JEE Physics Units and Dimensions MCQs Set 08

Answer: (b) (iv)

Explanation:
1. A scalar has only magnitude, so its value does not depend on the orientation of the axes.
2. Scalars need not be conserved (heat), can be negative (charge, work) and can vary with position (temperature).

Teacher's Note:
1. A scalar is unchanged by rotating the axes.
2. Conservation or sign are not part of the definition.

Answer: (d) Parallactic second

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

Teacher's Note:
1. Parsec is derived from parallax and measures distance.
2. Do not be misled by the word second.

Answer: (d) none of these

Explanation:
1. Temperature is a fundamental quantity in the SI system.
2. It cannot be written in terms of mass, length and time.

Teacher's Note:
1. Seven base quantities include temperature.
2. A derived quantity is built from base quantities, but temperature is itself a base quantity.

Answer: (a) only numerically correct

Explanation:
1. Numerically, the distance in the n-th second is u + a(2n - 1)/2, with n = t seconds.
2. Dimensionally, the term (2t - 1) subtracts a number from a time, which is not allowed. So it is not dimensionally correct.

Teacher's Note:
1. The constant 1 hides a time unit of one second.
2. A dimensionally correct equation cannot add a number to a time.

Answer: (c) strain, specific gravity, angle

Explanation:
1. Strain, specific gravity and angle are ratios of like quantities.
2. Work, couple and frequency are not dimensionless.

Teacher's Note:
1. Ratios of two like quantities are dimensionless.
2. Work and energy have dimensions ML²T⁻².

Answer: (c) Planck constant and energy

Explanation:
1. Planck constant h = E/ν has dimensions ML²T⁻¹.
2. Energy has dimensions ML²T⁻².
3. The other pairs have identical dimensions (the dimensions of the last two pairs are both zero).

Teacher's Note:
1. Use E = hν to get the dimensions of h.
2. Compare the powers of T.

Answer: (b) \( \gamma^2/\alpha\beta^2 \)

Explanation:
1. 1 J = 1 kg m² s⁻².
2. 1 kg = 1/α new mass units, 1 m = 1/β new length units and 1 s = 1/γ new time units.
3. 1 J = (1/α)(1/β²)(γ²) = γ²/(αβ²).

Teacher's Note:
1. Convert each base unit separately.
2. Time appears with a negative power, so γ ends up in the numerator.

Answer: (c) \( m=\frac{m_0}{\sqrt{1-v^2/c^2}} \)

Explanation:
1. The quantity under the root must be dimensionless, so v must be divided by c.
2. Then m and m₀ have the same dimensions.

Teacher's Note:
1. 1 - v² is not dimensionally valid.
2. Only v²/c² is dimensionless.

Answer: (c) \( ML^{5}T^{-2} \)

Explanation:
1. a/V² has the dimensions of P.
2. a = P V² = (ML⁻¹T⁻²)(L⁶) = ML⁵T⁻².

Teacher's Note:
1. Terms added together must share dimensions.
2. Multiply pressure by volume squared.

Answer: (c) \( M^{-1}L^3T^{-2} \)

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

Teacher's Note:
1. Make G the subject of the law.
2. Substitute the dimensions of F, r and m.

Answer: (b) \( [ML^{-1}T^0] \)

Explanation:
1. μ = F p²/(4ℓ²f²).
2. [μ] = (MLT⁻²)/(L²T⁻²) = ML⁻¹.
3. So μ is the mass per unit length.

Teacher's Note:
1. Solve for μ.
2. The result is the mass per unit length of the string.

Answer: (c) \( ML^{-1}T^{-1} \)

Explanation:
1. Viscous force F = ηA(dv/dx), so η = F/(A·dv/dx).
2. [η] = MLT⁻²/(L²·T⁻¹) = ML⁻¹T⁻¹.

Teacher's Note:
1. dv/dx has dimensions T⁻¹.
2. Divide force by area and velocity gradient.

Answer: (a) \( [M^{-3}L^{-2}A^4T^8] \)

Explanation:
1. [X] = M⁻¹L⁻²T⁴A² and [Z] = MT⁻²A⁻¹.
2. [Y] = [X]/[Z]² = M⁻¹L⁻²T⁴A²/(M²T⁻⁴A⁻²) = M⁻³L⁻²T⁸A⁴.

Teacher's Note:
1. Divide X by Z squared.
2. Subtract exponents carefully.

Answer: (b) momentum and Planck's constant

Explanation:
1. Momentum = MLT⁻¹ while Planck's constant = ML²T⁻¹.
2. The other pairs have identical dimensions.

Teacher's Note:
1. Check powers of L in each pair.
2. 1/√(μ₀ε₀) is the speed of light.

Answer: (a) \( T=k\sqrt{\rho r^3/S} \)

Explanation:
1. Let T = kρᵃrᵇSᶜ. S has dimensions MT⁻².
2. M: a + c = 0; L: -3a + b = 0; T: -2c = 1, so c = -1/2, a = 1/2, b = 3/2.
3. T = k√(ρr³/S).

Teacher's Note:
1. Match the powers of M, L and T on both sides.
2. Solve the three linear equations.

Answer: (b) \( [Ev^{-2}T^{-2}] \)

Explanation:
1. Surface tension S = MT⁻² (force per length).
2. E = Mv², so M = Ev⁻².
3. S = Ev⁻²T⁻².

Teacher's Note:
1. Express M through E and v.
2. Keep time as T.

Answer: (c) x and A

Explanation:
1. Cz is an angle, so C has dimensions of 1/z.
2. B tan(Cz) has the dimensions of x, so [B] = [x]. Also [A] = [x]/[y].
3. So x and A differ unless y is dimensionless.

Teacher's Note:
1. The argument of tan must be dimensionless.
2. Each term in x has the dimensions of x.

Answer: (d) \( \text{has dimensions }T^{-2} \)

Explanation:
1. The exponent αt² must be dimensionless.
2. So [α] = T⁻².

Teacher's Note:
1. The exponent of e is always dimensionless.
2. Invert the dimension of t².

Answer: (c) \( M^2LT^{-5}A^{-2} \)

Explanation:
1. [E] = MLT⁻³A⁻¹ and [B] = MT⁻²A⁻¹.
2. [E×B] = M²LT⁻⁵A⁻².

Teacher's Note:
1. Multiply the dimensions of the two fields.
2. Add exponents for each base unit.

Answer: (d) angle

Explanation:
1. EJ² = (ML²T⁻²)(M²L⁴T⁻²) = M³L⁶T⁻⁴.
2. M⁵G² = M⁵·M⁻²L⁶T⁻⁴ = M³L⁶T⁻⁴.
3. The ratio is dimensionless, like an angle.

Teacher's Note:
1. Compute numerator and denominator separately.
2. The ratio has no dimensions.

Answer: (b) \( M^{-1}L^{-3}T^4A^2 \)

Explanation:
1. F = q₁q₂/(4πε₀r²) gives ε₀ = q²/(F r²).
2. [ε₀] = (AT)²/(MLT⁻²·L²) = M⁻¹L⁻³T⁴A².

Teacher's Note:
1. Use Coulomb's law.
2. Charge has dimensions AT.

Answer: (d) henry (H)

Explanation:
1. Henry = ML²T⁻²A⁻² = ML²/(AT)² = ML²/Q².
2. So the unit is the henry.

Teacher's Note:
1. Charge Q = AT.
2. Rewrite the dimension of inductance with Q.