JEE Mains · Physics · STD 11 - 6. system of particles 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
- A \(9\)
- B \(6\)
- C \(60\)
- D \(7\)
Answer & Solution
Correct Answer
(B) \(6\)
Step-by-step Solution
Detailed explanation
Let mass of meter scale be \(m\). Balancing torque about knife edge \((0.02 \;g ) \times\left(30 \times 10^{-2}\right)= mg \times\left(10 \times 10^{-2}\right)\) \(m =0.06\; kg =6 \times 10^{-2} \;kg\)
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