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Q. If two springs A and B with spring constants 2k and k, are stretched separately by same suspended weight, then the ratio between the work done in stretching A and B is

KEAMKEAM 2011

Solution:

$ mg=2\,K{{x}_{A}} $ $ mg=K{{x}_{B}} $
$ \frac{{{x}_{A}}}{{{x}_{B}}}=\frac{1}{2} $
$ W=Fx $
So, $ \frac{{{W}_{A}}}{{{W}_{B}}}=\frac{F{{x}_{A}}}{F{{x}_{B}}}=\frac{1}{2} $