NCERT QUESTIONS
1.1 Fill in the blanks.
(a) The volume of a cube of side 1 cm is equal to …..m3
(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:
(a) Volume of cube, V = (1 cm)3 = (10-2 m)3 = 10-6 m3
(b) Surface area = curved area + area on top/base = 2πrh + 2πr2 = 2πr (h + r)
r = 2 cm = 20 mm
h = 10 cm = 100 mm
Surface area = 2πr (h + r) = 2 x 3.14 x 20 (100 + 20) = 15072 mm2
Hence, the answer is 15072 mm2
(c) Speed of vehicle = 18 km/h
1 km = 1000 m
1 hr = 60 x 60 = 3600 s
1 km/hr = 1000 m/3600 s = 5/18 m/s
18 km/h = = (18 x 1000)/3600
= 5 m/s
Distance travelled by the vehicle in 1 s = 5 m
(d) The relative density of lead is 11.3 g cm-3
=> 11.3 x 103 kg m-3 [1 kilogram = 103g, 1 meter = 102 cm]
=> 11.3 x 103 kg m-4
1.2 Fill in the blanks by suitable conversion of units.
(a) 1 kg m2 s–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 = …. (cm)3s–2 g–1
Answer:
(a) 1 kg m2 s–2 = ….g cm2 s–2
1 kg m2 s-2 = 1kg x 1m2 x 1s -2
We know that,
1kg = 103
1m = 100cm = 102cm
When the values are put together, we get
1kg x 1m2 x 1s-2 = 103g x (102cm)2 x 1s-2 = 103g x 104 cm2 x 1s-2 = 107 gcm2s-2
=>1kg m2 s-2 = 107 gcm2s-2
(b) 1 m = ….. ly
Using the formula,
Distance = speed x time
Speed of light = 3 x 108 m/s
Time = 1 yr = 365 days = 365 x 24 hr = 365 x 24 x 60 x 60 sec
Put these values in the formula mentioned above, and we get
One light year distance = (3 x 108 m/s) x (365 x 24 x 60 x 60) = 9.46×1015m
9.46 x 1015 m = 1ly
So that, 1m = 1/9.46 x 1015ly
=> 1.06 x 10-16ly
=>1 meter = 1.06 x 10-16ly
(c) 3.0 m s–2 = …. km h–2
1 km = 1000m so that 1m = 1/1000 km
3.0 m s-2 = 3.0 (1/1000 km) (1/3600 hour) -2 = 3.0 x 10-3 km x ((1/3600)-2h-2)
= 3 x 10-3km x (3600)2 hr-2 = 3.88 x 104 km h-2
=> 3.0 m s-2 = 3.88 x 104 km h-2
(d) G = 6.67 × 10–11 N m2 (kg)–2 = …. (cm)3s–2 g–1
G = 6.67 x 10-11 N m2 (kg)-2
We know that,
1N = 1kg m s-2
1 kg = 103 g
1m = 100cm= 102 cm
Put the values together, we get
=> 6.67 x 10-11 Nm2 kg-2 = 6.67 x 10-11 x (1kg m s -2) (1m2) (1kg-2)
Solve the following, and cancelling out the units, we get
=> 6.67 x 10-11 x (1kg -1 x 1m3 x 1s-2)
Put the above values together to convert kg to g and m to cm.
=> 6.67 x 10-11 x (103g)-1 x (102 cm)3 x (1s-2)
=> 6.67 x 10-8 cm3 s-2 g -1
=>G = 6.67 x 10-11 Nm2(kg)-2 = 6.67 x 10-8 (cm)3 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 m2s–2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, and the unit of time is γ s. Show that a calorie has a magnitude of 4.2 α–1 β–2 γ2 in terms of the new units.
Solution :
Given that,
1 Calorie=4.2 J = 4.2 Kg m2 s-2 …… (i)
As new unit of mass = α Kg
∴ 1 Kg = 1/α new unit of mass
Similarly, 1 m = β-1 new unit of length
1 s = γ-1 new unit of time
Putting these values in (i), we get
1 calorie = 4.2 (α-1 new unit of mass) (β-1 new unit of length)2 (γ-1 new unit of time)-2
= 4.2 α-1 β-2 γ2 new unit of energy (Proved)
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.
(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) In comparison with a soccer ball, atoms are very small
(b) When compared with a bicycle, a jet plane travels at high speed.
(c) When compared with the mass of a cricket ball, the mass of Jupiter is very large.
(d) As compared with the air inside a lunch box, the air inside the room has a large number of molecules.
(e) A proton is massive when compared with an electron.
(f) Like comparing the speed of a bicycle and a jet plane, the speed of light is more than the speed of sound.
1.5: A new unit of length is chosen such that the speed of light in a 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:
Distance between them = Speed of light x Time taken by light to cover the distance
Speed of light = 1 unit
Time taken = 8 x 60 + 20 = 480 + 20 = 500s
The distance between Sun and Earth = 1 x 500 = 500 units
1.6: Which of the following is the most precise device for measuring length?
(a) a vernier calliper 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 :
a. Least count of this vernier callipers = 1SD – 1 VD = 1 SD – 19/20 SD = 1/20 SD
= 1.20 mm = 1/200 cm = 0.005 cm
b. Least count of screw gauge = Pitch/Number of divisions = 1/1000 = 0.001 cm.
c. Wavelength of light, λ ≈ 10-5 cm = 0.00001 cm
Hence, it can be inferred that an optical instrument is the most suitable device to measure length.
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 the hair?
Solution:
Magnification of the microscope = 100
The average width of the hair in the field of view of the microscope = 3.5 mm
Actual thickness of hair =3.5 mm/100 = 0.035 mm
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 arbitrarily increase the accuracy of the screw gauge 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?
Solution:
(a) The thread should be wrapped around a pencil a number of times to form a coil with its turns touching each other closely. Measure the length of this coil with a metre scale. If L be the length of the coil and n be the number of turns of the coil, then the diameter of the thread is given by the relation
Diameter = L/n.
(b) Least count of the screw gauge = Pitch/number of divisions on the circular scale
So, theoretically, when the number of divisions on the circular scale is increased, the least count of the screw gauge will decrease. Hence, the accuracy of the screw gauge will increase. However, this is only a theoretical idea. Practically, there will be many other difficulties when the number of turns is increased.
(c) The probability of making random errors can be reduced to a greater extent in 100 observations than in the case of 5 observations.
1.9: The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected onto a screen, and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement?
Solution:
Arial Magnification = Area of the image/Area of the object
= 1.55/1.75 x 104
= 8.857x 103
Linear Magnification = √Arial magnification
= √8.857x 103
= 94. 1
1.10: State the number of significant figures in the following:
(a) 0.007 m2
(b) 2.64 × 1024 kg
(c) 0.2370 g cm–3
(d) 6.320 J
(e) 6.032 N m–2
(f) 0.0006032 m2
Solution:
(a) 0.007 m2
The given value is 0.007 m2.
Only one significant digit. It is 7.
(b) 2.64 × 1024 kg
Solution:
The value is 2.64 × 1024 kg
For the determination of significant values, the power of 10 is irrelevant. The digits 2, 6, and 4 are significant figures. The number of significant digits is 3.
(c) 0.2370 g cm–3
Solution:
The value is 0.2370 g cm–3
For the given value with decimals, all the numbers 2, 3, 7, and 0 are significant. The 0 before the decimal point is not significant
(d) All the numbers are significant. The number of significant figures here is 4.
(e) 6, 0, 3, and 2 are significant figures. Therefore, the number of significant figures is 4.
(f) 6, 0, 3, and 2 are significant figures. The number of significant figures is 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:
Area of the rectangular sheet = length x breadth
= 4.234 x 1.005 = 4.255 m2= 4.3 m2
The volume of the rectangular sheet = length x breadth x thickness = 4.234 x 1.005 x 2.01 x 10-2 = 8.55 x 10-2 m3.
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:
The mass of the box = 2.30 kg
and the mass of the first gold piece = 20.15 g
The mass of the second gold piece = 20.17 g
The total mass = 2.300 + 0.2015 + 0.2017 = 2.7032 kg
Since 1 is the least number of decimal places, the total mass = 2.7 kg.
The mass difference = 20.17 – 20.15 = 0.02 g
Since 2 is the least number of decimal places, the total mass = 0.02 g.
1.13: A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 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 = m0 / (1-v2)1/2 Guess where to put the missing c.
Solution:
Given the relation,
m = m0 / (1-v2)1/2
Dimension of m = M1 L0 T0
Dimension of m0 = M1 L0 T0
Dimension of v = M0 L1 T–1
Dimension of v2 = M0 L2 T–2
Dimension of c = M0 L1 T–1
The given formula will be dimensionally correct only when the dimension of L.H.S is the same as that of R.H.S. This is only possible when the factor, (1-v2)1/2 is dimensionless i.e., (1 – v2) is dimensionless. This is only possible if v2 is divided by c2. Hence, the correct relation is
m = m0 / (1 – v2/c2)1/2
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 m3 of a mole of hydrogen atoms?
Solution:
Radius of hydrogen atom, r = 0.5 Å = 0.5 × 10-10 m
Volume of hydrogen atom = (4/3) π r3
= (4/3) × (22/7) × (0.5 × 10-10)3
= 0.524 × 10-30 m3
1 mole of hydrogen contains 6.023 × 1023 hydrogen atoms.
∴ Volume of 1 mole of hydrogen atoms = 6.023 × 1023 × 0.524 × 10–30
= 3.16 × 10–7 m3
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 the hydrogen molecule to be about 1 Å)? Why is this ratio so large?
Solution:
Radius of hydrogen atom, r = 0.5 Å = 0.5 × 10-10 m
Volume of hydrogen atom = (4/3) π r3
= (4/3) × (22/7) × (0.5 × 10-10)3
= 0.524 × 10-30 m3
Now, 1 mole of hydrogen contains 6.023 × 1023 hydrogen atoms.
∴ Volume of 1 mole of hydrogen atoms, Va = 6.023 × 1023 × 0.524 × 10–30
= 3.16 × 10–7 m3
Molar volume of 1 mole of hydrogen atoms at STP,
Vm = 22.4 L = 22.4 × 10–3 m3
Hence, the molar volume is 7.08 × 104 times higher than the atomic volume. For this reason, the inter-atomic separation in hydrogen gas is much larger than the size of a hydrogen atom.
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 (hilltops, 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:
An imaginary line which joins the object and the observer’s eye is called the line of sight. When we observe the nearby objects, they move fast in the opposite direction as the line of sight changes constantly, whereas the distant objects seem to be stationary as the line of sight does not change rapidly.
1.17: The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 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 × 1030 kg, radius of the Sun = 7.0 × 108 m.
Solution :
Mass of the Sun, M = 2.0 × 1030 kg
Radius of the Sun, R = 7.0 × 108 m
Volume V = 4/3πr3
The density of the Sun is in the density range of solids and liquids. This high density is attributed to the intense gravitational attraction of the inner layers on the outer layer of the Sun.