Question

A circular table of radius 0.5 m has a smooth diametrical groove. A ball of mass 90 g is placed inside the groove along with a spring of spring constant ${10}^{2}Nc{m}^{-1}$. One end of the spring is tied to the edge of the table and the other end to the ball. The ball is at a distance of 0.1 m from the center when the table is at rest. On rotating the table with a constant angular frequency of ${10}^{2}rad{s}^{-1}$, the ball moves away from the center by a distance nearly equal to

• A. ${10}^{-1}m$
• B. ${10}^{-2}m$
• C. ${10}^{-3}m$
• D. $2\times {10}^{-1}m$

Answer 1

The correct option is B ${10}^{-2}m$

At equilibrium, balancing the centripetal force and the force experienced due to the spring:

$Kx=mr{\omega }^2\Rightarrow$

Given

$K={10}^{2}Nc{m}^{-1}={10}^{4}Nm-1\Rightarrow {10}^4\times x=\frac{90}{1000}\times (0.1+x){10}^4\Rightarrow 100x=0.9+9x\Rightarrow 91x=0.9\Rightarrow x=\frac{0.9}{91}=\frac{0.1}{90}\Rightarrow x=\frac{0.9}{91}\approx \frac{0.9}{90}=0.01m$