Question

The time taken by a body to slide down the smooth inclined plane is \(4\text{ sec}\). The time taken by a body to slide \(1/4^{\text{th}}\) of the length of the plane is

• A. \(1\text{ sec}\)
• B. \(2\text{ sec}\)
• C. \(3\text{ sec}\)
• D. \(0.5\text{ sec}\)

Hint:

For motion from rest with constant acceleration, distance is proportional to square of time: \(s\propto t^2\).

Answer 1

The Correct Option is B

The body slides down a smooth inclined plane from rest. For motion from rest with constant acceleration: \[ s=\frac{1}{2}at^2. \] This means: \[ s\propto t^2. \] Let the total length of the inclined plane be: \[ L. \] The time taken to slide the full length \(L\) is: \[ T=4\text{ sec}. \] So, \[ L=\frac{1}{2}aT^2. \] Now, let the time taken to slide \(\frac{1}{4}\) of the length be \(t\). Then: \[ \frac{L}{4}=\frac{1}{2}at^2. \] Now divide this equation by the full-length equation: \[ \frac{\frac{L}{4}}{L}=\frac{\frac{1}{2}at^2}{\frac{1}{2}aT^2}. \] Cancel common terms: \[ \frac{1}{4}=\frac{t^2}{T^2}. \] Taking square root: \[ \frac{1}{2}=\frac{t}{T}. \] Therefore: \[ t=\frac{T}{2}. \] Since \[ T=4\text{ sec}, \] we get: \[ t=\frac{4}{2}=2\text{ sec}. \] Hence, the required time is: \[ 2\text{ sec}. \]