Download JEE 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.
Chapter-wise Objective Questions: Units and Measurements
Access the complete set of multiple-choice questions for Units and Measurements below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.
Multiple Choice Questions
(a) \( [ML^{-3}T^{-1}] \)
(b) \( [ML^{-2}T^{-1}] \)
(c) \( [M^0L^{-1}T^{-3}] \)
(d) \( [ML^2T^{-1}] \)
Show Answer & Explanation
Answer: (d) \( [ML^2T^{-1}] \)
Explanation:
1. From \( E=h\nu \), \( h=E/\nu \).
2. Energy is \( [ML^2T^{-2}] \) and frequency is \( [T^{-1}] \).
3. So \( [h]=[ML^2T^{-1}] \), the same as angular momentum.
Teacher's Note:
1. P
2. l
3. a
4. n
5. c
6. k
7. '
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9.
10. c
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79. ν
80. .
(a) \( [ML^0T^{-2}A^{-1}] \)
(b) \( [ML^2T^{-2}A^{-1}] \)
(c) \( [ML^2T^{-1}A^3] \)
(d) \( [M^0L^{-2}T^{-2}A^{-2}] \)
Show Answer & Explanation
Answer: (b) \( [ML^2T^{-2}A^{-1}] \)
Explanation:
1. Magnetic flux \( \phi=BA \) and \( B=F/(IL)=[MT^{-2}A^{-1}] \).
2. Multiplying by area \( [L^2] \): \( [\phi]=[ML^2T^{-2}A^{-1}] \).
Teacher's Note:
1. F
2. l
3. u
4. x
5.
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85.
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89. I
90. L
91. .
(a) \( [ML^4T^{-1}] \)
(b) \( [ML^2T^{-2}] \)
(c) \( [ML^5T^{-3}] \)
(d) \( [ML^5T^{-2}] \)
Show Answer & Explanation
Answer: (d) \( [ML^5T^{-2}] \)
Explanation:
1. \( a/V^2 \) must have the dimensions of pressure, \( [ML^{-1}T^{-2}] \).
2. \( [a]=[P][V^2]=ML^{-1}T^{-2}\times L^6 \).
3. So \( [a]=[ML^5T^{-2}] \).
Teacher's Note:
1. O
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84. .
(a) \( [ML^2T^{-2}] \)
(b) \( [ML^{-1}T^{-1}] \)
(c) \( [M^{-1}L^3T^{-2}] \)
(d) \( [None of these] \)
Show Answer & Explanation
Answer: (c) \( [M^{-1}L^3T^{-2}] \)
Explanation:
1. From \( F=\dfrac{Gm_1m_2}{r^2} \), \( G=\dfrac{Fr^2}{m^2} \).
2. \( [G]=\dfrac{MLT^{-2}\times L^2}{M^2}=M^{-1}L^3T^{-2} \).
Teacher's Note:
1. U
2. s
3. e
4.
5. F
6.
7. =
8.
9. G
10. m
11. ₁
12. m
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18.
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45. g
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47. ²
48. .
(a) tension and surface tension
(b) stress and pressure
(c) Planck's constant and angular momentum
(d) angle and strain
Show Answer & Explanation
Answer: (a) tension and surface tension
Explanation:
1. Tension is a force, \( [MLT^{-2}] \), while surface tension is force per unit length, \( [MT^{-2}] \).
2. Stress and pressure are both force per area, Planck's constant and angular momentum are both \( [ML^2T^{-1}] \), and angle and strain are dimensionless.
3. So the pair in (a) differs.
Teacher's Note:
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97. .
(a) 9.378 m²
(b) 9.37 m²
(c) 9.378248 m²
(d) 9.3782 m²
Show Answer & Explanation
Answer: (a) 9.378 m²
Explanation:
1. Area = 3.124 × 3.002 = 9.378248 m².
2. Rounded to four significant figures this is 9.378 m².
Teacher's Note:
1. R
2. o
3. u
4. n
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119. u
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121. e
122. s
123. .
(a) \( [ML^3T^{-3}] \)
(b) \( [ML^{-1}T^{-1}] \)
(c) \( [ML^2T^{-2}] \)
(d) \( [MT^{-2}] \)
Show Answer & Explanation
Answer: (c) \( [ML^2T^{-2}] \)
Explanation:
1. Energy is work, force × distance.
2. \( [E]=MLT^{-2}\times L=ML^2T^{-2} \).
Teacher's Note:
1. W
2. o
3. r
4. k
5. ,
6.
7. e
8. n
9. e
10. r
11. g
12. y
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56. .
(a) \( \sqrt{\dfrac{1}{\varepsilon_0\mu_0}} \)
(b) \( \sqrt{\varepsilon_0/\mu_0} \)
(c) \( \varepsilon_0/\mu_0 \)
(d) \( \varepsilon_0\mu \)
Show Answer & Explanation
Answer: (a) \( \sqrt{\dfrac{1}{\varepsilon_0\mu_0}} \)
Explanation:
1. Maxwell's theory gives \( c=\dfrac{1}{\sqrt{\mu_0\varepsilon_0}} \).
2. Check: \( \mu_0\varepsilon_0 \) has dimensions \( [L^{-2}T^2] \), so its inverse square root is \( [LT^{-1}] \), a velocity.
Teacher's Note:
1. c
2.
3. =
4.
5. 1
6. /
7. √
8. (
9. μ
10. ₀
11. ε
12. ₀
13. )
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69. o
70. n
71. f
72. i
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74. m
75. .
(a) \( [M^{-1}LT^{-2}A] \)
(b) \( [ML^{-2}T^{-2}A^{-1}] \)
(c) \( [MLT^{-2}A^{-2}] \)
(d) \( [MLT^{-1}A^{-1}] \)
Show Answer & Explanation
Answer: (c) \( [MLT^{-2}A^{-2}] \)
Explanation:
1. From \( F/L=\dfrac{\mu_0I^2}{2\pi d} \), so the force per unit length, F/L, has the same dimensions as \( \mu_0I^2/d \) and \( [\mu_0]=[F]/[I^2] \).
2. \( [\mu_0]=\dfrac{MLT^{-2}}{A^2}=MLT^{-2}A^{-2} \).
Teacher's Note:
1. T
2. h
3. e
4.
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7. p
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55.
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75. N
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77. A
78. ⁻
79. ²
80. .
(a) \( [ML^{-2}T^2A] \)
(b) \( [M^{-1}L^{-3}T^4A^2] \)
(c) \( [MLT^{-2}A] \)
(d) \( [ML^2T^{-1}A^2] \)
Show Answer & Explanation
Answer: (b) \( [M^{-1}L^{-3}T^4A^2] \)
Explanation:
1. From \( F=\dfrac{q_1q_2}{4\pi\varepsilon_0r^2} \), \( \varepsilon_0=\dfrac{q^2}{Fr^2} \).
2. With \( q=AT \): \( [\varepsilon_0]=\dfrac{A^2T^2}{MLT^{-2}L^2}=M^{-1}L^{-3}T^4A^2 \).
Teacher's Note:
1. P
2. e
3. r
4. m
5. i
6. t
7. t
8. i
9. v
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11. t
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25. ²
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28. ⁻
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35. (
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37. a
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40. d
41.
42. p
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44. r
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60.
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69. =
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72. /
73. (
74. μ
75. ₀
76. ε
77. ₀
78. )
79. .
(a) time
(b) distance
(c) frequency
(d) angular acceleration
Show Answer & Explanation
Answer: (b) distance
Explanation:
1. A parsec is the distance at which an astronomical unit subtends an angle of one second of arc.
2. It is a unit of distance (about 3.26 light years).
Teacher's Note:
1. P
2. a
3. r
4. s
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101. .
(a) \( [ML^2T^{-3}A^{-1}] \)
(b) \( [ML^2T^{-3}A^{-2}] \)
(c) \( [ML^3T^{-3}A^{-2}] \)
(d) \( [M^{-1}L^{-3}T^3A^2] \)
Show Answer & Explanation
Answer: (b) \( [ML^2T^{-3}A^{-2}] \)
Explanation:
1. \( R=V/I \) with \( V=W/q=\dfrac{ML^2T^{-2}}{AT}=ML^2T^{-3}A^{-1} \).
2. Dividing by current: \( [R]=ML^2T^{-3}A^{-2} \).
Teacher's Note:
1. R
2. e
3. s
4. i
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6. t
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64. .
(a) \( [LA] \)
(b) \( [L^2A] \)
(c) \( [LT^{-1}A] \)
(d) \( [L^2T^{-1}A] \)
Show Answer & Explanation
Answer: (b) \( [L^2A] \)
Explanation:
1. Magnetic moment of a current loop is \( \mu=IA \), current times area.
2. So \( [\mu]=[A][L^2]=[L^2A] \).
Teacher's Note:
1. M
2. a
3. g
4. n
5. e
6. t
7. i
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80. .
(a) pressure and stress
(b) tension and surface tension
(c) strain and angle
(d) energy and work
Show Answer & Explanation
Answer: (b) tension and surface tension
Explanation:
1. Pressure and stress are force per area, strain and angle are dimensionless, and energy and work are \( [ML^2T^{-2}] \).
2. Tension (a force, \( [MLT^{-2}] \)) and surface tension (force per length, \( [MT^{-2}] \)) are different.
Teacher's Note:
1. B
2. e
3.
4. c
5. a
6. r
7. e
8. f
9. u
10. l
11. :
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51. a
52. c
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61. n
62.
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65.
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71.
72. p
73. e
74. r
75.
76. l
77. e
78. n
79. g
80. t
81. h
82. .
(a) \( [ML^2T^{-3}A^{-2}] \)
(b) \( [M^{-1}L^{-2}T^3A^2] \)
(c) \( [ML^3T^{-3}A^{-2}] \)
(d) \( [M^{-1}L^{-3}T^3A^2] \)
Show Answer & Explanation
Answer: (a) \( [ML^2T^{-3}A^{-2}] \)
Explanation:
1. Impedance is measured in ohm like resistance.
2. So \( [Z]=ML^2T^{-3}A^{-2} \).
Teacher's Note:
1. R
2. e
3. s
4. i
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(a) \( G^{1/2}h^{1/2}c^{-5/2} \)
(b) \( G^{-1/2}h^{1/2}c^{1/2} \)
(c) \( G^{1/2}h^{1/2}c^{-3/2} \)
(d) \( G^{1/2}h^{1/2}c^{1/2} \)
Show Answer & Explanation
Answer: (a) \( G^{1/2}h^{1/2}c^{-5/2} \)
Explanation:
1. Let \( T\propto G^xh^yc^z \) with \( [G]=M^{-1}L^3T^{-2} \), \( [h]=ML^2T^{-1} \), \( [c]=LT^{-1} \).
2. Mass: −x + y = 0. Length: 3x + 2y + z = 0. Time: −2x − y − z = 1.
3. Solving, x = y = 1/2 and z = −5/2, so \( T\propto G^{1/2}h^{1/2}c^{-5/2} \) (the Planck time).
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(a) \( [MT^2] \)
(b) \( [ML^{-3}T^{-4}] \)
(c) \( [ML^{-1}T^{-2}] \)
(d) \( [ML^{-1}T^{-1}] \)
Show Answer & Explanation
Answer: (d) \( [ML^{-1}T^{-1}] \)
Explanation:
1. Stokes' law gives \( F=6\pi\eta rv \), so \( \eta=\dfrac{F}{rv} \).
2. \( [\eta]=\dfrac{MLT^{-2}}{L\cdot LT^{-1}}=ML^{-1}T^{-1} \).
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