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# The steel wire can withstand a load up to 2940 N. A load of 150 kg is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position so that the wire does not break(A) $30^{\circ}$ (B) $60^{\circ}$ (C) $80^{\circ}$ (D) ${{85}^{\circ }}$

Last updated date: 20th Jun 2024
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Hint
We should know that velocity is defined as the rate change of displacement per unit time. Speed in a specific direction is also known as velocity. Velocity is equal to displacement divided by time. Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance which is a scalar quantity per time ratio. On the other hand, velocity is a vector quantity; it is direction-aware. An object which moves in the negative direction has a negative velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion in this case. Based on this we have to solve this question.

While crossing the equilibrium position $T=m g+\dfrac{m v^{2}}{r}=m g+\dfrac{m}{l}[2 g l(1-\cos \theta)]$
$=m g+2 m g-2 m g \cos \theta=m g[3-2 \cos \theta]$
In equilibrium, $\mathrm{T} \cos \theta=\mathrm{mg}$
$\Rightarrow \cos \theta=\dfrac{150 \times 9.8}{2940}$
$\Rightarrow \cos \theta=0.5$
$\Rightarrow \theta=60^{\circ}$