Practice MCQs for JEE Physics Units and Measurements
Access targeted multiple-choice questions for Units and Measurements designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.
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Multiple Choice Questions
(a) \( \sqrt{\mu_0/\varepsilon_0} \)
(b) \( \mu_0/\varepsilon_0 \)
(c) \( \sqrt{\varepsilon_0/\mu_0} \)
(d) \( \varepsilon_0/\mu_0 \)
Show Answer & Explanation
Answer: (a) \( \sqrt{\mu_0/\varepsilon_0} \)
Explanation:
1. \( \sqrt{\mu_0/\varepsilon_0} = \mu_0 c \), since \( c = 1/\sqrt{\mu_0\varepsilon_0} \).
2. Now \( \mu_0 \) has dimensions \( \mathrm{MLT^{-2}A^{-2}} \) and \( c \) has \( \mathrm{LT^{-1}} \), so \( \mu_0 c = \mathrm{ML^2T^{-3}A^{-2}} \).
3. This is the dimension of resistance (volt per ampere).
Teacher's Note:
1. \( \sqrt{\mu_0/\varepsilon_0} \) is the impedance of free space, about 377 ohm.
2. Remember it as an ohm-valued constant.
(a) \( [\mathrm{M^{-1}L^{-2}T^{4}A^{2}}] \)
(b) \( [\mathrm{M^{-2}L^{0}T^{-4}A^{-2}}] \)
(c) \( [\mathrm{M^{-3}L^{-2}T^{8}A^{4}}] \)
(d) \( [\mathrm{M^{-2}L^{-2}T^{6}A^{3}}] \)
Show Answer & Explanation
Answer: (c) \( [\mathrm{M^{-3}L^{-2}T^{8}A^{4}}] \)
Explanation:
1. Capacitance: \( [X] = \mathrm{M^{-1}L^{-2}T^{4}A^{2}} \). Magnetic field from \( F = BIl \): \( [Z] = \mathrm{MT^{-2}A^{-1}} \).
2. \( [Y] = [X]/[Z]^2 = \mathrm{M^{-1}L^{-2}T^{4}A^{2}} / \mathrm{M^{2}T^{-4}A^{-2}} \).
3. \( [Y] = \mathrm{M^{-3}L^{-2}T^{8}A^{4}} \).
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(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) \( S^{1/2}I^{3/2}h^{-1} \)
Show Answer & Explanation
Answer: (c) \( S^{1/2}I^{1/2}h^{0} \)
Explanation:
1. Let \( p = k\,h^aS^bI^c \). Dimensions: \( [p]=\mathrm{MLT^{-1}} \), \( [h]=\mathrm{ML^2T^{-1}} \), \( [S]=\mathrm{MT^{-2}} \), \( [I]=\mathrm{ML^2} \).
2. Mass: \( a+b+c=1 \). Length: \( 2a+2c=1 \). Time: \( -a-2b=-1 \).
3. Solving gives \( a=0,\ b=\tfrac12,\ c=\tfrac12 \), so \( p \propto S^{1/2}I^{1/2}h^0 \).
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(a) \( [\mathrm{A^{-1}TML^{3}}] \)
(b) \( [\mathrm{AT^{2}M^{-1}L^{-1}}] \)
(c) \( [\mathrm{AT^{-3}ML^{3/2}}] \)
(d) \( [\mathrm{A^{2}T^{3}M^{-1}L^{-2}}] \)
Show Answer & Explanation
Answer: (d) \( [\mathrm{A^{2}T^{3}M^{-1}L^{-2}}] \)
Explanation:
1. \( \sqrt{\varepsilon_0/\mu_0} = 1/(\text{impedance of free space}) \), which has the dimension of 1/resistance (conductance).
2. Conductance \( = \mathrm{M^{-1}L^{-2}T^{3}A^{2}} \).
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(a) \( [\mathrm{LA^{2}}] \)
(b) \( [\mathrm{LTA}] \)
(c) \( [\mathrm{A^{-1}}] \)
(d) \( [\mathrm{LA^{-2}}] \)
Show Answer & Explanation
Answer: (c) \( [\mathrm{A^{-1}}] \)
Explanation:
1. \( L/R \) has the dimension of time, and \( RC \) also has the dimension of time, so \( l/r \) and \( rc \) both reduce to T.
2. \( \dfrac{l}{rcv} = \dfrac{l/r}{c\,v} = \dfrac{T}{Q} \), since \( cv = Q \).
3. \( T/Q = T/(AT) = A^{-1} \).
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(a) \( [\mathrm{V^{-4}A^{-2}F}] \)
(b) \( [\mathrm{V^{-2}A^{2}F^{-2}}] \)
(c) \( [\mathrm{V^{-4}A^{2}F}] \)
(d) \( [\mathrm{V^{-4}A^{2}F^{2}}] \)
Show Answer & Explanation
Answer: (c) \( [\mathrm{V^{-4}A^{2}F}] \)
Explanation:
1. From \( V=LT^{-1} \) and \( A=LT^{-2} \): \( T=V/A \) and \( L=V^2/A \). Then \( M = F/(LT^{-2}) = F/A \).
2. Young's modulus \( = \mathrm{ML^{-1}T^{-2}} = \dfrac{F}{A}\cdot\dfrac{A}{V^2}\cdot\dfrac{A^2}{V^2} \).
3. So \( Y = V^{-4}A^{2}F \).
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(a) \( [\mathrm{MLT^{-2}}] \)
(b) \( [\mathrm{M^{0}L^{2}T^{-4}}] \)
(c) \( [\mathrm{M^{2}LT^{-4}}] \)
(d) \( [\mathrm{M^{2}L^{2}T^{-2}}] \)
Show Answer & Explanation
Answer: (c) \( [\mathrm{M^{2}LT^{-4}}] \)
Explanation:
1. The exponent is dimensionless, so \( [\alpha] = [x^2]/[kT] = \mathrm{L^2}/\mathrm{ML^2T^{-2}} = \mathrm{M^{-1}T^{2}} \).
2. \( [\beta] = [F]/[\alpha] = \mathrm{MLT^{-2}}/\mathrm{M^{-1}T^{2}} = \mathrm{M^{2}LT^{-4}} \).
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(a) 40
(b) 16
(c) 640
(d) 410
Show Answer & Explanation
Answer: (a) 40
Explanation:
1. One new length unit = 0.25 m and one new mass unit = 0.05 kg.
2. Density in new units \( = 128 \times \dfrac{(0.25)^3}{0.05} \).
3. \( = 128\times 0.015625/0.05 = 40 \).
Teacher's Note:
1. A larger length unit makes the numerical density larger (it enters as a cube), while a larger mass unit makes it smaller.
2. Convert units of the type mass/length³ by multiplying by (unit of length)³ and dividing by (unit of mass).
(a) \( \sqrt{Gh/c^5} \)
(b) \( \sqrt{hc^5/G} \)
(c) \( \sqrt{c^3/(Gh)} \)
(d) \( \sqrt{Gh/c^3} \)
Show Answer & Explanation
Answer: (a) \( \sqrt{Gh/c^5} \)
Explanation:
1. Let \( t = G^ah^bc^d \), with \( [G]=\mathrm{M^{-1}L^3T^{-2}} \), \( [h]=\mathrm{ML^2T^{-1}} \), \( [c]=\mathrm{LT^{-1}} \).
2. Mass: \( -a+b=0 \); length: \( 3a+2b+d=0 \); time: \( -2a-b-d=1 \).
3. Solving: \( a=b=\tfrac12,\ d=-\tfrac52 \), so \( t\propto\sqrt{Gh/c^5} \) (the Planck time).
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(a) \( [\varepsilon_0]=[\mathrm{M^{-1}L^{-3}T^{2}A}] \)
(b) \( [\varepsilon_0]=[\mathrm{M^{-1}L^{-3}T^{4}A^{2}}] \)
(c) \( [\varepsilon_0]=[\mathrm{M^{-2}L^{2}T^{-1}A^{-2}}] \)
(d) \( [\varepsilon_0]=[\mathrm{M^{-1}L^{2}T^{-1}A^{2}}] \)
Show Answer & Explanation
Answer: (b) \( [\varepsilon_0]=[\mathrm{M^{-1}L^{-3}T^{4}A^{2}}] \)
Explanation:
1. Coulomb's law gives \( \varepsilon_0 = q^2/(4\pi F r^2) \).
2. \( [\varepsilon_0] = \dfrac{(AT)^2}{MLT^{-2}\,L^2} = \mathrm{M^{-1}L^{-3}T^{4}A^{2}} \).
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(a) Pressure, Young's modulus, stress
(b) Emf, potential difference, electric potential
(c) Heat, work done, energy
(d) Dipole moment, electric flux, electric field
Show Answer & Explanation
Answer: (d) Dipole moment, electric flux, electric field
Explanation:
1. Pressure, Young's modulus and stress all have \( \mathrm{ML^{-1}T^{-2}} \). Emf, potential difference and potential all have \( \mathrm{ML^2T^{-3}A^{-1}} \). Heat, work and energy all have \( \mathrm{ML^2T^{-2}} \).
2. Dipole moment \( = \mathrm{LTA} \), electric flux \( = \mathrm{ML^3T^{-3}A^{-1}} \) and electric field \( = \mathrm{MLT^{-3}A^{-1}} \) are all different.
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(a) \( [\mathrm{M^{0}L^{2}T^{0}}] \)
(b) \( [\mathrm{ML^{2}T}] \)
(c) \( [\mathrm{ML^{0}T^{-1}}] \)
(d) \( [\mathrm{M^{0}L^{2}T^{-1}}] \)
Show Answer & Explanation
Answer: (a) \( [\mathrm{M^{0}L^{2}T^{0}}] \)
Explanation:
1. The exponent is dimensionless: \( [\alpha] = [k\theta]/[Z] = \mathrm{ML^2T^{-2}}/\mathrm{L} = \mathrm{MLT^{-2}} \).
2. \( [\beta] = [\alpha]/[p] = \mathrm{MLT^{-2}}/\mathrm{ML^{-1}T^{-2}} = \mathrm{L^2} \).
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(a) resistance
(b) charge
(c) voltage
(d) current
Show Answer & Explanation
Answer: (d) current
Explanation:
1. \( \varepsilon_0 \) has the dimension of \( q/(V\,L) \) (from \( C = \varepsilon_0 A/d \)), so \( \varepsilon_0 L V \) has the dimension of charge.
2. Dividing by time gives charge/time, which is current.
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(a) \( [\mathrm{ML^{-1}T^{-2}}] \)
(b) \( [\mathrm{ML^{2}T^{-1}}] \)
(c) \( [\mathrm{MLT^{-2}}] \)
(d) \( [\mathrm{ML^{2}T^{-2}}] \)
Show Answer & Explanation
Answer: (a) \( [\mathrm{ML^{-1}T^{-2}}] \)
Explanation:
1. \( \tfrac12\varepsilon_0E^2 \) is the energy stored per unit volume of an electric field.
2. Energy/volume \( = \mathrm{ML^2T^{-2}}/\mathrm{L^3} = \mathrm{ML^{-1}T^{-2}} \).
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
A. Boltzmann's constant B. Coefficient of viscosity C. Planck's constant D. Thermal conductivity
p. \( \mathrm{ML^2T^{-1}} \) q. \( \mathrm{ML^{-1}T^{-1}} \) r. \( \mathrm{MLT^{-3}K^{-1}} \) s. \( \mathrm{ML^2T^{-2}K^{-1}} \)
(a) A-s, B-q, C-p, D-r
(b) A-q, B-s, C-r, D-p
(c) A-s, B-p, C-q, D-r
(d) A-r, B-q, C-p, D-s
Show Answer & Explanation
Answer: (a) A-s, B-q, C-p, D-r
Explanation:
1. Boltzmann's constant is energy per kelvin: \( \mathrm{ML^2T^{-2}K^{-1}} \) (s).
2. Viscosity \( = F/(A\,dv/dx) = \mathrm{ML^{-1}T^{-1}} \) (q). Planck's constant \( = E/\nu = \mathrm{ML^2T^{-1}} \) (p).
3. Thermal conductivity \( = Q\,l/(A\,\Delta T\,t) = \mathrm{MLT^{-3}K^{-1}} \) (r).
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(1) \( l=\sqrt{\dfrac{nq^2}{\varepsilon k_BT}} \) (2) \( l=\sqrt{\dfrac{\varepsilon k_BT}{nq^2}} \) (3) \( l=\sqrt{\dfrac{q^2}{\varepsilon n^{2/3}k_BT}} \) (4) \( l=\sqrt{\dfrac{q^2}{\varepsilon n^{1/3}k_BT}} \)
(a) (1) and (2) only
(b) (2) and (4) only
(c) (3) and (4) only
(d) (1) and (3) only
Show Answer & Explanation
Answer: (b) (2) and (4) only
Explanation:
1. \( [\varepsilon] = q^2/(F L^2) \) and \( [k_BT] = FL \), so \( [\varepsilon k_BT] = q^2/L \). Also \( [n] = L^{-3} \).
2. (2): \( \varepsilon k_BT/(nq^2) = L^{-1}/L^{-3} = L^2 \), so \( l \) has the dimension of length. (4): \( q^2/(\varepsilon n^{1/3}k_BT) = L/L^{-1} = L^2 \), correct.
3. (1) gives \( L^{-2} \) and (3) gives \( L^{3} \) under the root, so they fail.
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(1) \( M\propto\sqrt c \) (2) \( M\propto\sqrt G \) (3) \( L\propto\sqrt h \) (4) \( L\propto\sqrt G \)
(a) (1), (3) and (4) only
(b) (2) and (3) only
(c) (1) and (2) only
(d) (1), (2) and (4) only
Show Answer & Explanation
Answer: (a) (1), (3) and (4) only
Explanation:
1. Solving the dimensions gives \( M=\sqrt{hc/G} \) and \( L=\sqrt{hG/c^3} \).
2. So \( M\propto\sqrt c \) (true), \( M\propto 1/\sqrt G \) (so (2) is false), \( L\propto\sqrt h \) (true) and \( L\propto\sqrt G \) (true).
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(1) \( \mu_0I^2=\varepsilon_0V^2 \) (2) \( \varepsilon_0I=\mu_0V \) (3) \( I=\varepsilon_0cV \) (4) \( \mu_0cI=\varepsilon_0V \)
(a) (1) and (3) only
(b) (2) and (4) only
(c) (1) and (4) only
(d) (2) and (3) only
Show Answer & Explanation
Answer: (a) (1) and (3) only
Explanation:
1. \( \mu_0I^2 \) has the dimension of force, and \( \varepsilon_0V^2 \) (energy per unit length) also has the dimension of force, so (1) is correct.
2. \( \varepsilon_0 V L \) is a charge, so \( \varepsilon_0 V c = \varepsilon_0VL/T \) is a current: (3) is correct.
3. (2) and (4) fail on dimensions (for example \( \mu_0 c I \) has the dimension of electric field, not of \( \varepsilon_0 V \)).
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(a) weber/ampere
(b) volt-second/ampere
(c) joule/(ampere)²
(d) All of these
Show Answer & Explanation
Answer: (d) All of these
Explanation:
1. \( \Phi = LI \) gives \( L = \) weber/ampere.
2. \( \mathcal E = L\,dI/dt \) gives \( L = \) volt-second/ampere. \( U=\tfrac12LI^2 \) gives \( L = \) joule/ampere².
3. All three forms are correct.
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(1) \( [\varepsilon_0]=[\mathrm{M^{-1}L^{-3}T^{2}I}] \) (2) \( [\varepsilon_0]=[\mathrm{M^{-1}L^{-3}T^{4}I^{2}}] \) (3) \( [\mu_0]=[\mathrm{MLT^{-2}I^{-2}}] \) (4) \( [\mu_0]=[\mathrm{ML^{2}T^{-1}I}] \)
(a) (1) and (3) only
(b) (2) and (3) only
(c) (2) and (4) only
(d) (1) and (4) only
Show Answer & Explanation
Answer: (b) (2) and (3) only
Explanation:
1. \( \varepsilon_0=q^2/(4\pi Fr^2) \Rightarrow [\varepsilon_0]=\mathrm{M^{-1}L^{-3}T^{4}I^{2}} \).
2. \( F/l=\mu_0I^2/(2\pi d) \Rightarrow [\mu_0]=\mathrm{MLT^{-2}I^{-2}} \).
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(1) Reynolds number and coefficient of friction (2) Curie and frequency of a light wave (3) Latent heat and gravitational potential (4) Planck's constant and torque
(a) (1), (2) and (3) only
(b) (2), (3) and (4) only
(c) (1) and (4) only
(d) (1), (2), (3) and (4)
Show Answer & Explanation
Answer: (a) (1), (2) and (3) only
Explanation:
1. Reynolds number and the coefficient of friction are both dimensionless. A curie (decays per second) and frequency both have \( \mathrm{T^{-1}} \). Latent heat and gravitational potential both have \( \mathrm{L^2T^{-2}} \).
2. Planck's constant is \( \mathrm{ML^2T^{-1}} \) while torque is \( \mathrm{ML^2T^{-2}} \), so pair (4) differs.
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
(1) Torque and work (2) Angular momentum and work (3) Energy and Young's modulus (4) Light year and wavelength
(a) (1) and (4) only
(b) (2) and (3) only
(c) (1) and (2) only
(d) (3) and (4) only
Show Answer & Explanation
Answer: (a) (1) and (4) only
Explanation:
1. Torque and work are both \( \mathrm{ML^2T^{-2}} \). A light year and a wavelength are both lengths.
2. Angular momentum is \( \mathrm{ML^2T^{-1}} \), not that of work; energy is \( \mathrm{ML^2T^{-2}} \) but Young's modulus is \( \mathrm{ML^{-1}T^{-2}} \).
Teacher's Note:
1. Check the dimensions or units term by term before choosing an option.
2. Work with the standard base quantities so that each step can be verified.
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