Question

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 to the same extension, work done on $S_1$, will be more than that on $S_2$

Statement-2: $k_1<k_2$

Choose the correct option.

• A. Statement-1 is true, Statement-2 is true and Statement-2 is not the correct explanation of Statement-1
• B. Statement-1 is false, Statement-2 is true
• C. Statement-1 is true, Statement-2 is false
• D. Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of statement-1.

Answer 1

The correct option is B Statement-1 is false, Statement-2 is true

As given,
Both springs are stretched by same force
$F=k_1{x}_1=k_2{x}_2$
$\frac{k_1}{k_2}=\frac{x_2}{x_1}...\left(1\right)$
$W_1=\frac{1}{2}{k}_1{x}_1^2$ and $W_2=\frac{1}{2}{k}_2{x}_2^2$
$W_1$ & $W_2$ are work done on spring (1) & (2)
as given in question,$W_1>W_2,\frac{W_1}{W_2}>1$
so, $\frac{\frac{1}{2}{k}_1{x}_1^2}{\frac{1}{2}{k}_2{x}_2^2}>1\because k_1{x}_1=k_2{x}_2$
$\frac{x_1}{x_2}>1,\frac{k_2}{k_1}>1${$\because$ eq (1)}
Therefore $k_2>k_1$, statement (2) is correct.
As, If $x_1=x_2=x$ statement (1)
$\frac{W_1}{W_2}=\frac{\frac{1}{2}{k}_1{x}^2}{\frac{1}{2}{k}_2{x}^2}$
$\frac{W_1}{W_2}=\frac{k_1}{k_2}<1$ {From statement (2)}

$W_1<W_2$ but in question $(W_1>W_2)$.
Hence Statement 1 is false.