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Laws of Motion question

2011 · Shift 1 · Q49
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  5. /2011 · Shift 1 · Q49

Laws of Motion question

2011 · Shift 1 · Q49

JEE AdvancedPhysicsLaws of MotionMCQ+3 / −0.75
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 vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in radian/s) is IIT-JEE 2011 Paper 1 Offline Physics - Laws of Motion Question 25 English
  1. A
    9
  2. B
    18
  3. C
    27
  4. D
    36
View written solutionFree

Correct answer: D

  1. Identify the force providing centripetal force

For a ball moving in a horizontal circle of radius LLL, the tension in the string provides the centripetal force:

T=mω2LT = m\omega^2 LT=mω2L

At the maximum possible angular velocity, the tension reaches its maximum allowed value:

Tmax⁡=324 NT_{\max} = 324\text{ N}Tmax​=324 N

  1. Substitute the given values

Given:

  • m=0.5 kgm = 0.5\text{ kg}m=0.5 kg
  • L=0.5 mL = 0.5\text{ m}L=0.5 m
  • Tmax⁡=324 NT_{\max} = 324\text{ N}Tmax​=324 N

So,

324=(0.5)ω2(0.5)324 = (0.5)\omega^2(0.5)324=(0.5)ω2(0.5)

324=0.25ω2324 = 0.25\omega^2324=0.25ω2

  1. Solve for ω\omegaω

ω2=3240.25=1296\omega^2 = \frac{324}{0.25} = 1296ω2=0.25324​=1296

ω=1296=36 rad/s\omega = \sqrt{1296} = 36\text{ rad/s}ω=1296​=36 rad/s

  1. Match with the options
  • A: 999
  • B: 181818
  • C: 272727
  • D: 363636

Hence, the correct option is:

D: 36\boxed{\text{D: }36}D: 36​

  1. Verification with stored answer

Stored correct answer is D, and our derived answer is also D. So they agree.

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