📘 Units and Dimensions — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Units and Dimensions for JEE Main with previous year questions on dimensional formulas, dimensional analysis, and their applications. Each solution is explained in a simple, step-by-step manner to boost your score.
⚛️
Units and Dimensions — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (7 April Shift 2)
The dimension of \(\sqrt{\frac{\mu_0}{\epsilon_0}}\) is equal to that of :
(\(\mu_0\) = Vacuum permeability and \(\epsilon_0\) = Vacuum permittivity)
(1) Voltage (2) Capacitance (3) Inductance (4) Resistance
(\(\mu_0\) = Vacuum permeability and \(\epsilon_0\) = Vacuum permittivity)
(1) Voltage (2) Capacitance (3) Inductance (4) Resistance
✅ Answer & Explanation:
Correct Option: (4)
The expression \(\sqrt{\frac{\mu_0}{\epsilon_0}}\) represents the intrinsic impedance of free space. Impedance has the same dimensions as Resistance: \([ML^2T^{-3}A^{-2}]\).
The expression \(\sqrt{\frac{\mu_0}{\epsilon_0}}\) represents the intrinsic impedance of free space. Impedance has the same dimensions as Resistance: \([ML^2T^{-3}A^{-2}]\).
Q2. JEE Main 2025 (4 April Shift 2)
Given below are two statements :
Statement (I) : The dimensions of Planck's constant and angular momentum are same.
Statement (II) : In Bohr's model electron revolve around the nucleus only in those orbits for which angular momentum is integral multiple of Planck's constant.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1) Both Statement I and Statement II are correct
(2) Statement I is incorrect but Statement II is correct
(3) Statement I is correct but Statement II is incorrect
(4) Both Statement I and Statement II are incorrect
Statement (I) : The dimensions of Planck's constant and angular momentum are same.
Statement (II) : In Bohr's model electron revolve around the nucleus only in those orbits for which angular momentum is integral multiple of Planck's constant.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1) Both Statement I and Statement II are correct
(2) Statement I is incorrect but Statement II is correct
(3) Statement I is correct but Statement II is incorrect
(4) Both Statement I and Statement II are incorrect
✅ Answer & Explanation:
Correct Option: (3)
- Statement I: True. Both Planck's constant (\(h = E/f\)) and angular momentum (\(L = mvr\)) have dimensions \([ML^2T^{-1}]\).
- Statement II: False. According to Bohr's quantization rule, angular momentum is an integral multiple of \(h/2\pi\), not just \(h\).
- Statement I: True. Both Planck's constant (\(h = E/f\)) and angular momentum (\(L = mvr\)) have dimensions \([ML^2T^{-1}]\).
- Statement II: False. According to Bohr's quantization rule, angular momentum is an integral multiple of \(h/2\pi\), not just \(h\).
Q3. JEE Main 2025 (4 April Shift 2)
In an electromagnetic system, a quantity defined as the ratio of electric dipole moment and magnetic dipole moment has dimension of \([M^P L^Q T^R A^S]\). The value of P and Q are :
(1) -1, 0 (2) -1, 1
(3) 1, -1 (4) 0, -1
(1) -1, 0 (2) -1, 1
(3) 1, -1 (4) 0, -1
✅ Answer & Explanation:
Correct Option: (4)
- Electric dipole moment \(p_e = q \cdot L = [A \cdot T \cdot L]\)
- Magnetic dipole moment \(p_m = I \cdot Area = [A \cdot L^2]\)
- Ratio \(p_e/p_m = \frac{[LTA]}{[AL^2]} = [L^{-1}T]\).
Thus, \(P = 0\) and \(Q = -1\).
- Electric dipole moment \(p_e = q \cdot L = [A \cdot T \cdot L]\)
- Magnetic dipole moment \(p_m = I \cdot Area = [A \cdot L^2]\)
- Ratio \(p_e/p_m = \frac{[LTA]}{[AL^2]} = [L^{-1}T]\).
Thus, \(P = 0\) and \(Q = -1\).
Q4. JEE Main 2025 (4 April Shift 1)
In an electromagnetic system, the quantity representing the ratio of electric flux and magnetic flux has dimension of \([M^P L^Q T^R A^S]\), where value of 'Q' and 'R' are:
(1) 3, -5 (2) -2, 2
(3) -2, 1 (4) 1, -1
(1) 3, -5 (2) -2, 2
(3) -2, 1 (4) 1, -1
✅ Answer & Explanation:
Correct Option: (4)
- Electric Flux \(\phi_e = E \cdot A = [ML^3T^{-3}A^{-1}]\)
- Magnetic Flux \(\phi_m = B \cdot A = [ML^2T^{-2}A^{-1}]\)
- Ratio \(\phi_e/\phi_m = [LT^{-1}]\) (the unit of velocity).
Thus, \(Q = 1\) and \(R = -1\).
- Electric Flux \(\phi_e = E \cdot A = [ML^3T^{-3}A^{-1}]\)
- Magnetic Flux \(\phi_m = B \cdot A = [ML^2T^{-2}A^{-1}]\)
- Ratio \(\phi_e/\phi_m = [LT^{-1}]\) (the unit of velocity).
Thus, \(Q = 1\) and \(R = -1\).
Q5. JEE Main 2025 (3 April Shift 1)
Match the LIST-I with LIST-II:
Choose the correct answer from the options given below :
(1) A-IV, B-III, C-II, D-I (2) A-III, B-II, C-I, D-IV
(3) A-II, B-IV, C-III, D-I (4) A-I, B-III, C-IV, D-II
| LIST-I | LIST-II |
|---|---|
| A. Gravitational constant | I. \([LT^{-2}]\) |
| B. Gravitational potential energy | II. \([L^2 T^{-2}]\) |
| C. Gravitational potential | III. \([ML^2 T^{-2}]\) |
| D. Acceleration due to gravity | IV. \([M^{-1} L^3 T^{-2}]\) |
Choose the correct answer from the options given below :
(1) A-IV, B-III, C-II, D-I (2) A-III, B-II, C-I, D-IV
(3) A-II, B-IV, C-III, D-I (4) A-I, B-III, C-IV, D-II
✅ Answer & Explanation:
Correct Option: (1)
- (A) G: From \(F = G \frac{m_1 m_2}{r^2} \implies [M^{-1}L^3T^{-2}]\) [IV]
- (B) Energy: \([ML^2T^{-2}]\) [III]
- (C) Potential: Energy per unit mass \(= [L^2T^{-2}]\) [II]
- (D) Acceleration: \([LT^{-2}]\) [I]
- (A) G: From \(F = G \frac{m_1 m_2}{r^2} \implies [M^{-1}L^3T^{-2}]\) [IV]
- (B) Energy: \([ML^2T^{-2}]\) [III]
- (C) Potential: Energy per unit mass \(= [L^2T^{-2}]\) [II]
- (D) Acceleration: \([LT^{-2}]\) [I]
Q6. JEE Main 2025 (29 Jan Shift 2)
Match List-I with List-II:
Choose the correct answer from the options given below :
(1) (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
(2) (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
(3) (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
(4) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
| List - I | List - II |
|---|---|
| (A) Young's Modulus | (I) \(ML^{-1} T^{-1}\) |
| (B) Torque | (II) \(ML^{-1} T^{-2}\) |
| (C) Coefficient of Viscosity | (III) \(M^{-1} L^3 T^{-2}\) |
| (D) Gravitational Constant | (IV) \(ML^2 T^{-2}\) |
Choose the correct answer from the options given below :
(1) (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
(2) (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
(3) (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
(4) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
✅ Answer & Explanation:
Correct Option: (3)
- (A) Young's Modulus: Same as stress/pressure \([ML^{-1}T^{-2}]\) [II]
- (B) Torque: \(Force \times Distance = [ML^2T^{-2}]\) [IV]
- (C) Viscosity: \([ML^{-1}T^{-1}]\) [I]
- (D) G Constant: \([M^{-1}L^3T^{-2}]\) [III]
- (A) Young's Modulus: Same as stress/pressure \([ML^{-1}T^{-2}]\) [II]
- (B) Torque: \(Force \times Distance = [ML^2T^{-2}]\) [IV]
- (C) Viscosity: \([ML^{-1}T^{-1}]\) [I]
- (D) G Constant: \([M^{-1}L^3T^{-2}]\) [III]
Q7. JEE Main 2025 (29 Jan Shift 1)
The expression given below shows the variation of velocity (v) with time (t), \(v = At^2 + \frac{Bt}{C+t}\). The dimension of ABC is :
(1) \([M^0 L^1 T^{-3}]\) (2) \([M^0 L^2 T^{-2}]\)
(3) \([M^0 L^1 T^{-2}]\) (4) \([M^0 L^2 T^{-3}]\)
(1) \([M^0 L^1 T^{-3}]\) (2) \([M^0 L^2 T^{-2}]\)
(3) \([M^0 L^1 T^{-2}]\) (4) \([M^0 L^2 T^{-3}]\)
✅ Answer & Explanation:
Correct Option: (4)
By principle of homogeneity:
- \([C] = [t] = [T]\)
- \([At^2] = [v] = [LT^{-1}] \implies [A] = [LT^{-3}]\)
- \([\frac{Bt}{T}] = [v] = [LT^{-1}] \implies [B] = [LT^{-1}]\)
Dimension of \(ABC = [LT^{-3}] \cdot [LT^{-1}] \cdot [T] = [L^2 T^{-3}]\).
By principle of homogeneity:
- \([C] = [t] = [T]\)
- \([At^2] = [v] = [LT^{-1}] \implies [A] = [LT^{-3}]\)
- \([\frac{Bt}{T}] = [v] = [LT^{-1}] \implies [B] = [LT^{-1}]\)
Dimension of \(ABC = [LT^{-3}] \cdot [LT^{-1}] \cdot [T] = [L^2 T^{-3}]\).
Q8. JEE Main 2025 (28 Jan Shift 1)
In a measurement, it is asked to find modulus of elasticity per unit torque applied on the system. The measured quantity has dimension of \([M^a L^b T^c]\). If \(b = -3\), the value of \(c\) is ________.
✅ Answer & Explanation:
Correct Answer: 0
- Modulus of Elasticity \([E] = [ML^{-1}T^{-2}]\)
- Torque \([\tau] = [ML^2T^{-2}]\)
- Ratio \(E/\tau = \frac{[ML^{-1}T^{-2}]}{[ML^2T^{-2}]} = [L^{-3}]\).
The dimension is \([M^0 L^{-3} T^0]\). Comparing with \([M^a L^b T^c]\), we find \(c = 0\).
- Modulus of Elasticity \([E] = [ML^{-1}T^{-2}]\)
- Torque \([\tau] = [ML^2T^{-2}]\)
- Ratio \(E/\tau = \frac{[ML^{-1}T^{-2}]}{[ML^2T^{-2}]} = [L^{-3}]\).
The dimension is \([M^0 L^{-3} T^0]\). Comparing with \([M^a L^b T^c]\), we find \(c = 0\).
Q9. JEE Main 2025 (23 Jan Shift 2)
Match List - I with List - II:
Choose the correct answer from the options given below :
(1) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(2) (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
(3) (A)-(I), (B)-(IV), (C)-(II), (D)-(III)
(4) (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
| List - I | List - II |
|---|---|
| (A) Permeability of free space | (I) \([ML^2 T^{-2}]\) |
| (B) Magnetic field | (II) \([MT^{-2} A^{-1}]\) |
| (C) Magnetic moment | (III) \([MLT^{-2} A^{-2}]\) |
| (D) Torsional constant | (IV) \([L^2 A]\) |
Choose the correct answer from the options given below :
(1) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(2) (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
(3) (A)-(I), (B)-(IV), (C)-(II), (D)-(III)
(4) (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
✅ Answer & Explanation:
Correct Option: (2)
- (A) \(\mu_0\): \([MLT^{-2}A^{-2}]\) [III]
- (B) Magnetic field: \([MT^{-2}A^{-1}]\) [II]
- (C) Magnetic moment: \(I \cdot Area = [L^2A]\) [IV]
- (D) Torsional constant: \(\tau/\theta = [ML^2T^{-2}]\) [I]
- (A) \(\mu_0\): \([MLT^{-2}A^{-2}]\) [III]
- (B) Magnetic field: \([MT^{-2}A^{-1}]\) [II]
- (C) Magnetic moment: \(I \cdot Area = [L^2A]\) [IV]
- (D) Torsional constant: \(\tau/\theta = [ML^2T^{-2}]\) [I]
Q10. JEE Main 2025 (22 Jan Shift 1)
If \(B\) is magnetic field and \(\mu_0\) is permeability of free space, then the dimensions of \((B/\mu_0)\) is:
(1) \([ML^2 T^{-2} A^{-1}]\) (2) \([MT^{-2} A^{-1}]\)
(3) \([L^{-1} A]\) (4) \([LT^{-2} A^{-1}]\)
(1) \([ML^2 T^{-2} A^{-1}]\) (2) \([MT^{-2} A^{-1}]\)
(3) \([L^{-1} A]\) (4) \([LT^{-2} A^{-1}]\)
✅ Answer & Explanation:
Correct Option: (3)
Using Ampere's Law: \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I \implies \frac{B}{\mu_0} = \frac{I}{L}\).
Dimension of \(B/\mu_0 = [AL^{-1}]\).
Using Ampere's Law: \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I \implies \frac{B}{\mu_0} = \frac{I}{L}\).
Dimension of \(B/\mu_0 = [AL^{-1}]\).
Q11. JEE Main 2025 (2 April Shift 2)
If \(\mu_0\) and \(\epsilon_0\) are the permeability and permittivity of free space, respectively, then the dimension of \((\frac{1}{\mu_0 \epsilon_0})\) is :
(1) \(L/T^2\) (2) \(L^2/T^2\) (3) \(T^2/L\) (4) \(T^2/L^2\)
(1) \(L/T^2\) (2) \(L^2/T^2\) (3) \(T^2/L\) (4) \(T^2/L^2\)
✅ Answer & Explanation:
Correct Option: (2)
From Maxwell's equations, the speed of light is \(c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}\).
Therefore, \(\frac{1}{\mu_0 \epsilon_0} = c^2\).
The dimensions of \(c^2\) are \([LT^{-1}]^2 = [L^2 T^{-2}]\).
From Maxwell's equations, the speed of light is \(c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}\).
Therefore, \(\frac{1}{\mu_0 \epsilon_0} = c^2\).
The dimensions of \(c^2\) are \([LT^{-1}]^2 = [L^2 T^{-2}]\).
Q12. JEE Main 2025 (2 April Shift 1)
Match List-I with List-II:
Choose the correct answer from the options given below :
(1) (A)-(I), (B)-(IV), (C)-(III), (D)-(II)
(2) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(3) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
(4) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
| List-I | List-II |
|---|---|
| (A) Coefficient of viscosity | (I) \([ML^0 T^{-3}]\) |
| (B) Intensity of wave | (II) \([ML^{-2} T^{-2}]\) |
| (C) Pressure gradient | (III) \([M^{-1}LT^2]\) |
| (D) Compressibility | (IV) \([ML^{-1} T^{-1}]\) |
Choose the correct answer from the options given below :
(1) (A)-(I), (B)-(IV), (C)-(III), (D)-(II)
(2) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(3) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
(4) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
✅ Answer & Explanation:
Correct Option: (2)
- (A) Viscosity: \([ML^{-1}T^{-1}]\) [IV]
- (B) Intensity: Power/Area \(= [MT^{-3}]\) [I]
- (C) Pressure gradient: \(dP/dx = [ML^{-2}T^{-2}]\) [II]
- (D) Compressibility: 1/Bulk Modulus \(= [M^{-1}LT^2]\) [III]
- (A) Viscosity: \([ML^{-1}T^{-1}]\) [IV]
- (B) Intensity: Power/Area \(= [MT^{-3}]\) [I]
- (C) Pressure gradient: \(dP/dx = [ML^{-2}T^{-2}]\) [II]
- (D) Compressibility: 1/Bulk Modulus \(= [M^{-1}LT^2]\) [III]
Q13. JEE Main 2024 (05 Apr Shift 1)
If \(G\) be the gravitational constant and \(u\) be the energy density then which of the following quantity have the dimensions as that of the \(\sqrt{uG}\) :
(1) pressure gradient per unit mass (2) Gravitational potential
(3) Energy per unit mass (4) Force per unit mass
(1) pressure gradient per unit mass (2) Gravitational potential
(3) Energy per unit mass (4) Force per unit mass
✅ Answer & Explanation:
Correct Option: (4)
- \([u] = [ML^{-1}T^{-2}]\)
- \([G] = [M^{-1}L^3T^{-2}]\)
- \([uG] = [ML^{-1}T^{-2}] \cdot [M^{-1}L^3T^{-2}] = [L^2 T^{-4}]\)
- \(\sqrt{uG} = [LT^{-2}]\).
This corresponds to acceleration, which is also Force per unit mass.
- \([u] = [ML^{-1}T^{-2}]\)
- \([G] = [M^{-1}L^3T^{-2}]\)
- \([uG] = [ML^{-1}T^{-2}] \cdot [M^{-1}L^3T^{-2}] = [L^2 T^{-4}]\)
- \(\sqrt{uG} = [LT^{-2}]\).
This corresponds to acceleration, which is also Force per unit mass.
Q14. JEE Main 2024 (04 Apr Shift 1)
The equation of stationary wave is: \(y = 2a \sin(\frac{2\pi nt}{\lambda}) \cos(\frac{2\pi x}{\lambda})\). Which of the following is NOT correct :
(1) The dimensions of \(n/\lambda\) is [T]
(2) The dimensions of \(n\) is \([LT^{-1}]\)
(3) The dimensions of \(x\) is [L]
(4) The dimensions of \(nt\) is [L]
(1) The dimensions of \(n/\lambda\) is [T]
(2) The dimensions of \(n\) is \([LT^{-1}]\)
(3) The dimensions of \(x\) is [L]
(4) The dimensions of \(nt\) is [L]
✅ Answer & Explanation:
Correct Option: (1)
Arguments of sine and cosine are dimensionless.
- \([2\pi x / \lambda] = [1] \implies [\lambda] = [x] = [L]\).
- \([2\pi nt / \lambda] = [1] \implies [n] = [LT^{-1}]\).
Checking (1): \([n/\lambda] = [LT^{-1}]/[L] = [T^{-1}]\). So (1) is incorrect.
Arguments of sine and cosine are dimensionless.
- \([2\pi x / \lambda] = [1] \implies [\lambda] = [x] = [L]\).
- \([2\pi nt / \lambda] = [1] \implies [n] = [LT^{-1}]\).
Checking (1): \([n/\lambda] = [LT^{-1}]/[L] = [T^{-1}]\). So (1) is incorrect.
Q15. JEE Main 2022 (27 Jul Shift 2)
An expression of energy density is given by \(u = \frac{\alpha}{\beta} \sin(\frac{\alpha x}{kt})\), where \(\alpha, \beta\) are constants, \(x\) is displacement, \(k\) is Boltzmann constant and \(t\) is the temperature. The dimensions of \(\beta\) will be :
(1) \([ML^2 T^{-2} \theta^{-1}]\) (2) \([M^0 L^2 T^{-2}]\)
(3) \([M^0 L^0 T^0]\) (4) \([M^0 L^2 T^0]\)
(1) \([ML^2 T^{-2} \theta^{-1}]\) (2) \([M^0 L^2 T^{-2}]\)
(3) \([M^0 L^0 T^0]\) (4) \([M^0 L^2 T^0]\)
✅ Answer & Explanation:
Correct Option: (4)
Argument of sine is dimensionless: \([\alpha x / kt] = [1]\).
Since \([kt] = [Energy] = [ML^2T^{-2}]\), then \([\alpha] = \frac{[ML^2T^{-2}]}{[L]} = [MLT^{-2}]\).
Now, \([u] = [\alpha/\beta] \implies [\beta] = \frac{[\alpha]}{[u]} = \frac{[MLT^{-2}]}{[ML^{-1}T^{-2}]} = [L^2]\).
Dimension of \(\beta = [M^0 L^2 T^0]\).
Argument of sine is dimensionless: \([\alpha x / kt] = [1]\).
Since \([kt] = [Energy] = [ML^2T^{-2}]\), then \([\alpha] = \frac{[ML^2T^{-2}]}{[L]} = [MLT^{-2}]\).
Now, \([u] = [\alpha/\beta] \implies [\beta] = \frac{[\alpha]}{[u]} = \frac{[MLT^{-2}]}{[ML^{-1}T^{-2}]} = [L^2]\).
Dimension of \(\beta = [M^0 L^2 T^0]\).
Q16. JEE Main 2021 (26 Aug Shift 2)
Match List - I with List - II :
Choose the most appropriate answer from the options given below :
(1) (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
(2) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
(3) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
(4) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
| List - I | List - II |
|---|---|
| (a) Magnetic induction | (i) \(ML^2 T^{-2} A^{-1}\) |
| (b) Magnetic flux | (ii) \(M^0 L^{-1} A\) |
| (c) Magnetic permeability | (iii) \(MT^{-2} A^{-1}\) |
| (d) Magnetization | (iv) \(MLT^{-2} A^{-2}\) |
Choose the most appropriate answer from the options given below :
(1) (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
(2) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
(3) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
(4) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
✅ Answer & Explanation:
Correct Option: (2)
- (a) Induction (B): \([MT^{-2}A^{-1}]\) [iii]
- (b) Flux (\(\phi\)): \(B \cdot A = [ML^2T^{-2}A^{-1}]\) [i]
- (c) Permeability (\(\mu\)): \([MLT^{-2}A^{-2}]\) [iv]
- (d) Magnetization (M): Magnetic moment per unit volume \(= [L^{-1}A]\) [ii]
- (a) Induction (B): \([MT^{-2}A^{-1}]\) [iii]
- (b) Flux (\(\phi\)): \(B \cdot A = [ML^2T^{-2}A^{-1}]\) [i]
- (c) Permeability (\(\mu\)): \([MLT^{-2}A^{-2}]\) [iv]
- (d) Magnetization (M): Magnetic moment per unit volume \(= [L^{-1}A]\) [ii]
Q17. JEE Main 2020 (08 Jan Shift 1)
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:
(1) \(h^{1/3} G^{2/3} c^{1/3} A^{-1}\) (2) \(h^0 c^5 G^{-1} A^{-1}\)
(3) \(h^{-2/3} c^{-1/3} G^{4/3} A^{-1}\) (4) \(h^2 G^{3/2} c^{1/3} A^{-1}\)
(1) \(h^{1/3} G^{2/3} c^{1/3} A^{-1}\) (2) \(h^0 c^5 G^{-1} A^{-1}\)
(3) \(h^{-2/3} c^{-1/3} G^{4/3} A^{-1}\) (4) \(h^2 G^{3/2} c^{1/3} A^{-1}\)
✅ Answer & Explanation:
Correct Option: (2)
By comparing dimensions on both sides or using dimensional analysis, it is found that stopping potential relates to these constants as shown in option (2). Note that power of \(h\) being zero implies independence from Planck's constant in this specific proportional arrangement.
By comparing dimensions on both sides or using dimensional analysis, it is found that stopping potential relates to these constants as shown in option (2). Note that power of \(h\) being zero implies independence from Planck's constant in this specific proportional arrangement.
Q18. JEE Main 2020 (04 Sep Shift 2)
A quantity \(x\) is given by \((I F v^2 / W L^4)\) in terms of moment of inertia \(I\), force \(F\), velocity \(v\), work \(W\) and length \(L\). The dimensional formula for \(x\) is same as that of :
(1) Planck's constant (2) force constant
(3) energy density (4) coefficient of viscosity
(1) Planck's constant (2) force constant
(3) energy density (4) coefficient of viscosity
✅ Answer & Explanation:
Correct Option: (3)
\([x] = \frac{[ML^2] \cdot [MLT^{-2}] \cdot [LT^{-1}]^2}{[ML^2T^{-2}] \cdot [L^4]}\)
\(= \frac{M^2 L^5 T^{-4}}{M L^6 T^{-2}} = [ML^{-1}T^{-2}]\).
This matches the dimension of Energy Density (Energy/Volume) or Pressure.
\([x] = \frac{[ML^2] \cdot [MLT^{-2}] \cdot [LT^{-1}]^2}{[ML^2T^{-2}] \cdot [L^4]}\)
\(= \frac{M^2 L^5 T^{-4}}{M L^6 T^{-2}} = [ML^{-1}T^{-2}]\).
This matches the dimension of Energy Density (Energy/Volume) or Pressure.
📘 Exam Preparation Tip for Units and Dimensions
- Memorize Key Dimensions: Make a table of dimensions for common physical quantities (e.g., force, energy, power, pressure, electric field, etc.). This will save time in the exam.
- Principle of Homogeneity: This is a powerful tool. In any correct equation, the dimensions of all terms must be the same. Use this to find the dimension of unknown constants.
- Applications of Dimensional Analysis: Practice problems where you need to derive the relation between different physical quantities or find the dimension of an expression.
- Common Constants: Know the dimensions of common constants like \(G\), \(h\), \(\epsilon_0\), \(\mu_0\), and the Boltzmann constant \(k\).
- Practice Matching Questions: Matching questions from JEE Main test your recall of dimensions for various quantities. Practice them regularly.