac{\mathrm{D} \lambda}{\mathrm{d}}$

$\mathrm{B}_{2}>\beta_{1}$

$\mathrm{y}=\mathrm{m}_{1} \frac{\mathrm{D} \lambda_{1}}{\mathrm{d}}=\mathrm{m}_{2} \frac{\mathrm{D} \lambda_{2}}{\mathrm{d}}$

$\frac{\mathrm{nD} \times \lambda_{2}}{\mathrm{d}}=\left(\mathrm{n}^{\prime}+\frac{1}{2}\right) \frac{\mathrm{D} \lambda_{1}}{\mathrm{d}} \Rightarrow 600 \mathrm{n}=\left(\mathrm{n}^{\prime}+\frac{1}{2}\right) \times 4$

Q. A Young’s double slit interference arrangement with slits $S_{1}$ and $S_{2}$ is immersed in water (refractive index $=4 / 3$ ) as shown in the figure. The positions of maxima on the surface of water are given by $x^{2}=p^{2} m^{2} \lambda^{2}-d^{2},$ where $\lambda$ is the wavelength of light in air (refractive index $=1$, $2 d$ is the separation between the slits and $m$ is an integer. The value of p is.

Sol. 3

Q. While conducting the Young’s double slit experiment, a student replaced the two slits with a large opaque plate in the x-y plane containing two small holes that act as two coherent point sources $\left(\mathrm{S}_{1}, \mathrm{S}_{2}\right)$ emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to the x-z plane (for z > 0) at a distance D = 3m from the mid-point of $\mathrm{S}_{1} \mathrm{S}_{2}$, as shown schematically in the figure. The distance between the sources d = 0.6003 mm. The origin O is at the intersection of the screen and the line joining $\mathrm{S}_{1} \mathrm{S}_{2}$. Which of the following is (are) true of the intensity pattern on the screen ?

(A) Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction

(B) Semi circular bright and dark bands centered at point O

(C) The region very close to the point O will be dark

(D) Straight bright and dark bands parallel to the x-axis

[JEE-Mains 2016]

Sol. (B,C)

Path difference at point O = d = .6003 mm = 600300 nm

$=\frac{2001}{2}(600 \mathrm{nm})=1000 \lambda+\frac{\lambda}{2}$

$\Rightarrow$ minima form at point $\mathrm{O}$

Line $S_{1} S_{2}$ and screen are $\perp$ to each other so fringe pattern is circular (semi-circular because only half of screen is available)

Q. Two coherent monochromatic  point sources $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ of wavelength $\lambda$ = 600 nm are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance d = 1.8mm. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is $\Delta \theta$. Which of the following options is/are correct ?

(A) A dark spot will be formed at the point $\mathrm{P}_{2}$

(B) The angular separation between two consecutive bright spots decreases as we move from $\mathrm{P}_{1}$ to $\mathrm{P}_{2}$ along the first quadrant

(C) At $\mathrm{P}_{2}$ the order of the fringe will be maximum

(D) The total number of fringes produced between $P_{1}$ and $\mathrm{P}_{2}$ in the first quadrant is close to 3000

Sol. (C,D)

Liquid Solution – JEE Main Previous Year Question of with Solutions

Class 9-10, JEE & NEET

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JEE Main Previous Year Question of Chemistry with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Chemistry will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Chemistry 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.

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.

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Previous Years AIEEE/JEE Mains Questions

Q. A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the folloowing statements is correct regarding the behaviour of the solution ?

(1) The solution is non-ideal, showing –ve deviation from Raoult’s law

(2) n-heptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s law

(3) The solution formed is an ideal solution.

(4) The solutionis non-ideal, showing +ve deviation from Raoult’s law

[AIEEE-2009]

Sol. (4)

(A) n–heptone : Non Polar

(B) Ethanol : Polar

$\mathrm{F}_{\mathrm{A}-\mathrm{B}}<\mathrm{F}_{\mathrm{A}-\mathrm{A}}, \mathrm{F}_{\mathrm{B}-\mathrm{B}} \Rightarrow+$ deviation

Q. Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectively :-

(1) 400 and 600 (2) 500 and 600 (3) 200 and 300 (4) 300 and 400

[AIEEE-2009]

Sol. (1)

$550=\mathrm{P}_{\mathrm{A}}^{\circ} \times \frac{1}{4}+\mathrm{P}_{\mathrm{B}}^{\circ} \times \frac{3}{4}$

$560=\mathrm{P}_{\mathrm{A}}^{\circ} \times \frac{1}{5}+\mathrm{P}_{\mathrm{B}}^{\circ} \times \frac{4}{5}$

$\mathrm{P}_{\mathrm{A}}^{\circ}=400 \quad \mathrm{P}_{\mathrm{B}}^{\circ}=600$ torr

Q. On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of

the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively.

Vapour pressure of the solution obtained by mixing 25.0 of heptane and 35 g of octane

will be (molar mass of heptane = 100 g $\mathrm{mol}^{-1}$ and of octane = 114 g $\mathrm{mol}^{-1}$) :-

(1) 144.5 kPa           (2) 72.0 kPa             (3) 36.1 kPa            (4) 96.2 kPa

[AIEEE-2010]

Sol. (2)

$\mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{X}_{\mathrm{A}}=105 \times \frac{1 / 4}{1 / 4+0.307}$

$=105 \times 0.449=47.13 \mathrm{K} \mathrm{Pa}$

$\mathrm{P}_{\mathrm{B}}=\mathrm{P}_{\mathrm{B}}^{\circ} \mathrm{X}_{\mathrm{B}}=45 \times 0.551=24.795$

$\mathrm{P}_{\mathrm{T}}=\mathrm{P}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}=71.925 \mathrm{atm}$

Q. If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water $\left(\Delta \mathrm{T}_{\mathrm{f}}\right)$, when 0.01 mol of sodium sulphate isdissolved in 1 kg of water, is $\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right):$ :-

(1) 0.0186 K            (2) 0.0372 K              (3) 0.0558 K              (4) 0.0744 K

[AIEEE-2010]

Sol. (3)

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{k}_{\mathrm{f}} \cdot \mathrm{m}$

$=3 \times 1.86 \times 0.01 / 1$

$=0.0558 \mathrm{K}$

Q. The molality of a urea solution in which 0.0100g of urea, $\left.\left[\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\right]$ is added to 0.3000 $\mathrm{dm}^{3}$ of water at STP is :-

(1) 0.555 m

(2) $5.55 \times 10^{-4} \mathrm{m}$

(3) 33.3 m

(4) $3.33 \times 10^{-2} \mathrm{m}$

[AIEEE-2011]

Sol. (2)

$\mathrm{m}=\frac{\mathrm{n}}{\mathrm{W}(\mathrm{kg})}=\frac{0.01 / 60}{0.3 \mathrm{kg}}=5.55 \times 10^{-4} \mathrm{mol} / \mathrm{kg}$

Q. A 5% solution of cane sugar (molar mass 342) is isotonic with 1% of a solution of an unknown solute. The molar mass of unknown solute in g/mol is :-

(1) 136.2           (2) 171.2           (3) 68.4            (4) 34.2

[AIEEE-2011]

Sol. (3)

$\pi_{\mathrm{c.s}}=\pi_{\mathrm{Unk}}$

$\left(\frac{\mathrm{n}}{\mathrm{V}}\right)_{\mathrm{c.s.}} \mathrm{RT}=\left(\frac{\mathrm{n}}{\mathrm{V}}\right)_{\mathrm{unk} .} \mathrm{RT}$

$\frac{5 \times 10}{342}=\frac{1 \times 10}{\mathrm{M}}$

M = 68.4 gm/mol

Q. Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at – $6^{\circ} \mathrm{C}$ will be :

$\left(\mathrm{K}_{\mathrm{f}} \text { for water }=1.86 \mathrm{K} \mathrm{kgmol}^{-1}, \text { and molar mass of ethylene glycol }=62 \mathrm{gmol}^{-1}\right)$

(1) 400.00 g             (2) 304.60 g            (3) 804.32 g            (4) 204.30 g

[AIEEE-2011]

Sol. (3)

$6=1.86 \times \frac{\mathrm{w} / 62}{4} \Rightarrow \mathrm{w}=800 \mathrm{gm}$

Q. The degree of dissociation () of a weak electrolyte, AxBy is related to van’t Hoff factor (i) by the expression :-

$(1) \alpha=\frac{\mathrm{x}+\mathrm{y}-1}{\mathrm{i}-1}$

(2) $\alpha=\frac{\mathrm{x}+\mathrm{y}+1}{\mathrm{i}-1}$

(3) $\alpha=\frac{\mathrm{i}-1}{(\mathrm{x}+\mathrm{y}-1)}$

(4) $\alpha=\frac{\mathrm{i}-1}{\mathrm{x}+\mathrm{y}+1}$

[AIEEE-2011]

Sol. (3)

Q. $\mathrm{K}_{\mathrm{f}}$ for water is 1.86 K kg $\mathrm{mol}^{-1}$. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol $\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)$ must you add to get the freezing point of the solution lowered to –$-2.8^{\circ} \mathrm{C} ?$

(1) 27 g            (2) 72 g            (3) 93 g             (4) 39 g

[AIEEE-2012]

Sol. (3)

$2.8=1.86 \times \frac{\mathrm{w} / 62}{1} \Rightarrow \mathrm{w}=93.33 \mathrm{gm}$

Q. A solution containing 0.85 g of $\mathrm{ZnCl}_{2}$ in 125.0 g of water freezes at $-0.23^{\circ} \mathrm{C}$ . The apparent degree of dissociation of the salt is :

($\mathbf{k}_{f}$ for water = 1.86 K kg $\mathrm{mol}^{-1}$, atomic mass ; Zn = 65.3 and Cl = 35.5)

(1) 1.36%            (2) 2.47%              (3) 73.5%             (4) 7.35%

[Jee (Main)-2012 online]

Sol. (3)

$0.23=(1+2 \alpha) \times 1.86 \times \frac{0.85 / 134.5}{0.125}$

$\alpha=0.735=73.5 \%$

Q. Liquids A and B form an ideal solution. At $30^{\circ}$C, the total vapour pressure of a solution containing 1 mol of A and 2 moles of B is 250 mm Hg. The total vapour pressure becomes 300 mm Hg when 1 more mol of A is added to the first solution. The vapour pressures of pure A and B at the same temperature are

(1) 450, 150 mm Hg

(2) 250, 300 mm Hg

(3) 125, 150 mm Hg

(4) 150, 450 mm Hg

[Jee (Main)-2012 online]

Sol. (1)

$250=\mathrm{P}_{\mathrm{A}}^{0} \times \frac{1}{3}+\mathrm{P}_{\mathrm{B}}^{0} \times \frac{2}{3}$

$300=\mathrm{P}_{\mathrm{A}}^{0} \times \frac{1}{2}+\mathrm{P}_{\mathrm{B}}^{0} \times \frac{1}{2}$

$\mathrm{P}_{\mathrm{A}}^{0}=450 \mathrm{mm}$

$\mathrm{P}_{\mathrm{B}}^{0}=150 \mathrm{mm}$

Q. The freezing point of a 1.00 m aqueous solution of HF is found to be $-1.91^{\circ} \mathrm{C}$. The

freezing point constant of water, $\mathrm{K}_{\mathrm{f}}$, is 1.86 K kg $\mathrm{mol}^{-1}$. The percentage dissociation of HF at this concentration is

(1) 2.7%             (2) 30%            (3) 10%             (4) 5.2%

[Jee (Main)-2012 online]

Sol. (1)

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \mathrm{m}$

$1.91=(1+\alpha) \times 1.86 \times 1$

$\alpha=0.027$

Q. How many grams of methyl alcohol should be added to 10 litre tank of water to prevent its freezing at 268 K ?

$\left(\mathrm{K}_{f} \text { for water is } 1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right)$

(1) 899.04 g              (2) 886.02 g            (3) 868.06 g                 (4) 880.07 g

[Jee (Main)-2013 online]

Sol. (2)

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{f}}^{0}-\mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times \mathrm{m}$

$273.15-268=1.86 \times \frac{\mathrm{w} / 32}{10}$

$\mathrm{w}=886.02 \mathrm{g}$

Q. Vapour pressure of pure benzene is 119 torr and that of toluene is 37.0 torr at the same temperature. Mole fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene 0.50, will be :

(1) 0.137           (2) 0.205            (3) 0.237            (4) 0.435

[Jee (Main)-2013 online]

Sol. (3)

Benzen $\rightarrow 4$

Toluene $\rightarrow B$

y $_{B}=\frac{P_{B}^{0} \times X_{B}}{P_{B}^{0} X_{B}+P_{A}^{0} X_{A}}=\frac{37 \times 0.5}{37 \times 0.5+119 \times 0.5}=0.237$

Q. A molecule M associates in a given solvent according to the equation M  $(\mathrm{M})_{\mathrm{n}}$. For a certain concentration of M, the van’t Hoff factor was found to be 0.9 and the fraction of associated molecules was 0.2. The value of n is :

(1) 2              (2) 4               (3) 5               (4) 3

[Jee (Main)-2013 online]

Sol. (1)

$\mathrm{M}=\mathrm{M}_{\mathrm{n}}$

$1-0.2 \quad 0.2 / \mathrm{n}$

$0.9=\frac{1-0.2+0.2 / \mathrm{n}}{1}$

$0.9=0.8+\frac{0.2}{\mathrm{n}}$

$0.1=\frac{0.2}{\mathrm{n}}$

$\mathrm{n}=2$

Q. 12g of a nonvolatile solute dissolved in 108g of water produces the relative lowering of vapour pressure of 0.1. The molecular mass of the solute is :

(1) 60           (2) 80             (3) 40            (4) 20

[Jee (Main)-2013 online]

Sol. (4)

$\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}=0.1=\frac{12 / \mathrm{m}}{108 / 18} \Rightarrow \mathrm{m}=20$

Q. The molarity of a solution obtained by mixing 750 mL of 0.5(M)HCl with 250 mL of 2(M)HCl will be :-

(1) 0.875 M           (2) 1.00 M            (3) 1.75 M            (4) 0.975 M

[Jee (Main)-2013]

Sol. (1)

$\mathrm{M}_{\mathrm{f}}=\frac{\mathrm{M}_{1} \mathrm{V}_{1}+\mathrm{M}_{2} \mathrm{V}_{2}}{\mathrm{V}_{1}+\mathrm{V}_{2}}=0.875 \mathrm{M}$

Q. The observed osmotic pressure for a 0.10 M solution of Fe$\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2}$ at $25^{\circ} \mathrm{C}$ is 10.8 atm. The expected and experimental (observed) values of Van’t Hoff factor (i) will be respectively : $\left(\mathrm{R}=0.082 \mathrm{L} \mathrm{atm} \mathrm{k}^{-} \mathrm{mol}^{-1}\right)$

(1) 3 and 5.42          (2) 5 and 3.42           (3) 4 and 4.00           (4) 5 and 4.42

[Jee (Main)-2014 online]

Sol. (4)

$\pi_{\mathrm{ob}}=\mathrm{i} \frac{\mathrm{n}}{\mathrm{V}} \mathrm{RT}$

$10.8=\mathrm{i} \times 0.1 \times 0.082 \times 298$

$\mathrm{i}=4.42$

Q. For an ideal Solution of two components A and B, which of the following is true ?

(1) $\Delta \mathrm{H}_{\text {mixing }}<0$ (zero)

(2) $\mathrm{A}-\mathrm{A}, \mathrm{B}-\mathrm{B}$ and $\mathrm{A}-\mathrm{B}$ interactions are identical

(3) $\mathrm{A}-\mathrm{B}$ interaction is stronger than $\mathrm{A}-\mathrm{A}$ and $\mathrm{B}-\mathrm{B}$ interactions

(4) $\Delta \mathrm{H}_{\text {mixing }}>0$ (zero)

[Jee(Main)-2014 online]

Sol. (2)

$\Delta \mathrm{H}_{\operatorname{mix}}=0$

Q. Consider separate solution of $0.500 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq}), 0.100 \mathrm{MM} \mathrm{g}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{aq}), 0.250 \mathrm{M} \mathrm{KBr}(\mathrm{aq})$ and 0.125 M $\mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq})$ at $25^{\circ} \mathrm{C}$. Which statement is true about these solutions, assuming all salts to be strong electrolytes ?

(1) 0.125 M $\mathrm{Na}_{3} \mathrm{PO}_{4}$ (aq) has the highest osmotic pressure.

(2) 0.500 M $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ (aq) has the highest osmotic pressure.

(3) They all have the same osmotic pressure.

(4) 0.100 M $\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}$ (aq) has the highest osmotic pressure.

[Jee (Main)-2014]

Sol. (3)

Q. Determination of the molar mass of acetic acid in benzene using freezing point depression is affected by :

(1) association

(2) dissociation

(3) complex formation

(4) partial ionization

[Jee (Main)-2015 online]

Sol. (1)

Acetic acid in non polar solvent (benzene) associates.

Q. A solution at $20^{\circ} \mathrm{C}$ is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are 74.7 torr and 22.3 torr, respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively :

(1) 38.0 torr and 0.589

(2) 30.5 torr and 0.389

(3) 35.8 torr and 0.280

(4) 35.0 torr and 0.480

[Jee (Main)-2015 online]

Sol. (1)

\begin{aligned} \mathrm{P}_{\mathrm{T}} &=\mathrm{P}_{\mathrm{A}}^{0} \mathrm{X}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^{0} \mathrm{X}_{\mathrm{B}} \\ &=747 \times \frac{1.5}{5}+22.3 \times \frac{3.5}{5} \\ &=38 \mathrm{torr} \end{aligned}

Q. The vapour pressure of acetone at $20^{\circ}$C is 185 torr. When 1.2 g of non-volatile substance was dissolved in 100 g of acetone at 20^{\circ}20^{\circ}C, its vapour pressure was 183 torr. The molar mass \left(\mathrm{g} \mathrm{mol}^{-1}\right) of the substance is : (1) 128 (2) 488 (3) 32 (4) 64 [Jee (Main)-2015] Sol. (4) \begin{array}{rl}{\frac{185-183}{185}} & {=\frac{1.2 / \mathrm{m}}{100 / 58}} \\ {\mathrm{m}=64} & {\mathrm{gm} / \mathrm{mol}}\end{array} Q. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point ? (1) \left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \cdot \mathrm{H}_{2} \mathrm{O} (2) \left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl} .2 \mathrm{H}_{2} \mathrm{O} (3) \left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] \cdot 3 \mathrm{H}_{2} \mathrm{O} (4) \left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3} [Jee (Main)-2018] Sol. (3) Kinematics 1D- JEE Advanced Previous Year Questions with Solutions Download India's Best Exam Preparation App Class 9-10, JEE & NEET 1,00,000+ learners 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. Download eSaral app for free study material and video tutorials. Simulator Previous Years JEE Advanced Questions Q. A train is moving along a straight line with a constant acceleration ‘a’. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of 60^{\circ} to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train, in \mathrm{m} / \mathrm{s}^{2}, is [IIT-JEE 2011] Sol. 5 Q. A rocket is moving in a gravity free space with a constant acceleration of 2 \mathrm{ms}^{-2} along + x direction (see figure). The length of a chamber inside the rocket is 4m. A ball is thrown from the left end of the chamber in + x direction with a speed of 0.3 \mathrm{ms}^{-1} relative to the rocket. At the same time, another ball is thrown in –x direction with a speed of 0.2 \mathrm{ms}^{-1} from its right end relative to the rocket. The time in seconds when the two balls hit each other is [JEE Advanced 2014] Sol. 8 or 2 Assuming open chamber \mathrm{V}_{\text {relative }}=0.5 \mathrm{m} / \mathrm{s} \mathrm{S}_{\text {relative }}=4 \mathrm{m} time =\frac{4}{0.5}=8 \mathrm{m} / \mathrm{s} Alternate Assuming closed chamber In the frame of chamber : Maximum displacement of ball A from its left end is \frac{\mathrm{u}_{\mathrm{A}}^{2}}{2 \mathrm{a}}=\frac{(0.3)^{2}}{2(2)}=0.0225 \mathrm{m} This is negligible with respect to the length of chamber i.e. 4m. So, the collision will be verym close to the left end. Hence, time taken by ball B to reach left end will be given by \mathrm{S}=\mathrm{u}_{\mathrm{B}} \mathrm{t}+\frac{1}{2} \mathrm{at}^{2} 4=(0.2)(\mathrm{t})+\frac{1}{2}(2)(\mathrm{t})^{2} Solving this, we get \mathrm{t} \approx 2 \mathrm{s} Q. Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30^{\circ} and 60^{\circ} with respect to the horizontal respectively as shown in figure. The speed of A is \mathrm{ms}^{-1}. At time t = 0 s, an observer in A finds B at a distance of 500 m. This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t = \mathrm{t}_{0}, A just escapes being hit by B, \mathrm{t}_{0} in seconds is [JEE Advanced-2014] Sol. 5 As observed from A, B moves perpendicular to line of motion of A. It means velocity of B along A is equal to velocity of A \mathrm{V}_{\mathrm{B}} \cos 30=100 \sqrt{3} \mathrm{V}_{\mathrm{B}}=200 If A is observer A remains stationary therefore \mathrm{t}=\frac{500}{\mathrm{V}_{\mathrm{B}} \sin 30}=\frac{500}{100}=5 Q. Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is that the instantaneous density \rho remains uniform throughout the volume. the rate of fractional change in density is \left(\frac{1}{\rho} \frac{d \rho}{d t}\right) constant. the velocity v of any point on the surface of the expanding sphere is proportional to (A) R^{3} (B) \frac{1}{R} (C) \mathrm{R} (D) R^{2 / 3} [JEE Advanced-2017] Sol. (C) Vector- JEE Advanced Previous Year Questions with Solutions Download India's Best Exam Preparation App Class 9-10, JEE & NEET 1,00,000+ learners 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. Download eSaral app for free study material and video tutorials. Simulator Previous Years JEE Advanced Questions Q. Three vectors \overrightarrow{\mathrm{P}}, \overrightarrow{\mathrm{Q}} and \overrightarrow{\mathrm{R}} are shown in the figure. Let S be any point on the vector \overrightarrow{\mathrm{R}}. The distance between the points P and S is b |\overrightarrow{\mathrm{R}}|. The general relation among vectors \overrightarrow{\mathrm{P}}, \overrightarrow{\mathrm{Q}} and \overrightarrow{\mathrm{S}} is : (\mathrm{A}) \overrightarrow{\mathrm{S}}=(1-\mathrm{b}) \overrightarrow{\mathrm{P}}+\mathrm{b}^{2} \overrightarrow{\mathrm{Q}} (B) \overrightarrow{\mathrm{S}}=(b-1) \overrightarrow{\mathrm{P}}+b \overrightarrow{\mathrm{Q}} (C) \overrightarrow{\mathrm{S}}=(1-\mathrm{b}) \overrightarrow{\mathrm{P}}+\mathrm{b} \overrightarrow{\mathrm{Q}} (\mathrm{D}) \overrightarrow{\mathrm{S}}=\left(1-\mathrm{b}^{2}\right) \overrightarrow{\mathrm{P}}+\mathrm{b} \overrightarrow{\mathrm{Q}} [JEE Advanced – 2017] Sol. (C) Let vector from point P to point S be \overrightarrow{\mathrm{c}} \Rightarrow \overrightarrow{\mathrm{c}}=\mathrm{b}|\overrightarrow{\mathrm{R}}| \hat{\mathrm{R}}=\mathrm{b}|\overrightarrow{\mathrm{R}}|\left(\frac{\overrightarrow{\mathrm{R}}}{|\overrightarrow{\mathrm{R}}|}\right)=\mathrm{b} \overrightarrow{\mathrm{R}}=\mathrm{b}(\overrightarrow{\mathrm{Q}}-\overrightarrow{\mathrm{P}}) from triangle rule of vector addition \overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{S}} \overrightarrow{\mathrm{P}}+\mathrm{b}(\overrightarrow{\mathrm{Q}}-\overrightarrow{\mathrm{P}})=\overrightarrow{\mathrm{S}} \Rightarrow \overrightarrow{\mathrm{S}}=(1-\mathrm{b}) \overrightarrow{\mathrm{P}}+\mathrm{b} \overrightarrow{\mathrm{Q}} Q. Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density  remains uniform throughout the volume. The rate of fractional change in density \left(\frac{1}{\rho} \frac{\mathrm{d} \rho}{\mathrm{dt}}\right) is constant. The velocity v of any point on the surface of the expanding sphere is proportional to : (A) \mathrm{R}^{3} (B) \frac{1}{\mathrm{R}} (C) R (D) \mathrm{R}^{2 / 3} [JEE Advanced – 2017] Sol. (C) Density of sphere is \rho=\frac{\mathrm{m}}{\mathrm{v}}=\frac{3 \mathrm{m}}{4 \pi \mathrm{R}^{3}} \Rightarrow \frac{1}{\rho} \frac{\mathrm{d} \rho}{\mathrm{dt}}=-\frac{3}{\mathrm{R}} \frac{\mathrm{d} \mathrm{R}}{\mathrm{dt}} since \Rightarrow \frac{1}{\rho} \frac{\mathrm{d} \rho}{\mathrm{dt}} is constant \therefore \frac{\mathrm{d} \mathrm{R}}{\mathrm{dt}} \propto \mathrm{R} Velocity of any point on the circumfrence V is equal to \frac{\mathrm{d} \mathrm{R}}{\mathrm{dt}} (rate of change of radius of outer layer) Ray Optics – JEE Advanced Previous Year Questions with Solutions Download India's Best Exam Preparation App Class 9-10, JEE & NEET 1,00,000+ learners 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. Download eSaral app for free study material and video tutorials. Simulator Previous Years JEE Advanced Questions Q. A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is 12.8 m above the water surface, the fish sees the speed of ball as \left[\mathrm{g}=10 \mathrm{m} / \mathrm{s}^{2} .\right] (A) 9 m/s (B) 12 m/s (C) 16 m/s (D) 21.33 m/s [IIT-JEE 2009] Sol. (C) Q. A student performed the experiment of determination of focal length of a concave mirror by u-v method using an optical bench of length 1.5 m. The focal length of the mirror used is 24 cm. The maximum error in the location of the image can be 0.2 cm. The 5 sets of (u, v) values recorded by the student (in cm) are : (42, 56), (48, 48), (60, 40), (66, 33), (78, 39). The data set(s) that cannot come from experiment and is (are) incorrectly recorded, is (are) : (A) (42, 56) (B) (48, 48) (C) (66, 33) (D) (78, 39) [IIT-JEE 2009] Sol. (C,D) \mathrm{V}=\frac{\mathrm{uf}}{\mathrm{u}-\mathrm{f}} \quad by sustituting the value of u and \mathrm{f} \mathrm{V}_{42} \Rightarrow \frac{24 \times 42}{18} \Rightarrow 56 ; \quad \mathrm{V}_{60}=\frac{24 \times 60}{36} \Rightarrow 40 \mathrm{v}_{48} \Rightarrow \frac{48 \times 24}{24} \Rightarrow 48 ; \mathrm{V}_{66}=\frac{66 \times 24}{42} \Rightarrow 37.71 \mathrm{V}_{78} \Rightarrow \frac{78 \times 24}{54} \Rightarrow 34.3 (66, 33) ; (78, 39) Q. A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is – (A) virtual and at a distance of 16 cm from the mirror (B) real and at a distance of 16 cm from the mirror (C) virtual and at a distance of 20 cm from the mirror (D) real and at a distance of 20 cm from the mirror [IIT-JEE 2010] Sol. (B) \frac{1}{15}=\frac{1}{v}-\frac{1}{10} Q. A ray OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle of 60^{\circ} (see figure). If the refractive index of the material of the prism is \sqrt{3}, which of the following is (are) correct? (A) The ray gets totally internally reflected at face CD (B) The ray comes out through face AD (C) The angle between the incident ray and the emergent ray is 90^{\circ} (D) The angle between the incident ray and the emergent ray is 120^{\circ} [IIT-JEE 2010] Sol. (A,B,C) Q. Two transparent media of refractive indices \mu_{1} and \mu_{3} have a solid lens shaped transparent material of refractive index \mu_{2} between them as shown in figures in Column II. A ray traversing these media is also shown in the figures. In Column I different relationships between \mu_{1}, \mu_{2} and \mu_{3} are given. Match them to the ray diagrams shown in Column II. [IIT-JEE 2010] Sol. (\mathrm{A})-\mathrm{pr},(\mathrm{B})-\mathrm{qst},(\mathrm{C})-\mathrm{prt},(\mathrm{D})-\mathrm{qs} Q. The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its image changes from \mathrm{m}_{25} to \mathrm{m}_{50} . The ratio \frac{m_{25}}{m_{50}} is – [IIT-JEE 2010] Sol. 6 \mathrm{m}=\frac{\mathrm{f}}{\mathrm{f}+\mathrm{u}} Q. Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to move from \frac{25}{3} m to \frac{50}{7} m in 30 seconds. What is the speed of the object in km per hour ? [IIT-JEE 2010] Sol. 3 \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \Rightarrow \frac{1}{10}=\frac{1}{u}+\frac{3}{25} \Rightarrow u=-50 \mathrm{m} \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}} \Rightarrow \frac{1}{10}=\frac{1}{\mathrm{u}}+\frac{7}{50} \Rightarrow \mathrm{u}=-25 \mathrm{m} Speed =\frac{25}{30} \times \frac{18}{5}=3 Q. A large glass slab \left(\mu=\frac{5}{3}\right) of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R? [IIT-JEE 2010] Sol. 6 r=\frac{h}{\sqrt{\mu^{2}-1}}=\frac{8}{\sqrt{\left(\frac{5}{3}\right)^{2}-1}}=6 \mathrm{cm} Q. A light ray traveling in glass medium is incident on glass-air interface at an angle of incidence \theta. The reflected (R) and transmitted (T) intensities, both as function of \theta, are plotted. The correct sketch is – [IIT-JEE 2011] Sol. (C) When \theta=0^{\circ}, maximum light is transmitted. At \theta>\theta_{\mathrm{C}} (critical angle), no further light is transmitted Q. Water (with refractive index = \frac{4}{3}) in a tank is 18 cm deep. Oil of refractive index \frac{7}{4} lies on water making a convex surface of radius of curvature ‘R = 6 cm’ as shown. Consider oil to act as a thin lens. An object ‘S’ is placed 24 cm above water surface. The location of its image is at ‘x’ cm above the bottom of the tank. Then ‘x’ is [IIT-JEE 2011] Sol. 2 First refraction [ Lens-air interface] \frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R} where \mu_{1}=1, \mathrm{u}=-24, \mu_{2}=\frac{7}{4}, \mathrm{R}=+6 After solving v = 21 Now for second refraction [Lens-water interface] \frac{4 / 3}{v_{2}}-\frac{7 / 4}{21}=0 \Rightarrow v_{2}=h=16 \mathrm{cm} So, from bottom 18-16=2 \Rightarrow \mathrm{x}=2 Q. A biconvex lens is formed with two thin plano-convex lenses as shown in the figure, Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvature R = 14 cm. For this biconvex lens, for an object distance of 40 cm, the image distance will be :- (A) –280.0 cm (B) 40.0 cm (C) 21.5 cm (D) 13.3 cm [IIT-JEE 2012] Sol. (B) \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_{1}}+\frac{1}{\mathrm{f}_{2}} \frac{1}{\mathrm{f}}=\left(\frac{\mu_{1}-1}{\mathrm{R}_{1}}\right)+\left(\frac{\mu_{2}-1}{\mathrm{R}_{2}}\right) \Rightarrow \frac{1}{\mathrm{f}}=\frac{5}{14}+\frac{2}{14} f = 20 c.m. \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \Rightarrow \frac{1}{\mathrm{v}}-\frac{1}{-40}=\frac{1}{20}=40 \mathrm{c.m} Paragraph for Questions 12 and 13 Most materials have the refractive index, n>1. So, when a light ray from air enters a naturally occurring material, then by Snell’s law, \frac{\sin \theta_{1}}{\sin \theta_{2}}=\frac{n_{2}}{n_{1}} , it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, n=\left(\frac{c}{v}\right)=\pm \sqrt{\varepsilon_{r} \mu_{r}} , where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, \varepsilon_{\mathrm{r}} and \mu_{\mathrm{r}} are the relative permittivity and permeability of the medium respectively. In normal materials, both \varepsilon_{\mathrm{r}} and \mu_{\mathrm{r}} are positive, implying positive n for the medium. When both \varepsilon_{\mathrm{r}} and \mu_{\mathrm{r}} are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called meta-materials. They exhibit significantly different optical behaviour, without violating any physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta materials. Q. For light incident from air on a meta-material, the appropriate ray diagram is – [IIT-JEE 2012] Sol. (C) _{1} \mu_{2}=\frac{\mu_{2}}{\mu_{1}}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}} \frac{(-)}{1}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}} sini = sin(–r) Q. Choose the correct statement. (A) The speed of light in the meta-material is \mathrm{v}=\mathrm{c}|\mathrm{n}| (B) The speed of light in the meta-material is v = \frac{c}{|n|} (C) The speed of light in the meta-material is v = c. (D) The wavelength of the light in the meta-material \left(\lambda_{\mathrm{m}}\right) is given by \lambda_{\mathrm{m}}=\lambda_{\mathrm{air}}|\mathrm{n}|, where \lambda_{\mathrm{air}} is the wavelength of the light in air. [IIT-JEE 2012] Sol. (B) \mu=\frac{\mathrm{c}}{\mathrm{v}} \Rightarrow \mathrm{v}=\frac{\mathrm{c}}{\mu}=\frac{\mathrm{c}}{\mathrm{n}} Q. A ray of light travelling in the direction \frac{1}{2}(\hat{i}+\sqrt{3} \hat{j}) is incident on a plane mirror. After reflection, it travels along the direction \frac{1}{2}(\hat{i}-\sqrt{3} \hat{j}). The angle of incidence is :- (A) 30^{\circ} (B) 45^{\circ} (C) 60^{\circ} (D) 75^{\circ} [JEE-Advance-2013] Sol. (A) Here normal is along \hat{j} Angle between incident ray and normal \cos \theta=\frac{\frac{1}{2}(\hat{i}+\sqrt{3} \hat{j}) \cdot \hat{j}}{(1)(1)}=\frac{\sqrt{3}}{2} \Rightarrow \theta=30^{\circ} Q. The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is \frac{2}{3} times the wavelength in free space. The radius of the curved surface of the lens is : (A) 1 m (B) 2 m (C) 3 m (D) 6 m [JEE-Advance-2013] Sol. (C) \mathrm{m}=-\frac{1}{3}=\frac{\mathrm{v}}{\mathrm{u}} v = 8m u = – 24 m \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \rightarrow \frac{1}{8}-\frac{1}{-24}=\frac{1}{\mathrm{f}} \Rightarrow \mathrm{f}=6 \mathrm{m} \mathrm{f}=\frac{\mathrm{R}}{(\mu-1)} \mu=\frac{\lambda_{\text {vaceum }}}{\lambda_{\text {medium }}}=\frac{3}{2} 6 \mathrm{m}=\left(\frac{\mathrm{R}}{3 / 2-1}\right) \Rightarrow \mathrm{R}=3 \mathrm{m} Q. A right angled prism of refractive index \mu_{1} is placed in a rectangular block of refractive index \mu_{2}, which is surrounded by a medium of refractive index \mu_{3}, as shown in the figure. A ray of light ‘e’ enters the rectangular block at normal incidence. Depending upon the relationships between \mu_{1}, \mu_{2}, and \mu_{3}, it takes one of the four possible paths ‘ef’, ‘eg’, ‘eh’ or ‘ei’. Match the paths in List I with conditions of refractive indices in List II and select the correct answer using the codes given below the lists : [JEE-Advance-2013] Sol. (D) (P) at prism surface ray moving towards normal so \left(\mu_{2}>\mu_{1}\right)at block surface ray moving away from normal so \left(\mu_{3}<\mu_{2}\right) (Q) No deflection of ray on both surface; so \mu_{1}=\mu_{2}=\mu_{3} (R) At prism surface ray moving away from normal so \mu_{2}<\mu_{1}. At block surface ray movign away from normal so µ3 < µ2 but since on total internal reflection not takes place on prism surface \mu_{1} \sin 45^{\circ}<\mu_{2} \sin 90^{\circ} \Rightarrow \mu_{1}<\sqrt{2} \mu_{2} (S) Total internal reflection takes place so \mu_{1} \sin 45^{\circ}>\mu_{2} \sin 90^{\circ} \Rightarrow \mu_{1}>\sqrt{2} \mu_{2} Q. A transparent thin film of uniform thickness and refractive index \mathrm{n}_{1}=1.4 is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index \mathrm{n}_{2}=1.5, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance \mathrm{f}_{1} from the film, while rays of light traversing from glass to air get focused at distance \mathrm{f}_{2} from the film. Then (\mathrm{A})\left|\mathrm{f}_{1}\right|=3 \mathrm{R} (B) \left|\mathrm{f}_{1}\right|=2.8 \mathrm{R} (C) \left|\mathrm{f}_{2}\right|=2 \mathrm{R}b (\mathrm{D})\left|\mathrm{f}_{2}\right|=1.4 \mathrm{R} [JEE-Advance-2014] Sol. (A,C) When rays are moving from air to glass, \frac{1.5}{\mathrm{f}_{1}}=\frac{(1.4-1)}{+\mathrm{R}}+\frac{(1.5-1.4)}{+\mathrm{R}} \frac{1.5}{\mathrm{f}_{1}}=\frac{0.4}{\mathrm{R}}+\frac{0.1}{\mathrm{R}}=\frac{0.5}{\mathrm{R}} \left|\mathrm{f}_{1}\right|=3 \mathrm{R} When rays are moving from glass to air, \frac{1}{\mathrm{F}_{2}}=\frac{(1-1.4)}{-\mathrm{R}}+\frac{(1.4-1.5)}{-\mathrm{R}}=\frac{0.5}{\mathrm{R}} \left|\mathrm{f}_{2}\right|=2 \mathrm{R} Q. A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is :- (A) 1.21 (B) 1.30 (C) 1.36 (D) 1.42 [JEE-Advance-2014] Sol. (C) From the given situation, at critical angle, \tan \theta=\frac{(\mathrm{d} / 2)}{\mathrm{h}}=\frac{5.77}{10} \therefore \sin \theta_{\mathrm{C}} \approx \frac{1}{2} \mu_{\text {denser }} \sin \theta_{\mathrm{C}}=\mu_{\text {raver }} \sin (\pi / 2) \Rightarrow 2.72 \times \frac{1}{2}=\mu_{\mathrm{r}} \Rightarrow \mu_{\mathrm{r}}=1.36 Q. Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is 1.5. Match lens combinations in List I with their focal length in List II and select the correct answer using the code given below the lists. [JEE-Advance-2014] Sol. (B) Q. Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification \mathrm{M}_{1}. When the set-up is kept in a medium of refractive index 7/6 the magnification becomes \mathrm{M}_{2}. The magnitude \left|\frac{M_{2}}{M_{1}}\right| is. [JEE-Advance-2015] Sol. 7 For reflectionfrom concave mirror, \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \Rightarrow \frac{1}{\mathrm{v}}-\frac{1}{15}=\frac{-1}{10} \frac{1}{\mathrm{v}}=\frac{1}{15}-\frac{1}{10}=\frac{-1}{30} \therefore \mathrm{v}=-30 magnification \left(\mathrm{m}_{1}\right)=-\frac{\mathrm{v}}{\mathrm{u}}=-2 Now for refraction from lens, \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20} \therefore magnification \left(\mathrm{m}_{2}\right)=\frac{\mathrm{v}}{\mathrm{u}}=-1 \therefore \mathrm{M}_{1}=\mathrm{m}_{1} \mathrm{m}_{2}=2 Now when the set-up is immersed in liquid, no effect for the image formed by mirror. we have \left(\mu_{\mathrm{L}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)=\frac{1}{10} \Rightarrow\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)=\frac{1}{5} when lens is immersed in liquid, \frac{1}{\mathrm{f}_{\mathrm{lens}}}=\left(\frac{\mu_{\mathrm{L}}}{\mu_{\mathrm{S}}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)=\frac{2}{7} \times \frac{1}{5}=\frac{2}{35} \therefore \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}_{\mathrm{Liquid}}} \Rightarrow \frac{1}{\mathrm{v}}=\frac{2}{35}-\frac{1}{20}=\frac{8-7}{140}=\frac{1}{140} \therefore magnification =-\frac{140}{20}=-7 \therefore \mathrm{M}_{2}=2 \times 7=14 \therefore\left|\frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}\right|=7 Q. Two identical glass rods \mathrm{S}_{1} and \mathrm{S}_{2} (refractive index = 1.5) have one convex end of radius of curvature 10 cm. They are placed with the curved surfaces at a distance d as shown in the figure, with their axes (shown by the dashded line) aligned. When a point source of light P is placed inside rod \mathrm{S}_{1} on its axis at a distance of 50 cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside \mathrm{S}_{2}. The distance d is : (A) 60 cm (B) 70 cm (C) 80 cm (D) 90 cm [JEE-Advance-2015] Sol. (B) For first surface \frac{1}{V}-\frac{1.5}{-50}=\frac{1-1.5}{-10} V = 50 cm for second surface \frac{1.5}{\infty}-\frac{1}{-(\mathrm{d}-50)}=\frac{1.5-1}{10} d = 70 cm \therefore(\mathrm{B}) Q. A monochromatic beam of light is incident at 60^{\circ} on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle \theta(n) with the normal (see the figure). For n = \sqrt{3} the value of \theta is 60^{\circ} and \frac{\mathrm{d} \theta}{\mathrm{dn}}=\mathrm{m}. The value of m is. [JEE-Advance-2015] Sol. 2 By snell’s law 1 \sin 60=n \sin r_{1} \Rightarrow \sin r_{1}=\frac{1}{2} r_{1}=30^{\circ} \quad \ldots \ldots(i) By differentiating ‘w.r.t’ n \mathrm{O}=\sin \mathrm{r}_{1}+\mathrm{n} \cos \mathrm{r}_{1}\left(\frac{\mathrm{dr}_{1}}{\mathrm{dn}}\right) =\frac{1}{2}+\sqrt{3}(\sqrt{\frac{3}{2}}) \frac{\mathrm{dr}_{1}}{\mathrm{dn}} \frac{\mathrm{d} \mathrm{r}_{1}}{\mathrm{dn}}=\frac{1}{3} ….(ii) By applying snell’s law n \sin r_{2}=1 \sin \theta n \sin \left(60-r_{1}\right)=1 \sin \theta\left[\therefore A=r_{1}+r_{2}\right] By diffrentiating ‘w.r.t’ n \sin \left(60-r_{1}\right)-n \cos \left(60-r_{1}\right) \frac{d r_{1}}{d n}=\cos \theta \frac{d \theta}{d n}\sin \left(60-r_{1}\right)-n \cos \left(60-r_{1}\right) \frac{d r_{1}}{d n}=\cos \theta \frac{d \theta}{d n}

By substituting value of $\mathrm{r}_{1}^{\prime}$ and $\frac{\mathrm{dr}_{1}}{\mathrm{dn}}$ from ( 1) and ( 2)

$\frac{\mathrm{d} \theta}{\mathrm{dn}}=2$

Paragraph for Question No. 23 and 24

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index $\mathbf{n}_{1}$ surrounded by a medium of lower refractive index $\mathrm{n}_{2}$. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media $\mathbf{n}_{1}$ and $\mathrm{n}_{2}$ as shown in the figure. All rays with the angle of incidence i less than a particular value of $\dot{l}_{\mathrm{m}}$ are confined in the medium of refractive index $\mathrm{n}_{1}$. The numerical aperture (NA) of the structure is defined as sin $i_{\mathrm{m}}$.

Q. For two structures namely $\mathrm{S}_{1}$ with $\mathrm{n}_{1}=\sqrt{45} / 4$ and $\mathrm{n}_{2}=3 / 2,$ and $\mathrm{S}_{2}$ with $\mathrm{n}_{1}=8 / 5$ and $\mathrm{n}_{2}=7 / 5$ and

taking the refractive index of water to be $4 / 3$ and that of air to be $1,$ the correct option(s) is (are)

(A) NA of $\mathrm{S}_{1}$ immersed in water is the same as that of $\mathrm{S}_{2}$ immersed in liquid of refractive index $\frac{16}{3 \sqrt{15}}$

(B) NA of $\mathrm{S}_{1}$ immersed in liquid of refractive index $\frac{6}{\sqrt{15}}$ is the same as that of $\mathrm{S}_{2}$ immersed in water.

(C) NA of $\mathrm{S}_{1}$ placed in air is the same as that of $\mathrm{S}_{2}$ immersed in liquid of refractive index .

(D) NA of $\mathrm{S}_{1}$ placed in air is the same as that of $\mathrm{S}_{2}$ placed in water.

Sol. (A,C)

Let the whole structure is placed in a medium of refractive index n’, then

$n^{\prime} \sin i=n_{1} \sin (90-\theta)$

$\mathrm{n}^{\prime} \sin \mathrm{i}=\mathrm{n}_{1} \cos \theta \quad \ldots(\mathrm{i})$

Here for $\mathrm{i}_{\mathrm{m}} ; \quad \theta=\mathrm{C}$ and $\sin \mathrm{C}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}$

from eq. (i), $\mathrm{n}^{\prime} \sin \mathrm{i}_{\mathrm{m}}=\mathrm{n}_{1} \sqrt{\frac{1-\mathrm{n}_{2}^{2}}{\mathrm{n}_{1}^{2}}}=\sqrt{\mathrm{n}_{1}^{2}-\mathrm{n}_{2}^{2}}$

$\Rightarrow \sin i_{m}=\frac{\sqrt{n_{1}^{2}-n_{2}^{2}}}{n^{\prime}}$

Now, for $(\mathrm{A})(\mathrm{NA})_{\mathrm{s} 1}=\frac{3}{4} \sqrt{\frac{45}{16}-\frac{9}{4}}=\frac{3}{4} \times \frac{3}{4}=\frac{9}{16}$

$(\mathrm{NA})_{\mathrm{s} 2}=\frac{3 \sqrt{15}}{16} \sqrt{\frac{64}{25}-\frac{49}{25}}=\frac{3 \sqrt{15}}{16} \frac{1}{5} \sqrt{15}=\frac{9}{16}$

For (B) $\quad(\mathrm{NA})_{\mathrm{s} 1}=\frac{\sqrt{15}}{6} \times \frac{3}{4}=\frac{\sqrt{15}}{8}$

$(\mathrm{NA})_{\mathrm{s} 2}=\frac{3}{4}=\frac{\sqrt{15}}{5}$ Not equal

For $(\mathrm{C})(\mathrm{NA})_{\mathrm{s} 1}=1 \times \frac{3}{4}=\frac{3}{4}$

$(\mathrm{NA})_{\mathrm{s} 2}=\frac{\sqrt{15}}{4} \times \frac{\sqrt{15}}{5}=\frac{15}{4 \times 5}=\frac{3}{4}$

For (D) $(\mathrm{NA})_{\mathrm{s} 1}=\frac{3}{4}$

$(\mathrm{NA})_{\mathrm{s} 2}=\frac{3}{4} \frac{\sqrt{15}}{5}$ Not equal

Q. If two structures of same cross-sectional area, but different numerical apertures $\mathrm{NA}_{1}$ and $\mathrm{NA}_{2}$ $\left(\mathrm{NA}_{2}<\mathrm{NA}_{1}\right)$ are joined longitudinally, the numerical aperture of the combined structure is

(A) $\frac{\mathrm{NA}_{1} \mathrm{NA}_{2}}{\mathrm{NA}_{1}+\mathrm{NA}_{2}}$

$(\mathrm{B}) \mathrm{NA}_{1}+\mathrm{NA}_{2}$

$(\mathrm{C}) \mathrm{NA}_{1}$

$(\mathrm{D}) \mathrm{NA}_{2}$

Sol. (D)

$\begin{array}{ll}{\text { It is given that }} & {\mathrm{NA}_{2}<\mathrm{NA}_{1}} \\ {\Rightarrow \mathrm{i}_{\mathrm{m} 2}<\mathrm{i}_{\mathrm{m} 1}}\end{array}$

Hence if the combination can be placed both ways i.e. 1st structure & then 2nd structure and then reversed also, then the condition of TIR is satisfied for lower $\dot{\mathbf{1}}_{\mathrm{m}}$ then it can be satisfied for all other less angler as well.

Hence $\mathrm{NA}_{2}$ will be the numerical aperture of the combined structure.

Q. A parallel beam of light is incident from air at an angle  on the side PQ of a right angled triangular prism of refractive index $\mathrm{n}=\sqrt{2}$. Light undergoes total internal reflection in the prism at the face PR when  has a minimum value of $45^{\circ}$. The angle $\theta$ of the prism is :

(A) $15^{\circ}$

(B) $22.5^{\circ}$

(C) $30^{\circ}$b

(D) $45^{\circ}$

Sol. (A)

$1 \sin 45^{\circ}=\sqrt{2} \sin \mathrm{r}_{1}$

$\mathrm{r}_{2}-\mathrm{r}_{1}=\theta$

$\theta=45^{\circ}-30^{\circ}$

$\Rightarrow \theta=15^{\circ}$

Q. A transparent slab of thickness d has a refractive index n(z) that increases with z. Here z is the vertical distance inside the slab, measured from the top. The slab is placed between two media with uniform refractive indices $\mathrm{n}_{1}$ and $\mathrm{n}_{2}\left(>\mathrm{n}_{1}\right)$, as shown in the figure. A ray of light is incident with angle $\theta_{\mathrm{i}}$ from medium 1 and emerges in medium 2 with refraction angle $\theta_{\mathrm{f}}$ with a lateral displacement . Which of the following statement(s) is(are) true ?

(A) $\ell$ is independent of $\mathrm{n}_{2}$

(B) $\ell$ is dependent on $\mathrm{n}(\mathrm{z})$

(C) $\mathrm{n}_{1} \sin \theta_{\mathrm{i}}=\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right) \sin \theta_{\mathrm{f}}$

(D) $\mathrm{n}_{1} \sin \theta_{\mathrm{i}}=\mathrm{n}_{2} \sin \theta_{\mathrm{f}}$

Sol. (A,B,D)

Q. A plano-convex lens is made of a material of refractive index n. When a small object is placed 30 cm away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of 10 cm away from the lens. Which of the following statement(s) is(are) true?

(A) The refractive index of the lens is 2.5

(B) The radius of curvature of the convex surface is 45 cm

(C) The faint image is erect and real

(D) The focal length of the lens is 20 cm.

Sol. (A,D)

Q. A small object is placed 50 cm to the left of thin convex lens of focal length 30 cm. A convex spherical mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm. The mirror is tilted such that the axis of the mirror is at an angle  = $30^{\circ}$ to the axis of the lens, as shown in the figure. If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the point (x, y) at which the image is formed are :

(A) $(25,25 \sqrt{3})$

$(\mathrm{B})\left(\frac{125}{3}, \frac{25}{\sqrt{3}}\right)$

(C) $(50-25 \sqrt{3}, 25)$

(D) (0, 0)

Sol. (A)

For lens $\mathrm{V}=\frac{(-50)(30)}{-50+30}=75$

For mirror $\mathrm{V}=\frac{\left(\frac{25 \sqrt{3}}{2}\right)(50)}{\frac{25 \sqrt{3}}{2}-50}=\frac{-50 \sqrt{3}}{4-\sqrt{3}}$

$\mathrm{m}=-\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{h}_{2}}{\mathrm{h}_{1}} \Rightarrow \mathrm{h}_{2}=-\left(\frac{\frac{-50 \sqrt{3}}{4-\sqrt{3}}}{\frac{25 \sqrt{3}}{2}}\right) \cdot \frac{25}{2}$

$\mathrm{h}_{2}=\frac{+50}{4-\sqrt{3}}$

The x coordinate of the images $=50-v \cos 30+h_{2} \cos 60 \approx 25$

The y coordinate of the images $=v \sin 30+h_{2} \sin 60 \approx 25 \sqrt{3}$

Q. For an isosceles prism of angle A and refractive index µ, it is found that the angle of minimum deviation $\delta_{\mathrm{m}}$ = A. Which of the following options is/are correct ?

(A) At minimum deviation, the incident angle $\dot{\mathbf{1}}_{1}$ and the refracting angle $\mathrm{r}_{1}$ at the first refracting surface are related by $\mathbf{r}_{1}=\left(\mathbf{i}_{1} / 2\right)$

(B) For this prism, the refractive index µ and the angle of prism A are related as $\mathrm{A}=\frac{1}{2} \cos ^{-1}\left(\frac{\mu}{2}\right)$

(C) For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $\mathrm{i}_{1}=\sin ^{-1}[\sin \mathrm{A} \sqrt{4 \cos ^{2} \frac{\mathrm{A}}{2}-1}-\cos \mathrm{A}]$

(D) For the angle of incidence $\mathbf{i}_{1}$ = A, the ray inside the prism is parallel to the base of the prism.

Sol. (A,C,D)

i = e (for minimum deviation)

$\mathrm{r}_{1}+\mathrm{r}_{2}=\mathrm{A}, \mathrm{r}_{1}=\mathrm{r}_{2}$

(A) $\delta_{\mathrm{m}}=2 \mathrm{i}-\mathrm{A}=\mathrm{A}$ (given)

$\Rightarrow \mathrm{i}=\mathrm{A}$

$\Rightarrow \mathrm{r}_{1}=\frac{\mathrm{A}}{2}=\frac{\mathrm{i}}{2}$

(B) $\mu=\frac{\sin (\mathrm{A})}{\sin (\mathrm{A} / 2)}=2 \cos \frac{\mathrm{A}}{2} \Rightarrow \mathrm{A}=2 \cos ^{-1}\left(\frac{\mu}{2}\right)$

(C) $\mu \sin \left(\mathrm{r}_{2}\right)=1$

$\sin \left(\mathrm{r}_{2}\right)=\frac{1}{\mu}$

$\mathrm{r}_{1}+\mathrm{r}_{2}=\mathrm{A}$

$\mathrm{r}_{1}=\mathrm{A}-\mathrm{r}_{2}$

$=\mathrm{A}-\sin ^{-1}\left[\frac{1}{\mu}\right]$

$\sin (\mathrm{i})=\mu \sin \left(\mathrm{r}_{1}\right)$

$\mathrm{i}=\sin ^{-1}\left[\mu \sin \left[\mathrm{A}-\sin ^{-1}\left[\frac{1}{\mu}\right]\right]\right.$

$\mathrm{i}_{\mathrm{g}}=\sin ^{-1}[\sqrt{\mu^{2}-1} \sin \mathrm{A}-\cos \mathrm{A}]=\sin ^{-1}\left[\mu \sin \left(\mathrm{A}-\theta_{\mathrm{C}}\right)\right]$

(Here $\left.\mu=2 \cos \frac{\mathrm{A}}{2}\right)$

(D) Condition of min. deviation $\mathrm{i}=\mathrm{e} \& \mathrm{r}_{1}=\mathrm{r}_{2}=\frac{\mathrm{A}}{2}$

Rays will be parallel to base.

Q. A monochromatic light is travelling in a medium of refractive index n = 1.6. It enters a stack of glass layers from the bottom side at an angle $\theta=30^{\circ}$. The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as $\mathrm{n}_{\mathrm{m}}=\mathrm{n}-\mathrm{m} \Delta \mathrm{n}$, where $\mathrm{n}_{\mathrm{m}}$ is the refractive index of the mth slab and $\Delta \mathrm{n}=0.1$ (see the figure). The ray is refracted out parallel to the interface between the (m – 1)th and mth slabs from the right side of the stack. What is the value of m ?

Sol. 8

Applying snell’s law between entry & exit surfaces,

n $\sin \theta=\mu \sin \left(\frac{\pi}{2}\right)$

Q. Sunlight of intensity $1.3 \mathrm{kW} \mathrm{m}^{-2}$ is incident normally on a thin convex lens of focal length 20 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, in kW m–2, at a distance 22 cm from the lens on the other side is

Sol. 130

$\frac{\mathrm{r}}{\mathrm{R}}=\frac{2}{20}=\frac{1}{10}$

$\therefore$ Ratio of area $=\frac{1}{100}$

Let energy incident on lens be E.

$\therefore$ Given $\frac{\mathrm{E}}{\mathrm{A}}=1.3$

So final, $\frac{\mathrm{E}}{\mathrm{a}}=? ?$

E = A × 1.30

Also $\frac{\mathrm{a}}{\mathrm{A}}=\frac{1}{100}$

$\therefore$ Average intensity of light at $22 \mathrm{cm}=\frac{\mathrm{E}}{\mathrm{a}}=\frac{\mathrm{A} \times 1.3}{\mathrm{a}}=100 \times 1.3=130 \mathrm{kW} / \mathrm{m}^{2}$

Q. A wire is bent in the shape of a right angled triangle and is placed in front of a concave mirror of focal length f, as shown in the figure. Which of the figures shown in the four options qualitatively represent(s) the shape of the image of the bent wire ? (These figures are not to scale.) ?

Sol. (D)

Distance of point A is f/2

Let A’ is the image of A from mirror, for this image

$\frac{1}{\mathrm{v}}+\frac{1}{-\mathrm{f} / 2}=\frac{1}{-\mathrm{f}}$

$\frac{1}{\mathrm{v}}=\frac{2}{\mathrm{f}}-\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}}$

image of line AB should be perpendicular to the principle axis & image of F will form at infinity, therefor correct image diagram is

Newtons Law of Motion-JEE Advanced Previous Year Questions with Solutions

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Q. A piece of wire is bent in the shape of a parabola y = $\mathrm{Kx}^{2}$ (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is

[IIT-JEE-2009]

(A) $\frac{\mathrm{a}}{\mathrm{gk}}$            (B) $\frac{\mathrm{a}}{2 \mathrm{gk}}$            (C) $\frac{2 \mathrm{a}}{\mathrm{gk}}$                  (D) $\frac{\mathrm{a}}{4 \mathrm{gk}}$

Sol. (B)

$\operatorname{ma} \cos \theta=\operatorname{mg} \sin \theta$

$\mathrm{a}=\mathrm{g} \tan \theta$

$\frac{a}{g}=\tan \theta$

$\frac{a}{g}=2 k x$

$\frac{a}{2 g k}=x$

Ray Optics – JEE Main Previous Year Questions with Solutions

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Communication System – JEE Main Previous Year Questions with Solutions

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Previous Years AIEEE/JEE Mains Questions

Q. This questions has Statement-1 and Statement-2. Of the four choice given after the statements, choose the one that best describes the two statements.

Statement-1 : Sky wave signals are used for long distance radio communication. These signals are in general, less stable than ground wave signals.

Statement-2 : The state of ionosphere varies from hour to hour, day to day and season to season.

(1) Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of statement-1

(2) Statement-1 is true, Statement-2 is true, Statrment-2 is not the correct explanation of Statement-1

(3) Statement-1 is false, Statement-2 is true

(4) Statement-1 is true, Statement-2 is false

[AIEEE-2011]

Sol. (2)

Q. A radar has a power of 1 kW and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m. The maximum distance up to which it can detect object located on the surface of the earth (Radius of earth $\left.=6.4 \times 10^{6} \mathrm{m}\right)$ is :

(1) 40 km (2) 64 (3) 80 km (4) 16km

[AIEEE-2012]

Sol. (3)

$\mathrm{d}=\sqrt{2 \mathrm{Rh}}=\sqrt{2 \times 6400 \times 0.5}=80 \mathrm{km}$

Q. A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance 100 kilo ohm. Find the maximum modulated frequency which could be detected by it.

(1) 10.62kHz (2) 5.31 MHz (3) 5.31 kHz (4) 10.62MHz

[JEE Main-2013]

Sol. (3)

$\mathrm{f}_{\mathrm{max}} \leq \frac{\sqrt{\frac{1}{\mathrm{m}^{2}}-1}}{2 \pi \mathrm{RC}}$

$\mathrm{f}_{\max } \leq \frac{8}{6 \times 2 \pi \times 100 \times 10^{3} \times 250 \times 10^{-12}}$

$\mathrm{f}_{\max } \leq \frac{8 \times 10^{6}}{12 \pi \times 25}$

$\mathrm{f}_{\max } \leq 8.4925 \mathrm{kHz}$

Q. A single of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are –

(1) 2005 kHz, 2000 kHz and 1995 kHz

(2) 2000 kHz and 1995 kHz

(3) 2 MHz only

(4) 2005 kHz and 1995 kHz

[JEE Main-2015]

Sol. (1)

Frequency present after modulation

$\mathrm{f}_{\mathrm{c}}, \mathrm{f}_{\mathrm{c}} \pm \mathrm{f}_{\mathrm{s}}$

$\Rightarrow 2000 \mathrm{KHz}, 2005 \mathrm{KHz}$ and $1995 \mathrm{KHz}$

Q. Choose the correct statement :

(1) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal.

(2) In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

(3) In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

(4) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

[JEE Main-2016]

Sol. (2)

Q. In amplitude modulation, sinusoidal carrier frequency used is denoted by $\omega_{\mathrm{c}}$ and the signal frequency is denoted by $\omega_{\mathrm{m}}$. The bandwidth $\left(\Delta \omega_{\mathrm{m}}\right)$ of the signal is such that $\Delta \omega_{\mathrm{m}}<<\omega_{\mathrm{c}}$. Which of the following frequencies is not contained in the modulated wave ?

[JEE Main-2017]

Sol. (3)

Refer NCERT Page No. 526 Three frequencies are contained $\omega_{\mathrm{m}}+\omega_{\mathrm{c}}, \omega_{\mathrm{c}}-\omega_{\mathrm{m}} \& \omega_{\mathrm{c}}$

Q. A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz ?

[JEE Main-2018]

Sol. (2)

Since the carrier frequency is distributed as band width frequency, so 10% of 10 GHz = n × 5 kHz

where n = no of channels $\frac{10}{100} \times 10 \times 10^{9}=n \times 5 \times 10^{3} \mathrm{n}=2 \times 10^{5}$ telephonic channels

Kinematics 2D – JEE Main Previous Year Questions with Solutions

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Previous Years AIEEE/JEE Mains Questions

Q. A particle is moving with velocity $\overrightarrow{\mathrm{v}}=\mathrm{K}(\mathrm{y} \hat{\mathrm{i}}+\mathrm{x} \hat{\mathrm{j}})$ where K is a constant. The general equation for its path is :

(1) $\mathrm{y}^{2}=\mathrm{x}^{2}+$ constant

(2) $\mathrm{y}=\mathrm{x}^{2}+$ constant

(3) $\mathrm{y}^{2}=\mathrm{x}+$ constant

(4) xy = constant

[AIEEE – 2010]

Sol. (1)

$\overrightarrow{\mathrm{v}}=\mathrm{k}(\mathrm{y} \hat{\mathrm{i}}+\mathrm{x} \hat{\mathrm{j}})$

$\overrightarrow{\mathrm{v}}=\mathrm{v}_{\mathrm{x}} \hat{\mathrm{i}}+\mathrm{v}_{\mathrm{y}} \hat{\mathrm{j}}$

$\overrightarrow{\mathrm{v}}=\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}} \hat{\mathrm{i}}+\frac{\mathrm{dy}}{\mathrm{dt}} \hat{\mathrm{j}}$

$\frac{d x}{d t}=k y \& \frac{d y}{d t}=k x$

$\Rightarrow \frac{d x}{d y}=\frac{y}{x}$

$\Rightarrow \mathrm{y}^{2}=\mathrm{x}^{2}+$ constant

Q. A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is :-

(1) $\frac{\pi}{2} \frac{v^{4}}{g^{2}}$

(2) $\pi \frac{\mathrm{v}^{2}}{\mathrm{g}^{2}}$

( 3)$\pi \frac{\mathrm{v}^{2}}{\mathrm{g}}$

( 4)$\pi \frac{\mathrm{v}^{4}}{\mathrm{g}^{2}}$

[AIEEE – 2011]

Sol. (4)

$\mathrm{r}=\mathrm{R}_{\max }=\frac{\mathrm{v}^{2}}{\mathrm{g}}$

area $=\pi \mathrm{r}^{2}$

$=\pi\left(\frac{\mathrm{v}^{2}}{\mathrm{g}}\right)^{2}$

$=\pi \frac{\mathrm{v}^{4}}{\mathrm{g}^{2}}$

Q. A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force $\mathrm{F}(\mathrm{t})=\mathrm{F}_{0} \mathrm{e}^{-\mathrm{bt}}$ in the x direction. Its speed v(t) is depicted by which of the following curves ?

[AIEEE – 2012]

Sol. (3)

$\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=\frac{\mathrm{F}_{0}}{\mathrm{m}} \mathrm{e}^{-\mathrm{bt}}$

$\int_{0}^{\mathrm{v}} \mathrm{d} \mathrm{v}=\frac{\mathrm{F}_{0}}{\mathrm{m}} \int_{0}^{\mathrm{t}} \mathrm{e}^{-\mathrm{bt}} \mathrm{dt}$

$\mathrm{v}=\frac{-\mathrm{F}_{0}}{\mathrm{m} \mathrm{b}}\left[\mathrm{e}^{-\mathrm{bt}}-1\right]$

$=\frac{\mathrm{F}_{0}}{\mathrm{m} \mathrm{b}}\left[1-\mathrm{e}^{-\mathrm{bt}}\right]$

Q. A projectile is given an initial velocity of $(\hat{i}+2 \hat{j}) m / s$ where $\hat{\mathbf{i}}$ is along the ground and $\hat{j}$ is along the vertical. If g = 10 m/s2, the equation of its trajectory is :

(1) $\mathrm{y}=\mathrm{x}-5 \mathrm{x}^{2}$

(2) $\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}$

(3) $4 y=2 x-5 x^{2}$

(4) $4 y=2 x-25 x^{2}$

[AIEEE – 2013]

Sol. (2)

$\mathrm{u}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}$

$\mathrm{u}_{\mathrm{x}}=1$

$\mathrm{u}_{\mathrm{y}}=2$

$\tan \theta=\frac{2}{1}$

$\mathrm{y}=\mathrm{x} \tan \theta-\frac{1}{2} \mathrm{g} \frac{\mathrm{x}^{2}}{\mathrm{u}^{2} \cos ^{2} \theta}$

$\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}$

Q. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first ?

(Assume stones do not rebound after hitting the ground and neglect air resistance, take

g = 10 $\mathrm{M} / \mathrm{S}^{2}$) (The figure are schematic and not drawn to scale)

[JEE Mains – 2015]

Sol. (1)

$\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}$

For particle 2

$-240=40 \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2}$

$5 t^{2}-40 t-240=0$

$\mathrm{t}_{2}=12 \mathrm{sec}$

for $0<\mathrm{t}<8$ sec $\rightarrow \mathrm{a}_{\text {rel }}=0$

straight line x-t graph

for $8<\mathrm{t}<12$ sec $\rightarrow \mathrm{a}_{\mathrm{rel}}=-\mathrm{g}$

downward parabola

for $\mathrm{t}>12 \mathrm{sec} \rightarrow$ Both particles comes to rest

Kinematics 2D – JEE Main Previous Year Questions with Solutions

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Q. A train is moving along a straight line with a constant acceleration ‘a’. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of $60^{\circ}$ to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train, in $\mathrm{m} / \mathrm{s}^{2}$, is –

[IIT-JEE 2011]

Sol. 5

Q. A rocket is moving in a gravity free space with a constant acceleration of 2 $\mathrm{ms}^{-2}$along + x direction (see figure). The length of a chamber inside the rocket is 4m. A ball is thrown from the left end of the chamber in + x direction with a speed of 0.3 $\mathrm{ms}^{-1}$ relative to the rocket. At the same time, another ball is thrown in –x direction with a speed of 0.2 $\mathrm{ms}^{-1}$ from its right end relative to the rocket. The time in seconds when the two balls hit each other is –

Sol. 8 or 2

Assuming open chamber

$\mathrm{V}_{\text {relative }}=0.5 \mathrm{m} / \mathrm{s}$

$S_{\text {relative }}=4 \mathrm{m}$

time $=\frac{4}{0.5}=8 \mathrm{s}$

Alternate

Assuming closed chamber

In the frame of chamber :

Maximum displacement of ball A from its left end is $\frac{\mathrm{u}_{\mathrm{A}}^{2}}{2 \mathrm{a}}=\frac{(0.3)^{2}}{2(2)}=0.0225 \mathrm{m}$

This is negligible with respect to the length of chamber i.e. 4m. So, the collision will be very close to the left end.

Hence, time taken by ball B to reach left end will be given by

$\mathrm{S}=\mathrm{u}_{\mathrm{B}} \mathrm{t}+\frac{1}{2} \mathrm{at}^{2}$

$4=(0.2)(t)+\frac{1}{2}(2)(t)^{2}$

Solving this, we get

$\mathrm{t} \approx 2 \mathrm{s}$

Q. Airplanes A and B are flying with constant velocity in the same vertical plane at angles $30^{\circ}$ and $60^{\circ}$ with respect to the horizontal respectively as shown in figure. The speed of A is $\mathrm{ms}^{-1}$. At time t = 0 s, an observer in A finds B at a distance of 500 m. This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t = $t_{0}$, A just escapes being hit by B, $t_{0}$ in seconds is

Sol. 5

As observed from A, B moves perpendicular to line of motion of A. It means velocity of B along A is equal to velocity of A

$\mathrm{V}_{\mathrm{B}} \cos 30=100 \sqrt{3}$

$\mathrm{V}_{\mathrm{B}}=200$

If A is observer A remains stationary therefore

$\mathrm{t}=\frac{500}{\mathrm{V}_{\mathrm{B}} \sin 30}=\frac{500}{100}=5$

Q. A ball is projected from the ground at an angle of $45^{\circ}$ with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of $30^{\circ}$ with the horizontal surface. The maximum height it reaches after the bounce, in metres, is

Sol. 30

$\mathrm{H}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} 45}{2 \mathrm{g}}=120$

$\Rightarrow \frac{\mathrm{u}^{2}}{4 \mathrm{g}}=120$ ….(i)

when half of kinetic energy is lost $\mathrm{v}=\frac{\mathrm{u}}{\sqrt{2}}$

$\mathrm{H}_{2}=\frac{\left(\frac{\mathrm{u}}{\sqrt{2}}\right)^{2} \sin ^{2} 30}{2 \mathrm{g}}=\frac{\mathrm{u}^{2}}{16 \mathrm{g}}$

from (i) & (ii)

$\mathrm{H}_{2}=\frac{\mathrm{H}_{1}}{4}=30 \mathrm{m}$ on 30.00

Physics Revision series for JEE, NEET, Class 11 and 12

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Revise complete Physics on eSaral YouTube Channel. Get everything you are looking for JEE and NEET Preparation in this Physics Revision Series by Saransh Gupta Sir. Here you will get to know about important key concepts and all the formulae that are very important to solve conceptual problems that will come in main Exam.

All the Revision Videos and mind maps are very important concepts including some tricks and tips to solve Multiple Choice questions of JEE & NEET in less time.

### Class XI

TOPIC Revision Videos Mind Maps
Unit & Dimensions Revise Mindmap
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JEE Advanced Previous Year Papers Chapter-wise with Solution

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Here you will find JEE Advanced Previous Year Papers which are divided chapter-wise. All the JEE Advanced Previous Year papers are provided with complete and detailed Solution. You can download them in the PDF format or you can also read them online. eSaral also provides JEE Main Topic-wise Previous Year Question Papers with Solutions.

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.

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Physics Topic-wise JEE Advanced Previous Year Questions with Solutions

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Practicing JEE Advanced Previous Year Questions 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 papers questions with solutions.
These JEE Advanced previous Year Question for Physics plays an important role in IIT-JEE preparation. We are providing IIT-JEE Mains Previous Year Question Papers with detailed Solution.

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 Physics Topic-wise Previous year question for JEE Main:

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.

Download eSaral app to study from free video tutorials for JEE & NEET by top IITian Faculties of Kota.

Work, Power & Energy – JEE Main Previous Year Questions with Solutions

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JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their 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.

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Previous Years AIEEE/JEE Main Questions

Q. The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $\mathrm{U}(\mathrm{x})=\frac{\mathrm{a}}{\mathrm{x}^{12}}-\frac{\mathrm{b}}{\mathrm{x}^{6}}$ , where a and b are constant and x is the distance between the atoms. if the dissociation energy of the molecule is $\left[\mathrm{U}(\mathrm{x}=\infty)-\mathrm{U}_{\text {at equilibrium }}\right], \mathrm{D}$ is :

( 1)$\frac{b^{2}}{6 a}$

( 2)$\frac{\mathrm{b}^{2}}{2 \mathrm{a}}$

(3) $\frac{b^{2}}{12 a}$

(4) $\frac{\mathrm{b}^{2}}{4 \mathrm{a}}$

[AIEEE-2010]

Sol. (4)

$\mathrm{U}(\mathrm{x}=\infty)=0,$ at equilibrium

$\frac{\mathrm{d} \mathrm{U}}{\mathrm{dx}}=0 ;-\frac{12 \mathrm{a}}{\mathrm{x}^{13}}+\frac{6 \mathrm{b}}{\mathrm{x}^{7}}=0 \Rightarrow \mathrm{x}^{6}=\frac{2 \mathrm{a}}{\mathrm{b}}$

Q. At time t = 0s particle starts moving along the x-axis. If its kinetic energy increases uniformly with time ‘t’, the net force acting on it must be proportional to :-

(1) $\sqrt{t}$

(2) constant

(3) t

(4) $\frac{1}{\sqrt{t}}$

[AIEEE-2011]

Sol. (4)

$\frac{1}{2} \mathrm{mv}^{2}=\mathrm{kt}$

$\mathrm{v}=\sqrt{\frac{2 \mathrm{kt}}{\mathrm{m}}}$

$\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=\sqrt{\frac{\mathrm{k}}{2 \mathrm{mt}}}$

$\mathrm{F}=\mathrm{m} \cdot \frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}$

$\mathrm{F} \propto \frac{1}{\sqrt{\mathrm{t}}}$

Q. This question has Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

If two springs $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ of force constants $\mathrm{k}_{1}$ and $\mathrm{k}_{2}$, respectively, are stretched by the same force, it is found that more work is done on spring $\mathrm{S}_{1}$ than on spring $\mathrm{S}_{2}$.

Statement-1: If stretched by the same amount, work done on $\mathrm{S}_{1}$, will be more than that on $\mathrm{S}_{2}$

Statement-2 : $\mathrm{k}_{1}$ < $\mathrm{k}_{2}$

(1) Statement-1 is true, Statement-2 is true and Statement-2 is not the correct explanation of Statement-1.

(2) Statement-1 is false, Statement-2 is true

(3) Statement-1 is true, Statement-2 is false

(4) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of statement-1.

[AIEEE-2012]

Sol. (2)

Given same force $\mathrm{F}=\mathrm{k}_{1} \mathrm{x}_{1}=\mathrm{k}_{2} \mathrm{x}_{2}$

$\Rightarrow \frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}=\frac{\mathrm{x}_{2}}{\mathrm{x}_{1}}$

$\mathrm{W}_{1}=\frac{1}{2} \mathrm{k}_{1} \mathrm{x}_{1}^{2} \& \mathrm{W}_{2}=\frac{1}{2} \mathrm{k}_{2} \mathrm{x}_{2}^{2}$

As $\frac{\mathrm{W}_{1}}{\mathrm{W}_{2}}>1$ so $\frac{\frac{1}{2} \mathrm{k}_{1} \mathrm{x}_{1}^{2}}{\frac{1}{2} \mathrm{k}_{2} \mathrm{x}_{2}^{2}}>1$

$\Rightarrow \frac{\mathrm{Fx}_{1}}{\mathrm{Fx}_{2}}>1 \Rightarrow \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}>1$

$\therefore \mathrm{k}_{2}>\mathrm{k}_{1}$ statement 2 is true

OR

if $\mathrm{x}_{1}=\mathrm{x}_{2}=\mathrm{x}$

$\frac{\mathrm{W}_{1}}{\mathrm{W}_{2}}=\frac{\frac{1}{2} \mathrm{K}_{1} \mathrm{x}^{2}}{\frac{1}{2} \mathrm{K}_{2} \mathrm{x}^{2}}=\frac{\mathrm{K}_{1}}{\mathrm{K}_{2}}$

$\therefore \frac{\mathrm{W}_{1}}{\mathrm{W}_{2}}=\frac{\mathrm{K}_{1}}{\mathrm{K}_{2}}<1$

$\therefore \mathrm{W}_{1}<\mathrm{W}_{2}$

statement 1 is false

Q. When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude $\mathrm{F}=\mathrm{ax}+\mathrm{bx}^{2}$ where a and b are constants. The work done in stretching the unstretched rubber-band by L is:-

(1) $\frac{\mathrm{aL}^{2}}{2}+\frac{\mathrm{bL}^{3}}{3}$

(2) $\frac{1}{2}\left(\frac{\mathrm{aL}^{2}}{2}+\frac{\mathrm{b} \mathrm{L}^{3}}{3}\right)$

(3) aL $^{2}+\mathrm{bL}^{3}$

(4) $\frac{1}{2}\left(\mathrm{aL}^{2}+\mathrm{bL}^{3}\right)$

[JEE-Mains-2014]

Sol. (1)

Work done $=\int_{0}^{\mathrm{L}} \mathrm{Fdx}$

$=\int_{0}^{\mathrm{L}}\left(\mathrm{ax}+\mathrm{bx}^{2}\right) \mathrm{d} \mathrm{x}$

$=\frac{\mathrm{aL}^{2}}{2}+\frac{\mathrm{bL}^{3}}{3}$

Q. A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $3.8 \times 10^{7}$ J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 $\mathrm{ms}^{-2}$ :-

(1) $12.89 \times 10^{-3} \mathrm{kg}$

(2) $2.45 \times 10^{-3} \mathrm{kg}$

(3) $6.45 \times 10^{-3} \mathrm{kg}$

(4) $9.89 \times 10^{-3} \mathrm{kg}$

[JEE-Mains-2016]

Sol. (1)

Work done against gravity = (mgh) 1000

in lifting 1000 times

$=10 \times 9.8 \times 10^{3}$

$=9.8 \times 10^{4}$ Joule

$20 \%$ efficiency is to converts fat into energy.

$\left[20 \% \text { of } 3.8 \times 10^{7} \mathrm{J}\right] \times(\mathrm{m})=9.8 \times 10^{4}$

(Where m is mass)

$\mathrm{m}=12.89 \times 10^{-3} \mathrm{kg}$

Q. A point particle of mass, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals . The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and PR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction $\mu$ and the distance x(=QR) are, respecitvely close to :

[JEE-Mains-2016]

Sol. (4)

Q. A body of mass m $=10^{-2}$ kg is moving in a medium and experiences a frictional force F = $-\mathrm{K} \mathrm{V}^{2}$. Its intial speed is $\mathrm{v}_{0}$ = 10 $\mathrm{ms}^{-1}$. If, after 10 s, its energy is , the value of k will be :-

(1) $10^{-4} \mathrm{kg} \mathrm{m}^{-1}$

(2) $10^{-1} \mathrm{kg} \mathrm{m}^{-1} \mathrm{s}^{-1}$

(3) $10^{-3} \mathrm{kg} \mathrm{m}^{-1}$

(4) $10^{-3} \mathrm{kg} \mathrm{s}^{-1}$

[JEE MAINS-2017]

Sol. (1)

$\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}=\frac{1}{8} \mathrm{mv}_{0}^{2}$

$\mathrm{v}_{\mathrm{f}}=\frac{\mathrm{v}_{0}}{2}=5 \mathrm{m} / \mathrm{s}$

$\left(10^{-2}\right) \frac{d V}{d t}=-k v^{2}$

$\int_{10}^{5} \frac{\mathrm{dv}}{\mathrm{v}^{2}}=-100 \mathrm{k} \int_{0}^{10} \mathrm{dt}$

$\frac{1}{5}-\frac{1}{10}=100 \mathrm{k}(10)$

$\mathrm{k}=10^{-4}$

Q. A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be :

(1) 9 J            (2) 18 J              (3) 4.5 J            (4) 22 J

[JEE MAINS-2017]

Sol. (3)

F = 6t = ma

$\Rightarrow \mathrm{a}=6 \mathrm{t}$

$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=6 \mathrm{t}$

$\int_{0}^{\mathrm{v}} \mathrm{d} \mathrm{v}=\int_{0}^{1} 6 \mathrm{t} \mathrm{dt}$

$\mathrm{v}=\left(3 \mathrm{t}^{2}\right)_{0}^{1}=3 \mathrm{m} / \mathrm{s}$

From work energy theorem

$\mathrm{W}_{\mathrm{F}}=\Delta \mathrm{K.E}=\frac{1}{2} \mathrm{m}\left(\mathrm{v}^{2}-\mathrm{u}^{2}\right)$

$=\frac{1}{2}(1)(9-0)=4.5 \mathrm{J}$

Q. A particle is moving in a circular path of radius a under the action of an attractive potential $\mathrm{U}=-\frac{\mathrm{k}}{2 \mathrm{r}^{2}} .$ Its total energy is :-.

(1) $\frac{\mathrm{k}}{2 \mathrm{a}^{2}}$

(2) Zero

(3) $-\frac{3}{2} \frac{\mathrm{k}}{\mathrm{a}^{2}}$

$(4)-\frac{\mathrm{k}}{4 \mathrm{a}^{2}}$

[JEE MAINS-2018]

Sol. (2)

$\mathrm{F}=-\frac{\partial \mathrm{u}}{\partial \mathrm{r}}=\frac{\mathrm{K}}{\mathrm{r}^{3}}$

Since it is performing circular motion

$\mathrm{F}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\frac{\mathrm{K}}{\mathrm{r}^{3}}$

$\mathrm{mv}^{2}=\frac{\mathrm{K}}{\mathrm{r}^{2}}$

$\Rightarrow \mathrm{K.E.}=\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{K}}{2 \mathrm{r}^{2}}$

Total energy = P.E. + K.E.

$=-\frac{\mathrm{K}}{2 \mathrm{r}^{2}}+\frac{\mathrm{K}}{2 \mathrm{r}^{2}}=\mathrm{Zero}$

Radioactivity – JEE Main Previous Year Questions with Solutions

Class 9-10, JEE & NEET

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JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their 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.

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.

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Previous Years AIEEE/JEE Mains Questions

Q. The half life of a radioactive substance is 20 minutes. The approximate time interval $\left(\mathfrak{t}_{2}-\mathfrak{t}_{1}\right)$ between the time $t_{2}$ when $\frac{2}{3}$ of it has decayed and time $t_{1}$ when $\frac{1}{3}$ of it had decayed is :-

(1) 20 min (2) 28 min (3) 7 min (4) 14 min

[AIEEE – 2011]

Sol. (1)

\because \frac{\mathrm{N}}{\mathrm{N}_{0}}=\left[\frac{1}{2}\right]^{t / T} \quad \therefore \frac{1}{3}=\left[\frac{1}{2}\right]^{\frac{t_{2}}{T}} \& \frac{2}{3}=\left[\frac{1}{2}\right]^{\frac{t_{1}}{T}}

\Rightarrow \frac{1}{2}=\left[\frac{1}{2}\right]^{\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right) \frac{1}{\mathrm{T}}} \Rightarrow 1=\frac{\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)}{\mathrm{T}} \Rightarrow \mathrm{T}=\mathrm{t}_{2}-\mathrm{t}_{1}

\Rightarrow \mathrm{t}_{2}-\mathrm{t}_{1}=20 \mathrm{min}

Q. Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have equal number of nuclei. After 80 minutes , the ratio of decayed numbers of A and B nuclei will be :-

(1) 5 : 4 (2) 1 : 16 (3) 4 : 1 (4) 1 : 4

[JEE-Mains – 2016]

Sol. (1)

\mathrm{t}=80 \mathrm{min}=4 \mathrm{T}_{\mathrm{A}}=2 \mathrm{T}_{\mathrm{B}}

\therefore \text { no. of nuclei of } \mathrm{A} \text { decayed }=\mathrm{N}_{0}-\frac{\mathrm{N}_{0}}{2^{4}}=\frac{15 \mathrm{N}_{0}}{16}

\therefore \text { no. of nuclei of } \mathrm{B} \text { decayed }=\mathrm{N}_{0}-\frac{\mathrm{N}_{0}}{2^{2}}=\frac{3 \mathrm{N}_{0}}{4}

required ratio $=\frac{5}{4}$

Q. A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by :

(1) $\mathrm{t}=\mathrm{T} \log (1.3)$

$(2) \mathrm{t}=\frac{\mathrm{T}}{\log (1.3)}$

(3) $\mathrm{t}=\frac{\mathrm{T}}{2} \frac{\log 2}{\log 1.3}$

(4) \mathrm{t}=\mathrm{T} \frac{\log 1.3}{\log 2}

[JEE-Mains – 2017]

Sol. (4)

At time t

$\frac{\mathrm{N}_{\mathrm{B}}}{\mathrm{N}_{\mathrm{A}}}=.3 \Rightarrow \mathrm{N}_{\mathrm{B}}=.3 \mathrm{N}_{\mathrm{A}}$

also let initially there are total $\mathrm{N}_{0}$ number of nuclei

$\mathrm{N}_{\mathrm{A}}+\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{0}$

$\mathrm{N}_{\mathrm{A}}=\frac{\mathrm{N}_{0}}{1.3}$

Also as we know

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

$\frac{\mathrm{N}_{0}}{1.3}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$

$\frac{1}{1.3}=\mathrm{e}^{-\lambda t} \Rightarrow \ell \mathrm{n}(1.3)=\lambda \mathrm{t}$ or $\mathrm{t}=\frac{\ell \mathrm{n}(1.3)}{\lambda}$

$\mathrm{t}=\frac{\ell \mathrm{n}(1.3)}{\frac{\ell \mathrm{n}(2)}{\mathrm{T}}}=\frac{\ell \mathrm{n}(1.3)}{\ell \mathrm{n}(2)} \mathrm{T}$

Photoelectric Effect – JEE Main Previous Year Questions with Solutions

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JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their 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.

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.

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Previous Years AIEEE/JEE Mains Questions

Q. The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is : (hc = 1240 eV-nm)

(1) 1.51 eV            (2) 1.68 eV          (3) 3.09 eV              (4) 1.41 eV

[AIEEE – 2009]

Sol. (4)

$\mathrm{E}_{\mathrm{k}}=\frac{\mathrm{hc}}{\lambda}-\phi_{0} \Rightarrow 1.68=\frac{12400}{4000}-\phi_{0}$

By solving it $\phi_{0}=1.42 \mathrm{eV}$

Q. Statement-1 : When ultraviolet light is incident on a photocell, its stopping potential is V0 and the maximum kinetic energy of the photoelectrons is $\mathrm{K}_{\mathrm{max}}$. When the ultraviolet light is replaced by X-rays, both $\mathrm{V}_{0}$ and $\mathrm{K}_{\mathrm{max}}$ increase.

Statement-2 : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light

(1) Statement–1 is true, Statement–2 is false

(2) Statement–1 is true, Statement–2 is true; Statement–2 is the correct explanation of Statement– 1

(3) Statement–1 is true, Statement–2 is true; Statement–2 is not the correct explanation of Statement– 1

(4) Statement–1 is false, Statement–2 is true

[AIEEE – 2010]

Sol. (1)

Speed of emitted electrons is independent of frequency of incident light.

Q. This question has Statememtn-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

Statement–1 : A metallic surface is irradiated by a monochromatic light of frequency $\mathrm{v}>\mathrm{v}_{0}$ (the threshold frequency). The maximum kinetic energy and the stopping potential are $\mathrm{K}_{\max }$ and $\mathrm{V}_{0}$ respectively. If the frequency incident on the surface is doubled, both the $\mathrm{K}_{\max }$ and $\mathrm{V}_{0}$ are also boubled.

Statement-2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.

(1) Statement–1 is true, Statement–2 is true, Statement–2 is not the correct explanationof Statement– 1

(2) Statement–1 is false, Statement–2 is true

(3) Statement–1 is true, Statement–2 is false

(4) Statement–1 is true, Statement–2 is true, Statement–2 is the correct explanation of Statement– 1

[AIEEE-2011]

Sol. (2)

Q. The anode voltage of photocell is kept fixed. The wavelength $\lambda$ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows

[AIEEE-2013]

Sol. (4)

For constant intensity as wavelength decreases energy of photons increases and number of photons decreases. So it may seem that current should decrease. But the probability that a photon will be successful in emitting an electron will also increase. So as wavelength decreases current increases

Q. Radiation of wavelength $\lambda$, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength of changed to $\frac{3 \lambda}{4}$, the speed of the fastest emitted electron will be :-

[JEE Main-2016]

Sol. (2)

$\mathrm{E}=(\mathrm{KE})_{\max }+\mathrm{f}$

$\left[\frac{\mathrm{hc}}{\lambda}=(\mathrm{KE})_{\max }+\phi\right] \ldots .$ (1)

\frac{4}{3} \frac{\mathrm{hc}}{\lambda}=\left(\frac{4}{3} \mathrm{KE}_{\max }+\frac{\phi}{3}\right)+\phi

Nuclear Physics – JEE Main Previous Year Questions with Solutions

Class 9-10, JEE & NEET

1,00,000+ learners

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their 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.

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.

Simulator

Previous Years AIEEE/JEE Mains Questions

Q. The above is a plot of binding energy per nucleon $\mathrm{E}_{\mathrm{b}}$, against the nuclear mass M; A, B, C, D, E, F correspond to different nuclei. Consider four reactions :

(i) $\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}+\varepsilon$

(ii) $\mathrm{C} \rightarrow \mathrm{A}+\mathrm{B}+\varepsilon$

(iii) $\mathrm{D}+\mathrm{E} \rightarrow \mathrm{F}+\varepsilon$

$(\mathrm{iv}) \mathrm{F} \rightarrow \mathrm{D}+\mathrm{E}+\varepsilon$

where $\varepsilon$ is the energy released ? In which reactions is $\varepsilon$ positive ?

(1) (ii) and (iv) (2) (ii) and (iii) (3) (i) and (iv) (4) (i) and (iii)

[AIEEE – 2009]

Sol. (3)

Q. The speed of daughter nuclei is :-

(1) $\mathrm{c} \sqrt{\frac{\Delta \mathrm{m}}{\mathrm{M}+\Delta \mathrm{m}}}$

(2) $\mathrm{c} \frac{\Delta \mathrm{m}}{\mathrm{M}+\Delta \mathrm{m}}$

(3) $\mathrm{c} \sqrt{\frac{2 \Delta \mathrm{m}}{\mathrm{M}}}$

(4) $\mathrm{c} \sqrt{\frac{\Delta \mathrm{m}}{\mathrm{M}}}$

[AIEEE-2010]

Sol. (3)

Total kinetic energy of products

$=$ Total energy released $\frac{\mathrm{p}^{2}}{2 \mathrm{m}}+\frac{\mathrm{p}^{2}}{2 \mathrm{m}}$

$\left.=(\text { mass defect }) \mathrm{c}^{2} \text { (where } \mathrm{m}=\frac{\mathrm{M}}{2} \text { given }\right)$

$\Rightarrow 2\left(\frac{\mathrm{p}^{2}}{2 \mathrm{m}}\right)=\left[(\mathrm{M}+\Delta \mathrm{m})-\left(\frac{\mathrm{M}}{2}+\frac{\mathrm{M}}{2}\right)\right] \times \mathrm{c}^{2}$

$\Rightarrow 2 \times\left[\frac{\mathrm{p}^{2}}{2\left(\frac{\mathrm{M}}{2}\right)}\right]=(\Delta \mathrm{m}) \mathrm{c}^{2}$

$\Rightarrow \frac{2\left(\frac{\mathrm{M}}{2} \mathrm{v}\right)^{2}}{\mathrm{M}}=(\Delta \mathrm{m}) \mathrm{c}^{2} \Rightarrow \mathrm{v}=\mathrm{c} \sqrt{\frac{2 \Delta \mathrm{m}}{\mathrm{M}}}$

Q. The binding energy per nucleon for the parent nucleus is $E_{1}$ an that for the daughter nuclei is $\mathrm{E}_{2}$. Then:-

(1) $\mathrm{E}_{1}=2 \mathrm{E}_{2}$

$(2) \mathrm{E}_{2}=2 \mathrm{E}_{1}$

(3) $\mathrm{E}_{1}>\mathrm{E}_{2}$

$(4) \mathrm{E}_{2}>\mathrm{E}_{1}$

[AIEEE – 2010]

Sol. (4)

Because energy is releasing $\Rightarrow$ Binding energy per nucleon of product> that of parent $\Rightarrow \mathrm{E}_{2}>$

$\mathrm{E}_{1}$

Q. A radioactive nucleus (initial mass number A and atomic number Z) emits 3 -particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be:-

(1) $\frac{\mathrm{A}-\mathrm{Z}-4}{\mathrm{Z}-2}$

(2) $\frac{\mathrm{A}-\mathrm{Z}-8}{\mathrm{Z}-4}$

(3) $\frac{\mathrm{A}-\mathrm{Z}-4}{\mathrm{Z}-8}$

(4) $\frac{\mathrm{A}-\mathrm{Z}-12}{\mathrm{Z}-4}$

[AIEEE – 2010]

Sol. (3)

Q. After absorbing a slowly moving neutron of mass $\left.\mathrm{m}_{\mathrm{N}} \text { (momentum } \sim 0\right)$ a nucleus of mass M breaks into two nuclei of masses $\mathrm{m}_{1}$ and $5 \mathrm{m}_{1}\left(6 \mathrm{m}_{1}=\mathrm{M}+\mathrm{m}_{\mathrm{N}}\right)$, respectively. If the de Broglie wavelength of the nucleus with mass $\mathrm{m}_{1}$ is $\lambda$, then de Broglie wavelength of the other nucleus will be:-

(1) $25 \lambda$

(2) $5 \lambda$

(3) $\frac{\lambda}{5}$

( 4)$\lambda$

[AIEEE – 2011]

Sol. (4)

Q. Statement-1: A nucleus having energy $\mathrm{E}_{1}$ decays be $\beta^{-}$ emissionto daughter nucleus having energy $E_{2}$, but the – rays are emitted with a continuous energy spectrum having end point energy $\mathrm{E}_{1}-\mathrm{E}_{2}$.

Statement-1: To conserve energy and momentum in -decay at least three particles must take part in the transformation.

(1) Statement-1 is incorrect, statement-2 is correct

(2) Statement-1 is correct, statement-2 is incorrect

(3) Statement-1 is correct, statement-2 correct; statement-2 is the correct explanation of statement-1

(4) Statement-1 is correct, statement-2 is correct; statement -2 is not the correct explanation of statement-1.

[AIEEE – 2011]

Sol. (3)

Q. Assume that a neutron breaks into a proton and an electron. The energy released during this process is :

(Mass of neutron $=1.6725 \times 10^{-27} \mathrm{kg}$

Mass of proton $=1.6725 \times 10^{-27} \mathrm{kg}$

Mass of electron $\left.=9 \times 10^{-31} \mathrm{kg}\right)$

(1) 5.4 MeV (2) 0.73 MeV (3) 7.10 MeV (4) 6.30 MeV

[AIEEE – 2012]

Sol. (2)

Released energy

$=\left[1.6747 \times 10^{-27}-1.6725 \times 10^{-27}-9 \times 10^{-31}\right]$

$\times\left(3 \times 10^{8}\right)^{2} \mathrm{J}=0.73 \mathrm{MeV}$

Atomic Structure – JEE Main Previous Year Questions with Solutions

Class 9-10, JEE & NEET

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JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their 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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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.

Simulator

Previous Years AIEEE/JEE Mains Questions

Q. The transistion from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from :-

(1) $4 \rightarrow 2$

(2) $5 \rightarrow 4$

(3) $2 \rightarrow 1$

(4) $3 \rightarrow 2$

[AIEEE – 2009]

Sol. (2)

Use Rydberg equation to find the required transition.

Q. Energy required for the electron excitation in $\mathrm{Li}^{++}$ from the first to the third Bohr orbit is:

(1) 108.8 eV

(2) 122.4 eV

(3) 12.1 eV

(4) 36.3 eV

[AIEEE-2011]

Sol. (1)

Rquired energy $=13.6(Z)^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$ eV

$=13.6(3)^{2}\left[\frac{1}{1^{2}}-\frac{1}{3^{2}}\right]=108.8 \mathrm{eV}$

Q. Hydrogen atom is excited from ground state to another state with principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be :-

(1) 6 (2) 2 (3) 3 (4) 5

[AIEEE-2012]

Sol. (1)

Number of lines $=^{n} C_{2}=^{4} C_{2}=\frac{(4)(3)}{(2)}=6$

Q. In a hydrogen like atom electron makes transition from an energy level with quantum number n to another with quantum number $(n-1) .$ If $n>>1$, the frequency of radiation emitted is proportional to :

( 1)$\frac{1}{\mathrm{n}}$

( 2)$\frac{1}{\mathrm{n}^{2}}$

(3) $\frac{1}{\mathrm{n}^{3 / 2}}$

(4) $\frac{1}{n^{3}}$

[JEE Main-2013]

Sol. (4)

Energy of $\mathrm{E}=\mathrm{h} \theta=\mathrm{E}_{0} \mathrm{z}^{2}\left[\frac{1}{(\mathrm{n}-1)^{2}}-\frac{1}{\mathrm{n}^{2}}\right]$

$=\mathrm{E}_{0} \mathrm{z}^{2}\left[\frac{2 \mathrm{n}-1}{\mathrm{n}^{2}(\mathrm{n}-1)^{2}}\right]$

$\mathrm{h} \theta \approx \mathrm{E}_{0} \mathrm{z}^{2}\left[\frac{2 \mathrm{n}}{\mathrm{n}^{4}}\right] \Rightarrow \mathrm{v} \propto \frac{1}{\mathrm{n}^{3}}$

Q. As an electron makes a transition from an excited state to the ground state of a hydrogen – like atom/ion :

(1) kinetic energy decreases, potential energy increases but total energy remains same

(2) kinetic energy and total energy decrease but potential energy increases

(3) its kinetic energy increases but potential energy and total energy decreases

(4) kinetic energy, potential energy and total energy decrease.

[JEE Main-2015]

Sol. (3)

$\mathrm{K}=+13.6 \frac{\mathrm{z}^{2}}{\mathrm{n}^{2}}$ as $\mathrm{n}$ decreases $\mathrm{k}$ increases

Q. Match List-I (Fundament Experiment) with List-II (its conclusion) and select the correct option from the choices given below the list :

(1) A-ii, B-i, C-iii

(2) A-iv, B-iii, C-ii

(3) A-i, B-iv, C-iii

(4) A-ii, B-iv, C-iii

[JEE Main-2015]

Sol. (1)

Self Explanatory/Theory

(A) Franck-Hertz experiment explains disrete energy levels of atom

(B) Photo-electric experiment explain particle nature of light

(C) Davison Germer experiment explain wave nature of electron.

Q. Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths $r=\lambda_{1} / \lambda_{2}$, is given by :

(1) $\mathrm{r}=\frac{3}{4}$

(2) $\mathrm{r}=\frac{1}{3}$

(3) $\mathrm{r}=\frac{4}{3}$

(4) $r=\frac{2}{3}$

[JEE Main-2017]

Sol. (2)

using $\Delta \mathrm{E}=\frac{\mathrm{hC}}{\lambda}$

for $\lambda_{1} \quad-\mathrm{E}-(-2 \mathrm{E})=\frac{\mathrm{hC}}{\lambda_{1}}$

$\lambda_{1}=\frac{\mathrm{hC}}{\mathrm{E}}$

for $\lambda_{2} \quad-\mathrm{E}-\left(-\frac{4 \mathrm{E}}{3}\right)=\frac{\mathrm{hC}}{\lambda_{2}}$

$\lambda_{2}=\frac{3 \mathrm{h} \mathrm{C}}{\mathrm{E}}$

$\frac{\lambda_{1}}{\lambda_{2}}=\mathrm{r}=\frac{1}{3}$

Q. A particle A of mass m and initial velocity v collides with a particle B of mass which is at rest. The collision is head on, and elastic. The ratio of the de–Broglie wavelengths $\lambda_{\mathrm{A}}$ to $\lambda_{\mathrm{B}}$after the collision is :

(1) $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{2}{3}$

(2) $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{1}{2}$

(3) $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{1}{3}$

(4) $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=2$

[JEE Main-2017]

Sol. (4)

By conservation of linear momentum

$\mathrm{mv}=\mathrm{mv}_{1}+\frac{\mathrm{m}}{2} \mathrm{v}_{2}$

$2 \mathrm{v}=2 \mathrm{v}_{1}+\mathrm{v}_{2}$ ….(1)

by law of collision

$\mathrm{e}=\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{u}_{1}-\mathrm{u}_{2}}$

$\mathrm{u}=\mathrm{v}_{2}-\mathrm{v}_{1}$ …(2)

By equation (1) and (2)

Option (4)

Q. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let $\lambda_{\mathrm{n}}, \lambda_{\mathrm{g}}$ be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)

$(1) \Lambda_{\mathrm{n}} \approx \mathrm{A}+\mathrm{B} \lambda_{\mathrm{n}}$

$(2) \Lambda_{\mathrm{n}}^{2} \approx \mathrm{A}+\mathrm{B} \lambda_{\mathrm{n}}^{2}$

(3) $\Lambda_{\mathrm{n}}^{2} \approx \lambda$

(4) $\Lambda_{\mathrm{n}} \approx \mathrm{A}+\frac{\mathrm{B}}{\lambda_{\mathrm{n}}^{2}}$

[JEE Main-2018]

Sol. (4)

$\lambda_{\mathrm{n}}=\frac{\mathrm{h}}{\mathrm{mu}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}_{\mathrm{n}}}}$

$\Rightarrow \mathrm{k}_{\mathrm{n}}=\frac{\mathrm{h}^{2}}{2 \mathrm{m} \lambda_{\mathrm{n}}^{2}} ; \mathrm{k}_{\mathrm{g}}=\frac{\mathrm{h}^{2}}{2 \mathrm{m} \lambda_{\mathrm{g}}^{2}}$

$\Rightarrow \mathrm{k}_{\mathrm{g}}-\mathrm{k}_{\mathrm{n}}=\frac{\mathrm{h}^{2}}{2 \mathrm{m}}\left[\frac{1}{\lambda_{\mathrm{g}}^{2}}-\frac{1}{\lambda_{\mathrm{n}}^{2}}\right]$

$\mathrm{E}_{\mathrm{n}}=-\mathrm{k}_{\mathrm{n}}$

for emitted photon

$\frac{\mathrm{hc}}{\Lambda_{\mathrm{n}}}=\mathrm{E}_{\mathrm{n}}-\mathrm{E}_{\mathrm{g}}=\mathrm{K}_{\mathrm{g}}-\mathrm{K}_{\mathrm{n}}$

$\frac{1}{\Lambda_{\mathrm{n}}}=\frac{\mathrm{K}_{\mathrm{g}}-\mathrm{K}_{\mathrm{n}}}{\mathrm{hc}}$

$\Lambda_{\mathrm{n}}=\frac{\mathrm{hc}}{\mathrm{K}_{\mathrm{g}}-\mathrm{K}_{\mathrm{n}}} \Rightarrow \Lambda_{\mathrm{n}}=\frac{\mathrm{hc}}{\frac{\mathrm{h}^{2}}{2 \mathrm{m}}\left[\frac{1}{\lambda_{\mathrm{g}}^{2}}-\frac{1}{\lambda_{\mathrm{n}}^{2}}\right]}$

$\Lambda_{\mathrm{n}}=\frac{2 \mathrm{mc}}{\mathrm{h}\left(\frac{\lambda_{\mathrm{n}}^{2}-\lambda_{\mathrm{g}}^{2}}{\lambda_{\mathrm{g}}^{2} \lambda_{\mathrm{n}}^{2}}\right)}$

$\Lambda_{\mathrm{n}}=\frac{2 \mathrm{mc} \lambda_{\mathrm{g}}^{2} \lambda_{\mathrm{n}}^{2}}{\mathrm{h}\left(\lambda_{\mathrm{n}}^{2}-\lambda_{\mathrm{g}}^{2}\right)}$

as $\lambda_{\mathrm{n}} \propto \mathrm{n}$

$\lambda_{\mathrm{n}}>>\lambda_{\mathrm{g}}$

$\Lambda_{\mathrm{n}}=\frac{2 \mathrm{mc} \lambda_{\mathrm{g}}^{2}}{\mathrm{h}}\left[1-\left(\frac{\lambda_{\mathrm{g}}}{\lambda_{\mathrm{n}}}\right)^{2}\right]^{-1}$

Q. If the series limit frequency of the Lyman series is $\mathrm{v}_{\mathrm{L}}$, then the series limit frequency of the Pfund series is :

(1) $16 \mathrm{v}_{\mathrm{L}}$

(2) $\mathrm{v}_{\mathrm{L}} / 16$

(3) $\mathrm{v}_{\mathrm{L}} / 25$

(4) $25 \mathrm{v}_{\mathrm{L}}$

[JEE Main-2018]

Sol. (3)

Magnetism – JEE Main Previous Year Questions with Solutions

Class 9-10, JEE & NEET

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JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their 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.

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.

Simulator

Previous Years AIEEE/JEE Mains Questions

Q. Due to the presence of the current $\mathrm{I}_{1}$ at the origin:-

(1) The magnitude of the net force on the loop is given by $\frac{\mathrm{I}_{1} \mathrm{I}}{4 \pi} \mu_{0}\left[2(\mathrm{b}-\mathrm{a})+\frac{\pi}{3}(\mathrm{a}+\mathrm{b})\right]$

(2) The magnitude of the net force on the loop is given by $\frac{\mu_{0} \mathrm{I} \mathrm{I}_{1}}{24 \mathrm{ab}}(\mathrm{b}-\mathrm{a})$

(3) The forces on $\mathrm{AB}$ and $\mathrm{DC}$ are zero

(4) The forces on $\mathrm{AD}$ and $\mathrm{BC}$ are zero

[AIEEE – 2009]

Sol. (4)

For AD and BC, $\overrightarrow{\mathrm{B}} \square \overrightarrow{\mathrm{dL}}$. Hence force on AD and BC is zero.

Q. Two short bar magnets of length 1 cm each have magnetic moments $1.20 \mathrm{Am}^{2}$ and $1.00 \mathrm{Am}^{2}$ respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the midpoint O of the line joining their centres is close to :-

(Horizontal component of earth’s magnetic induction is $\left.3.6 \times 10^{-5} \mathrm{Wb} / \mathrm{m}^{2}\right)$

(1) $3.6 \times 10^{-5} \mathrm{Wb} / \mathrm{m}^{2}$

(2) $2.56 \times 10^{-4} \mathrm{Wb} / \mathrm{m}^{2}$

(3) $3.50 \times 10^{-4} \mathrm{Wb} / \mathrm{m}^{2}$

(4) $5.80 \times 10^{-4} \mathrm{Wb} / \mathrm{m}^{2}$

[JEE(Mains) – 2013]

Sol. (2)

Q. The coercivity of a small magnet where the ferromagnet gets demagnetized is $3 \times 10^{3} \mathrm{A} \mathrm{m}^{-1}$. The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is :

(1) 3A          (2) 6 A            (3) 30 mA             (4) 60 mA

[JEE(Mains) – 2014]

Sol. (1)

Q. Hysteresis loops for two magnetic materials A and B are given below :

These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use ;

(1) B for electromagnets and transformers.

(2) A for electric generators and transformers.

(3) A for electromagnets and B for electric transformers.

(4) A for transformers and B for electric generators.

[JEE(Mains) – 2016]

Sol. (1)

For electromagnet and transformers, we require the core that can be magnitised and demagnetised quickly when subjected to alternating current. From the given graphs, graph B is suitable.

Q. A magnetic needle of magnetic moment $6.7 \times 10^{-2} \mathrm{Am}^{2}$ and moment of inertia $7.5 \times 10^{-6} \mathrm{kg} \mathrm{m}^{2}$ is performing simple harmonic oscillations in a magnetic field of 0.01 T.

Time taken for 10 complete oscillations is :

(1) 6.98 s

(2) 8.76 s

(3) 6.65 s

(4) 8.89 s

[JEE(Mains) – 2017]

Sol. (3)