A small sphere is given vertical velocity of magnitude v0=5 m s -1 and it swings in a vertical plane about the end of a massless string. The angle θ with the vertical at which string will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere, is

Question

A small sphere is given vertical velocity of magnitude v0=5 m s -1 and it swings in a vertical plane about the end of a massless string. The angle θ with the vertical at which string will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere, is (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="image" src="https://cdn.tardigrade.in/img/question/physics/9155c97bc52edbc8bab112b549575a13-.png" /> v2=v02+2 g l cos θ T=m g cos θ+(m v2/l) Given T=2 m g 2 m g=m g cos θ+(m/l) (v02+2 g l cos θ) ⇒ cos θ=(1/4)

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

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

$6 0^{\circ}$

$3 0^{\circ}$

$v^{2} = v_{0}^{2} + 2 g l cos θ$
$T = m g cos θ + \frac{m v ^{2}}{l}$
Given $T = 2 m g$
$2 m g = m g cos θ + \frac{m}{l}$
$(v_{0}^{2} + 2 g l cos θ)$
$\Rightarrow cos θ = \frac{1}{4}$

Q. A small sphere is given vertical velocity of magnitude v0​=5ms−1 and it swings in a vertical plane about the end of a massless string. The angle θ with the vertical at which string will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere, is

cos−132​

cos−1(41​)

60∘

30∘

v2=v02​+2glcosθ T=mgcosθ+lmv2​ Given T=2mg 2mg=mgcosθ+lm​ (v02​+2glcosθ) ⇒cosθ=41​

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

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

$6 0^{\circ}$

$3 0^{\circ}$

$v^{2} = v_{0}^{2} + 2 g l cos θ$
$T = m g cos θ + \frac{m v ^{2}}{l}$
Given $T = 2 m g$
$2 m g = m g cos θ + \frac{m}{l}$
$(v_{0}^{2} + 2 g l cos θ)$
$\Rightarrow cos θ = \frac{1}{4}$