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6.System of Particles and Rotational Motion
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A metre scale is balanced on a knife edge at its centre. When two coins, each of mass $10\, g$ are put one on the top of the other at the $10.0\, cm$ mark the scale is found to be balanced at $40.0\, cm$ mark. The mass of the metre scale is found to be $x \times 10^{-2}$ $kg$. The value of $x$ is
A
$9$
B
$6$
C
$60$
D
$7$
(JEE MAIN-2022)
Solution

Let mass of meter scale be $m$.
Balancing torque about knife edge
$(0.02 \;g ) \times\left(30 \times 10^{-2}\right)= mg \times\left(10 \times 10^{-2}\right)$
$m =0.06\; kg =6 \times 10^{-2} \;kg$
Standard 11
Physics
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