Water and mercury are filled in two cylindrical vessels upto same height. Both vessels have a hole in the wall near the bottom. The velocity of water and mercury coming out of the holes are $v_{1}$ and $v_{2}$ respectively. Thus

v_{1} = v_{2}

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v_{1} = 13.6 v_{2}

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v_{1} = \frac{v_{2}}{13.6}

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v_{1} = \sqrt{13.6} v_{2}

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Solution

The correct option is A $v_{1} = v_{2}$
Since velocity of liquid coming out of tank is irrespective of density of material & given as velocity = $\sqrt{2 g h}$ . Hence both will come out at same velocity.

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Water and mercury are filled in two cylindrical vessels upto same height. Both vessels have a hole in the wall near the bottom. The velocity of water and mercury coming out of the holes are v1 and v2 respectively. Thus

The correct option is A v1=v2 Since velocity of liquid coming out of tank is irrespective of density of material & given as velocity = √2gh. Hence both will come out at same velocity.

Water and mercury are filled in two cylindrical vessels upto same height. Both vessels have a hole in the wall near the bottom. The velocity of water and mercury coming out of the holes are $v_{1}$ and $v_{2}$ respectively. Thus

The correct option is A $v_{1} = v_{2}$
Since velocity of liquid coming out of tank is irrespective of density of material & given as velocity = $\sqrt{2 g h}$ . Hence both will come out at same velocity.