Question

If the ratio of lengths, radii and Young's Moduli of steel and brass wires in the figure are $ a $, $ b $, and $ c $ respectively, then the corresponding ratio of increase in their lengths would be:

• A. \( \frac{a}{3bc} \)
• B. \( \frac{3a}{2bc} \)
• C. \( \frac{2a}{3bc} \)
• D. \( \frac{2ab^2}{c} \)

Hint:

When dealing with problems involving Young's Modulus and changes in length, remember the formula \( \Delta L = \frac{F L}{A Y} \), where the length increase depends on the applied force, original length, cross-sectional area, and Young’s Modulus.

Answer 1

The Correct Option is C

Let the increase in the length of the brass wire be \( \Delta L_1 \) and the increase in the length of the steel wire be \( \Delta L_2 \). The general formula for the increase in the length of a wire under a force \( F \) is given by: \[ \Delta L = \frac{F L}{A Y} \] where \( F \) is the applied force, \( L \) is the original length, \( A \) is the cross-sectional area, and \( Y \) is Young's Modulus. 
For both the brass and steel wires, the force applied is the same. 
We are given the ratios of lengths, radii, and Young's Moduli as \( a \), \( b \), and \( c \) respectively. 
The corresponding increases in lengths for brass and steel wires will be: 
For the brass wire: \[ \Delta L_1 = \frac{F a}{b c} \] For the steel wire: \[ \Delta L_2 = \frac{F 2a}{b c} \] 
The total increase in length is the sum of \( \Delta L_1 \) and \( \Delta L_2 \): \[ \Delta L_{\text{total}} = \Delta L_1 + \Delta L_2 = \frac{F a}{b c} + \frac{F 2a}{b c} = \frac{3F a}{b c} \] Thus, the ratio of the increase in lengths is: \[ \frac{\Delta L_{\text{total}}}{\Delta L_1} = \frac{\frac{3F a}{b c}}{\frac{F a}{b c}} = \frac{3}{1} = 3 \] The final ratio of the increase in lengths is: \[ \frac{2a}{3bc} \] Thus, the correct answer is (C).