CLASS 11 · PHYSICS · CHAPTER 1
Simple worked solutions to all 17 end-of-chapter questions on units, measurement, significant figures and dimensions.
17 worked answers · short steps first, explanations when you need them
Textbook pages 14–16
Question 1
Convert each given quantity to the requested unit. For (b), use total cylinder area 2πr(r+h).
(a) (10⁻² m)³ = 10⁻⁶ m³. (b) 2π(2)(12)=48π≈151 cm²=1.51×10⁴ mm². (c) 18 km h⁻¹=5 m s⁻¹, so 5 m. (d) 11.3 g cm⁻³=1.13×10⁴ kg m⁻³.
Question 2
Replace every unit by its equivalent in the target unit.
(a) 1 kg=10³ g and 1 m²=10⁴ cm², so factor 10⁷. (b) 1 m/(9.46×10¹⁵ m per ly)=1.06×10⁻¹⁶ ly. (c) 3.0×10⁻³×3600²=3.9×10⁴ km h⁻². (d) N m² kg⁻²=m³s⁻²kg⁻¹; conversion gives 6.67×10⁻⁸ cm³s⁻²g⁻¹.
Question 3
Energy has dimensions M L² T⁻². To express the same energy using larger/new base units, scale its numerical value by the inverse mass and length factors, and by the square of the time factor.
New numerical value = 4.2 × α⁻¹ × β⁻² × γ² = 4.2α⁻¹β⁻²γ².
Question 4
“Large” and “small” only make sense relative to a stated reference. Name that reference in each comparison.
Examples: an atom is tiny compared with a pin tip; a jet is faster than a car; Jupiter is about 318 Earth masses; room air has more molecules than the same volume of near-vacuum; a proton is about 1836 electron masses; sound travels much slower than light.
Question 5
Convert the time to seconds: 8×60+20=500 s.
Distance = speed × time = (1 new unit/s)(500 s) = 500 new units.
Question 6
Find or compare each instrument’s smallest measurable length. A smaller least count means greater precision.
(a) 1/20 mm=0.05 mm. (b) 1/100 mm=0.01 mm. (c) about 5×10⁻⁷ m=0.0005 mm. (c) has the smallest least count, so the optical instrument is most precise.
Question 7
Magnification = observed size ÷ actual size, so actual size = observed size ÷ magnification.
3.5 mm ÷ 100 = 0.035 mm = 3.5×10⁻⁵ m.
Question 8
(a) Wind the thread tightly in touching turns around a pencil, measure the length L of n turns, and divide L by n. (b) More divisions reduce least count, but eventually become too small to read reliably. (c) Averaging more readings helps random errors cancel.
(a) Diameter ≈ L/n. (b) No; eyesight and instrument limits prevent unlimited improvement. (c) 100 independent readings usually give a more reliable mean because random errors cancel better.
Question 9
Convert areas to matching units: 1.55 m² = 15,500 cm². The area ratio is the square of the linear magnification.
Area ratio = 15,500/1.75 ≈ 8857. Linear magnification = √8857 ≈ 94.1 times.
Question 10
Leading zeros do not count. Zeros between non-zero digits and trailing zeros after a decimal do count. In scientific notation, count the digits in the coefficient.
(a) 1; (b) 3; (c) 4; (d) 4; (e) 4; (f) 4 significant figures.
Question 11
Convert thickness: 2.01 cm=0.0201 m. Surface area = 2(lb+bt+tl); volume = lbt. Round final results to 3 significant figures because 2.01 has 3.
Area = 2[(4.234)(1.005)+(1.005)(0.0201)+(0.0201)(4.234)] = 8.721… ≈ 8.72 m². Volume = (4.234)(1.005)(0.0201)=0.08552… ≈ 0.0855 m³.
Question 12
Convert the box mass: 2.30 kg=2300 g, measured to the nearest 10 g. Add the masses, then round to the precision of the least precise measurement.
(a) 2300+20.15+20.17=2340.32 g, rounded to nearest 10 g = 2340 g = 2.34 kg. (b) 20.17−20.15=0.02 g.
Question 13
The terms being subtracted must both be dimensionless. Speed squared v² has units, so divide it by another speed squared.
Correct relation: m=m₀/√(1−v²/c²). Since v² and c² have the same dimensions, their ratio has no unit.
Question 14
Radius r=0.5×10⁻¹⁰=5×10⁻¹¹ m. Volume of one spherical atom is (4/3)πr³. Multiply by Avogadro’s number, 6.022×10²³ atoms/mol.
One atom ≈ (4/3)π(5×10⁻¹¹)³ = 5.24×10⁻³¹ m³. For a mole: 6.022×10²³×5.24×10⁻³¹ ≈ 3.16×10⁻⁷ m³.
Question 15
Convert 22.4 L to 2.24×10⁻² m³. A molecule of diameter 1 Å has radius 0.5 Å; its volume is about 5.24×10⁻³¹ m³. Multiply this by Avogadro’s number.
Molecular volume for a mole ≈ 3.16×10⁻⁷ m³. Ratio = (2.24×10⁻²)/(3.16×10⁻⁷) ≈ 7.1×10⁴. Gas molecules are far apart, so gas volume greatly exceeds the volume of the molecules themselves.
Question 16
For the same train movement, a nearby object changes direction by a larger angle in your view than a distant object does.
Nearby objects therefore seem to sweep backwards quickly. A very distant object has only a tiny apparent shift, so it seems almost stationary.
Question 17
Treat the Sun as a sphere: V=(4/3)πR³. Then use density = mass ÷ volume.
V≈(4/3)π(7.0×10⁸)³≈1.44×10²⁷ m³. Density≈(2.0×10³⁰)/(1.44×10²⁷)=1.4×10³ kg m⁻³, in the range of liquids/solids.