 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2010

Let $C$ be the capacitance of a capacitor discharging through a resistor $R.$ Suppose ${t_1}$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and ${t_2}$ is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio ${t_1}/{t_2}$ will be
A
$1$
B
${1 \over 2}$
C
${1 \over 4}$
D
$2$

Explanation

Initial energy of capacitor, ${E_1} = {{q_1^2} \over {2C}}$

Final energy of capacitor, ${E_2} = {1 \over 2}{E_1} = {{q_1^2} \over {4C}} = {\left( {{{{{{q_1}} \over {\sqrt 2 }}} \over {2C}}} \right)^2}$

$\therefore$ ${t_1}=$ time for the charge to reduce to ${1 \over {\sqrt 2 }}$ of its initial value

and ${t_2} =$ time for the charge to reduce to ${1 \over 4}$ of its initial value

We have, ${q_2} = {q_1}{e^{ - t/CR}}$

$\Rightarrow \ln \left( {{{{q_2}} \over {{q_1}}}} \right) = - {t \over {CR}}$

$\therefore$$\ln \left( {{1 \over {\sqrt 2 }}} \right) = {{ - {t_1}} \over {CR}}...\left( 1 \right)$

and $\ln \left( {{1 \over 4}} \right) = {{ - {t_2}} \over {CR}}\,\,...\left( 2 \right)$

By $(1)$ and $(2),$ ${{{t_1}} \over {{t_2}}} = {{\ln \left( {{1 \over {\sqrt 2 }}} \right)} \over {\ln \left( {{1 \over 4}} \right)}}$

$= {1 \over 2}{{\ln \left( {{1 \over 2}} \right)} \over {2\ln \left( {{1 \over 2}} \right)}} = {1 \over 4}$
2

AIEEE 2008

A $5V$ battery with internal resistance $2\Omega$ and a $2V$ battery with internal resistance $1\Omega$ are connected to a $10\Omega$ resistor as shown in the figure. The current in the $10\Omega$ resistor is

A
$0.27A{P_2}\,\,to\,\,{P_1}$
B
$0.03A{P_1}\,\,to\,\,{P_2}$
C
$0.03A{P_2}\,\,to\,\,{P_1}$
D
$0.27A{P_1}\,\,to\,\,{P_2}$

Explanation

Applying kirchoff's loop law in $AB\,{P_2}{P_1}A$ we get

$- 2i + 5 - 10\,{i_1} = 0\,\,\,\,\,\,\,\,...\left( i \right)$ Again applying kirchoffs loop law in ${P_2}$ $CD{P_1}{P_2}$ we get,

$10{i_1} + 2 - i + {i_1} = 0\,\,\,\,\,\,...\left( {ii} \right)$

From $\left( i \right)$ and $\left( {ii} \right)$ $11{i_1} + 2 - \left[ {{{5 - 10{i_1}} \over 2}} \right] = 0$

$\Rightarrow {i_1} = {1 \over {32}}$ A from ${P_2}$ to ${P_1}$
3

AIEEE 2008

Consider a block of conducting material of resistivity $'\rho '$ shown in the figure. Current $'I'$ enters at $'A'$ and leaves from $'D'$. We apply superposition principle to find voltage $'\Delta V'$ developed between $'B'$ and $'C'$. The calculation is done in the following steps:
(i) Take current $'I'$ entering from $'A'$ and assume it to spread over a hemispherical surface in the block.
(ii) Calculate field $E(r)$ at distance $'r'$ from A by using Ohm's law $E = \rho j,$ where $j$ is the current per unit area at $'r'$.
(iii) From the $'r'$ dependence of $E(r)$, obtain the potential $V(r)$ at $r$.
(iv) Repeat (i), (ii) and (iii) for current $'I'$ leaving $'D'$ and superpose results for $'A'$ and $'D'.$ $\Delta V$ measured between $B$ and $C$ is

A
${{\rho I} \over {\pi a}} - {{\rho I} \over {\pi \left( {a + b} \right)}}$
B
${{\rho I} \over a} - {{\rho I} \over {\left( {a + b} \right)}}$
C
${{\rho I} \over {2\pi a}} - {{\rho I} \over {2\pi \left( {a + b} \right)}}$
D
${{\rho I} \over {2\pi \left( {a - b} \right)}}$

Explanation

Let $j$ be the current density.

Then $j \times 2\pi {r^2} = I \Rightarrow j = {I \over {2\pi {r^2}}}$

$\therefore$ $E = \rho j = {{\rho I} \over {2\pi {r^2}}}$

Now, $\Delta V{'_{BC}} =$ $- \int\limits_{a + b}^a {\overrightarrow E .\,\overrightarrow {dr} }$ $= - \int\limits_{a + b}^a {{{\rho I} \over {2\pi {r^2}}}} dr$

$= - {{\rho I} \over {2\pi }}\left[ { - {1 \over r}} \right]_{a + b}^a$

$= {{\rho I} \over {2\pi a}} - {{\rho I} \over {2\pi \left( {a + b} \right)}}$

On applying superposition as mentioned we get

$\Delta {V_{BC}} = 2 \times \Delta {V_{BC}} = {{\rho I} \over {\pi a}} - {{\rho I} \over {\pi \left( {a + b} \right)}}$
4

AIEEE 2008

Consider a block of conducting material of resistivity $'\rho '$ shown in the figure. Current $'I'$ enters at $'A'$ and leaves from $'D'$. We apply superposition principle to find voltage $'\Delta V'$ developed between $'B'$ and $'C'$. The calculation is done in the following steps:
(i) Take current $'I'$ entering from $'A'$ and assume it to spread over a hemispherical surface in the block.
(ii) Calculate field $E(r)$ at distance $'r'$ from A by using Ohm's law $E = \rho j,$ where $j$ is the current per unit area at $'r'$.
(iii) From the $'r'$ dependence of $E(r)$, obtain the potential $V(r)$ at $r$.
(iv) Repeat (i), (ii) and (iii) for current $'I'$ leaving $'D'$ and superpose results for $'A'$ and $'D'.$ For current entering at $A,$ the electric field at a distance $'r'$ from $A$ is

A
${{\rho I} \over {8\pi {r^2}}}$
B
${{\rho I} \over {{r^2}}}$
C
${{\rho I} \over {2\pi {r^2}}}$
D
${{\rho I} \over {4\pi {r^2}}}$

Explanation

As shown above $E = {{\rho I} \over {2\pi {r^2}}}$