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

6   Which of the following is the most precise device for measuring length:$\\$ a vernier callipers with $20$ divisions on the sliding scale$\\$ a screw gauge of pitch 1 mm and 100 divisions on the circular scale$\\$ an optical instrument that can measure length to within a wavelength of light ?

Solution :

A device with minimum count is the most suitable to measure length. Least count of vernier callipers$\\$ $= 1$ standard division (SD) - $1$ vernier division (VD)$\\$ $=1-\dfrac{9}{10}=\dfrac{1}{10}=0.01cm$$\\$ Least count of screw gauge =$ \dfrac{\text{pitch}}{\text{Number of divisions}}$$\\$ $=\dfrac{1}{1000}=0.001 cm $$\\$ Least count of an optical device = Wavelength of light$\\$ $= 0.00001 cm$$\\$ Hence, it can be inferred that an optical instrument is the most suitable device to measure length.

7   A student measures the thickness of a human hair by looking at it through a microscope of magnification $100$. He makes $20$ observations and finds that the average width of the hair in the field of view of the microscope is $3.5 mm$. What is the estimate on the thickness of hair?

Solution :

Magnification of the microscope = $100$$\\$ Average width of the hair in the field of view of the microscope = $3.5 mm$$\\$ $\therefore $ Actual thickness of the hair is $ \dfrac{3.4}{100}=0.035 mm$

8   Answer the following:$\\$ You are given a thread and a metre scale How will you estimate the diameter of the thread?$\\$ A screw gauge has a pitch of $1.0 mm$ and $200$ divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?$\\$ The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of $100$ measurements of the diameter expected to yield a more reliable estimate than a set of $5$ measurements only?

Solution :

Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a metre scale. The diameter of the thread is given by the relation.$\\$ Diameter =$\dfrac{\text{Lenght of thread}}{\text{ Number of turns }}$$\\$ It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number divisions of the circular scale will increase its accuracy to a certain extent only.$\\$ A set of $100$ measurements is more reliable than a set of $5$ measurements because random errors involved in the former are very less as compared to the latter.

9   The photograph of a house occupies an area of $1.75cm ^2$ on a $35mm$ slide. The slide is projected on to a screen, and the area of the house on the screen $1.55m ^2$ is the linear magnification of the projector-screen arrangement?

Solution :

Area of the house on the slide $= 1.75cm ^2$$\\$ Area of the image of the house formed on the screen $=1.55m ^2 = 1.55 * 10^ 4 cm^ 2$$\\$ Arial magnification $m _a=\dfrac{\text{Area of image}}{\text{Area of object}}=\dfrac{1.55}{1.75}*10^4$$\\$ $\therefore $ Linear magnifications, $m_1=\sqrt{m_a}$$\\$ $=\sqrt{\dfrac{1.55}{1.75}*10^4}=94.11$

10   State the number of significant figures in the following:$\\$ $0.007 m ^2\\ 2.64 *10 ^{24} kg\\ 0.2370 g cm^{-3}\\ 6.320 J\\ 6.032 Nm^{- 2}\\ 0.0006032 m ^2$

Solution :

The given quantity is $0.007m ^2$$\\$ If the number is less than one, then all zeros on the right of the decimal point (but left to the first non-zero) are insignificant. This means that here, two zeros after the decimal are not significant. Hence, only $7$ is a significant figure in this quantity.$\\$ The given quantity is $2.64 * 10^{ 24} kg .$$\\$ Here, the power of $10$ is irrelevant for the determination of significant figures. Hence, all digits i.e., $2, 6$ and $4 $are significant figure$\\$ The given quantity is $0.2370 g cm ^{- 3}$$\\$ For a number with decimals, the trailing zeroes are significant. Hence, besides digits $2, 3$ and $7, 0$ that appears after the decimal point is also a significant figure. The given quantity is $6.320J$$\\$ For a number with decimals, the trailing zeroes are significant. Hence, all four digits appearing in the given quantity are significant figures.$\\$ The given quantity is $6.032 Nm^{-2}$$\\$ All zeroes between two non-zero digits are always significant$\\$ The given quantity is $0.0006032m ^2$$\\$ If the number is less than one, then the zeroes on the right of the decimal point (but left to the first non-zero) are insignificant. Hence, all three zeroes appearing before $6$ are not significant figures. All zeros between two non-zero digits are always significant. Hence the remaining four digits are significant figures.

11   .

Solution :