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

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

A $5$ metre long wire is fixed to the ceiling. A weight of $10 k g$ is hung at the lower end and is $1$ metre above the floor. The wire was elongated by $1 m m$ . The energy stored in the wire due to stretched is

The correct option is B $0.05$ joule
$W = \frac{1}{2} \times F \times l = \frac{1}{2} m g l$
$= \frac{1}{2} \times 10 \times 10 \times 1 \times 10^{- 3} = 0.05 J$