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### JEE Main 2016 (Online) 9th April Morning Slot

Figure shows elliptical path abcd of a planet around the sun S such that the area of triangle csa is ${1 \over 4}$ the area of the ellipse. (See figure) With db as the semimajor axis, and ca as the semiminor axis. If t1 is the time taken for planet to go over path abc and t2 for path taken over cda then :

A
t1 = t2
B
t1 = 2t2
C
t1 = 3t2
D
t1 = 4t2

## Explanation

Let the area of ellipse = A

$\therefore$   Area of abcSa = ${A \over 2} + {A \over 4}$ = Area of half of the ellipse + Area of the triangle = ${{3A} \over 4}$

Area of adcSa = ${A \over 2} - {A \over 4}$ = ${A \over 4}$

$\therefore$   ${{{t_1}} \over {{t_2}}}$ = ${{{{3A} \over 4}} \over {{A \over 4}}}$ = 3

$\therefore$   t1 = 3t2
2

### JEE Main 2016 (Online) 10th April Morning Slot

An astronaut of mass m is working on a satellite orbiting the earth at a distance h from the earth’s surface. The radius of the earth is R, while its mass is M. The gravitational pull FG on the astronaut is :
A
Zero since astronaut feels weightless
B
0 < FG < ${{GMm} \over {{R^2}}}$
C
${{GMm} \over {{{\left( {R + h} \right)}^2}}}$ < FG < ${{GMm} \over {{R^2}}}$
D
FG = ${{GMm} \over {{{\left( {R + h} \right)}^2}}}$

## Explanation

The gravitional pull on the astronaut is,

FG  =  ${{GMm} \over {{{\left( {R + h} \right)}^2}}}$

The satellite is moving so fast around the earth that whenever it trying to fall on the earch it is missing the earth and it will keep going arround the earth. On the satellite the astronaut feels weightless because the satellite is constantly freefalling.

If there was no gravity then the satellite will move in straight line in the space. Then also the astronaut will feel weightless. But the reason in this case is here is no gravitional force on the satellite in free space.
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### JEE Main 2017 (Offline)

The variation of acceleration due to gravity $g$ with distance d from centre of the earth is best represented by (R = Earth’s radius):
A
B
C
D

## Explanation

When d < R means distance of a point is d from the center of the circle where d is inside the earth surface. In this case the value of acceleration, $g = -{{GMd} \over {{R^3}}}$

$\therefore$ $g \propto d$

So inside of the earth surface g - d graph is straight line.

When d > R means distance of a point is d from the center of the circle where d is outside of the earth's surface. In this case the value of acceleration, $g = - {{GM} \over {{d^2}}}$

$\therefore$ $g \propto {1 \over {{d^2}}}$

So outside of the earth surface g - d graph is hyperbolic.
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### JEE Main 2017 (Online) 8th April Morning Slot

If the Earth has no rotational motion, the weight of a person on the equator is W. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh ${3 \over 4}$ W. Radius of the Earth is 6400 km and g=10 m/s2.
A
1.1 $\times$ 10−3 rad/s
B
0.83 $\times$ 10−3 rad/s
C
0.63 $\times$ 10−3 rad/s
D
0.28 $\times$ 10−3 rad/s

## Explanation

Initially when earth is not rotating then weight of the person is $w$.

When earth rotares about it's axis then weight = ${{3\omega } \over 4}$

Then,

g' = g $-$ $\omega$2R cos2$\theta$

$\Rightarrow$ $\,\,\,$ ${{3g} \over 4}$ = g $-$ $\omega$2R cos2 0o

$\Rightarrow$ $\,\,\,$ $\omega$2R = ${g \over 4}$

$\,\,\,$ $\omega$ = $\sqrt {{g \over {4R}}}$

$\Rightarrow$ $\omega = \sqrt {{{10} \over {4 \times 6400 \times {{10}^3}}}}$

= 0.63 $\times$ 10$-$3 rad/s