6.System of Particles and Rotational Motion
medium

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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