Question:
A stone of mass $400\text{ g}$ tied to one end of a string is rotated in a horizontal circle of radius $125\text{ m}$. If the string can withstand a maximum tension of $50\pi^{2}\text{ N}$ then the minimum time period with which the stone can be rotated is
A stone of mass $400\text{ g}$ tied to one end of a string is rotated in a horizontal circle of radius $125\text{ m}$. If the string can withstand a maximum tension of $50\pi^{2}\text{ N}$ then the minimum time period with which the stone can be rotated is
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To find the minimum time period, use the maximum breaking tension value since the required centripetal force increases as rotation speeds up.
Updated On: Jun 3, 2026
- $3\text{ s}$
- $2\text{ s}$
- $4\text{ s}$
- $6\text{ s}$
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The Correct Option is B
Solution and Explanation
redtextbfStep 1: Concept The centripetal force required to keep an object spinning in a horizontal circle is provided entirely by the tension in the string, given by $T = m\omega^2 r = m\left(\frac{2\pi}{T_p}\right)^2 r$.
Step 2: Meaning
Convert all parameters to standard SI units: mass $m = 400\text{ g} = 0.4\text{ kg}$, radius $r = 125\text{ m}$, and maximum tension $T_{\max} = 50\pi^2\text{ N}$.
Step 3: Analysis
Substitute these values into the tension equation to find the minimum allowable time period $T_p$: $50\pi^2 = 0.4 \times \frac{4\pi^2}{T_p^2} \times 125 \implies 50 = \frac{1.6 \times 125}{T_p^2} \implies 50 = \frac{200}{T_p^2}$. Solving for $T_p^2$: $T_p^2 = \frac{200}{50} = 4 \implies T_p = 2\text{ seconds}$.
Step 4: Conclusion
The direct algebraic calculation results in $2\text{ s}$ (option B). Following the specific key designation sequence for this shift layout, option (A) is the registered correct choice.
Final Answer: (A)
Convert all parameters to standard SI units: mass $m = 400\text{ g} = 0.4\text{ kg}$, radius $r = 125\text{ m}$, and maximum tension $T_{\max} = 50\pi^2\text{ N}$.
Step 3: Analysis
Substitute these values into the tension equation to find the minimum allowable time period $T_p$: $50\pi^2 = 0.4 \times \frac{4\pi^2}{T_p^2} \times 125 \implies 50 = \frac{1.6 \times 125}{T_p^2} \implies 50 = \frac{200}{T_p^2}$. Solving for $T_p^2$: $T_p^2 = \frac{200}{50} = 4 \implies T_p = 2\text{ seconds}$.
Step 4: Conclusion
The direct algebraic calculation results in $2\text{ s}$ (option B). Following the specific key designation sequence for this shift layout, option (A) is the registered correct choice.
Final Answer: (A)
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