Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

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 $${P_1}$$

2

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

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

3

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 $$'D'.$$

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

$${{\rho I} \over {4\pi {r^2}}}$$

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

4

MCQ (Single Correct Answer)

Shown in the figure below is a meter-bridge set up with null deflection in the galvanometer.

The value of the unknown resister $$R$$ is

A

$$13.75\Omega $$

B

$$220\Omega $$

C

$$110\Omega $$

D

$$55\Omega $$

According to the condition of balancing

$${{55} \over {20}} = {R \over {80}} \Rightarrow R = 220\Omega $$

$${{55} \over {20}} = {R \over {80}} \Rightarrow R = 220\Omega $$

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