(A) The energy stored in a stretched wire is given by the formula for elastic potential energy: $U = \frac{1}{2} \times F \times \Delta L$.Given:Force

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(A) The energy stored in a stretched wire is given by the formula for elastic potential energy: $U = \frac{1}{2} \times F \times \Delta L$.Given:Force $F = 20 \ N$Extension $\Delta L = 1.0 \ mm = 1.0 \times 10^{-3} \ m$.Substituting the values into the formula:$U = \frac{1}{2} \times 20 \times 1.0 \times 10^{-3} \ J$.$U = 10 \times 10^{-3} \ J$.$U = 0.01 \ J$.

Class 11 Physics Mechanical Properties of Solids Work Done in Stretching a Wire

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$A$ wire is suspended by one end. At the other end,a weight equivalent to $20 \ N$ force is applied. If the increase in length is $1.0 \ mm$,the increase in energy of the wire will be ....... $J$.

A
$0.01$
B
$0.02$
C
$0.04$
D
$1.00$

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Mechanical Properties of Solids

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Work Done in Stretching a Wire

The elastic potential energy of a strained body is

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$A$ wire suspended vertically from one end is stretched by attaching a weight $200 \,N$ to the lower end. The weight stretches the wire by $1 \,mm$. The elastic potential energy gained by the wire is ....... $J$

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If the work done in stretching a wire by $1 \ mm$ is $2 \ J$,the work necessary for stretching another wire of same material but with double radius of cross-section and half the length by $1 \ mm$ is $.... \ J$.

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$A$ wire of length $50\, cm$ and cross-sectional area of $1\, mm^2$ is extended by $1\, mm$. The required work done will be $(Y = 2 \times 10^{10}\, N/m^2)$.

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The work done per unit volume to stretch a wire by $1\%$ of its length,having a cross-sectional area of $1\,mm^2$,is: $[Y = 9 \times 10^{11}\,N/m^2]$

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$A$ wire is suspended by one end. At the other end,a weight equivalent to $20 \ N$ force is applied. If the increase in length is $1.0 \ mm$,the increase in energy of the wire will be ....... $J$.

Similar Questions

The elastic potential energy of a strained body is

$A$ wire suspended vertically from one end is stretched by attaching a weight $200 \,N$ to the lower end. The weight stretches the wire by $1 \,mm$. The elastic potential energy gained by the wire is ....... $J$

If the work done in stretching a wire by $1 \ mm$ is $2 \ J$,the work necessary for stretching another wire of same material but with double radius of cross-section and half the length by $1 \ mm$ is $.... \ J$.

$A$ wire of length $50\, cm$ and cross-sectional area of $1\, mm^2$ is extended by $1\, mm$. The required work done will be $(Y = 2 \times 10^{10}\, N/m^2)$.

The work done per unit volume to stretch a wire by $1\%$ of its length,having a cross-sectional area of $1\,mm^2$,is: $[Y = 9 \times 10^{11}\,N/m^2]$

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