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Home » A metre stick is balanced on a knife-edge at its centre. When two coins, each of mass are put one on top of the other at the mark, the stick is found to be balanced at . What is the mass of the metre stick?
A metre stick is balanced on a knife-edge at its centre. When two coins, each of mass 5 \mathrm{~g} are put one on top of the other at the 12.0 \mathrm{~cm} mark, the stick is found to be balanced at 45.0 \mathrm{~cm}. What is the mass of the metre stick?
CBSE | Class 11 | NCERT | Physics | System of Particles and Rotational Motion
A metre stick is balanced on a knife-edge at its centre. When two coins, each of mass 5 \mathrm{~g} are put one on top of the other at the 12.0 \mathrm{~cm} mark, the stick is found to be balanced at 45.0 \mathrm{~cm}. What is the mass of the metre stick?

When two coins are inserted at the 12 cm point, the metre rule’s centre of mass shifts to 45 cm from 50 cm.

Mass of two coins is given as m=10 \mathrm{~g}

The distance at which the coins are placed from the new support is given as \mathrm{d}_{1}=45-12=33 \mathrm{~cm}

Distance of the center of mass from the new support is given as 50-45=5 \mathrm{~cm}

We should use the balancing moments to find the mass of the scale,

Moment due to the coins can be calculated by r_{1}=m x d_{1}

Moment due to the mass of the scale can be calculated as r_{2}=M \times d_{2}

We know,

r_{1}=r_{2}

m \times d_{1}=M \times d_{2}

On substituting we get,

10 \times 33=M \times d_{2}

M=(10 \times 33) / 5=330 / 5=66 \mathrm{~g}

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