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Circular Motion question

2017 · 9 Apr · Shift 1 · Q48
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  5. /2017 · 9 Apr · Shift 1 · Q48

Circular Motion question

2017 · 9 Apr · Shift 1 · Q48

JEE MainPhysicsCircular MotionMCQ+4 / −1
A conical pendulum of length 1 m makes an angle θ\thetaθ = 45o w.r.t. Z-axis and moves in a circle in the XY plane. The radius of the circle is 0.4 m and its center is vertically below O. The speed of the pendulum, in its circular path, will be: (Take g = 10 ms−2 ) JEE Main 2017 (Online) 9th April Morning Slot Physics - Circular Motion Question 63 English
  1. A
    0.4 m/s
  2. B
    4 m/s
  3. C
    0.2 m/s
  4. D
    2 m/s
View written solutionFree

Correct answer: D

  1. Model the conical pendulum

For a conical pendulum of string length lll, making angle θ\thetaθ with the vertical (here Z-axis), the bob moves in a horizontal circle of radius rrr.

The forces on the bob are:

  • Tension TTT along the string
  • Weight mgmgmg downward

Resolving tension:

Tcos⁡θ=mgT\cos\theta = mgTcosθ=mg Tsin⁡θ=mv2rT\sin\theta = \frac{mv^2}{r}Tsinθ=rmv2​

  1. Eliminate TTT

Divide the second equation by the first:

Tsin⁡θTcos⁡θ=mv2/rmg\frac{T\sin\theta}{T\cos\theta} = \frac{mv^2/r}{mg}TcosθTsinθ​=mgmv2/r​

tan⁡θ=v2rg\tan\theta = \frac{v^2}{rg}tanθ=rgv2​

So,

v2=rgtan⁡θv^2 = rg\tan\thetav2=rgtanθ

  1. Substitute the given values

Given:

  • r=0.4 mr = 0.4\,\text{m}r=0.4m
  • g=10 m/s2g = 10\,\text{m/s}^2g=10m/s2
  • θ=45∘⇒tan⁡45∘=1\theta = 45^\circ \Rightarrow \tan 45^\circ = 1θ=45∘⇒tan45∘=1

Hence,

v2=0.4×10×1=4v^2 = 0.4 \times 10 \times 1 = 4v2=0.4×10×1=4

v=2 m/sv = 2\,\text{m/s}v=2m/s

  1. Check with options
  • A: 0.4 m/s0.4\,\text{m/s}0.4m/s
  • B: 4 m/s4\,\text{m/s}4m/s
  • C: 0.2 m/s0.2\,\text{m/s}0.2m/s
  • D: 2 m/s2\,\text{m/s}2m/s

Therefore, the correct option is:

D: 2 m/s\boxed{\text{D: } 2\,\text{m/s}}D: 2m/s​

  1. Comparison with stored answer

Stored correct answer: D\text{D}D

Our derived answer is also D\text{D}D, so they agree.

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