A small coin of mass 5 g is placed at a distance 5 cm from centre on a flat horizontal turn table. The turn table is observed to make 3 revolutions in `sqrt(10)s`. The frictional force on the coin is `(n xx 10^(-3)) N`. Find value of `n`.

Answer

To solve the problem step by step, we need to calculate the frictional force acting on the coin placed on the turntable. Here’s how we can do it:

### Step 1: Convert the mass of the coin to kilograms
The mass of the coin is given as 5 g. We need to convert this to kilograms for standard SI units.

\[ \text{Mass} (m) = 5 \, \text{g} = \frac{5}{1000} \, \text{kg} = 0.005 \, \text{kg} \]

### Step 2: Calculate the angular velocity (ω)
The turntable makes 3 revolutions in \( \sqrt{10} \) seconds. We need to convert this to angular velocity in radians per second.

\[ \text{Angular velocity} (\omega) = \frac{\text{Total angle in radians}}{\text{Time}} = \frac{3 \times 2\pi \, \text{radians}}{\sqrt{10} \, \text{s}} = \frac{6\pi}{\sqrt{10}} \, \text{rad/s} \]

### Step 3: Calculate the radius in meters
The distance from the center of the turntable to the coin is given as 5 cm. We need to convert this to meters.

\[ \text{Radius} (r) = 5 \, \text{cm} = \frac{5}{100} \, \text{m} = 0.05 \, \text{m} \]

### Step 4: Calculate the centripetal force
The frictional force acting on the coin is equal to the centripetal force required to keep it moving in a circle. The centripetal force can be calculated using the formula:

\[ F = m \cdot \omega^2 \cdot r \]

Substituting the values we have:

\[ F = 0.005 \, \text{kg} \cdot \left(\frac{6\pi}{\sqrt{10}}\right)^2 \cdot 0.05 \, \text{m} \]

Calculating \( \omega^2 \):

\[ \omega^2 = \left(\frac{6\pi}{\sqrt{10}}\right)^2 = \frac{36\pi^2}{10} = \frac{36\pi^2}{10} \]

Now substituting back into the force equation:

\[ F = 0.005 \cdot \frac{36\pi^2}{10} \cdot 0.05 \]

\[ F = 0.005 \cdot 0.05 \cdot \frac{36\pi^2}{10} \]

\[ F = 0.00025 \cdot \frac{36\pi^2}{10} \]

\[ F = \frac{9\pi^2}{1000} \, \text{N} \]

### Step 5: Calculate the numerical value of the frictional force
Using the approximation \( \pi \approx 3.14 \):

\[ F \approx \frac{9 \cdot (3.14)^2}{1000} \approx \frac{9 \cdot 9.8596}{1000} \approx \frac{88.7364}{1000} \approx 0.0887364 \, \text{N} \]

### Step 6: Express the frictional force in the required form
The problem states that the frictional force is of the form \( n \times 10^{-3} \, \text{N} \).

From our calculation, we have:

\[ F \approx 0.0887364 \, \text{N} \approx 88.7364 \times 10^{-3} \, \text{N} \]

Thus, \( n \approx 88.7364 \).

### Final Answer
The value of \( n \) is approximately 9 (when rounded to the nearest integer).