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. $3\text{ s}$
• B. $2\text{ s}$
• C. $4\text{ s}$
• D. $6\text{ s}$

Hint:

To find the minimum time period, use the maximum breaking tension value since the required centripetal force increases as rotation speeds up.

Answer 1

The Correct Option is B

Step 1: Know what pulls the stone in.
For a stone going in a circle, the string tension supplies the needed centre-pulling force. So tension equals the centripetal force.

Step 2: Write tension with the time period.
Centripetal force is $m\omega^{2}r$, and $\omega = \dfrac{2\pi}{T_{p}}$. \[ T_{\text{string}} = m\left(\frac{2\pi}{T_{p}}\right)^{2} r = \frac{4\pi^{2} m r}{T_{p}^{2}} \]
Step 3: Convert to SI units.
Mass $m = 400$ g $= 0.4$ kg, radius $r = 125$ m. Greatest tension allowed is $50\pi^{2}$ N.

Step 4: The smallest time needs the largest tension.
A smaller $T_{p}$ means faster spin and more tension. So the minimum time period happens when tension is at its limit $50\pi^{2}$.

Step 5: Plug in and solve.
\[ 50\pi^{2} = \frac{4\pi^{2}(0.4)(125)}{T_{p}^{2}} \]The $\pi^{2}$ cancels, giving $50 = \dfrac{200}{T_{p}^{2}}$.

Step 6: Find $T_{p}$.
\[ T_{p}^{2} = \frac{200}{50} = 4 \quad\Rightarrow\quad T_{p} = 2 \text{ s} \]This is option 2.
\[ \boxed{T_{p} = 2 \text{ s}} \]