A uniform metre rod is capable of turning free about a knife edge at 70 cm mark. A mass of 10 g is suspended at 15 cm mark and a mass of 150 g is suspended at 85 cm mark makes the rod exactly horizontal. Find the weight of the metre rod.

Suppose the mass of the metre rod is 'm'.
According to the conservation of net torque for rotational equilibrium,
10 g(70-15) +mg(70-50) -150g (85-70)=0550 g+20 mg -2250g=020 mg=2250g-550 g=1700or W=mg=1700 /20=85 gThus, the weight of the rod is 85 g.

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