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Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A straight rod of length L extends from x = a to x = L + a. The gravitational force it exerts on a point mass 'm' at x = 0, if the mass per unit length of the rod is A + Bx^{2} , is given by :

A

$$Gm\left[ {A\left( {{1 \over a} - {1 \over {a + L}}} \right) - BL} \right]$$

B

$$Gm\left[ {A\left( {{1 \over a} - {1 \over {a + L}}} \right) + BL} \right]$$

C

$$Gm\left[ {A\left( {{1 \over {a + L}} - {1 \over a}} \right) + BL} \right]$$

D

$$Gm\left[ {A\left( {{1 \over {a + L}} - {1 \over a}} \right) - BL} \right]$$

dm = (A + Bx

dF = $${{GMdm} \over {{x^2}}}$$

F = $$\int_a^{a + L} {{{GM} \over {{x^2}}}} $$ (A + Bx

= GM$$\left[ { - {A \over x} + Bx} \right]_a^{a + L}$$

= GM$$\left[ {A\left( {{1 \over a} - {1 \over {a + L}}} \right) + BL} \right]$$

2

MCQ (Single Correct Answer)

The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be :

A

$${{\sqrt 3 } \over 2}$$ s

B

$${3 \over 2}$$ s

C

$${2 \over {\sqrt 3 }}$$ s

D

$$2\sqrt 3 $$ s

$$ \because $$ g = $${{GM} \over {{R^2}}}$$

$${{{g_p}} \over {{g_e}}}$$ = $${{{M_e}} \over {{M_e}}}{\left( {{{{{\mathop{\rm R}\nolimits} _e}} \over {{R_p}}}} \right)^2}$$ = 3$${\left( {{1 \over 3}} \right)^2}$$ = $${{1 \over 3}}$$

Also T $$ \propto $$ $${1 \over {\sqrt g }}$$

$$ \Rightarrow $$ $${{{T_p}} \over {{T_e}}}$$ = $$\sqrt {{{{g_e}} \over {{g_p}}}} $$ = $$\sqrt 3 $$

$$ \Rightarrow $$ T_{p} = 2$$\sqrt 3 $$ s

$${{{g_p}} \over {{g_e}}}$$ = $${{{M_e}} \over {{M_e}}}{\left( {{{{{\mathop{\rm R}\nolimits} _e}} \over {{R_p}}}} \right)^2}$$ = 3$${\left( {{1 \over 3}} \right)^2}$$ = $${{1 \over 3}}$$

Also T $$ \propto $$ $${1 \over {\sqrt g }}$$

$$ \Rightarrow $$ $${{{T_p}} \over {{T_e}}}$$ = $$\sqrt {{{{g_e}} \over {{g_p}}}} $$ = $$\sqrt 3 $$

$$ \Rightarrow $$ T

3

MCQ (Single Correct Answer)

A satellite is revolving in a circular orbit at a height h form the earth surface, such that h < < R where R is the earth. Assuming that the effect of earth's atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is :

A

$$\sqrt {gR} \left( {\sqrt 2 - 1} \right)$$

B

$$\sqrt {2gR} $$

C

$$\sqrt {gR} $$

D

$${{\sqrt {gR} } \over 2}$$

v_{0} = $$\sqrt {g(R + h)} \approx \sqrt {gR} $$

v_{e} = $$\sqrt {2g(R + h)} \approx \sqrt {2gR} $$

$$\Delta $$v=v_{e} $$-$$ v_{0} = $$\left( {\sqrt 2 - 1} \right)\sqrt {gR} $$

v

$$\Delta $$v=v

4

MCQ (Single Correct Answer)

Two stars of masses 3 $$ \times $$ 10^{31} kg each, and at distance 2 $$ \times $$ 10^{11} m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the starâ€™s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is - (Take Gravitational constant; G = 6.67 $$ \times $$ 10^{â€“11} Nm^{2} kg^{â€“2})

A

2.4 $$ \times $$ 10^{4} m/s

B

1.4 $$ \times $$ 10^{5} m/s

C

3.8 $$ \times $$ 10^{4} m/s

D

2.8 $$ \times $$ 10^{5} m/s

By energy convervation between 0 & $$\infty $$.

$$ - {{GMm} \over r} + {{ - GMm} \over r} + {1 \over 2}m{V^2} = 0 + 0$$

[M is mass of star m is mass of meteroite)

$$ \Rightarrow $$ v $$ = \sqrt {{{4GM} \over r}} = 2.8 \times {10^5}$$m/s

$$ - {{GMm} \over r} + {{ - GMm} \over r} + {1 \over 2}m{V^2} = 0 + 0$$

[M is mass of star m is mass of meteroite)

$$ \Rightarrow $$ v $$ = \sqrt {{{4GM} \over r}} = 2.8 \times {10^5}$$m/s

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Units & Measurements

Motion

Laws of Motion

Work Power & Energy

Simple Harmonic Motion

Center of Mass and Collision

Rotational Motion

Gravitation

Properties of Matter

Heat and Thermodynamics

Waves

Vector Algebra

Circular Motion

Ray & Wave Optics

Electrostatics

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Capacitor

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Atoms and Nuclei

Dual Nature of Radiation

Electronic Devices

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