NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurement

Q1.1 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 ... $\small (mm)^2$

(c) A vehicle moving with a speed of $\small 18\ km h^{-1}$ covers....m in 1 s

(d) The relative density of lead is 11.3. Its density is .... $\small g cm^{-3}$ or .... $\small kg m^{-3}$ .

Answer:

(a) We know, $\small 1 cm = .01 m$ (Tip: Divide by 100 to convert cm to m)

The volume of a cube of side a = $\small a^3\ m^3$

Volume of cube of side 1 cm (i.e, .01 m ) = $\small (.01)^3 = 10^{-6}\ m^3$

(b) We know, $1 cm = 10 mm$ (Tip: Multiply by 10 to convert cm to mm)

The surface area of a solid cylinder of radius r and height h = $2(\pi r^2) + 2 \pi rh = 2\pi r(r + h)$

Required area = $2\pi (20)(\ 20 + 100) mm^3 = 40 \pi (120) mm^3 = 4800 \pi\ mm^3 = 1.5 \times 10^{4 }\ mm^3$

(c) (Tip: multiply by 5/18 to convert $km h^{-1}\ to\ m s^{-1}$ )

Ditance covered = $Speed \times time = (18 \times 5/18)ms^{-1} \times (1s) = 5 m$

(d) Density = Relative Density $\times$ Density of water

(Density of water = $1 \ gcm^{-3} = 1000\ kgm^{-3}$ )

Density of lead= $(11.3)\times1\ gcm^{-3} = 11.3\ gcm^{-3} \\ $

$\text{Or, Density of lead} = (11.3)\times1000\ kgm^{-3} = 11300\ kgm^{-3} = 1.13 \times 10^4\ kgm^{-3}$

Q 1.2 : Fill in the blanks by suitable conversion of units

(a) $1 kg m^2s^{-2} = \ \ \ \ \ \ \ \ gcm^2s^{-2}$

b) $1 m = \ \ \ \ ly$

(c) $3.0 \ m s^{-2} = \ \ \ \ km h^{-2}$

(d) $G = 6.67 \times 10^{-11} Nm^2(kg)^{-2} = \ \ \ \ \ (cm)^3 s^{-2}g^{-1}$

Answer:

(a) $(1 kg \rightarrow 1000g ;\ 1m\rightarrow 100 cm)$

$1 kgm^2s^{-2} = (1000g)(100cm)^2s^{-2} = (10^3 \times10^{4})gcm^2s^{-2}$

$\Rightarrow 1 kgm^2s^{-2}= 10^{7}\ gcm^2s^{-2}$

(b) $1 m = 1.057\times10^{-16} ly$ (1 ly = Distance travelled by light in 1 year )

(c) $(1 m \rightarrow 10^{-3}km ;\ (60\times60 =) 3600 s \rightarrow 1hr)$

$\\3.0 ms^{-2} = 3.0 (10^{-3}km)(1/3600 hr)^{-2} = 3.0 \times (3600)^2 / 1000\ kmhr^{-2} \\ $

$\Rightarrow 3.0 ms^{-2} = 3.9 \times 10^4 \ kmhr^{-2}$

(d) $(N \rightarrow J.s = kgms^{-2};\ 1 m \rightarrow 100cm;\ 1 kg\rightarrow 1000g)$

$G = 6.67 \times 10^{-11} (kgms^{-2})m^2kg^{-2} = (6.67 \times 10^{-11})(m)^3 s^{-2}kg^{-1}$

$\Rightarrow G = 6.67 \times 10^{-8}cm^3 s^{-2}g^{-1}$

Q 1.3 : A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where $1J = 1 kg m^2 s^{-2}$ . 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.

Answer:

Given,

$1 Cal = 4.2 (kg)(m)^{2}(s)^{-2}$

Given new unit of mass = $\alpha kg$ (In old unit 1kg corresponded to a unit mass, but in new unit $\alpha kg$ corresponds to a unit mass)

$\therefore$ In terms of the new unit, $1 kg = 1/\alpha = \alpha ^{-1}$

Similarly in terms of new units , $1 m = 1/\beta = \beta^{-1}$ and $1 s = 1/\gamma = \gamma ^{-1}$

$\therefore$ $1 Cal = 4.2 (kg)(m)^{2}(s)^{-2} = $

$ 1Cal = 4.2 (\alpha ^{-1})(\beta ^{-1})^{2}(\gamma ^{-1})^{-2} = 4.2 \alpha ^{-1}\beta ^{-2}\gamma ^{2}$

Q 1.4 : 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 :

(a) atoms are very small objects

(b) a jet plane moves with great speed

(c) the mass of Jupiter is very large

(d) the air inside this room contains a large number of molecules

(e) a proton is much more massive than an electron

(f) the speed of sound is much smaller than the speed of light

Answer:

The given statement is true. A dimensional quantity may be small with respect to one reference and maybe large with respect to another reference. Hence, we require a standard reference to judge for comparison.

(a) An atom is a very small object with respect to a tennis ball. (but larger than an electron!)

(b) A jet plane moves with great speed with respect to a train.

(c) The mass of Jupiter is very large as compared to an apple.

(d) The air inside this room contains a large number of molecules as compared to in your lungs.

(e) A proton is much more massive than an electron

(f) The speed of sound is less than the speed of light

Q 1.5 : 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 terms of the new unit if light takes 8 min and 20 s to cover this distance?

Answer:

Distance between Sun and Earth = Speed of light x Time taken by light to cover the distance

Speed of light = 1 unit

Time taken by light to reach earth is 8 minute 20 seconds

Time taken = $(8\times60) + 20 = 500 s$

The distance between Sun and Earth = 1 x 500 = 500 units.

Q 1.6 : Which of the following is the most precise device for measuring length :

(a) a vernier callipers with 20 divisions on the sliding scale

(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale

(c) an optical instrument that can measure length to within a wavelength of light?

Answer:

To judge which tool is more precise, we have to find out their least count. Least count defines the margin of error and hence the precision. Hence the instrument with lower least count will be more precise.

(a) Least count = $1MSD - 1VSD = 1 - (19/20)= 1/20 = 0.05\ cm$ (Taking 1 MSD as 1 mm)

(b) Least count = pitch/ number of divisions = $1/10000 = 0.001 cm$

(c) least count = wavelength of light = 400nm to 700nm, that is in the range of $10^{-7}$ m

Therefore, the optical instrument is the most precise device used to measure length.

Q 1.7 : 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?

Answer:

Given,

Magnification of Microscope = 100

The average width of hair under the microscope = 3.5 mm

(20 observations were made to calculate the average i.e. 3.5 mm as an experimental procedure. No need in our calculations.)

(Note: When magnified, the width is 3.5 mm. Hence actual width will be less by a factor of magnification value)

The average thickness of hair = $\frac{3.5 mm}{100} = 0.035 \ mm.$

Q 1.8 Answer the following :

(a)You are given a thread and a metre scale. How will you estimate the diameter of the thread?

(b)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?

(c) 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?

Answer:

(a) Take the thread and wrap it around the metre scale. Make sure the coils are packed closely without any space in between. If the diameter of the thread is d and number of turns obtained are n, then (n x d) corresponds to the marking on the metre scale, l.

Therefore, the diameter of the thread would be, d = l/n

(b) Theoretically, by increasing the number of divisions on the circular divisions, the value of least count decreases and hence accuracy increases.(Lower the value of least count, better will be the reading)

But practically, the number of divisions can be increased only up to a certain limit. (Also two adjacent divisions cannot be separated by a distance less than the human eye resolution!)

(c) With an increase in the number of observations, the accuracy of the experiment increases as the error is now distributed over a large range. Hence, a set of 100 measurements of the diameter is expected to yield a more reliable estimate than a set of 5 measurements.

Q 1.9 : The photograph of a house occupies an area of 1.75 $cm^2$ on a 35 $mm$ slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 $m^2$ . What is the linear magnification of the projector-screen arrangement?

Answer:

Given,

Area of the house in the photo = $1.75\ cm^{2}$

Area of the house on the screen = $1.55 m^{2} = 1.55 \times10^{4} cm^2$

$\therefore$ Arial magnification, $m_{a}$ = Area on the screen / area on photo = $\frac{1.55\times 10^{4}}{1.75} = 0.886 \times 10^4$

Linear magnification of the projector- screen arrangement $m_{l} = \sqrt{m_{a}} = \sqrt{.886 \times 10^{4}}= 94.1$

Q 1.10 : State the number of significant figures in the following :
(a)$0.007 \ m^2$

(b) $2.64 \times 10^{24} kg$

(c) $0.2370 \ g cm^{-3}$

(d) $6.320 \ J$

(e) $6.032 \ Nm^{-2}$

(f) $0.0006032 \ m^{2}$

Answer:

(a) The given value is $0.007 \ m^2$ .

Since, the number is less than 1 , the zeros on the right to the decimal before the first non-zero integer is insignificant. So, the number 7 is the only significant digit.

$\therefore$ It has 1 significant digit.

(b) The value is $2.64 \times 10^{24} kg$

For the determination of significant values, we do not consider the power of 10 (Number is not less than 1). The digits 2, 6, and 4 are significant figures.

$\therefore$ It has 3 significant digits.

(c)The value is $0.2370 \ g cm^{-3}$ .

For the given value with decimals, all the numbers 2, 3, 7 and 0 are significant.

$\therefore$ It has 4 significant digits.

(d) The value is $6.320 \ J$ .

It has 4 significant digits.

(e) The value is $6.032 \ Nm^{-2}$ .

All the four digits are significant as the zeros in between two non-zero values are also significant.

$\therefore$ It has 4 significant digits.

(f) The value is $0.0006032 \ m^{2}$

Same as (a), first three zeroes after the decimal is insignificant. Only 6, 0, 3, 2 are significant.

$\therefore$ It has 4 significant digits.

Q 1.11 : The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Answer:

Given,

Length, l = 4.234 m ; Breadth, b = 1.005 m; Height, h = 2.01 cm = 0.0201 m

The length has 4 significant figures

The breadth has 4 significant figures

The height has 3 significant figures (Since the number is less than 1, hence zeroes after decimal before the first non-zero integer is insignificant)

We know,

Surface area of a cuboid, S = 2(l x b + b x h + h x l)

$\Rightarrow$ S= $2(4.234 \times 1.005 + 1.005 \times 0.0201 + 0.0201 \times 4.234)$

$\Rightarrow$ S= $2(4.255 + 0.020 + 0.085)$

$\Rightarrow$ S= $8.72 \ m^{2}$

(Note: For addition/subtraction , the number of places after the decimal point in the answer is less than or equal to the number of decimal places in every term in the sum; i.e decided by the numbers after the decimal. )

Volume, V = l x b x h

$\Rightarrow$ V= $4.234 \times 1.005 \times 0.0201$ = $0.0855 \ m^3$ ( $\therefore$ 3 significant digits)

(Note: For Multiplication The LEAST number of significant digits in any number determines the number of significant figures in the answer; i.e decided by the number of significant digits )

The area has three significant values 2, 7 and 8.

The volume has three significant values 5, 5 and 8.

Q 1.12 : The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?

Answer:

Given,

The mass of the box = 2.30 kg

and the mass of the first gold piece = 20.15 g = 0.02015 kg

The mass of the second gold piece = 20.17 g = 0.02017 kg

(a) The total mass = 2.30 + 0.02015 + 0.02017 = 2.34032 kg

Since one is the least number of decimal places, the total mass = 2.3 kg.

(Note: For addition/subtraction , the number of places after the decimal point in the answer is less than or equal to the number of decimal places in every term in the sum; i.e decided by the numbers after the decimal. )

(b) Difference in masses $=20.17-20.15=0.02 \mathrm{~g}$

In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal place

Q 1.13 : A famous relation in physics relates ‘moving mass’ $m$ to the ‘rest mass’ $m_{o}$ 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 due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes : $m = \frac{m_{o}}{(1-v^2)^{1/2}}$

Guess where to put the missing c.

Answer:

The relation given is $m = \frac{m_{o}}{(1-v^2)^{1/2}}$

Divide both sides by $m_{o}$ ;

$\therefore$ L.H.S becomes $m / m_{o}$ which is dimensionless.

Hence, R.H.S must be dimensionless too. (After Dividing by $m_{o}$ !)

$\frac{1}{(1-v^2)^{1/2}}$ can be dimensionless only when $v \rightarrow (v/c)$

Therefore, the dimensional equation is $m = \frac{m_{o}}{(1-(\frac{v}{c})^2)^{1/2}}$

Q 1.14 : The unit of length convenient on the atomic scale is known as an angstrom and is denoted by: 1Å = 10^{-10} m$ . The size of a hydrogen atom is about 0.5Å. What is the total atomic volume in $m^3$ of a mole of hydrogen atoms?

Answer:

Radius of an Hydrogen atom = 0.5 $ Å$ = 0.5 x 10 -10 m

Volume, V = $\frac{4}{3} \pi r^3$

$\Rightarrow$ V = $4/3 \times 22/7 \times (0.5 \times 10^{-10})^3$

$\Rightarrow$ V = $0.524 \times 10^{-30} m^3$

1 hydrogen mole contains $6.023 \times 10^{23}$ hydrogen atoms.

The volume of 1 mole of hydrogen atom, V' = $6.023 \times 10^{23} \times 0.524 \times 10^{-30}$

$\Rightarrow$ V' = $3.16 \times 10^{-7} m^3 \approx 3 \times 10^{-7} m^3$ .

Q1.15 : One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large?

Answer:

Radius of hydrogen atom = 0.5 $Å$ = 0.5 x 10 -10 m (Size here refers to Diameter!)

Volume occupied by the hydrogen atom, V= $\frac{4}{3} \pi r^3$

$\Rightarrow$ V= $4/3 \times 22/7 \times ( 0.5 \times 10^{-10})^3$

$\Rightarrow$ V= $0.524 \times 10^{-30} m^3$

1 mole of hydrogen contains 6.023 x 10 23 hydrogen atoms.

Volume of 1 mole of hydrogen atom, V' = 6.023 x 10 23 x 0.524 x 10 -30

$\Rightarrow$ V' = 3.16 x 10^-7 m³

$V_{m} = 22.4 L = 22.4 \times 10^{-3} m^3$

$\frac{V_{m}}{V'} = \frac{22.4\times10^{-3}}{3.16\times10^{-7}} = 7.09 \times 10^4$

The molar volume is $7.09 \times 10^4$ times greater than the atomic volume.

Hence, intermolecular separation in gas is much larger than the size of a molecule.

Q 1.16 Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

Answer:

Our eyes detect angular velocity, not absolute velocity. An object far away makes a lesser angle than an object which is close. That's why the moon (which is so far away!) does not seem to move at all angularly and thus seems to follow you while driving.

In other words, while in a moving train, or for that matter in any moving vehicle, a nearby object moves in the opposite direction while the distant object moves in the same direction. !

Q 1.17: The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding $10^7 K$ , and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data: mass of the Sun = $2.0 \times 10^{30} kg$ , radius of the Sun = $7.0 \times 10^8 m$

Answer:

Given,

Mass of the Sun, m = $3 \times10^{30}\ kg$

The radius of the Sun, r = $8 \times10^8\ m$

$\therefore$ Volume V = $\frac{4}{3}\pi r^3$

$\Rightarrow$ V= $\frac{4}{3} \times \frac{22}{7} \times (8 \times 10^8)^3 = 2145.52 \times10^{24} m^3$

Density = Mass/Volume

$\Rightarrow$ Density = $\frac{3\times10^{30}}{2145.52\times10^{24}} = 1.39 \times 10^3\ kgm^{-3}$

Therefore, the density of the sun is in the range of solids and liquids and not gases. This high density arises due to inward gravitational attraction on outer layers due to inner layers of the Sun. (Imagine layers and layers of gases stacking up like a pile!)