QJEEM23Ph7779: JEE Main 2023 Physics - Class 11 JEEM23Ph7779 Solution: Two trains 'A' and 'B' of length ' ' and ' ' are travelling…

Let's start by converting the velocities of both trains to m/s:

Train A: $108 \frac{km}{h} \times \frac{1000 m}{km} \times \frac{1 h}{3600 s} = 30 \frac{m}{s}$

Train B: $72 \frac{km}{h} \times \frac{1000 m}{km} \times \frac{1 h}{3600 s} = 20 \frac{m}{s}$

To cross the tunnel, Train A has to cover a distance equal to the length of the tunnel plus its own length: $L + l$

Similarly, Train B has to cover a distance equal to the length of the tunnel plus its own length: $L + 4l$

We are given that Train A takes 35 seconds less time than Train B to cross the tunnel. Let's denote the time taken by Train A as $t_A$ and the time taken by Train B as $t_B$ . Then, we have:

$t_B = t_A + 35$

Using the formula distance = velocity × time, we can write the equations for both trains:

Train A: $(L + l) = 30t_A$

Train B: $(L + 4l) = 20t_B$

Now, we can substitute $t_B = t_A + 35$ in the equation for Train B:

$(L + 4l) = 20(t_A + 35)$

We have two equations and two unknowns ( $L$ and $t_A$ ). We can solve this system of equations by eliminating one of the unknowns. Let's eliminate $t_A$ by expressing it from the equation for Train A:

$t_A = \frac{L + l}{30}$

Now, substitute this expression for $t_A$ in the equation for Train B:

$(L + 4l) = 20\left(\frac{L + l}{30} + 35\right)$

Multiplying both sides by 30 to get rid of the fraction:

$30(L + 4l) = 20(L + l) + 20 \times 35 \times 30$

Expanding the equation:

$30L + 120l = 20L + 20l + 21,000$

Simplify:

$10L + 100l = 21,000$

Since we are given that $L = 60l$ , substitute this into the equation:

$10(60l) + 100l = 21,000$

Solve for $l$ :

$700l = 21,000$

$l = 30$

Now, substitute the value of $l$ back into the equation for $L$ :

$L = 60l = 60 \times 30 = 1800$

So, the length of the tunnel is:

1800 m