Question

If the ratio of lengths, radii, and Young's moduli of steel and brass wires are \(a\), \(b\), and \(c\) respectively, their respective loads are in the ratio 5:3, then the corresponding ratio of increase in their lengths would be:

• A. \(\frac{3a^2c}{b}\)
• B. \(\frac{5a}{3b^2c}\)
• C. \(\frac{3ac}{5b^2}\)
• D. \(\frac{5c}{3ab^2}\)

Hint:

When calculating the elongation of different materials under varying loads, remember that the elongation is directly proportional to the load and the length, and inversely proportional to the cross-sectional area and Young's modulus.

Answer 1

The Correct Option is B

Step 1: Understand the relationship between elongation, applied load, length, cross-sectional area, and Young's modulus. 
The elongation \(\Delta L\) in a wire under a load \(F\) can be calculated by Hooke's Law for materials, which is given by: \[ \Delta L = \frac{FL}{AE}, \] where \(L\) is the initial length, \(A\) is the cross-sectional area (\(\pi r^2\) if the wire is cylindrical), and \(E\) is Young's modulus. 
Step 2: Apply the given ratios to the formula. For steel (S) and brass (B) wires: \[ \Delta L_S = \frac{F_S L_S}{A_S E_S}, \quad \Delta L_B = \frac{F_B L_B}{A_B E_B}. \] Given: \(L_S = aL_B, \quad r_S = br_B, \quad E_S = cE_B, \quad F_S = 5F_B\), \[ A_S = \pi r_S^2 = \pi (br_B)^2 = b^2 A_B, \] \[ \Delta L_S = \frac{5F_B aL_B}{b^2 A_B cE_B} = \frac{5a}{b^2c} \Delta L_B. \]