An object of mass 100 kg falls from point A to B as shown in figure. The change in its weight, corrected to the nearest integer is ($R_E$ is the radius of the earth):

Question

Two identical particles each of mass $ m $ go around a circle of radius $ a $ under the action of their mutual gravitational attraction. The angular speed of each particle will be:

Answer 1

$Mg' = Mg \frac{R^2}{(R+h)^2}$

$Mg' = Mg \frac{R^2}{(R+2R)^2} = \frac{Mg}{9}$

$Mg' = Mg \frac{R^2}{(R+\frac{3}{2}R)^2} = \frac{Mg}{25}$

$Mg \left( \frac{4}{25} - \frac{1}{9} \right) = 49 \, \text{N}$

An object of mass 100 kg falls from point A to B as shown in figure. The change in its weight, corrected to the nearest integer is ($R_E$ is the radius of the earth):

The Correct Option is A

Solution and Explanation

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