Question

In searle' experiment, diameter of wire was measured 0.05 cm by *** 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' 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