(N/A) Given: Voltage $V = 200 \ V$, Power $P = 100 \ W$, Number of bulbs $n = 5$, Time $t = 4 \ hours$.
$(a)$ We know that $P = \frac{V^2}{R}$, therefore, the resistance of each bulb is:
$R = \frac{V^2}{P} = \frac{(200)^2}{100} = \frac{40000}{100} = 400 \ \Omega$.
$(b)$ Electrical energy consumed by one bulb in $4 \ hours$ is:
$E_1 = P \times t = 100 \ W \times 4 \ h = 400 \ Wh$.
Total energy consumed by $5$ bulbs in $4 \ hours$ is:
$E_{total} = 5 \times 400 \ Wh = 2000 \ Wh = 2 \ kWh$.
$(c)$ Cost of consumed electricity:
Since $1 \ unit = 1 \ kWh$ and the rate is $50 \ paise = ₹ 0.50$ per unit:
$Cost = 2 \ units \times ₹ 0.50/unit = ₹ 1.00$.