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*Simulator*

**Previous Years JEE Advanced Questions**

(A) the current I through the battery is 7.5 mA

(B) the potential difference across $\mathrm{R}_{\mathrm{L}}$ is 18V

(C) ratio of powers dissipated in $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ is 3

(D) if $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ are interchanged, magnitude of the power dissipated in $\mathrm{R}_{\mathrm{L}}$ will decrease by a factor of 9

**[IIT-JEE 2009]**

**Sol.**(A,D)

**–**

$(\mathrm{A}) \frac{1}{\mathrm{R}_{100}}=\frac{1}{\mathrm{R}_{40}}+\frac{1}{\mathrm{R}_{60}}$

$(\mathrm{B}) \mathrm{R}_{100}=\mathrm{R}_{40}+\mathrm{R}_{60}$

(C) $\mathrm{R}_{100}>\mathrm{R}_{60}>\mathrm{R}_{40}$

$(\mathrm{D}) \frac{1}{\mathrm{R}_{100}}>\frac{1}{\mathrm{R}_{60}}>\frac{1}{\mathrm{R}_{40}}$

**[IIT-JEE 2010]**

**Sol.**(D)

**[IIT-JEE 2010]**

**Sol.**(C)

(A) directly proportional to L

(B) directly proportional to t

(C) independent of L

(D) independent of t

**[IIT-JEE 2010]**

**Sol.**(C)

$\mathrm{R}=\frac{\rho \mathrm{L}}{\mathrm{A}}=\frac{\rho \mathrm{L}}{\mathrm{Lt}}=\frac{\rho}{\mathrm{t}} \Rightarrow$ independent of $\mathrm{L}$

**[IIT-JEE 2010]**

**Sol.**4

$J_{1}=\left(\frac{2 \varepsilon}{R+2}\right)^{2} R$ and $J_{2}=\left(\frac{\varepsilon}{R+1 / 2}\right)^{2} R$ as $\frac{J_{1}}{J_{2}}=2.25$ so $\frac{4 \varepsilon^{2}}{(R+2)}=2.25 \frac{4 \varepsilon^{2}}{(1+2 R)^{2}} \Rightarrow R=4 \Omega$

**[IIT-JEE 2011]**

**Sol.**5

$E_{A B}=\frac{\frac{E_{1}}{R_{1}}+\frac{E_{2}}{R_{2}}}{\frac{1}{R_{1}}+\frac{1}{R_{2}}}=\frac{\frac{6}{1}+\frac{3}{2}}{\frac{1}{1}+\frac{1}{2}}=\frac{\frac{15}{2}}{\frac{3}{2}}=5$ volt

(A) 10.2 ohm (B) 10.6 ohm (C) 10.8 ohm (D) 11.1 ohm

**[IIT-JEE 2011]**

**Sol.**(B)

Apply condition of wheatstone bridge, $\frac{x}{52+1}=\frac{10}{48+2} \Rightarrow x=\frac{10}{50} \times 53 \Rightarrow \mathrm{x}=10.6 \Omega$

(A) the current through PQ is zero

(B) $\mathrm{I}_{1}=3 \mathrm{A}$

(C) The potential at S is less than that at Q

(D) $\mathrm{I}_{2}=2 \mathrm{A}$

**[IIT-JEE 2012]**

**Sol.**(A,B,C,D)

Since resitances in two arms upper and lower are in same ratio.

Current through PQ is zero.

$I_{2}=\frac{12}{6}=2 A$, $I_{1}=\frac{12}{6}+\frac{12}{12}=3 A$ and Potential of $\mathrm{S}$ is less than that at $\mathrm{Q}$

(A) 4 if wires are in parallel

(B) 2 if wires are in series

(C) 1 if wires are in series

(D) 0.5 if wires are in parallel

**[JEE Advanced 2014]**

**Sol.**(B,D)

Resistance of heater $1, \mathrm{R}=\frac{4 \rho \mathrm{L}}{\pi \mathrm{d}^{2}}$

Resistance of heater $2, R_{1}=\frac{R}{4}, R_{2}=\frac{R}{4}$

Series $\mathrm{R}_{\mathrm{net}}=\frac{\mathrm{R}}{2}$

Power $=2 \frac{\mathrm{V}^{2}}{\mathrm{R}}$

$\Rightarrow \quad$ power is twice, hence time is $\frac{1}{2}$

time $=\frac{1}{2}$ of $4 \min =2 \min$

Parallel $\mathrm{R}_{\mathrm{Net}}=\frac{\mathrm{R}}{8}$

Power = 8 times

time $=\frac{\mathrm{R}}{8}$ times $=\frac{\mathrm{R}}{8} \times 4 \min =0.5 \mathrm{sec}$

(A) $\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{R}_{1}=\mathrm{R}_{2}=\mathrm{R}_{3}$

(B) $\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{R}_{1}=2 \mathrm{R}_{2}=\mathrm{R}_{3}$

(C) $\mathrm{V}_{1}=2 \mathrm{V}_{2}$ and $2 \mathrm{R}_{1}=2 \mathrm{R}_{2}=\mathrm{R}_{3}$

(D) $2 \mathrm{V}_{1}=\mathrm{V}_{2}$ and $2 \mathrm{R}_{1}=\mathrm{R}_{2}=\mathrm{R}_{3}$

**[JEE Advanced 2014]**

**Sol.**(A,B,D)

since current through $\mathrm{R}_{2}$ is zero

Hence $\left[\frac{\mathrm{V}_{1}}{\mathrm{R}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{R}_{3}}\right]$

The above equation is satisfied by options (A, B, D)

**[JEE Advanced 2014]**

**Sol.**(5)

(A) $60 \pm 0.15 \Omega$

(B) $135 \pm 0.56 \Omega$

(C) $60 \pm 0.25 \Omega$

(D) $135 \pm 0.23 \Omega$

** [JEE Advanced 2014]**

**Sol.**(C)

For meter bridge,

$\mathrm{R}=\mathrm{X}\left(\frac{\ell}{100-\ell}\right)$

$=90\left(\frac{\ell}{100-\ell}\right)$

$=90\left(\frac{40}{60}\right)$v

$=60 \Omega$

error in R : (erro = least count)

$\frac{\mathrm{d} \mathrm{R}}{\mathrm{R}}=\frac{\mathrm{d} \mathrm{x}}{\mathrm{x}}+\frac{\mathrm{d} \ell}{\ell}+\frac{\mathrm{d}(100-\ell)}{(100-\ell)}$

$=0+\frac{\operatorname{lmm}}{40 \mathrm{cm}}+\frac{1 \mathrm{mm}}{60 \mathrm{cm}}$

subsequent variation of the magnitude of current density j(t) at any point in the material?

**[JEE Advanced 2016]**

**Sol.**(A)

This is the problem of RC circuit where the product RC is a constant.

So due to leakage current, charge & current density will exponentially decay & will become zero at infinite time. So correct answer is (A)

for any small element

Resistance $\mathrm{R}=\frac{\mathrm{dr}}{\sigma(2 \pi r \ell)}$

Capacitance $\mathrm{C}=\frac{\in 2 \pi \mathrm{r} \ell}{\mathrm{dr}}$

Product $\mathrm{R} \times \mathrm{C}=\frac{\epsilon}{\sigma}=\mathrm{constant}$

$\mathrm{q}=\mathrm{q}_{0} \mathrm{e}^{-\left(\frac{\mathrm{t} \sigma}{\epsilon}\right)}$

$\mathrm{I}=\frac{\mathrm{d} q}{\mathrm{dt}}=\frac{\mathrm{q}_{0} \sigma}{\epsilon} \mathrm{e}^{-\left(\frac{\mathrm{t} \sigma}{\epsilon}\right)}$

(A) The temperature distribution over the filament is uniform

(B) The resistance over small sections of the filament decreases with time

(C) The filament emits more light at higher band of frequencies before it breaks up

(D) The filament consumes less electrical power towards the end of the life of the bulb

**[JEE Advanced 2016]**

**Sol.**(C,D)

Because of non-uniform evaporation at different section, area of cross-section would be different at different sections.

Region of highest evaporation rate would have rapidly reduced area and would become break up cross-section.

Resistance of the wire as whole increases with time.

Overall resistance increases hence power decreases. At break up junction temperature would be highest, thus light of highest band frequency would be emitted at those cross-section.

(A) The maximum voltage range is obtained when all the components are connected in series

(B) The maximum voltage range is obtained when the two resistors and one galvanometer are connected in series, and the second galvanometer is connected in parallel to the first galvanometer

(C) The maximum current range is obtained when all the components are connected in parallel

(D) The maximum current range is obtained when the two galvanometers are connected in series and the combination is connected in parallel with both the resistors.

**[JEE Advanced 2016]**

**Sol.**(B,C)

Range $=\mathrm{i}_{\mathrm{g}}\left(2 \mathrm{R}_{\mathrm{C}}+2 \mathrm{R}\right)$

Range $=\mathrm{i}_{\mathrm{g}}+4 \mathrm{i}_{\mathrm{g}} \frac{\mathrm{R}_{\mathrm{C}}}{\mathrm{R}}$

$=2 \mathrm{i}_{\mathrm{g}}\left[\frac{1}{2}+\frac{2 \mathrm{R}_{\mathrm{c}}}{\mathrm{R}}\right]$

$=2 \mathrm{i}_{\mathrm{g}}\left(\frac{1}{2}+\frac{\mathrm{R}_{\mathrm{C}}}{\mathrm{R}}+\frac{\mathrm{R}_{\mathrm{C}}}{\mathrm{R}}\right)$

So (C)

**Paragraph 2**

Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r <<h. Now a high voltage source (HV) is connected

across the conducting plates such that the bottom plate is at $+V_{0}$ and the top plate at

$-\mathbf{V}_{0}$. Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collision between the balls and the interaction between them is negligible.

(Ignore gravity)

Which of the following statements is correct ?

(A) The balls will bounce back to the bottom plate carrying the opposite charge they went up with

(B) the balls will execute simple harmonic motion between the two plates

(C) The balls will bounce back to the bottom plate carrying the same charge they went up with

(D) The balls will stick to the top plate and remain there

**[JEE Advanced 2016]**

**Sol.**(A)

Balls placed on +ve plate become positive charge and move upward due to electric field.

These balls on colliding with negative plate become negatively charged and move opposite to the direction of electric field.

(A) Proportional to $\mathrm{V}_{0}^{1 / 2}$

(B) Proportional to $V_{0}^{2}$

(C) Proportional to the potential $\mathrm{V}_{0}$

(D) Zero

**[JEE Advanced 2016]**

**Sol.**(B)