question_answer

When a person stands on a weighing balance, workingon the principle of Hooke's law, it shows a reading of 60 kg after a long time and the spring gets compressed by2.5 cm. If the person jumps on the balance from a height of 10 cm, the maximum reading of the balance will be

A) 60 kg    

B) 120kg

C) 180 kg

D) 240kg

Correct Answer: D

Solution :

[d] initially, \[60g=kx=k(2.5)\]                 (i) Let x' be the maximum compression when the person jumps on the balance, then \[\frac{1}{2}k{x}'=60g({{x}^{'2}}+10)\] \[\Rightarrow \frac{1}{2}\left[ \frac{60g}{2.5} \right]x{{'}^{2}}=60g(x'+10)\] \[\Rightarrow x{{'}^{2}}=5x'+50\] \[\Rightarrow x{{'}^{2}}-5x{{'}^{2}}-50=0\] Solving for x', we get \[x'=10\,\,cm\] if m kg is the reading, then \[Mg=k(10)\]                                  (ii) From Eqs. (i) and (ii), we get \[m=240\text{ }kg\]