Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Two identical charged spheres suspended from a common point by two massless strings of length $$l$$ are initially a distance $$d\left( {d < < 1} \right)$$ apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result charges approach each other with a velocity $$v$$. Then as a function of distance $$x$$ between them,

A

$$v\, \propto \,{x^{ - 1}}$$

B

$$y\, \propto \,{x^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}$$

C

$$v\, \propto \,x$$

D

$$v\, \propto \,{x^{ - {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}$$

At any instant

$$T\cos \theta = mg\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

$$T\sin \theta = {F_e}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

$$ \Rightarrow {{\sin \theta } \over {\cos \theta }} = {{{F_e}} \over {mg}} \Rightarrow {F_e} = mg\,\tan \theta $$

$$ \Rightarrow {{k{q^2}} \over {{x^2}}} = mg\,\tan \theta \Rightarrow {q^2} \propto {x^2}\tan \theta $$

$$\sin \theta = {\textstyle{x \over {2l}}}$$

For small $$\theta ,\,\sin \theta \approx \tan \theta $$

$$\therefore$$ $${q^2} \propto {x^3}$$

$$ \Rightarrow q{{dq} \over {dt}} \propto {x^2}{{dx} \over {dt}}$$

$$\therefore$$ $${{dq} \over {dt}} = const.$$

$$\therefore$$ $$q \propto {x^2}.v \Rightarrow {x^{3/2}}\alpha {x^2}.v\,\,$$ $$\,\,\,\,\,$$ $$\left[ {\,\,} \right.$$ as $$\left. {{q^2} \propto {x^3}\,\,} \right]$$

$$ \Rightarrow v \propto {x^{ - 1/2}}$$

$$T\cos \theta = mg\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

$$T\sin \theta = {F_e}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

$$ \Rightarrow {{\sin \theta } \over {\cos \theta }} = {{{F_e}} \over {mg}} \Rightarrow {F_e} = mg\,\tan \theta $$

$$ \Rightarrow {{k{q^2}} \over {{x^2}}} = mg\,\tan \theta \Rightarrow {q^2} \propto {x^2}\tan \theta $$

$$\sin \theta = {\textstyle{x \over {2l}}}$$

For small $$\theta ,\,\sin \theta \approx \tan \theta $$

$$\therefore$$ $${q^2} \propto {x^3}$$

$$ \Rightarrow q{{dq} \over {dt}} \propto {x^2}{{dx} \over {dt}}$$

$$\therefore$$ $${{dq} \over {dt}} = const.$$

$$\therefore$$ $$q \propto {x^2}.v \Rightarrow {x^{3/2}}\alpha {x^2}.v\,\,$$ $$\,\,\,\,\,$$ $$\left[ {\,\,} \right.$$ as $$\left. {{q^2} \propto {x^3}\,\,} \right]$$

$$ \Rightarrow v \propto {x^{ - 1/2}}$$

2

MCQ (Single Correct Answer)

A thin semi-circular ring of radius $$r$$ has a positive charges $$q$$ distributed uniformly over it. The net field $$\overrightarrow E $$ at the center $$O$$ is

A

$${q \over {4{\pi ^2}{\varepsilon _0}{r^2}}}\,j$$

B

$$ - {q \over {4{\pi ^2}{\varepsilon _0}{r^2}}}\,j$$

C

$$ - {q \over {2{\pi ^2}{\varepsilon _0}{r^2}}}\,j$$

D

$$ {q \over {2{\pi ^2}{\varepsilon _0}{r^2}}}\,j$$

Let us consider a differential element $$dl.$$ charge on this element.

$$dq = \left( {{q \over {\pi r}}} \right)dl$$

$$ = {q \over {\pi r}}\left( {rd\theta } \right)\,\,\,\,\,$$ (as $$dl = rd\theta $$)

$$ = \left( {{q \over \pi }} \right)d\theta $$

Electric field at $$O$$ due to $$dq$$ is

$$dE = {1 \over {4\pi { \in _0}}}.{{dq} \over {{r^2}}}$$

$$ = {1 \over {4\pi { \in _0}}}.{q \over {\pi {r^2}}}d\theta $$

The component $$dE\cos \theta $$ will be counter balanced by another element on left portion.

Hence resultant field at $$O$$ is the resultant of the component $$dE\sin \theta $$ only.

$$\therefore$$ $$E = \int {dE\sin \theta = \int\limits_0^\pi {{q \over {4{\pi ^2}{r^2}{ \in _0}}}} } \sin \theta d\theta $$

$$ = {q \over {4{\pi ^2}{r^2}{ \in _0}}}\left[ { - \cos \theta } \right]_0^\pi $$

$$ = {q \over {4{\pi ^2}{r^2}{ \in _0}}}\left( { + 1 + 1} \right)$$

$$ = {q \over {2{\pi ^2}{r^2}{ \in _0}}}$$

The directions of $$E$$ is towards negative $$y$$-axis.

$$\therefore$$ $$\overrightarrow E = - {q \over {2{\pi ^2}{r^2}{ \in _0}}}\widehat j$$

3

MCQ (Single Correct Answer)

Let there be a spherically symmetric charge distribution with charge density varying as $$\rho \left( r \right) = {\rho _0}\left( {{5 \over 4} - {r \over R}} \right)$$ upto $$r=R,$$ and $$\rho \left( r \right) = 0$$ for $$r>R,$$ where $$r$$ is the distance from the erigin. The electric field at a distance $$r\left( {r < R} \right)$$ from the origin is given by

A

$${{{\rho _0}r} \over {4{\varepsilon _0}}}\left( {{5 \over 3} - {r \over R}} \right)$$

B

$${{4\pi {\rho _0}r} \over {3{\varepsilon _0}}}\left( {{5 \over 3} - {r \over R}} \right)$$

C

$${{4{\rho _0}r} \over {4{\varepsilon _0}}}\left( {{5 \over 4} - {r \over R}} \right)$$

D

$${{{\rho _0}r} \over {3{\varepsilon _0}}}\left( {{5 \over 4} - {r \over R}} \right)$$

Let us consider a spherical shell of radius $$x$$ and thickness $$dx.$$

Charge on this shell

$$dq = \rho .4{\pi ^2}dx = {\rho _0}\left( {{5 \over 4} - {x \over R}} \right).4\pi {x^2}dx$$

$$\therefore$$ Total charge in the spherical region from center to $$r$$ $$\left( {r < R} \right)$$ is

$$q = \int {dq = 4\pi {\rho _0}\int\limits_0^r {\left( {{5 \over 4} - {x \over R}} \right)} } {x^2}dx$$

$$ = 4\pi {\rho _0}\left[ {{5 \over 4}.{{{r^3}} \over 3} - {1 \over R}.{{{r^4}} \over 4}} \right]$$

$$ = \pi {\rho _0}{r^3}\left( {{5 \over 3} - {r \over R}} \right)$$

$$\therefore$$ Electric field at $$r,$$ $$E = {1 \over {4\pi { \in _0}}}.{q \over {{r^2}}}$$

$$ = {1 \over {4\pi { \in _0}}}.{{\pi {\rho _0}{r^3}} \over {{r^2}}}\left( {{5 \over 3} - {r \over R}} \right)$$

$$ = {{{\rho _0}r} \over {4{ \in _0}}}\left( {{5 \over 3} - {r \over R}} \right)$$

4

MCQ (Single Correct Answer)

Two points $$P$$ and $$Q$$ are maintained at the potentials of $$10$$ $$V$$ and $$-4$$ $$V$$, respectively. The work done in moving $$100$$ electrons from $$P$$ to $$Q$$ is :

A

$$9.60 \times {10^{ - 17}}J$$

B

$$ - 2.24 \times {10^{ - 16}}J$$

C

$$ 2.24 \times {10^{ - 16}}J$$

D

$$- 9.60 \times {10^{ - 17}}J$$

$$ {{{W_{PQ}}} \over q} = \left( {{V_Q} - {V_P}} \right)$$

$$ \Rightarrow {W_{PQ}} = q\left( {{V_Q} - {V_P}} \right)$$

$$ = \left( { - 100 \times 1.6 \times {{10}^{ - 19}}} \right)\left( { - 4 - 10} \right)$$

$$ = + 2.24 \times {10^{ - 16}}J$$

$$ \Rightarrow {W_{PQ}} = q\left( {{V_Q} - {V_P}} \right)$$

$$ = \left( { - 100 \times 1.6 \times {{10}^{ - 19}}} \right)\left( { - 4 - 10} \right)$$

$$ = + 2.24 \times {10^{ - 16}}J$$

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