Two springs have their force constant as $k_{1}$ and $k_{2} (k_{1} > k_{2})$ . When they are stretched by the same constant force up to equilibrium -

No work is done by this force in case of both the springs

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Equal work is done by this force in case of both the springs

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More work is done by this force in case of second spring

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More work is done by this force in case of first spring

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Solution

The correct option is C More work is done by this force in case of second spring
Since force in both springs is same: $k_{1} x_{1} = k_{2} x_{2}$
If $k_{1} > k_{2}$ then $x_{1} < x_{2}$
Work= $0.5 k x^{2}$
Hence, work by second spring is greater than first

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Two springs have their force constant as k1 and k2(k1>k2). When they are stretched by the same constant force up to equilibrium -

The correct option is C More work is done by this force in case of second springSince force in both springs is same: k1x1=k2x2If k1>k2 then x1<x2Work= 0.5kx2Hence, work by second spring is greater than first

Two springs have their force constant as $k_{1}$ and $k_{2} (k_{1} > k_{2})$ . When they are stretched by the same constant force up to equilibrium -

The correct option is C More work is done by this force in case of second spring
Since force in both springs is same: $k_{1} x_{1} = k_{2} x_{2}$
If $k_{1} > k_{2}$ then $x_{1} < x_{2}$
Work= $0.5 k x^{2}$
Hence, work by second spring is greater than first