To solve this problem, we need to use Torricelli’s law to determine the instantaneous velocity of discharge through a small opening at the base of the tank. The law is derived from Bernoulli’s principle and is given by:
v=2ghv = \sqrt{2gh}v=2gh
where:
- vvv is the instantaneous velocity of discharge,
- ggg is the acceleration due to gravity,
- hhh is the height of the liquid column above the opening.
In this problem, we have two cases to consider:
- A tank filled with three layers of liquids with different densities.
- A tank filled with a single liquid of density ρ\rhoρ only.
Step 1: Instantaneous velocity of discharge for the tank with three liquid layers
For this case, the tank is filled with three liquids:
- Liquid 1 has density ρ\rhoρ and height h6\frac{h}{6}6h,
- Liquid 2 has density 2ρ2\rho2ρ and height h3\frac{h}{3}3h,
- Liquid 3 has density 3ρ3\rho3ρ and height h2\frac{h}{2}2h.
The pressure at the base is the sum of the pressures exerted by each liquid layer. The pressure due to a liquid column is given by:
P=ρghP = \rho g hP=ρgh
For the three layers:
- The pressure from the first liquid (density ρ\rhoρ, height h6\frac{h}{6}6h) is: P1=ρg(h6)P_1 = \rho g \left( \frac{h}{6} \right)P1=ρg(6h)
- The pressure from the second liquid (density 2ρ2\rho2ρ, height h3\frac{h}{3}3h) is: P2=2ρg(h3)P_2 = 2\rho g \left( \frac{h}{3} \right)P2=2ρg(3h)
- The pressure from the third liquid (density 3ρ3\rho3ρ, height h2\frac{h}{2}2h) is: P3=3ρg(h2)P_3 = 3\rho g \left( \frac{h}{2} \right)P3=3ρg(2h)
Total pressure at the base is the sum of these pressures:
Ptotal=P1+P2+P3=ρg(h6)+2ρg(h3)+3ρg(h2)P_{\text{total}} = P_1 + P_2 + P_3 = \rho g \left( \frac{h}{6} \right) + 2\rho g \left( \frac{h}{3} \right) + 3\rho g \left( \frac{h}{2} \right)Ptotal=P1+P2+P3=ρg(6h)+2ρg(3h)+3ρg(2h)
Simplifying:
Ptotal=ρg(h6+2h3+3h2)P_{\text{total}} = \rho g \left( \frac{h}{6} + \frac{2h}{3} + \frac{3h}{2} \right)Ptotal=ρg(6h+32h+23h)
To combine the terms, find a common denominator (6):
Ptotal=ρg(h6+4h6+9h6)=ρg×14h6=7ρgh3P_{\text{total}} = \rho g \left( \frac{h}{6} + \frac{4h}{6} + \frac{9h}{6} \right) = \rho g \times \frac{14h}{6} = \frac{7 \rho g h}{3}Ptotal=ρg(6h+64h+69h)=ρg×614h=37ρgh
Now, the velocity of discharge is:
v1=2Ptotalρ=2×7ρgh3ρ=14gh3v_1 = \sqrt{\frac{2 P_{\text{total}}}{\rho}} = \sqrt{\frac{2 \times \frac{7 \rho g h}{3}}{\rho}} = \sqrt{\frac{14 g h}{3}}v1=ρ2Ptotal=ρ2×37ρgh=314gh
Step 2: Instantaneous velocity of discharge for the tank with liquid of density ρ\rhoρ only
In this case, the tank is filled with a single liquid of density ρ\rhoρ and height hhh. The pressure at the base is:
Ptotal=ρghP_{\text{total}} = \rho g hPtotal=ρgh
The velocity of discharge is:
v2=2Ptotalρ=2ρghρ=2ghv_2 = \sqrt{\frac{2 P_{\text{total}}}{\rho}} = \sqrt{\frac{2 \rho g h}{\rho}} = \sqrt{2 g h}v2=ρ2Ptotal=ρ2ρgh=2gh
Step 3: Ratio of the velocities
Now, we calculate the ratio of the velocity in the first case (with three layers of liquids) to the velocity in the second case (with a single liquid of density ρ\rhoρ):
Ratio=v1v2=14gh32gh=143×2=146=73≈1.53\text{Ratio} = \frac{v_1}{v_2} = \frac{\sqrt{\frac{14 g h}{3}}}{\sqrt{2 g h}} = \sqrt{\frac{14}{3 \times 2}} = \sqrt{\frac{14}{6}} = \sqrt{\frac{7}{3}} \approx 1.53Ratio=v2v1=2gh314gh=3×214=614=37≈1.53
Final Answer:
The ratio of the instantaneous velocity of discharge through a small opening at the base of the tank in the first case (three liquid layers) to that in the second case (single liquid of density ρ\rhoρ) is approximately 1.53.