JEE Physics Units and Measurements MCQs Set 12

Answer: (a) length, time and velocity

Explanation:
1. Fundamental quantities must be independent of each other.
2. Velocity equals length divided by time, so length, time and velocity are not independent. The other sets are independent.

Teacher's Note:
1. Independence means none of the chosen quantities can be formed from the others.
2. Check each set for a relation among its members.

Answer: (d) may have a unit

Explanation:
1. Angle is dimensionless but has units (radian, degree); strain and refractive index are dimensionless without units.
2. So a dimensionless quantity may or may not have a unit.

Teacher's Note:
1. Dimension and unit are different ideas: dimension is the nature of the quantity, a unit is a chosen standard.
2. Plane angle and solid angle are the standard examples of dimensionless quantities with units.

Answer: (c) \( v^2=kg\lambda \)

Explanation:
1. Let \( v=k\lambda^ag^b\rho^c \). Then \( LT^{-1}=L^a(LT^{-2})^b(ML^{-3})^c \).
2. Mass: \( c=0 \). Time: \( -2b=-1\Rightarrow b=\tfrac12 \). Length: \( a+b-3c=1\Rightarrow a=\tfrac12 \).
3. So \( v=k\sqrt{g\lambda} \), that is \( v^2=kg\lambda \).

Teacher's Note:
1. The absence of mass in the velocity dimension forces the density exponent to be zero.
2. Equate powers of M, L and T.

Answer: (c) 6.67 × 10⁻⁸

Explanation:
1. \( 1\ \text{N}=10^5\ \text{dyne},\ 1\ \text{m}^2=10^4\ \text{cm}^2,\ 1\ \text{kg}^{-2}=10^{-6}\ \text{g}^{-2} \).
2. \( G=6.67\times10^{-11}\times10^{5}\times10^{4}\times10^{-6}=6.67\times10^{-8}\ \text{dyne cm}^2\text{g}^{-2} \)

Teacher's Note:
1. Convert each unit separately and multiply the factors.
2. Cross-check with [G] = M⁻¹L³T⁻²: length factor 10⁶ (cm³ per m³), mass factor 10⁻³ (g per kg, inverted), giving 10³.

Answer: (b) doubled

Explanation:
1. The unit of g is length/time², so the new unit is \( \frac{2}{2^2}=\frac12 \) of the old unit.
2. A smaller unit gives a larger number: the numerical value of g doubles.

Teacher's Note:
1. Numerical value is inversely proportional to the size of the unit.
2. Work out how the compound unit changes first.

Answer: (c) (1)-Q-c, (2)-S-a, (3)-P-b, (4)-R-d

Explanation:
1. G: \( \left[M^{-1}L^3T^{-2}\right] \), N m²/kg². Torque: \( \left[ML^2T^{-2}\right] \), N m. Momentum: \( \left[MLT^{-1}\right] \), N s. Pressure: \( \left[ML^{-1}T^{-2}\right] \), pascal.

Teacher's Note:
1. Derive each dimension from its defining formula, then match the unit.
2. N s is the unit of impulse, which equals momentum.

Answer: (b) \( \left[M^0L^0T^0\right] \) and \( \left[M^0L^1T^{-1}\right] \)

Explanation:
1. Arguments of sine and cosine are dimensionless, so \( [C]=T^{-1} \) and \( [D]=L^{-1} \).
2. A and B each have the dimension of force, so A/B is dimensionless.
3. \( \frac{C}{D}=\frac{T^{-1}}{L^{-1}}=LT^{-1} \)

Teacher's Note:
1. Trigonometric arguments must be dimensionless.
2. Terms added together must have the same dimension.

Answer: (b) \( \left[ML^2T^{-3}A^{-2}\right] \)

Explanation:
1. \( R=\frac{V}{I}=\frac{W/q}{I}=\frac{ML^2T^{-2}}{(AT)A}=ML^2T^{-3}A^{-2} \)

Teacher's Note:
1. Resistance is potential difference divided by current.
2. Potential difference is work per unit charge, and charge is current times time.

Answer: (c) \( \left[M^0L^0T^0I^0\right] \)

Explanation:
1. \( \frac{e^2}{4\pi\varepsilon_0} \) has the dimension of energy × length, \( ML^3T^{-2} \).
2. \( c\hbar=LT^{-1}\cdot ML^2T^{-1}=ML^3T^{-2} \).
3. The ratio is dimensionless (the fine-structure constant).

Teacher's Note:
1. Coulomb's law gives e²/(4πε₀) = force × distance².
2. Compare with the dimension of cħ.

Answer: (b) \( y=ax^b \)

Explanation:
1. Take logs of \( y=ax^b \): \( \log y=\log a+b\log x \), a straight line with intercept log a and slope b.
2. The other relations do not give a straight line between log x and log y.

Teacher's Note:
1. A straight line on a log-log plot indicates a power law.
2. The slope is the exponent and the intercept is log of the coefficient.

Answer: (c) magnetic flux

Explanation:
1. Faraday's law: \( \varepsilon=\frac{d\phi}{dt} \), so \( \phi=\varepsilon\,t \) has unit volt second (weber).

Teacher's Note:
1. Induced emf is the rate of change of magnetic flux.
2. Volt-second is the same as weber.

Answer: (b) \( \sqrt{\frac{Gm}{r^3}} \)

Explanation:
1. Let \( \omega\propto m^xr^yG^z \), with \( [G]=M^{-1}L^3T^{-2} \).
2. Time: \( -2z=-1\Rightarrow z=\tfrac12 \). Mass: \( x-z=0\Rightarrow x=\tfrac12 \). Length: \( y+3z=0\Rightarrow y=-\tfrac32 \).

Teacher's Note:
1. Match the time dimension first; it fixes the power of G.
2. Then use mass and length.

Answer: (b) (i), (ii) and (iii) only

Explanation:
1. Statements (i), (ii) and (iii) restate the independence of base quantities, so they are true.
2. (iv) is false: for example velocity has zero dimension in mass, and angle has zero dimension in all base quantities.

Teacher's Note:
1. A base quantity appears only in itself, so its dimension in any other base quantity is zero.
2. Derived quantities can have zero power in some base quantities.

Answer: (d) (i), (ii) and (iv) only

Explanation:
1. Dimensional consistency is necessary but not sufficient: a consistent equation can still be wrong by a numerical factor (ii), and it may be right (i).
2. An equation that is not dimensionally homogeneous cannot be physically correct (iv); (iii) is false.

Teacher's Note:
1. Dimensional analysis can reject an equation but cannot prove one.
2. Numerical constants are invisible to dimensional analysis.

Answer: (a) PV²

Explanation:
1. a/V² must have the dimension of pressure, so \( [a]=[P][V^2] \).

Teacher's Note:
1. Only like quantities can be added.
2. The term a/V² is added to P.

Answer: (a) \( \left[ML^8T^{-2}\right] \)

Explanation:
1. \( [a]=[PV^2]=ML^{-1}T^{-2}L^6=ML^5T^{-2} \) and \( [b]=[V]=L^3 \).
2. \( [ab]=ML^5T^{-2}\cdot L^3=ML^8T^{-2} \)

Teacher's Note:
1. Find a and b separately, then multiply.
2. b has the dimension of volume.

Answer: (b) (volt)(coulomb)(metre)

Explanation:
1. \( [GM_eM_s]=M^{-1}L^3T^{-2}\cdot M^2=ML^3T^{-2} \).
2. (volt)(coulomb)(metre) = joule × metre = \( ML^2T^{-2}\cdot L=ML^3T^{-2} \): same.
3. The other two have dimension L²T⁻².

Teacher's Note:
1. Gravitational potential energy is GMm/r, so GMm has dimension of energy × length.
2. Volt × coulomb is energy.

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

Explanation:
1. Boltzmann constant: energy per kelvin, \( ML^2T^{-2}K^{-1} \) (4).
2. Viscosity: \( ML^{-1}T^{-1} \) (2).
3. Planck constant: \( ML^2T^{-1} \) (1).
4. Thermal conductivity: \( \frac{\text{heat flow/area}}{\text{temperature gradient}}=\frac{MT^{-3}}{KL^{-1}}=MLT^{-3}K^{-1} \) (3).

Teacher's Note:
1. Use E = kT for the Boltzmann constant and E = hν for Planck's constant.
2. Thermal conductivity is power per unit length per kelvin.

Answer: (a) (i), (iii) and (iv) only

Explanation:
1. Planck mass: \( M=\sqrt{\frac{hc}{G}} \) so \( M\propto\sqrt c \), and \( M\propto G^{-1/2} \) (not √G).
2. Planck length: \( L=\sqrt{\frac{hG}{c^3}} \) so \( L\propto\sqrt h \) and \( L\propto\sqrt G \).

Teacher's Note:
1. Derive M and L in terms of h, c and G by matching dimensions.
2. M is inversely proportional to √G.

Answer: (a) (i) and (iii) only

Explanation:
1. (i) \( [\mu_0I^2]=MLT^{-2} \) and \( [\varepsilon_0V^2]=M^{-1}L^{-3}T^4A^2\cdot M^2L^4T^{-6}A^{-2}=MLT^{-2} \): correct.
2. (iii) \( \varepsilon_0c=\frac{1}{\mu_0c} \) has the dimension of 1/resistance, so \( \varepsilon_0cV \) is a current: correct.
3. (ii) and (iv) fail the dimension check.

Teacher's Note:
1. Use c² = 1/(μ₀ε₀) to relate the dimensions of μ₀ and ε₀.
2. Check each equation by reducing it to base units.

Answer: (b) \( R \)

Explanation:
1. Mass is constant: \( \rho R^3=\text{const} \), so \( \frac{1}{\rho}\frac{d\rho}{dt}=-\frac{3}{R}\frac{dR}{dt} \).
2. This is constant, so \( v=\frac{dR}{dt}\propto R \).

Teacher's Note:
1. Differentiate ρR³ = constant with respect to time.
2. Constant fractional rate means v/R is constant.

Answer: (a) \( [\mu_0]=[\varepsilon_0]^{-1}[L]^{-2}[T]^2 \)

Explanation:
1. \( c^2=\frac{1}{\mu_0\varepsilon_0}\Rightarrow\mu_0=\frac{1}{\varepsilon_0c^2} \).
2. \( [\mu_0]=[\varepsilon_0]^{-1}[L]^{-2}[T]^2 \)

Teacher's Note:
1. The speed of light connects μ₀ and ε₀.
2. Substitute [c] = L/T.

Answer: (d) henry (H)

Explanation:
1. Inductance: \( L=\frac{\phi}{I}=\frac{ML^2T^{-1}Q^{-1}}{QT^{-1}}=\frac{ML^2}{Q^2} \).
2. Hence the unit is the henry.

Teacher's Note:
1. Flux has dimension ML²T⁻¹Q⁻¹ and current has QT⁻¹.
2. Energy divided by current squared also gives ML²Q⁻².

Answer: (b) \( MT^{-1}C^{-1} \)

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

Teacher's Note:
1. Use F = qvB.
2. Length cancels out.

Answer: (a) 9

Explanation:
1. Weight scales as volume: \( W\propto L^3 \); the area of the leg scales as \( L^2 \).
2. \( \text{stress}=\frac{W}{A}\propto L\Rightarrow\text{factor}=9 \)

Teacher's Note:
1. Mass scales as the cube of linear size and cross-section as the square.
2. Stress therefore grows in proportion to linear size.