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Question

If work done in stretching a wire by 1 mm is 2J. Then the work necessary for stretching another wire of same material but with double the radius and half the length by 1 mm in joule is


A
1/4
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B
4
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C
8
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D
16
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Solution

The correct option is D 16
The stretching force $$F=\dfrac{YA\Delta l}{l}$$
where $$Y=$$Young's modulus, $$A=$$Area of cross-section of wire, $$l=$$actual length of wire, $$\Delta l=$$increase in length.
$$F=\dfrac{Y\pi r^{2}\Delta l}{l}$$
As the material is same $$Y$$ does not change.
$$\dfrac{F_1}{F_2}=\dfrac{\dfrac{r_1^{2}\Delta l_1}{l_1}}{\dfrac{r_2^{2}\Delta l_2}{l_2}}$$
Here $$\Delta l_1=1mm$$
$$\Delta l_2=1mm$$
$$l_2=\dfrac{1}{2}l_1$$
$$r_2=2r_1$$
$$\dfrac{F_1}{F_2}=\dfrac{\dfrac{r_1^{2}\times 1mm}{l_1}}{\dfrac{4r_1^{2}\times 1mm}{\dfrac{1}{2}l_1}}$$
$$\dfrac{F_1}{F_2}=\dfrac{1}{8}$$
The work done in stretching wire by amount $$\Delta l$$ is $$W=\dfrac{1}{2} F\Delta l$$
Hence $$\dfrac{W_1}{W_2}=\dfrac{F_1}{F_2}=\dfrac{1}{8}$$
As $$F_1=2$$
$$F_2=2\times 8=16$$
Hence the correct option is (D).

Physics

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