Question

An old light bulb draws only 54.3 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what factor is the diameter of the filament reduced, assuming uniform thinning along its length? Neglect any effects caused by temperature differences.

Answer 1

Step-by-step explanation:

Po = 60 W

P = 54.3 W

Let the initial diameter of the filament is do and the final diameter of the filament is d.

Let the voltage is V and the initial resistance is Ro and the final resistance is R.

The formula for power is given by

P = V²/R

The resistance of the filament is inversely proportional to the square of the diameter of the filament. As voltage is constant so the power is

Power α diameter²

So, initial power is

Po α do² ..... (1)

Final power is

P α d² ..... (2)

Divide equation (2) by equation (1), we get

P / Po = d² / do²

54.3 / 60 = d² / do²

d² / do² = 0.905

d = 0.95 d

Thus, the diameter of the filament is reduced to a factor of 0.95 .

Answer 2

The factor of the diameter is 0.95.

Step-by-step explanation:

Given that,

Power of old light bulb = 54.3 W

Power = 60 W

We know that,

The resistance is inversely proportional to the diameter.

`R\propto(1)/(D)`

The power is inversely proportional to the resistance.

`P\propto(1)/(R)`

`P\propto D^2`

We need to calculate the factor of the diameter of the filament reduced

Using relation of power and diameter

`(P_(i))/(P_(f))=(D_(i)^2)/(D_(f)^2)`

Put the value into the formula

`(D_(i)^2)/(D_(f)^2)=(54.3)/(60)`

`(D_(i))/(D_(f))=0.95`

`D_(i)=0.95 D_(f)`

Hence, The factor of the diameter is 0.95.