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Question

If the work done is stretching a wire by $$1\ mm$$ is $$W$$. Then, the work required to stretch another wire of same material but with half the radius of cross-section and double the length by $$2\ mm$$ is.


A
4W
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B
W
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C
14W
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D
12W
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Solution

The correct option is D $$\dfrac{1}{2}W$$
Given,
    $${ \quad W }_{ 1 }=W\\ \triangle { l }_{ 1 }=1mm={ 10 }^{ -3 }m\\ { r }_{ 2 }=\frac { { r }_{ 1 } }{ 2 } \\ { l }_{ 2 }=2{ l }_{ 1 }\\ \triangle { l }_{ 2 }=2mm=2\times { 10 }^{ -3 }mm$$
We know that 
   $$W=\frac { 1 }{ 2 } \times F\times \triangle l$$
Here we have substitute $$F$$ 
  $$F=\frac { YA\triangle l }{ l } $$
We can write
  $$W\quad \propto \frac { A\triangle l\times \triangle l }{ l } $$
  $$\frac { { W }_{ 1 } }{ { W }_{ 2 } } =\frac { { A }_{ 1 }{ \left( \triangle { l }_{ 1 } \right)  }^{ 2 } }{ { l }_{ 1 } } \times \frac { { l }_{ 2 } }{ { A }_{ 2 }{ \left( \triangle { l }_{ 2 } \right)  }^{ 2 } } =\frac { { \left( { r }_{ 1 } \right)  }^{ 2 }{ \left( \triangle { l }_{ 1 } \right)  }^{ 2 } }{ { l }_{ 1 } } \times \frac { { l }_{ 2 } }{ { \left( { r }_{ 2 } \right)  }^{ 2 }{ \left( \triangle { l }_{ 2 } \right)  }^{ 2 } } $$
Substitute all value in above equation
$$\frac { { W }_{ 1 } }{ { W }_{ 2 } } ={ \left( \frac { { r }_{ 1 } }{ { r }_{ 2 } }  \right)  }^{ 2 }\times { \left( \frac { \triangle { l }_{ 1 } }{ \triangle { l }_{ 2 } }  \right)  }^{ 2 }\times \left( \frac { { l }_{ 2 } }{ { l }_{ 1 } }  \right) \\ \quad \quad \quad \quad ={ 2 }^{ 2 }\times { \left( \frac { 1 }{ 2 }  \right)  }^{ 2 }\times 2\\ \quad \quad \quad \quad =4\times \frac { 1 }{ 4 } \times 2\\ { W }_{ 2 }=\frac { 1 }{ 2 } W$$
Correct option is D
  
    

Physics

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