Question

A vertical disc of diameter 10 cm makes 20 revolutions per second about a horizontal axis through its centre. A uniform magnetic field \(10^{-1}\) T acts perpendicular to the plane. If potential difference between centre and rim is \(\frac{\pi}{x} \times 10^{-p}\) Volt, find \(x/p\).

Hint:

Rewrite numbers smartly (e.g., \(5 \times 10^{-3} = 0.5 \times 10^{-2}\)) to match given form quickly.

Answer 1

Correct Answer: 1

Concept: EMF between centre and rim of rotating disc: \[ V = \frac{1}{2} B \omega R^2 \]

Step 1: Given data \[ R = 5\,\text{cm} = 5 \times 10^{-2}\,\text{m} \] \[ f = 20 \Rightarrow \omega = 2\pi f = 40\pi \] \[ B = 10^{-1}\,\text{T} \]

Step 2: Substitute \[ V = \frac{1}{2} \times 10^{-1} \times 40\pi \times (5 \times 10^{-2})^2 \] \[ = \frac{1}{2} \times 10^{-1} \times 40\pi \times 25 \times 10^{-4} \]

Step 3: Simplify \[ V = 20 \times 25 \times \pi \times 10^{-5} = 500\pi \times 10^{-5} = 5\pi \times 10^{-3} \]

Step 4: Compare with given form \[ V = \frac{\pi}{x} \times 10^{-p} \] \[ 5\pi \times 10^{-3} = \frac{\pi}{x} \times 10^{-p} \] Cancel \(\pi\): \[ 5 \times 10^{-3} = \frac{1}{x} \times 10^{-p} \] Rewrite: \[ 5 \times 10^{-3} = 0.5 \times 10^{-2} \] \[ = \frac{1}{2} \times 10^{-2} \]

Step 5: Identify values \[ \frac{1}{x} = \frac{1}{2} \Rightarrow x = 2 \] \[ p = 2 \] Final: \[ {\frac{x}{p} = \frac{2}{2} = 1} \]