Answer
The power dissipated by each bulb is increased by a factor of $1.04$
Work Step by Step
Let $I_0$ be the original current when 50 bulbs are functioning. We can write an expression for the power:
$P = I_0^2~R$
We know that the original current is $I_0 = \frac{V}{R_{eq}} = \frac{V}{50~R}$. If one of the bulbs connected in series is removed, then the total resistance is reduced. We can find the new current:
$I = \frac{V}{R_{eq}}$
$I = \frac{V}{49~R}$
$I = \frac{50}{49} \times \frac{V}{50~R}$
$I = \frac{50}{49} \times I_0$
We can find an expression for the new power dissipated by each bulb:
$P = I^2~R$
$P = (\frac{50}{49}~I_0)^2~R$
$P = 1.04\times I_0^2~R$
The power dissipated by each bulb is increased by a factor of $1.04$
