\[(a)\;10 \quad (b)\;2 \quad (c)\;5 \quad(d)\;infinite \]

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The relative velocity of one train relative to another is $30-(-30)=60 km/hr$

As they are $60\; km$ apart they will crash in 1 hr

Relative velocity of bird with respect to train towards which it is moving is $90\; km/hr$

So time taken for $1^{st}$ trip $t_1=\large\frac{60}{90}=\frac{2}{3}$$hr$

In this time the train has moved $\large\frac{2}{3} $$\times 60=40 \;km$. So remaining distance for the bird to fly for $2^{nd}$ trip is $60-40=20 km$

Time taken for II trip$= \large\frac{20}{90}=\frac{2}{3^2}$$ hr$

Similarly for the $n^{th}\; trip =\large\frac{2}{3^n} $$hr$

Since total time is 1 hr

Therefore $ \large\frac{2}{3}+\frac{2}{3^2}+.....+\frac{2}{3^n}=1$

$\large\frac{2}{3} \bigg[1+\frac{1}{3}+.....+\frac{1}{3^{n-1}}\bigg]=1$

$\large\frac{2}{3} \bigg[\frac{1-(\Large\frac{1}{3})^n}{1-\Large\frac{1}{3}}\bigg]=1$

$1-\bigg(\large\frac{1}{3}\bigg)^n$$=1$

$3^n=\infty$

$n=\infty$

bird makes infinite trip

Hence d is the correct answer.

Hence d is the correct answer.

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