Question
A tank open at the top with a water level of 1 m, as shown in the figure, has a hole at a height of 0.5 m. A free jet leaves horizontally from the smooth hole. The distance X (in m) where the jet strikes the floor is
Options :
0.5
1.0
2.0
4.0
Answer :
1.0
Solution :
Let free jet velocity is x displacement
t = X/V
As per Torricelli’s formula, V = √(2gh)
t = X/√(2gh) ..(i)
By second law of motion equation
s = ut+ gt2/2
For free fall u=0, s= gt2/2
t = √(2s/g) ..(i) is the time of free fall of object
By equation (i) and (ii)
X/√(2gh) = √(2s/g)
X = √(4hs) = √(4*0.5*0.5) = 1m
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