A pendulum of mass 1 kg and length l - 1 m is released from rest at angle - - 60 - - The power delivered by all the forces acting on the bob at the angle - - 30 - will be g - 10 m-s2-

The correct option is A 13.4 watt h=l(cos 30 cos 60)=1(0.86 0.5)=0.36m v=√(2gh)√(2× 10× 0.36)=2.68m/s p=mgsinθ =1 × 10 ×sin 30 × 2.68 ...

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Standard X

Physics

Power

A pendulum of...

Question

A pendulum of mass $1 k g$ and length $l = 1 m$ is released from rest at angle $θ = 60^{\circ}$ . The power delivered by all the forces acting on the bob at the angle $θ = 30^{\circ}$ will be $(g = 10 m/ s^{2})$ .

13.4 w a t t

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Solution

The correct option is A $13.4 w a t t$
$h = l (cos 30 - cos 60) = 1 (0.86 - 0.5) = 0.36 m$

$v = \sqrt{2 g h} \sqrt{2 \times 10 \times 0.36} = 2.68 m / s$

$p = m g sin θ$

$= 1 \times 10 \times sin 30 \times 2.68$

$= 13.4 w a t t$

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A pendulum of mass 1 kg and length l=1m is released from rest at angle θ=60∘ . The power delivered by all the forces acting on the bob at the angle θ=30∘ will be (g=10m/s2).

The correct option is A 13.4 watth=l(cos30−cos60)=1(0.86−0.5)=0.36mv=√2gh√2×10×0.36=2.68m/sp=mgsinθ=1×10×sin30×2.68=13.4watt

A pendulum of mass $1 k g$ and length $l = 1 m$ is released from rest at angle $θ = 60^{\circ}$ . The power delivered by all the forces acting on the bob at the angle $θ = 30^{\circ}$ will be $(g = 10 m/ s^{2})$ .

The correct option is A $13.4 w a t t$
$h = l (cos 30 - cos 60) = 1 (0.86 - 0.5) = 0.36 m$

$v = \sqrt{2 g h} \sqrt{2 \times 10 \times 0.36} = 2.68 m / s$

$p = m g sin θ$

$= 1 \times 10 \times sin 30 \times 2.68$

$= 13.4 w a t t$