Question

A metal string A is suspended from a rigid support and its free end is attached to a block of mass M. Second block having mass 2M is suspended at the bottom of the first block using a string B. The area of cross sections of strings A and B are same. The ratio of lengths of strings of A to B is 2 and the ratio of their Young's moduli ($Y_A / Y_B$) is 0.5. The ratio of elongations in A to B is _______.

• A. 1
• B. 4
• C. 8
• D. 6

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Two strings suspend successive masses. The tension in the top string bears the weight of all suspended masses below it, while the tension in the bottom string supports only the bottom mass. We apply Hooke's Law relating tension, area, length, and Young's modulus to find individual elongations.

Step 2: Key Formula or Approach:
Young's Modulus definition: $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{T / A}{\Delta L / L}$
Rearranging for elongation: $\Delta L = \frac{TL}{AY}$

Step 3: Detailed Explanation:
Calculate the tension in each string:
String B supports only the lower block of mass 2M.
$T_B = 2Mg$.
String A supports both the upper block of mass M and the lower block of mass 2M.
$T_A = (M + 2M)g = 3Mg$.
Using the elongation formula for both strings:
$\Delta L_A = \frac{T_A L_A}{A_A Y_A}$
$\Delta L_B = \frac{T_B L_B}{A_B Y_B}$
We are given the following ratios:
Areas are the same: $\frac{A_B}{A_A} = 1$.
Length ratio: $\frac{L_A}{L_B} = 2$.
Young's moduli ratio: $\frac{Y_A}{Y_B} = 0.5 \implies \frac{Y_B}{Y_A} = 2$.
Tension ratio (calculated): $\frac{T_A}{T_B} = \frac{3Mg}{2Mg} = \frac{3}{2}$.
Divide the elongation formulas to find their ratio:
$\frac{\Delta L_A}{\Delta L_B} = \left( \frac{T_A}{T_B} \right) \cdot \left( \frac{L_A}{L_B} \right) \cdot \left( \frac{A_B}{A_A} \right) \cdot \left( \frac{Y_B}{Y_A} \right)$
$\frac{\Delta L_A}{\Delta L_B} = \left( \frac{3}{2} \right) \times (2) \times (1) \times (2) = \frac{3 \times 4}{2} = 6$.

Step 4: Final Answer:
The ratio of elongations is 6.