Chapter 1: Units and Measurements

Q2.12Written answer

The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is

(a) the total mass of the box,

(b) the difference in the masses of the pieces to correct significant figures?

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Worked solution

The mass of the box = 2.30 kg

and the mass of the first gold piece = 20.15 g

The mass of the second gold piece = 20.17 g

The total mass = 2.300 + 0.2015 + 0.2017 = 2.7032 kg

Since 1 is the least number of decimal places, the total mass = 2.7 kg.

The mass difference = 20.17 – 20.15 = 0.02 g

Since 2 is the least number of decimal places, the total mass = 0.02 g.

Q2.13Written answer

A physical quantity P is related to four observables a, b, c and d as follows:

P =

\(\begin{array}{l}\frac{a^{3}b^{2}}{\sqrt{c}d}\end{array} \)

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?

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Worked solution

\(\begin{array}{l}\frac{a^{3}b^{2}}{\sqrt{c}d}\end{array} \)

\(\begin{array}{l}\frac{\Delta P}{P}\end{array} \)

\(\begin{array}{l}\frac{3\Delta a}{a}\end{array} \)

\(\begin{array}{l}\frac{2\Delta b}{b}\end{array} \)

\(\begin{array}{l}\frac{1}{2}\end{array} \)

\(\begin{array}{l}\frac{\Delta c}{c}\end{array} \)

\(\begin{array}{l}\frac{\Delta d}{d}\end{array} \)

(

\(\begin{array}{l}\frac{\Delta P}{P}\end{array} \)

\(\begin{array}{l}\frac{\Delta a}{a}\end{array} \)

\(\begin{array}{l}\frac{\Delta b}{b}\end{array} \)

\(\begin{array}{l}\frac{1}{2}\end{array} \)

\(\begin{array}{l}\frac{\Delta c}{c}\end{array} \)

\(\begin{array}{l}\frac{\Delta d}{d}\end{array} \)

= 3 x 1 + 2 x 3 +

\(\begin{array}{l}\frac{1}{2}\end{array} \)

= 3 + 6 + 2 + 2 = 13 %

P = 4.235

\(\begin{array}{l}\Delta P\end{array} \)

=

\(\begin{array}{l}\frac{13P}{100}\end{array} \)

=

\(\begin{array}{l}\frac{13\times 4.235}{100}\end{array} \)

= 0.55

The error lies in the first decimal point, so the value of p = 4.3

Q2.14Written answer

A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion:

(a) y = a sin (

\(\begin{array}{l}\frac{2\pi t}{T}\end{array} \)

)

(b) y = a sin vt

(c) y =

\(\begin{array}{l}\frac{a}{T}\end{array} \)

sin
\(\begin{array}{l}\frac{t}{a}\end{array} \)

(d) y =

\(\begin{array}{l}a\sqrt{2}\end{array} \)

( sin
\(\begin{array}{l}\frac{2\pi t}{T}\end{array} \)
+ cos
\(\begin{array}{l}\frac{2\pi t}{T}\end{array} \)
)

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Worked solution

(a) y = a sin

\(\begin{array}{l}\frac{2\pi t}{T}\end{array} \)

Dimension of y = M⁰ L¹T⁰

The dimension of a = M⁰ L¹ T⁰

Dimension of sin

\(\begin{array}{l}\frac{2\pi t}{T}\end{array} \)

⁰⁰⁰

Since the dimensions on both sides are equal, the formula is dimensionally correct.

(b) It is dimensionally incorrect, as the dimensions on both sides are not equal.

(c) It is dimensionally incorrect, as the dimensions on both sides are not equal.

(d) y =

\(\begin{array}{l}a\sqrt{2}\end{array} \)

\(\begin{array}{l}\frac{2\pi t}{T}\end{array} \)

Dimension of y = M⁰ L¹ T⁰

The dimension of a = M⁰ L¹ T⁰

Dimension of

\(\begin{array}{l}\frac{t}{T}\end{array} \)

⁰⁰⁰

The formula is dimensionally correct.

Q2.15Written answer

A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m_o of a particle in terms of its speed v and the speed of light, c (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :

m =

\(\begin{array}{l}\frac{m_{0}}{\sqrt{1 – \nu ^{2}}}\end{array} \)

Guess where to put the missing c.

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Worked solution

The relation given is

\(\begin{array}{l}\frac{m_{0}}{\sqrt{1 – \nu ^{2}}}\end{array} \)

We can get,

\(\begin{array}{l}\frac{m_{0}}{m}\end{array} \)

\(\begin{array}{l}\sqrt{1-\nu ^{2}}\end{array} \)

\(\begin{array}{l}\frac{m_{0}}{m}\end{array} \)

To satisfy this,

\(\begin{array}{l}\sqrt{1-\nu ^{2}}\end{array} \)

\(\begin{array}{l}\sqrt{1-\frac{\nu ^{2}}{c^{2}}}\end{array} \)

Thus, m =

\(\begin{array}{l}m_{0}\sqrt{1-\frac{\nu ^{2}}{c^{2}}}\end{array} \)

Q2.16Written answer

The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10^–10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m³ of a mole of hydrogen atoms?

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Worked solution

hydrogen atom radius = 0.5 A = 0.5 x 10^-10m

Volume =

\(\begin{array}{l}\frac{4}{3}\pi r^{3}\end{array} \)

=

\(\begin{array}{l}\frac{4}{3}\end{array} \)

\(\begin{array}{l}\frac{22}{7}\end{array} \)

^-10³

= 0.524 x 10^-30 m³

1 hydrogen mole contains 6.023 x 10²³ hydrogen atoms.

Volume of 1 mole of hydrogen atom = 6.023 x 10²³ x 0.524 x 10^-30

= 3.16 x 10^-7 m³.

Q2.17Written answer

One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen (Take the size of the hydrogen molecule to be about 1 Å)? Why is this ratio so large?

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Worked solution

Radius = 0.5 A = 0.5 x 10^-10 m

Volume =

\(\begin{array}{l}\frac{4}{3}\pi r^{3}\end{array} \)

=

\(\begin{array}{l}\frac{4}{3}\end{array} \)

\(\begin{array}{l}\frac{22}{7}\end{array} \)

^-10³

= 0.524 x 10^-30 m³

1 hydrogen mole contains 6.023 x 10²³ hydrogen atoms.

Volume of 1 mole of hydrogen atom = 6.023 x 10²³ x 0.524 x 10^-30

= 3.16 x 10^-7 m³

V_m = 22.4 L = 22.4 x 10^-3 m³

\(\begin{array}{l}\frac{V_{m}}{V_{a}}\end{array} \)

\(\begin{array}{l}\frac{22.4\times 10^{-3}}{3.16 \times 10^{-7}}\end{array} \)

⁴

The molar volume is 7.1 x 10⁴times more than the atomic volume. Hence, the inter-atomic separation in hydrogen gas is larger than the size of the hydrogen atom.

Q2.18Written answer

Explain this common observation clearly: If you look out of the window of a fast-moving train, the nearby trees, houses etc., seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hilltops, the Moon, the stars etc.) seem to be stationary (In fact, since you are aware that you are moving, these distant objects seem to move with you).

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Worked solution

An imaginary line which joins the object and the observer’s eye is called the line of sight. When we observe the nearby objects, they move fast in the opposite direction as the line of sight changes constantly, whereas the distant objects seem to be stationary as the line of sight does not change rapidly.

Q2.19Written answer

The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the 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 such a long baseline, they show parallax only of the order of 1” (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1” (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres?

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Worked solution

Diameter of Earth’s orbit = 3 × 10¹¹ m

Radius of Earth’s orbit r = 1.5 × 10¹¹ m

Let the distance parallax angle be θ=1″ (s) = 4.847 × 10^–6 rad.

Let the distance of the star be D.

Parsec is defined as the distance at which the average radius of the Earth’s orbit subtends an angle of 1″

Therefore, D = 1.5 × 10¹¹ /4.847 × 10^–6= 0.309 x 10¹⁷

Hence 1 parsec ≈ 3.09 × 10¹⁶ m.

Q2.20Written answer

2. 20. The nearest star to our solar system is 4.29 light-years away. How much is this
distance in terms of parsecs? How much parallax would this star (named Alpha
Centauri) show when viewed from two locations of the Earth six months apart in its
orbit around the Sun?

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Worked solution

1 light year is the distance travelled by light in a year.

1 light year = 3 x 108 x 365 x 24 x 60 x 60 = 9.46 x 10¹⁵ m

Therefore, distance travelled by light in 4.29 light years = 4.29 x 9.46 x 10¹⁵ = 4.058 x 10¹⁶ m

Parsec is also a unit of distance
1 parsec = 3.08 x 10¹⁶ m
Therefore, the distance travelled by light in parsec is given as

4.29 light years =4.508 x 10¹⁶/3.80 x 10¹⁶ = 1.318 parsec = 1.32 parsec

Using the relation,
θ = d / D
here,
d is the diameter of Earths orbit, d = 3 × 10¹¹ m
D is the distance of the star from the earth, D = 405868.32 × 10¹¹ m
∴ θ = 3 × 10¹¹ / 405868.32 × 10¹¹ = 7.39 × 10^-6 rad
But the angle covered in 1 sec = 4.85 × 10^–6 rad
∴ 7.39 × 10^-6 rad = 7.39 × 10^-6 / 4.85 × 10^-6= 1.52″

Q2.21Written answer

Precise measurements of physical quantities are a need for science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc., are needed. Also, wherever you can, give a quantitative idea of the precision needed.

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Worked solution

Precise measurement is essential for the development of science. The ultra-short laser pulse is used for measurement of time intervals. X-ray spectroscopy is used to find the interatomic separation. To measure the mass of atoms, the mass spectrometer is developed.

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