The rest distance between Lucknow and Bhopal is 1000 km . A nonstop...

$\Delta t = \dfrac{L_0}{v} = \dfrac{10^6}{100} = 10^4 \text{ s}$

Converting this time into minutes:
$\Delta t = \dfrac{10000}{60} \text{ min} = \dfrac{500}{3} \text{ min}$

The time elapsed in the train frame is the proper time $\Delta t_0$, given by the time dilation formula:
$\Delta t_0 = \Delta t \sqrt{1 - \dfrac{v^2}{c^2}}$

The difference in time between the two frames is:
\Delta t - \Delta t_0 = \Delta t \left(1 - \sqrt{1 - \dfrac{v^2}{c^2}}\right)<br><br>Sincev \ll c$, using the binomial expansion$(1 - x)^{1/2} $\approx$1 - $\dfrac{x}{2}:<br>\Delta$t - $\Delta$t_0 $\approx \Delta$t \left(\dfrac{v^2}{2c^2}\right)$<br><br>Substituting the given values ($c = 3 \times 10^8 \text{ m s}^{-1}$):<br>$\Delta t - \Delta t_0 = 10^4 \times \dfrac{(100)^2}{2 \times (3 \times 10^8)^2}$<br><br>$\Delta t - \Delta t_0 = 10^4 \times \dfrac{10^4}{18 \times 10^{16}}$<br><br>$\Delta t - \Delta t_0 = \dfrac{1}{18 \times 10^8} \text{ s}$<br><br>$\Delta t - \Delta t_0 = \dfrac{10}{18} \times 10^{-9} \text{ s} = \dfrac{5}{9} \text{ ns} \approx 0.56 \text{ ns}$<br><br>Thus, the time elapsed in the train frame is$0.56 \text{ ns}$less than$\dfrac{500}{3} \text{ min}$.$