Q134 / 150 · MHT CET 2026 (13 April Shift 1) · Physics
A string of length 'L' fixed at one end carries a mass 'm' at the other end. The string makes $\dfrac{3}{\pi}$ r.p.s. around the vertical axis through fixed end. The tension in the string is
- A$72 mL$
- Correct answerB$36 mL$
- C$18 mL$
- D$9 mL$
Solution
For a mass rotating in a horizontal circle as a conical pendulum, the horizontal component of the tension provides the necessary centripetal force.
$T \sin \theta = m \omega^2 r$
From the geometry of the pendulum, the radius of the circular path is $r = L \sin \theta$. Substituting this into the equation gives:
$T \sin \theta = m \omega^2 (L \sin \theta)$
$T = m \omega^2 L$
The frequency of revolution is given as $f = \dfrac{3}{\pi}$. The angular velocity $\omega$ is:
$\omega = 2 \pi f = 2 \pi \left( \dfrac{3}{\pi} \right) = 6$
Substituting the value of $\omega$ into the expression for tension:
$T = m (6)^2 L = 36 mL$
Answer: $36 mL$
$T \sin \theta = m \omega^2 r$
From the geometry of the pendulum, the radius of the circular path is $r = L \sin \theta$. Substituting this into the equation gives:
$T \sin \theta = m \omega^2 (L \sin \theta)$
$T = m \omega^2 L$
The frequency of revolution is given as $f = \dfrac{3}{\pi}$. The angular velocity $\omega$ is:
$\omega = 2 \pi f = 2 \pi \left( \dfrac{3}{\pi} \right) = 6$
Substituting the value of $\omega$ into the expression for tension:
$T = m (6)^2 L = 36 mL$
Answer: $36 mL$
Hence, the correct answer is option: 'B'.