Question

A train running at the speed of \(60\ \text{km/hr}\) crosses a pole in \(9\) seconds. What is the length of the train?

• A. \(120\ \text{m}\)
• B. \(180\ \text{m}\)
• C. \(324\ \text{m}\)
• D. \(150\ \text{m}\)

Hint:

Always make units consistent before applying \( \text{Distance = \text{Speed \times \text{Time\).
Use the quick conversion: \( \text{km/hr \to \text{m/s \) by multiplying with \( \frac{5{18 \); inverse is \( \times \frac{18{5 \).

Answer 1

The Correct Option is D

Step 1 (What does “crosses a pole” mean?)
When a train crosses a stationary point-sized object (like a pole), the distance covered relative to the pole} equals the entire length of the train.
Therefore, Length of train \(L = \text{Distance covered in the given time}\).
Step 2 (Write down given speed and time with units).
Speed \(v = 60\ \text{km/hr}\).
Time \(t = 9\ \text{s}\).
Step 3 (Convert speed to m/s so units match metres and seconds).
Use \(1\ \text{km/hr} = \dfrac{5}{18}\ \text{m/s}\).
\[ v = 60 \times \frac{5}{18}\ \text{m/s} = \frac{300}{18}\ \text{m/s} = \frac{50}{3}\ \text{m/s} \ (\approx 16.67\ \text{m/s})
\] Step 4 (Apply the basic formula \( \text{Distance} = \text{Speed} \times \text{Time}\)).
\[ L = v \times t = \left(\frac{50}{3}\ \text{m/s}\right)\times (9\ \text{s}) = \frac{50}{3}\times 9\ \text{m} = 150\ \text{m}
\] Step 5 (Alternative method without decimals — keep everything in km and hr, then convert).
Convert time to hours: \(9\ \text{s} = \dfrac{9}{3600}\ \text{hr} = \dfrac{1}{400}\ \text{hr}\).
Distance in km \(= 60 \times \dfrac{1}{400} = \dfrac{60}{400} = \dfrac{3}{20}\ \text{km}\).
Convert to metres: \(\dfrac{3}{20}\ \text{km} = \dfrac{3}{20}\times 1000 = 150\ \text{m}\).
Step 6 (Sanity check).
At \(60\ \text{km/hr}\) \(= 16.67\ \text{m/s}\), in \(9\ \text{s}\) the train should cover about \(16.67 \times 9 \approx 150\ \text{m}\).
This is consistent, so the calculation is correct.
Step 7 (Common pitfalls to avoid).
(i) Forgetting to convert km/hr to m/s leads to wrong answers by a factor of \(3.6\).
(ii) Using only the engine length—remember “crossing a pole” implies full train length.
\[ \boxed{150\ \text{metres (Option (d)}} \]