JEE Physics Units and Measurements MCQs Set 06

Answer: (b) A-IV, B-III, C-I, D-II

Explanation:
1. Surface tension = force/length = N/m = kg s⁻² (IV).
2. Pressure = force/area = kg m⁻¹ s⁻² (III).
3. Viscosity coefficient = force/(area x velocity gradient) = kg m⁻¹ s⁻¹ (I).
4. Impulse = force x time = momentum = kg m s⁻¹ (II).

Teacher's Note:
1. Convert each quantity to base units using its defining formula.
2. Newton = kg m s⁻².

Answer: (b) A-III, B-I, C-II, D-IV

Explanation:
1. Young's modulus is a stress: \( [ML^{-1}T^{-2}] \) (III).
2. Viscosity: \( [ML^{-1}T^{-1}] \) (I).
3. Planck's constant \( h=\frac{E}{\nu} \): \( [ML^2T^{-1}] \) (II).
4. Work function is an energy: \( [ML^2T^{-2}] \) (IV).

Teacher's Note:
1. Planck's constant has the dimensions of angular momentum.
2. Work function and energy share the same dimensions.

Answer: (c) A-III, B-II, C-IV, D-I

Explanation:
1. Pressure gradient = pressure/length: \( \frac{ML^{-1}T^{-2}}{L}=ML^{-2}T^{-2} \) (III).
2. Energy density = energy/volume: \( ML^{-1}T^{-2} \) (II).
3. Electric field = force/charge: \( MLT^{-3}A^{-1} \) (IV).
4. Latent heat = energy/mass: \( L^2T^{-2} \) (I).

Teacher's Note:
1. Energy density has the same dimensions as pressure.
2. Latent heat is energy per unit mass.

Answer: (b) \( A=[LT],\ B=[L^{-1}T^{-1}] \)

Explanation:
1. Each term inside the brackets must have the dimension of length: \( [At]=[L]\Rightarrow[A]=\frac{L}{T^{-1}}=[LT] \).
2. \( \left[\frac tB\right]=[L]\Rightarrow[B]=\frac{T^{-1}}{L}=[L^{-1}T^{-1}] \).

Teacher's Note:
1. x and y have the dimension of length, so everything subtracted from them must too.
2. Use the given dimension of t.

Answer: (b) \( \text{Nm}^2\text{C}^{-1};\ \text{Nm}^3\text{C}^{-1} \)

Explanation:
1. The unit of E is N C⁻¹.
2. \( A=E\,x^2\Rightarrow\text{N C}^{-1}\text{m}^2 \) and \( B=E\,y^3\Rightarrow\text{N C}^{-1}\text{m}^3 \).

Teacher's Note:
1. Multiply the unit of E by the unit of the denominator.
2. x² gives m² and y³ gives m³.

Answer: (a) 220

Explanation:
1. Pitch = 0.5 mm, so the least count \( =\frac{0.5}{100}=0.005\ \text{mm} \).
2. Zero error \( =+6\times0.005=+0.03\ \text{mm} \) (zero below the reference line).
3. Reading \( =4\times0.5+46\times0.005=2.23\ \text{mm} \).
4. Diameter \( =2.23-0.03=2.20\ \text{mm}=220\times10^{-2}\ \text{mm} \).

Teacher's Note:
1. A zero error found with the studs in contact must be subtracted when it is positive.
2. The linear scale reading is 4 pitch divisions = 2.0 mm.

Answer: (d) A-IV, B-I, C-III, D-II

Explanation:
1. Torque = force x distance = kg m² s⁻² (IV).
2. Energy density = energy/volume = kg m⁻¹ s⁻² (I).
3. Pressure gradient = pressure/length = kg m⁻² s⁻² (III).
4. Impulse = force x time = kg m s⁻¹ (II).

Teacher's Note:
1. Torque and energy have the same dimensions but different physical meaning.
2. Energy density has the same unit as pressure.

Answer: (b) A-III, B-I, C-IV, D-II

Explanation:
1. Angular momentum \( =mvr \): \( ML^2T^{-1} \) (III).
2. Torque: \( ML^2T^{-2} \) (I).
3. Stress = force/area: \( ML^{-1}T^{-2} \) (IV).
4. Pressure gradient: \( ML^{-2}T^{-2} \) (II).

Teacher's Note:
1. Use the defining formula of each quantity.
2. Stress and pressure have identical dimensions.

Answer: (c) Compressibility

Explanation:
1. \( [\alpha]=[PV^2] \) (same as P) and \( [b]=[V] \).
2. \( \left[\frac{b^2}{\alpha}\right]=\frac{V^2}{PV^2}=\frac1P \).
3. The reciprocal of pressure (or of bulk modulus) is compressibility.

Teacher's Note:
1. Terms added to P must have the dimension of pressure.
2. (V - b) shows that b has the dimension of volume.

Answer: (d) \( \left[h^{\frac12}c^{\frac12}G^{-\frac12}\right] \)

Explanation:
1. Let \( M=c^aG^bh^k \) with \( [c]=LT^{-1},\ [G]=M^{-1}L^3T^{-2},\ [h]=ML^2T^{-1} \).
2. Mass: \( -b+k=1 \). Length: \( a+3b+2k=0 \). Time: \( -a-2b-k=0 \).
3. Adding the last two: \( b+k=0 \), so \( k=-b \) and \( b=-\frac12,\ k=\frac12,\ a=\frac12 \).

Teacher's Note:
1. Write each fundamental constant in terms of M, L and T and solve for the three exponents.
2. This combination is the Planck mass.