1

### 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
3

### JEE Main 2019 (Online) 12th January Morning Slot

A satellite of mass M is in a circular orbit of radius R about the centre of the earth. A meteorite of the same mass, falling towards the earth, collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before the collision. The subsequent motion of the combined body will be :
A
in the same circular orbit of radius R
B
such that it escapes to infinity
C
in a circular orbit of a different radius
D
in an elliptical orbit

## Explanation mv$\widehat i$ + mv$\widehat j$

= 2m${\overrightarrow v ^1}$

$\overrightarrow v$ = ${1 \over {\sqrt 2 }} \times \sqrt {{{GM} \over R}}$
4

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