Question

A body weighing $ 700gm-wt $ on the surface of the earth. How much will be the weight of the body on the surface of a planet whose mass is $ \dfrac{1}{7} $ and radius half of the earth
 $ \begin{align}
  & A.200gm-wt \\
 & B.400gm-wt \\
 & C.500gm-wt \\
 & D.300gm-wt \\
\end{align} $

Answer 1

Hint: The acceleration due to gravity is given as the product of gravitational constant and mass of the earth which is divided by the square of the radius of the earth. As the mass of the body has been changed as well as the radius is also changed, there will be change in the weight of the object. Weight of an object is found out by taking the product of the mass of the object and acceleration due to gravity. These all help you to solve this question.

Complete step-by-step answer:
The acceleration due to gravity is the acceleration provided by the earth when there is a fall due to gravitational force acting on a body.
It can be written as the formula,
 $ g=\dfrac{GM}{{{R}^{2}}} $
Where $ G $ be the gravitational constant, $ M $ be the mass of the earth and $ R $ be the radius of the earth.
The mass of the planet is given as the $ \dfrac{1}{7} $ mass of earth. This can be written as,
 $ {M}'=\dfrac{M}{7} $
And the radius of the planet is found to be half of that of earth. This can be written as,
 $ {R}'=\dfrac{R}{7} $
Substituting this in the equation of acceleration due to gravity,
\[g=\dfrac{4GM}{7{{R}^{2}}}\]
Now as we all know, the weight of a body on the surface can be found by the formula,
\[W=mg\]
Here it is mentioned that the mass of the body is given as,
\[m=700g\]
Therefore the weight of the body will be,
\[W=\dfrac{700\times 4GM}{7{{R}^{2}}}=400gm-wt\]
Therefore the correct answer is obtained.

So, the correct answer is “Option B”.

Note: Gravitational acceleration is given as the free fall acceleration of a body in the space without any external force. This is uniform gain in speed because of the gravitational force of attraction only. This is a vector quantity also.