When a man stands on a moving escalator he goes up in $50 s e c .$ and when he walks up the moving escalator he goes up in $30 s e c .$ Then the man walks up the stationary escalator in a time of $- - - - s e c$

60

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75

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90

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18.75

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Solution

The correct option is B $75$
Suppose the escalator is $x$ meter long.
When man stand then his net velocity will be $o n l y$ that of $e s c a l a t o r$ i.e. $v_{e}$ so $x = v_{e} \times 50$ .....(1)

When the man walks on the escalator then the net velocity will be sum of both i.e $v_{e} + v_{m}$ so $x = (v_{e} + v_{m}) \times 30$ .....(2)

If the man only walks and the escalator is stationary then the net velocity is $v_{m}$ so time being $t$ give us $x = v_{m} \times t$ .....(3)
Equation-1 $s u b s t r a c e d$ from equation-2 we get $20 v_{e} = 30 v_{m}$
putting it in equation-1 and put $x = 50 v_{e}$ in eqaution-3
we will get $x = \frac{2 t \times v_{e}}{3}$ , comparing it with eqaution-1
we get $t = 75 s$
Option B is correct.

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