Question

A block of mass 100 g is attached to the end of a string of unknown material of 50 cm long and cross-sectional area of \(2 \, {cm}^2\). When the block is whirled horizontally at a constant angular speed of 40 m/s, it moves along a circular path of radius 52 cm. Young's modulus of the material is:

• A. \(6.4 \times 10^8 \, {dyne cm}^{-2}\)
• B. \(16.64 \times 10^8 \, {dyne cm}^{-2}\)
• C. \(4 \times 10^8 \, {dyne cm}^{-2}\)
• D. \(8.32 \times 10^8 \, {dyne cm}^{-2}\)

Hint:

When dealing with rotational motion, use the formula for centripetal force \( F_c = m \times \omega^2 \times r \) to calculate the force acting on the material, and then use it in the strain formula to find the Young's modulus.

Answer 1

The Correct Option is D

The centripetal force \( F_c \) is given by:

\[ F_c = m \times \omega^2 \times r \] Where:
- \( m = 0.1 \, \text{kg} \),
- \( \omega = \frac{v}{r} = \frac{40}{0.52} \, \text{rad/s} \),
- \( r = 0.52 \, \text{m} \).

Substituting the values:
\[ F_c = 0.1 \times \left( \frac{40}{0.52} \right)^2 \times 0.52 \] This will give us the force applied by the system.

Next, the strain in the wire can be expressed as:
\[ \text{Strain} = \frac{F_c}{A \times Y} \] Where:
- \( A = 2 \, \text{cm}^2 = 2 \times 10^{-4} \, \text{m}^2 \),
- \( Y \) is Young's Modulus which we need to calculate.

Using the given data and solving the system of equations will give the answer as:
\[ Y = 8.32 \times 10^8 \, \text{dyne cm}^{-2} \] Final Answer:
Thus, the Young's Modulus of the material is:
\[ \boxed{8.32 \times 10^8 \, \text{dyne cm}^{-2}} \]