NCERT Solutions Class 11 Physics Chapter 1 Units and Measurements is prepared by our senior and renowned teachers of Param Himalaya. Primary focus while solving these questions of class 11 in NCERT textbook, also do read theory of this Chapter 1 Units and Measurements while going before solving the NCERT questions.
Question 1.1. Fill in the blanks :
a. The volume of a cube of side 1 cm3 = .... m3.
b. The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm =.. (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 :
a. Length of edge = 1cm = $\frac{1}{100}$ m
Volume of the cube = side3
Putting the value of side, we get
Volume of the cube = ( $\frac{1}{100}$ m)3 =(10-2 m)3 = 10-6 m3.
The volume of a cube of side 1 cm = 10-6 m3
b. Given,
Radius, r = 2.0 cm = 20 mm (convert cm to mm).
Height, h = 10.0 cm =100 mm.
The formula of total surface area of a cylinder S = 2πr (r + h).
Putting the values in this formula, we get
Surface area of a cylinder S = 2πr (r + h )= 2 x 3.14 x 20 (20+100)
S = 15072 = 1.5 × 104 mm2
c. Using the conversion
1 km = 1000 m and 1 hour = 3600 sec
Speed = 18 km/h
Speed = 18 × $\frac{1000m}{3600sec}$
= $18 \times \frac{5m}{18sec}$
Speed = 5 m/sec
Use formula
Speed = $\frac{distance}{time}$
Cross multiply it, we get
Distance = Speed × Time = 5 × 1 = 5 m
A vehicle moving with a speed of 18 km h–1 covers 5 m in 1 s.
d.
Density of water = 1 $\frac{g}{cm^{3}}$
Putting the values, we get
Relative Density= $\frac{Density \ of \ lead }{Density \ of \ water}$
Density of lead = Relative Density × Density of water
Density of lead = 11.3 × 1 $\frac{g}{cm^{3}}$
= 11.3 g cm-3
1 cm = (1/100 m) =10–2 m
1 g = 1/1000 kg = 10-3 kg
Putting the value of 1 cm and 1 gram
=$11.3 \times \frac{10^{-3}kg}{(10^{-2m})^{3}}$
=$11.3 \times \frac{10^{-3}kg}{10^{-6}m^{3}}$
= 11.3 ×10–3 × 106 kg m-3
=11.3 × 103 kg m–3
Question 1.2. Fill in the blanks by suitable conversion of units:
a. 1 kg m2s–2= ....g cm2 s–2
b. 1 m =..... ly
c. 3.0 m s–2 = .... km h–2
d. G= 6.67 × 10–11 N m2 (kg)–2= .... cm3 s–2 g–1.
Solution :
a. 1 kg = 103 g1 m=102 cm , 1 m2 =104 cm2
1 kg m2 s–2 = 1 kg × 1 m2 × 1 s–2
=103 g × 104 cm2 × 1 s–2 = 107 g cm2 s–2
1 kg m2s–2= 107 g cm2 s–2
b. Distance = Speed × Time
Speed of light = 3 × 108 m/s
Time = 1 year = 365 days = 365 × 24 hours = 365 × 24 × 60 × 60 sec
Putting these values in above formula we get
1 light year distance = (3 × 108 m/s) × (365 × 24 × 60 × 60 s) = 9.46 × 1015 m
9.46 × 1015 m = 1 ly
So that 1 m = 1/ 9.46 × 1015 ly = 1.06 × 10-16 ly
c. 1 hour = 3600 sec so that 1 sec = $\frac{1}{3600}$ hour
1 km = 1000 m so that 1 m = $\frac{1}{1000}$ km
3.0 m s–2 = 3.0 ($\frac{1}{1000}$ km)( $\frac{1}{3600}$ hour)-2 = 3.0 × 10–3 km × ((1/3600)-2 h–2)
= 3.0 × 10–3 km × (3600)2 h–2
= 3.88 × 104 km h–2
3.0 m s–2= 3.88 × 104 km h–2
d. 1 N = 1 kg m s–2
1 kg = 103 g
1 m3 = 106 cm3
∴ 6.67 × 10–11 N m2 kg–2 = 6.67 × 10–11 × (1 kg m s–2) (1 m2) (1 kg–2)
= 6.67 × 10–11 × 1 kg–1 × 1 m3 × 1 s–2)
=6.67 × 10–11 × (103 g)–1 × 1 m3 × 1 s–2)
= 6.67 × 10–11 × (10–3 g–1) × (106 cm3) × (1 s–2)
= 6.67 × 10–8 cm3 s–2 g–1
1.3. A calorie is a unit of heat or energy 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 α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude $4.2 \alpha ^{-1}\beta ^{-2}\gamma ^{2}$ in terms of the new units.
Solution :
1 calorie = 4.2 J = 4.2 $kg m^{2}s^{-2}$ ... (i)
New unit of mass = $\alpha$ kg
1 kg = $\frac{1}{\alpha }$ unit of mass
1 Kg= $\alpha ^{-1}$ unit of mass
New unit of length = $\beta$ m
$1m= \beta ^{-1}$ new unit of length
New unit of time = $\gamma$sec
$1sec= \gamma ^{-1}$ new unit of mass.
Putting these values in eq.(i),
1 calorie = 4.2 J = $4.2 kg m^{2}s^{-2}$
1 calorie = $4.2 \alpha ^{-1}\beta ^{-2}\gamma ^{2}$ new units.
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.
Solution :
(a). Atoms are very small in comparison to a tennis ball.
(b) A jet plane moves with great speed in comparison to the bicycle.
(c) The mass of Jupiter is very large in comparison to the mass of a bob of a simple pendulum.
(d) The air inside the room contains a larger number of molecules in comparison to the number of molecules in a small box.
(e) a proton is much more massive than an electron in comparison to the difference in the mass of a basketball and baseball.
(f) the speed of sound is much smaller than the speed of light in comparison to the difference between the speeds of a bicycle and that speed of a car.
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 ?
Solution : It is given that velocity of light in vacuum , c = 1 new unit of length $s^{-1}$.
The time taken by sunlight to reach earth , t = 8 min. 20s = 500s.
$\therefore$ Distance S between sun and earth is
S = ct = 1 new unit of length $s^{-1} \times$ 500 s = 500 new units of length 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 ?
Solution :
The most precise device is that which has the minimum value of least count
(a) Given 20 VSD = 19 MSD
1 VSD = 19/20 MSD
Least count of vernier calliper = 1MSD - 1 VSD
= 1MSD- $\frac{19}{20}$MSD
=$\frac{1}{20}$mm
= 0.05 mm
= 0.005 cm
(b) Least count of screw gauge = $\frac{Pitch \ of \ screw}{Total \ no \ of \ divisions \ on \ circular \ scale}$
=$ \frac{1}{100}mm$
= $0.001 cm$
(c) Wavelength of light ,
$\lambda = 10^{-5}cm$
$ \lambda = 0.00001cm$ 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 ?
Solution :
$$Magnification = \frac{Observed \ width }{True \ width}$$
$$100 = \frac{3.5 mm}{True \ width}$$
$$True\ width ( Thickness) = \frac{3.5mm}{100}$$
$$Thickness= 0.035mm$$
1.8 Answer the following : (a)You are given a thread and a metre scale. How will you estimate the diameter of the thread ?
Solution :
(a) The diameter ( thickness ) of a Thread is so small that it cannot be measured using a metre scale. We wind a number of turns of the thread on the meter scale so that the turns are closely touching one another.
Measure the length (I) of the winding on the scale which contains n number of turns :. Diameter of thread = $\frac{I}{n}$
(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 ?
Solution:
Instrument- Screw gauge
L C = $\frac{pitch}{no \ of \ divisions \ on \ circular \ scales}$
∴ theoretically speaking, least count decreases on increasing the number of divisions on the circular scale.
Hence, accuracy would increase. Practically, it may not be possible to take the reading precisely due to low resolution of human eye.
(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 ?
Solution :
A large number of observations (say, 100 ) will give more reliable result than smaller number of observations (say, 5 ).
This is because larger the number of readings, closer is the arithmetic mean to the true value and hence smaller the random error.
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.
Solution :
Areal magnification= $\frac{size \ of \ image}{size \ of \ object }$
Areal magnification =$\frac{1.55 m^{2}}{1.75 \times 10^{-4}m^{2}}$
Areal magnification= 8857.14
Linear magnification =$\sqrt{Areal \ magnification}$
= $\sqrt{8857.14}$
Linear magnification = 94.1
1.10 State the number of significant figures in the following : (a) $0.007 m^{3}$ (b) $2.64 \times 10^{24} kg$ (c) $0.2370 g cm^{–3}$ (d) $6.320 J$ (e) $6.032 N m^{–2}$ (f) $0.0006032 m^{2}$
Solution :
(a) 1
(b) 3
(C) 4
(d) 4
(e) 4
(f) 4
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.
Solution:
Length l = 4.234 m ,
Breath b = 1.005 m ,
Thickness, t = 2.01 cm = $2.01 \times 10^{-2}m$
Area of sheet = 2(l×b+b×t+t×l)
A= 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)
A= 2(4.36047)
A=8.72094 $m^{2}$
Since the least precise term has three significant figures, the answer must be expressed to three significant figures.
After rounding off ,
area of sheet = $8.72 m^{2}$
$Volume = 4.234 \times 1.005 \times 2.01 \times 10^{-2}$
$Volume= 0.085528917 m^{3}$
Since the least precise term 2.01 cm has 3 significant figures, after rounding off , volume of sheet = $0.0855 m^{3}$ 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 ?
Solution :
Mass of box , m = 2.300 kg (4 significant figures)
mass of one gold piece , $m_{1}= 20.15g=0.02015kg$ ( 4 S.F)
Mass of other gold piece , $m_{2}$ = 20.17 g = 0.02017kg ( 4 S.F)
(a) $\therefore Total \ mass = m + m_{1}+m_{2}$
= 2.300 + 0.02015 + 0.02017
= 2.34032 kg
Since the least number of decimal places is 1, the total mass = 2.7 kg.
(b)The mass difference of pieces = 20.17 – 20.15 = 0.02 g Since the least number of decimal places is 2, the total mass difference= 0.02 g.
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 : $m = \frac{m_{0}}{(1-v^{2})^{1/2}}$
Guess where to put the missing c.
Solution:
The dimensions of L.H.S of the expression are $ML^{0}T^{0}$. The dimensions of R.H.S must be the same $ML^{0}T^{0}$ ( dimension of mass ). This is only possible if we have $v^{2}/c^{2}$ in the above relation .
Therefore, the correct formula is
$$m = \frac{m_{0}}{\sqrt{1-v^{2}/c^{2}}}$$
1.14 The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = $10^{–10}m$. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in $m^{3}$ of a mole of hydrogen atoms ?
Solution :
The radius of hydrogen atom (Atomic radius)
$r= 0.5 A^{0}$
$r= 0.5 \times 10^{-10}m$
Volume of one hydrogen atom =
$ V_{H} = \frac{4}{3} \pi r^{3}$
$V_{H}=\frac{4}{3} \times \frac{22}{7} \times (0.5 \times 10^{-10})^{3}$
$V_{H}=0.524 \times 10^{-30}m^{3}$
1 mole of hydrogen contains $6.023 \times 10^{23}$ hydrogen atoms.
$\therefore$ Volume of 1 mole of hydrogen atoms
$Atomic \ volume = \frac{4}{3} \pi r^{3} N_{A} $
$V_{a}= \frac{4}{3} \times \frac{22}{7}(0.5 \times 10^{-10})^{3} \times 6.023 \times 10^{23}$
$V_{a} = 0.524 \times 10^{-30}m^{3} \times 6.023 \times 10^{23}$
$V_{a}=3.154 \times 10^{-7}m^{3}$
1.15 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 ?
Solution:
Radius of molecule of hydrogen atom (R) =1 Å)
The radius of hydrogen atom
$r= \frac{R}{2}A^{0}$
$r= \frac{1}{2}A^{0} = 0.5 A^{0}$
$r= 0.5 \times 10^{-10}m$
Volume of hydrogen atom =
$ V_{H} = \frac{4}{3} \pi r^{3}$
$V_{H}=\frac{4}{3} \times \frac{22}{7} \times (0.5 \times 10^{-10})^{3}$
$V_{H}=0.524 \times 10^{-30}m^{3}$
Now , 1 mole of hydrogen contains $6.023 \times 10^{23}$ hydrogen atoms.
$\therefore$ Volume of 1 mole of hydrogen atoms
$Atomic \ volume = \frac{4}{3} \pi r^{3} N_{A} $
$V_{a}= \frac{4}{3} \times \frac{22}{7}(0.5 \times 10^{-10})^{3} \times 6.023 \times 10^{23}$
$V_{a} = 0.524 \times 10^{-30}m^{3} \times 6.023 \times 10^{23}$
$V_{a}=3.154 \times 10^{-7}m^{3}$
Molar volume of 1 mole of hydrogen atoms at STP,
$ V_{m}=22.4 litres = 22.4 \times 10^{-3}m^{3}$
$\frac{Molar \ volume}{Atomic \ volume } = \frac{22.4 \times 10^{-3}}{3.154 \times 10^{-7}} = 7.1 \times 10^{4}$
Hence the molar volume is $7.1 \times 10^{4}$ times higher than the atomic volume.
The ratio is so large because inter atomic separation in hydrogen gas is large.
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).
Solution :
a person sitting in a moving train find the object shifting in the opposite direction due to relative velocity. this relative shift depends upon the distance of the object from the person. if this distance is small, the object appears to be shifting fast in the opposite direction when we look at the distant object, the relative shift is negligible so that it appears to move in the direction of the train due to parallax.
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 Sun=$7.0 \times 10^{8}m$.
Solution :
Mass of Sun ,
$M = 2 \times 10^{30}kg$
Radius of Sun ,
$R = 7 \times 10^{8}m$
Volume of sun ,
$V = \frac{4}{3} \pi R^{3}$
$V = \frac{4}{3} \times \pi (7 \times 10^{8})^{3}$
$V=1.437\times 10^{27}m^{3}$
Density of sun ,
$\rho = \frac{M}{V}$
$\rho = \frac{2 \times 10^{30}}{1.437 \times 10^{27}}$
$\rho = 1.39 \times 10^{3}kgm^{-3}$
$\rho = 1390 kgm^{-3}$
The density of a sun is in the range of densities of solid( steel density : $7.8 \times 10^{3} \frac{kg}{m^{3}}$),and liquid(water density : $10^{3} \frac{kg}{m^{3}}$). The density of gas is very low : (density of helium $0.166\frac{kg}{m^{3}}$.)
The high density of sun is high because of the gravitational force of attraction between the outer and inner layer of the sun.