A circular disc of mass 10 kg is suspend | Class 11 Physics Chapter Oscillations, Oscillations NCERT Solutions

Question:

A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire.

(Torsional spring constant α is defined by the relation J = -α ø, where J is the restoring couple and ø¸ the angle of twist).

Answer:

Mass of the circular disc, m = 10 kg

Radius of the disc, r = 15 cm = 0.15 m

The torsional oscillations of the disc has a time period, T = 1.5 s

The moment of inertia of the disc is:

l  =  1/2 mr2

= 1/2 x 10 x (0.15)2

= 0.1125 kg m2

Time period,  T  = 2π underroot 1 / α

α is the torsional constant.

α = 4 π2 l / T2

= 4 x π2 x 0.1125  / (1.5)2

= 1.972 Nm/rad

Hence, the torsional spring constant of the wire is 1.972 Nm rad-1.


Study Tips for Answering NCERT Questions:

NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:

  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about Oscillations.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
  • Practice answering questions within the time limit to improve your speed and accuracy.
  • Discuss your answers with your teachers or peers to get feedback and improve your understanding.

Comments

Comment(s) on this Question

Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 23: A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by ....