Two springs have their force constants as $K_{1}$ and $K_{2}$ ( $K_{1}$ > $K_{1}$ ). The work done when both are stretched by same amount of length will be :

equal

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greater for

K_{1}

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greater for

K_{2}

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given data is incomplete

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Solution

The correct option is A greater for $K_{1}$
When the two springs are structed by same amount say $x$ , then
$\frac{W_{1}}{W_{2}} = \frac{\frac{1}{2} K_{1} x^{2}}{\frac{1}{2} K_{2} x^{2}} = \frac{K_{1}}{K_{2}}$

given $K_{1} > K_{2}$

hence $\frac{W_{1}}{W_{2}} = \frac{K_{1}}{K_{2}} > 1 \Rightarrow \frac{W_{1}}{W_{2}} > 1$

$\Rightarrow W_{1} > W_{2}$ so, $(B)$ is correct option
so greater for $K_{1}$ spring

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Two springs have their force constants as K1 and K2 (K1>K1). The work done when both are stretched by same amount of length will be :

The correct option is A greater for K1When the two springs are structed by same amount say x, thenW1W2=12K1x212K2x2=K1K2given K1>K2hence W1W2=K1K2>1 ⇒ W1W2>1⇒ W1>W2 so, (B) is correct optionso greater for K1 spring

Two springs have their force constants as $K_{1}$ and $K_{2}$ ( $K_{1}$ > $K_{1}$ ). The work done when both are stretched by same amount of length will be :