JEE Physics Units and Dimensions MCQs Set 06

Answer: (c) impulse × distance

Explanation:
1. Action \( = ML^2T^{-2}\cdot T = ML^2T^{-1} \).
2. Impulse × distance \( = MLT^{-1}\cdot L = ML^2T^{-1} \).
3. Power and force × velocity are both \( ML^2T^{-3} \).

Teacher's Note:
1. Action has the dimensions of Planck's constant.
2. It also equals angular momentum.

Answer: (a) wave frequency

Explanation:
1. The argument \( bt \) is dimensionless, so \( [b] = T^{-1} \), the dimension of frequency.

Teacher's Note:
1. In the standard form sin(ωt − kx), b is ω.
2. Then b/c is the wave speed.

Answer: (d) wave velocity

Explanation:
1. \( [b] = T^{-1} \) and \( [c] = L^{-1} \), so \( [b/c] = LT^{-1} \).

Teacher's Note:
1. Ratio of angular frequency to wave number is the wave speed.
2. Wave number c has dimension L⁻¹.

Answer: (b) \( \alpha = ML^{-1}T^0,\ \beta = T^{-1} \)

Explanation:
1. \( \beta t \) is dimensionless, so \( [\beta] = T^{-1} \).
2. \( [\alpha] = \frac{MLT^{-2}}{L^2T^{-2}} = ML^{-1} \).

Teacher's Note:
1. The sine is dimensionless.
2. Divide the dimensions of force by those of velocity squared.

Answer: (d) the unit of momentum is doubled

Explanation:
1. Time \( = \frac{\text{length}}{\text{velocity}} \): factor \( \frac22 = 1 \). Mass \( = \frac{FT^2}{L} \): factor \( \frac{2}{2} = 1 \).
2. Momentum \( = FT \): factor 2. Energy \( = FL \): factor 4.

Teacher's Note:
1. Express each derived unit through the three doubled ones.
2. Energy is force × length, so it becomes four times.

Answer: (a) \( \rho a^4f^{-6} \)

Explanation:
1. \( T = f^{-1} \), \( L = aT^2 = af^{-2} \), \( M = \rho L^3 = \rho a^3f^{-6} \).
2. \( F = MLT^{-2} = \rho a^3f^{-6}\cdot af^{-2}\cdot f^2 = \rho a^4f^{-6} \).

Teacher's Note:
1. Express M, L and T in terms of the new base quantities first.
2. Frequency is the inverse of time.

Answer: (c) \( MT^{-2} \)

Explanation:
1. \( [\alpha] = \frac{[K]}{[s^2]} = \frac{ML^2T^{-2}}{L^2} = MT^{-2} \).

Teacher's Note:
1. Divide by L² to remove the length from energy.
2. α is like a spring constant.

Answer: (c) \( ML^{7/2}T^{-2} \)

Explanation:
1. \( \beta \) is added to y, so \( [\beta] = L \).
2. \( [\alpha] = [U]\frac{[y+\beta]}{[\sqrt y]} = ML^2T^{-2}\cdot\frac{L}{L^{1/2}} = ML^{5/2}T^{-2} \).
3. \( [\alpha\beta] = ML^{7/2}T^{-2} \).

Teacher's Note:
1. The constant beta takes the dimensions of y.
2. Then find alpha from the energy dimension.

Answer: (a) 10

Explanation:
1. 1 poiseuille \( = 1\ \frac{\text{kg}}{\text{m s}} = \frac{1000\text{ g}}{100\text{ cm}\cdot\text{s}} = 10\ \frac{\text{g}}{\text{cm s}} \).
2. So 1 poiseuille = 10 poise.

Teacher's Note:
1. Replace kg by 1000 g and m by 100 cm.
2. The ratio 1000/100 gives 10.

Answer: (b) moment of inertia and moment of a force

Explanation:
1. Moment of inertia \( = ML^2 \), moment of force \( = ML^2T^{-2} \): different.
2. Angular momentum and Planck's constant: \( ML^2T^{-1} \). Impulse and momentum: \( MLT^{-1} \). Work and torque: \( ML^2T^{-2} \).

Teacher's Note:
1. Work and torque share dimensions but not the same nature.
2. Moment of inertia has no time dependence.

Answer: (d) the energy delivered by radiation to a target

Explanation:
1. The rad is a unit of absorbed dose: 1 rad = 0.01 J of energy absorbed per kg of target.
2. Ion production is exposure (roentgen), biological effect is dose equivalent (rem) and decay rate is activity (becquerel or curie).

Teacher's Note:
1. Absorbed dose measures energy deposited per unit mass.
2. The SI unit of absorbed dose is the gray (1 Gy = 100 rad).

Answer: (a) henry (H)

Explanation:
1. \( \text{H} = \frac{\text{V s}}{\text{A}} = \frac{ML^2T^{-3}A^{-1}\cdot T}{A} \); with \( Q = AT \), \( \text{H} = \frac{ML^2}{Q^2} \).
2. Wb/m\(^2\) is the tesla (\( MT^{-1}Q^{-1} \)) and Wb is \( ML^2T^{-1}Q^{-1} \).

Teacher's Note:
1. Express inductance through the energy U = ½LI².
2. Then L = 2U/I² = ML²T⁻²/(Q²T⁻²) = ML²/Q².

Answer: (c) 17.6 kg m s\(^{-1}\)

Explanation:
1. \( p = 3.513\times5.00 = 17.565 \).
2. The speed has three significant figures, so the product is limited to three: 17.6.

Teacher's Note:
1. In multiplication the result has as many significant figures as the least precise factor.
2. Round 17.565 to 17.6.

Answer: (c) \( [MT^{-1}C^{-1}] \)

Explanation:
1. \( B = \frac{F}{qv} \), so \( [B] = \frac{MLT^{-2}}{C\cdot LT^{-1}} = MT^{-1}C^{-1} \).

Teacher's Note:
1. The length cancels between force and velocity.
2. In SI this is kg s⁻¹ C⁻¹, i.e. tesla.

Answer: (a) one minute

Explanation:
1. 1 VSD \( = \frac{29}{30} \) MSD. Least count \( = 1\text{ MSD} - 1\text{ VSD} = \frac{1}{30}\text{ MSD} \).
2. \( = \frac{0.5^\circ}{30} = \frac{1}{60}^\circ = 1' \).

Teacher's Note:
1. Least count = value of one main scale division divided by the number of vernier divisions.
2. 1/60 of a degree is one minute of arc.

Answer: (b) \( [\varepsilon_0] = M^{-1}L^{-3}T^4A^2 \)

Explanation:
1. From \( F = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{r^2} \), \( \varepsilon_0 = \frac{q^2}{Fr^2} \).
2. \( [\varepsilon_0] = \frac{A^2T^2}{MLT^{-2}\cdot L^2} = M^{-1}L^{-3}T^4A^2 \).

Teacher's Note:
1. Use Coulomb's law, with charge = current × time.
2. The SI unit is C² N⁻¹ m⁻² (farad per metre).