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CBSE Class 11 Physics · 10 questions · 24 marks
Before you can solve a single problem in Class 11, you need a language for saying how big things are and how sure you are about it. This chapter sets up the SI base units, the algebra of dimensions, and the rules that decide how many digits of an answer you are actually allowed to write down. Dimensional analysis alone will let you catch wrong formulas and even guess correct ones.
A physical quantity Q has the dimensional formula [M L² T⁻³]. Which of the following is Q?
Answer
Power — power is energy per unit time, so its dimensions are [M L² T⁻²] / [T] = [M L² T⁻³]. Momentum is [M L T⁻¹], pressure is [M L⁻¹ T⁻²] and angular momentum is [M L² T⁻¹], so none of those match.
The measured length of a metal strip is recorded as 0.03060 m. How many significant figures does this reading contain?
Answer
4 — the two zeros before 3 are only place markers and are not significant, while the zero between 3 and 6 is a captive zero and the final zero is a trailing zero after the decimal point. So the significant digits are 3, 0, 6 and 0, giving four significant figures.
The radius of a sphere is measured with a 2% error. What is the maximum percentage error in the calculated volume?
Answer
6% — volume V = (4/3)πr³, so ΔV/V = 3(Δr/r). With Δr/r = 2%, the volume error is 3 × 2% = 6%. The numerical factor 4π/3 is exact and contributes no error.
In the relation P = a/(b + V), where P is pressure and V is volume, what are the dimensions of the constant a?
Answer
[M L² T⁻²] — by homogeneity b must have the dimensions of volume, [L³], so (b + V) is [L³]. Then a = P × (b + V) = [M L⁻¹ T⁻²][L³] = [M L² T⁻²], which is the dimension of energy.
Assertion (A): A dimensionally correct equation is not necessarily physically correct. Reason (R): Dimensional analysis cannot determine dimensionless constants or distinguish between quantities that share the same dimensional formula.
Answer
Both A and R are true and R is the correct explanation of A — because dimensions are blind to pure numbers and cannot tell apart work from torque (both [M L² T⁻²]), an equation such as kinetic energy = mv² passes the dimensional test while being physically wrong by a factor of ½. So the limitation stated in R is exactly why A holds.
The time period of a simple pendulum is measured five times as 2.03 s, 2.06 s, 2.01 s, 2.09 s and 2.06 s. Find the mean value and the mean absolute error.
Answer
Mean value T = (2.03 + 2.06 + 2.01 + 2.09 + 2.06)/5 = 10.25/5 = 2.05 s. Absolute errors: |2.03 − 2.05| = 0.02, |2.06 − 2.05| = 0.01, |2.01 − 2.05| = 0.04, |2.09 − 2.05| = 0.04, |2.06 − 2.05| = 0.01. Mean absolute error ΔT = (0.02 + 0.01 + 0.04 + 0.04 + 0.01)/5 = 0.12/5 = 0.024 ≈ 0.02 s. So T = (2.05 ± 0.02) s.
A rectangular sheet has length 8.24 cm measured to ±0.01 cm and breadth 3.5 cm measured to ±0.1 cm. Calculate its area and express the result with the correct number of significant figures and percentage error.
Answer
Area A = 8.24 × 3.5 = 28.84 cm². The breadth has only 2 significant figures, so the area must be rounded to 2 significant figures: A ≈ 29 cm². Percentage error: ΔA/A = Δl/l + Δb/b = 0.01/8.24 + 0.1/3.5 = 0.00121 + 0.02857 = 0.02978, i.e. about 3.0%. So A = 29 cm² with a maximum error of roughly 3%, which is about ±0.9 cm². The dominant contribution comes from the crudely measured breadth.
Using the method of dimensions, derive an expression for the frequency f of a stretched wire, assuming it depends on the length L of the wire, the tension F in it, and the mass per unit length μ. Also state two limitations of the method you used.
Answer
Assume f ∝ L^a F^b μ^c, that is f = k L^a F^b μ^c where k is a dimensionless constant. Dimensions: [f] = [T⁻¹], [L] = [L], [F] = [M L T⁻²], [μ] = [M L⁻¹]. Substituting: [M⁰ L⁰ T⁻¹] = [L]^a [M L T⁻²]^b [M L⁻¹]^c = [M^(b+c) L^(a+b−c) T^(−2b)]. Comparing powers: For M: b + c = 0 For T: −2b = −1, so b = ½, and therefore c = −½ For L: a + b − c = 0, so a + ½ + ½ = 0, giving a = −1. Hence f = k L⁻¹ F^(1/2) μ^(−1/2) = (k/L)√(F/μ). Experiment shows k = ½ for the fundamental mode, so f = (1/2L)√(F/μ). Limitations: (i) the dimensionless constant k cannot be obtained from dimensional analysis and must come from experiment or a full derivation; (ii) the method fails for relations involving sums of terms, trigonometric, exponential or logarithmic functions, and it cannot be used when a quantity depends on more than three independent dimensional quantities in this simple power-law way.
(a) State the principle of homogeneity of dimensions. (b) Check whether the equation v² = u² + 2as is dimensionally consistent. (c) A student writes the escape speed of a body from a planet of mass M and radius R as v = √(GM/R²). Use dimensions to show this is wrong and write the dimensionally correct form.
Answer
(a) The principle of homogeneity states that in any valid physical equation, every term on both sides must have exactly the same dimensional formula. Quantities of different dimensions can never be added, subtracted or equated. (b) [v²] = [L T⁻¹]² = [L² T⁻²]. [u²] = [L² T⁻²]. [2as] = [L T⁻²][L] = [L² T⁻²] (the 2 is dimensionless). All three terms have dimensions [L² T⁻²], so the equation is dimensionally consistent. (c) The dimensional formula of G is [M⁻¹ L³ T⁻²], since from F = Gm₁m₂/r², G = F r²/(m₁m₂) = [M L T⁻²][L²]/[M²] = [M⁻¹ L³ T⁻²]. For the student's expression: GM/R² = [M⁻¹ L³ T⁻²][M]/[L²] = [L T⁻²]. Its square root is [L^(1/2) T⁻¹], which is not [L T⁻¹], so the formula is dimensionally wrong. Trying GM/R instead: [M⁻¹ L³ T⁻²][M]/[L] = [L² T⁻²], whose square root is [L T⁻¹] — the dimension of speed. So the dimensionally correct form is v = k√(GM/R), and the full derivation gives k = √2, i.e. v = √(2GM/R).
Read the following and answer the questions that follow: A school lab team is calibrating a screw gauge to measure the diameter of a thin copper wire. The gauge has a pitch of 0.5 mm and 50 divisions on its circular scale. On closing the jaws with no wire in between, the zero of the circular scale sits 3 divisions above the reference line. With the wire in place the main scale reads 2 divisions of 0.5 mm each and the circular scale reads 27. (a) Calculate the least count of the screw gauge. (b) State the sign and magnitude of the zero error and the corresponding zero correction. (c) Find the corrected diameter of the wire in mm. (d) If the same wire is measured with a metre scale of least count 1 mm, comment on which instrument is appropriate and why.
Answer
(a) Least count = pitch / number of circular scale divisions = 0.5 mm / 50 = 0.01 mm. (b) The zero of the circular scale lies above the reference line, so the reading has been shortened: this is a negative zero error of magnitude 3 × 0.01 = 0.03 mm. The zero correction is therefore +0.03 mm, which must be added to every observed reading. (c) Observed reading = main scale reading + (circular scale reading × least count) = (2 × 0.5) + (27 × 0.01) = 1.00 + 0.27 = 1.27 mm. Corrected diameter = 1.27 + 0.03 = 1.30 mm. (d) The screw gauge is appropriate. Its least count of 0.01 mm is about one hundred and thirty times smaller than the quantity being measured, giving a relative error near 0.8%, whereas a metre scale with least count 1 mm could not even resolve the 1.30 mm diameter reliably and would give a relative error of nearly 80%. An instrument should have a least count much smaller than the quantity being measured.
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