Young's moduli of two wires A and B are in the ratio 7:4. Wire A is 2 m long and has radius R. Wire B is 1.5 m long and has radius 2 mm. If the two wires stretch by the same length for a given load, then the value of R is close to?

Correct option is A. 1.7 mm. Given Y_A / Y_B = 7/4, L_A = 2 m, A_A = π R^2, L_B = 1.5 m, A_B = π (2 mm)^2, F and ΔL are the same. We know that Young's modulus Y is given by:

Y = (F/A) / (ΔL/L) = FL / (AΔL)

For wire A:

Y_A = FL_A / (A_A ΔL) = F(2m) / (πR²ΔL)

For wire B:

Y_B = FL_B / (A_B ΔL) = F(1.5m) / (π(2mm)²ΔL)

Since the wires stretch by the same length (ΔL) for a given load (F), we can write the ratio of Young's moduli as:

Y_A / Y_B = (F(2m) / (πR²ΔL)) / (F(1.5m) / (π(2mm)²ΔL))

Simplifying, we get:

Y_A / Y_B = (2m * π(2mm)²) / (1.5m * πR²)

We are given that Y_A / Y_B = 7/4. Substituting this, we have:

7/4 = (2m * π(2mm)²) / (1.5m * πR²)

Solving for R²:

R² = (4 * 2m * π(2mm)²) / (7 * 1.5m * π)

R² = (8 * (2mm)²) / (10.5)

R² ≈ 3.05 mm²

Taking the square root:

R ≈ √3.05 mm² ≈ 1.75 mm

Therefore, the value of R is close to 1.7 mm.