A long plying record revolves with a speed of `33 (1)/(3) "rev min"^(-1)`, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the coefficient of friction between the coins and the record is 0.15, which of the two coins will revolve with the record ? Take `g = 9.8 ms^(-2)`.

Explanation

To determine which of the two coins will revolve with the long playing record, we need to analyze the forces acting on the coins and the conditions for circular motion.

### Step-by-Step Solution:

1. **Convert the speed of the record to radians per second:**
The speed of the record is given as \( 33 \frac{1}{3} \) revolutions per minute. First, convert this to revolutions per second:
\[
v = 33 \frac{1}{3} \text{ rev/min} = \frac{100}{3} \text{ rev/min}
...