JEE Physics Units and Dimensions MCQs Set 02

Practice MCQs for JEE Physics Units and Dimensions

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Multiple Choice Questions

Question 1: If the error in measuring the diameter of a circle is 4%, the error in the circumference of the circle would be
(a) 2%
(b) 8%
(c) 4%
(d) 1%
Show Answer & Explanation

Answer: (c) 4%

Explanation:
1. Circumference = πd, which is proportional to d, so the fractional error is the same, 4%.

Teacher's Note:
1. A constant factor does not change the fractional error.
2. Area would show 8%.

Question 2: The external and internal radii of a hollow cylinder are measured to be (4.23 ± 0.01) cm and (3.89 ± 0.01) cm. The thickness of the wall of the cylinder is
(a) (0.34 ± 0.02) cm
(b) (0.17 ± 0.02) cm
(c) (0.17 ± 0.01) cm
(d) (0.34 ± 0.01) cm
Show Answer & Explanation

Answer: (a) (0.34 ± 0.02) cm

Explanation:
1. Thickness = 4.23 − 3.89 = 0.34 cm.
2. For a difference, absolute errors add: 0.01 + 0.01 = 0.02 cm.

Teacher's Note:
1. Add absolute errors for both sum and difference.
2. The thickness is the difference of radii, not half of it.

Question 3: The heat generated in a circuit depends on the resistance, current and time for which the current flows. If the errors in measuring these are 1%, 2% and 1% respectively, the maximum error in the heat will be
(a) 2%
(b) 3%
(c) 6%
(d) 1%
Show Answer & Explanation

Answer: (c) 6%

Explanation:
1. H = I²Rt, so error = 2 × 2 + 1 + 1 = 6%.

Teacher's Note:
1. The current appears squared.
2. Add all the percentage contributions.

Question 4: The percentage errors in the measurement of mass and speed are 2% and 3% respectively. The maximum error in the estimate of kinetic energy obtained by measuring mass and speed will be
(a) 11%
(b) 8%
(c) 5%
(d) 1%
Show Answer & Explanation

Answer: (b) 8%

Explanation:
1. KE = ½mv², so error = 2 + 2 × 3 = 8%.

Teacher's Note:
1. Speed is squared, so its error doubles.
2. Constants like ½ have no error.

Question 5: The pressure on a square plate is measured by measuring the force on the plate and the length of its sides. If the maximum errors in the measurement of force and length are 4% and 2% respectively, the maximum error in the measurement of pressure is
(a) 1%
(b) 2%
(c) 6%
(d) 8%
Show Answer & Explanation

Answer: (d) 8%

Explanation:
1. P = F/ℓ², so the error = 4 + 2 × 2 = 8%.

Teacher's Note:
1. Area is the square of the side.
2. Errors of a quotient add.

Question 6: The error in measuring the side of a cube is ±1%. The error in the calculation of the volume of the cube will be about
(a) ±0.001%
(b) ±1%
(c) ±6%
(d) ±3%
Show Answer & Explanation

Answer: (d) ±3%

Explanation:
1. V = a³, so the error is 3 × 1% = 3%.

Teacher's Note:
1. Cube: multiply by 3.
2. Square: multiply by 2.

Question 7: The resistance R of a wire is given by R = ρℓ/(πr²). The percentage errors in the measurement of ρ, ℓ and r are 1%, 2% and 3% respectively. The percentage error in the measurement of R is
(a) 6%
(b) 9%
(c) 8%
(d) 10%
Show Answer & Explanation

Answer: (b) 9%

Explanation:
1. Error = 1 + 2 + 2 × 3 = 9%.

Teacher's Note:
1. The radius is squared in the denominator.
2. Denominator terms also add.

Question 8: Which of the following has the highest number of significant figures?
(a) 0.007 m²
(b) 2.64 × 10²⁴ kg
(c) 0.0006032 m²
(d) 6.3200 J
Show Answer & Explanation

Answer: (d) 6.3200 J

Explanation:
1. 0.007 has 1, 2.64 × 10²⁴ has 3, 0.0006032 has 4, and 6.3200 has 5 significant figures.

Teacher's Note:
1. Leading zeros are not significant.
2. Trailing zeros after the decimal point are significant.

Question 9: In a vernier callipers, N divisions of the vernier scale coincide with (N − 1) divisions of the main scale (in which 1 division represents 1 mm). The least count of the instrument in cm should be
(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, so LC = 1 MSD − 1 VSD = 1/N mm = 1/(10N) cm.

Teacher's Note:
1. LC = 1 MSD ÷ number of vernier divisions.
2. Convert mm to cm.

Question 10: A vernier callipers has 20 divisions on the vernier scale which coincide with 19 divisions on the main scale. The least count of the instrument is 0.1 mm. The main scale divisions are of
(a) 0.5 mm
(b) 1 mm
(c) 2 mm
(d) 1/4 mm
Show Answer & Explanation

Answer: (c) 2 mm

Explanation:
1. LC = MSD/20 = 0.1 mm, so MSD = 2 mm.

Teacher's Note:
1. LC = 1 MSD/N for N vernier divisions equal to N − 1 main divisions.
2. Check: 2 mm/20 = 0.1 mm.

Question 11: A student measured the diameter of a wire using a screw gauge with least count 0.001 cm and listed the measurements. The correct measurement is
(a) 5.3 cm
(b) 5.32 cm
(c) 5.320 cm
(d) 5.3200 cm
Show Answer & Explanation

Answer: (c) 5.320 cm

Explanation:
1. With a least count of 0.001 cm a reading must be quoted to three decimal places.
2. So 5.320 cm is the correct form.

Teacher's Note:
1. The number of decimal places must match the least count.
2. 5.3200 would imply a finer instrument.

Question 12: The velocity v of a particle at time t is given by v = at + b/(t + c), where a, b and c are constants. The dimensions of a, b and c are respectively
(a) LT⁻², L and T
(b) L², T and LT²
(c) LT², LT and L
(d) L, LT and T²
Show Answer & Explanation

Answer: (a) LT⁻², L and T

Explanation:
1. c is added to t, so [c] = T.
2. [b/(t + c)] = [LT⁻¹], so [b] = [L].
3. [at] = [LT⁻¹] gives [a] = [LT⁻²].

Teacher's Note:
1. Only quantities of equal dimension can be added.
2. Work out c first.

Question 13: The dimensions of electrical resistance are
(a) [ML²T⁻³A⁻¹]
(b) [ML²T⁻³A⁻²]
(c) [ML³T⁻³A⁻²]
(d) [ML⁻¹L³T³A²]
Show Answer & Explanation

Answer: (b) [ML²T⁻³A⁻²]

Explanation:
1. R = V/I and V = work/charge = [ML²T⁻²]/[AT] = [ML²T⁻³A⁻¹].
2. R = [ML²T⁻³A⁻¹]/[A] = [ML²T⁻³A⁻²].

Teacher's Note:
1. Ohm = V/A = kg m² s⁻³ A⁻².
2. Build up from work and charge.

Question 14: Which two of the following five physical parameters have the same dimensions? (a) energy density, (b) refractive index, (c) dielectric constant, (d) Young's modulus, (e) magnetic field
(a) (a) and (d)
(b) (a) and (e)
(c) (b) and (d)
(d) (c) and (e)
Show Answer & Explanation

Answer: (a) (a) and (d)

Explanation:
1. Energy density = energy/volume = [ML⁻¹T⁻²], which is also the dimension of Young's modulus (stress).
2. Refractive index and dielectric constant are dimensionless.

Teacher's Note:
1. Energy density and pressure share dimensions.
2. Ratios are dimensionless.

Question 15: If the dimensions of a physical quantity are given by MᵃLᵇTᶜ, then the physical quantity will be
(a) Force if a = 0, b = −1, c = −2
(b) Pressure if a = 1, b = −1, c = −2
(c) Velocity if a = 1, b = 0, c = −1
(d) Acceleration if a = 1, b = 1, c = −2
Show Answer & Explanation

Answer: (b) Pressure if a = 1, b = −1, c = −2

Explanation:
1. [Pressure] = [ML⁻¹T⁻²], which matches a = 1, b = −1, c = −2.
2. The other options give wrong dimensions (force needs a = 1, b = 1, c = −2; velocity [LT⁻¹]; acceleration [LT⁻²]).

Teacher's Note:
1. Compare powers with the known formulae.
2. Pressure = force/area.

Question 16: The dimensions of (1/2)ε₀E², where ε₀ is the permittivity of free space and E is the electric field, are
(a) [MLT⁻¹]
(b) [ML²T⁻²]
(c) [ML⁻¹T⁻²]
(d) [ML²T⁻¹]
Show Answer & Explanation

Answer: (c) [ML⁻¹T⁻²]

Explanation:
1. (1/2)ε₀E² is the energy density of the electric field = energy/volume.
2. = [ML²T⁻²]/[L³] = [ML⁻¹T⁻²].

Teacher's Note:
1. Energy per unit volume has the dimension of pressure.
2. Recognise the formula for field energy density.

Question 17: The dimensions of magnetic permeability are
(a) [MLT²A⁻²]
(b) [ML⁻¹T⁻²A⁻²]
(c) [ML⁻²T²A²]
(d) [MLT⁻²A⁻²]
Show Answer & Explanation

Answer: (d) [MLT⁻²A⁻²]

Explanation:
1. μ₀ = F/(I²) per unit length ratio: [μ₀] = [MLT⁻²A⁻²].

Teacher's Note:
1. Same as H m⁻¹ = N A⁻².
2. Check using F = μ₀I₁I₂l/(2πd).

Question 18: The dimensions of (μ₀ε₀)^(−1/2) are
(a) [L^(1/2)T^(−1/2)]
(b) [L⁻¹T]
(c) [LT⁻¹]
(d) [L^(−1/2)T^(1/2)]
Show Answer & Explanation

Answer: (c) [LT⁻¹]

Explanation:
1. c = 1/√(μ₀ε₀) is the speed of light, which has dimensions [LT⁻¹].

Teacher's Note:
1. Maxwell's relation: c² = 1/(μ₀ε₀).
2. Speed of light in vacuum.

Question 19: The dimensions of the magnetic field B in terms of C (coulomb), M, L and T are
(a) [M¹L¹T⁻²C]
(b) [M¹L⁰T⁻¹C⁻¹]
(c) [M¹L⁰T⁻²C]
(d) [M¹L⁰T⁻¹C]
Show Answer & Explanation

Answer: (b) [M¹L⁰T⁻¹C⁻¹]

Explanation:
1. From F = qvB, [B] = [MLT⁻²]/([C][LT⁻¹]) = [MT⁻¹C⁻¹].

Teacher's Note:
1. Use the Lorentz force law.
2. Unit: tesla = N s C⁻¹ m⁻¹.

Question 20: If the voltage across a bulb rated 220 V, 100 W drops by 2.5% of its rated value, the percentage by which the power decreases is
(a) 5%
(b) 10%
(c) 20%
(d) 2.5%
Show Answer & Explanation

Answer: (a) 5%

Explanation:
1. P = V²/R with constant R, so the percentage change in P = 2 × 2.5% = 5%.

Teacher's Note:
1. Power depends on the square of voltage.
2. Resistance is assumed fixed.

Question 21: In an experiment four quantities a, b, c and d are measured with percentage errors 1%, 2%, 3% and 4% respectively. The quantity P is calculated as P = a³b²/(cd). The percentage error in P is
(a) 4%
(b) 14%
(c) 10%
(d) 7%
Show Answer & Explanation

Answer: (b) 14%

Explanation:
1. Error = 3 × 1 + 2 × 2 + 3 + 4 = 14%.

Teacher's Note:
1. Add all weighted errors.
2. Denominator terms also add.

Question 22: The dimensional formula of angular momentum is
(a) [M¹L²T⁻¹]
(b) [M²L²T⁻²]
(c) [M¹L²T⁻³]
(d) [M¹L¹T⁻¹]
Show Answer & Explanation

Answer: (a) [M¹L²T⁻¹]

Explanation:
1. L = mvr = [M][LT⁻¹][L] = [ML²T⁻¹].

Teacher's Note:
1. Angular momentum has the same dimensions as Planck's constant.
2. L = Iω also works.

Question 23: If energy (E), velocity (V) and time (T) are chosen as the 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 = [MT⁻²]. E/V² = [ML²T⁻²]/[L²T⁻²] = [M].
2. So [MT⁻²] = [EV⁻²T⁻²].

Teacher's Note:
1. Express M through E and V.
2. Then attach T⁻².

Question 24: If the dimension of the critical velocity v_c of a liquid flowing through a tube is expressed as (ηˣρʸrᶻ), where η, ρ and r are the coefficient of viscosity, density of the liquid and radius of the tube, the values of x, y and z are
(a) 1, 1, 1
(b) 1, −1, −1
(c) −1, −1, 1
(d) −1, 1, 1
Show Answer & Explanation

Answer: (b) 1, −1, −1

Explanation:
1. [η] = [ML⁻¹T⁻¹], [ρ] = [ML⁻³], [r] = [L].
2. Time: −x = −1 so x = 1. Mass: x + y = 0 so y = −1. Length: −x − 3y + z = 1 so z = −1.

Teacher's Note:
1. Match powers of M, L and T on both sides.
2. Verify with the known formula v_c = Re·η/(ρr).

Question 25: The dimensions of inductance are
(a) [ML²T⁻²A⁻²]
(b) [ML²T⁻¹A⁻¹]
(c) [MLT⁻¹A⁻¹]
(d) [MLTA⁻²]
Show Answer & Explanation

Answer: (a) [ML²T⁻²A⁻²]

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

Teacher's Note:
1. Henry = J A⁻².
2. Use the energy stored in an inductor.

Question 26: Planck's constant (h), the speed of light in vacuum (c) and Newton's gravitational constant (G) are three fundamental constants. Which of the following combinations has the dimension of length?
(a) √(hc/G)
(b) √(Gc/h^(3/2))
(c) √(hG/c³)
(d) √(hG/c^(5/2))
Show Answer & Explanation

Answer: (c) √(hG/c³)

Explanation:
1. [hG/c³] = [ML²T⁻¹][M⁻¹L³T⁻²]/[L³T⁻³] = [L²].
2. So √(hG/c³) has dimension [L] (the Planck length).

Teacher's Note:
1. Combine constants so that M and T cancel.
2. Check each option's powers.

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