JEE PYQs Units and Measurements (Set-3) Class 11 Physics provides a carefully selected set of previous year solved questions to strengthen your understanding of this fundamental chapter. These questions focus on key concepts such as dimensional analysis, significant figures, and error calculation. Practicing this set will help improve accuracy, enhance problem-solving speed, and give you better insight into JEE Main exam patterns. Practice more with Units and Measurements JEE Main PYQs Set-1 and JEE Main PYQs Set-2 to strengthen your concepts and improve problem-solving accuracy.
Units and Measurements – JEE Main PYQs Previous Year Solved Questions
Problem : Dimensional formula for thermal conductivity is (Here, $K$ denotes the temperature)
(a) $[MLT^{-2}K]$
(b) $[MLT^{-2}K^{-2}]$
(c) $[MLT^{-3}K^{-1}]$
(d) $[MLT^{-3}K]$
[2020, 4 Sep Shift-I]
Solution: Ans. (c)
For conduction of heat,
$$\frac{dQ}{dt} = KA\frac{dT}{dx}$$
$$K = \frac{\left(\frac{dQ}{dt}\right)}{A\left(\frac{dT}{dx}\right)} = \frac{dQ \times dx}{A \times dt \times dT}$$
$$= \frac{\text{joule} \times \text{metre}}{(\text{metre})^2 \times \text{second} \times \text{kelvin}}$$
$$= \frac{\frac{\text{kilogram} \times (\text{metre})^2}{(\text{second})^2} \times \text{metre}}{(\text{metre})^2 \times \text{second} \times \text{kelvin}}$$
$$= \frac{\text{kg} \cdot \text{m}^2 \cdot \text{m}}{\text{s}^2 \cdot \text{m}^2 \cdot \text{s} \cdot \text{K}} = \frac{\text{kg} \cdot \text{m}}{\text{s}^3 \cdot \text{K}}$$
$$[K] = \frac{[M^1][L^1]}{[T^3][K^1]}$$
$$\Rightarrow [K] = [M^1L^1T^{-3}K^{-1}]$$
Hence, correct option is (c).
Practice JEE Physics PYQs to understand real exam patterns and improve accuracy.
Problem : A quantity $x$ is given by $x = \dfrac{IFv^2}{WL^4}$ where $I$ is moment of inertia, $F$ is force, $v$ is velocity, $W$ is work and $L$ is length. The dimensional formula for $x$ is same as that of
(a) Planck’s constant
(b) force constant
(c) coefficient of viscosity
(d) energy density
[2020, 4 Sep Shift-II]
Solution: Ans. (d)
Given that, $x = \dfrac{IFv^2}{WL^4}$
Dimensionally,
$$[x] = \frac{[I][F][v]^2}{[W][L]^4}$$
$$= \frac{[M^1L^2][M^1L^1T^{-2}][L^1T^{-1}]^2}{[M^1L^2T^{-2}][L^1]^4}$$
$$= \frac{[M^1L^2][M^1L^1T^{-2}][L^2T^{-2}]}{[M^1L^2T^{-2}][L^4]}$$
$$= [M^1L^{-1}T^{-2}] \quad \dots(i)$$
On checking the alternatives:
(a) Planck’s constant $\Rightarrow [h] = [M^1L^2T^{-1}]$
doesn’t match with dimensional formula of $x$.
(b) Force constant $\Rightarrow [K] = [M^1T^{-2}]$
doesn’t match with dimensional formula of $x$.
(c) Coefficient of viscosity $\Rightarrow [\eta] = [M^1L^{-1}T^{-1}]$
doesn’t match with dimensional formula of $x$.
(d) Energy density $\Rightarrow [E_d] = [M^1L^{-1}T^{-2}]$
matches with dimensional formula of $x$.
Hence, option (d) is correct.
Problem : The quantities $x = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}}$, $y = \dfrac{E}{B}$ and $z = \dfrac{l}{CR}$ are defined, where $C$ is capacitance, $R$ is resistance, $l$ is length, $E$ is electric field, $B$ is magnetic field, $\varepsilon_0$ is free space permittivity and $\mu_0$ is permeability, respectively. Then,
(a) $x$, $y$ and $z$ have the same dimension
(b) Only $x$ and $z$ have the same dimension
(c) Only $x$ and $y$ have the same dimension
(d) Only $y$ and $z$ have the same dimension
[2020, 5 Sep Shift-II]
Solution: Ans. (a)
$x = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}} = \text{speed of light in vacuum}$
$\therefore \text{Dimension of } x, [x] = [M^0 L^1 T^{-1}]$
$y = \dfrac{E}{B} = \text{speed of EM wave}$
$\therefore \text{Dimension of } y, [y] = [M^0 L^1 T^{-1}]$
$z = \dfrac{l}{RC} = \dfrac{l}{\tau} = \frac{\text{length}}{\text{time}}$
$\therefore \text{Dimension of } z, [z] = [M^0 L^1 T^{-1}]$
Thus, all quantities have same dimensions i.e., of velocity.
Hence, correct option is (a).
Problem : The dimension of $\dfrac{B^2}{2\mu_0}$, where $B$ is magnetic field and $\mu_0$ is the magnetic permeability of vacuum, is
(a) $[ML^{-1}T^{-2}]$
(b) $[MLT^{-2}]$
(c) $[ML^2T^{-1}]$
(d) $[ML^2T^{-2}]$
[2020, 7 Jan Shift-II]
Solution: Ans. (a)
As, $\dfrac{B^2}{2\mu_0} = \text{energy density of magnetic field}$
$$\dfrac{B^2}{2\mu_0} = \dfrac{\text{Energy}}{\text{Volume}}$$
So, $\dfrac{B^2}{2\mu_0} = [\text{Energy} / \text{Volume}]$
$$\dfrac{B^2}{2\mu_0} = \dfrac{[ML^2T^{-2}]}{[L^3]}$$
$$\dfrac{B^2}{2\mu_0} = [ML^{-1}T^{-2}]$$
Hence, correct option is (a).
Problem : The dimension of stopping potential $V_0$ in photoelectric effect in units of Planck’s constant $h$, speed of light $c$ and gravitational constant $G$ and ampere $A$ is
(a) $h^{-2/3} c^{-1/3} G^{4/3} A^{-1}$
(b) $h^{1/3} G^{2/3} c^{1/3} A^{-1}$
(c) $h^2 G^{3/2} c^{1/3} A^{-1}$
(d) $h^{2/3} c^{5/3} G^{1/3} A^{-1}$
[2020, 8 Jan Shift-I]
Solution: Ans. (*)
Let $V_0 = (h)^a \cdot (c)^b \cdot (G)^c \cdot (A)^d \quad \dots(i)$
Then, $[V_0] = [\text{potential}]$
$$[V_0] = \left[ \frac{\text{potential energy}}{\text{charge}} \right]$$
$$[V_0] = \frac{[ML^2T^{-2}]}{[AT]} = [ML^2T^{-3}A^{-1}]$$
$[h] = \left[ \frac{\text{Energy}}{\text{Frequency}} \right] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$
$[c] = [\text{Speed}] = [LT^{-1}]$
$[G] = \left[ \frac{\text{Force} \times (\text{Distance})^2}{(\text{Mass})^2} \right]$
$$[G] = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$$
Substituting the dimensions of $V_0, h, c, G$ and $A$ in Eq. (i) and equating dimensions on both sides, we get:
$[ML^2T^{-3}A^{-1}] = [ML^2T^{-1}]^a \times [LT^{-1}]^b \times [M^{-1}L^3T^{-2}]^c \times [A]^d$
$$\Rightarrow a – c = 1 \quad \dots(ii)$$
$$2a + b + 3c = 2 \quad \dots(iii)$$
$$-a – b – 2c = -3 \quad \dots(iv)$$
$$d = -1 \quad \dots(v)$$
On solving above equations, we get:
$a = 0, b = 5, c = -1, d = -1$
Substituting these values in Eq. (i), we get:
$$V_0 = h^0 \cdot c^5 \cdot G^{-1} \cdot A^{-1}$$
None of the given options matches with the result.
Problem : A quantity $f$ is given by $f = \sqrt{hc^5 / G}$, where $c$ is speed of light, $G$ universal gravitational constant and $h$ is the Planck’s constant. Dimension of $f$ is that of
(a) area
(b) volume
(c) momentum
(d) energy
[2020, 9 Jan Shift-I]
Solution: Ans. (d)
Dimensions of quantity $f$ are
$$[f] = \frac{[h]^{1/2} [c]^{5/2}}{[G]^{1/2}} \quad \dots(i)$$
[Note: To produce dimensions of different constants, just remember/recall nearest formulae containing these constants.]
As, $h = \frac{E}{\nu}; [h] = [ML^2T^{-2}][T] = [ML^2T^{-1}]$
$c = [LT^{-1}]$ and $G = \frac{F \cdot r^2}{m^2}$
$$\Rightarrow [G] = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$$
So, dimensions of $f$ using Eq. (i),
$$[f] = \frac{[ML^2T^{-1}]^{\frac{1}{2}} [LT^{-1}]^{\frac{5}{2}}}{[M^{-1}L^3T^{-2}]^{\frac{1}{2}}}$$
$$[f] = \left[ M^{\frac{1}{2} + \frac{1}{2}} L^{\frac{2}{2} + \frac{5}{2} – \frac{3}{2}} T^{-\frac{1}{2} – \frac{5}{2} + \frac{2}{2}} \right]$$
$$[f] = [ML^2T^{-2}]$$
Thus, it is the dimensions of energy.
Hence, option (d) is correct.
Problem : In SI units, the dimensions of $\sqrt{\dfrac{\varepsilon_0}{\mu_0}}$ is
(a) $[A^{-1}TML^3]$
(b) $[AT^2M^{-1}L^{-1}]$
(c) $[AT^{-3}ML^{3/2}]$
(d) $[A^2T^3M^{-1}L^{-2}]$
[2019, 8 April Shift-I]
Solution: Ans.(d)
Dimensions of $\varepsilon_0$ (permittivity of free space) are
$$[\varepsilon_0] = [M^{-1}L^{-3}T^4A^2]$$
As, $c = \text{speed of light}$.
$$\therefore \text{Dimension of } [c] = [LT^{-1}]$$
So, dimensions of $\sqrt{\dfrac{\varepsilon_0}{\mu_0}}$ are
$$\left[\sqrt{\dfrac{\varepsilon_0}{\mu_0}}\right] = \left[\sqrt{\dfrac{\varepsilon_0^2}{\varepsilon_0\mu_0}}\right] = [\varepsilon_0c] \quad \left[\because c^2 = \dfrac{1}{\mu_0\varepsilon_0}\right]$$
$$\left[\sqrt{\dfrac{\varepsilon_0}{\mu_0}}\right] = [M^{-1}L^{-3}T^4A^2][LT^{-1}]$$
$$\left[\sqrt{\dfrac{\varepsilon_0}{\mu_0}}\right] = [M^{-1}L^{-2}T^3A^2]$$
Hence, correct option is (d).
Problem : If surface tension ($S$), moment of inertia ($I$) and Planck’s constant ($h$), were to be taken as the fundamental units, the dimensional formula for linear momentum would be
(a) $S^{1/2} I^{1/2} h^{-1}$
(b) $S^{3/2} I^{1/2} h^0$
(c) $S^{1/2} I^{1/2} h^0$
(d) $S^{1/2} I^{3/2} h^{-1}$
[2019, 8 April Shift-II]
Solution: Ans. (c)
Suppose, linear momentum ($p$) depends upon the Planck’s constant ($h$) raised to the power ($a$), surface tension ($S$) raised to the power ($b$) and moment of inertia ($I$) raised to the power ($c$).
Then, $p \propto (h)^a (S)^b (I)^c$ or $p = k h^a S^b I^c$
where, $k$ is a dimensionless proportionality constant.
Thus, $[p] = [h]^a [S]^b [I]^c \quad \dots(i)$
Then, the respective dimensions of the given physical quantities, i.e.
$[p] = [\text{mass} \times \text{velocity}] = [MLT^{-1}]$
$[I] = [\text{mass} \times \text{distance}^2] = [ML^2T^0]$
$[S] = [\frac{\text{force}}{\text{length}}] = [MT^{-2}]$
$[h] = [ML^2T^{-1}]$
Then, substituting these dimensions in Eq. (i), we get
$[MLT^{-1}] = [ML^2T^{-1}]^a [MT^{-2}]^b [ML^2]^c$
For dimensional balance, the dimensions on both sides should be same. Thus, equating dimensions, we have:
$$a + b + c = 1$$
$$2(a + c) = 1 \quad \text{or} \quad a + c = \frac{1}{2}$$
$$-a – 2b = -1 \quad \text{or} \quad a + 2b = 1$$
Solving these three equations, we get
$$a = 0, b = \frac{1}{2}, c = \frac{1}{2}$$
$$\therefore p = h^0 S^{1/2} I^{1/2} \quad \text{or} \quad p = S^{1/2} I^{1/2} h^0$$
Hence, correct option is (c).
Problem : In the formula $X = 5YZ^2$, $X$ and $Z$ have dimensions of capacitance and magnetic field, respectively. What are the dimensions of $Y$ in SI units?
(a) $[M^{-1}L^{-2} T^4A^2]$
(b) $[M^{-2}L^{0} T^{-4}A^{-2}]$
(c) $[M^{-3}L^{-2} T^8A^4]$
(d) $[M^{-2}L^{-2} T^6A^3]$
[2019, 10 April Shift-II]
Solution: Ans. (c)
To find dimensions of capacitance in the given relation, we can use the formula for energy.
Capacitor’s energy is $U = \frac{1}{2}CV^2$
So, dimensionally,
$\Rightarrow [C] = \left[ \dfrac{U}{V^2} \right]$
As, $V = \text{potential} = \dfrac{\text{potential energy}}{\text{charge}}$
We have,
$[C] = \dfrac{[U]}{\left[ \dfrac{U^2}{q^2} \right]} = \dfrac{[q^2]}{[U]} = \left[ \dfrac{A^2T^2}{ML^2T^{-2}} \right]$
$\therefore X = [M^{-1}L^{-2}A^2T^4]$
To get dimensions of magnetic field, we use force on a current carrying conductor in magnetic field,
$F = BIl \Rightarrow [B] = \dfrac{[F]}{[I][l]} = \left[ \dfrac{MLT^{-2}}{AL} \right]$
$\therefore Z = [M^1L^0T^{-2}A^{-1}]$
Now, using given relation,
$X = 5YZ^2$
$[Y] = \dfrac{[X]}{[Z^2]} = \left[ \dfrac{M^{-1}L^{-2}A^2T^4}{(M^1L^0T^{-2}A^{-1})^2} \right]$
$ [Y] = \dfrac{M^{-1}L^{-2}A^2T^4}{M^2T^{-4}A^{-2}}$
$\therefore [Y] = [M^{-3}L^{-2}T^8A^4]$
Problem : Which of the following combinations has the dimension of electrical resistance ($\varepsilon_0$ is the permittivity of vacuum and $\mu_0$ is the permeability of vacuum)?
(a) $\sqrt{\dfrac{\mu_0}{\varepsilon_0}}$
(b) $\dfrac{\mu_0}{\varepsilon_0}$
(c) $\sqrt{\dfrac{\varepsilon_0}{\mu_0}}$
(d) $\dfrac{\varepsilon_0}{\mu_0}$
[2019, 12 April Shift-I]
Solution: Ans. (a)
A formula is valid only, if the dimensions of LHS and RHS are same. So, we need to balance dimensions of given options with the dimension of electrical resistance.
Let dimensions of resistance $R$, permittivity $\varepsilon_0$ and permeability $\mu_0$ are $[R], [\varepsilon_0]$ and $[\mu_0]$, respectively.
So, $[R] = [\varepsilon_0]^\alpha [\mu_0]^\beta \dots(i)$
$[R] = [M^1L^2T^{-3}A^{-2}]$,
$[\varepsilon_0] = [M^{-1}L^{-3}T^4A^2]$,
$[\mu_0] = [M^1L^1T^{-2}A^{-2}]$
Now, from Eq. (i), we get
$[M^1L^2T^{-3}A^{-2}] = [M^{-1}L^{-3}T^4A^2]^\alpha [M^1L^1T^{-2}A^{-2}]^\beta$
$[M^1L^2T^{-3}A^{-2}] = M^{-\alpha+\beta} L^{-3\alpha+\beta} T^{4\alpha-2\beta} A^{2\alpha-2\beta}$
On comparing both sides, we get
$-\alpha + \beta = 1 \dots(ii)$
$-3\alpha + \beta = 2 \dots(iii)$
$4\alpha – 2\beta = -3 \dots(iv)$
$2\alpha – 2\beta = -2 \dots(v)$
Value of $\alpha$ and $\beta$ can be found using any two Eqs. from (ii) to (v).
On subtracting Eq. (iii) from Eq. (ii), we get
$(-\alpha + \beta) – (-3\alpha + \beta) = 1 – 2$
$\Rightarrow 2\alpha = -1$ or $\alpha = \frac{-1}{2}$
Put the value of $\alpha$ in Eq. (ii), we get
$\beta = + \frac{1}{2}$
$\therefore [R] = [\varepsilon_0]^{-1/2} [\mu_0]^{1/2} = \sqrt{\dfrac{\mu_0}{\varepsilon_0}}$
Problem : In form of $G$(universal gravitational constant), $h$(Planck constant) and $c$ (speed of light), the time period will be proportional to
(a) $\sqrt{\frac{Gh}{c^5}}$
(b) $\sqrt{\frac{hc^5}{G}}$
(c) $\sqrt{\frac{c^3}{Gh}}$
(d) $\sqrt{\frac{Gh}{c^3}}$
[2019, 9 Jan Shift-II]
Solution: Ans. (a)
According to dimensional analysis, if a physical quantity (let $y$) depends on another physical quantities (let $A, B, C$), then the dimension formula is given by
$$y = A^a B^b C^c$$
(where, $a, b, c$ are the power of physical quantity and $k = \text{constant}$)
For the given question, the time($t$) depends on the $G$ (gravitational constant), $h$ (Planck’s constant) and $c$ (velocity of light), then according to dimensional analysis
$$t = k(G)^a (h)^b (c)^c \quad \dots(i)$$
For calculating the values of $a, b$ and $c$, compare the dimensional formula for both side.
LHS $t = \text{time} = [M^0 L^0 T^1]$
RHS $G = \dfrac{F \cdot r^2}{m_1 m_2}$
$G = \dfrac{\text{kg ms}^{-2} \times \text{m}^2}{(\text{kg})^2} = [M^{-1} L^3 T^{-2}]^a$
$h = [M^1 L^2 T^{-1}]^b$
$$c = \frac{d}{t} = [M^0 L^1 T^{-1}]^c$$
Compare both side for powers of M, L and T,
$$M \Rightarrow 0 = -a + b \quad \dots(i)$$
$$L \Rightarrow 0 = 3a + 2b + c \quad \dots(ii)$$
$$T \Rightarrow 1 = -2a – b – c \quad \dots(iii)$$
Solving Eqs. (i), (ii), (iii), we get
$$a = \frac{1}{2}, b = \frac{1}{2} \text{ and } c = \frac{-5}{2}$$
So, put these values in Eq. (i), we get
$$t = k G^{1/2} h^{1/2} c^{-5/2}$$
$$t = k \sqrt{\frac{Gh}{c^5}}$$
Problem : The force of interaction between two atoms is given by
$F = \alpha \beta \exp \left( -\frac{x^2}{\alpha kT} \right)$; where $x$ is the distance, $k$ is the Boltzmann constant and $T$ is temperature and $\alpha$ and $\beta$ are two constants. The dimension of $\beta$ is
(a) $[MLT^{-2}]$
(b) $[M^0L^2T^{-4}]$
(c) $[M^2LT^{-4}]$
(d) $[M^2L^2T^{-2}]$
[2019, 11 Jan Shift-I]
Solution: Ans. (c)
Force of interaction between two atoms is given as
$$F = \alpha \beta \exp(-x^2 / \alpha kT)$$
As we know, exponential terms are always dimensionless, so
$$\text{dimensions of } \left( \dfrac{-x^2}{\alpha kT} \right) = [M^0 L^0 T^0]$$
$\Rightarrow \text{Dimensions of } \alpha = \text{Dimension of } (x^2 / kT)$
Now, substituting the dimensions of individual term in the given equation, we get
$\text{Dimensions of } \alpha = \dfrac{[M^0 L^2 T^0]}{[M^1 L^2 T^{-2}]}$ $\{\because \text{Dimensions of } kT$ equivalent to the dimensions of energy $= [M^1 L^2 T^{-2}] \} $
$\text{Dimensions of } \alpha = [M^{-1} L^0 T^2] \quad \dots(i)$
Now from given equation, we have dimensions of
$$F = \text{dimensions of } \alpha \times \text{dimensions of } \beta$$
$\Rightarrow \text{Dimensions of } \beta = \text{Dimensions of } \left( \frac{F}{\alpha} \right)$
$$\text{Dimensions of } \beta = \dfrac{[M^1 L^1 T^{-2}]}{[M^{-1} L^0 T^2]} \quad [\because \text{using Eq. (i)}]$$
$$\text{Dimensions of } \beta = [M^2 L^1 T^{-4}]$$
Problem : If speed ($v$), acceleration ($A$) and force ($F$) are considered as fundamental units, the dimension of Young’s modulus will be
(a) $[v^{-4} A^{-2} F]$
(b) $[v^{-2} A^2 F^2]$
(c) $[v^{-2} A^2 F^{-2}]$
(d) $[v^{-4} A^2 F]$
[2019, 11 Jan Shift-II]
Solution: Ans. (d)
Dimensions of speed are, $[v] = [LT^{-1}]$
Dimensions of acceleration are, $[A] = [LT^{-2}]$
Dimensions of force are, $[F] = [MLT^{-2}]$
Dimension of Young modulus is, $[Y] = [ML^{-1} T^{-2}]$
Let dimensions of Young’s modulus is expressed in terms of speed, acceleration and force as;
$$[Y] = [v]^\alpha [A]^\beta [F]^\gamma \quad \dots(i)$$
Then substituting dimensions in terms of $M, L$ and $T$ we get,
$$[ML^{-1} T^{-2}] = [LT^{-1}]^\alpha [LT^{-2}]^\beta [MLT^{-2}]^\gamma$$
$$[ML^{-1} T^{-2}] = [M^\gamma L^{\alpha + \beta + \gamma} T^{-\alpha – 2\beta – 2\gamma}]$$
Now comparing powers of basic quantities on both sides we get,
$\quad \gamma = 1$
$\alpha + \beta + \gamma = -1$
and $-\alpha – 2\beta – 2\gamma = -2$
Solving these, we get
$$\alpha = -4, \beta = 2, \gamma = 1$$
Substituting $\alpha, \beta$, and $\gamma$ in Eq. (i) we get;
$$[Y] = [v^{-4} A^2 F^1]$$
Problem : Let $l, r, c$, and $v$ represent inductance, resistance, capacitance and voltage, respectively. The dimension of $\dfrac{l}{rcv}$ in SI units will be
(a) $[LT^2]$
(b) $[LTA]$
(c) $[A^{-1}]$
(d) $[LA^{-2}]$
[2019, 12 Jan Shift-II]
Solution : Ans. (c)
Dimensions of given quantities are
$l = \text{inductance} = [M^1 L^2 T^{-2} A^{-2}]$
$r = \text{resistance} = [M^1 L^2 T^{-3} A^{-2}]$
$c = \text{capacitance} = [M^{-1} L^{-2} T^4 A^2]$
$v = \text{voltage} = [M^1 L^2 T^{-3} A^{-1}]$
So, dimensions of $\dfrac{l}{rcv}$ are
$$\left[ \dfrac{l}{rcv} \right] = \dfrac{[ML^2 T^{-2} A^{-2}]}{[M^1 L^2 T^{-2} A^{-1}]} = [A^{-1}]$$
Problem : Let $[\varepsilon_0]$ denotes the dimensional formula of the permittivity of vacuum. If $M = \text{mass}, L = \text{length}, T = \text{time and } A = \text{electric current}$, then
(a) $[\varepsilon_0] = [M^{-1} L^{-3} T^2 A]$
(b) $[\varepsilon_0] = [M^{-1} L^{-3} T^4 A^2]$
(c) $[\varepsilon_0] = [M^{-2} L^2 T^{-1} A^{-2}]$
(d) $[\varepsilon_0] = [M^{-1} L^2 T^{-1} A^2]$
[JEE Main 2013]
Solution : Ans. (b)
Electrostatic force between two charges,
$$F = \dfrac{1}{4\pi\varepsilon_0} \dfrac{q_1q_2}{R^2} \Rightarrow \varepsilon_0 = \dfrac{q_1q_2}{4\pi FR^2}$$
Substituting the units.
Hence, $\varepsilon_0 = \dfrac{C^2}{N \text{-} m^2} = \dfrac{[AT]^2}{[MLT^{-2}][L^2]}$
$$\varepsilon_0 = [M^{-1} L^{-3} T^4 A^2]$$
Problem : The dimensions of magnetic field in $M, L, T$ and $C$ (coulomb) is given as
(a) $[MLT^{-1} C^{-1}]$
(b) $[MT^2 C^{-2}]$
(c) $[MT^{-1} C^{-1}]$
(d) $[MT^2 C^{-1}]$
[AIEEE 2008]
Solution : Ans. (c)
From the relation $F = qvB$
$\Rightarrow [MLT^{-2}] = [C][LT^{-1}][B]$
$\Rightarrow [B] = [MC^{-1} T^{-1}]$
Problem : Which of the following units denotes the dimensions $[ML^2 / Q^2]$, where $Q$ denotes the electric charge?
(a) $Wb/m^2$
(b) $henry (H)$
(c) $H/m^2$
(d) $weber (Wb)$
[AIEEE 2006]
Solution : Ans. (b)
Magnetic energy $= \frac{1}{2} L I^2 = \dfrac{L q^2}{2 t^2} \left[ \text{as } I = \dfrac{q}{t} \right]$
where $L = \text{inductance}, I = \text{current}$
Energy has the dimensions $= [ML^2 T^{-2}]$
Equate the dimensions, we have
$$[ML^2 T^{-2}] = [henry] \times \dfrac{[Q^2]}{[T^2]}$$
$$\Rightarrow [henry] = \dfrac{[ML^2]}{[Q^2]}$$
Problem : Which one of the following represents the correct dimensions of the coefficient of viscosity?
(a) $[ML^{-1} T^{-2}]$
(b) $[MLT^{-1}]$
(c) $[ML^{-1} T^{-1}]$
(d) $[ML^{-2} T^{-2}]$
[AIEEE 2004]
Solution : Ans. (c)
By Newton’s formula, $\eta = \dfrac{F}{A(\Delta v_x / \Delta z)}$
$\therefore$ Dimensions of $\eta$
$$= \dfrac{\text{Dimensions of force}}{\text{Dimensions of area} \times \text{Dimensions of velocity gradient}}$$
$$= \dfrac{[MLT^{-2}]}{[L^2][T^{-1}]} = [ML^{-1} T^{-1}]$$
Problem : Dimensions of $1/\mu_0 \varepsilon_0$, where symbols have their usual meaning, are
(a) $[L^{-1} T]$
(b) $[L^2 T^2]$
(c) $[L^2 T^{-2}]$
(d) $[LT^{-1}]$
[AIEEE 2003]
Solution : Ans. (c)
As we know that formula of velocity is
$$v = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}}$$
$$\Rightarrow v^2 = \dfrac{1}{\mu_0 \varepsilon_0} = [LT^{-1}]^2$$
$$\therefore \dfrac{1}{\mu_0 \varepsilon_0} = [L^2 T^{-2}]$$
Problem : The physical quantities not having same dimensions are
(a) torque and work
(b) momentum and Planck’s constant
(c) stress and Young’s modulus
(d) speed and $(\mu_0 \varepsilon_0)^{-1/2}$
[AIEEE 2003]
Solution : Ans. (b)
Planck’s constant (in terms of unit)
$$h = J\text{-}s = [ML^2 T^{-2}][T] = [ML^2 T^{-1}]$$
Momentum $(p) = kg\text{-}ms^{-1}$
$$Momentum (p) = [M][L][T^{-1}]$$
$$ (p) = [MLT^{-1}]$$
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