A ball of mass 0.1 Kg. is whirled in a horizontal circle of radius 1 m . by means of a string at an initial speed of 10 R.P.M. Keeping the radius constant, the tension in the string is reduced to one quarter of its initial value. The new speed is

Options

A. 5r.p.m.
B. 10 r.p.m.
C. 20 r.p.m
D. 14 r.p.m.

Answer

To solve the problem step by step, we will follow the process outlined in the video transcript.

### Step 1: Calculate the initial speed in meters per second
The initial speed of the ball is given as 10 R.P.M. (revolutions per minute). We need to convert this to meters per second.

1. The distance covered in one complete revolution (circumference of the circle) is given by:
\[ \text{Distance in one revolution} = 2\pi r \]
where \( r = 1 \) m.
\[ \text{Distance in one revolution} = 2\pi \times 1 = 2\pi \text{ meters} \]

2. In 10 revolutions, the distance covered is:
\[ \text{Distance in 10 revolutions} = 10 \times 2\pi = 20\pi \text{ meters} \]

3. To find the speed in meters per second, we divide the distance by the time taken (60 seconds):
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{20\pi}{60} = \frac{\pi}{3} \text{ m/s} \]

### Step 2: Set up the equation for tension and centripetal force
The tension in the string provides the centripetal force required to keep the ball moving in a circle. The tension \( T \) can be expressed as:
\[ T = \frac{m v^2}{r} \]
where:
- \( m = 0.1 \) kg (mass of the ball)
- \( v = \frac{\pi}{3} \) m/s (initial speed)
- \( r = 1 \) m (radius)

### Step 3: Calculate the initial tension
Substituting the values into the tension formula:
\[ T = \frac{0.1 \left(\frac{\pi}{3}\right)^2}{1} = 0.1 \times \frac{\pi^2}{9} = \frac{0.1\pi^2}{9} \text{ N} \]

### Step 4: Determine the new tension
According to the problem, the new tension \( T' \) is one quarter of the initial tension:
\[ T' = \frac{T}{4} = \frac{1}{4} \times \frac{0.1\pi^2}{9} = \frac{0.1\pi^2}{36} \text{ N} \]

### Step 5: Set up the equation for the new speed
The new tension can also be expressed in terms of the new speed \( v' \):
\[ T' = \frac{m v'^2}{r} \]
Substituting for \( T' \):
\[ \frac{0.1\pi^2}{36} = \frac{0.1 v'^2}{1} \]

### Step 6: Solve for the new speed \( v' \)
Cancelling \( 0.1 \) from both sides:
\[ \frac{\pi^2}{36} = v'^2 \]
Taking the square root:
\[ v' = \frac{\pi}{6} \text{ m/s} \]

### Step 7: Convert the new speed back to R.P.M.
To find the new speed in revolutions per minute:
1. The distance covered in one revolution is still \( 2\pi \) meters.
2. The time taken to cover \( 2\pi \) meters at the new speed \( v' = \frac{\pi}{6} \) m/s is:
\[ \text{Time} = \frac{2\pi}{\frac{\pi}{6}} = 12 \text{ seconds} \]
3. Therefore, the number of revolutions per minute is:
\[ \text{Revolutions per minute} = \frac{60 \text{ seconds}}{12 \text{ seconds}} = 5 \text{ R.P.M.} \]

### Final Answer
The new speed of the ball is 5 R.P.M.

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