Units and Measurement

Class 11 NCERT Physics

NCERT

1   Fill in the blanks$\\$ (a) The volume of a cube of side $1 cm$ is equal to.... $m ^3$$\\$ (b) The surface area of a solid cylinder of radius $2.0 cm$ and height $10.0 cm$ is equal to ... $(mm ^2)$$\\$ (c) A vehicle moving with a speed of $18 km \ h^{- 1}$ covers....m in $1 s$$\\$ (d) The relative density of lead is $11.3.$ Its density is ....$g cm ^ 3$ or .$ ...kg \ m ^{- 3} .$

Solution :

$ 1 cm = \dfrac{1}{100} m $$\\$ Volume of the cube $= 1cm ^3$$\\$ But, $1 cm^ 3 = 1 cm * 1 cm * 1 cm =\\ (\dfrac{1}{100})m*(\dfrac{1}{100})m*(\dfrac{1}{100})m$$\\$ $\therefore 1 cm^3=10^{-6} m^3$$\\$ Hence, the volume of a cube of side $1 cm$ is equal to $10^{-6} m^ 3$$\\$ The total surface area of a cylinder of radius $r$ and height h is $S = 2 \pi r ( r + h )$$\\$ Give that $r =2 cm = 2 * 1 cm = 2 * 10 mm = 20 mm\\ h = 10 cm = 10 = 10 mm = 100 mm\\ \therefore S = 2 * 3.14 * 20 ( 20 + 100 )= 15072 =1.5 * 10^ 4 mm ^2$$\\$ Using the conversion,$\\$ $ 1 km / hr=\dfrac{5}{18}m/s\\ 18 km/hr=18*\dfrac{5}{18}=5 m/s$$\\$ Therefore, distance can be obtained using the relation: Distance = Speed × Time =$ 5 * 1 = 5m$$\\$ Hence, the vehicle covers $5 m$ in $1 s.$$\\$ Relative density of a substance is given by the relation,$\\$ Relative density =$\dfrac{\text{Density of substance}}{\text{ Density of water}} $$\\$ Density of water = $kg / cm ^3$$\\$ Density of lead = Relative density of lead * Density of water$\\$ $=11.3*1=11.3g /cm^3$$\\$ Again,$ 1g=\dfrac{1}{1000}kg\\ 1 cm^3=10^{-6} m^3\\ 1g.cm^3=\dfrac{10^{-3}}{10^{-6}} lg/m^3\\ =10^3 kg/m^3\\ \therefore 11.3 g /cm^3\\ =11.3*10^3 kg/m^3$

2   Fill in the blanks by suitable conversion of units:$\\$ $1 k g/m ^2 s ^{-2}= .... g\ cm ^2 s ^{- 2}\\ 1 m = ....1 y\\ 3.0 ms^{-2}- .... km /h^{-2}\\ G = 6.67 * 10^{-11} Nm ^2 ( kg)^{-2}= .... ( cm )^3s^{-2 } g{- 1}$

Solution :

$1 kg = 10^3 g\\ 1 m^2=10^4 cm^2\\ 1 kg/m^2 s^{-2}=1 kg * 1 m^2*1s^{-2}\\ =10^3 g * 10^4 cm^2*1s^{-2}=10^7 g/cm^2 s^{-2} $Light year is the total distance travelled by light in one year.$1\text{ly= Speed of light * One year}\\ =(3*10^8m/s)*(365*24*60*60s)\\ =9.46*10^{15}m\\ \therefore 1m=\dfrac{1}{9.46*10^{15}}=1.057*10^{-16}\text{ly}\\ 1 m=10^{-3}km$ Again,$1s=\dfrac{1}{3600}h\\ 1s^{-1}=3600h^{-1}\\ 1s^{-1}=(3600)^2h^{-2}\\ \therefore 3ms^{-2}\\ =(3*10^{-3 }km)*((36000)^2h^{-2})\\ =3.88*10^{-4} km/h^{-2}\\ 1N=1 kg/ms^{-2}\\ 1 kg =10^{-3} g^{-1}\\ 1 m^3=10^6 cm^3\\ \therefore 6.67*10^{-11}Nm^2 kg^{-2}\\ =6.67*10^{-11}*(1 kg /ms^{-2})(1 m^2)(1 s^{-2})\\ \therefore 6.67*10^{-11}*(1 kg *1 m^3 * 1 s^{-2})\\ =6.67 * 10^{-11}*(10^{-3} g^{-1})*(10^6 cm^3)*(1 s^{-2})\\ =6.67 * 10^{-8} cm^3 s^{-2}g^{-1}$

3   A calorie is a unit of heat or energy and it equals about $4.2 J$ where $1J = 1 kg/ m$ Suppose we employ a system of units in which the unit of mass equals $\alpha kg,$ the unit of length equals $\beta m$, the unit of time is $\gamma s$. Show that a calorie has a magnitude $4.2 \alpha ^{-1} \beta ^{-2 } \gamma^ 2$ in terms of the new units.

Solution :

Given that,$\\$ $1 \text{calorie}=4.2(1 kg)(1 m^2)(1 s^{-2})$$\\$ New unit of mass = $\alpha kg$$\\$ Hence, in terms of the new unit,$ 1 kg=\dfrac{1}{\alpha }=\alpha^{-1}$$\\$ In terms of the new unit of length,$\\$ $1 m \dfrac{1}{\beta}=\beta^{-1} \ or \ 1 m^2 =\beta ^{-2}$$\\$ And, in terms of the new unit of time,$\\$ $ 1s=\dfrac{1}{\gamma}=\gamma^{-1}\\ 1 s^2=\gamma^{-2}\\ 1 s^{-2}=\gamma^2\\ \therefore 1 \text{Calories}=4.2(1 \alpha^{-1})(1 \beta^{-2})(1 \gamma^2)\\ =4.2 \alpha^{-1}\beta^{-2}\gamma^2$

4   Explain this statement clearly:$\\$ “To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. $\\$In view of this, reframe the following statements wherever necessary:$\\$ $\bullet $ atoms are very small objects$\\$ $\bullet $ a jet plane moves with great speed$\\$ $\bullet $ the mass of Jupiter is very large$\\$ $\bullet $ the air inside this room contains a large number of molecules$\\$ $\bullet $ a proton is much more massive than an electron$\\$ $\bullet $ the speed of sound is much smaller than the speed of light.

Solution :

The given statement is true because a dimensionless quantity may be large or small in comparison to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction.$\\$ $\bullet $An atom is a very small object in comparison to a soccer ball.$\\$ $\bullet $A jet plane moves with a speed greater than that of a bicycle.$\\$ $\bullet $Mass of Jupiter is very large as compared to the mass of a cricket ball.$\\$ $\bullet $The air inside this room contains a large number of molecules as compared to that present in a geometry box.$\\$ $\bullet $A proton is more massive than an electron.$\\$ $\bullet $Speed of sound is less than the speed of light.

5   A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in tem of the new unit if light takes $8$ min and $20$ s to cover this distance?

Solution :

Distance between the Sun and the Earth:$\\$ = Speed of light × Time taken by light to cover the distance$\\$ Given that in the new unit, speed of light = 414 unit$\\$ Time taken, $t = 8$ min $20 s = 500 s$$\\$ $\therefore $ Distance between the Sun and the Earth = $1 × 500 = 500 $ units