Question

If the reading of the spring balance is same as that at depth $d$ below the earths surface at poles, the value of d will be

• A. $\frac{{\omega }^2{R}^2}{g}$
• B. $\frac{{\omega }^2{R}^2}{2g}$
• C. $\frac{2{\omega }^2{R}^2}{g}$
• D. $\frac{2{\omega }^2{R}^2}{3g}$

Answer 1

The correct option is A $\frac{{\omega }^2{R}^2}{g}$

Weight at the equator $=m(g-{\omega }^{2}R)$
Weight at a depth $d$ at poles $=mg(1-\frac{d}{R})$
Equating these two, we get
$m(g-{\omega }^{2}R)=mg(1-\frac{d}{R})$
$\Rightarrow g-{\omega }^R=g-\frac{gd}{R}$
$\Rightarrow d=\frac{{\omega }^2{R}^2}{g}$