MCQEasyJEE 2023Circular Motion Dynamics

JEE Physics 2023 Question with Solution

A stone of mass 1kg1 \, \text{kg} is tied to the end of a massless string of length 1m1 \, \text{m}. If the breaking tension of the string is 400N400 \, \text{N}, then maximum linear velocity the stone can have without breaking the string, while rotating in horizontal plane, is:

  • A

    20m/s20 \, \text{m/s}

  • B

    40m/s40 \, \text{m/s}

  • C

    400m/s400 \, \text{m/s}

  • D

    10m/s10 \, \text{m/s}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: mass of stone m=1kgm = 1 \, \text{kg}, string length r=1mr = 1 \, \text{m}, maximum tension T=400NT = 400 \, \text{N}.

Find: the maximum linear velocity of the stone without breaking the string.

At the limiting condition, the maximum velocity occurs when the centripetal force equals the maximum tension in the string.

Fc=mv2rF_c = \frac{mv^2}{r}

Setting centripetal force equal to tension:

mv2r=T\frac{mv^2}{r} = T

Substitute the given values:

1×v21=400\frac{1 \times v^2}{1} = 400v2=400v^2 = 400v=20m/sv = 20 \, \text{m/s}

Therefore, the maximum linear velocity is 20m/s20 \, \text{m/s}. The solution states the correct option is B, but this value matches option A, so the source has an option-label discrepancy.

Using the tension limit

Given: the stone moves in a horizontal circle and the string can sustain at most 400N400 \, \text{N} tension.

Find: the highest possible speed before the string breaks.

The string provides the required centripetal force. Hence the largest allowed speed is obtained when the required centripetal force becomes exactly equal to the breaking tension.

Tmax=mv2rT_{\max} = \frac{mv^2}{r}

With m=1kgm = 1 \, \text{kg} and r=1mr = 1 \, \text{m}:

400=1v21400 = \frac{1 \cdot v^2}{1}v2=400v^2 = 400v=400=20m/sv = \sqrt{400} = 20 \, \text{m/s}

Therefore, the correct numerical value is 20m/s20 \, \text{m/s}, which corresponds to option A in the listed options.

Common mistakes

  • Using the wrong force relation, such as F=mrωF = mr\omega or a linear-force guess, is incorrect because circular motion requires centripetal force. Use mv2r\frac{mv^2}{r} or equivalently mrω2mr\omega^2.

  • Choosing 40m/s40 \, \text{m/s} by forgetting to take the square root is wrong because from v2=400v^2 = 400, the speed is v=20m/sv = 20 \, \text{m/s}, not 400400 or 4040.

  • Treating the breaking tension as an extra force in addition to centripetal force is incorrect because the tension itself provides the centripetal force here. Set the maximum tension equal to the required centripetal force.

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