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### JEE Main 2019 (Online) 10th January Evening Slot

Two stars of masses 3 $\times$ 1031 kg each, and at distance 2 $\times$ 1011 m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star’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 $\times$ 10–11 Nm2 kg–2)
A
2.4 $\times$ 104 m/s
B
1.4 $\times$ 105 m/s
C
3.8 $\times$ 104 m/s
D
2.8 $\times$ 105 m/s

## Explanation

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
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### JEE Main 2019 (Online) 11th January Morning Slot

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

## Explanation

v0 = $\sqrt {g(R + h)} \approx \sqrt {gR}$

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

$\Delta$v=ve $-$ v0 = $\left( {\sqrt 2 - 1} \right)\sqrt {gR}$
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### JEE Main 2019 (Online) 11th January Evening Slot

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

## Explanation

$\because$    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 $\propto$ ${1 \over {\sqrt g }}$

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

$\Rightarrow$  Tp = 2$\sqrt 3$ s
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### JEE Main 2019 (Online) 12th January Morning Slot

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 + Bx2 , 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
$Gm\left[ {A\left( {{1 \over {a + L}} - {1 \over a}} \right) - BL} \right]$

## Explanation dm = (A + Bx2)dx

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

F = $\int_a^{a + L} {{{GM} \over {{x^2}}}}$ (A + Bx2)dx

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

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

NEET