Given that, two trains cross each other in $14$ seconds.
Let the length of the slower train be $\text{L}$ meter.
The faster train crosses a lamp pole in $12$ seconds.
$S_{\text{faster train}} = \frac{160}{12} \;\text{m/sec}=\frac{40}{3}\;\text{m/sec}$
The speed of slower train $ =S _{\text{faster train}} = \frac{40}{3}\;\text{m/sec} – 6 \times \frac{5}{18}\;\text{m/sec}$
$\qquad \qquad = \left(\frac{40}{3} – \frac{5}{3}\right)\;\text{m/sec}$
$\qquad \qquad = \left(\frac{40-5}{3}\right)\;\text{m/sec}$
$\qquad \qquad=\frac{35}{3}\;\text{m/sec}$
We know that,
- When two trains are moving in opposite directions, then their speed will be added.
- When two trains are moving in the same direction, then their speed will be subtracted.
Now, $\frac{L+160}{14}= \frac{35}{3} + \frac{40}{3}$
$\Rightarrow \frac{L+160}{14}= \frac{75}{3}$
$\Rightarrow \frac{L+160}{14}= 25$
$\Rightarrow L+160= 25 \times 14$
$\Rightarrow L+160= 350$
$\Rightarrow L= 350-160$
$\Rightarrow \boxed{L= 190\;\text{meters}}$
$\therefore$ The length of slower train is $190\;\text{meters.}$
Correct Answer $:\text{A}$