Dimensions — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Dimensions MCQs with step-by-step solutions (35 questions). Part of Units and Measurements. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Dimensions · medium
If force $[F]$, acceleration $[a]$ and time $[T]$ are chosen as the fundamental physical quantities. Find the dimensions of energy.
A. $[F][a][T]$
B. $[F][a][T^2]$ ✓ Correct
C. $[F][a][T^{-1}]$
D. $[F][a^{-1}][T]$
Solution: Let $[E]=[F]^a[a]^b[T]^c$, so $[ML^2T^{-2}]=[MLT^{-2}]^a[LT^{-2}]^b[T]^c=[M^a L^{a+b} T^{-2a-2b+c}]$. Comparing powers: $a=1$; $a+b=2\Rightarrow b=1$; $-2a-2b+c=-2\Rightarrow c=2$. Hence energy $=[F][a][T^2]$.
Q2 — Dimensions · medium
If $E$ and $G$ respectively denote energy and gravitational constant, then $\dfrac{E}{G}$ has the dimensions of
A. $[M^2][L^{-1}][T^0]$ ✓ Correct
B. $[M][L^{-1}][T^{-1}]$
C. $[M][L^0][T^0]$
D. $[M^2][L^{-2}][T^{-1}]$
Solution: $[E]=[F]\cdot[d]=[MLT^{-2}][L]=[ML^2T^{-2}]$. From $F=\dfrac{GM_1M_2}{r^2}$, $G=\dfrac{Fr^2}{M_1M_2}=\dfrac{[MLT^{-2}][L^2]}{[M^2]}=[M^{-1}L^3T^{-2}]$. Thus $\dfrac{E}{G}=\dfrac{[ML^2T^{-2}]}{[M^{-1}L^3T^{-2}]}=[M^2L^{-1}T^0]$.
Q3 — Dimensions · medium
Dimensions of stress are
A. $[ML^2T^{-2}]$
B. $[ML^0T^{-2}]$
C. $[ML^{-1}T^{-2}]$ ✓ Correct
D. $[MLT^{-2}]$
Solution: Stress $=\dfrac{\text{Force}}{\text{Area}}=\dfrac{[MLT^{-2}]}{[L^2]}=[ML^{-1}T^{-2}]$.
Q4 — Dimensions · medium
A physical quantity of the dimensions of length that can be formed out of $c$, $G$ and $\dfrac{e^2}{4\pi\varepsilon_0}$ is [$c$ is velocity of light, $G$ is universal constant of gravitation and $e$ is charge]
A. $\dfrac{1}{c^2}\left[G\dfrac{e^2}{4\pi\varepsilon_0}\right]^{1/2}$ ✓ Correct
B. $c^2\left[G\dfrac{e^2}{4\pi\varepsilon_0}\right]^{1/2}$
C. $\dfrac{1}{c^2}\left[\dfrac{e^2}{G\,4\pi\varepsilon_0}\right]^{1/2}$
D. $\dfrac{1}{c}G\dfrac{e^2}{4\pi\varepsilon_0}$
Solution: As $F=\dfrac{e^2}{4\pi\varepsilon_0 r^2}\Rightarrow\dfrac{e^2}{4\pi\varepsilon_0}=r^2 F$, so $\left[\dfrac{e^2}{4\pi\varepsilon_0}\right]=[ML^3T^{-2}]$. Also $G=[M^{-1}L^3T^{-2}]$ and $\dfrac{1}{c^2}=[L^{-2}T^2]$. Checking option (a): $\dfrac{1}{c^2}\left[G\dfrac{e^2}{4\pi\varepsilon_0}\right]^{1/2}=[L^{-2}T^2][L^6T^{-4}]^{1/2}=[L^{-2}T^2][L^3T^{-2}]=[L]$.
Q5 — Dimensions · medium
If energy $(E)$, velocity $(v)$ and time $(T)$ are chosen as the fundamental quantities, the dimensional formula of surface tension will be
A. $[Ev^{-2}T^{-1}]$
B. $[Ev^{-1}T^{-2}]$
C. $[Ev^{-2}T^{-2}]$ ✓ Correct
D. $[E^{-2}v^{-1}T^{-3}]$
Solution: Surface tension $S=\dfrac{\text{Force}}{\text{Length}}=\dfrac{[MLT^{-2}]}{[L]}=[ML^0T^{-2}]$. Let $S\propto E^a v^b T^c$ with $[E]=[ML^2T^{-2}]$, $[v]=[LT^{-1}]$. Then $[ML^0T^{-2}]=[M^a L^{2a+b} T^{-2a-b+c}]$. Equating: $a=1$; $2a+b=0\Rightarrow b=-2$; $-2a-b+c=-2\Rightarrow c=-2$. So $S=[Ev^{-2}T^{-2}]$.
Q6 — Dimensions · medium
If dimensions of critical velocity $v_c$ of a liquid flowing through a tube are expressed as $[\eta^x\rho^y r^z]$, where $\eta$, $\rho$ and $r$ are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of $x$, $y$ and $z$ are given by
A. $1, -1, -1$ ✓ Correct
B. $-1, -1, 1$
C. $-1, -1, -1$
D. $1, 1, 1$
Solution: With $[\eta]=[ML^{-1}T^{-1}]$, $[\rho]=[ML^{-3}]$, $[r]=[L]$ and $v_c=[M^0LT^{-1}]$: $[M^0LT^{-1}]=[ML^{-1}T^{-1}]^x[ML^{-3}]^y[L]^z$. Comparing exponents of M, L, T: $x+y=0$, $-x-3y+z=1$, $-x=-1$. Solving gives $x=1$, $y=-1$, $z=-1$.
Q7 — Dimensions · medium
If force $(F)$, velocity $(v)$ and time $(T)$ are taken as fundamental units, then the dimensions of mass are
A. $[FvT^{-1}]$
B. $[FvT^{-2}]$
C. $[Fv^{-1}T^{-1}]$
D. $[Fv^{-1}T]$ ✓ Correct
Solution: From $F=ma=\dfrac{mv}{t}$, we get $m=\dfrac{Ft}{v}$, so $[M]=\dfrac{[F][T]}{[v]}=[Fv^{-1}T]$.
Q8 — Dimensions · medium
The dimensions of $(\mu_0 \varepsilon_0)^{-1/2}$ are
A. $[L^{1/2}T^{-1/2}]$
B. $[L^{-1}T]$
C. $[LT^{-1}]$ ✓ Correct
D. $[L^{1/2}T^{1/2}]$
Solution: $(\mu_0\varepsilon_0)^{-1/2}$ is the expression for the velocity of light, since $c=\dfrac{1}{\sqrt{\mu_0\varepsilon_0}}$. So its dimensions are $[LT^{-1}]$.
Q9 — Dimensions · medium
The dimensions of $\dfrac{1}{2}\varepsilon_0 E^2$, where $\varepsilon_0$ is permittivity of free space and $E$ is electric field, are
A. $[ML^2T^{-2}]$
B. $[ML^{-1}T^{-2}]$ ✓ Correct
C. $[ML^2T^{-1}]$
D. $[MLT^{-1}]$
Solution: $\dfrac{1}{2}\varepsilon_0 E^2$ is the energy density of an electric field. With $[\varepsilon_0]=[M^{-1}L^{-3}T^4A^2]$ and $[E]=[MLT^{-3}A^{-1}]$, $\left[\dfrac{1}{2}\varepsilon_0E^2\right]=[M^{-1}L^{-3}T^4A^2]\times[MLT^{-3}A^{-1}]^2=[ML^{-1}T^{-2}]$.
Q10 — Dimensions · medium
If the dimensions of a physical quantity are given by $[M^a L^b T^c]$, then the physical quantity will be
A. pressure if $a=1, b=-1, c=-2$ ✓ Correct
B. velocity if $a=1, b=0, c=-1$
C. acceleration if $a=1, b=1, c=-2$
D. force if $a=0, b=-1, c=-2$
Solution: Dimensions of velocity $=[M^0L^1T^{-1}]$, acceleration $=[M^0L^1T^{-2}]$, force $=[M^1L^1T^{-2}]$ and pressure $=[M^1L^{-1}T^{-2}]$. Thus $a=1, b=-1, c=-2$ corresponds to pressure.
Q11 — Dimensions · medium
Which two of the following five physical parameters have the same dimensions? (i) Energy density (ii) Refractive index (iii) Dielectric constant (iv) Young's modulus (v) Magnetic field
A. (ii) and (iv)
B. (iii) and (v)
C. (i) and (iv) ✓ Correct
D. (i) and (v)
Solution: Energy density $u=\dfrac{\text{Energy}}{\text{Volume}}=\dfrac{[ML^2T^{-2}]}{[L^3]}=[ML^{-1}T^{-2}]$. Young's modulus $=\dfrac{\text{stress}}{\text{strain}}=[ML^{-1}T^{-2}]$. Refractive index and dielectric constant are dimensionless, and magnetic field $=\dfrac{F}{qv}=[MT^{-2}A^{-1}]$. Hence (i) and (iv) have the same dimensions.
Q12 — Dimensions · medium
Dimensions of resistance in an electrical circuit, in terms of dimension of mass $M$, of length $L$, of time $T$ and of current $I$, would be
A. $[ML^2T^{-3}I^{-1}]$
B. $[ML^2T^{-2}]$
C. $[ML^2T^{-1}I^{-1}]$
D. $[ML^2T^{-3}I^{-2}]$ ✓ Correct
Solution: By Ohm's law $R=\dfrac{V}{I}=\dfrac{W}{qI}=\dfrac{\text{work}}{\text{charge}\times\text{current}}=\dfrac{[ML^2T^{-2}]}{[IT][I]}=[ML^2T^{-3}I^{-2}]$.
Q13 — Dimensions · medium
The velocity $v$ of a particle at time $t$ is given by $v=at+\dfrac{b}{t+c}$, where $a$, $b$ and $c$ are constants. The dimensions of $a$, $b$ and $c$ are respectively
A. $[LT^{-2}], [L]$ and $[T]$ ✓ Correct
B. $[L^2], [T]$ and $[LT^2]$
C. $[LT^2], [LT]$ and $[L]$
D. $[L], [LT]$ and $[T^2]$
Solution: By the principle of homogeneity $[a][t]=[v]$, so $[a]=\dfrac{[LT^{-1}]}{[T]}=[LT^{-2}]$. Also $[c]=[t]=[T]$, and $[b]=[v][t+c]=[LT^{-1}][T]=[L]$.
Q14 — Dimensions · medium
The ratio of the dimensions of Planck's constant and that of the moment of inertia is the dimension of
A. frequency ✓ Correct
B. velocity
C. angular momentum
D. time
Solution: From $E=h\nu$, $[h]=[ML^2T^{-1}]$, and moment of inertia $[I]=[ML^2]$. So $\dfrac{[h]}{[I]}=\dfrac{[ML^2T^{-1}]}{[ML^2]}=[T^{-1}]$, which is the dimension of frequency.
Q15 — Dimensions · medium
The dimensions of universal gravitational constant are
A. $[M^{-1}L^3T^{-2}]$ ✓ Correct
B. $[ML^2T^{-3}]$
C. $[M^{-2}L^2T^{-2}]$
D. $[M^{-2}L^2T^{-1}]$
Solution: From Newton's law of gravitation $F=\dfrac{Gm_1m_2}{r^2}$, so $G=\dfrac{Fr^2}{m_1m_2}=\dfrac{[MLT^{-2}][L^2]}{[M]^2}=[M^{-1}L^3T^{-2}]$.
Q16 — Dimensions · medium
Planck's constant has the dimensions of
A. linear momentum
B. angular momentum ✓ Correct
C. energy
D. power
Solution: From $E=h\nu$, $[h]=\dfrac{E}{\nu}=\dfrac{[ML^2T^{-2}]}{[T^{-1}]}=[ML^2T^{-1}]$, which is the dimension of angular momentum.
Q17 — Dimensions · medium
A pair of physical quantities having same dimensional formula is
A. force and torque
B. work and energy ✓ Correct
C. force and impulse
D. linear momentum and angular momentum
Solution: Work $=$ Force $\times$ displacement $=[MLT^{-2}][L]=[ML^2T^{-2}]$ and energy $=\dfrac{1}{2}mv^2=[M][LT^{-1}]^2=[ML^2T^{-2}]$. Thus work and energy have the same dimensional formula. (Torque is also $[ML^2T^{-2}]$ but is not energy; the intended pair is work and energy.)
Q18 — Dimensions · medium
The dimensional formula for magnetic flux is
A. $[ML^2T^{-2}A^{-1}]$ ✓ Correct
B. $[ML^3T^{-2}A^{-2}]$
C. $[M^0L^{-2}T^2A^{-2}]$
D. $[ML^2T^{-1}A^2]$
Solution: Magnetic flux $\phi=BA$. Since $F=BIl$, $B=\dfrac{F}{Il}$, so $\phi=\dfrac{F}{Il}A=\dfrac{[MLT^{-2}][L^2]}{[A][L]}=[ML^2T^{-2}A^{-1}]$.
Q19 — Dimensions · medium
The force $F$ on a sphere of radius $r$ moving in a medium with velocity $v$ is given by $F=6\pi\eta rv$. The dimensions of $\eta$ are
A. $[ML^{-3}]$
B. $[MLT^{-2}]$
C. $[MT^{-1}]$
D. $[ML^{-1}T^{-1}]$ ✓ Correct
Solution: Viscous force on a sphere of radius $r$ is $F=6\pi\eta rv\Rightarrow\eta=\dfrac{F}{6\pi rv}$. So $[\eta]=\dfrac{[F]}{[r][v]}=\dfrac{[MLT^{-2}]}{[L][LT^{-1}]}=[ML^{-1}T^{-1}]$.
Q20 — Dimensions · medium
Which of the following will have the dimensions of time?
A. $LC$
B. $\dfrac{R}{L}$
C. $\dfrac{L}{R}$ ✓ Correct
D. $\dfrac{C}{L}$
Solution: $\dfrac{L}{R}$ is the time constant of an $R$-$L$ circuit, so its dimensions are the same as that of time. Alternatively, $\dfrac{\text{Dimensions of }L}{\text{Dimensions of }R}=\dfrac{[ML^2T^{-2}A^{-2}]}{[ML^2T^{-3}A^{-2}]}=[T]$.
Q21 — Dimensions · medium
An equation is given as $\left(p+\dfrac{a}{V^2}\right)=b\dfrac{\theta}{V}$, where $p$ = pressure, $V$ = volume and $\theta$ = absolute temperature. If $a$ and $b$ are constants, then dimensions of $a$ will be
A. $[ML^5T^{-2}]$ ✓ Correct
B. $[M^{-1}L^5T^2]$
C. $[ML^{-5}T^{-1}]$
D. $[ML^5T]$
Solution: From the principle of homogeneity, dimensions of $p$ = dimensions of $\dfrac{a}{V^2}$. So $a=pV^2=[ML^{-1}T^{-2}][L^3]^2=[ML^5T^{-2}]$.
Q22 — Dimensions · medium
Which of the following is a dimensional constant?
A. Refractive index
B. Poisson's ratio
C. Relative density
D. Gravitational constant ✓ Correct
Solution: A quantity which has dimensions and also a constant value is called a dimensional constant. Refractive index, Poisson's ratio and relative density are dimensionless, while the gravitational constant $G$ has dimensions and a fixed value, so it is a dimensional constant.
Q23 — Dimensions · medium
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 $\eta=\dfrac{p(r^2-x^2)}{4vl}$ where $v$ is the velocity of oil at distance $x$ from the axis of the tube. The dimensions of $\eta$ are
A. $[M^0L^0T^0]$
B. $[MLT^{-1}]$
C. $[ML^2T^{-2}]$
D. $[ML^{-1}T^{-1}]$ ✓ Correct
Solution: Pressure $p=\dfrac{\text{Force}}{\text{Area}}=\dfrac{[MLT^{-2}]}{[L^2]}=[ML^{-1}T^{-2}]$ and velocity $v=[LT^{-1}]$. By homogeneity, $r^2$ and $x^2$ have the same dimensions, so $\eta=\dfrac{[ML^{-1}T^{-2}][L^2]}{[LT^{-1}][L]}=[ML^{-1}T^{-1}]$.
Q24 — Dimensions · medium
The time dependence of physical quantity $p$ is given by $p=p_0\exp(-\alpha t^2)$, where $\alpha$ is a constant and $t$ is the time. The constant $\alpha$
A. is dimensionless
B. has dimensions $[T^{-2}]$ ✓ Correct
C. has dimensions $[T^2]$
D. has dimensions of $p$
Solution: Since the power of an exponential is dimensionless, $\alpha t^2$ is dimensionless, i.e. $\alpha t^2=[M^0L^0T^0]$. Thus $\alpha=\dfrac{1}{t^2}=\dfrac{1}{[T^2]}=[T^{-2}]$.
Q25 — Dimensions · medium
If $p$ represents radiation pressure, $c$ represents speed of light and $S$ represents radiation energy striking unit area per sec. The non-zero integers $x,y,z$ such that $p^x S^y c^z$ is dimensionless are
A. $x=1,\;y=1,\;z=1$
B. $x=-1,\;y=1,\;z=1$
C. $x=1,\;y=-1,\;z=1$ ✓ Correct
D. $x=1,\;y=1,\;z=-1$
Solution: Radiation pressure $p=[ML^{-1}T^{-2}]$, velocity of light $c=[LT^{-1}]$, and $S=\dfrac{[ML^2T^{-2}]}{[L^2T]}=[MT^{-3}]$. For $p^x S^y c^z$ to be dimensionless, $[M^0L^0T^0]=[M]^{x+y}[L]^{-x+z}[T]^{-2x-3y-z}$, giving $x+y=0$, $-x+z=0$, $-2x-3y-z=0$. Solving, $x=1,\;y=-1,\;z=1$.
Q26 — Dimensions · medium
The dimensional formula for permeability of free space, $\mu_0$ is
A. $[MLT^{-2}A^{-2}]$ ✓ Correct
B. $[ML^{-1}T^2A^{-2}]$
C. $[ML^{-1}T^{-2}A^2]$
D. $[MLT^{-2}A^{-1}]$
Solution: From the Biot-Savart law, $dB=\dfrac{\mu_0}{4\pi}\dfrac{Idl\sin\theta}{r^2}$, so $\mu_0=\dfrac{4\pi r^2(dB)}{Idl\sin\theta}=\dfrac{[L^2][MT^{-2}A^{-1}]}{[A][L]}=[MLT^{-2}A^{-2}]$.
Q27 — Dimensions · medium
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. $x=\dfrac{1}{2},\;y=\dfrac{1}{2}$
B. $x=-\dfrac{1}{2},\;y=-\dfrac{1}{2}$
C. $x=\dfrac{1}{2},\;y=-\dfrac{1}{2}$
D. $x=-\dfrac{1}{2},\;y=\dfrac{1}{2}$ ✓ Correct
Solution: Since $f=Cm^x k^y$, $[T^{-1}]=[M]^x[MT^{-2}]^y$ (with $k=\dfrac{\text{force}}{\text{length}}$). By homogeneity, $x+y=0$ and $-2y=-1$, giving $y=\dfrac{1}{2}$ and $x=-\dfrac{1}{2}$.
Q28 — Dimensions · medium
According to Newton, the viscous force acting between liquid layers of area $A$ and velocity gradient $\dfrac{\Delta v}{\Delta z}$ is given by $F=-\eta A\dfrac{dv}{dz}$, where $\eta$ is a constant called
A. $[ML^{-2}T^{-2}]$
B. $[M^0L^0T^0]$
C. $[ML^2T^{-2}]$
D. $[ML^{-1}T^{-1}]$ ✓ Correct
Solution: As $F=-\eta A\dfrac{dv}{dz}\Rightarrow\eta=-\dfrac{F}{A\left(\dfrac{dv}{dz}\right)}$. With $F=[MLT^{-2}]$, $A=[L^2]$, $dv=[LT^{-1}]$, $dz=[L]$, $\eta=\dfrac{[MLT^{-2}][L]}{[L^2][LT^{-1}]}=[ML^{-1}T^{-1}]$.
Q29 — Dimensions · medium
The dimensional formula of pressure is
A. $[MLT^{-2}]$
B. $[ML^{-1}T^2]$
C. $[ML^{-1}T^{-2}]$ ✓ Correct
D. $[MLT^{-2}]$
Solution: Pressure $=\dfrac{\text{Force}}{\text{Area}}=\dfrac{F}{A}=\dfrac{[MLT^{-2}]}{[L^2]}=[ML^{-1}T^{-2}]$.
Q30 — Dimensions · medium
The dimensional formula of torque is
A. $[ML^2T^{-2}]$ ✓ Correct
B. $[MLT^{-2}]$
C. $[ML^{-1}T^{-2}]$
D. $[ML^{-2}T^{-2}]$
Solution: Torque $\tau=\mathbf{r}\times\mathbf{F}$, so dimensions of $\tau=[L][MLT^{-2}]=[ML^2T^{-2}]$.