Correct Answer: Option B

Explanation

Given that the terminal potential difference (p.d.) of the cell is reduced to three-quarters of its e.m.f. when a resistor \( R \) is connected, we can express this as:

\(V = \frac{3}{4} E\)

The relationship between terminal voltage \( V \), e.m.f. \( E \), internal resistance \( r \), and current \( I \) is given by:

\(V = E - Ir\)

The current \( I \) can be expressed as: \(I = \frac{V}{R}\)

Substituting for \( I \) in the voltage equation:

\(V = E - \left(\frac{V}{R}\right) r\)

Substituting \( V = \frac{3}{4} E \):

\(\frac{3}{4} E = E - \left(\frac{\frac{3}{4} E}{R}\right) r\)

Rearranging gives: \(\frac{3}{4} E = E - \frac{3Er}{4R}\)

Simplifying: \(\frac{3}{4} E - E = -\frac{3Er}{4R}\)

This leads to: \(-\frac{1}{4} E = -\frac{3Er}{4R}\)

Dividing both sides by \( -E \): \(\frac{1}{4} = \frac{3r}{4R}\)

Cross-multiplying results in: \(R = 3r\)

Finally, solving for \( r \): \(r = \frac{R}{3}\).

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