Work Energy And Power Ques 3
3. Two similar springs $\mathrm{P}$ and $\mathrm{Q}$ have spring constants $\mathrm{K} _{\mathrm{P}}$ and $\mathrm{K} _{\mathrm{Q}}$, such that $\mathrm{K} _{\mathrm{P}} >\mathrm{K} _{\mathrm{Q}}$. They are stretched, first by the same amount (case a,) then by the same force (case b). The work done by the springs $\mathrm{W} _{\mathrm{P}}$ and $\mathrm{W} _{\mathrm{Q}}$ are related as, in case (a) and case (b), respectively
[2015]
(a) $\mathrm{W} _{\mathrm{P}}=\mathrm{W} _{\mathrm{Q}} ; \mathrm{W} _{\mathrm{P}}=\mathrm{W} _{\mathrm{Q}}$
(b) $\mathrm{W} _{\mathrm{P}}>\mathrm{W} _{\mathrm{Q}} ; \mathrm{W} _{\mathrm{Q}}>\mathrm{W} _{\mathrm{P}}$
(c) $\mathrm{W} _{\mathrm{P}}<\mathrm{W} _{\mathrm{Q}} ; \mathrm{W} _{\mathrm{Q}}<\mathrm{W} _{\mathrm{P}}$
(d) $\mathrm{W} _{\mathrm{P}}=\mathrm{W} _{\mathrm{Q}} ; \mathrm{W} _{\mathrm{P}}>\mathrm{W} _{\mathrm{Q}}$
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Answer:
Correct Answer: 3.(b)
Solution: (b) Case (a): Suppose the two springs are stretched by the same distance $x$. Then
$ \frac{W_P}{W_Q}=\frac{\frac{1}{2} k_p x^2}{\frac{1}{2} k_Q x^2}=\frac{k_p}{k_Q} $
As $\mathrm{k} _{\mathrm{p}}>\mathrm{k} _{\mathrm{Q}}$ so, $\mathrm{W} _{\mathrm{p}}>\mathrm{W} _{\mathrm{Q}}$
Case (b): Suppose the two springs stretched by distance $x_P$ and $x_Q$ by the same force $\mathrm{F}$.
Then,
$ \mathrm{F}=k_p x_p=k_Q x_Q $
$\frac{W_p}{W_Q}=\frac{\frac{1}{2} k_p x_p^2}{\frac{1}{2} k_Q x_Q^2}=\frac{k_p \cdot x_p \cdot x_p}{k_Q \cdot x_Q \cdot x_Q}=\frac{F x_p}{F x_Q}=\frac{k_Q}{k_p} $
$\text { As } k_P>k_Q \quad \therefore \quad \mathrm{W} _{\mathrm{Q}}>\mathrm{W} _{\mathrm{P}}$
BETA
