NCERT Solutions for Class 11 Physics Chapter 2 Units And Measurement

Ans : (a)\(1 \mathrm{cm}=\frac{1}{100} \mathrm{m}\)
Volume of cube = 1 cm³
\(\mathrm{But}, 1 \mathrm{cm} 3=1 \mathrm{cm} \times 1 \mathrm{cm} \times 1 \mathrm{cm}=\left(\frac{1}{100}\right) \mathrm{m} \times\left(\frac{1}{100}\right) \mathrm{m} \times\left(\frac{1}{100}\right) \mathrm{m}\)
\(\begin{array}{l}{ 1 \mathrm{cm} 3=10-6 \mathrm{m} 3} \\ {\text { Hence, the volume of a cube of side } 1 \mathrm{cm} \text { is equal to } 10-6 \mathrm{m} 3 \text { . }}\end{array}\)
\(\begin{array}{l}{\text { (b) The total surface area of a cylinder of radius } r \text { and height } h \text { is }} \\ {S=2 \pi r(r+h)} \\ {\text { Given that, }} \\ {r=2 \mathrm{cm}=2 \times 1 \mathrm{cm}=2 \times 10 \mathrm{mm}=20 \mathrm{mm}} \\ {h=10 \mathrm{cm}=10 \times 10 \mathrm{mm}=100 \mathrm{mm}} \\ {\therefore S=2 \times 3.14 \times 20 \times(20+100)=15072=1.5 \times 104 \mathrm{mm} 2}\end{array}\)
\(\begin{array}{l}{\text { (c) Using the conversion, }} \\ {1 \mathrm{km} / \mathrm{h}=\frac{5}{18} \mathrm{m} / \mathrm{s}} \\ {18 \mathrm{km} \mathrm{h}=18 \times \frac{5}{18}=5 \mathrm{m} / \mathrm{s}}\end{array}\)
\(\begin{array}{l}{\text { Therefore, distance can be obtained using the relation: }} \\ {\text { Distance }=\text { Speed } \times \text { Time }=5 \times 1=5 \mathrm{m}} \\ {\text { Hence, the vehicle covers } 5 \mathrm{m} \text { in } 1 \mathrm{s} \text { . }}\end{array}\)
(d) Relative density of a substance is given by the relation, 
Relative density = \(\frac{\text { Density of substance }}{\text { Density of water }}\)
Density of water = 1 g/cm³
\(\begin{aligned} \text { Density of lead } &=\text { Relative density of lead } \times \text { Density of water } \\ &=11.3 \times 1=11.3 \mathrm{g} / \mathrm{cm}^{3} \end{aligned}\)
\(1 \mathrm{g}=\frac{1}{1000} \mathrm{kg}\)
1 cm³  = 10⁻⁶  m³
\(1\mathrm{g} / \mathrm{cm}^ 3=\frac{10^{-3}}{10^{-6}} \mathrm{kg} / \mathrm{m}^{3}=10^{3} \mathrm{kg} / \mathrm{m}^{3}\)
11.3 g/cm³  = 11.3 x 10. kg/m³ 

Ans : \(\begin{aligned}(a) 1 \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-2} &=\frac{1 \mathrm{kg} \mathrm{m}^{2}}{s^{2}}=\frac{1 \times 1000 \times\left(10^{2}\right)^{2}}{s^{2}} \mathrm{g} \mathrm{cm}^{2} \\ &=10^{7} \mathrm{g} \mathrm{cm}^{2} \mathrm{s}^{-2} \end{aligned}\)
\(1 \mathrm{m} \quad=\frac{1}{9.46 \times 10^{5}} \mathrm{ly} \approx \frac{1}{10^{16}} \mathrm{ly}=10^{-16} \mathrm{ly}\)
\(\begin{aligned}(b) 3 \mathrm{ms}^{-2} &=\frac{3 \times 10^{-3} \mathrm{km}}{\left(\frac{1}{3600}\right)^{2} \mathrm{h}^{2}}=3 \times 3600 \times 3600 \times 10^{-3} \mathrm{km} \mathrm{h}^{-2} \\ &=3.888 \times 10^{4} \mathrm{km} \mathrm{h}^{-2} \end{aligned}\)
\(\begin{aligned}(c) G &=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{kg}^{-2}=6.67 \times 10^{-11} \frac{\mathrm{kg} \mathrm{m}}{\mathrm{s}^{2}} \mathrm{m}^{2} \mathrm{kg}^{-2} \\ &=6.67 \times 10^{-11} \mathrm{kg}^{-1} \mathrm{m}^{3} \mathrm{s}^{-2} \\ &=6.67 \times 10^{-11} \frac{\mathrm{m}^{3}}{\mathrm{kgs}^{2}}=\frac{6.67 \times 10^{-11} \times\left(10^{2}\right)^{3}}{\left(10^{3}\right)^{2}} \\ &=6.67 \times 10^{-8} \mathrm{cm}^{-3} \mathrm{s}^{-2} \mathrm{g}^{-1} \end{aligned}\) 

Ans : \(1 \mathrm{cal}=4.2 \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-2}\)
Hence in terms of new unit ,  = \(1 \mathrm{kg}=\frac{1}{\mathrm{a}}=\alpha^{-1}\)
In terms of the new unit length,
\(\mathrm{I} \mathrm{m}=\frac{1}{\beta}=\beta^{-1} \text { or } \mathrm{Im}^{2}=\beta^{-2}\)
And, in terms of the new unit of time, 
\(\begin{array}{l}{1 \mathrm{s}=\frac{1}{\gamma}=\gamma^{-1}} \\ {1 \mathrm{s}^{2}=\gamma^{-2}} \\ {1 \mathrm{s}^{-2}=\gamma^{2}}\end{array}\)
Calorie = \(4.2(1 \mathrm{a}-1)(1 \beta-2)(1 \mathrm{y} 2)=4.2 \mathrm{a}-1 \beta-2 \mathrm{y}^2\) 

Ans : Physical quantities are called large or small depending on the unit (standard) of measurement For example, the distance between two cities on earth is measured in kilometres but the distance between stars or intergalactic distances are measured in parsec The later standard parsec is equal to \(3.08 \times 10^{16} \mathrm{m} \text { or } 3.08 \times 10^{12} \) km is certainly larger than metre or kilometre Therefore, the inter-stellar or intergalactic distances are certainly larger than the distances between two cities on earth. 
(a) The size of an atom is much smaller than even the sharp tip of a pin. 
(b) A Jet plane moves with a speed greater than that of a super fast train. 
(c) The mass of Jupiter is very large compared to that of the earth. 
(d) The air inside this room contains more number of molecules than in one mole of air. 
(e) This is a correct statement. 
(f) This is a correct statement. 

Ans : \(\begin{array}{l}{\text { Distance between Sun and Earth }} \\ {=\text { Speed of light in vacuum } x \text { time taken by light to travel from Sim to Earth }=3 \times 10^{8} \mathrm{m} / \mathrm{s} \times 8} \\ {\min 20 \mathrm{s}=3 \times 10^{8} \mathrm{m} / \mathrm{s} \times 500 \mathrm{s}=500 \times 3 \times 10^{8} \mathrm{m} \text { . }} \\ {\text { In the new system, the speed of light in vacuum is unity. So, the new unit of length is } 3 \times 10^{8} \mathrm{m} \text { . }} \\ {\therefore \text { distance between Sun and Earth }=500 \text { new units. }}\end{array}\)