Page No 36: - Chapter 2 Units & Measurements class 11 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 2.11:

The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Answer:

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:

Hence, area and volume both must have least significant figures i.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 2.12:

The mass of a box measured by a grocer’s balance is 2.300 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?

Answer:

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 places as there are in the number with the least decimal places. Hence, the total mass of 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 decimal places as there are in the number with the least decimal places.

Question 2.13:

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 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?

Answer:

Percentage error in P = 13 %

Value of P is given as 3.763.

By rounding off the given value to the first decimal place, we get P = 3.8.

Question 2.14:

A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic 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 out the wrong formulas on dimensional grounds.

Answer:

(a) Answer: Correct

Dimension of y = M⁰ L¹ T⁰

Dimension of a = M⁰ L¹ T⁰

Dimension of  = M⁰ L⁰ T⁰

Dimension of L.H.S = Dimension of R.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 be dimensionless, which is not so in the given case. Hence, the given formula is dimensionally incorrect.

(c) Answer: Incorrect

Dimension of y = M⁰L¹T⁰

Dimension of  = M⁰L¹T^–1

Dimension of = M⁰ L^–1 T¹

But the argument of the trigonometric function must be dimensionless, which is not so in the given case. Hence, the formula is dimensionally incorrect.

(d) Answer: Correct

Dimension of y = M⁰ L¹ T⁰

Dimension of a = M⁰ L¹ T⁰

Dimension of   = M⁰ L⁰ T⁰

Since the argument of the trigonometric function must be dimensionless (which is true in the given case), the dimensions of y and a are the same. Hence, the given formula is dimensionally correct.

Question 2.15:

A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m₀ 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:

Answer:

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

The given formula will be dimensionally correct only when the dimension of L.H.S is the same as that of R.H.S. This is only possible when the factor,   is dimensionless i.e., (1 – v²) is dimensionless. This is only possible if v² is divided by c². Hence, the correct relation is

.

Question 2.16:

The unit of length convenient on the atomic scale is known as an angstrom and is denoted by . The size of a hydrogen atom is about what is the total atomic volume in m³ of a mole of hydrogen atoms?

Answer:

Radius of hydrogen atom, r = 0.5  = 0.5 × 10^–10 m

Volume of hydrogen atom =

1 mole of hydrogen contains 6.023 × 10²³ hydrogen atoms.

∴ Volume of 1 mole of hydrogen atoms = 6.023 × 10²³ × 0.524 × 10^–30

= 3.16 × 10^–7 m³

Question 2.17:

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 hydrogen molecule to be about 1 ). Why is this ratio so large?

Answer:

Radius of hydrogen atom, r = 0.5  = 0.5 × 10^–10 m

Volume of hydrogen atom =

Now, 1 mole of hydrogen contains 6.023 × 10²³ hydrogen atoms.

∴ Volume of 1 mole of hydrogen atoms, V_a = 6.023 × 10²³ × 0.524 × 10^–30

= 3.16 × 10^–7 m³

Molar volume of 1 mole of hydrogen atoms at STP,

V_m = 22.4 L = 22.4 × 10^–3 m³

Hence, the molar volume is 7.08 × 10⁴ times higher than the atomic volume. For this reason, the inter-atomic separation in hydrogen gas is much larger than the size of a hydrogen atom.

Question 2.18:

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 (hill tops, 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).

Answer:

Line of sight is defined as an imaginary line joining an object and 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 in the 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 sight does not change its direction rapidly.

Question 2.19:

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 meters?

Answer:

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 of the Earth’s orbit subtends an angle of .

∴ We have

Hence, 1 parsec ≈ 3.09 × 10¹⁶ m.