KVPY 2020-21 SA Exam Previous Papers

Question : 29 of 80

Marks: +1, -0

The ratio of gravitational force and electrostatic repulsive force between two electrons is approximately (gravitational constant

=6.7 \×10^{-11} {Nm}^{2} \/ {Kg}^{2}

, mass of an electron

=9.1 \×10^{-31} {kg}

, chargeon an electron

=

1.6 \×10^{19} {C})

Solution:

F_{G}={G m_{1} m_{2}}/{r^{2}}

={6.7 \×10^{-11} \× 9.1 \×10^{-31} \×9.1 \×10^{-31}}/{r^{2}}

F_{C}={k q_{1} q_{2}}/{r^{2}}

={9 \×10^{9} \×1.6 \×10^{-19} \×1.6 \×10^{-19}}/{r^{2}}

{F_{G}}/{F_{C}}={6.7 \×(9.1)^{2} \×10^{-73}}/{9 \×(1.6)^{2} \×10^{-29}}

{F_{G}}/{F_{C}}={554.8 \×10^{-44}}/{23.04}=24 \×10^{-44}

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