Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A mass is supported on a frictionless horizontal surface. It is attached to a string and rotates about a fixed centre at an angular velocity $\omega_0$ . If the length of the string and angular velocity are doubled, the tension in the string which was initially $T_0$ is now:

$T_0$

$T_0/2$

$4T_0$

$8T_0$

Step-by-Step Solution

Identify the Force: For a mass moving in a horizontal circle on a frictionless surface, the tension ( $T$ ) in the string provides the necessary centripetal force ( $F_c$ ) [Source 61, 66].
Formula: The centripetal force in terms of angular velocity ( $\omega$ ) and radius ( $r$ ) is given by $F_c = m \omega^2 r$ . Therefore, the initial tension is: $T_0 = m \omega_0^2 l$ (where $l$ is the initial length of the string) [Source 42, 61].
New Conditions:

New length, $l' = 2l$
New angular velocity, $\omega' = 2\omega_0$

Calculate New Tension ( $T'$ ): $T' = m (\omega')^2 l'$ $T' = m (2\omega_0)^2 (2l)$ $T' = m (4\omega_0^2) (2l)$ $T' = 8 (m \omega_0^2 l)$ $T' = 8 T_0$

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started