Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A ball of mass 0.1 kg is whirled in a horizontal circle of radius 1 m by means of a string at an initial speed of 10 rpm. Keeping the radius constant, the tension in the string is reduced to one quarter of its initial value. The new speed is:

5 rpm

10 rpm

20 rpm

14 rpm

Step-by-Step Solution

Identify the Force: For a ball whirled in a horizontal circle, the tension ( $T$ ) in the string provides the necessary centripetal force ( $F_c$ ) to maintain the circular motion.
Formula: The centripetal force is given by $F_c = m \omega^2 R$ , where $m$ is mass, $\omega$ is angular velocity, and $R$ is radius [Source 44, 64]. $T = m \omega^2 R$
Proportionality: Since the mass ( $m$ ) and radius ( $R$ ) are kept constant, the tension is directly proportional to the square of the angular speed. $T \propto \omega^2$
Calculation:

Initial Tension: $T_1 \propto \omega_1^2$
Final Tension: $T_2 \propto \omega_2^2$
Given: $T_2 = \frac{1}{4} T_1$ $\frac{T_2}{T_1} = \frac{\omega_2^2}{\omega_1^2}$ $\frac{1}{4} = \left( \frac{\omega_2}{10} \right)^2$ Taking the square root of both sides: $\frac{1}{2} = \frac{\omega_2}{10}$ $\omega_2 = 5 \text{ rpm}$

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