Question

A block takes \(t\) time to slide down a plane inclined at \(45^\circ\) to the horizontal. If the surface is made smooth (frictionless), the block takes time \( \frac{t}{2} \) to slide down the plane. The coefficient of friction between the block and the inclined plane is \( \left(\frac{\alpha}{100}\right) \). The value of \( \alpha \) is _____.

Answer 1

Correct Answer: 60

Step 1: Understanding the Question:
We are comparing the sliding time on a rough inclined plane versus a smooth one to find the friction coefficient \(\mu\).
Step 2: Key Formula or Approach:
For a distance \(s\) from rest: \(s = \frac{1}{2}at^2 \Rightarrow t \propto \frac{1}{\sqrt{a}}\).
- Smooth plane acceleration: \(a_s = g \sin\theta\).
- Rough plane acceleration: \(a_r = g(\sin\theta - \mu \cos\theta)\).
Ratio of times: \(\frac{t_r}{t_s} = \sqrt{\frac{a_s}{a_r}}\).
Step 3: Detailed Explanation:
1. Given \(\theta = 45^{\circ}\), \(t_r = t\) and \(t_s = t/2\). So, \(\frac{t_r}{t_s} = 2\).
\[ 2 = \sqrt{\frac{g \sin 45^{\circ}}{g (\sin 45^{\circ} - \mu \cos 45^{\circ})}} \]
\[ 4 = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} - \mu \frac{1}{\sqrt{2}}} = \frac{1}{1 - \mu} \]
\[ 1 - \mu = \frac{1}{4} = 0.25 \]
\[ \mu = 0.75 \]
2. Given \(\mu = \frac{\alpha}{100}\):
\[ \frac{\alpha}{100} = 0.75 \Rightarrow \alpha = 75 \]
Step 4: Final Answer:
The value of \(\alpha\) is 75.