This set of Fluid Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Fluid Properties – 1”.
1. Which one of the following is the unit of mass density?
a) kg = m3
b) kg = m2
c) kg = m
d) kg = ms
a) kg = m3
b) kg = m2
c) kg = m
d) kg = ms
Answer: a
Explanation: Mass Density(p) is defined as the mass(m) per unit volume(V ), i.e., p = m ⁄v
Thus, the unit of p is kg = m3.
Explanation: Mass Density(p) is defined as the mass(m) per unit volume(V ), i.e., p = m ⁄v
Thus, the unit of p is kg = m3.
2. The specific gravity of a liquid has
a) the same unit as that of mass density
b) the same unit as that of weight density
c) the same unit as that of specific volume
d) no unit
a) the same unit as that of mass density
b) the same unit as that of weight density
c) the same unit as that of specific volume
d) no unit
Answer: d
Explanation: The specific gravity of a liquid is the ratio of two similar quantities (densities) which makes it unitless.
Explanation: The specific gravity of a liquid is the ratio of two similar quantities (densities) which makes it unitless.
3. The specific volume of a liquid is the reciprocal of
a) weight density
b) mass density
c) specific weight
d) specific volume
a) weight density
b) mass density
c) specific weight
d) specific volume
Answer: b
Explanation: Specific volume(v) is defined as the volume(V ) per unit mass(m).
v = v⁄m = 1 / m⁄v = 1⁄p
where p is the mass density.
Explanation: Specific volume(v) is defined as the volume(V ) per unit mass(m).
v = v⁄m = 1 / m⁄v = 1⁄p
where p is the mass density.
4. Which one of the following is the unit of specific weight?
a) N = m3
b) N = m2
c) N = m
d) N = ms
a) N = m3
b) N = m2
c) N = m
d) N = ms
Answer: a
Explanation: Specific weight(γ) is defined as the weight(w) per unit volume(V ), i.e.,
γ = w / v
Thus, unit of is N = m3.
Explanation: Specific weight(γ) is defined as the weight(w) per unit volume(V ), i.e.,
γ = w / v
Thus, unit of is N = m3.
5. Which one of the following is the dimension of mass density?
a) [M1 L-3 T0].
b) [M1 L3 T0].
c) [M0 L-3 T0].
d) [M0 L3 T0]
a) [M1 L-3 T0].
b) [M1 L3 T0].
c) [M0 L-3 T0].
d) [M0 L3 T0]
Answer: a
Explanation: Mass Density(p) is defined as the mass(m) per unit volume(V ), i.e.,
[p] = [m]/[v] = [m] /[L3] = [ML-3].
Explanation: Mass Density(p) is defined as the mass(m) per unit volume(V ), i.e.,
[p] = [m]/[v] = [m] /[L3] = [ML-3].
6. Which one of the following is the dimension of specific gravity of a liquid?
a) [M1 L-3 T0].
b) [M1 L0 T0].
c) [M0 L-3 T0].
d) [M0 L0 T0].
a) [M1 L-3 T0].
b) [M1 L0 T0].
c) [M0 L-3 T0].
d) [M0 L0 T0].
Answer: d
Explanation: The specific gravity of a liquid is the ratio of two similar quantities (densities) which makes it dimensionless.
Explanation: The specific gravity of a liquid is the ratio of two similar quantities (densities) which makes it dimensionless.
7. Which one of the following is the dimension of specific volume of a liquid?
a) [M1 L-3 T0].
b) [M-1 L3 T0].
c) [M-1 L-3 T0].
d) [M0 L3 T0].
a) [M1 L-3 T0].
b) [M-1 L3 T0].
c) [M-1 L-3 T0].
d) [M0 L3 T0].
Answer: b
Explanation: Specific volume(v) is defined as the volume(V ) per unit mass(m). Thus,
[v] = [V]/[m] = [L3]/[M] = [M-1L3].
Explanation: Specific volume(v) is defined as the volume(V ) per unit mass(m). Thus,
[v] = [V]/[m] = [L3]/[M] = [M-1L3].
8. Which one of the following is the dimension of specific weight of a liquid?
a) [ML-3 T -2].
b) [ML3 T-2].
c) [ML-2 T-2].
d) [ML2 T-2].
a) [ML-3 T -2].
b) [ML3 T-2].
c) [ML-2 T-2].
d) [ML2 T-2].
9. Two fluids 1 and 2 have mass densities of p1 and p2 respectively. If p1 > p2, which one of the following expressions will represent the relation between their specific volumes v1 and v2?
a) v1 > v2
b) v1 < v2
c) v1 = v2
d) Cannot be determined due to insufficient information.
a) v1 > v2
b) v1 < v2
c) v1 = v2
d) Cannot be determined due to insufficient information.
Answer: b
Explanation: Specific volume(v) is defined as the volume(V ) per unit mass(m).
v = v⁄m = 1 / m⁄v = 1⁄p
where p is the mass density. Thus, if p1 > p2, the relation between the specific volumes v1 and v2
will be represented by v1 < v2.
Explanation: Specific volume(v) is defined as the volume(V ) per unit mass(m).
v = v⁄m = 1 / m⁄v = 1⁄p
where p is the mass density. Thus, if p1 > p2, the relation between the specific volumes v1 and v2
will be represented by v1 < v2.
10. A beaker is filled with a liquid up to the mark of one litre and weighed. The weight of the liquid is found to be 6.5 N. The specific weight of the liquid will be
a) 6:5 kN = m3
b) 6:6 kN = m3
c) 6:7 kN = m3
d) 6:8 kN = m3
a) 6:5 kN = m3
b) 6:6 kN = m3
c) 6:7 kN = m3
d) 6:8 kN = m3
Answer: a
Explanation: Specific weight(γ) is defined as the weight(w) per unit volume(V ), i.e.,
γ = w⁄V
Thus, γ = 6:5 ⁄10-3 N ⁄ m3 = 6:5 kN/m3.
Explanation: Specific weight(γ) is defined as the weight(w) per unit volume(V ), i.e.,
γ = w⁄V
Thus, γ = 6:5 ⁄10-3 N ⁄ m3 = 6:5 kN/m3.
11. A beaker is filled with a liquid up to the mark of one litre and weighed. The weight of the liquid is found to be 6.5 N. The specific gravity of the liquid will be
a) 0.65
b) 0.66
c) 0.67
d) 0.68
a) 0.65
b) 0.66
c) 0.67
d) 0.68
Answer: b
Explanation: Specific gravity(S) of a liquid is defined as the ratio of the density of the liquid(pl) to that of water(pw).
Thus, S = 0:66.
Explanation: Specific gravity(S) of a liquid is defined as the ratio of the density of the liquid(pl) to that of water(pw).
Thus, S = 0:66.
12. A beaker is filled with a liquid up to the mark of one litre and weighed. The weight of the liquid is found to be 6.5 N. The specific volume of the liquid will be
a) 1 l =kg
b) 1:5 l =kg
c) 2 l =kg
d) 2:5 l =kg
a) 1 l =kg
b) 1:5 l =kg
c) 2 l =kg
d) 2:5 l =kg