Question

A body hanging from a massless spring stretches it by $3$cm on Earth's surface. At a place $800$km above the Earth's surface, the same body will stretch the spring by : (Radius of Earth $=6400$km)

• A. $\left(\frac{34}{27}\right)$cm
• B. $\left(\frac{64}{27}\right)$cm
• C. $\left(\frac{27}{64}\right)$cm
• D. $\left(\frac{27}{34}\right)$cm
• E. $\left(\frac{35}{81}\right)$cm

Answer 1

The correct option is B $\left(\frac{64}{27}\right)$cm

Acceleration due to gravity,
$g=\frac{GM}{r^2}$
$\therefore g\propto \frac{1}{r^2}$
When the body is hanged on a spring
$F=-kx=mg$ ............(i)
Spring force $F$ acts on it, where x is extension in spring. Let $800$km above the Earth's surface, the stretch in the length of spring is $x$' and value of $g$ is $g^{\prime }$.
$g=\frac{GM}{R^2}$ and $g^{\prime }=\frac{GM}{(R+h{)}^2}$
So, $g^{\prime }=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}$ .........(ii)
where, R=radius of Earth
From Eq. (i), we get
$kx=mg$
$\Rightarrow x\propto g$ .........(iii)
Therefore, $\frac{x^{\prime }}{x}=\frac{g^{\prime }}{g}$
From Eqs. (ii) and (iii), we get
$\frac{x^{\prime }}{x}=\frac{g}{g{\left(1+\frac{h}{R}\right)}^2}=\frac{g}{g{(1+\frac{800\times {10}^3}{6400\times {10}^3})}^2}$
$=\frac{g}{g{(1+\frac{1}{8})}^2}$
$=\frac{g}{g{\left(\frac{g}{8}\right)}^2}=\frac{g\times 64}{g\times 81}=\frac{64}{81}$
$x^{\prime }=\frac{64}{81}x$ $[x=3cm]$
$\Rightarrow x^{\prime }=\frac{64}{81}\times 3=\frac{64}{27}$
$\Rightarrow x^{\prime }=\frac{64}{27}$cm.