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
B
8×102J
C
20×102J
D
11×103J

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 $$1.02mm$$$$\Delta W={ W }_{ 2 }-{ W }_{ 1 }=\left( 24.99-8.961 \right) \times { 10 }^{ -3 }\simeq 16\times { 10 }^{ -3 }J$$Physics

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