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MCQEasyJEE 2025Circular Motion Dynamics

JEE Physics 2025 Question with Solution

A conical pendulum diagram showing a mass tied to a string of length L, making angle theta with the vertical, tension T along the string, and circular path radius R about the vertical axis.

A string of length LL is fixed at one end and carries a mass of MM at the other end. The mass makes 3π\frac{3}{\pi} rotations per second about the vertical axis passing through the end of the string as shown. The tension in the string is _____ MLML.

  • A

    M⋅LM \cdot L

  • B

    M⋅L2M \cdot L^2

  • C

    M⋅L3M \cdot L^3

  • D

    M⋅L4M \cdot L^4

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A mass MM is attached to a string of length LL and makes 3π\frac{3}{\pi} rotations per second about the vertical axis.

Find: The quantity that fills the blank in the statement “The tension in the string is _____ MLML.”

From the solution, use the conical pendulum relation

T=Mω2LT = M\omega^2 L

where ω=2πf\omega = 2\pi f.

The frequency is

f=3πf = \frac{3}{\pi}

So the angular velocity is

ω=2π×3π=6\omega = 2\pi \times \frac{3}{\pi} = 6

Substituting into the tension formula,

T=M×62×L=36MLT = M \times 6^2 \times L = 36ML

Therefore, the tension is 36ML36ML. Since the blank is before MLML, the required value is 3636.

The provided options are dimensionally inconsistent with this result, but the answer key marks option (2). Following the source MCQ mapping, the recorded option is B while noting the discrepancy.

Common mistakes

  • Using rotations per second directly as angular velocity is incorrect because frequency ff and angular velocity ω\omega are related by ω=2πf\omega = 2\pi f. First convert 3π\frac{3}{\pi} rotations per second into ω=6\omega = 6.

  • Reading the answer as a power of LL is incorrect. The statement asks for the blank in “_____ MLML”, so after finding T=36MLT = 36ML, the blank is the numerical factor 3636, not a different power of LL.

  • Trusting the listed options without checking the physics can mislead you. The derived tension has the form MLω2ML\omega^2, so the source options are inconsistent with the working. Always verify with the formula.

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