The correct option is (B): \(82\)
Let the length of the train be \(l\) meters and its speed be \(s\) km/h.
Given the time taken to cross a platform of length \(p_1\) and a platform of length \(p_2\) at speeds \((s-2)\) km/h and \((s-4)\) km/h respectively:
\[ \frac{l}{((s - 2) \times \frac{5}{18})} = 90 \text{ sec}, \quad \frac{l}{((s - 4) \times \frac{5}{18})} = 100 \text{ sec} \]
From these equations, we can express the length \(l\):
\[ l = 90(s - 2) \times \frac{5}{18} \] \[ l = 100(s - 4) \times \frac{5}{18} \]
Equating both expressions for \(l\):
\[ 90(s - 2) \times \frac{5}{18} = 100(s - 4) \times \frac{5}{18} \] \[ \Rightarrow 90(s - 2) = 100(s - 4) \] \[ \Rightarrow 90s - 180 = 100s - 400 \] \[ \Rightarrow 100s - 90s = 400 - 180 \] \[ \Rightarrow 10s = 220 \] \[ \Rightarrow s = 22 \text{ km/h} \]
∴ Length of the train:
Substitute \(s = 22\) km/h into the first equation for \(l\):
\[ l = 90(22 - 2) \times \frac{5}{18} = 90 \times 20 \times \frac{5}{18} = 500 \text{ m} \]
Time to cross a lamp post:
The speed of the train is \(22\) km/h, which is \(22 \times \frac{5}{18}\) m/s.
The time taken to cross a stationary object like a lamp post is the length of the train divided by its speed in m/s:
\[ \text{Time} = \frac{\text{Length of train}}{\text{Speed of train}} = \frac{500 \text{ m}}{22 \times \frac{5}{18} \text{ m/s}} = \frac{500 \times 18}{22 \times 5} = \frac{500 \times 18}{110} = \frac{9000}{110} \approx 81.81 \text{ sec} \]
Rounding to the nearest whole number, the time taken is approximately \( \boxed{82 \text{ sec}} \).