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Standard XII

Physics

Newton's Laws for System of Particles

A uniform rod...

Question

A uniform rod of mass 1 kg is hanging from a thread attached at the midpoint of the rod. A block of mass $m$ =3kg hangs at the left end of the rod and block of mass $M$ hangs at the right side at 80 cm from block $m$ . If system is in equilibrium. Calculate $M$ .

4 kg

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5 kg

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6 kg

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8 kg

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9 kg

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Solution

The correct option is B 5 kg
Given : $m = 3 k g$
As the rod is in equilibrium, thus net torque about the point O is zero i.e $τ_{o} = 0$
$∴$ $m g (0.5) - M g (0.3) = 0$

$⟹$ $M = \frac{m (0.5)}{0.3} = \frac{3 \times 0.5}{0.3} = 5 k g$

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A uniform rod of mass 1 kg is hanging from a thread attached at the midpoint of the rod. A block of mass m=3kg hangs at the left end of the rod and block of mass M hangs at the right side at 80 cm from block m. If system is in equilibrium. Calculate M.

The correct option is B 5 kgGiven : m=3kg As the rod is in equilibrium, thus net torque about the point O is zero i.e τo=0∴ mg(0.5)−Mg(0.3)=0⟹ M=m(0.5)0.3=3×0.50.3=5kg

A uniform rod of mass 1 kg is hanging from a thread attached at the midpoint of the rod. A block of mass $m$ =3kg hangs at the left end of the rod and block of mass $M$ hangs at the right side at 80 cm from block $m$ . If system is in equilibrium. Calculate $M$ .

The correct option is B 5 kg
Given : $m = 3 k g$
As the rod is in equilibrium, thus net torque about the point O is zero i.e $τ_{o} = 0$
$∴$ $m g (0.5) - M g (0.3) = 0$

$⟹$ $M = \frac{m (0.5)}{0.3} = \frac{3 \times 0.5}{0.3} = 5 k g$