Question

Two circular holes, one larger than the other, are cut in the side of a large water tank whose top is open to the atmosphere. Hole 1 is the larger hole, and hole 2 is the smaller hole. The center of one of these holes is located 6 times as far beneath the surface of the water as the other. The volume flow rate of the water coming out of the holes is the same. (a) Decide which hole (1 or 2) is located nearest the surface of the water. (b) Calculate the ratio of the radius of the larger hole to the radius of the smaller hole, rA/rB.

Answer 1

Answer:$\frac{r_1}{r_2}=1.565$

Explanation:

Given

two holes are made with different sizes

Hole 1 is large in size and hole 2 is small

If the volume flow rate of water is same for both the hole then small hole must be below the large hole because for same flow rate, velocity of water is large while cross-sectional area is small so it compensate to give same flow for both the holes.

Now for radius apply Bernoulli's theorem at hole 1 and 2

$P_1+\rho gh_1=P_{atm}+\frac{1}{2}\rho v_1^2$

$P_2+\rho g6h_2=P_{atm}+\frac{1}{2}\rho v_2^2$

if hole 1 is h distance below water surface then $h_2=6h$

and $P_1=P_2=P_{atm}$

Also $v_1=\sqrt{2gh}$

$v_2=\sqrt{2g(6h)}$

and $Q=A_1v_1=A_2v_2$

$A=\pi r^2$

thus $\dfrac{r_1}{r_2}=\sqrt{\dfrac{v_2}{v_1}}$

$\dfrac{r_1}{r_2}=\sqrt{\dfrac{\sqrt{6h}}{\sqrt{h}}}$

$\frac{r_1}{r_2}=1.565$