JEE MAIN - Physics (2012 - No. 21)

This question has Statement $$1$$ and Statement $$2.$$ Of the four choices given after the Statements, choose the one that best describes the two Statements.

If two springs $${S_1}$$ and $${S_2}$$ of force constants $${k_1}$$ and $${k_2}$$, respectively, are stretched by the same force, it is found that more work is done on spring $${S_1}$$ than on spring $${S_2}$$.

STATEMENT 1: If stretched by the same amount work done on $${S_1}$$, Work done on $${S_1}$$ is more than $${S_2}$$
STATEMENT 2: $${k_1} < {k_2}$$

Statement 1 is false, Statement 2 is true
Statement 1 is true, Statement 2 is false
Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1
Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1

Explanation

We know force (F) = kx

$$W = {1 \over 2}k{x^2}$$

$$W =$$ $${{{{\left( {kx} \right)}^2}} \over {2k}}$$ $$\,\,\,$$

$$\therefore$$ $$W = {{{F^2}} \over {2k}}$$ [ as $$F=kx$$ ]

When force is same then,

$$W \propto {1 \over k}$$

Given that, $${W_1} > {W_2}$$

$$\therefore$$ $${k_1} < {k_2}$$

Statement-2 is true.

For the same extension, x1 = x2 = x

Work done on spring S1 is W1 = $${1 \over 2}{k_1}x_1^2 = {1 \over 2}{k_1}{x^2}$$

Work done on spring S2 is W2 = $${1 \over 2}{k_2}x_2^2 = {1 \over 2}{k_2}{x^2}$$

$$ \therefore $$ $${{{W_1}} \over {{W_2}}} = {{{k_1}} \over {{k_2}}}$$

As $${k_1} < {k_2}$$ then $${W_1} < {W_2}$$

So, Statement-1 is false.

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