A) \[m,\frac{l}{x}\]
B) \[m,\frac{l}{{{x}^{2}}}\]
C) \[m,x\]
D) \[m,{{x}^{2}}\]
Correct Answer: A
Solution :
Let, 'l' and 'M' be the length and mass of rod for equilibrium, taking moment about X, we get \[m\times x=M\times (\frac{l}{2}-x)\] this can be written as, \[m=(\frac{Ml}{2})\frac{l}{x}-M\] comparing with straight line equation, \[y=mx+c\] we can say that, \[m\propto \frac{l}{x}\]You need to login to perform this action.
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