JEE Physics Units and Measurements MCQs Set 07

Answer: (c) force

Explanation:
1. Force = mass × acceleration = kg × m/s².
2. So kg m s⁻² is the newton, the unit of force.

Teacher's Note:
1. Check by building the unit: kg m s⁻² is exactly the newton.
2. Energy would be kg m² s⁻², momentum kg m s⁻¹.

Answer: (c) 10⁻⁷ cm

Explanation:
1. 1 nm = 10⁻⁹ m.
2. 1 m = 100 cm, so 1 nm = 10⁻⁹ × 10² cm = 10⁻⁷ cm.
3. Also 1 nm = 10⁻⁶ mm.

Teacher's Note:
1. Nano = 10⁻⁹.
2. Convert the base unit step by step to avoid errors.

Answer: (d) fermi

Explanation:
1. Nuclear radii are of the order of 10⁻¹⁵ m.
2. The fermi (1 fm = 10⁻¹⁵ m) is the convenient unit.

Teacher's Note:
1. An angstrom (10⁻¹⁰ m) is suited to atomic sizes.
2. A fermi is also called a femtometre.

Answer: (a) 1

Explanation:
1. \( 1000\ \text{kg m}^{-3}=\frac{1000\times10^{3}\ \text{g}}{10^{6}\ \text{cm}^3}=1\ \text{g cm}^{-3} \)

Teacher's Note:
1. 1000 kg/m³ is the density of water, which is 1 g/cm³.
2. To convert kg/m³ to g/cm³, divide by 1000.

Answer: (c) The unit of momentum is doubled

Explanation:
1. Time = length/velocity: 2/2 = 1, so the unit of time is unchanged.
2. Mass = force × time²/length: 2 × 1/2 = 1, so the unit of mass is unchanged.
3. Momentum = mass × velocity becomes 1 × 2 = 2 times.
4. Energy = force × length becomes 2 × 2 = 4 times.

Teacher's Note:
1. Express each derived unit through the base units, then substitute the factors.
2. Only momentum doubles.

Answer: (c) coefficient of friction

Explanation:
1. Angle is dimensionless but has a unit (radian).
2. Luminous intensity (candela) and current (ampere) have dimensions and units.
3. The coefficient of friction is a ratio of two forces, so it has neither.

Teacher's Note:
1. A ratio of two like quantities is a pure number.
2. Refractive index, strain and relative density are other examples.

Answer: (d) energy

Explanation:
1. Energy (or work) = force × distance = MLT⁻² × L = ML²T⁻².
2. Pressure is ML⁻¹T⁻², momentum MLT⁻¹ and power ML²T⁻³.

Teacher's Note:
1. ML²T⁻² is shared by work, energy, torque and heat.
2. Power has an extra T⁻¹.

Answer: (a) torque and energy

Explanation:
1. Torque = force × distance = ML²T⁻²; energy = ML²T⁻².
2. Angular momentum is ML²T⁻¹, pressure is ML⁻¹T⁻², and light year (length) differs from solar day (time).

Teacher's Note:
1. Torque and work share dimensions but are different physical quantities.
2. Torque is a vector, whereas work is a scalar.

Answer: (c) [M⁰L¹T⁻³]

Explanation:
1. \( f=\frac{3S}{t^3} \) has dimensions \( \frac{L}{T^3}=LT^{-3} \).

Teacher's Note:
1. The factor 1/3 is a number and has no dimensions.
2. f has the dimensions of distance divided by t³.

Answer: (a) [ML⁻¹T⁻¹]

Explanation:
1. \( [\eta]=\frac{[F]}{[a][v]}=\frac{MLT^{-2}}{L\cdot LT^{-1}}=ML^{-1}T^{-1} \).

Teacher's Note:
1. 6π is a pure number.
2. η is the coefficient of viscosity.

Answer: (b) [M⁰L¹T⁻¹]

Explanation:
1. [ω] = T⁻¹ and [k] = L⁻¹.
2. [ω/k] = T⁻¹ × L = LT⁻¹, which is velocity (the wave speed).

Teacher's Note:
1. ω/k is the phase (wave) velocity.
2. Dimensions match speed.

Answer: (c) [M⁰L⁰T⁻¹]

Explanation:
1. The exponent at + 3 must be dimensionless.
2. So [a][T] = 1 and [a] = T⁻¹.

Teacher's Note:
1. A number such as 3 is dimensionless, so at must be too.
2. Exponent rule again.

Answer: (d) LT⁻²

Explanation:
1. \( \left[\frac{d^2y}{dx^2}\right]=\frac{L}{T^2}=LT^{-2} \).

Teacher's Note:
1. A derivative divides the dimension of the numerator by that of the denominator.
2. The second derivative divides by the square of the denominator.

Answer: (b) 5

Explanation:
1. Zeros between non-zero digits count.
2. Trailing zeros after a decimal point also count.
3. So 2, 0, 0, 4, 0 are all significant: 5 digits.

Teacher's Note:
1. Trailing zeros are significant only if there is a decimal point.
2. Leading zeros never count.

Answer: (c) 4.4 cm

Explanation:
1. In addition the result keeps as many decimal places as the least precise number.
2. B has one decimal place, so round 4.431 to 4.4 cm.

Teacher's Note:
1. For addition/subtraction count decimal places; for multiplication/division count significant figures.
2. The least precise measurement limits the result.

Answer: (b) 55.4

Explanation:
1. 17.8 has three significant figures; 3.1143 has five.
2. The product keeps the smaller number: three significant figures.
3. 55.4354 → 55.4.

Teacher's Note:
1. For multiplication keep the fewest significant figures.
2. Never quote more precision than the least precise measurement.

Answer: (d) (22.10 ± 0.10) cm

Explanation:
1. Sum = 11.05 + 11.05 = 22.10 cm.
2. Absolute errors add: 0.05 + 0.05 = 0.10 cm.

Teacher's Note:
1. In sums and differences, absolute errors add.
2. Keep the same number of decimals as the data.

Answer: (b) 0.75%

Explanation:
1. P = K V^(−3/2).
2. ΔP/P = −(3/2) ΔV/V = −(3/2)(−0.5%) = +0.75%.

Teacher's Note:
1. A decrease in volume raises pressure.
2. Take the exponent with its sign.

Answer: (b) 0.05 mm

Explanation:
1. \( \text{LC}=\frac{\text{pitch}}{\text{number of divisions}}=\frac{5\ \text{mm}}{100}=0.05\ \text{mm} \).

Teacher's Note:
1. Pitch is the distance moved in one full rotation.
2. Least count = pitch ÷ number of circular divisions.

Answer: (c) (P − Q)/R

Explanation:
1. P and Q have different dimensions, so P − Q is not possible.
2. PQ − R is meaningful if PQ has the same dimensions as R, and PR − Q² is meaningful if PR and Q² match, so those can exist.

Teacher's Note:
1. Subtraction needs equal dimensions.
2. Products and ratios of different quantities are always allowed.

Answer: (d) M⁰L⁻¹T⁰, M⁰L⁰T⁻¹, M⁰L⁻¹T¹

Explanation:
1. αx is dimensionless, so [α] = L⁻¹.
2. β(t − Bx) is dimensionless and t is a time, so [β] = T⁻¹.
3. Bx has the dimension of time, so [B] = T/L = L⁻¹T.

Teacher's Note:
1. The exponent and the sine argument are dimensionless.
2. Inside the bracket, t and Bx must have the same dimension.