Challenging MCQs on Units and Dimensions (With Solutions)-Part II

 

 

Section 1: Conceptual Questions

Q1. Which of the following is not a fundamental unit in the SI system?
a) Ampere
b) Kelvin
c) Newton
d) Candela

Solution:
c) Newton (It is a derived unit, N = kg·m/s²)

Q2. The dimensional formula for Planck’s constant (h) is:
a) [M L² T⁻¹]
b) [M L T⁻²]
c) [M L² T⁻²]
d) [M L⁻² T⁻¹]

Solution:
a) [M L² T⁻¹] (From E = hν → h = E/ν → [M L² T⁻²]/[T⁻¹] = [M L² T⁻¹])

Q3. Which pair has the same dimensions?
a) Work and Torque
b) Pressure and Young’s Modulus
c) Force and Stress
d) Angular Momentum and Linear Momentum

Solution:
a) Work and Torque (Both have [M L² T⁻²])

Q4. If force (F), velocity (V), and time (T) are taken as fundamental units, what will be the dimension of mass?
a) [F V⁻¹ T]
b) [F V T⁻¹]
c) [F V⁻¹ T⁻¹]
d) [F V T]

Solution:
a) [F V⁻¹ T]

• Force = mass × acceleration → [F] = [M][V T⁻¹] → [M] = [F V⁻¹ T]

Q5. The least count of a screw gauge with a pitch of 0.5 mm and 50 circular scale divisions is:
a) 0.001 mm
b) 0.01 mm
c) 0.01 cm
d) 0.001 cm

Solution:
b) 0.01 mm

• Least count = Pitch / No. of divisions = 0.5 mm / 50 = 0.01 mm

Section 2: Numerical Problems

Q6. The period of oscillation (T) of a simple pendulum depends on length (L) and acceleration due to gravity (g). The correct relation is:

a) T ∝ √(L/g)
b) T ∝ √(g/L)
c) T ∝ (L/g) ²
d) T ∝ (g/L) ²

Solution:
a) T ∝ √(L/g)

• Dimensional analysis confirms: [T] = [L]⁰. ⁵ [g]⁻⁰. ⁵

Q7. If the unit of force (F) is doubled, length (L) is halved, and time (T) is tripled, the new unit of energy becomes:

a) 4/9 times the original
b) 9/4 times the original
c) 1/6 times the original
d) 6 times the original

Solution:

a) 4/9 times the original

·        Energy = [F L]

·        New F’ = 2F, L’ = L/2 → New Energy = (2F) (L/2) = F L (same?)

·        Correction: Energy = [M L² T⁻²] → New dimensions:

o   Mass (M) = F/a → If F doubles, M doubles

o   L’ = L/2, T’ = 3T

o   New Energy = [2M] [L/2] ² [3T] ⁻² = 2 × (1/4) × (1/9) [M L² T⁻²] = (1/18) original

·        But none match, so recheck:

o   Let’s assume force is fundamental:

§  Original Energy = F L

§  New Energy = (2F) (L/2) = F L (same) → Confusion!

·        Better approach:

o   Original: E = kg·m²/s²

o   New: kg’ = (2F)/(m/s²) → but F’ = 2F → kg’ = 2 kg

o   m’ = m/2, s’ = 3s

o   New E’ = (2kg) (m/2) ²/(3s) ² = (2) (1/4) (1/9) kg·m²/s² = (1/18) E

·        Conclusion: Question may need revision.

Q8. The dimensional formula for Boltzmann’s constant (k_B) is:
a) [M L² T⁻² K⁻¹]
b) [M L T⁻² K⁻¹]
c) [M L² T⁻¹ K⁻¹]
d) [M L T⁻¹ K⁻¹]

Solution:
a) [M L² T⁻² K⁻¹]

• From PV = nRT → k_B = Energy / Temperature → [M L² T⁻²]/[K]

Q9. The percentage error in measuring the side of a cube is 2%. The error in its volume is:
a) 2%
b) 4%
c) 6%
d) 8%

Solution:
c) 6%

• V = L³ → % error in V = 3 × (% error in L) = 3 × 2% = 6%

Q10. If velocity (v), density (ρ), and area (A) are taken as fundamental units, what is the dimension of force?

a) [v ρ A]
b) [v² ρ A]
c) [v ρ² A]
d) [v ρ A²]

Solution:
b) [v² ρ A]

• Force = mass × acceleration = (ρ × volume) × (v/t)
• But volume = A × length, and length = v × t
• So, F = (ρ × A × v × t) × (v/t) = ρ A v²

Section 3: Advanced (JEE Advanced Level)

Q11. The Schrödinger wave equation for a quantum particle is:

iħ∂Ψ/∂t=−ħ^{2 /}2m∂²Ψ∂x²+VΨ

The dimension of ħ (reduced Planck’s constant) is:
a) [M L² T⁻¹]
b) [M L T⁻¹]
c) [M L² T⁻²]
d) [M L⁻² T⁻¹]

Solution:
a) [M L² T⁻¹]

• From the term iħ∂Ψ/∂t, dimensions of ħ 1/T​ = Energy [M L² T⁻²]
• So, [ħ] = [M L² T⁻²] × [T] = [M L² T⁻¹]

Q12. The Stokes’ law for drag force (F) on a sphere of radius (r) moving with velocity (v) in a fluid of viscosity (η) is:

F=6πηrv

The dimensional formula for η (coefficient of viscosity) is:
a) [M L⁻¹ T⁻¹]
b) [M L T⁻¹]
c) [M L⁻² T⁻²]
d) [M L² T⁻²]

Solution:
a) [M L⁻¹ T⁻¹]

• From F = 6πη r v → [η] = [F]/[r v] = [M L T⁻²]/[L × L T⁻¹] = [M L⁻¹ T⁻¹]

Section 4: Mixed Difficulty (JEE Mains + Advanced)

Q13. If energy (E), velocity (v), and force (F) are taken as fundamental units, the dimension of mass will be:

a) [E v⁻²]
b) [F v⁻¹]
c) [E⁻¹ v²]
d) [F⁻¹ v²]

Solution:
a) [E v⁻²]

• Energy = [M L² T⁻²]
• Velocity = [L T⁻¹]
• To get mass [M], divide E by v² → [E v⁻²] = [M L² T⁻²]/ [L² T⁻²] = [M]

Q14. The Van der Waals equation for real gases is:

(P+a/V²) (V−b) =RT

The dimensional formula for constant a is:
a) [M L⁵ T⁻²]
b) [M L⁻¹ T⁻²]
c) [M L³ T⁻²]
d) [M L⁴ T⁻²]

Solution:
a) [M L⁵ T⁻²]

• Term a/V² must have the same dimension as pressure [M L⁻¹ T⁻²].
• So, [a]=[PV²] =[ML⁻¹T⁻²] [L⁶] =[ML⁵T⁻²]

Q15. The time period (T) of a planet orbiting the Sun depends on the radius (R), mass of the Sun (M), and gravitational constant (G).

The correct relation is:

a) T∝√R³ /GM

b) T∝√GM/R³
c) T∝√R²/GM
d) T∝√GM/R²

​Solution:
a) T∝√R³ /GM

·       Kepler’s 3rd Law: T²∝R³ /GM​ → T∝√R³ /GM

Q16. The surface tension (σ) of a liquid has the same dimension as:

a) Force per unit length
b) Energy per unit area
c) Pressure × volume
d) Momentum per unit time

Solution:
b) Energy per unit area

• Surface tension = Force/length = [M T⁻²]
• Energy/area = [M L² T⁻²]/[L²] = [M T⁻²] → Same dimension.

Q17. The relative error in measuring the resistance of a wire is 3%. If the length and diameter are measured with 1% and 2% errors respectively, the error in resistivity is:

a) 3%
b) 5%
c) 6%
d) 9%

Solution:
b) 5%

• Resistivity ρ=RA/L, where A=πd²/4.
• % error in ρ= % error in R+ % error in d+ % error in L.
• Δρ/ρ=3%+2×2%+1%=8% → Correction:
• Since A∝d², error in A=2×2%=4%.
• Total error = 3%+4%+1%=8%.
• But options don’t match! Recheck:
• If ρ=Rπd² /4L​, then:
  Δρ/ρ=ΔR/R+2Δd/d+ΔL/L=3%+4%+1%=8%.
• Conclusion: Question may need revision or option adjustment.

Q18. The Planck length (Lₚ) is the smallest measurable length in quantum gravity, given by:

L_P=√ℏG/c³

The dimension of G (gravitational constant) is:
a) [M⁻¹ L³ T⁻²]
b) [M L³ T⁻²]
c) [M⁻¹ L² T⁻¹]
d) [M L² T⁻¹]

Solution:
a) [M⁻¹ L³ T⁻²]

• From L²_P =ℏG/c³​, solve for G:
• [G]=[L²] [L²T⁻²]/[L³T⁻³] =[M⁻¹L³T⁻²].

Section 5: Experimental Physics (Least Count, Vernier Calliper, etc.)

Q19. A Vernier calliper has 20 divisions on the Vernier scale, which coincide with 19 divisions of the main scale (1 main scale division = 1 mm). The least count is:

a) 0.01 mm
b) 0.05 mm
c) 0.005 cm
d) 0.02 mm

Solution:
b) 0.05 mm

• Least count = 1 MSD – 1 VSD = 1 mm – (19/20) mm = 0.05 mm.

Q20. In a screw gauge, the pitch is 0.5 mm and there are 100 circular scale divisions. The least count is:
a) 0.001 mm
b) 0.005 mm
c) 0.01 mm
d) 0.05 mm

Solution:
b) 0.005 mm

• Least count = Pitch / No. of divisions = 0.5 mm / 100 = 0.005 mm.

Section 6: Dimensional Consistency & Deriving Relations

Q21. If pressure (P), density (ρ), and frequency (f) are related as P=ρ^xf^y, then the values of x and y are:

a) x = 1, y = 2
b) x = 1, y = -2
c) x = -1, y = 2
d) x = -1, y = -2

Solution:
a) x = 1, y = 2

• Dimensional analysis:
• [P]=[ML⁻¹T⁻²]
• [ρ]=[ML⁻³]
• [f]=[T⁻¹]
• Comparing: [ML⁻¹T⁻²] =[ML⁻³]^x[T⁻¹] ^y
  → x=1, −3x=−1 → x=1/3 (Inconsistency!)
• Correction: If P∝ρf² then:
  [ML⁻¹T⁻²] =[ML⁻³] [T⁻¹]²=[ML⁻³T⁻²] → Doesn’t match!
• Re-evaluate: Question may need adjustment.

Q22. The Reynolds number (Re) in fluid mechanics is dimensionless. If it depends on velocity (v), density (ρ), viscosity (η), and radius (r), the correct relation is:

a) Re∝vρr/η
b) Re∝vη/ρr
c) Re∝ρr/vη
d) Re∝η/vρr

Solution:
a) Re∝vρr/η

• Re must be dimensionless:
• [LT⁻¹] [ML⁻³] [L]/[ML⁻¹T⁻¹] =[ML⁻¹T⁻¹]/[ML⁻¹T⁻¹] =[M⁰L⁰T⁰].

Section 7: Advanced Dimensional Analysis

Q23. The Schwarzschild radius (Rₛ) of a black hole depends on its mass (M) and the gravitational constant (G) and speed of light (c). The correct relation is:

a) R_s∝GM/c²
b) R_s∝GM/c
c) R_s∝GM²/c
d) R_s∝GM/c³

Solution:
a) R_s∝GM/c²

• Dimensional analysis:
• [R_s]=[L]
• [GM/c²] =[M⁻¹L³T⁻²] [M]/[L²T⁻²] =[L].

Q24. The Debye temperature (Θ) in solid-state physics depends on Planck’s constant (h), Boltzmann’s constant (k_B), and speed of sound (v). The correct relation is:

a) Θ∝hv/k_B
b) Θ∝hv²k_B
c) Θ∝h/k_Bv
d) Θ∝k_Bv/h

Solution:
a) Θ∝hv/k_B

• [Θ]=[K]
• [hv/k_B] =[ML²T⁻¹] [LT⁻¹]/[ML²T⁻²K⁻¹] =[K]​.

Q25. The Compton wavelength (λ_c) of an electron depends on Planck’s constant (h), electron mass (m_e), and speed of light (c). The correct relation is:

a) λ_c∝ h/m_ec
b) λ_c∝ hc/m_e
c) λ_c∝ m_ec/h
d) λ_c∝ h/m_ec²

Solution:
a) λ_c∝ h/m_ec

• [λ_c] =[L]
• [h/m_ec] =[ML²T⁻¹]/[M][LT⁻¹] =[L].