Physics Multiple Choice Questions on “Fluids Mechanical Properties – Bernoulli’s Equation”
1. Consider a tank of height 20m filled with liquid of density 100kg/m3. The area of tank is 10m2. If the tank has a hole of area 2m2 at the bottom, find the speed of the liquid flowing out through the hole when the height of liquid in the tank is 10m . Assume speed of liquid descending at top of tank is 5m/s.
a) 20m/s
b) 14.14m/s
c) 15m/s
d) 20.615m/s
Answer: c
Clarification: We can’t consider the speed of efflux to be (sqrt{2gh}) as the areas are comparable. So, we use Bernoulli theorem between the top of the tank and the hole. Pressure at the top of tank and hole will be same, equal to P0, since both are exposed to the atmosphere.
P0 + ρgH + 1/2ρ (v_{1}^{2}) = P0+ 1/2ρ(v_{2}^{2})
∴ v2 = (sqrt{2(gh + v_{1}^{2}/2)})
= (sqrt{2(10 * 10 + 25/2)})
= 15m/s.
2. In which of the following conditions can the Bernoulli equation not be used?
a) Viscous flow
b) incompressible fluid
c) steady flow
d) laminar flow
Answer: a
Clarification: Bernoulli’s equation can be used for non-viscous, incompressible and steady laminar flow.
3. The speed of efflux, in case of a tank with a hole at the bottom, depends upon which of the following factors? Assume that the area of the tank is > > area of hole.
a) area of tank
b) density of liquid
c) height of hole from liquid
d) atmospheric pressure value
Answer: c
Clarification: Given that the area of the tank is > > area of hole, we can use the formula v = (sqrt{2gh}). This shows that speed depends only on acceleration due to gravity and height of hole from the surface.
4. A cylindrical tank of Height H has a hole on its side. It is kept on a flat surface. Assuming that hole’s area is much smaller than the area of the tank, what should be the distance of the hole below the top surface so that water coming out of the hole travels the maximum horizontal distance at the instant when the height of water is H?
a) H
b) H/3
c) H/2
d) same for all positions
Answer: c
Clarification: Let the hole be at a height x below the top surface.
The distance of hole from the ground will be H-x.
Speed of efflux ‘v’ = (sqrt{2gx}).
Let the time taken for water to reach the ground be ’t’. H-x = 1/2gt2.
∴ t = (sqrt{2/g(H – x)}).
For maximum range, v*t should be maximum.
∴ d((sqrt{2/g(H – x)})*(sqrt{2gx}))/dx = 0
∴ d((sqrt{Hx – x^2}))/dx = 0
∴ (H-2x)/2 (sqrt{Hx – x^2}) = 0
The numerator should be zero but denominator should be non-zero.
∴ H ≠ x (for denominator to be non-zero)
∴ H = 2x OR x = H/2.
5. A cylindrical tank, filled with water, has an area of 10m2. A piston covers its entire top surface. A uniformly distributed load of 1000N is applied on the piston from top. A hole of area 0.001m2is at the bottom of the tank. Find the speed of efflux when the height of water level is 0.1m .
a) 24.5m/s
b) 1.48m/s
c) 0.2m/s
d) 2m/s
Answer: b
Clarification: We should apply Bernoulli theorem between the top surface of water and the hole. The area of hole is very small when compared to the area of the tank. So, 1/2ρv2 at the top surface of the tank can be neglected.
P0 + F/AP + ρgh = P0 + 1/2ρv2
∴ 1000/10 + 1000*10*0.1 = 0.5*1000 *v2
∴ v = 1.48m/s
6. A cylindrical tank of area 5m2 contains water filled to a height of 10cm. A hole of area 0.005m2 is present at the bottom. What is the time required for the water level to become half?
a) 100s
b) 10s
c) 1min
d) 10min
Answer: a
Clarification: The speed of efflux when height of water is ’h’ = (sqrt{2gh}). The volume of water flowing out in time dt is Avdt = 0.005*(sqrt{2gh})*dt. Therefore the decrease in height of water level (dh) in time dt is vol flowing out / area of tank = 0.005(sqrt{2gh})dt/5.
dh = (sqrt{2gh}) (0.005)dt/5
(int_{0}^{0.05}dh/sqrt{h} = int_{0}^{t}sqrt{2g})0.001 dt
∴ 2(sqrt{0.05}) = (sqrt{20})0.001 t
∴ t = (sqrt{0.2})*1000/(sqrt{20}) = 100s.