Published by:
CGP EDU Academic Team
Published on: September 12, 2026
A mass is kept on a horizontal frictionless surface. It is attached to a string and rotates about a fixed center at an angular velocity
0 . If the length of the string and angular velocity are doubled, find the tension in the string which was initially T 0
Text Solution
Verified by ExpertsThe correct answer is:
A
Step 1: The tension in the string when the mass rotates is given by the formula: T = m r \omega^2, where m is the mass, r is the length of the string, and \omega is the angular velocity.
Step 2: Initially, the tension is T_0 = m r_0 \omega_0^2.
Step 3: When the length of the string and angular velocity are both doubled, we have:
T_new = m (2r_0) (2\omega_0)^2 = m (2r_0) (4\omega_0^2) = 8 m r_0 \omega_0^2 = 8 T_0.
Therefore, the new tension in the string is 8T_0.
Step 2: Initially, the tension is T_0 = m r_0 \omega_0^2.
Step 3: When the length of the string and angular velocity are both doubled, we have:
- New length of the string: r = 2r_0
- New angular velocity: \omega = 2\omega_0
T_new = m (2r_0) (2\omega_0)^2 = m (2r_0) (4\omega_0^2) = 8 m r_0 \omega_0^2 = 8 T_0.
Therefore, the new tension in the string is 8T_0.
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