A ball of mass m=0.5 kg is attached to the end of a string having length L=0.5 m . The ball is rotated on a horizontal circular path about the vertical axis. The maximum tension that the string can bear is 324 N . The maximum possible value of the angular velocity of ball (in rad s- 1 ) is

Question

A ball of mass m=0.5 kg is attached to the end of a string having length L=0.5 m . The ball is rotated on a horizontal circular path about the vertical axis. The maximum tension that the string can bear is 324 N . The maximum possible value of the angular velocity of ball (in rad s- 1 ) is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-up6o5bsch4zqfqvl.jpg" /> (A) 9 (B) 18 (C) 27 (D) 36

Tardigrade Admin · Accepted Answer

textitT textsinθ =mω 2r textitT textsinθ =mω 2L textsinθ <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-up6o5bsch4zqp59b.jpg" /> ⇒ T=mω 2L ⇒ 324=(1/2)((ω )2)(1/2) Therefore ω =36 rad s- 1

$9$

$18$

$27$

$36$

$T sin θ = m ω^{2} r$
$T sin θ = m ω^{2} L sin θ$

$\Rightarrow T = m ω^{2} L$
$\Rightarrow 324 = \frac{1}{2} ((ω)^{2}) \frac{1}{2}$
Therefore $ω = 36 r a d s^{- 1}$

Q. A ball of mass m=0.5kg is attached to the end of a string having length L=0.5m . The ball is rotated on a horizontal circular path about the vertical axis. The maximum tension that the string can bear is 324N . The maximum possible value of the angular velocity of ball (in rads−1 ) is

9

18

27

36

T sinθ=mω2r T sinθ=mω2Lsinθ ⇒T=mω2L ⇒324=21​((ω)2)21​ Therefore ω=36rads−1

$9$

$18$

$27$

$36$

$T sin θ = m ω^{2} r$
$T sin θ = m ω^{2} L sin θ$

$\Rightarrow T = m ω^{2} L$
$\Rightarrow 324 = \frac{1}{2} ((ω)^{2}) \frac{1}{2}$
Therefore $ω = 36 r a d s^{- 1}$