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Standard XII

Physics

Young's Modulus of Elasticity

A 5 metre lon...

Question

A 5 metre long wire is fixed to the ceiling. A weight of 10 kg is hung at the lower end and is 1 metre above the floor. The wire was elongated by 1 mm. The energy stored in the wire due to stretching is

Zero

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0.05 joule

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100 joule

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500 joule

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Solution

The correct option is B
0.05 joule

$W = \frac{1}{2} \times F \times l = \frac{1}{2} m g l$
= $\frac{1}{2} \times 10 \times 10 \times 1 \times 10^{- 3} = 0.05 J$

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A 5 metre long wire is fixed to the ceiling. A weight of 10 kg is hung at the lower end and is 1 metre above the floor. The wire was elongated by 1 mm. The energy stored in the wire due to stretching is

The correct option is B 0.05 joule W=12×F×l=12mgl =12×10×10×1×10−3=0.05J

The correct option is B
0.05 joule

$W = \frac{1}{2} \times F \times l = \frac{1}{2} m g l$
= $\frac{1}{2} \times 10 \times 10 \times 1 \times 10^{- 3} = 0.05 J$