## Unit 1 Physical World And Measurement Chapter 1 Measurement And Dimension Of Physical Quantity Long Answer Type Questions

**Question 1. Name a dimensionless physical quantity and show that it is dimensionless.**

**Answer:**

‘Angle’ is a dimensionless quantity.

As per the definition, the angle subtended at the center of a circle by an arc has a value equal to the ratio of the arc and the radius of the circle.

Thus, angle = \(\frac{\text { arc }}{\text { radius }} \text {. }\). Arc and radius both denote lengths.

Hence, dimension of angle = \(\frac{\mathrm{L}}{\mathrm{L}}=\mathrm{L}^{1-1}=\mathrm{L}^0=1\)

Hence, the angle is a dimensionless quantity.

**Question 2. What are**

- Light year,
- Astronomical unit, and
- Parsec?

**Answer:**

Each of them is a unit of length.

- 1 light year (ly) is the distance traveled by light in a vacuum in 1 year.
- 1 ly = (365.25 x 24 x 60 x 60 s) x (3 x 10
^{8}m/s) ≈ 9.47 x 10^{15}m - It is often rounded off as 1 ly ~ 10
^{16}m.

- 1 ly = (365.25 x 24 x 60 x 60 s) x (3 x 10
- 1 astronomical unit (AU) is the mean distance between the sun and the earth. 1AU ≈ 1.5 x 10
^{11}m. - 1 parsec (pc) is the distance at which 1 AU subtends an angle of 1 arc-second (1″). 1pc ≈ 3.1 x 10
^{16}m. Light year and parsec are used as very large units of length. Astronomical distances are usually expressed in ly or pc. 1 pc ≈ 3.26 ly

**Question 3. Is a light year a fundamental unit or a derived unit?**

**Answer:**

The distance covered by light in 1 year in a vacuum is one light year. This is a unit of length and hence, fundamental.

**Question 4. Units of three physical quantities X, Y, and Z are g · cm ^{2} · s^{-5}, g · s^{-1} and cm · s^{-2}. Find the relationship between X Y and Z.**

**Answer:**

Units of three physical quantities X, Y, and Z are g · cm^{2} · s^{-5}, g · s^{-1} and cm · s^{-2}.

Let X = kY^{a}Z^{b}, where k is a dimensionless constant, and a and b are numeric indices. From the given units, dimensions of X = ML²T^{-5}, those of Y and Z are MT^{-1} and LT^{-2} respectively.

∴ \(\mathrm{ML}^2 \mathrm{~T}^{-5}=\left(\mathrm{MT}^{-1}\right)^a\left(\mathrm{LT}^{-2}\right)^b\)

= \(\mathrm{M}^a \mathrm{~L}^b \mathrm{~T}^{-a-2 b}\)

Equating powers of the same base, a = 1, b = 2

∴ X = kYZ².

**Question 5. If force (F), Length (L), and time (T) have the bask units, what would be the dimension of mass?**

**Answer:**

If force (F), Length (L), and time (T) have the bask units,

From Newton’s second law of motion, force mass = \(\frac{\text { force }}{\text { acceleration }}\)

Dimension of force = F (as given)

Dimension of acceleration = LT^{-2}

∴ Dimensional of mass = \(\frac{\mathrm{F}}{\mathrm{LT}^{-2}}=\mathrm{FL}^{-1} \mathrm{~T}^2 .\)

**Question 6. A famous relation in physics relates ‘moving mass’ m with the ‘rest mass’ m _{0} 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 by Albert Einstein. A boy recalls the relation almost correctly but forgets where to put the constant c. He writes: \(m=\frac{m_0}{\left(1-v^2\right)^{1 / 2}}\). Guess where to put the missing c.**

**Answer:**

A famous relation in physics relates ‘moving mass’ m with the ‘rest mass’ m_{0} 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 by Albert Einstein. A boy recalls the relation almost correctly but forgets where to put the constant c. He writes: \(m=\frac{m_0}{\left(1-v^2\right)^{1 / 2}}\).

1 is a dimensionless number, but v² has a dimension L²T^{-2}, because [v] = LT^{-1}.

So the expression (1 – v²) does not have dimensional homogeneity.

Here, v² should be replaced by a dimensionless quantity. As [c] = LT^{-1} and [c²] = L²T^{-2}, we note that \(\frac{v^2}{c^2}\) is dimensionless.

So the correct relation should be: \(m=\frac{m_0}{\left(1-\frac{v^2}{c^2}\right)^{1 / 2}}\)