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Question: When the load on a wire is increased from 3 kg wt to 5 kg wt the elongation increases from 0.61 mm t...

When the load on a wire is increased from 3 kg wt to 5 kg wt the elongation increases from 0.61 mm to 1.02 mm. The required work down during the extension of the wire is

A

16×10−3 J16 \times 10^{- 3}\text{ J}

B

8×10−2 J8 \times 10^{- 2}\text{ J}

C

20×10−2 J20 \times 10^{- 2}\text{ J}

D

11×10−3 J11 \times 10^{- 3}\text{ J}

Answer

16×10−3 J16 \times 10^{- 3}\text{ J}

Explanation

Solution

: work done in stretching the wire through 0.61 mm under the load of 3 kg wt.

W1=12W_{1} = \frac{1}{2}Stretching force × extension]

=12×3×9.8×0.61×10−3= \frac{1}{2} \times 3 \times 9.8 \times 0.61 \times 10^{- 3}

=8.967×10−3J= 8.967 \times 10^{- 3}J

Work done in stretching the wire through 1.02 mm under the load of 5 kg wt.

W2=12×5×9.8×1.02×10−3W_{2} = \frac{1}{2} \times 5 \times 9.8 \times 1.02 \times 10^{- 3}

=24.99×10−3J= 24.99 \times 10^{- 3}J

Hence the work done in stretching the wire from 0.61 mm to 1.02 mm.

ΔW=W2−W1=(24.99−8.961)×10−3\Delta W = W_{2} - W_{1} = (24.99 - 8.961) \times 10^{- 3}

=16×10−3J= 16 \times 10^{- 3}J