JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
To mop-clean a floor, a cleaning machine presses a circular mop of radius \(R\) vertically down with a total force \(F\) and rotates it with a constant angular speed about its axis. If the force \(F\) is distributed uniformly over the mop and the floor is \(\mu \), the torque, applied by the machine on the mop is
- A \(\mu FR/3\)
- B \(\mu FR/6\)
- C \(\mu FR/2\)
- D \(\frac{2}{3}\,\mu FR\)
Answer & Solution
Correct Answer
(D) \(\frac{2}{3}\,\mu FR\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l} d\tau = \left( {\mu dN} \right)x\\ \,\,\,\,\,\,\, = \mu \left( {\frac{F}{{\pi {R^2}}} \times 2\pi \times dx} \right)x\\ \,\,\,\,\tau = \int\limits_0^R {d\tau } \end{array}\)
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