Question

Two smooth inclined planes A and B each of height 20 m have angles of inclination $30^\circ$ and $60^\circ$ respectively. If $t_1$ and $t_2$ are the times taken by two blocks to reach the bottom of the planes A and B from the top, then find the value of $t_1 - t_2$. (Acceleration due to gravity $g = 10\, m/s^2$)

• A. \(\frac{\sqrt{3} - 1}{\sqrt{3}}\, s\)
• B. \(3(\sqrt{3} - 1)\, s\)
• C. \(4 \left(\frac{\sqrt{3} - 1}{\sqrt{3}}\right) s\)
• D. \((3\sqrt{3} - 2) s\)

Hint:

For blocks sliding down frictionless inclined planes, use kinematic equations with acceleration as \(g \sin \theta\) and distance as the length of the incline to find time taken.

Answer 1

The Correct Option is C

Step 1: Height \(h = 20\, m\) for both planes.
Step 2: Length of incline for plane A: \(L_1 = \frac{h}{\sin 30^\circ} = \frac{20}{1/2} = 40\, m\)
Length of incline for plane B: \(L_2 = \frac{h}{\sin 60^\circ} = \frac{20}{\sqrt{3}/2} = \frac{40}{\sqrt{3}}\, m\)
Step 3: Acceleration down the incline due to gravity component: \[ a = g \sin \theta \] For plane A: \(a_1 = 10 \times \sin 30^\circ = 5\, m/s^2\)
For plane B: \(a_2 = 10 \times \sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\, m/s^2\)
Step 4: Time taken to travel down the incline starting from rest: Using \(s = \frac{1}{2} a t^2\), solve for \(t\): \[ t = \sqrt{\frac{2s}{a}} \] For plane A: \[ t_1 = \sqrt{\frac{2 \times 40}{5}} = \sqrt{16} = 4\, s \] For plane B: \[ t_2 = \sqrt{\frac{2 \times \frac{40}{\sqrt{3}}}{5 \sqrt{3}}} = \sqrt{\frac{80/\sqrt{3}}{5 \sqrt{3}}} = \sqrt{\frac{80}{5 \times 3}} = \sqrt{\frac{80}{15}} = \frac{4}{\sqrt{3}}\, s \] Step 5: Calculate difference \(t_1 - t_2\): \[ 4 - \frac{4}{\sqrt{3}} = 4 \left(1 - \frac{1}{\sqrt{3}}\right) = 4 \left(\frac{\sqrt{3} - 1}{\sqrt{3}}\right) s \]