Question

A body hanging from a massless spring, stretches it by \[3\,{\text{cm}}\] on Earth’s surface. At a place \[800\,{\text{km}}\] above the Earth’s surface, the same body will stretch the spring by: (Radius of Earth=\[6400\,{\text{km}}\])
A. \[\left( {\dfrac{{34}}{{27}}} \right)\,{\text{cm}}\]
B. \[\left( {\dfrac{{64}}{{27}}} \right)\,{\text{cm}}\]
C. \[\left( {\dfrac{{27}}{{64}}} \right)\,{\text{cm}}\]
D. \[\left( {\dfrac{{27}}{{34}}} \right)\,{\text{cm}}\]
E. \[\left( {\dfrac{{35}}{{81}}} \right)\,{\text{cm}}\]

Answer 1

Hint: Use the formula for the spring force on the spring. Derive the relation between the stretching of the spring and acceleration due to gravity. Use the relation for the acceleration due to gravity in terms of the radius of the planet and mass of the planet.

Formula used:
The spring force \[F\] is given by
\[F = kx\]
Here, \[k\] is the spring constant and \[x\] is the displacement of the spring from the mean position.
The expression for the acceleration due to gravity \[g\] is
\[g = \dfrac{{GM}}{{{R^2}}}\]
Here, \[G\] is a universal gravitational constant, \[M\] is the mass of the planet and \[R\] is the radius of the planet.

Complete step by step answer:
It is given that a mass \[m\] is hanging freely from a massless spring.
The spring force of the spring is balanced by the weight of the mass attached to the spring.
\[kx = mg\]
From the above equation, it can be concluded that the displacement or stretching \[x\] of the spring is directly proportional to the acceleration due to gravity \[g\] at that place.
\[x \propto g\] …… (3)
Rewrite equation (2) for the acceleration due to gravity \[g\] on the surface of the Earth.
\[g = \dfrac{{GM}}{{{R^2}}}\]
Here, \[M\] is the mass of the Earth and \[R\] is the radius of the Earth.
Rewrite equation (2) for the acceleration due to gravity \[g'\] at a height \[h\] from the surface of the Earth.
\[g' = \dfrac{{GM}}{{{{\left( {R + h} \right)}^2}}}\]
Rewrite equation (3) for the stretching of the spring on the Earth and above the Earth.
\[\dfrac{{x'}}{x} = \dfrac{{g'}}{g}\]
Substitute \[\dfrac{{GM}}{{{R^2}}}\] for \[g\] and \[\dfrac{{GM}}{{{{\left( {R + h} \right)}^2}}}\] for \[g'\] in the above equation.
\[\dfrac{{x'}}{x} = \dfrac{{\dfrac{{GM}}{{{{\left( {R + h} \right)}^2}}}}}{{\dfrac{{GM}}{{{R^2}}}}}\]
\[ \Rightarrow x' = x\dfrac{{{R^2}}}{{{{\left( {R + h} \right)}^2}}}\]
\[ \Rightarrow x' = x\dfrac{1}{{{{\left( {1 + \dfrac{h}{R}} \right)}^2}}}\]
Substitute \[800\,{\text{km}}\] for \[h\] and \[6400\,{\text{km}}\] for \[R\] in the above equation.
\[ \Rightarrow x' = x\dfrac{1}{{{{\left( {1 + \dfrac{{800\,{\text{km}}}}{{6400\,{\text{km}}}}} \right)}^2}}}\]
\[ \Rightarrow x' = \dfrac{{64}}{{81}}x\]
The stretching of the spring on the surface of the Earth is \[3\,{\text{cm}}\].
Substitute for in the above equation.
\[ \Rightarrow x' = \dfrac{{64}}{{81}}\left( {3\,{\text{cm}}} \right)\]
\[ \Rightarrow x' = \left( {\dfrac{{64}}{{27}}} \right)\,{\text{cm}}\]
Therefore, the body will stretch the spring by \[\left( {\dfrac{{64}}{{27}}} \right)\,{\text{cm}}\].

Hence, the correct option is B.

Note:
One may think of converting the unit of the stretching of spring on the surface of the Earth in the SI system of units. But there is no need for unit conversion. The unit of the height of the spring from the surface of earth and the radius of the Earth gets cancelled and final stretching of the spring will be in centimeter.