Work Energy And Power Question 152

Question: 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

Options:

A) 60 kg

B) 120kg

C) 180 kg

D) 240kg

Show Answer

Answer:

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}[ \frac{60g}{2.5} ]x{{’}^{2}}=60g(x’+10) $

$ \Rightarrow x{{’}^{2}}=5x’+50 $

$ \Rightarrow x{{’}^{2}}-5x{{’}^{2}}-50=0 $

Solving for x’, we get $ x’=10cm $ if m kg is the reading, then $ Mg=k(10) $ (ii)

From Eqs. (i) and (ii), we get $ m=240kg $



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