Verified Solution • 2025–26 Session moderate

QJEEM12Ph8f42: JEE Main 2012 Physics - Class 11 JEEM12Ph8f42 Solution: This question has Statement and Statement Of the four choices…

This question has Statement 11 and Statement 2.2. Of the four choices given after the Statements, choose the one that best describes the two Statements.

If two springs S1{S_1} and S2{S_2} of force constants k1{k_1} and k2{k_2} , respectively, are stretched by the same force, it is found that more work is done on spring S1{S_1} than on spring S2{S_2} .

STATEMENT 1: If stretched by the same amount work done on S1{S_1} , Work done on S1{S_1} is more than S2{S_2}
STATEMENT 2: {k_1} < {k_2}

Multiple Choice Options

  • A Statement 1 is false, Statement 2 is true
  • B Statement 1 is true, Statement 2 is false
  • C Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1
  • D Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1

Step-by-Step Verified Solution

We know force (F) = kx

W=12kx2W = {1 \over 2}k{x^2}

W=W = (kx)22k{{{{\left( {kx} \right)}^2}} \over {2k}}    \,\,\,

∴\therefore W=F22kW = {{{F^2}} \over {2k}} [ as F=kxF=kx ]

When force is same then,

W∝1kW \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 = 12k1x12=12k1x2{1 \over 2}{k_1}x_1^2 = {1 \over 2}{k_1}{x^2}

Work done on spring S2 is W2 = 12k2x22=12k2x2{1 \over 2}{k_2}x_2^2 = {1 \over 2}{k_2}{x^2}

∴\therefore W1W2=k1k2{{{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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