Question 450: Centre of Mass and Rotational Motion

Question

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 $$	imes$$ 10
$$-$$2
kg. The value of x is ___________.

Accepted Answer

Let $\mathrm{M}$ be the mass of the meter scale.
The weight $\mathrm{Mg}$ of the scale acts at $50 \mathrm{~cm}$ mark.
Finally after putting two coins on the meter scale, balancing the torques about the knife edge, we get,
$20 g 	imes 30=\mathrm{Mg} 	imes 10$
$$
\begin{aligned}
M & =60 \mathrm{~g} \\
& =60 	imes 10^{-3} \mathrm{~kg} \\
& =6 	imes 10^{-2} \mathrm{~kg} \\
& =x 	imes 10^{-2} \mathrm{~kg}
\end{aligned}
$$
On comparing, we get $x=6$

Centre of Mass and Rotational Motion

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 ___________.

Integer / Numeric Answer Type