Question

In searle's experiment, diameter of wire was measured 0.05 cm by screw gauge of least count 0.001 cm. The length of wire was measured 110 cm by metes scale of least count 0.1 cm. When a weight of 50 N is suspended from the wire, extension is measured to be 0.125 cm by a micrometer of lecst count 0.001 cm. Find maximum error in the measurement of Young's modulus of the wire material. 

(A) 1.09 × 10¹² 

(B) 1.09 × 10¹⁰ 

(C) 3.09 × 10¹⁰ 

(D) 3.09 × 10¹²

Answer 1

The correct option is (B) 1.09 × 10¹⁰.

Explanation:

Y = {(stress) / (strain)} = {(F/A) / (ℓ/L)} 

= (FL / Aℓ) = [FL / {ℓ(πd² / 4)}] 

∴ Y = {(4FL) / (πd²ℓ)} 

∴ Y = constant × {L / (d² ℓ)} ---- (4F / π) is constant 

∴ (ΔY / Y) = (ΔL / L) + {(2Δd) / d} + (Δℓ / ℓ) 

= {(0.1) / (110)} + 2{(0.001) / (0.050)} + {(0.001) / (0.125)} 

(ΔY / Y) = {1 / (1100)} + (2 / 50) + {1 / (125)} 

∴ {(%ΔY) / Y} = [100 × {1 / (1100)}] + [100 × {2 / 50}] + [100 × {1 / (125)}] 

= (1 / 11) + 4 + 0.8 

= 4.89% 

Now 

Y = {(4FL) / (πd² ℓ)} 

= [{4 × 50 × 1.1} / {π × (0.05 × 10^–2)² × 0.125 × 10^–2}] 

∴ Y = 2.24 × 10¹¹ N/m² 

As (ΔY / Y) = 4.89% 

Hence 

ΔY = 2.24 × 10¹¹ × 4.89 × 10^–2 

ΔY = 1.09 × 10¹⁰ N/m²