JEE MAIN - Physics (2022 - 28th June Evening Shift - No. 6)
Two objects of equal masses placed at certain distance from each other attracts each other with a force of F. If one-third mass of one object is transferred to the other object, then the new force will be :
$${2 \over 9}$$ F
$${16 \over 9}$$ F
$${8 \over 9}$$ F
F
Explanation
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Let the masses are m and distance between them is l, then $$F = {{G{m^2}} \over {{l^2}}}$$.
When 1/3rd mass is transferred to the other then masses will be $${{4m} \over 3}$$ and $${{2m} \over 3}$$. so new force will be
$$F' = {{G{{4m} \over 3} \times {{2m} \over 3}} \over {{l^2}}} = {8 \over 9}{{G{m^2}} \over {{l^2}}} = {8 \over 9}F$$
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