A uniform bar of length `12 L` and mass `48 m` is supported horizontally on two smooth tables as shown in the figure. A small moth (an insect) of mass `8m` is sitting on end `A` of the rod and a spider (an insect) of mass `16 m` is sitting on the other end `B`. Both the insects start moving towards each other along the rod with moth moving at speed `2v` and the spider at half of this speed. They meet at a point `P` on the rod and the spider eats the moth. After this the spider moves with a velocity `v//2` relative to the rod towards the end `A`. The spider takes negligible time in eating the insect. Also, let `v = L//T`, where `T` is a constant having value `4 sec`. The speed of the bar after the spider eats up the moth and moves towards `A` isA. `v//2`B. `v`C. `v//6`D. `2v`

A uniform bar of length `12 L` and mass `48 m` is supported horizontal

1.	A uniform bar of length `12 L` and mass `48 m` is supported horizontally on two smooth tables as shown in the figure. A small moth (an insect) of mass `8m` is sitting on end `A` of the rod and a spider (an insect) of mass `16 m` is sitting on the other end `B`. Both the insects start moving towards each other along the rod with moth moving at speed `2v` and the spider at half of this speed. They meet at a point `P` on the rod and the spider eats the moth. After this the spider moves with a velocity `v//2` relative to the rod towards the end `A`. The spider takes negligible time in eating the insect. Also, let `v = L//T`, where `T` is a constant having value `4 sec`. The speed of the bar after the spider eats up the moth and moves towards `A` isA. `v//2`B. `v`C. `v//6`D. `2v`
Answer» Correct Answer - C After spider eats the moth let speed of rod becomes `v_(0)` (towards right) `therefore` Speed of spider relative to rod `= (v)/(2)` (left ward) From com `24m ((v)/(2)-v_(0)) = 48 mv_(0) , v_(0) = v//6`

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