Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A proton and an $$\alpha $$-particle (with their masses in the ratio of 1 : 4 and charges in the ratio of 1 : 2) are accelerated from rest through a potential difference V. If a uniform magnetic field (B) is set up perpendicular to their velocities, the ratio of the radii r_{p} : r_{$$\alpha $$} of the circular paths described by them will be ;

A

$$1:\sqrt 3 $$

B

1 : 3

C

$$1:\sqrt 2 $$

D

1 : 2

KE = q$$\Delta $$V

r = $${{\sqrt {2mq\Delta V} } \over {qB}}$$

r $$ \propto $$ $$\sqrt {{m \over q}} $$

$${{{r_p}} \over {{r_ \propto }}}$$ = $${1 \over {\sqrt 2 }}$$

r = $${{\sqrt {2mq\Delta V} } \over {qB}}$$

r $$ \propto $$ $$\sqrt {{m \over q}} $$

$${{{r_p}} \over {{r_ \propto }}}$$ = $${1 \over {\sqrt 2 }}$$

2

MCQ (Single Correct Answer)

As shown in the figure, two infinitely long, identical wires are bent by 90^{o} and placed in such a way that the segments LP and QM are along the x-axis, while segments PS and QN are parallel to the y-axis. If OP = OQ = 4cm, and the magnitude of the magnetic field at O is 10^{–4} T, and the two wires carry equal
currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at O will be ($$\mu $$_{0} = 4$$\pi $$ $$ \times $$ 10^{–7} NA^{–2}) :

A

40 A, perpendicular into the page

B

40 A, perpendicular out of the page

C

20 A, perpendicular into the page

D

40 A, perpendicular out of the page

Magnetic field at 'O' will be done to 'PS' and 'QN' Only

i.e. B_{0} = B_{PS} + B_{QN} $$ \to $$ Both inwards

Let current in each wire = i

$$ \therefore $$ B_{0} = $${{{\mu _0}i} \over {4\pi d}} + {{{\mu _0}i} \over {4\pi d}}$$

or 10^{$$-$$4} = $${{{\mu _0}i} \over {2\pi d}}$$ = $${{2 \times {{10}^{ - 7}} \times i} \over {4 \times {{10}^{ - 2}}}}$$

$$ \therefore $$ i = 20 A

i.e. B

Let current in each wire = i

$$ \therefore $$ B

or 10

$$ \therefore $$ i = 20 A

3

MCQ (Single Correct Answer)

A particle of mass m and charge q is in an electric and magnetic field given by

$$\overrightarrow E = 2\widehat i + 3\widehat j;\,\,\,\overrightarrow B = 4\widehat j + 6\widehat k.$$

The charged particle is shifted from he origin to the point P(x = 1; y = 1) along a straight path. The magnitude of the total work done is :

$$\overrightarrow E = 2\widehat i + 3\widehat j;\,\,\,\overrightarrow B = 4\widehat j + 6\widehat k.$$

The charged particle is shifted from he origin to the point P(x = 1; y = 1) along a straight path. The magnitude of the total work done is :

A

(2.5) q

B

(0.35) q

C

(0.15) q

D

5 q

$${\overrightarrow F _{net}} = q\overrightarrow E + q\left( {\overrightarrow v \times \overrightarrow B } \right)$$

$$ = \left( {2q\widehat i + 3q\widehat j} \right) + q\left( {\overrightarrow v \times \overrightarrow B } \right)$$

$$W = {\overrightarrow F _{net}}.\overrightarrow S $$

$$=$$ 2q + 3q

$$=$$ 5q

$$ = \left( {2q\widehat i + 3q\widehat j} \right) + q\left( {\overrightarrow v \times \overrightarrow B } \right)$$

$$W = {\overrightarrow F _{net}}.\overrightarrow S $$

$$=$$ 2q + 3q

$$=$$ 5q

4

MCQ (Single Correct Answer)

The region between y = 0 and y = d contains a magnetic field $$\overrightarrow B = B\widehat z$$. A particle of mass m and charge q enters the region with a velocity $$\overrightarrow v = v\widehat i.$$ If d $$=$$ $${{mv} \over {2qB}},$$ the acceleration of the charged particle at the point of its emergence at the other side is :

A

$${{qvB} \over m}\left( -{{{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)$$

B

$${{qvB} \over m}\left( {{1 \over 2}\widehat i - {{\sqrt 3 } \over 2}\widehat j} \right)$$

C

$${{qvB} \over m}\left( {{{ - \widehat j + \widehat i} \over {\sqrt 2 }}} \right)$$

D

$${{qvB} \over m}\left( {{{\widehat j + \widehat i} \over {\sqrt 2 }}} \right)$$

Here R = $${{mv} \over {qB}}$$ = 2d

cos $$\theta $$ = $${{{R \over 2}} \over R}$$ = $${1 \over 2}$$

$$ \Rightarrow $$ $$\theta $$ = 60

Acceleration of the charged particle at the point of its emergence,

$$\overrightarrow {{a_c}} = {a_{{c_x}}}\left( { - \widehat i} \right) + {a_{{c_y}}}\left( { - \widehat j} \right)$$

= $${a_c}\cos 30^\circ \left( { - \widehat i} \right) + {a_c}\sin 30^\circ \left( { - \widehat j} \right)$$

= $${a_c}\left( {{{\sqrt 3 } \over 2}\left( { - \widehat i} \right) + {1 \over 2}\left( { - \widehat j} \right)} \right)$$

= $${{qvB} \over m}\left( { - {{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)$$

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