Question

A block ‘A’ takes 2 s to slide down a frictionless incline of 30° and length ‘l’, kept inside a lift going up with uniform velocity ‘v’. If the incline is changed to 45°, the time taken by the block, to slide down the incline, will be approximately

• A. 2.66 s
• B. 0.83 s
• C. 1.68 s
• D. 0.70 s

Answer 1

The Correct Option is C

To solve this problem, we need to calculate the time taken by a block to slide down a frictionless incline inside a lift moving with uniform velocity. Let's solve the problem step-by-step.

Step 1: Understand the Initial Condition

Given that the block takes 2 seconds to slide down a 30° incline of length l inside a lift moving up with velocity v. For such situations, the relative motion between the block and incline inside the lift needs to be analyzed. Since the lift is moving with constant velocity, its effect can be ignored on block's sliding time.

Step 2: Calculate the Acceleration on the Incline

For a block sliding down an incline due to gravity without friction:

\(a = g \sin \theta\)

Where \(g\) is the acceleration due to gravity (9.8 m/s²), and \(\theta\) is the angle of the incline.

For \(\theta = 30^\circ\): \(a_{30} = g \sin 30^\circ = \frac{g}{2}\)

Step 3: Determine the Length of the Incline

We are given the time \(t = 2 s\) to slide this length initially. Using the equation of motion:

\(l = \frac{1}{2} a t^2\)

Substitute \(a = \frac{g}{2}\):

\(l = \frac{1}{2} \cdot \frac{g}{2} \cdot (2^2) = g\)

Step 4: Use Similar Logic for the 45° Incline

For a 45° incline:

\(a_{45} = g \sin 45^\circ = \frac{g}{\sqrt{2}}\)

Step 5: Calculate the Time for the New Incline

Using the equation \(l = \frac{1}{2} a t^2\) again, where l remains the same at \(g\):

\(g = \frac{1}{2} \cdot \frac{g}{\sqrt{2}} \cdot t^2\)

Solving for \(t\):

\(t^2 = \frac{2g}{g/\sqrt{2}} = 2 \sqrt{2}\)

\(t = \sqrt{2 \sqrt{2}} \approx 1.68 \, \text{s}\)

The time taken for the block to slide down the 45° incline is approximately 1.68 seconds.

Conclusion: The correct answer is 1.68 s.