Rotational Motion

JEE Main level, numerical answer, medium

116 of 280

Question 116 of 280

A metre scale is balanced on a knife edge at its centre. When two coins, each of mass 10g10g are put one on the top of the other at the 10.0cm10.0cm mark the scale is found to be balanced at 40.0cm40.0cm mark. The mass of the metre scale is found to be x×10−2kgx\times {10}^{-2}kg. The value of xx is _____ .

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Answer: 6

Let the mass of the scale be mm.
The centre of mass of the scale will be at 50cm50cm.
Applying torque balance,
0=mg×(50−40)−2×10g×(40−10)⇒m=60g=6×10−2kg0=mg\times \left(50-40\right)-2\times 10g\times \left(40-10\right) \Rightarrow m=60g=6\times {10}^{-2}kg

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