Question 2.1: Fill in the blanks
(a) The volume of a cube of side 1 cm is equal to ________ m3.
Solution:
The side of the cube is 1 cm.
Volume \( V \) of a cube = side3
\[
V = (1 \, \text{cm})^3 = 1 \, \text{cm}^3
\]
Converting \( \text{cm}^3 \) to \( \text{m}^3 \):
\[
1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3
\]
Therefore, the volume of a cube of side 1 cm is \( 1 \times 10^{-6} \, \text{m}^3 \).
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ________ (mm)2.
Solution:
Radius \( r \) = 2.0 cm
Height \( h \) = 10.0 cm
Surface area \( A \) of a cylinder = \( 2\pi r(h + r) \)
\[
A = 2\pi \times 2.0 \, \text{cm} \times (10.0 \, \text{cm} + 2.0 \, \text{cm}) = 2\pi \times 2.0 \, \text{cm} \times 12.0 \, \text{cm}
\]
\[
A = 48\pi \, \text{cm}^2
\]
Converting \( \text{cm}^2 \) to \( \text{mm}^2 \):
\[
1 \, \text{cm}^2 = 100 \, \text{mm}^2 \quad \Rightarrow \quad 48\pi \, \text{cm}^2 = 4800\pi \, \text{mm}^2
\]
Therefore, the surface area of a solid cylinder is \( 4800\pi \, \text{mm}^2 \approx 15079.64 \, \text{mm}^2 \).
(c) A vehicle moving with a speed of 18 km h-1 covers ________ m in 1 s.
Solution:
Speed \( v \) = 18 km/h
Converting speed to m/s:
\[
18 \, \text{km/h} = 18 \times \frac{1000 \, \text{m}}{3600 \, \text{s}} = 5 \, \text{m/s}
\]
Distance covered in 1 s = speed × time
\[
\text{Distance} = 5 \, \text{m/s} \times 1 \, \text{s} = 5 \, \text{m}
\]
Therefore, a vehicle moving with a speed of 18 km h-1 covers \( 5 \, \text{m} \) in 1 s.
(d) The relative density of lead is 11.3. Its density is ________ g cm-3 or ________ kg m-3.
Solution:
Relative density (R.D.) = 11.3
Density of water = 1 g/cm3
\[
\text{Density of lead} = 11.3 \times 1 \, \text{g/cm}^3 = 11.3 \, \text{g/cm}^3
\]
Converting \( \text{g/cm}^3 \) to \( \text{kg/m}^3 \):
\[
1 \, \text{g/cm}^3 = 1000 \, \text{kg/m}^3 \quad \Rightarrow \quad 11.3 \, \text{g/cm}^3 = 11300 \, \text{kg/m}^3
\]
Therefore, the density of lead is \( 11.3 \, \text{g/cm}^3 \) or \( 11300 \, \text{kg/m}^3 \).
Question 2.2: Fill in the blanks by suitable conversion of units
(a) 1 kg m2 s-2 = ________ g cm2 s-2
Solution:
We need to convert kg to g and m to cm:
\[
1 \, \text{kg} = 1000 \, \text{g}
\]
\[
1 \, \text{m} = 100 \, \text{cm}
\]
Therefore,
\[
1 \, \text{kg} \, \text{m}^2 \, \text{s}^{-2} = 1000 \, \text{g} \times (100 \, \text{cm})^2 \, \text{s}^{-2}
\]
\[
= 1000 \, \text{g} \times 10000 \, \text{cm}^2 \, \text{s}^{-2} = 10^7 \, \text{g} \, \text{cm}^2 \, \text{s}^{-2}
\]
Therefore, 1 kg m2 s-2 = \( 10^7 \) g cm2 s-2.
(b) 1 m = ________ ly
Solution:
1 light year (ly) is the distance light travels in one year:
\[
1 \, \text{ly} = 9.461 \times 10^{15} \, \text{m}
\]
Therefore,
\[
1 \, \text{m} = \frac{1}{9.461 \times 10^{15}} \, \text{ly} \approx 1.057 \times 10^{-16} \, \text{ly}
\]
Therefore, 1 m = \( 1.057 \times 10^{-16} \) ly.
(c) 3.0 m s-2 = ________ km h-2
Solution:
We need to convert m/s2 to km/h2:
1 m = 0.001 km
\[
1 \, \text{s}^2 = \left( \frac{1}{3600} \, \text{h} \right)^2 = \frac{1}{12960000} \, \text{h}^2
\]
Therefore,
\[
3.0 \, \text{m/s}^2 = 3.0 \times 0.001 \, \text{km} \times 12960000 \, \text{h}^{-2} = 38880 \, \text{km/h}^2
\]
Therefore, 3.0 m s-2 = 38880 km h-2.
(d) G = 6.67 x 10-11 N m2 (kg)-2 = ________ (cm)3 s-2 g-1.
Solution:
We need to convert N to dynes, m to cm, and kg to g:
1 N = 105 dynes
\[
1 \, \text{m} = 100 \, \text{cm}
\]
\[
1 \, \text{kg} = 1000 \, \text{g}
\]
Therefore,
\[
G = 6.67 \times 10^{-11} \, \text{N} \, \text{m}^2 \, \text{(kg)}^{-2} \times 10^5 \, \text{dynes} \times (100 \, \text{cm})^2 \times (1000 \, \text{g})^{-2}
\]
\[
= 6.67 \times 10^{-11} \times 10^5 \times 10^4 \times 10^{-6} \, \text{(cm)}^3 \, \text{s}^{-2} \, \text{g}^{-1}
\]
\[
= 6.67 \times 10^{-8} \, \text{(cm)}^3 \, \text{s}^{-2} \, \text{g}^{-1}
\]
Therefore, G = \( 6.67 \times 10^{-8} \) (cm)3 s-2 g-1.
Question 2.3: Conversion of units for a calorie
A calorie is a unit of heat or energy and it equals about 4.2 J where 1 J = 1 kg m2 s-2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, and the unit of time is γ s. Show that a calorie has a magnitude 4.2 α-1 β-2 γ2 in terms of the new units.
Solution:
Given: \( 1 \, \text{cal} = 4.2 \, \text{J} \)
\[
1 \, \text{J} = 1 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}
\]
In the new system of units:
1 unit of mass = \( \alpha \, \text{kg} \)
1 unit of length = \( \beta \, \text{m} \)
1 unit of time = \( \gamma \, \text{s} \)
We need to express 1 J in terms of the new units:
\[
1 \, \text{J} = 1 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} = \left( \frac{1 \, \text{kg}}{\alpha \, \text{kg}} \right) \cdot \left( \frac{1 \, \text{m}}{\beta \, \text{m}} \right)^2 \cdot \left( \frac{1 \, \text{s}}{\gamma \, \text{s}} \right)^{-2}
\]
\[
= \alpha^{-1} \cdot \beta^{-2} \cdot \gamma^{2}
\]
Therefore,
\[
4.2 \, \text{J} = 4.2 \cdot \alpha^{-1} \cdot \beta^{-2} \cdot \gamma^{2}
\]
Thus, a calorie has a magnitude of \( 4.2 \alpha^{-1} \beta^{-2} \gamma^{2} \) in terms of the new units.