Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Two conductors have the same resistance at $${0^ \circ }C$$ but their temperature coefficients of resistance are $${\alpha _1}$$ and $${\alpha _2}.$$ The respective temperature coefficients of their series and parallel combinations are nearly

A

$${{{\alpha _1} + {\alpha _2}} \over 2},\,{\alpha _1} + {\alpha _2}$$

B

$${\alpha _1} + {\alpha _2},\,{{{\alpha _1} + {\alpha _2}} \over 2}$$

C

$${\alpha _1} + {\alpha _2},\,{{{\alpha _1}{\alpha _2}} \over {{\alpha _1} + {\alpha _2}}}$$

D

$${{{\alpha _1} + {\alpha _2}} \over 2},\,{{{\alpha _1} + {\alpha _2}} \over 2}$$

$${R_1} = {R_0}\left[ {1 + {\alpha _1}\Delta t} \right];$$

$${R_2} = {R_0}\left[ {1 + {\alpha _2}\Delta t} \right]$$

$$R = {R_1} + {R_2}$$

$$ = {R_0}\left[ {2 + \left( {{\alpha _1} + {\alpha _2}} \right)\Delta t} \right]$$

$$ = 2{R_0}\left[ {1 + \left( {{{{\alpha _1} + {\alpha _2}} \over 2}} \right)\Delta t} \right]$$

$${\alpha _{eq}} = {{{\alpha _1} + {\alpha _2}} \over 2}$$

In Parallel, $${1 \over R} = {1 \over {{R_1}}} + {1 \over {{R_2}}}$$

$$ = {1 \over {{R_0}\left[ {1 + {\alpha _1}\Delta t} \right]}} + {1 \over {{R_0}\left[ {1 + {\alpha _2}\Delta t} \right]}}$$

$$ \Rightarrow {1 \over {{{{R_0}} \over 2}\left( {1 + {\alpha _{eq}}\Delta t} \right)}} = {1 \over {{R_0}\left( {1 + {\alpha _1}\Delta t} \right)}} + {1 \over {{R_0}\left( {1 + {\alpha _2}\Delta t} \right)}}$$

$$2\left( {1 - {a_{eq}}\Delta t} \right) = \left( {1 - {\alpha _1}\Delta t} \right)\left( {1 - {\alpha _2}\Delta t} \right)$$

$$\therefore$$ $${\alpha _{eq}} = {{{\alpha _1} + {\alpha _2}} \over 2}$$

$${R_2} = {R_0}\left[ {1 + {\alpha _2}\Delta t} \right]$$

$$R = {R_1} + {R_2}$$

$$ = {R_0}\left[ {2 + \left( {{\alpha _1} + {\alpha _2}} \right)\Delta t} \right]$$

$$ = 2{R_0}\left[ {1 + \left( {{{{\alpha _1} + {\alpha _2}} \over 2}} \right)\Delta t} \right]$$

$${\alpha _{eq}} = {{{\alpha _1} + {\alpha _2}} \over 2}$$

In Parallel, $${1 \over R} = {1 \over {{R_1}}} + {1 \over {{R_2}}}$$

$$ = {1 \over {{R_0}\left[ {1 + {\alpha _1}\Delta t} \right]}} + {1 \over {{R_0}\left[ {1 + {\alpha _2}\Delta t} \right]}}$$

$$ \Rightarrow {1 \over {{{{R_0}} \over 2}\left( {1 + {\alpha _{eq}}\Delta t} \right)}} = {1 \over {{R_0}\left( {1 + {\alpha _1}\Delta t} \right)}} + {1 \over {{R_0}\left( {1 + {\alpha _2}\Delta t} \right)}}$$

$$2\left( {1 - {a_{eq}}\Delta t} \right) = \left( {1 - {\alpha _1}\Delta t} \right)\left( {1 - {\alpha _2}\Delta t} \right)$$

$$\therefore$$ $${\alpha _{eq}} = {{{\alpha _1} + {\alpha _2}} \over 2}$$

2

MCQ (Single Correct Answer)

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

$$2$$

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

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

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

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

3

MCQ (Single Correct Answer)

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

$$0.27A{P_1}\,\,to\,\,{P_2}$$

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 $${P_1}$$

$$ - 2i + 5 - 10\,{i_1} = 0\,\,\,\,\,\,\,\,...\left( i \right)$$

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 $${P_1}$$

4

MCQ (Single Correct Answer)

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 $$'D'.$$

(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

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

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

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

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

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

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