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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2021 (Online) 1st September Evening Shift

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 $$\times$$ 103 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 ________

Answer

Correct Answer is 3

Explanation


T1 = 1 hour

$$\Rightarrow$$ $$\omega$$1 = 2$$\pi$$ rad/hour

T2 = 8 hours

$$\Rightarrow$$ $$\omega$$2 = $${\pi \over 4}$$ rad/hour

R1 = 2 $$\times$$ 103 km

As T2 $$\propto$$ R3

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

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


V1 = $$\omega$$1R1 = 4$$\pi$$ $$\times$$ 103 km/h

V2 = $$\omega$$2R2 = 2$$\pi$$ $$\times$$ 103 km/h

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

JEE Main 2021 (Online) 27th August Morning Shift

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 ________

Answer

Correct Answer is 2

Explanation

Energy is maximum when mass is split equally so $${M \over m} = 2$$
3

JEE Main 2021 (Online) 27th July Morning Shift

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]

Your Input ________

Answer

Correct Answer is 4

Explanation


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

JEE Main 2021 (Online) 17th March Morning Shift

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 ________

Answer

Correct Answer is 64

Explanation

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

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

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

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

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