Solveeit Logo
Login

Question

Physics Question on mechanical properties of solids

When the load on a wire is increased from 3 kg wt3 \,kg \,wt to 5 kg wt5 \,kg \,wt the elongation increases from 0.61 mm0.61\, mm to 1.02 mm 1.02\, mm. The work done during the extension of the wire is

A

16×16\times 10−3 J10^{-3} \,J

B

8×8\times 10−2 J10^{-2} \,J

C

20×20\times 10−2 J10^{-2} \,J

D

11×11\times 10−3 J10^{-3} \,J

Answer

16×16\times 10−3 J10^{-3} \,J

Explanation

Solution

Work done in stretching the wire through 0.61 mm0.61 \,mm under the load of 3 kg wt3\, kg \,wt. W1W_{1} =12=\frac{1}{2} stretching force ×\timesextension =12×3×9.8×0.61×10−3=\frac{1}{2}\times3\times9.8\times0.61\times10^{-3} =8.967×10−3J=8.967\times10^{-3} J Work done in stretching the wire through 1.02 mm1.02 \,mm under the load of 5 kg wt5 \,kg\, wt. W2W_{2} =12×5×9.8×1.02×10−3=\frac{1}{2}\times5\times9.8\times1.02\times10^{-3} =24.99×10−3J=24.99\times10^{-3}J. Hence the work done in stretching the wire from 0.61 mm0.61 \,mm to 1.02 mm1.02 \,mm. ΔW\Delta W =W2−W1=W_{2}-W_{1} =(24.99−8.967)×10−3=\left(24.99-8.967\right)\times10^{-3} ≃\simeq 16×10−3J16\times10^{-3}J