MCQ on Unit and Dimension chapter for different exams like JELET, NEET, JEE, CUET, board exams and first year engineering physics
1. The dimensional formula of pressure is
a. [MLT-2]
b. [ML-1T2]
c. [ML-1T-2]
d. [MLT2]
option c
Pressure=\frac{Force}{Area} \\
Pressure=\frac{Mass\times Acceleration}{Area} \\
[Pressure]=\frac{[Mass]\times [Acceleration]}{[Area]} \\
[Pressure]=\frac{[M]\times [LT^{-2}]}{[L^{2}]}\\
The dimension of pressure is
[Pressure]=[ML^{-1}T^{-2}]
2. The dimensional formula for angular momentum is
a. [M0L-2T-2]
b. [ML2T-1]
c. [MLT-1]
d. [ML2T-2]
option b
Angular momentum,L = mvr
[L] = [m][v][r]
[L] = [M][LT-1][L]
[L] = [ML2T-1]
3. The dimensional formula of torque is
a. [ML2T-2]
b. [MLT-2]
c. [ML-1T-2]
d. [ML-2T-2]
option a
Torque = Force \times {Distance}
[Torque] = [Force] \times [{Distance}]
[Torque] = [MLT^{-2}] \times [{L}]
[Torque] = [ML^{2}T^{-2}]
4. The dimension of universal gravitation constant is
a. [M-2L2T-1]
b. [M-1L3T-2]
c. [ML2T-1]
d. [M-2L3T-2]
option b
The gravitational force, F is given by
F=G\frac{m_{1}m_{2}}{r^{2}}
G=F\frac{r^{2}}{m_{1}m_{2}}
The dimension of gravitational constant is
[G]=[F]\frac{[r^{2}]}{[m_{1}][m_{2}]}
[G]=[MLT^{-2}]\frac{[L^{2}]}{[M][M]}
[G]=[M^{-1}L^{3}T^{-2}]
So, option (b) is the correct dimension of gravitational constant.
5. According to Newton, the viscous force acting between liquid layers of area A and velocity gradient \frac{\Delta V}{\Delta Z} is given by F=-\eta A\frac{\Delta V}{\Delta Z} where \eta is constant called coefficient of viscosity, the dimensional formula of coefficient of viscosity \eta is
a. [ML-2T-2]
b. [M0L0T0]
c. [ML2T-2]
d. [ML-1T-1]
option d
To find the dimension of coefficient of viscosity, write the equation
F=-\eta A\frac{\Delta V}{\Delta Z} where \eta is the coefficient of viscosity
The dimension of coefficient of viscosity can be found from the above equation
[\eta ] = \frac{[F]}{[A]} \frac{[\Delta Z]}{[\Delta V]}
[\eta ] = \frac{[MLT^{-2}]}{[L^{2}]} \frac{[L]}{[LT^{-1}]}
So, the dimension of coefficient of viscosity is
[\eta ] = [ML^{-1}T^{-1}]
The right answer is option d
6. The frequency of vibration f of a mass m suspended from a spring of spring constant k is given by a relation of the type f=cm^{x}k^{y} where c is a dimensionless constant. The values of x and y are
a. x=\frac{1}{2}, y=\frac{1}{2}
b. x=-\frac{1}{2}, y=-\frac{1}{2}
c. x=\frac{1}{2}, y=-\frac{1}{2}
d. x=-\frac{1}{2}, y=\frac{1}{2}
option d
7. Turpentine oil is flowing through a tube of length l and radius r. The pressure difference between the two ends of the tube is P. The viscosity of oil is given by \eta =P \frac{(r^{2}-x^{2})}{4vl} , where v is the velocity of oil at a distance x from the axis of the tube. The dimension of coefficient of viscosity \eta is
a. [M0L0T0]
b. [MLT-1]
c. [ML2T-2]
d. [ML-1T-1]
option d
The given equation is
\eta =P \frac{(r^{2}-x^{2})}{4vl}
[\eta] =[P] \frac{[(r^{2}-x^{2})]}{4[v][l]}
The dimension of pressure is [P] = [ML-1T-2]
The dimension of [(r^{2}-x^{2})] = [L^{2}]
Dimension of velocity is [v]=[LT^{-1}]
The dimension of length is, [l]=[L]
So, the dimension of coefficient of viscosity is
[\eta] =[ML^{-1}T^{-2}] \frac{[L^{2}])}{[LT^{-1}][L]}
[\eta] =[ML^{-1}T^{-1}]
8. The velocity v of a particle at time t is given by v =at+\frac{b}{t+c} , where a, b and c are constant. The dimensions of a, b and c are respectively
a. L^{2},T and LT^{2}
b. LT^{2},LT and L
c. L,LT and T^{2}
d. LT^{-2},L and T
option d
9. The dimension of (\mu_{0}\epsilon _{0})^{-\frac{1}{2}} is
a. L^{\frac{1}{2}}T^{-\frac{1}{2}}
b. L^{-1}T
c. LT^{-1}
d. L^{\frac{1}{2}}T^{\frac{1}{2}}
option c
The velocity of light in free space or vacuum is c = (\mu_{0}\epsilon _{0})^{-\frac{1}{2}}
So, the dimension of velocity = the dimension of (\mu_{0}\epsilon _{0})^{-\frac{1}{2}} = LT^{-1}
10. If x = at+bt^{2} , where x is the distance travelled by the body in kilometers while t is the time in seconds, then the unit of b is
a. Km/s
b. kms
c. km/s2
d.kms2
option d
x = at+bt^{2}
Now the dimension of [x]= [at] = [bt^{2}]
[x] = [bt^{2}]
[x] = [b][t^{2}]
[b]=[\frac{x}{t^{2}}]
So, the unit of b is = [\frac{unit of x}{(unit of time)^{2}}]
The unit of b is = [\frac{km}{(s)^{2}}]
So, option C is the right answer.
11. A force is given by F = at+bt^{2} , where t is the time. The dimension of a and b are
a. MLT^{-4} and MLT
b. MLT^{-1} and MLT^{0}
c. MLT^{-3} and MLT^{-4}
d. MLT^{-3} and MLT^{-0}
option c
The dimension of [F]= [at] = [bt^{2}]
[F]=[at]
[a]=\frac{[F]}{[t]}
[a]=\frac{[MLT^{-2}]}{[T]}
[a]=MLT^{-3}
Again, [F]= [bt^{2}]
[b]=[\frac{F}{t^{2}}]
[b]=[\frac{MLT^{-2}}{T^{2}}]
[b]=MLT^{-4}
12. In the relation p=\frac{\alpha }{\beta }e^{-\frac{\alpha z}{k\theta }} , p is pressure, z is distance, k is Boltzmann constant and θ is the temperature. The dimensional formula of β will be
a. M^{0}L^{2}T^{0}
b. ML^{3}T
c. ML^{0}T^{-1}
d. M^{0}L^{2}T^{-1}
option a
13. \frac{\alpha }{t^{2} } =Fv+\frac{\beta }{v^{2}} , Find dimension formula for \alpha and \beta
a. ML^{3}T^{-1} and ML^{4}T^{-2}
b. ML^{2}T^{-2} and ML^{4}T^{-3}
c. ML^{2}T^{-1} and ML^{4}T^{-3}
d. ML^{3}T^{-2} and ML^{4}T^{-2}
option c
14. For one mole of gas, Van der waals’ equation is (P+\frac{a}{v^{2}})(V-b)=RT , Find the dimensions of a and b.
a. ML^{5}T^{-2} and M^{0}L^{2}T^{0}
b. ML^{4}T^{-2} and M^{0}L^{3}T^{0}
c. ML^{4}T^{-2} and ML^{2}T^{-2}
d. ML^{5}T^{-2} and M^{0}L^{3}T^{0}
option d
(P+\frac{a}{v^{2}})(V-b)=RT
So, [P]=[\frac{a}{v^{2}}] and [V]=[b]
[a]=[PV^{2}]
[a]=[ML^{-1}T^{-2}][(L^{3})^{2}]
[a]=[ML^{5}T^{-2}]
Again, [b]=[V]
[b]=[M^{0}L^{3}T^{0}]
15. The equation of a wave is given by y=aSin\omega (\frac{x}{v}-k) where \omega is angular velocity and v is the linear velocity. The dimensions of k will be
a. T^{2}
b. T^{-1}
c. T
d. LT
option c
[k]=[\frac{x}{v}]
[k]=[\frac{L}{LT^{-1}}]
[k]=[T]
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