^{rd} December 2007. A few multiple choice questions (with solution) in this section were discussed in the posts dated 4^{th} and 5^{th} December 2007. You can access all posts related to fluid mechanics on this site by clicking on the

Today we will discuss a few more multiple choice questions on fluid mechanics:

**(1)** A water storage tank has a square hole of side 2 cm at its bottom. A plumber, unaware of the hole, admits water into the tank at a constant rate of √2 litre per second. Up to what height can water remain in the tank? (*g = *10 ms^{–2})

(a) 12.25 m

(b) 8.25 m

(c) 2.45 m

(d) 1.25 m

(e) 0.625 m

If *h* is the height up to which water can remain in the tank, the velocity *v *of water flowing out through the hole in the steady state is given by

*v* = √(2*gh*)

The volume of water flowing out in the steady state is *av* where *a* is the area of the hole.

This must be equal to the volume of incoming water which is given as √2 litre = √2 ×10^{–3}

m^{3}. Thus we have

*a*√(2*gh*) = √2 ×10^{–3}

Here *a =* 4 cm^{2} = 4×10^{–4}* *m^{2}.

Substituting, 4×10^{–4}×√(2×10×*h*) = √2 ×10^{–3}

Or, √(10*h*) = 2.5 from which *h =* 6.25/10 = 0.625 m.

**(2)** A piece of granite floats at the interface of mercury and water contained in a beaker (Fig.). If the densities of granite, water and mercury are *ρ*, *ρ*_{1} and *ρ*_{2} respectively, the ratio of the volume of granite in water to the volume in mercury is

(a) (*ρ*_{2}* *– *ρ*) /(*ρ *– *ρ*_{1})

(b) (*ρ*_{2}* *+ *ρ*) /(*ρ+* *ρ*_{1})

(c) *ρ*_{1}* ρ*_{2} /*ρ*

(d) *ρ*_{1} /*ρ*_{2}

(e) *ρ*_{2} /*ρ*_{1}

The weight of a floating body is equal to the weight of the displaced fluid. If *V* and *v* represent the total volume of the piece of granite and volume of granite in water respectively, we have

*V* *ρg = v ρ*_{1}*g + *(*V *–* v*)* ρ*_{2}*g*

Or, *v*(*ρ*_{1}* *– *ρ*_{2}) = *V*(*ρ *– *ρ*_{2})

Therefore, *v/V = *(*ρ *– *ρ*_{2}) /(*ρ*_{1}* *– *ρ*_{2})

The ratio required in the question is *v/* (*V**–**v*) and is given by

*v/* (*V**–**v*) = (*ρ *– *ρ*_{2}) /[(*ρ*_{1}* *– *ρ*_{2})* **–*(*ρ *– *ρ*_{2})]

Or, *v/* (*V**–**v*) = (*ρ *– *ρ*_{2}) /(*ρ*_{1}* *– *ρ*) = (*ρ*_{2}* *– *ρ*) /(*ρ *– *ρ*_{1})

**(3)** The pressure of water at the bottom of a water tank exceeds the atmospheric pressure by 10^{4} pascal. The velocity of efflux of water through an orifice at the bottom of the water tank will be (*g = *10 ms^{–2})

(a) √5 ms^{–1}

(b) √10 ms^{–1}

(c) √15 ms^{–1}

(d) √20 ms^{–1}

(e) √30 ms^{–1}

If *h* is the height of water column in the tank, we have

*h**ρg = *10^{4}^{}

Since the density (*ρ*) of water is 1000 kg m^{–3},

*h*×1000×10 = 10^{4} from which *h = *1m.

The velocity of efflux (Torricelli’s theorem) is √(2*gh*) = √(2×10×1) = √20 ms^{–1}

**(4) **In a wind tunnel the flow speeds (of air) on the upper and lower surfaces of the wing of a model airplane are *v*_{1} and *v*_{2} respectively (*v*_{1}> *v*_{2}). If the wing area is *A* and the density of air is *ρ*, the lift on the wing is

(a) *ρ** *(*v*_{1}^{2 }–* v*_{2}^{2})*A*

(b) ½ *ρ** *(*v*_{1}^{ }–* v*_{2})*A*

(c) ½ *ρ** *(*v*_{1}^{2 }–* v*_{2}^{2})*A*

(d) ½ (*v*_{1}^{2 }–* v*_{2}^{2})*A /**ρ*

(e) *ρ** *(*v*_{1}^{ }–* v*_{2})*A*

If *P*_{1} and *P*_{2} are the pressures on the upper and lower surfaces of the wing we have, by

*P*_{1} + ½ *ρ** **v*_{1}^{2} = *P*_{2} + ½ *ρ** **v*_{2}^{2}

[We have ignored the small height difference between the top and bottom of the wing so that the gravitational potential energy is treated as constant].

The pressure at the top is less than the pressure at the bottom since *v*_{1}> *v*_{2}.

The lift on the wing is (*P*_{2 }–* P*_{1})*A = ***½ ***ρ** ***( v_{1}^{2 }– v_{2}^{2})A**

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