JEE Physics Units And Measurements MCQs Set 18

Multiple Choice Questions (MCQs) for JEE Physics: Units and Measurements

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Practice Units and Measurements MCQs for JEE Physics

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Multiple Choice Questions

Question 1: Two resistors R₁ = (100 ± 3) Ω and R₂ = (200 ± 4) Ω are connected in series. The equivalent resistance is
(a) (300 ± 7) Ω
(b) (300 ± 1) Ω
(c) (100 ± 7) Ω
(d) none of these
Show Answer & Explanation

Answer: (a) (300 ± 7) Ω

Explanation:
1. R = R₁ + R₂ = 300 Ω.
2. ΔR = ΔR₁ + ΔR₂ = 3 + 4 = 7 Ω.

Teacher's Note:
1. Absolute errors add in a sum, even in a difference.

Question 2: In a vernier callipers, N divisions of the vernier scale coincide with (N − 1) divisions of the main scale, in which one division is 1 mm. The least count in cm is
(a) N
(b) N − 1
(c) 1/(10N)
(d) (1/N) − 1
Show Answer & Explanation

Answer: (c) 1/(10N)

Explanation:
1. 1 VSD = (N − 1)/N mm.
2. LC = 1 MSD − 1 VSD = 1 − (N−1)/N = 1/N mm = 1/(10N) cm.

Teacher's Note:
1. Least count of a vernier = value of one main scale division / number of vernier divisions.

Question 3: 1° (one degree) is equal to
(a) 17 rad
(b) 17.45 × 10⁻² rad
(c) 1.745 × 10⁻² rad
(d) 1.745 × 10² rad
Show Answer & Explanation

Answer: (c) 1.745 × 10⁻² rad

Explanation:
1. 1° = π/180 rad = 3.1416/180 = 1.745 × 10⁻² rad.

Teacher's Note:
1. Convert degrees to radians by multiplying with π/180.
2. Always keep track of the power of ten.

Question 4: The density of a material in the CGS system is 4 g/cm³. In a system in which the unit of length is 10 cm and the unit of mass is 100 g, the value of the density is
(a) 0.4
(b) 40
(c) 400
(d) 0.04
Show Answer & Explanation

Answer: (b) 40

Explanation:
1. n₂ = n₁ (M₁/M₂)(L₁/L₂)⁻³.
2. n₂ = 4 × (1/100) × (10)³ = 40.

Teacher's Note:
1. A larger unit gives a smaller number, so adjust the powers carefully.

Question 5: Which of the following is the correct dimension of the gravitational constant?
(a) [M⁻¹L³T⁻²]
(b) [MLT⁻¹]
(c) [ML⁻¹T⁻²]
(d) [ML⁻²T⁻²]
Show Answer & Explanation

Answer: (a) [M⁻¹L³T⁻²]

Explanation:
1. F = Gm₁m₂/r², so G = Fr²/(m₁m₂) = [MLT⁻²][L²]/[M²] = [M⁻¹L³T⁻²].

Teacher's Note:
1. G is a dimensional constant.

Question 6: Match the columns I and II.
Column IColumn II
(A) Practical unit(1) radian
(B) Base unit(2) light year
(C) Derived unit(3) kg m s⁻¹
(D) Complementary unit(4) second

(a) A-2, B-1, C-3, D-4
(b) A-2, B-4, C-3, D-1
(c) A-2, B-3, C-4, D-1
(d) A-4, B-2, C-3, D-1
Show Answer & Explanation

Answer: (b) A-2, B-4, C-3, D-1

Explanation:
1. Light year is a practical unit, second is a base unit, kg m s⁻¹ (momentum) is a derived unit, radian is a supplementary (complementary) unit.

Teacher's Note:
1. Classify units as base, derived, supplementary or practical.

Question 7: The wrong unit conversion among the following is
(a) 1 angstrom = 10⁻¹⁰ m
(b) 1 fermi = 10⁻¹⁵ m
(c) 1 light year = 9.46 × 10¹⁵ m
(d) 1 astronomical unit = 1.496 × 10⁻¹¹ m
Show Answer & Explanation

Answer: (d) 1 astronomical unit = 1.496 × 10⁻¹¹ m

Explanation:
1. 1 astronomical unit = 1.496 × 10¹¹ m (Earth-Sun distance).
2. The power of ten in option (d) has the wrong sign.

Teacher's Note:
1. Memorise the common conversions with their powers of ten.

Question 8: The diameter of a steel ball is measured with a vernier callipers having 0.1 cm main scale divisions, 10 vernier divisions matching 9 main scale divisions. Three readings (main scale, vernier divisions) are (0.5 cm, 8), (0.5 cm, 4), (0.5 cm, 6). The zero error is −0.03 cm. The mean corrected diameter is
(a) 0.52 cm
(b) 0.59 cm
(c) 0.56 cm
(d) 0.53 cm
Show Answer & Explanation

Answer: (b) 0.59 cm

Explanation:
1. LC = 0.1/10 = 0.01 cm.
2. Readings: 0.58, 0.54, 0.56 cm; mean 0.56 cm.
3. Zero error −0.03 cm, so the corrected value = 0.56 + 0.03 = 0.59 cm.

Teacher's Note:
1. Corrected reading = observed reading − zero error.

Question 9: The dimensions of the coefficient of self inductance are
(a) [ML²T⁻²A⁻²]
(b) [ML²T⁻²A⁻¹]
(c) [MLT⁻²A⁻²]
(d) [MLT⁻²A⁻¹]
Show Answer & Explanation

Answer: (a) [ML²T⁻²A⁻²]

Explanation:
1. U = ½LI², so L = 2U/I².
2. [L] = [ML²T⁻²]/[A²] = [ML²T⁻²A⁻²].

Teacher's Note:
1. Compare with the energy stored in the inductor.

Question 10: Which of the following is a dimensional constant?
(a) Refractive index
(b) Dielectric constant
(c) Relative density
(d) Gravitational constant
Show Answer & Explanation

Answer: (d) Gravitational constant

Explanation:
1. Refractive index, dielectric constant and relative density are ratios and have no dimension.
2. G has dimensions [M⁻¹L³T⁻²].

Teacher's Note:
1. Dimensional constants have dimensions but fixed values.

Question 11: Which of the following statements are correct? I. 1 calorie = 4.18 J. II. 1 Å = 10⁻¹⁰ m. III. 1 MeV = 1.6 × 10⁻¹³ J. IV. 1 newton = 10⁻⁵ dyne.
(a) I, II and III
(b) III and IV
(c) I only
(d) IV only
Show Answer & Explanation

Answer: (a) I, II and III

Explanation:
1. I, II and III are correct.
2. 1 N = 10⁵ dyne, so IV is wrong.

Teacher's Note:
1. 1 N = 10⁵ dyne and 1 J = 10⁷ erg are the CGS conversions.

Question 12: The physical quantity that does not have the dimensional formula [ML⁻¹T⁻²] is
(a) force
(b) pressure
(c) stress
(d) modulus of elasticity
Show Answer & Explanation

Answer: (a) force

Explanation:
1. Pressure, stress and modulus of elasticity all have dimensions of force per unit area, [ML⁻¹T⁻²].
2. Force has dimensions [MLT⁻²].

Teacher's Note:
1. Pressure, stress, Young's modulus and energy density share the same dimensions.

Question 13: One centimetre on the main scale of a vernier callipers is divided into 10 equal parts. If 10 divisions of the vernier coincide with 8 small divisions of the main scale, the least count is
(a) 0.01 cm
(b) 0.02 cm
(c) 0.05 cm
(d) 0.005 cm
Show Answer & Explanation

Answer: (b) 0.02 cm

Explanation:
1. 1 MSD = 0.1 cm. 10 VSD = 8 MSD, so 1 VSD = 0.08 cm.
2. LC = 0.1 − 0.08 = 0.02 cm.

Teacher's Note:
1. LC = 1 MSD − 1 VSD.

Question 14: 1 parsec is equal to
(a) 3.1 × 10⁻¹⁶ m
(b) 3.26 light year
(c) 6.3 × 10⁴ astronomical unit
(d) 1.496 × 10¹¹ m
Show Answer & Explanation

Answer: (b) 3.26 light year

Explanation:
1. 1 parsec = 3.08 × 10¹⁶ m and 1 ly = 9.46 × 10¹⁵ m.
2. Ratio = 3.08 × 10¹⁶ / 9.46 × 10¹⁵ = 3.26 ly.

Teacher's Note:
1. 1 parsec is about 3.26 light years, and about 2.06 × 10⁵ astronomical units.

Question 15: If L = 2.331 cm and B = 2.1 cm, then L + B is
(a) 4.4 cm
(b) 4 cm
(c) 4.43 cm
(d) 4.431 cm
Show Answer & Explanation

Answer: (a) 4.4 cm

Explanation:
1. The sum is 4.431 cm.
2. In addition the result keeps the least number of decimal places, here one (from 2.1).
3. So L + B = 4.4 cm.

Teacher's Note:
1. For addition and subtraction, count decimal places; for multiplication and division, count significant figures.

Question 16: Consider the statements: I. Light year and year both measure time. II. Both have the dimension of time. III. Light year measures length. Which is/are correct?
(a) I and II
(b) II and III
(c) II only
(d) III only
Show Answer & Explanation

Answer: (d) III only

Explanation:
1. A year measures time, but a light year is a distance.
2. So only statement III is correct.

Teacher's Note:
1. Always check what physical quantity a unit actually measures.

Question 17: Which of the following pairs has the same dimensions?
(a) Angular momentum and Planck's constant
(b) Dipole moment and electric field
(c) Both (a) and (b)
(d) None of these
Show Answer & Explanation

Answer: (a) Angular momentum and Planck's constant

Explanation:
1. Angular momentum and Planck's constant both have dimension [ML²T⁻¹].
2. Electric dipole moment is [ATL] but electric field is [MLT⁻³A⁻¹], which differ.

Teacher's Note:
1. Planck's constant has the dimensions of angular momentum, since E = hf.

Question 18: The error in the measurement of the radius of a sphere is 1%. The error in the measurement of its volume is
(a) 1%
(b) 5%
(c) 3%
(d) 8%
Show Answer & Explanation

Answer: (c) 3%

Explanation:
1. V ∝ r³, so ΔV/V = 3 Δr/r = 3 × 1% = 3%.

Teacher's Note:
1. The percentage error is multiplied by the power of the quantity.

Question 19: The density of a material in the CGS system is 8 g/cm³. In a system in which the unit of length is 5 cm and the unit of mass is 20 g, the density is
(a) 8
(b) 20
(c) 50
(d) 80
Show Answer & Explanation

Answer: (c) 50

Explanation:
1. n₂ = 8 × (5)³/20 = 8 × 125/20 = 50.

Teacher's Note:
1. The number gets bigger when the unit of length is larger because of the cube.

Question 20: The dimensions of pressure are the same as those of
(a) force per unit volume
(b) energy per unit volume
(c) force
(d) energy
Show Answer & Explanation

Answer: (b) energy per unit volume

Explanation:
1. Energy per unit volume = [ML²T⁻²]/[L³] = [ML⁻¹T⁻²], which is the dimension of pressure.
2. Force per unit volume would be [ML⁻²T⁻²].

Teacher's Note:
1. Pressure = energy density is a useful identity.

Question 21: When two quantities are added or subtracted, the absolute error in the result is
(a) the sum of the absolute errors of the quantities
(b) the difference of the absolute errors
(c) the product of the absolute errors
(d) the quotient of the absolute errors
Show Answer & Explanation

Answer: (a) the sum of the absolute errors of the quantities

Explanation:
1. For Z = A ± B, ΔZ = ΔA + ΔB, because errors can add up in the worst case.

Teacher's Note:
1. Even in subtraction the absolute errors add.

Question 22: The relative density of a metal is found with a spring balance. In air it reads (5.00 ± 0.05) N and in water (4.00 ± 0.05) N. The relative density is
(a) (5.00 ± 0.05)
(b) (5.00 ± 11%)
(c) (5.00 ± 0.10)
(d) (5.00 ± 6%)
Show Answer & Explanation

Answer: (b) (5.00 ± 11%)

Explanation:
1. Relative density = W_air/(W_air − W_water) = 5.00/1.00 = 5.00.
2. Error in the loss of weight = 0.05 + 0.05 = 0.10 N, which is 10% of 1.00 N.
3. Total = 1% + 10% = 11%.

Teacher's Note:
1. The difference of two close readings has a large percentage error.

Question 23: One unified atomic mass unit is equal to
(a) 12 times the mass of one carbon-12 atom
(b) 1/12 of the mass of 12 atoms of C-12
(c) 1/12 of the mass of one atom of C-12
(d) 12 times the mass of 12 atoms of C-12
Show Answer & Explanation

Answer: (c) 1/12 of the mass of one atom of C-12

Explanation:
1. By definition 1 u = (1/12) × mass of one C-12 atom = 1.66 × 10⁻²⁷ kg.

Teacher's Note:
1. Option (b) is wrong because it divides the mass of 12 atoms, which equals the mass of one atom.

Question 24: The frequency of vibration of a string is f = (n/2L)√(T/m), where T is the tension, L the length and n the harmonic number. The dimensional formula of m is
(a) [M⁰LT]
(b) [M L⁻¹T⁻¹]
(c) [M L⁻¹T⁰]
(d) [M⁰LT⁻¹]
Show Answer & Explanation

Answer: (c) [M L⁻¹T⁰]

Explanation:
1. f² = n²T/(4L²m), so m = T/(f²L²) (n is a number).
2. [m] = [MLT⁻²]/([T⁻²][L²]) = [ML⁻¹]. m is the mass per unit length.

Teacher's Note:
1. Square both sides of the formula before comparing dimensions.

Question 25: If energy (E), velocity (V) and time (T) are chosen as fundamental quantities, the dimensional formula of surface tension will be
(a) [EV⁻¹T⁻²]
(b) [EV⁻²T⁻²]
(c) [E⁻²V⁻¹T⁻³]
(d) [EV⁻²T⁻¹]
Show Answer & Explanation

Answer: (b) [EV⁻²T⁻²]

Explanation:
1. Surface tension = energy/area = E/L².
2. L = VT, so L² = V²T².
3. Surface tension = [EV⁻²T⁻²].

Teacher's Note:
1. Express length through V and T first.

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