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 \(x/p\).
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 \(x/p\).
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Rewrite numbers smartly (e.g., \(5 \times 10^{-3} = 0.5 \times 10^{-2}\)) to match given form quickly.
Updated On: Apr 17, 2026
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Correct Answer: 1
Solution and Explanation
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: \[ {\frac{x}{p} = \frac{2}{2} = 1} \]
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: \[ {\frac{x}{p} = \frac{2}{2} = 1} \]
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