Chapter 2 Units and Measurements Solutions
Question - 11 : - The length, breadth and thickness of a rectangular sheet of metalare 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of thesheet to correct significant figures.
Answer - 11 : -
Length of sheet, l = 4.234 m
Breadth of sheet, b = 1.005 m
Thickness of sheet, h = 2.01 cm = 0.0201 m
The given table lists the respective significant figures:
| Quantity | Number | Significant Figure |
| l | 4.234 | 4 |
| b | 1.005 | 4 |
| h | 0.0201 | 3 |
Hence, area and volume both must have least significant figuresi.e., 3.
Surface area of the sheet = 2 (l × b + b × h + h × l)
= 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)
= 2(4.25517 + 0.0202005 + 0.0851034)
= 2 × 4.36
= 8.72 m2
Volume of the sheet = l × b × h
= 4.234 × 1.005 × 0.0201
= 0.0855 m3
This number has only 3 significant figures i.e., 8, 5, and 5.
Question - 12 : - The mass of a box measured by a grocer’s balance is 2.300 kg. Twogold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) thetotal mass of the box, (b) the difference in the masses of the pieces tocorrect significant figures?
Answer - 12 : -
Mass of grocer’s box = 2.300 kg
Mass of gold piece I =20.15g = 0.02015 kg
Mass of gold piece II =20.17 g = 0.02017 kg
(a) Total mass of the box = 2.3 + 0.02015 + 0.02017 = 2.34032 kg
In addition, the final result should retain as many decimal placesas there are in the number with the least decimal places. Hence, the total massof the box is 2.3 kg.
(b) Difference in masses = 20.17 – 20.15 = 0.02 g
In subtraction, the final result should retain as many decimalplaces as there are in the number with the least decimal places.
Question - 13 : - A physical quantity P is related to four observables a,b, c and d as follows:
The percentage errors of measurement in a, b, c and d are1%, 3%, 4% and 2%, respectively. What is the percentage error in thequantity P? If the value of P calculated using theabove relation turns out to be 3.763, to what value should you round off theresult?
Answer - 13 : - 
Percentage error in P = 13 %
Value of P is given as 3.763.
By rounding off the given value to the first decimal place, weget P = 3.8.
A book with many printing errors contains four different formulasfor the displacement y of a particle undergoing a certainperiodic motion:
(a) 
(b) y = a sin vt
(c) 
(d) 
(a = maximum displacement of the particle, v =speed of the particle. T = time-period of motion). Rule outthe wrong formulas on dimensional grounds.
Answer - 14 : -
(a) Answer: Correct
Dimension of y = M0 L1 T0
Dimension of a = M0 L1 T0
Dimensionof 
= M0 L0 T0
Dimension of L.H.S = Dimension ofR.H.S
Hence, the given formula is dimensionally correct.
(b) Answer: Incorrect
y = a sin vt
Dimension of y = M0 L1 T0
Dimension of a = M0 L1 T0
Dimension of vt = M0 L1 T–1 ×M0 L0 T1 = M0 L1 T0
But the argument of the trigonometric function must bedimensionless, which is not so in the given case. Hence, the given formula isdimensionally incorrect.
(c) Answer: Incorrect
Dimension of y = M0L1T0
Dimensionof
= M0L1T–1
Dimensionof
= M0 L–1 T1But the argument of the trigonometric function must bedimensionless, which is not so in the given case. Hence, the formula isdimensionally incorrect.
(d) Answer: Correct
Dimension of y = M0 L1 T0
Dimension of a = M0 L1 T0
Dimensionof 
= M0 L0 T0
Since the argument of the trigonometric function must bedimensionless (which is true in the given case), the dimensions of y and a arethe same. Hence, the given formula is dimensionally correct.
Question - 15 : - A famous relation in physics relates ‘moving mass’ m tothe ‘rest mass’ m0 of a particle in terms ofits speed v and the speed of light, c. (Thisrelation first arose as a consequence of special relativity due to AlbertEinstein). A boy recalls the relation almost correctly but forgets where to putthe constant c. He writes:
Answer - 15 : -
Given the relation,
Dimension of m = M1 L0 T0
Dimension of 
= M1 L0 T0
Dimension of v = M0 L1 T–1
Dimension of v2 = M0 L2 T–2
Dimension of c = M0 L1 T–1
Thegiven formula will be dimensionally correct only when the dimension of L.H.S isthe same as that of R.H.S. This is only possible when the factor, 
is dimensionless i.e., (1 – v2) isdimensionless. This is only possible if v2 isdivided by c2. Hence, the correct relation is
.
Question - 16 : - The unit of length convenient on the atomic scale isknown as an angstrom and is denoted by
. The size of a hydrogen atom is about 
what is the total atomic volumein m3 of a mole of hydrogen atoms?
Answer - 16 : -
Radius of hydrogen atom, r = 0.5 
= 0.5 × 10–10 m
Volumeof hydrogen atom =
1 mole of hydrogen contains 6.023 × 1023 hydrogenatoms.
∴ Volume of 1 mole of hydrogenatoms = 6.023 × 1023 × 0.524 × 10–30
= 3.16 × 10–7 m3
One mole of an ideal gas at standard temperature and pressureoccupies 22.4 L (molar volume). What is the ratio of molar volume to the atomicvolume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1). Why is this ratio so large?
Answer - 17 : - Radius of hydrogen atom, r =0.5 = 0.5 × 10–10 m
Volumeof hydrogen atom =
Now, 1 mole of hydrogen contains 6.023 × 1023 hydrogenatoms.
∴ Volume of 1 mole of hydrogenatoms, Va = 6.023 × 1023 × 0.524 ×10–30
= 3.16 × 10–7 m3
Molar volume of 1 mole of hydrogen atoms at STP,
Vm =22.4 L = 22.4 × 10–3 m3
Hence, the molar volume is 7.08 × 104 times higherthan the atomic volume. For this reason, the inter-atomic separation inhydrogen gas is much larger than the size of a hydrogen atom.
Explain this common observation clearly : If you look out of thewindow of a fast moving train, the nearby trees, houses etc. seem to moverapidly in a direction opposite to the train’s motion, but the distant objects(hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, sinceyou are aware that you are moving, these distant objects seem to move withyou).
Answer - 18 : -
Line of sight is defined as an imaginary line joining an objectand an observer’s eye. When we observe nearby stationary objects such as trees,houses, etc. while sitting in a moving train, they appear to move rapidly inthe opposite direction because the line of sight changes very rapidly.
On the other hand, distant objects such as trees, stars, etc.appear stationary because of the large distance. As a result, the line of sightdoes not change its direction rapidly.
Question - 19 : - The principle of ‘parallax’ in section 2.3.1 is used in thedetermination of distances of very distant stars. The baseline AB isthe line joining the Earth’s two locations six months apart in its orbit aroundthe Sun. That is, the baseline is about the diameter of the Earth’s orbit ≈ 3 ×1011m. However, even the nearest stars are so distant that with sucha long baseline, they show parallax only of the order of 1” (second) of arc orso. A parsec is a convenient unit of length on theastronomical scale. It is the distance of an object that will show a parallaxof 1” (second) of arc from opposite ends of a baseline equal to the distancefrom the Earth to the Sun. How much is a parsec in terms of meters?
Answer - 19 : -
Diameter of Earth’s orbit = 3 × 1011 m
Radius of Earth’s orbit, r = 1.5 × 1011 m
Let the distance parallax angle be
= 4.847 × 10–6 rad.
Let the distance of the star be D.
Parsec is defined as the distance at which the average radius ofthe Earth’s orbit subtends an angle of.
∴ We have
Hence, 1 parsec ≈ 3.09 × 1016 m.
The nearest star to our solar system is 4.29 light years away. Howmuch is this distance in terms of parsecs? How much parallax would this star(named Alpha Centauri) show when viewed from two locations of theEarth six months apart in its orbit around the Sun?
Answer - 20 : -
Distance of the star from the solar system = 4.29 ly
1 light year is the distance travelled by light in one year.
1 light year = Speed of light × 1 year
= 3 × 108 × 365 × 24 × 60 × 60 = 94608 × 1011 m
∴4.29 ly = 405868.32 × 1011 m
∴1 parsec = 3.08 × 1016 m
∴4.29 ly =
= 1.32 parsecUsing the relation,
But, 1 sec = 4.85 × 10–6 rad
∴