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Question

Physics Question on Current electricity

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

2:1

Explanation

Solution

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 R100R_{100} and the resistance of the 25-W bulb as R25R_{25}.

The power consumed by a bulb can be written as:

P=V2RP = \frac{V^2}{R}

where PP is the power, VV is the voltage, and RR 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 d25d_{25} and the diameter of the 100-W bulb as d100d_{100}.

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

R100=kl(d100)2R_{100} = k \cdot \frac{l}{(d_{100})^2}

R25=kl(d25)2R_{25} = k \cdot \frac{l}{(d_{25})^2}

where kk is a constant and ll is the length of the filament.

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

V2R100=V2R25\frac{V^2}{R_{100}} = \frac{V^2}{R_{25}}

Substituting the expressions for R100R_{100} and R25R_{25} and simplifying, we get:

(d100)2(d25)2=10025\frac{(d_{100})^2}{(d_{25})^2} = \frac{100}{25}

d100d25=10025=4=2\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