JEE Physics Units and Dimensions MCQs Set 01

Answer: (a) SI

Explanation:
1. MKS, CGS and FPS use only mass, length and time as base quantities.
2. SI has seven base quantities including current, temperature, amount of substance and luminous intensity.

Teacher's Note:
1. Compare the number of base quantities in each system.
2. SI is the most general system.

Answer: (c) Displacement gradient

Explanation:
1. Displacement gradient = length/length, a pure ratio.
2. Velocity gradient has unit s⁻¹; the others have units of force per area or per length.

Teacher's Note:
1. Gradient = change of a quantity per unit length.
2. A ratio of two lengths has no unit.

Answer: (a) 10³

Explanation:
1. 1 micron = 10⁻⁶ m and 1 nm = 10⁻⁹ m.
2. Ratio = 10⁻⁶/10⁻⁹ = 10³.

Teacher's Note:
1. Know the SI prefixes: micro = 10⁻⁶, nano = 10⁻⁹.
2. Divide the powers of ten.

Answer: (d) None of these

Explanation:
1. Temperature is a base quantity in the SI system (kelvin).
2. It cannot be derived from length, mass and time.

Teacher's Note:
1. The seven SI base quantities cannot be obtained from each other.
2. Temperature measures a distinct thermal property.

Answer: (b) 3, 2, 4, 1

Explanation:
1. Stellar distances are in light years (3).
2. Inter-atomic spacing is about 1 Å (2); nuclear size is a few fermi (4); infrared wavelength is about a micron (1).
3. So A-3, B-2, C-4, D-1.

Teacher's Note:
1. Match each quantity with the unit of the right order of size.
2. Fermi = 10⁻¹⁵ m, angstrom = 10⁻¹⁰ m, micron = 10⁻⁶ m.

Answer: (d) Parallactic second

Explanation:
1. A parallactic second (parsec) is a unit of distance.
2. The other three are units of time.

Teacher's Note:
1. 1 parsec = 3.26 light years.
2. Light year and parsec are both distance units.

Answer: (d) 1 newton = 10⁻⁵ dynes

Explanation:
1. 1 N = 10⁵ dyne, not 10⁻⁵ dyne.
2. The other three relations are correct.

Teacher's Note:
1. 1 N = 1 kg m s⁻² = 10³ g × 10² cm s⁻² = 10⁵ dyne.
2. 1 eV = 1.6 × 10⁻¹⁹ J.

Answer: (b) Distance

Explanation:
1. A parsec is the distance at which one astronomical unit subtends an angle of one second of arc.

Teacher's Note:
1. 1 parsec ≈ 3.08 × 10¹⁶ m.
2. Used for stellar distances.

Answer: (a) Wave velocity

Explanation:
1. bt and cx are dimensionless, so [b] = [T⁻¹] and [c] = [L⁻¹].
2. [b/c] = [LT⁻¹], the dimension of velocity.

Teacher's Note:
1. The argument of a sine is dimensionless.
2. Velocity = ω/k.

Answer: (c) [M⁰L⁻¹T⁰]

Explanation:
1. Wave number = 1/wavelength, so its dimension is [L⁻¹].

Teacher's Note:
1. Wave number k = 2π/λ.
2. Its SI unit is m⁻¹.

Answer: (a) Potential gradient

Explanation:
1. Energy gradient = energy/length = force; weight is a force; dp/dt = force.
2. Potential gradient (electric) = volt/metre, which is electric field, not force.

Teacher's Note:
1. Gradient of potential energy is force, but of potential is field.
2. V/m = N/C.

Answer: (c) A dimensionally incorrect equation may be correct

Explanation:
1. A dimensionally incorrect equation must be wrong.
2. So 'may be correct' is the false statement.

Teacher's Note:
1. Dimensional correctness is necessary but not sufficient.
2. Missing numerical factors cannot be detected.

Answer: (c) The dimension of a derived quantity is never zero in any base quantity

Explanation:
1. A derived quantity can have zero power in a base quantity, e.g. velocity has zero power of mass.
2. So statement 3 is incorrect.

Teacher's Note:
1. Dimension of velocity = [M⁰L¹T⁻¹].
2. Zero powers are written as 0 or omitted.

Answer: (b) Torque and work

Explanation:
1. Torque (vector) and work (scalar) both have dimension [ML²T⁻²].
2. Work and energy are both scalars; impulse and momentum are both vectors.

Teacher's Note:
1. Check the vector/scalar nature of each pair.
2. Torque and work differ in direction behaviour.

Answer: (b) Has dimensions [T⁻²]

Explanation:
1. The exponent αt² is dimensionless, so [α] = [t²]⁻¹ = [T⁻²].

Teacher's Note:
1. Argument of an exponential is dimensionless.
2. Match α with the inverse square of time.

Answer: (a) [M⁰L⁰T⁻¹]

Explanation:
1. Angular velocity = angle/time, and angle is dimensionless, so the dimension is [T⁻¹].

Teacher's Note:
1. rad/s is its SI unit.
2. Radian has no dimension.

Answer: (a) [MLT⁻³] and [MLT⁻⁴]

Explanation:
1. [at] = [F] gives [a] = [MLT⁻²]/[T] = [MLT⁻³].
2. [bt²] = [F] gives [b] = [MLT⁻²]/[T²] = [MLT⁻⁴].

Teacher's Note:
1. Each term must have the dimension of force.
2. Divide by the power of t.

Answer: (a) −1/2, 1/2, 5/2

Explanation:
1. [G] = [M⁻¹L³T⁻²], [h] = [ML²T⁻¹], [c] = [LT⁻¹], [E] = [ML²T⁻²].
2. Mass: −p + q = 1. Length: 3p + 2q + r = 2. Time: −2p − q − r = −2.
3. Solving: p = −1/2, q = 1/2, r = 5/2.

Teacher's Note:
1. Equate powers of M, L and T on both sides.
2. Check by substituting back.

Answer: (a) Tension and surface tension

Explanation:
1. Tension is a force [MLT⁻²]; surface tension is force per length [MT⁻²].
2. The other pairs have identical dimensions.

Teacher's Note:
1. Do not confuse tension (force) with surface tension.
2. Planck's constant = angular momentum.

Answer: (c) [ML²T⁻²]

Explanation:
1. Torque = force × distance = [MLT⁻²][L] = [ML²T⁻²].

Teacher's Note:
1. Torque has the same dimensions as work and energy.
2. Its unit N m is not joule, to avoid confusion.

Answer: (c) Dimensionless

Explanation:
1. Relative density is the ratio of two densities, so it has no dimensions.

Teacher's Note:
1. Ratios of like quantities are dimensionless.
2. It also has no unit.

Answer: (d) [M⁻¹L³T⁻²]

Explanation:
1. F = Gm₁m₂/r² gives [G] = [F][L²][M⁻²] = [MLT⁻²][L²][M⁻²] = [M⁻¹L³T⁻²].

Teacher's Note:
1. Rearrange the law of gravitation.
2. Unit: N m² kg⁻².

Answer: (a) (αa + βb + γc)%

Explanation:
1. For a product of powers the maximum fractional errors add, each multiplied by the magnitude of its power.
2. So the percentage error is (aα + bβ + cγ)%.

Teacher's Note:
1. Maximum error: add the weighted fractional errors.
2. Subtraction of errors never occurs.

Answer: (a) ±13%

Explanation:
1. Percentage error = 1 + 2 × 3 + 3 × 2 = 13%.

Teacher's Note:
1. Powers multiply the percentage errors.
2. Add all the contributions.

Answer: (c) 2x + y

Explanation:
1. g = 4π²ℓ/T².
2. δg/g = δℓ/ℓ + 2δT/T = y + 2x.

Teacher's Note:
1. The power of T is 2, so its error is doubled.
2. Fractional errors add.

Answer: (a) 4%

Explanation:
1. Area = ℓ², so the fractional increase is doubled: 2 × 2% = 4%.

Teacher's Note:
1. For small changes, δA/A = 2δℓ/ℓ.
2. Linear expansion of copper is the same in both cases.