JEE Physics Units and Dimensions MCQs Set 05

Answer: (a) length, time and velocity

Explanation:
1. Fundamental quantities must be independent of each other.
2. Velocity has dimensions \( LT^{-1} \), which can be formed from length and time, so the set length, time, velocity is not independent.
3. The other sets are independent.

Teacher's Note:
1. Independent here means no quantity of the set can be derived from the others.
2. Mass, length and time form the usual set.

Answer: (c) never has a nonzero dimension

Explanation:
1. If a quantity has a nonzero dimension, it must be measured in some unit.
2. So a quantity that has no unit cannot have a nonzero dimension.

Teacher's Note:
1. A unitless quantity such as a ratio is also dimensionless.
2. The reverse is not true: a dimensionless quantity such as angle may have a unit (radian).

Answer: (a) may have a unit

Explanation:
1. Angle has no dimensions, yet it is measured in radian or degree.
2. So a dimensionless quantity may or may not have a unit.

Teacher's Note:
1. Strain has neither dimension nor unit.
2. Angle is the standard example of a dimensionless quantity with units.

Answer: (d) \( a/b \)

Explanation:
1. Quantities of different dimensions cannot be added or subtracted.
2. The exponent \( a/b \) in \( e^{a/b} \) would have to be dimensionless, which it is not.
3. Only the ratio \( a/b \) is a meaningful new quantity.

Teacher's Note:
1. Division and multiplication of unlike quantities give new quantities.
2. Addition and subtraction need identical dimensions.

Answer: (b) \( LT^{-2},\ L,\ T \)

Explanation:
1. \( c \) must have the dimensions of \( t \), so \( [c] = T \).
2. \( at \) has the dimensions of \( v \): \( [a] = LT^{-1}/T = LT^{-2} \).
3. \( \frac{b}{t + c} \) also has the dimensions of \( v \), so \( [b] = LT^{-1}\cdot T = L \).

Teacher's Note:
1. Every term added must have the dimensions of velocity.
2. Quantities added to \( t \) must themselves have dimensions of time.

Answer: (d) \( M^0LT^{-1} \) and \( T^{-1} \)

Explanation:
1. The exponent \( \alpha t \) is dimensionless, so \( [\alpha] = T^{-1} \).
2. \( x = V_0/\alpha \times \) (number), so \( [V_0] = [x][\alpha] = LT^{-1} \).

Teacher's Note:
1. The bracket \( 1 - e^{-\alphat} \) is a pure number.
2. \( V_{0} \) is therefore a velocity-like quantity.

Answer: (a) dimensions \( T^{-2} \)

Explanation:
1. The exponent \( \alpha t^2 \) must be dimensionless, so \( [\alpha] = T^{-2} \).

Teacher's Note:
1. The argument of any exponential must be a pure number.
2. \( P_{0} \) carries the dimensions of \( P \).

Answer: (c) (3) and (4) only

Explanation:
1. Work: \( ML^2T^{-2} \). Momentum: \( MLT^{-1} \).
2. Pressure: \( \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \). Energy per volume: \( \frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2} \).
3. So (3) and (4) match.

Teacher's Note:
1. Pressure and energy density have the same dimensions.
2. This is why the energy density of a field is sometimes called its pressure.

Answer: (c) \( Av^2\rho \)

Explanation:
1. Let \( F = A^av^b\rho^c \): \( MLT^{-2} = L^{2a}\,L^bT^{-b}\,M^cL^{-3c} \).
2. \( M \): \( c = 1 \). \( T \): \( b = 2 \). \( L \): \( 2a + b - 3c = 1 \) gives \( a = 1 \).
3. So \( F \propto Av^2\rho \).

Teacher's Note:
1. Equate the powers of \( M, L \) and \( T \) on both sides.
2. A is an area, so it contributes \( L^{2} \).

Answer: (a) 16 W

Explanation:
1. Power has dimensions \( ML^2T^{-3} \).
2. One unit of power \( = \frac{20\times10^2}{5^3} = \frac{2000}{125} = 16 W \).

Teacher's Note:
1. Substitute the size of each fundamental unit in the dimensional formula.
2. Mass 20 and length² 100 go in the numerator, time³ 125 in the denominator.

Answer: (d) metre

Explanation:
1. The term \( \frac{2kl}{ma} \) is dimensionless, so \( [k] = \frac{[m][a]}{[l]} \).
2. Then \( \frac{ma}{k} = l \), a length, and the bracket is a pure number.
3. So \( b \) has the unit of length, the metre.

Teacher's Note:
1. Expressions inside a root with 1 must be dimensionless.
2. This fixes the dimensions of \( k \).

Answer: (b) angle (it is dimensionless)

Explanation:
1. \( [E] = ML^2T^{-2} \), \( [J] = ML^2T^{-1} \), \( [G] = M^{-1}L^3T^{-2} \).
2. Numerator: \( ML^2T^{-2}\cdot M^2L^4T^{-2} = M^3L^6T^{-4} \).
3. Denominator: \( M^5\cdot M^{-2}L^6T^{-4} = M^3L^6T^{-4} \).
4. The ratio is dimensionless, matching an angle.

Teacher's Note:
1. An angle is dimensionless, so only a dimensionless result fits.
2. The listed alternatives (length, mass, time) all have dimensions.

Answer: (b) 2

Explanation:
1. \( MLT^{-2} = L^2\cdot ML^{-3}\cdot L^xT^{-x} = ML^{x-1}T^{-x} \).
2. Matching \( T \): \( x = 2 \). Matching \( L \): \( x - 1 = 1 \), which is also 2.

Teacher's Note:
1. Both \( L \) and \( T \) equations agree on \( x = 2 \).
2. This \( v^{2} \) law is the quadratic drag law.

Answer: (c) \( x = -\frac12,\ y = \frac12 \)

Explanation:
1. \( [f] = T^{-1} \) and \( [K] = \frac{MLT^{-2}}{L} = MT^{-2} \).
2. \( T^{-1} = M^xM^yT^{-2y} \) gives \( y = \frac12 \) and \( x + y = 0 \), so \( x = -\frac12 \).

Teacher's Note:
1. The force constant has dimensions of force per length.
2. The final result \( f ∝ \sqrt{K/m} \) is the SHM frequency.

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

Explanation:
1. Let \( v = k\lambda^ag^b\rho^c \). Mass appears only in \( \rho \), so \( c = 0 \).
2. \( LT^{-1} = L^{a+b}T^{-2b} \) gives \( b = \frac12 \) and \( a = \frac12 \).
3. So \( v^2 = kg\lambda \).

Teacher's Note:
1. Density introduces mass, which cannot cancel.
2. Hence the speed is independent of density.

Answer: (d) pressure

Explanation:
1. Torque \( = ML^2T^{-2} \), surface tension \( = MT^{-2} \), viscosity \( = ML^{-1}T^{-1} \).
2. Pressure \( = ML^{-1}T^{-2} \).

Teacher's Note:
1. Use the given relations τ = \( r \times F, S = F/l \) and \( F \) = 6πηrv.
2. Pressure is force per unit area.

Answer: (b) angular momentum

Explanation:
1. \( h = \frac{E}{\nu} = \frac{ML^2T^{-2}}{T^{-1}} = ML^2T^{-1} \).
2. Angular momentum \( = mvr = ML^2T^{-1} \).

Teacher's Note:
1. \( E \) = hν gives the dimensions of \( h \).
2. Angular momentum has the same dimensions.

Answer: (a) \( M^0L^0T^0 \) and \( M^0L^1T^{-1} \)

Explanation:
1. A and \( B \) each have the dimensions of force, so \( A/B \) is dimensionless.
2. Ct and Dx are dimensionless, so \( [C] = T^{-1} \) and \( [D] = L^{-1} \), giving \( C/D = LT^{-1} \).

Teacher's Note:
1. Arguments of sine and cosine are dimensionless.
2. \( C/D \) has the dimensions of a speed, like a wave speed \( \omega/k \).

Answer: (c) \( MLT^{-3} \) and \( MLT^{-4} \)

Explanation:
1. \( [at] = [F] \) gives \( [a] = MLT^{-2}/T = MLT^{-3} \).
2. \( [bt^2] = [F] \) gives \( [b] = MLT^{-4} \).

Teacher's Note:
1. Divide the dimensions of force by those of the variable multiplying the coefficient.
2. Each coefficient absorbs the missing powers of time.

Answer: (c) (2) and (3) only

Explanation:
1. (1) \( 2\pi t/T \) is dimensionless and \( y \) has length: correct.
2. (2) \( vt \) has the dimensions of length, not dimensionless: wrong.
3. (3) \( A/T \) is a velocity, not a length, and \( t/A \) is not dimensionless: wrong.
4. (4) \( \omega t \) is dimensionless and the factor keeps the length: correct.

Teacher's Note:
1. Check the argument of the trigonometric function and then the multiplying factor.
2. Both must be correct for the formula to be dimensionally valid.

Answer: (a) (1), (2) and (3) only

Explanation:
1. Statements (1), (2) and (3) are the basic idea of dimensions.
2. (4) is false: velocity \( LT^{-1} \) has zero dimension in mass, for example.

Teacher's Note:
1. A derived quantity may have a zero power in some base quantity.
2. Base quantities are independent of one another.

Answer: (d) joule K\(^{-1}\) mol\(^{-1}\)

Explanation:
1. From \( PV = nRT \), \( R = \frac{PV}{nT} = \frac{\text{energy}}{\text{mol}\cdot\text{K}} \).
2. In SI the unit is \( J K ^{-1} \) mol\(^{-1}\); erg is a CGS unit.

Teacher's Note:
1. PV has the dimensions of energy.
2. Watt is power, newton is force.

Answer: (b) linear momentum

Explanation:
1. Impulse \( = F\Delta t = \Delta p \), so its unit \( (N s) \) equals that of momentum (kg \( m/s \)).

Teacher's Note:
1. Impulse equals the change in momentum.
2. Rate of change of momentum is force.

Answer: (b) momentum

Explanation:
1. \( \frac{P}{t} \) has the dimensions of force, so \( [P] = MLT^{-2}\cdot T = MLT^{-1} \).
2. These are the dimensions of momentum.

Teacher's Note:
1. Multiply the dimensions of force by those of \( t \).
2. Check that \( \alphat \) also has dimensions of force: α is force per time.