JEE Physics Units And Measurements MCQs Set 17

Answer: (d) 8%

Explanation:
1. g = 4π²L/T².
2. Δg/g = 2% + 2 × 3% = 8%.

Teacher's Note:
1. Remember g ∝ L/T².

Answer: (d) taking a large number of observations and then the mean

Explanation:
1. Random errors have no fixed cause or sign.
2. Their effect reduces when the mean of many observations is taken.

Teacher's Note:
1. More observations reduce random error but cannot remove systematic error.

Answer: (d) A-3, B-1, C-4, D-2

Explanation:
1. Joule = ML²T⁻², newton = MLT⁻², hertz = T⁻¹, pascal = ML⁻¹T⁻².

Teacher's Note:
1. Derive each unit from its defining formula.

Answer: (d) has all of these properties

Explanation:
1. The caesium frequency is the same at every place and every time.
2. It is also independent of temperature and pressure, so it is a reliable standard.

Teacher's Note:
1. A good standard must be invariable and reproducible.
2. Atomic clocks are the most accurate time standard.

Answer: (c) A is true, but R is false.

Explanation:
1. Leading zeros in 0.005 are not significant, so it has one significant figure; the zeros after 5 in 0.500 are significant.
2. The Assertion is true but the Reason is false, because many zeros are significant.

Teacher's Note:
1. Zeros between digits and trailing zeros after a decimal point are significant.

Answer: (b) [MLT⁻¹]

Explanation:
1. Adding quantities of the same dimension does not change the dimension.

Teacher's Note:
1. Dimensions do not add; only the numerical values do.

Answer: (a) 5, 1 and 2

Explanation:
1. 23.023 has five (zeros between digits count).
2. 0.0003 has one.
3. 2.1 × 10⁻³ has two.

Teacher's Note:
1. Powers of ten do not affect the count.

Answer: (c) 1 mm and 50

Explanation:
1. Least count = pitch / number of divisions.
2. 1 mm / 50 = 0.02 mm, while 0.5/50 = 0.01 mm and 1/100 = 0.01 mm.

Teacher's Note:
1. Check every option by dividing pitch by divisions.

Answer: (d) A-4, B-1, C-2, D-3

Explanation:
1. Nucleus about 10⁻¹⁴ m, Sun-Earth about 10¹¹ m, Earth radius about 10⁷ m, proton about 10⁻¹⁵ m.

Teacher's Note:
1. Orders of magnitude help in estimation questions.

Answer: (d) Energy/work

Explanation:
1. Force/acceleration = mass, velocity/acceleration = time, volume/area = length.
2. Energy and work have the same dimensions, so their ratio is dimensionless.

Teacher's Note:
1. A ratio of two like quantities is always dimensionless.

Answer: (b) increases 16 times

Explanation:
1. Energy = force × length, so the unit becomes 4 × 4 = 16 times.

Teacher's Note:
1. Write the dimensions in terms of the quantities whose units change.

Answer: (b) [M⁻¹T²A]

Explanation:
1. [v] = [LT⁻¹] and [E] = [MLT⁻³A⁻¹].
2. Mobility = [LT⁻¹]/[MLT⁻³A⁻¹] = [M⁻¹T²A].

Teacher's Note:
1. Use E = F/q to get the dimension of electric field.

Answer: (a) A-3, B-1, C-2, D-4

Explanation:
1. Mean absolute error is Σ|Δaᵢ|/n, relative error is Δa_mean/a_mean, percentage error is that times 100, and an individual absolute error is the difference a_n − a_mean.

Teacher's Note:
1. Percentage error is relative error multiplied by 100.

Answer: (a) Torque and work

Explanation:
1. Torque and work are both [ML²T⁻²].
2. Stress is [ML⁻¹T⁻²], energy is [ML²T⁻²], force is [MLT⁻²].

Teacher's Note:
1. Torque and work are equal in dimensions but torque is a vector.

Answer: (c) 0.8%

Explanation:
1. Percentage error = (0.2/25) × 100 = 0.8%.

Teacher's Note:
1. Maximum error in a single reading equals the least count.

Answer: (d) e²a₀/(hc)

Explanation:
1. [C] = [M⁻¹L⁻²T⁴A²].
2. Write u = eᵃ a₀ᵇ hᶜ cᵈ and equate dimensions: a = 2, b = 1, c = −1, d = −1.
3. So u = e²a₀/(hc).

Teacher's Note:
1. Set up simultaneous equations for the exponents of M, L, T and A.

Answer: (c) A is true, but R is false.

Explanation:
1. Constants such as G and h have fixed values, so the Assertion is true.
2. They have dimensions, so the Reason is false.

Teacher's Note:
1. Dimensional constants (G, h, c) are not dimensionless constants like π.

Answer: (d) I, II and III

Explanation:
1. These are the standard rules for significant figures, so all three are correct.

Teacher's Note:
1. Writing 2.5 cm as 0.025 m keeps two significant figures.

Answer: (c) 7%

Explanation:
1. ΔR/R = ΔV/V + ΔI/I.
2. ΔV/V = 5% and ΔI/I = 0.2/10 = 2%.
3. Total = 7%.

Teacher's Note:
1. For a quotient, add the percentage errors of the numerator and denominator.

Answer: (c) 3

Explanation:
1. In division the result has as many significant figures as the factor with the least, here 2.51 has three.
2. So the result has 3 significant figures (1.72 g/cm³).

Teacher's Note:
1. Use the smallest number of significant figures among the factors.

Answer: (a) least count

Explanation:
1. The least count is the smallest division or resolution of the instrument.

Teacher's Note:
1. Least count error is a systematic limit of any instrument.

Answer: (c) [MT⁻¹C⁻¹]

Explanation:
1. F = qvB gives B = F/(qv).
2. [B] = [MLT⁻²]/([C][LT⁻¹]) = [MT⁻¹C⁻¹].

Teacher's Note:
1. Use the magnetic force F = qvB to find the dimensions of B.

Answer: (c) Angular displacement

Explanation:
1. Strain, Reynolds number and Poisson's ratio are pure numbers with no unit.
2. Angular displacement has the unit radian (or degree) but is dimensionless.

Teacher's Note:
1. Radian and steradian are dimensionless units.

Answer: (c) I, III and IV

Explanation:
1. I: six digits, correct.
2. II: leading zeros do not count, only 4289 so four, wrong.
3. III: trailing zeros without a decimal point are not counted, so three, correct.
4. IV: zeros between non-zero digits count, so five, correct.

Teacher's Note:
1. Leading zeros never count; zeros between digits always count.

Answer: (d) all of these

Explanation:
1. Dimensional analysis checks equations, derives relations (in simple cases) and converts units.

Teacher's Note:
1. It cannot give numerical constants or handle sums such as sin or exponential functions of dimensional arguments.