Units & Measurement
NCERT Solution
Part 3
Question 14: 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 `(2πt)/T`
Answer: Here, [LHS] = [y] = [L]
[RHS] = [a sin `(2πt)/T`] = [L sin `T/T`] = [L]
So, this formula is dimensionally correct.
(b) y = a sin vt
Answer: Here, [y] = [L]
[a sin vt] = [L sin (LT-1T)] = [L sin L]
This equation is wrong
(c) y = `(a/T)` sin `t/a`
Answer: Here, [y] = [L]
[`(a/T)` sin `t/a`] = [`L/T` sin `T/L`] = [LT-1 sin TL-1]
This equation is wrong
(d) y =( a√2) (sin `(2πt) / T` + cos `(2πt) / T` )
Answer: Here, [y] = [L] [a√2] = [L]
[sin`(2πt)/T` + cos `(2πt)/T`]
= [sin `T/T` + cos `T/T`] = dimensionless
This equation is correct.
(a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.
Question 15: A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo o f 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=(m_0)/((1-v^2)^(1/2))`
Guess where to put the missing c.
Answer: From given equation, `(m_0)/m=sqrt(1-v^2)`
LHS is dimensionless
So, RHS should also be dimensionless
It is possible only when `sqrt(1-v^2)` shall be `sqrt(1-(v^2)/(c^2))`
So, the correct formula is `m=m_0(1-(v^2)/(c^2))^(-1/2)`
Question 16: 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?
Answer: Volume of hydrogen atom = volume of sphere = 4/3 πr3
`=4/3× 3.14×(0.5×10^(-10))^3` m3
= 5.22 × 10(-30) m3
We know that as per Avogadro’s principle, 1 mole contains 6.023 × 1023 particles
So, volume of 1 mole of hydrogen atoms
= 6.023 × 1023 × 5.22 × 10-30 = 3.15 × 10-7 m3
Question 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 Å). Why is this ratio so large?
Answer: Volume of 1 mole of ideal gas Vg = 22.4 liter = 22.4 × 10-3 m3
As calculated in previous answer, volume of 1 mole hydrogen molecule = 3.15 × 10-7
Now, ratio of volume of ideal gas and volume of 1 mole of hydrogen
`(V_g)/(V_H)` = (22.4 × 10-3) ÷ (3.15 × 10-7)
= 7.1 × 104
Compared to the size of a hydrogen molecule, the interatomic separation in a gas is very large. Hence, this ratio is very large.
Question 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: The line of sight is the line joining an object and our eyes. The line of sight from a nearby object changes rapidly. On the other hand, the line of sight of a distant object changes not so rapidly. So, nearby object appear to move rapidly while distant object appears to be stationary.
Question 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 metres?
Answer: Parallax method gives the following formula:
`θ=b/D`
Here, b = baseline, D = distance of distant object
Given, θ = 1” = 4.85 × 10-6
and b = 1.5 × 1011 m (radius of earth)
So, D = b/θ
`=(1.5×10^(11))/(4.85×10^(-6))` m
= 3.09 × 1016 = 3 × 1016 m