Units and Dimensions JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Units and Dimensions, free to read — no sign-in needed. The full chapter has 18 questions; sign in to attempt the remaining 13 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   In a new system of units, the units of mass, length, time and current are $5$ kg, $5$ m, $5$ s and $5$ A, respectively. If $\mu_0$ and $\epsilon_0$ are the permeability and permittivity of free space, respectively, then in this new system of units, the magnitude of one SI unit of $\sqrt{\mu_0/\epsilon_0}$, is:
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   Answer: 25

   The dimensional formula for permeability of free space $\mu_0$ is $[M L T^{-2} A^{-2}]$. The dimensional formula for permittivity of free space $\epsilon_0$ is $[M^{-1} L^{-3} T^4 A^2]$. The dimensional formula for $\sqrt{\dfrac{\mu_0}{\epsilon_0}}$ is: $\left( \dfrac{[M L T^{-2} A^{-2}]}{[M^{-1} L^{-3} T^4 A^2]} \right)^{1/2} = \left( [M^2 L^4 T^{-6} A^{-4}] \right)^{1/2} = [M L^2 T^{-3} A^{-2}]$ Let the magnitude of $1$ SI unit in the new system be $n$. Using the principle of dimensional homogeneity: $n_1 u_1 = n_2 u_2$ $1 \times [M_1 L_1^2 T_1^{-3} A_1^{-2}] = n \times [M_2 L_2^2 T_2^{-3} A_2^{-2}]$ Given the new units are $M_2 = 5$ kg, $L_2 = 5$ m, $T_2 = 5$ s, and $A_2 = 5$ A, while the SI units are $M_1 = 1$ kg, $L_1 = 1$ m, $T_1 = 1$ s, and $A_1 = 1$ A. $n = \left( \dfrac{M_1}{M_2} \right) \left( \dfrac{L_1}{L_2} \right)^2 \left( \dfrac{T_1}{T_2} \right)^{-3} \left( \dfrac{A_1}{A_2} \right)^{-2}$ $n = \left( \dfrac{1}{5} \right) \left( \dfrac{1}{5} \right)^2 \left( \dfrac{1}{5} \right)^{-3} \left( \dfrac{1}{5} \right)^{-2}$ $n = 5^{-1} \times 5^{-2} \times 5^3 \times 5^2$ $n = 5^{-1 - 2 + 3 + 2} = 5^2 = 25$ Thus, the magnitude of one SI unit in the new system of units is $25$. Answer: $25$
2. Q2JEE Advanced Adv 2025 (Paper 2)
   A temperature difference can generate e.m.f. in some materials. Let $S$ be the e.m.f. produced per unit temperature difference between the ends of a wire, $\sigma$ the electrical conductivity and $\kappa$ the thermal conductivity of the material of the wire. Taking $M, L, T, I$ and $K$ as dimensions of mass, length, time, current and temperature, respectively, the dimensional formula of the quantity $Z=\frac{S^2 \sigma}{\kappa}$ is :-
   1. A.$\left[M^0 L^0 T^0 I^0 K^0\right]$
   2. B.$\left[M^0 L^0 T^0 I^0 K^{-1}\right]$
   3. C.$\left[M^1 L^2 T^{-2} I^{-1} K^{-1}\right]$
   4. D.$\left[M^1 L^2 T^{-4} I^{-1} K^{-1}\right]$
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   Answer: (B)

   $\begin{aligned} & \mathrm{S}=\text { emf per unit temperature difference } \\ & \sigma=\text { Electrical conductivity } \\ & \mathrm{k}=\text { Thermal conductivity } \\ & {[\mathrm{S}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{I}^{-1} \mathrm{~K}^{-1}\right]} \\ & {[\sigma]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^3 \mathrm{I}^2\right]} \\ & {[\mathrm{K}]=\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-3} \mathrm{~K}^{-1}\right]} \\ & {[\mathrm{Z}]=\frac{\mathrm{S}^2 \sigma}{\mathrm{~K}}=\frac{\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-3} \mathrm{~K}^{-2}\right]}{\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-3} \mathrm{~K}^{-1}\right]}}\end{aligned}$ $[\mathrm{Z}]=\left[\mathrm{K}^{-1}\right]$
3. Q3JEE Advanced Adv 2024 (Paper 1)
   A dimensionless quantity is constructed in terms of electronic charge $e$, permittivity of free space $\varepsilon_0$, Planck's constant $h$ and speed of light $c$. If the dimensionless quantity is written as $e^\alpha \varepsilon_0^\beta h^\gamma c^\delta$ and $n$ is a non-zero integer, then $(\alpha, \beta, \gamma, \delta)$ is given by
   1. A.$(2 n,-n,-n,-n)$
   2. B.$(n,-n,-2 n,-n)$
   3. C.$(n,-n,-n,-2 n)$
   4. D.$(2 n,-n,-2 n,-2 n)$
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   Answer: (A)

   For the quantity to be dimensionless $\begin{aligned} & \mathrm{e}^\alpha \varepsilon_0^\beta \mathrm{h}^\gamma \mathrm{c}^{\mathrm{d}}=\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \mathrm{~A}^0 \\ & \Rightarrow(\mathrm{AT})^\alpha\left(\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right)^\beta\left(\mathrm{ML}^2 \mathrm{~T}^{-1}\right)^\gamma\left(\mathrm{LT}^{-1}\right)^\delta=\mathrm{A}^0 \mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \\ & \therefore \alpha+2 \beta=0, \alpha+4 \beta-\gamma-\delta=0,-\beta+\gamma=0 ~\&~-3 \beta+2 \gamma+\delta=0 \\ & \therefore \alpha=-2 \beta, \beta=\gamma ~\&~ \gamma=\delta\end{aligned}$ $\therefore$ Option (1) satisfies the given condition
4. Q4JEE Advanced Adv 2023 (Paper 2)
   Young’s modulus of elasticity $Y$ is expressed in terms of three derived quantities, namely, the gravitational constant $G$, Planck’s constant $h$ and the speed of light $c$, as $Y={c}^{\alpha }{h}^{\beta }{G}^{\gamma }$. Which of the following is the correct option?
   1. A.$\alpha =7,\beta =-1,\gamma =–2$
   2. B.$\alpha =–7,\beta =–1,\gamma =–2$
   3. C.$\alpha =7,\beta =–1,\gamma =2$
   4. D.$\alpha =–7,\beta =1,\gamma =–2$
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   Answer: (A)

   Dimensions of Young's modulus: $\left[{M}^{1}{L}^{–1}T{}^{–2}\right]$ Dimensions of velocity of light: $\left[{M}^{0}{L}^{1}{T}^{–1}\right]$ Dimensions of Planck's constant: $\left[{M}^{1}{L}^{2}{T}^{–1}\right]$ Dimensions of Gravitational constant: $\left[{M}^{–1}{L}^{3}{T}^{–2}\right]$ Now, as given in the question $Y={c}^{\alpha }{h}^{\beta }{G}^{\gamma }$ $\Rightarrow \left[{M}^{1}{L}^{–1}T{}^{–2}\right]={\left[{M}^{0}{L}^{1}{T}^{–1}\right]}^{\alpha }{\left[{M}^{1}{L}^{2}{T}^{–1}\right]}^{\beta }{\left[{M}^{–1}{L}^{3}{T}^{–2}\right]}^{\gamma }$ Comparing exponent of the M, L & T both sides, we get $1=\beta -\gamma \Rightarrow \beta =1+\gamma ...\left(1\right)$ $-1=\alpha +2\beta +3\gamma ...\left(2\right)$ and $-2=-\alpha -\beta -2\gamma ...\left(3\right)$ From equation(1) and (2), we get $-1=\alpha +2\left(1+\gamma \right)+3\gamma \Rightarrow \alpha +5\gamma =-3...\left(4\right)$ From equation(1) and (3), we get $-2=-\alpha -\left(1+\gamma \right)-2\gamma \Rightarrow \alpha +3\gamma =1...\left(5\right)$ From equation(4) and equation(5), we get $\alpha =7,\gamma =–2$ and then $\beta =1+\gamma =-1$.
5. Q5JEE Advanced Adv 2022 (Paper 2)
   In a particular system of units, a physical quantity can be expressed in terms of the electric charge $e$, electron mass ${m}_{e}$. Planck's constant $h$, and Coulomb's constant $k=\dfrac{1}{4\pi {\epsilon }_{0}}$, where ${\epsilon }_{0}$ is the permittivity of vacuum. In terms of these physical constants, the dimension of the magnetic field is $\left[B\right]={\left[e\right]}^{\alpha }{\left[{m}_{e}\right]}^{\beta }{\left[h\right]}^{\gamma }{\left[k\right]}^{\delta }$. The value of $\alpha +\beta +\gamma +\delta$ is _______.
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   Answer: 4

   Given here: $\left[B\right]={\left[e\right]}^{\alpha }{\left[{m}_{e}\right]}^{\beta }{\left[h\right]}^{\gamma }{\left[k\right]}^{\delta }$ The dimensions of $\left[e\right]=\left[Current\times time\right]=\left[IT\right]$ and $\left[{m}_{e}\right]=\left[M\right]$, Planck's constant $\left[h\right]=\dfrac{Energy}{Frequency}=\dfrac{\left[M{L}^{2}{T}^{-2}\right]}{\left[{T}^{-1}\right]}=\left[M{L}^{2}{T}^{-1}\right]$ and $\left[k\right]=\dfrac{1}{\left[{\epsilon }_{0}\right]}={\left[{M}^{-1}{L}^{-3}{T}^{4}{I}^{2}\right]}^{-1}=\left[M{L}^{3}{T}^{-4}{I}^{-2}\right]$ Using Principle of homogeneity, $\left[{M}^{1}{T}^{-2}{I}^{-1}\right]={\left[IT\right]}^{\alpha }{\left[M\right]}^{\beta }{\left[M{L}^{2}{T}^{-1}\right]}^{\gamma }{\left[M{L}^{3}{T}^{-4}{I}^{-2}\right]}^{\delta }$ Comparing both sides, we get $\beta +\gamma +\delta =1...\left(i\right)$ $2\gamma +3\delta =0...\left(ii\right)$ $\alpha -\gamma -4\delta =-2...\left(iii\right)$ $\alpha -2\delta =-1...\left(iv\right)$ On solving above four equations, we get $\alpha =3,\beta =-4,\gamma =3$ and $\delta =2$ So, $\alpha +\beta +\gamma +\delta =4$