Question

A 100-W bulb and 4 25-W bulbs are made for the same voltage. They have filaments of the same length and material. The ratio of the diameter of the 100-W bulb to that of the 25-W bulb is:

• A. 4:1
• B. √2:1
• C. 2:1
• D. 1:√2

Answer 1

The Correct Option is C

The power consumed by a bulb is related to its resistance and voltage. Since both the 100-W and 25-W bulbs are made for the same voltage, their resistances must be different. Let's denote the resistance of the 100-W bulb as \( R_{100} \) and the resistance of the 25-W bulb as \( R_{25} \).

The power consumed by a bulb can be written as:

\[ P = \frac{V^2}{R} \]

where \( P \) is the power, \( V \) is the voltage, and \( R \) is the resistance.

Since the filaments of the bulbs are made of the same material and have the same length, the resistance is directly proportional to the length of the filament and inversely proportional to the cross-sectional area of the filament. Let's denote the diameter of the 25-W bulb as \( d_{25} \) and the diameter of the 100-W bulb as \( d_{100} \).

The cross-sectional area of the filament is proportional to the square of its diameter. Therefore, we can write:

\[ R_{100} = k \cdot \frac{l}{(d_{100})^2} \]

\[ R_{25} = k \cdot \frac{l}{(d_{25})^2} \]

where \( k \) is a constant and \( l \) is the length of the filament.

Since the bulbs are made for the same voltage, we can write:

\[ \frac{V^2}{R_{100}} = \frac{V^2}{R_{25}} \]

Substituting the expressions for \( R_{100} \) and \( R_{25} \) and simplifying, we get:

\[ \frac{(d_{100})^2}{(d_{25})^2} = \frac{100}{25} \]

\[ \frac{d_{100}}{d_{25}} = \sqrt{\frac{100}{25}} = \sqrt{4} = 2 \]

So, the ratio of the diameter of the 100-W bulb to that of the 25-W bulb is 2:1.

The correct answer is option (C) : 2:1