Question

A stone of mass \(10\,\text{g}\) is attached to a string of length \(80\,\text{cm}\) and is rotated in a horizontal circle at a height of \(h\) from the ground with a centripetal acceleration of \(125\,\text{m s}^{-2}\). At a moment the string breaks and the stone strikes the ground at a horizontal distance of \(10\,\text{m}\). Then the value of \(h\) is \[ (\text{Acceleration due to gravity}=10\,\text{m s}^{-2}) \]

• A. \(10\,\text{m}\)
• B. \(5\,\text{m}\)
• C. \(15\,\text{m}\)
• D. \(20\,\text{m}\)

Hint:

For a body projected horizontally, \[ \boxed{x=vt,\qquad h=\frac12gt^2.} \] First find the horizontal speed from centripetal motion and then use projectile motion equations.

Answer 1

The Correct Option is B

Step 1: Find the horizontal velocity of the stone. The centripetal acceleration is \[ a_c=\frac{v^2}{r}. \] Given, \[ a_c=125\,\text{m s}^{-2}, \qquad r=0.8\,\text{m}. \] Hence, \[ v^2=125\times0.8=100, \] \[ v=10\,\text{m s}^{-1}. \]

Step 2: Find the time of flight. After the string breaks, the stone is projected horizontally. The horizontal distance travelled is \[ x=vt. \] Given, \[ x=10\,\text{m}, \qquad v=10\,\text{m s}^{-1}, \] therefore, \[ t=\frac{10}{10}=1\,\text{s}. \]

Step 3: Calculate the height. For vertical motion, \[ h=\frac12gt^2. \] Substituting, \[ g=10\,\text{m s}^{-2}, \qquad t=1\,\text{s}, \] \[ h=\frac12(10)(1)^2=5\,\text{m}. \] Hence, \[ \boxed{h=5\,\text{m}.} \] Therefore, the correct option is \(\boxed{(B)}\).