Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

Numerical

Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hours respectively. The radius of the orbit of nearer satellite is 2 $$\times$$ 10^{3} km. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is $${\pi \over x}rad\,{h^{ - 1}}$$ where x is ____________.

Your Input ________

Correct Answer is **3**

T

$$\Rightarrow$$ $$\omega$$

T

$$\Rightarrow$$ $$\omega$$

R

As T

$$ \Rightarrow {\left( {{{{R_2}} \over {{R_1}}}} \right)^3} = {\left( {{{{T_2}} \over {{T_1}}}} \right)^2}$$

$$ \Rightarrow {{{R_2}} \over {{R_1}}} = {\left( {{8 \over 1}} \right)^{2/3}} = 4 \Rightarrow {R_2} = 8 \times {10^3}$$ km

V

V

Relative $$\omega$$ = $${{{V_1} - {V_2}} \over {{R_2} - {R_1}}} = {{2\pi \times {{10}^3}} \over {6 \times {{10}^3}}}$$

$$ = {\pi \over 3}$$ rad/hour

x = 3

2

Numerical

A body of mass (2M) splits into four masses (m, M $$-$$ m, m, M $$-$$ m}, which are rearranged to form a square as shown in the figure. The ratio of $${M \over m}$$ for which, the gravitational potential energy of the system becomes maximum is x : 1. The value of x is ............ .

Your Input ________

Correct Answer is **2**

Energy is maximum when mass is split equally so $${M \over m} = 2$$

3

Numerical

Suppose two planets (spherical in shape) in radii R and 2R, but mass M and 9M respectively have a centre to centre separation 8 R as shown in the figure. A satellite of mass 'm' is projected from the surface of the planet of mass 'M' directly towards the centre of the second planet. The minimum speed 'v' required for the satellite to reach the surface of the second planet is $$\sqrt {{a \over 7}{{GM} \over R}} $$ then the value of 'a' is ____________.

[Given : The two planets are fixed in their position]

[Given : The two planets are fixed in their position]

Your Input ________

Correct Answer is **4**

Acceleration due to gravity will be zero at P therefore,

$${1 \over 2}m{v^2} - {{GMm} \over R} - {{G9Mm} \over {7R}} = 0 - {{GMm} \over {2R}} - {{G9Mm} \over {6R}}$$

$$\therefore$$ $$V = \sqrt {{4 \over 7}{{GM} \over R}} $$

4

Numerical

The radius in kilometer to which the present radius of earth (R = 6400 km) to be compressed so that the escape velocity is increased 10 times is ___________.

Your Input ________

Correct Answer is **64**

$${V_{es}} = \sqrt {{{2GM} \over R}} $$

$${V_{es}}\sqrt R $$ = const

$$ \therefore $$ $${V_{es}}.\sqrt R = 10{V_{es}}\sqrt {R'} $$

$$ \Rightarrow $$ $$R' = {R \over {100}}$$ = 64 km

$${V_{es}}\sqrt R $$ = const

$$ \therefore $$ $${V_{es}}.\sqrt R = 10{V_{es}}\sqrt {R'} $$

$$ \Rightarrow $$ $$R' = {R \over {100}}$$ = 64 km

On those following papers in Numerical

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