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NEET 2015
Exam
NEET
Year
2015
Subject
Physics
Chapter
Work, Power and Energy

NEET 2015 · Physics · Work, Power and Energy

Two similar springs P and Q have spring constants K _ P and K _ Q ,...

Two similar springs P\text{P} and Q\text{Q} have spring constants KP{K}_{P} and KQ{K}_{Q}, such that KP>KQ{K}_{P} > {K}_{Q}. They are stretched, first by the same amount (case a), then by the same force (case b). the work done by the springs WP{W}_{P} and WQ{W}_{Q} are related as respectively, in case

Options

  1. AWP=WQ;WP>WQ{W}_{P} = {W}_{Q}; {W}_{P} > {W}_{Q}
  2. BWP=WQ;WP=WQ{W}_{P} = {W}_{Q}; {W}_{P} = {W}_{Q}
  3. CWP>WQ;WQ>WP{W}_{P} > {W}_{Q}; {W}_{Q} > {W}_{P}
  4. DWP<WQ;WQ<WP{W}_{P} < {W}_{Q}; {W}_{Q} < {W}_{P}

Correct answer

C. WP>WQ;WQ>WP{W}_{P} > {W}_{Q}; {W}_{Q} > {W}_{P}

Step-by-step solution

KP>KQ{K}_{P} > {K}_{Q}

Case (a): Stretched by same amount

WP=12KPx2{W}_{P} = \dfrac{1}{2}{K}_{P}{x}^{2}

WQ=12KQx2{W}_{Q} = \dfrac{1}{2}{K}_{Q}{x}^{2} (∵asxP=xQ=x)\left(∵ as{x}_{P} = {x}_{Q} = x\right)

∴WP>WQ∴ {W}_{P} > {W}_{Q}

Case (b): Stretched by same force,

F=KPxP=KQxQF = {K}_{P}{x}_{P} = {K}_{Q}{x}_{Q}

xP=FKP;xQ=FKQ{x}_{P} = \dfrac{F}{{K}_{P}}; {x}_{Q} = \dfrac{F}{{K}_{Q}}

WP=12KPxP2=12KPF2KP2=12F2KP{W}_{P} = \dfrac{1}{2}{K}_{P}{x}_{P}^{2} = \dfrac{1}{2}{K}_{P}\dfrac{{F}^{2}}{{K}_{P}^{2}} = \dfrac{1}{2}\dfrac{{F}^{2}}{{K}_{P}}

WQ=12KQxQ2=12KQF2KQ2=12F2KQ{W}_{Q} = \dfrac{1}{2}{K}_{Q}{x}_{Q}^{2} = \dfrac{1}{2}{K}_{Q}\dfrac{{F}^{2}}{{K}_{Q}^{2}} = \dfrac{1}{2}\dfrac{{F}^{2}}{{K}_{Q}}

∴WP<WQ∴ {W}_{P} < {W}_{Q}
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