A small sphere is given vertical velocity of magnitude v0=5m / s and it swings in a vertical plane about the end of the massless string. If maximum tension string can withstand is equal to twice of the weight of sphere, then angle θ with vertical at which string will break is

Question

A small sphere is given vertical velocity of magnitude v0=5m / s and it swings in a vertical plane about the end of the massless string. If maximum tension string can withstand is equal to twice of the weight of sphere, then angle θ with vertical at which string will break is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-didzzqjgkka9rysy.jpg" /> (A) (cos)- 1 ((2/3)) (B) (cos)- 1((1/4)) (C) 60°  (D) 30°

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-didzzqjgkka9xmsi.jpg" /> FBD of body at the position shown in diagram T-mgcosθ =(m v2/l) .....(i) Applying work energy theorem ⇒ mglcos θ =(1/2)mv2-(1/2)mv02 ....(ii) T=2mg ......(iii) By solving ∴ cos θ =(1/4)

Q. A small sphere is given vertical velocity of magnitude $v_{0} = 5 m / s$ and it swings in a vertical plane about the end of the massless string. If maximum tension string can withstand is equal to twice of the weight of sphere, then angle $θ$ with vertical at which string will break is

$(cos)^{- 1} (\frac{2}{3})$

$(cos)^{- 1} (\frac{1}{4})$

$6 0^{\circ}$

$3 0^{\circ}$

FBD of body at the position shown in diagram
$T - m g cos θ = \frac{m v ^{2}}{l}$ .....(i)
Applying work energy theorem
$\Rightarrow m g l cos θ = \frac{1}{2} m v^{2} - \frac{1}{2} m v_{0}^{2}$ ....(ii)
$T = 2 m g$ ......(iii)
By solving $∴ cos θ = \frac{1}{4}$

Q. A small sphere is given vertical velocity of magnitude v0​=5m/s and it swings in a vertical plane about the end of the massless string. If maximum tension string can withstand is equal to twice of the weight of sphere, then angle θ with vertical at which string will break is

(cos)−1(32​)

(cos)−1(41​)

60∘

30∘