(D) The energy stored in the stretched wire (elastic potential energy) is given by $U = \frac{1}{2} F \Delta l$,where $F = 20\, N$ is the applied forc

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(D) The energy stored in the stretched wire (elastic potential energy) is given by $U = \frac{1}{2} F \Delta l$,where $F = 20\, N$ is the applied force and $\Delta l = 1.0\, mm$ is the extension.The decrease in gravitational potential energy of the load as it moves downwards by $\Delta l = 1.0\, mm$ is given by $W_g = F \Delta l$.The ratio of the increase in energy of the wire to the decrease in gravitational potential energy is:Ratio $= \frac{U}{W_g} = \frac{\frac{1}{2} F \Delta l}{F \Delta l} = \frac{1}{2}$.

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

Medium

$A$ 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 ratio of the increase in energy of the wire to the decrease in gravitational potential energy when the load moves downwards by $1\, mm$ will be:

A
$1$
B
$1/4$
C
$1/3$
D
$1/2$

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Mechanical Properties of Solids

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Work Done in Stretching a Wire

$A$ wire $2 \,m$ in length suspended vertically stretches by $10 \,mm$ when a mass of $10 \,kg$ is attached to the lower end. The elastic potential energy gained by the wire is ...... $J$ (take $g=10 \,m/s^2$)

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Identical springs of steel and copper are equally stretched. On which more work will have to be done?

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

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$A$ wire fixed at the upper end stretches by length $l$ by applying a force $F$. The work done in stretching is

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When the load applied to a wire is increased from $5 \ kg$ wt to $8 \ kg$ wt,the elongation of the wire increases from $1 \ mm$ to $1.8 \ mm$. The work done during the elongation of the wire is (Acceleration due to gravity $= 10 \ m/s^2$)

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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 ratio of the increase in energy of the wire to the decrease in gravitational potential energy when the load moves downwards by $1\, mm$ will be:

Similar Questions

$A$ wire $2 \,m$ in length suspended vertically stretches by $10 \,mm$ when a mass of $10 \,kg$ is attached to the lower end. The elastic potential energy gained by the wire is ...... $J$ (take $g=10 \,m/s^2$)

Identical springs of steel and copper are equally stretched. On which more work will have to be done?

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

$A$ wire fixed at the upper end stretches by length $l$ by applying a force $F$. The work done in stretching is

When the load applied to a wire is increased from $5 \ kg$ wt to $8 \ kg$ wt,the elongation of the wire increases from $1 \ mm$ to $1.8 \ mm$. The work done during the elongation of the wire is (Acceleration due to gravity $= 10 \ m/s^2$)

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