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Standard XII

Physics

Centripetal Acceleration

A metallic ro...

Question

A metallic rod of mass 20 kg and of uniform thickness rests against a wall while the lower end of the rod is in contact with floor. The rod makes an angle of $60^{o}$ with floor. If the weight of rod produces a torque 150 Nm about its lower end, the length of the rod is (g = 10 $m s^{- 2}$ )

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Solution

The correct option is C
3m

$∴ τ = (m g) (\frac{1}{2} c o s 60^{o})$
$\Rightarrow 150 = 20 \times 10 \times \frac{L}{2} \times \frac{1}{2}$
$\Rightarrow L = 3 m$

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A metallic rod of mass 20 kg and of uniform thickness rests against a wall while the lower end of the rod is in contact with floor. The rod makes an angle of 60o with floor. If the weight of rod produces a torque 150 Nm about its lower end, the length of the rod is (g = 10ms−2)

The correct option is C 3m ∴τ=(mg)(12cos60o) ⇒150=20×10×L2×12 ⇒L=3m

A metallic rod of mass 20 kg and of uniform thickness rests against a wall while the lower end of the rod is in contact with floor. The rod makes an angle of $60^{o}$ with floor. If the weight of rod produces a torque 150 Nm about its lower end, the length of the rod is (g = 10 $m s^{- 2}$ )

The correct option is C
3m

$∴ τ = (m g) (\frac{1}{2} c o s 60^{o})$
$\Rightarrow 150 = 20 \times 10 \times \frac{L}{2} \times \frac{1}{2}$
$\Rightarrow L = 3 m$