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Question

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


A
16×103J
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B
8×102J
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C
20×102J
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D
11×103J
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Solution

The correct option is C $$16\times { 10 }^{ -3 }J$$
Given:
Apply Load:    $$M_1=3kg \ \ \ \ \Delta L= 0.61mm \\ M_2=5kg \ \ \ \Delta L=1.02mm$$    

Work done is stretching the wire through $$0.61mm$$ under the load of $$3kg$$ $$wt$$
$${ W }_{ 1 }=\cfrac { 1 }{ 2 } stretching\quad force\times extension=\cfrac { 1 }{ 2 } \times 3\times 9.8\times 0.61\times { 10 }^{ -3 }=8.967\times { 10 }^{ -3 }J$$
Work done in stretching the wire through $$1.02mm$$ under the load of $$5kg$$ $$wt$$
$${ W }_{ 2 }=\cfrac { 1 }{ 2 } \times 5\times 9.8\times 1.02\times { 10 }^{ -3 }=24.99\times { 10 }^{ -3 }J$$
Hence work done in stretching the wire from $$0.61mm$$ to $$1.02mm$$
$$\Delta W={ W }_{ 2 }-{ W }_{ 1 }=\left( 24.99-8.961 \right) \times { 10 }^{ -3 }\simeq 16\times { 10 }^{ -3 }J$$
1031875_936989_ans_e85dccce7c664c61b4a64f10f29eb961.PNG

Physics

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