1

JEE Main 2021 (Online) 1st September Evening Shift

Four particles each of mass M, move along a circle of radius R under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is :

A
${1 \over 2}\sqrt {{{GM} \over {R(2\sqrt 2 + 1)}}}$
B
${1 \over 2}\sqrt {{{GM} \over R}(2\sqrt 2 + 1)}$
C
${1 \over 2}\sqrt {{{GM} \over R}(2\sqrt 2 - 1)}$
D
$\sqrt {{{GM} \over R}}$

Explanation

${F_{net}} = {{M{V^2}} \over R}$

$\sqrt 2 F + {F_1} = {{M{V^2}} \over R}$

$\sqrt 2 {{GMM} \over {{{\left( {\sqrt 2 R} \right)}^2}}} + {{GMM} \over {{{(2R)}^2}}} = {{M{V^2}} \over R}$

${{GM} \over R}\left( {{1 \over {\sqrt 2 }} + {1 \over 4}} \right) = {V^2}$

${{GM} \over R}\left( {{{4 + \sqrt 2 } \over {4\sqrt 2 }}} \right) = {V^2}$

$V = \sqrt {{{GM\left( {4 + \sqrt 2 } \right)} \over {R4\sqrt 2 }}}$

$V = {1 \over 2}\sqrt {{{GM\left( {2\sqrt 2 + 1} \right)} \over R}}$

Option (b)
2

JEE Main 2021 (Online) 31st August Evening Shift

If RE be the radius of Earth, then the ratio between the acceleration due to gravity at a depth 'r' below and a height 'r' above the earth surface is : (Given : r < RE)
A
$1 - {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} - {{{r^3}} \over {R_E^3}}$
B
$1 + {r \over {{R_E}}} + {{{r^2}} \over {R_E^2}} + {{{r^3}} \over {R_E^3}}$
C
$1 + {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} + {{{r^3}} \over {R_E^3}}$
D
$1 + {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} - {{{r^3}} \over {R_E^3}}$

Explanation

${g_{up}} = {g \over {{{\left( {1 + {r \over R}} \right)}^2}}}$

${g_{down}} = g\left( {1 - {r \over R}} \right)$

${{{g_{down}}} \over {{g_{up}}}} = \left( {1 - {r \over R}} \right){\left( {1 + {r \over R}} \right)^2}$

$= \left( {1 - {r \over R}} \right)\left( {1 + {{2r} \over R} + {{{r^2}} \over {{R^2}}}} \right)$

$= 1 + {r \over R} - {{{r^2}} \over {{R^2}}} - {{{r^3}} \over {{R^3}}}$
3

JEE Main 2021 (Online) 31st August Morning Shift

The masses and radii of the earth and moon are (M1, R1) and (M2, R2) respectively. Their centres are at a distance 'r' apart. Find the minimum escape velocity for a particle of mass 'm' to be projected from the middle of these two masses :
A
$V = {1 \over 2}\sqrt {{{4G({M_1} + {M_2})} \over r}}$
B
$V = \sqrt {{{4G({M_1} + {M_2})} \over r}}$
C
$V = {1 \over 2}\sqrt {{{2G({M_1} + {M_2})} \over r}}$
D
$V = {{\sqrt {2G} ({M_1} + {M_2})} \over r}$

Explanation

${1 \over 2}m{V^2} - {{G{M_1}m} \over {r/2}} - {{G{M_2}m} \over {r/2}} = 0$

${1 \over 2}m{V^2} = {{2Gm} \over r}({M_1} + {M_2})$

$V = \sqrt {{{4G({M_1} + {M_2})} \over r}}$

Option (b)
4

JEE Main 2021 (Online) 27th August Evening Shift

A mass of 50 kg is placed at the centre of a uniform spherical shell of mass 100 kg and radius 50 m. If the gravitational potential at a point, 25 m from the centre is V kg/m. The value of V is :
A
$-$60 G
B
+2 G
C
$-$20 G
D
$-$4 G

Explanation

${V_A} = \left[ { - {{G{M_1}} \over r} - {{G{M_2}} \over R}} \right]$

$= \left[ { - {{50} \over {25}}G - {{100} \over {50}}G} \right]$

$= - 4G$