Q13 / 122 · NTA Abhyas (JEE Main) 2020 · Physics · Rotational Motion
A metre stick is balanced on a knife-edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass (in gram) of the metre stick?
Correct answer: 66
Solution
Let total mass of the metre stick be M gram

Distance between mid-point E and new centre of gravity (DE),
= 50 - 45 = 5 cm
Distance between 12 cm mark and new centre of gravity (CD),
= 45 - 12 = 33 cm
From principle of moments in equilibrium,
Mg × DE = (2 × 5g) × CD
M × 5 = 10 × 33
or M = 66 g
∴ Mass of the metre stick is 66 g.

Distance between mid-point E and new centre of gravity (DE),
= 50 - 45 = 5 cm
Distance between 12 cm mark and new centre of gravity (CD),
= 45 - 12 = 33 cm
From principle of moments in equilibrium,
Mg × DE = (2 × 5g) × CD
M × 5 = 10 × 33
or M = 66 g
∴ Mass of the metre stick is 66 g.
Hence, the correct answer is 66.