Q75 / 100 · BITSAT 2023 (Memory Based Paper 2) · Physics
A passenger sitting in a train A moving at $90 \mathrm{~km} /$ $h$ observes another train B moving in the opposite direction for $8 \mathrm{~s}$. If the velocity of the train B is $54 \mathrm{~km} / \mathrm{h}$, then length of train B is :
- A$80 \mathrm{~m}$
- B$200 \mathrm{~m}$
- C$120 \mathrm{~m}$
- Correct answerD$320 \mathrm{~m}$
Solution
Velocity of train A
\(\mathrm{V}_{\mathrm{A}}=90 \frac{\mathrm{km}}{\mathrm{hr}}=90 \times \frac{5}{18}=25 \mathrm{~m} / \mathrm{s}\)
Velocity of train B
\(\mathrm{V}_{\mathrm{B}}=54 \frac{\mathrm{km}}{\mathrm{hr}}=54 \times \frac{5}{18}=15 \mathrm{~m} / \mathrm{s}\)
Velocity of train B w.r.t. train
\(\begin{aligned}<br />A=\vec{V}_B & -\vec{V}_A \\<br />& =15-(-25) \mathrm{m} / \mathrm{s} \\<br />& =40 \mathrm{~m} / \mathrm{s}<br />\end{aligned}\)
Time of crossing \(=\frac{\text { length of train }}{\text { relative velocity }}\)
\((8)=\frac{\ell}{40}\)
\(\ell=8 \times 40=320\) meter.
\(\mathrm{V}_{\mathrm{A}}=90 \frac{\mathrm{km}}{\mathrm{hr}}=90 \times \frac{5}{18}=25 \mathrm{~m} / \mathrm{s}\)
Velocity of train B
\(\mathrm{V}_{\mathrm{B}}=54 \frac{\mathrm{km}}{\mathrm{hr}}=54 \times \frac{5}{18}=15 \mathrm{~m} / \mathrm{s}\)
Velocity of train B w.r.t. train
\(\begin{aligned}<br />A=\vec{V}_B & -\vec{V}_A \\<br />& =15-(-25) \mathrm{m} / \mathrm{s} \\<br />& =40 \mathrm{~m} / \mathrm{s}<br />\end{aligned}\)
Time of crossing \(=\frac{\text { length of train }}{\text { relative velocity }}\)
\((8)=\frac{\ell}{40}\)
\(\ell=8 \times 40=320\) meter.
Hence, the correct answer is option: 'D'.