Neet Solved Paper 2015 Question 6
Question: Two similar springs P and Q have spring constants $ K _{P} $ and $ K _{Q} $ , such that $ 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 $ W _{P} $ and $ W _{Q} $ are related as, in case (a) and case (b), respectively
Options:
A) $ W _{P}=W _{Q};W _{P}>W _{Q} $
B) $ W _{P}=W _{Q};W _{P}=W _{Q} $
C) $ W _{P}>W _{Q};W _{Q}>W _{P} $
D) $ W _{P}<W _{Q};W _{Q}<W _{P} $
Show Answer
Answer:
Correct Answer: C
Solution:
- Given, $ K _{P}>K _{Q} $ In case (a), the elongation is same i.e. $ x _1=x _2=x $ So, $ W _{P}=\frac{1}{2}K _{P}x^{2} $ and $ W _{Q}=\frac{1}{2}K _{Q}x^{2} $
$ \therefore $ $ \frac{W _{P}}{W _{Q}}=\frac{K _{P}}{K _{Q}}>1 $
$ \Rightarrow ,W _{P}>W _{Q} $ In case (b), the spring force is same i.e. $ F _1=F _2=F $ So, $ x _1=\frac{F}{K _{P}},x _2\frac{F}{K _{Q}} $
$ \therefore $ $ W _{P}=\frac{1}{2}K _{P}x_1^{2}=\frac{1}{2}K _{p}\frac{F^{2}}{K_P^{2}}=\frac{1}{2}\frac{F^{2}}{K _{P}} $ and $ W _{Q}=\frac{1}{2}K _{Q}x_2^{2}=\frac{1}{2}K _{Q}.\frac{F^{2}}{K_Q^{2}}=\frac{1}{2}\frac{F^{2}}{K _{Q}} $
$ \therefore $ $ \frac{W _{P}}{W _{Q}}=\frac{K _{Q}}{K _{P}}<1 $
$ \Rightarrow W _{P}<W _{Q} $
