Question

A circular turn table of radius 0.5 m has a smooth grrove as shown in fig. A ball of mass 90 g is placed inside the groove along with a spirng of spring constant ${10}^{2}N/cm$. the ball is at a distance of 0.1 m from the centre when the turn table is at rest. on rotating the turn table with a constant angular velocity of ${10}^2-rad-se{c}^{-1}$ the ball, moves away from the initial position 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 ${10}^{-2}m$

$\begin{array}{c}\text{Given, mass of ball}P\left(m\right)=90gm\\ \text{spring const.}\left(k\right)=100N/cm\\ \text{and radius of turn table}=0.5m\text{.}\end{array}$

$\text{Let, ball}P\text{moves '}x\text{'}cm\text{towards the spring.}$

$\begin{array}{cc}{F}_{\text{centrifugal}}& =m{\omega }^{2}r\\ & \Rightarrow m{\omega }^{2}r=kx\end{array}\text{where}\begin{array}{cc}r& =0.1m+x=10cm+xcm\end{array}$

$\begin{array}{cc}\Rightarrow & \frac{90}{1000}\times (100{)}^2\times \frac{(10+x)}{100}=100x\\ \therefore x& =\frac{90}{91}cm\approx 0.01={10}^{-2}m\end{array}$