Class 11 Physics Ch-Units and Measurement Solution-NCERT Answers

Understanding physical quantities begins with the basics—Class 11 Physics Ch: Units and Measurement Solution lays the foundation for all future topics by explaining how quantities are measured, the types of units used, and the importance of accuracy and precision. In this chapter, students learn about systems of units, significant figures, dimensional analysis, and error types.

These NCERT solutions are designed to help you grasp each concept easily, solve textbook questions step-by-step, and build a strong base for advanced physics.

Class 11 Physics Ch-Units and Measurement Solution

Question 2. 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.
Answer:  

(a) Side of cube = 1 cm
Convert to metre: 1 cm = 0.01 m

Volume of cube = side × side × side
= 0.01 × 0.01 × 0.01
= 0.000001 m³
= 1 × 10⁻⁶ m³

(b) Surface area = 2πrh + 2πr² = 2πr (h + r)
= 2 x 22/7 x 2 x 10 (10 x 10 + 2 x 10) mm² 

= 1.5 x 10⁴ mm² 

Hence, answer is 1.5 x 10⁴.

(c) Convert to m/s:
18 × (1000 / 3600) = 18 × (1 / 3.6) = 5 m/s

This means the vehicle covers 5 m in 1 second.

(d) Relative density of lead = 11.3
Density of water = 1 g cm⁻³ = 1000 kg m⁻³

Density of lead = Relative density × Density of water
= 11.3 × 1 g cm⁻³
= 11.3 g cm⁻³

Now convert to SI unit:
1 g cm⁻³ = 1000 kg m⁻³
So, 11.3 g cm⁻³ = 11.3 × 1000 = 11300 kg m⁻³ = 1.13 × 10⁴ kg m⁻³

Final Answer: The density of lead is 11.3 g cm⁻³ or 1.13 × 10⁴ kg m⁻³.

Question 2. 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 x 10^-11 N m² (kg)^-2 = …. (cm)³ s^-2 g^-1.
Answer:

Question 2. 3. A calorie is a unit of heat or energy and it equals about 4.2 J where 1 J = 1 kgm² s^-2. Suppose we employ a system of units in which the unit of mass equals a kg, the unit of length equals j8 m, the. unit of time is ys. Show that a calorie has a magnitude 4.2  ^{α-1 β-2 γ2} in terms of the new units.
Answer:

Question 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.

Answer:  Explanation:
A physical quantity can only be described as large or small when it is compared with some reference standard. For example, the length of 1 metre is very large compared to the size of an atom, but very small compared to the distance between two cities. Therefore, words like “large”, “small”, “fast”, or “slow” are meaningless unless a standard or frame of reference is given.

Reframed statements:

(a) Atoms are very small objects → The size of an atom is very small compared to the size of a cricket ball.

(b) A jet plane moves with great speed → A jet plane moves with much greater speed compared to an ordinary car.

(c) The mass of Jupiter is very large → The mass of Jupiter is very large compared to the mass of the Earth.

(d) The air inside this room contains a large number of molecules → The air inside this room contains a large number of molecules compared to the number of grains of sand in the same room.

(e) A proton is much more massive than an electron → The mass of a proton is about 1836 times the mass of an electron.

(f) The speed of sound is much smaller than the speed of light → The speed of sound in air (≈ 340 m/s) is much smaller than the speed of light (≈ 3 × 10⁸ m/s).

Question 2. 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: Time taken by light = 8 minutes 20 seconds
= (8 × 60) + 20
= 480 + 20
= 500 seconds

In this new system, speed of light = 1 unit per second

Distance = speed × time
= 1 × 500
= 500 units

Question 2.6
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:

• Vernier callipers:
  Least count = 1 main scale division / total vernier divisions
  = 1 mm / 20 = 0.05 mm = 5 × 10⁻⁵ m
• Screw gauge:
  Least count = pitch / number of circular scale divisions
  = 1 mm / 100 = 0.01 mm = 1 × 10⁻⁵ m
• Optical instrument:
  Can measure up to wavelength of light ≈ 5 × 10⁻⁷ m

Comparison:

• Vernier callipers → 5 × 10⁻⁵ m
• Screw gauge → 1 × 10⁻⁵ m
• Optical instrument → 5 × 10⁻⁷ m

Since 5 × 10⁻⁷ m is the smallest, the optical instrument is the most precise.

Final Answer: The most precise device is the optical instrument that can measure length to within a wavelength of light.

Question 2.7. 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: Magnified size of hair = 3.5 mm

Magnification of microscope = 100

Actual thickness of hair = magnified size ÷ magnification
= 3.5 mm ÷ 100
= 0.035 mm

Convert to metre:
0.035 mm = 0.035 × 10⁻³ m = 3.5 × 10⁻⁵ m

Final Answer: The estimated thickness of the human hair is 0.035 mm or 3.5 × 10⁻⁵ m.

Question 2. 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) You are given a thread and a metre scale. How will you estimate the diameter of the thread?
Procedure:

1. Fold the thread or place several (say n = 20 or 50) adjacent turns/strands side by side so that their combined thickness is easily measurable.
2. Hold the n strands together carefully and place them along the metre scale. Mark or read the length L occupied by these n strands.
3. Diameter of the thread ≈ L / n.
   Comments: use as large an n as practicable (so L is large compared with the least division of the metre scale) to reduce relative error. Repeat the measurement a few times and take the average.

(b) A screw gauge has pitch 1.0 mm and 200 divisions on the circular scale. Can accuracy be increased arbitrarily by increasing circular divisions?
No. The least count (theoretical resolution) = pitch / number of circular divisions = 1.0 mm / 200 = 0.005 mm. While increasing the number of divisions reduces the least count, accuracy cannot be increased arbitrarily because of practical limits:
• Mechanical imperfections — nonuniform pitch, play/backlash in threads, wear.
• Zero error and calibration errors.
• Difficulty in reading very fine divisions (human reading error).
• Thermal expansion of instrument and object.
• Surface roughness of the object and imperfect contact.
Thus making more divisions helps resolution up to a point, but true accuracy is limited by instrument construction, calibration, human error and environmental effects.

(c) Why is 100 measurements more reliable than 5 for mean diameter measured by vernier callipers?
• Random errors average out: the standard error of the mean decreases as 1/√n. So increasing n from 5 to 100 reduces the statistical uncertainty by a factor √(100/5)=√20 ≈ 4.47.
• More measurements make the mean less sensitive to occasional outliers or random fluctuations and let you detect systematic biases or inconsistent readings.
• With many readings you can compute standard deviation and estimate uncertainty reliably; with only 5 readings the uncertainty estimate is poor.
Note: increasing number of readings reduces random uncertainty but does not remove a systematic error (e.g., zero error); you must still check and correct systematic errors.

Question 2. 9. The photograph of a house occupies an area of 1.75 cm² 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 m². What is the linear magnification of the projector-screen arrangement?
Answer: Here area of the house on slide = 1.75 cm² = 1.75 x 10^-4 m² and area of the house of projector-screen = 1.55 m²
.•. Areal magnification =Area on screen/Area on slide = 1.55 m² / 1.75 x 10^-4 m² = 8.857 x 10³
.•. Linear magnification

Question 2. 10. State the number of significant figures in the following:
(a) 0.007 m² (b) 2.64 x 10⁴ kg
(c) 0.2370 g cm^-3 (d) 6.320 J
(e) 6.032 N m^-2 (f) 0.0006032 m²
Answer: (a) 1 (b) 3 (c) 4 (d) 4 (e) 4 (f) 4.

Question 2 .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:  Step 1: Convert all quantities to the same unit (metres).
Length = 4.234 m (4 significant figures)
Breadth = 1.005 m (4 significant figures)
Thickness = 2.01 cm = 0.0201 m (3 significant figures)

Step 2: Calculate area of sheet (length × breadth).
Area = 4.234 × 1.005
= 4.25517 m²

Both have 4 significant figures → result must have 4 significant figures.
Area = 4.255 m²

Step 3: Calculate volume (length × breadth × thickness).
Volume = 4.234 × 1.005 × 0.0201
= 0.085598 m³

Lowest significant figures among values = 3 (thickness has 3 s.f.).
So volume should be written with 3 significant figures.
Volume = 0.0856 m³

Question 2. 12. The mass of a box measured by a grocer’s balance is 2.3 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: Step 1: Convert all quantities to same unit (kg).
Box mass = 2.3 kg (2 significant figures)
First gold piece = 20.15 g = 0.02015 kg (4 significant figures)
Second gold piece = 20.17 g = 0.02017 kg (4 significant figures)

Step 2: Total mass.
Total mass = 2.3 + 0.02015 + 0.02017
= 2.34032 kg

Now, rule for addition: result is limited by least precise decimal place.

• 2.3 kg has 1 decimal place.
  So, final result must have 1 decimal place.
  Total mass = 2.3 kg

Step 3: Difference in masses of gold pieces.
Difference = 20.17 g – 20.15 g = 0.02 g

Rule: in subtraction, result is limited by least precise decimal place.
Both have 2 decimal places → answer should also have 2 decimal places.
Difference = 0.02 g

Question 2. 13. A physical quantity P is related to four observables a, b, c and d as follows:

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?
Answer:

As the error lies in first decimal place, the answer should be rounded off to first decimal place. Hence, we shall express the value of P after rounding it off as P = 3.8.

Question 2. 14. A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion:

(a = maximum displacement of the particle, v = speed of the particle, T = time-period of motion)Rule out the wrong formulas on dimensional grounds.

Answer: Given: a = maximum displacement → [a] = L
v = speed → [v] = L T⁻¹
T = time period → [T] = T
t = time → [t] = T

(a) y = a sin(2π t / T)
Argument of sin: 2π t / T → 2π is dimensionless and t/T has dimension T/T = 1 (dimensionless).
So sin(2π t / T) is dimensionless. Prefactor a has dimension L.
Therefore y has dimension L — correct on dimensional grounds.
Conclusion: formula (a) is dimensionally correct.

(b) y = a sin(v t)
Argument of sin: v t → [v][t] = (L T⁻¹)(T) = L (length).
The sine function requires a dimensionless argument, but vt has dimension L, not dimensionless.
Thus the formula is dimensionally inconsistent and therefore wrong.
Comment / possible correction: if the symbol v were intended to be an angular frequency ω (with [ω] = T⁻¹), then y = a sin(ω t) would be correct, where ω = 2π/T. Alternatively, if v truly means linear speed, the argument must be made dimensionless (for example vt/λ where λ is a length), but as written it is wrong.

(c) y = (a / T) sin(t / a)
Prefactor: a/T → [a/T] = L / T (velocity), not a length.
Argument: t / a → [t/a] = T / L, which is not dimensionless.
So (i) the argument of sine is not dimensionless (invalid), and (ii) even if the sine were dimensionless, the prefactor would give y dimension L/T, not L.
Therefore the formula is doubly wrong on dimensional grounds.
Comment / likely misprint and correction: a sensible form would be y = a sin(2π t / T) (prefactor a and argument 2π t/T dimensionless). As written, (c) is wrong.

(d) y = (a/√2) [ sin(2π t / T) + cos(2π t / T) ]
Each trig argument 2π t / T is dimensionless. Prefactor a/√2 has dimension L. Sum of two dimensionless trig functions is dimensionless, so whole RHS has dimension L.
Thus the formula is dimensionally correct. (Extra remark: amplitude of this combination = (a/√2)·√(1²+1²) = a, so the maximum displacement is indeed a.)

Question 2. 15. 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:

Guess where to put the missing c.
Answer:

Question 2. 16. The unit of length convenient on the atomic scale is known as an angstrom and is denoted by A: 1 A = 10^-10 m. The size of a hydrogen atom is about 0.5 A. What is the total atomic volume in m³ of a mole of hydrogen atoms?
Answer: Volume of one hydrogen atom = 4/3 πr3 (volume of sphere)
= 4/3 x 3.14 x (0.5 x 10^-10) m³ = 5.23 x 10^-31 m³
According to Avagadro’s hypothesis, one mole of hydrogen contains 6.023 x 10²³ atoms.
Atomic volume of 1 mole of hydrogen atoms
= 6.023 x 1023 x 5.23 x 10^-31 = 3.15 x 10^-7m³.

Question 2. 17. 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 A.) Why is this ratio so large?
Answer:  Volume of one mole of ideal gas, V_g
= 22.4 litre = 22.4 x 10^-3 m³
Radius of hydrogen molecule = 1A/2
= 0.5 A = 0.5 x 10^-10 m
Volume of hydrogen molecule = 4/3 πr³
=4/3 x 22/7 (0.5 x 10^-10)³ m³
= 0.5238 x 10^-30 m³
One mole contains 6.023 x 10²³ molecules.
Volume of one mole of hydrogen, VH
= 0.5238 x 10^-30 x 6.023 x 10²³ m³ = 3.1548 x 10^-7 m³
Now V_g/V_H=22.4 x 10^-3/3.1548 x 10^-7 =7.1 x 10⁴
The ratio is very large. This is because the interatomic separation in the gas is very large compared to the size of a hydrogen molecule.

Question 2. 18. 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: The line joining a given object to our eye is known as the line of sight. When a train moves rapidly, the line of sight of a passenger sitting in the train for nearby trees changes its direction rapidly. As a result, the nearby trees and other objects appear to run in a direction opposite to the train’s motion. However, the line of sight of distant and large size objects e.g., hill tops, the Moon, the stars etc., almost remains unchanged (or changes by an extremely small angle). As a result, the distant object seems to be stationary.

Question 2. 19. The principle of ‘parallax’ is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit =3 x 10 n m. However, even the nearest stars are so distant that with such a long baseline, they show parallel only of the order of 1″ (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1″ (second) of arc from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres?
Answer:  From parallax method we can say
θ=b/D,where b=baseline ,D = distance of distant object or star
Since, θ=1″ (s) and b=3 x 10¹¹ m
D=b/20=3 x 10¹¹/2 x 4.85 x 10^-6 m
or D=3 x 10¹¹/9.7 x 10^-6 m =30 x 10¹⁶/9.7 m
= 3.09 x 10¹⁶ m = 3 x 10¹⁶ m.

Question 2. 20. The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?
Answer:  As we know, 1 light year = 9.46 x 10¹⁵ m
.•. 4.29 light years = 4.29 x 9.46 x 10¹⁵ = 4.058 x 10¹⁶ m
Also, 1 parsec = 3.08 x 10¹⁶ m
.•. 4.29 light years =4.508 x 10¹⁶/3.80 x 10¹⁶ = 1.318 parsec = 1.32 parsec.
As a parsec distance subtends a parallax angle of 1″ for a basis of radius of Earth’s orbit around the Sun (r).In present problem base is the distance between two locations of the Earth six months apart in its orbit around the Sun = diameter of Earth’s orbit (b = 2r).
.•. Parallax angle subtended by 1 parsec distance at this basis = 2 second (by definition of parsec).
.•. Parallax angle subtended by the star Alpha Centauri at the given basis θ = 1.32 x 2 = 2.64″.

Question 2. 21. Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modem science where precise measurements of length, time, mass etc., are needed. Also, wherever you can, give a quantitative idea of the precision needed.
Answer:  Extremely precise measurements are needed in modem science. As an example, while launching a satellite using a space launch rocket system we must measure time to a precision of 1 micro second. Again working with lasers we require length measurements to an angstrom unit (1 A° = 10^-10m) or even a fraction of it. For estimating nuclear sizes we require a precision of 10^-15 m. To measure atomic masses using mass spectrograph we require a precision of 10^-30kg and so on.

Question 2.22. Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):
(a) the total mass of rain-bearing clouds over India during the Monsoon
(b) the mass of an elephant
(c) the wind speed during a storm
(d) the number of strands of hair on your head
(e) the number of air molecules in your classroom.
Answer: (a) The average rainfall of nearly 100 cm or 1 m is recorded by meteorologists, during Monsoon, in India. If A is the area of the country, then A = 3.3 million sq. km = 3.3 x 10⁶ (km)2 = 3.3 x 10⁶ x 10⁶m²= 3.3 x 10¹² m²
Mass of rain-bearing clouds
= area x height x density = 3.3 x 10¹² x 1 x 1000 kg = 3.3 x 10¹⁵ kg.
(b) Measure the depth of an empty boat in water. Let it be d1. If A be the base area of the boat, then volume of water displaced by boat, V1 = Ad2
Let d2 be the depth of boat in water when the elephant is moved into the boat. Volume of water displaced by (boat + elephant), V2 = Ad2 Volume of water displaced by elephant,
V = V2-V1 = A(d2 -d1)
If p be the density of water, then mass of elephant = mass of water displaced by it = A(d2 – d1) p.
(c) Wind speed can be estimated by floating a gas-filled balloon in air at a known height h. When there is no wind, the balloon is at A. Suppose the wind starts blowing to the right such that the balloon drifts to position B in 1 second.
Now, AB = d = hθ.

The value of d directly gives the wind speed.
(d) Let us assume that the man is not partially bald. Let us further assume that the hair on the head are uniformly distributed. We can estimate the area of the head. The thickness of a hair can be measured by using a screw gauge. The number of hair on the head is clearly the ratio of the area of head to the cross-sectional area of a hair.
Assume that the human head is a circle of radius 0.08 m i.e., 8 cm. Let us further assume that the thickness of a human air is 5 x 10^-5m.
Number of hair on the head
=Area of the head/Area of cross – section of a hair
=π (0.08)2/π(5 x 10^-5)=64 x 10-4/25 x 10^-10=2.56 x 10⁶
The number of hair on the human head is of the order of one million.
(e) We can determine the volume of the class-room by measuring its length, breadth and height. Consider a class room of size 10 m x 8 m x 4 m. Volume of this room is 320 m³. We know that 22.4l or 22.4 x 10^-3 m³ of air has 6.02 x 10²³ molecules (equal to Avogadro’s number).
Number of molecules of air in the class room
=(6.02 x 10²³ /22.4 x 10^-3 ) x 320 =8.6 x 10²⁷

Question 2. 23. 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 x 10³⁰ kg, radius of the Sun = 7.0 x 10⁸ m.
Answer: Given M = 2 x 10³⁰ kg, r = 7 x 10⁸ m
.-. Volume of Sun = 4/3πr³ x 3.14 x (7 x 10⁸)³  = 1.437 x 10²⁷ m³
As p = M/V,                .’. p = 2 x 10³⁰/1.437 x 10²⁷= 1391.8 kg m^-3 = 1.4 x 10³ kg m^-3
Mass density of Sun is in the range of mass densities of solids/liquids and not gases.

Question 2. 24. When the planet Jupiter is at a distance of 824.7 million kilometres from the Earth, its angular diameter is measured to be 35.72″ of arc. Calculate the diameter of Jupiter.
Answer: Given angular diameter θ = 35.72= 35.72 x 4.85 x 10^-6 rad
= 173.242 x 10^-6 = 1.73 x 10^-4 rad
Diameter of Jupiter D = θ x d = 1.73 x 10^-4 x 824.7 x 10⁹ m
=1426.731 x 10³ = 1.43 x 10⁸ m

Question 2. 25. A man walking briskly in rain with speed v must slant his umbrella forward making an angle θ with the vertical. A student derives the following relation between θ and v: tanθ = v and checks that the relation has a correct limit: as v—>θ, θ —>0, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct? If not, guess the correct relation.
Answer: According to principle of homogenity of dimensional equations,
Dimensions of L.H.S. = Dimensions of R.H.S.
Here, v = tan θ
i. e., [L¹ T^-1] = dimensionless, which is incorrect.
Correcting the L.H.S., we. get
v/u= tan θ, where u is velocity of rain.

Question 2. 26. It is claimed that two cesium clocks, if allowed to run for 180 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s?
Answer:  Total time = 100 years = 100 x 365 x 24 x 60 x 60 s
Error in 1 second=0.02/100 x 365 x 24 x 60 x 60
=6.34 x 10^-12 s
.•. Accuracy of 1 part in 10¹¹ to 10¹².

Question 2. 27. Estimate the average mass density of a sodium atom assuming its size to be about 2.5 A. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase: 970 kg m^3-. Are the two densities of the same order of magnitude? If so, why?
Answer: It is given that radius of sodium atom, R = 2.5 A = 2.5 x 10^-10 m
Volume of one mole atom of sodium, V = NA .4/3 π R³
V = 6.023 x 10²³ x –4/3 x 3.14 x (2.5 x 10^-10)³ m³ and mass of one mole atom of sodium, M = 23 g = 23 x 10^-3 kg
.•. Average mass density of sodium atom, p = M/V
=(23 x 10^-3/6.023 x 10²³ x 4/3 x 3.14 x (2.5 x 10^-10))
= 6.96 x 10² kg m^-3 = 0.7 x 10^-3 kg m^-3
The density of sodium in its crystalline phase = 970 kg m^-3
= 0.97 x 10³ kg m^-3
Obviously the two densities are of the same order of magnitude (= 10³ kg m^-3). It is on account of the fact that in solid phase atoms are tightly packed and so the atomic mass density is close to the mass density of solid.

Question 2. 28. The unit of length convenient on the nuclear scale is afermi: I f=10^-15 m. Nuclear sizes obey roughly the following empirical relation:
r = r0 A^1/3
where r is the radius of the nucleus, A its mass number, and r0 is a constant equal to about,1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise 2.27.
Answer:  Assume that the nucleus is spherical. Volume of nucleus
= 4/3 πr³ = 4/3 π [r₀ A^1/3]³ = 4/3 πr₀³A
Mass of nucleus = A
.•. Nuclear mass density = Mass of nucleus/Volume of nucleus
= A/(4/3πr₀³A) = 3/4πr₀³
Since r0 is a constant therefore the right hand side is a constant. So, the nuclear mass density is independent of mass number. Thus, nuclear mass density is constant for different nuclei.
For sodium, A = 23
.’. radius of sodium nucleus,
r = 1.2 x 10^-15 (23)^1/3 m = 1.2 x 2.844 x 10^-15 m =3.4128 x 10^-15

So, the nuclear mass density is nearly 50 million times more than the atomic mass density for a sodium atom.

Question 2. 29. A LASER is source of very intense, monochromatic, and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth?
Answer: We known that speed of laser light = c = 3 x 10⁸ m/s. If d be the distance of Moon from the earth, the time taken by laser signal to return after reflection at the Moon’s surface

Question 2. 30. A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water = 1450 m s^-1).
Answer:  Here speed of sound in water v = 1450 m s^-1 and time of echo t = 77.0 s.
If distance of enemy submarine be d, then t = 2d/v
.’. d=vt/2 =1450 x 77.0/2 =55825 m=55.8 x 10³ m or 55.8 km.

Question 2. 31. The farthest objects in our Universe discovered by modem astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?
Answer: The time taken by light from the quasar to the observer
t = 3.0 billion years = 3.0 x 10⁹ years As 1 ly = 9.46 x 10¹⁵ m
.•. Distance of quasar from the observer d = 3.0 x 10⁹  x 9.46 x 10¹⁵  m
= 28.38 x 10²⁴ m = 2.8 x 10²⁵ m or 2.8 x 10²² km.

Question 2. 32. It is a well known fact that during a total solar eclipse the disk of the Moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the Moon.
Answer: From examples 2.3 and 2.4, we get θ = 1920″ and S = 3.8452 x 10⁸ m. During the total solar eclipse, the disc of the moon completely covers the disc of the sun, so the angular diameter of both the sun and the moon must be equal. Angular diameter of the moon, θ= Angular diameter of the sun
= 1920″ = 1920 x 4.85 x 10^-6 rad [1″ = 4.85 x 10^-6 rad]
The earth-moon distance, S = 3.8452 x 10⁸ m .’. The diameter of the moon, D = θ x S
= 1920 x 4.85 x 10^-6 x 3.8452 x 10⁸ m = 35806.5024 x 10² m = 3581 x 10³ m 3581 km.

Question 2. 33. A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (-15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?
Answer: The values of different fundamental constants are given below:

We have to try to make permutations and combinations of the universal constants and see if there can be any such combination whose dimensions come out to be the dimensions of time. One such combination is:

QUESTIONS BASED ON SUPPLEMENTARY CONTENTS

Question 1. The radius of a sphere is measured as (2.1 ± 0.5) cm calculate its surface area with error limits.
Answer: Radius of the sphere = (2.1 ± 0.5) cm
.-. r = 2.1 and Ar = ± 0.5
S.A. = 4 π r2
= 4 x 3.14 x 2.1 x 2.1
=55.4 cm2

Question 2. The voltage across a lamp is (6.0 ± 0.1) volt and the current passing through it is (4.0 ± 0.2) ampere. Find the power consumed by the lamp.
Answer: Power P = V x I

Question 3. The length and breadth of a rectangular block are 25.2 cm and 16.8 cm, which have both been measured to an accuracy of 0.1 cm. Find the area of the rectangular block.
Answer:  Here l = (25.2 ± 0.1) cm
b = (16.8 ±0.1) cm
Area= l x b
= 25.2 x 16.8
= 423.4 cm²

Question 4. A force of (2500 ±5) N is applied over an area of (0.32 ± 0.02) m². Calculate the pressure exerted over the area.
Answer:

Question 5. To find the value of ‘g by using a simple pendulum, the following observations were made : Length of thread l = (100 ± 0.1) cm
Time period of oscillation T = (2 ± 0.1) sec
Calculate the maximum permissible error in measurement of ‘g’. Which quantity should be measured more accurately and why?
Answer:

Question 6. For a glass prism of refracting angle 60°, the minimum angle of deviation Dm is found to be 36° with a maximum error of 1.05°. When a beam of parallel light is incident on the prism, find the range of experimental value of refractive index ‘μ’. It is known that the refractive index ‘μ’ of the material of the prism is given by

Answer:

Question 7. The radius of curvature of a concave mirror, measured by a spherometer is given by R=l²/6h +h/2
The value of l and h are 4.0 cm and 0.065 cm respectively where l is measured by a metre scale and h by the spherometer. Find the relative error in the measurement of R.
Answer:  Given that l = 4 cm and Al = 0.1 cm (least count of the metre scale) here l is the distance between the legs of the spherometer.

Question 8. In Searle’s experiment, the diameter of the wire as measured by a screw guage of least count 0.001 cm, is 0.500 cm. The length, measured by a scale of least count 0.1 cm is 110.0 cm. When a weight of 40 N is suspended from the wire, its extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the Young’s modulus of the material of the wire from this data.
Answer: Young’s modulus of the material of the wire is given as

Question 9. A small error in the measurement of the quantity having the highest power (in a given formula) will contribute maximum percentage error in the value of the physical quantity to whom it is related. Explain why?
Answer:

Question 10. The two specific heat capacities of a gas are measured as Cp = (12.28 ± 0.2) units and Cv = (3.97 ± 0.3) units. Find the value of the gas constant R.
Answer:

MORE QUESTIONS SOLVED

I. Very Short Answer Type Questions
Question 1. What are the derived units?
Answer: Units of those physical quantities which are derived from the fundamental units are called derived units.

Question 2. What do you understand by fundamental physical quantities?
Answer:  Fundamental physical quantities are those quantities which are independent of each other. For example, mass, length, time, temperature, electric current, luminous intensity and amount of substance are seven fundamental physical quantities.

Question 3. Define parsec.
Answer: The distance at which a star would have annual parallax of 1 second of arc.
1 parsec = 3.08 x 10¹⁶ m

Question 4. Define Atomic mass unit (a.m.u.).
Answer: 1 a.m.u. = 1/12 th mass of carbon-12 atom, i.e., 1.66 x 10^-27 kg.

Question 5. Which is a bigger unit-light year or parsec?
Answer: Parsec is bigger unit than light year (1 parsec = 3.26 light year).

Question 6. Do A and A.U. stand for same length?
Answer:  No, 1 A = 10^-10 m
1 A.U. = 1.496 x 10¹¹ m

Question 7. Name two pairs of physical quantities whose dimensions are same.
Answer:  —> Stress and Young’s modulus.
—> Work and Energy.

Question 8. What is the order of precision of an atomic clock?
Answer:  About 1 in 10¹² to 10¹³ s.

Question 9. What does RADAR stand for?
Answer: RADAR stands for ‘Radio detection and ranging’.

Question 10. What does SONAR stand for?
Answer: SONAR stands for ‘sound navigation and ranging’.

Question 11. f= x², then what is the relative error in f?
Answer: 2Δx/x

Question 12. Name at least six physical quantities whose dimensions are ML² T^-2.
Answer: (i) Work (ii) Torque (iii) Moment of force (iv) Couple (v) Potential energy (vi) Kinetic energy.

Question 13. Name four units used in the measurement of extremely short distances.
Answer: 1 micron (1 p) = 10^-6 m
1 nano metre (1 nm) = 10^-9 m
1 angstrom (1 A) = 10^-10 m
1 fermi (1 f) = 10^-15 m.

Question 14. If x = a + bt + ct² where x is in metre and t in second, then what is the unit of e?
Answer: According to the principle of homogeneity of dimensions.
[ct²] = [L] or [c] = [LT^-2]
So, the unit of c is ms^-2.

Question 15. Do all physical quantities have dimensions? If no, name four physical quantities which are dimensionless.
Answer: No, all physical quantities do not possess dimensions. Angle, specific gravity, Poisson’s ratio and Strain are four examples of dimensionless quantities.

Question 16. Obtain the dimensions of relative density.
Answer: As relative density is defined as the ratio of the density of given substance and the density of standard distance (water), it is a dimensionless quantity.

Question 17. Obtain the dimensional formula for coefficient of viscosity.
Answer:

Question 18. Do specific heat and latent heat have the same dimensions?
Answer: No.

Question 19. Do mass and weight have the same dimensions?
Answer:  No.

Question 20. Given that the value of G in the CGS system as 6.67 x 10^-8dyne cm² g^-2 , find the value in MKS system.
Answer: 6.67 x 10^-8 dyne cm² g^-2 = 6.67 x 10¹¹ Nm² /kg² .

Question 21. Is Avogadro’s number a dimensionless quantity?
Answer: No, it has dimensions. In fact its dimensional formula is [mol^-1].

Question 22. Can a physical quantity have dimensions but still have no units?
Answer:  No, it is not possible.

Question 23.  Are all constants dimensionless?
Answer:  No, it is not true.

Question 24. What is N m^-1 s² equal to?
Answer: N m^-1 s² is nothing but SI unit of mass i.e., the kilogram.

Question 25. Express a joule in terms of fundamental units.
Ans. [Energy] = [M L² T^-2], hence 1 joule = 1 kg x 1 m² x 1 s^-2 = 1 kg m² s^-2.

Question 26. What is the dimensional formula for torque?
Answer: [M L² T^-2].

Question 27. Is nuclear mass density dependent on the mass number? (Given: r = r₀ A^1/3)
Answer: No, since density = Mass/Volume

Question 28. What does LASER stand for?
Answer: LASER stands for ‘Light Amplification by Stimulated Emission of Radiation’.

II. Short Answer Type Questions
Question 1. A body travels uniformly a distance of (13.8 ± 0.2) m in a time (4.0 ± 0.3) s. What is the velocity of the body within error limits?
Answer:

Question 2. What do you mean by order of magnitude? Explain.
Answer: The order of magnitude of a numerical quantity (N) is the nearest power of 10 to which its value can be written.For example. Order of magnitude of nuclear radius 1.5 x 10^-14 m is -14.

Question 3. A laser signal is beamed towards the planet Venus from Earth and its echo is received 8.2 minutes later. Calculate the distance of Venus from the Earth at that time.
Answer:  We know that speed of laser light, c = 3 x 10⁸m/ s Time of echo, t = 8.2 minutes = 8.2 x 60 seconds
If distance of Venus be d, then t = 2d/c
d = 1/2ct = 1/2 x 3 x 10⁸ x 8.2 x 60 m
= 7.38 x 10¹⁰ m
= 7.4 x 10¹⁰ m.

Question 4. The parallax of a heavenly body measured from two points diametrically opposite on earth’s equator is 60 second. If the radius of earth is 6.4 x 10⁶ m, determine the distance of the heavenly body from the centre of earth. Convert this distance in A.U. Given 1 A.U. = 1.5 x 10¹¹ m.
Answer:

Question 5. If the length and time period of an oscillating pendulum have errors of 1% and 2% respectively, what is the error in the estimate of g?
Answer:

Question 6. If x = at² + bt + c; where x is displacement as a function of time. Write the dimensions of a, b and c.
Answer: All the terms should have the same dimension

Question 7. The number of particles crossing per unit area perpendicular to x-axis in unit time N is given by N= -D(n₂-n₁/x₂-x₁), where n₁ and n₂ are the number of particles per unit volume at x₁ and x₂ respectively. Deduce the dimensional formula for D.
Answer:

Question 8. An experiment measured quantities a, b, c and then x is calculated by using the relation ab²x =ab²/c³. If the percentage errors in measurements of a, b and c are ± 1%, ±2% and ± 1.5% respectively, then calculate the maximum percentage error in value of x obtained.
Answer:

Question 9. If instead of mass, length and time as fundamental quantities, we choose velocity, acceleration and force as fundamental quantities and express their dimensions by V, A and F respectively, show that the dimensions of Young’s modulus can be expressed as [FA² V^-4].
Answer:  We know that the usual dimensions of Y are [MLT^–2]/[L²] i.e.,[M L^-2T^-2]
To express these in terms of F, A and V, we must express, M, L and T in terms of these new ‘fundamental’ quantities.
Now, [V] = [LT^–1], [A] = [LT^–2] and [F] = [MLT^–2]
It follow that M = FA-1, T = VA~X, L = V2 A-1
[Y] = [ML^-1T^–2]
= [FA^-1] [V² A^-1]^-1[VA^-1]^-2=FA²V^-4 Thus the ‘new’ dimensions of Young’s modulus are [FV^-4 A²]

Question 10. The density of a cylindrical rod was measured by using the formula ρ=4m/πD²l
The percentage errors in m, D and l are 1%, 1.5% and 0.5%. Calculate the percentage error in the calculated value of density.
Answer:

Question 11. The force experienced by a mass moving with a uniform speed v in a circular path of radius r experiences a force which depends on its mass, speed and radius. Prove that the relation is f=mv²/r.
Answer:

Question 12. The distance of the Sun from the Earth is 1.496 x 10¹¹ m (i.e., 1 A.U.). If the angular diameter of the Sun is 2000″, find the diameter of the Sun.
Answer: Here, θ=2000
=2000/3600 x π /180 rad
= 9.7 x 10^-3 rad d = 1.496 x 10¹¹ m
From the figure,
θ=D/d
D=θ d
= 9.7 x 10^-3 x 1.496 x 10¹¹
= 1.45 x 10⁹m

Question 13. Experiments show that the frequency (n) of a tuning fork depends upon the length (l) of the prong, the density (d) and Young’s modulus (Y) of its material. From dimensional considerations, find a possible relation for the frequency of a tuning fork.
Answer:

Question 14. Calculate focal length of a spherical mirror from the following observations: object distance u = (50.1 ± 0.5) cm and image distance v = (20.1 ± 0.2) cm.
Answer:

Question 15. The radius of the Earth is 6.37 x 10⁶ m and its mass is 5.975 x 10²⁴ kg. Find the Earth’s average density to appropriate significant figures.
Answer: Radius of the Earth (R) = 6.37 x 10⁶ m
Volume of the Earth (V) = 4/3 πR³m³ = 4/3 x (3.142) x (6.37 x 10⁶)³ m³
Average density (D)=Mass/Volume=M/V= 0.005517 x 10⁶ kg  m^-3
The density is accurate only up to three significant figures which is the accuracy of the least accurate factor, namely, the radius of the earth.

Question 16. The orbital velocity v of a satellite may depend on its mass m, distance r from the centre of Earth and acceleration due to gravity g. Obtain an expression for orbital velocity.
Answer:  Let orbital velocity of satellite be given by the relation
v = kma rb gc where k is a dimensionless constant and a, b and c are the unknown powers. Writing dimensions on two sides of equation, we have
[M° L¹ T^-1] = [M]^a [L]^b [L T^-2]^c = [M^a L^{b + c} T^-2c]
Applying principle of homogeneity of dimensional equation, we find that
a = 0 …(i)
b + c = 1 …(H)
– 2c = – 1 …(Hi)
On solving these equations, we find that
a=0,b=+(1/2) and c=+(1/2)

Question 17.  Check by the method of dimensional analysis whether the following relations are correct.

Answer:

Hence, the relation is correct.

Question 18. Given that the time period T of oscillation of a gas bubble from an explosion under water depends upon P, d and E where P is the static pressure, d the density of water and E is the total energy of explosion, find dimensionally a relation for T.
Answer: We are given that
T = f (P, d, E)
Assuming that T = k Pa db Ec and substituting dimensions of all the quantities involved, we have
[T] = [M L^-1 T^-2]^a [M L^-3]^b [M L² T^-2]^c  Equating powers of M, L and T on both sides,
we have a + b + c = 0
-a-3b + 2c = 0
-2 a-2 c = 1
Solving these equations, we get
a = -5/6 b = 1/2 c = 1/3

Question 19. The radius of curvature of a concave mirror measured by spherometer is given by R =l²/6h + h/2.The values of l and h are 4 cm and 0.065 cm respectively. Compute the error in measurement of radius of curvature.
Answer:

Question 20. The radius of the Earth is 6.37 x10⁶ m and its average density is 5.517 x 10³ kg m^-3. Calculate the mass of earth to correct significant figures.
Answer:  Mass = Volume x density
Volume of earth = 4/3π R³
= 4/3 x 3.142 x (6.37 x 10⁶)³m³
Mass of earth = — x 3.142 x (6.37 x 10⁶)³ x 5.517 x 10³ kg
= 5974.01 x 10²¹ kg = 5.97401 x 10²⁴ kg
The radius has three significant figures and the density has four. Therefore, the final result should be rounded up to three significant figures. Hence, mass of the earth = 5.97 x 10²⁴  kg.

Question 21. Find the dimensions of the following quantities
(i) Acceleration (ii) Angle (iii) Density
(iv) Kinetic energy (v) Gravitational constant (vi) Permeability
Answer:

Question 22. The length, breadth and thickness of a block of metal were measured with the help of a Vernier Callipers. The measurements are l = (5.250 ± 0.001) cm, b = (3.450 ± 0.001) cm,t = (1.740 ± 0.001) cm.Find the percentage error in volume of the block.
Answer: Volume of the block is given by
V = I b t
Relative error in the volume of block

Question 23. Find the value of 60 W on a system having 100 g, 20 cm and 1 minute as the fundamental units.
Answer: Here n₁ = 60 W. Obviously, the physical quantity is power whose dimensional formula is [M¹ L² T^3-]. The first system, in which unit of power is 1 watt, is SI system in which M₁ = 1 kg,L₁ = 1 m and T₁= ls in second system, M₂ = 100 g, L₂ = 20 cm and T₂ = 1 min = 60 s.

Question 24. By using the method of dimension, check the accuracy of the following formula: T =rhρg/2cos θ , where T is the surface tension, h is the height of the liquid in a capillary tube, p is the density of the liquid, g is the acceleration due to gravity, 6 is the angle of contact, and r is the radius of the capillary tube.
Answer:  In order to find out the accuracy of the given equation we shall compare the dimensions of T and rh ρg/2cos θ

III. Long Answer Type Questions
Question 1. P.A.M. Dirac, a great physicist of 20th century found that from the following basic constants, a number having dimensions of time can be constructed:

• (i) charge on electron (e),
• (ii) permittivity of free space (ε0),
• (iii) mass of electron (me),
• (iv) mass of proton (me)
• (v) speed of light (c),
• (vi) universal gravitational constant (G).
  

Obtain Dirac’s number, given that the desired number is proportional to mp-1 and me-2. What is the significance of this number?
Answer:

Substituting values of all known parameters we find that the value of x is nearly 15 billion years, which is approximately equal to the present estimate of the age of the universe.

Question 2. To determine acceleration due to gravity, the time of 20 oscillations of a simple pendulum of length 100 cm was observed to be 40 s. Calculate the value of g and maximum percentage error in the measured value of g.
Answer:

Here 0.1% is the error in the measurement of length, and 0.5% is the error in the measurement of time. Therefore, time needs more careful measurement.

Question 3. It is known that the period T of a magnet of magnetic moment M vibrating in a uniform magnetic field of intensity B depends upon M, B and I where I is the moment of inertia of the magnet about its axis of oscillations. Show that

Answer: We first note that the dimension of I are [ML²]. Also the magnetic moment has the units Am2 so that its dimensions can be written as [AL²] where A stands for the dimensions of the electric current. Finally the magnetic field vector B has the units newton (per ampere metre) so that its dimensions can be written as

Question 4. Briefly explain how you will estimate the molecular diameter of oleic acid.
Answer: Answer:

1. Make a very dilute solution of oleic acid by dissolving 1 milliliter in 20 milliliters of alcohol. Then take 1 milliliter of this solution and dissolve it again in 20 milliliters of alcohol. The final solution has 1/400th concentration of oleic acid.
2. Fill a trough with water and sprinkle a little lycopodium powder on the surface.
3. Put a few drops of the solution on water. The drops spread to form a thin circular film. Measure the diameter of the film to find its surface area.
4. Find the total volume of oleic acid in the drops.
5. Thickness of the film equals the total volume of oleic acid divided by the area of the film. This thickness is approximately the molecular diameter of oleic acid.

Result: The molecular diameter of oleic acid is about one nanometer (10 to the minus 9 meters).

Question 5. Obtain a relation between the distance travelled by a body in time t, if its initial velocity be u and acceleration.
Answer: Let the distance covered is S,

IV. Multiple Choice Questions
Question 1: The SI units of magnetic field is
Options:
(a) weber per metre²
(b) newton per coulomb per (metre per second)
(c) newton per ampere per metre
(d) all the above

Answer: (c) newton per ampere per metre
Explanation: Magnetic field (B) has SI unit tesla (T). 1 T = 1 N/(A·m).

Question 2: The dimensions of energy per unit volume are the same as those of
Options:
(a) pressure
(b) force
(c) modulus of elasticity
(d) all the above

Answer: (d) all the above
Explanation:

• Energy per unit volume = (energy)/(volume) = (ML²T⁻²)/(L³) = ML⁻¹T⁻².
• Pressure = Force/Area = (MLT⁻²)/(L²) = ML⁻¹T⁻²
• Modulus of elasticity has same dimension as pressure.

Question 3: The SI units of the universal gravitational constant G are
Options:
(a) kg m² s⁻²
(b) kg⁻¹ m³ s⁻²
(c) N kg² m⁻²
(d) N kg² m⁻²

Answer: (b) kg⁻¹ m³ s⁻²
Explanation: From Newton’s law of gravitation: F = G(m₁m₂/r²)

• [F] = MLT⁻², [m₁] = M, [m₂] = M, [r] = L
• So, [G] = [F][r²]/([m₁][m₂]) = (MLT⁻² L²)/M² = L³ M⁻¹ T⁻²

Question 4: The number of particles crossing per unit area perpendicular to X-axis in unit time is N = -D (n₂ – n₁)/(x₂ – x₁)
Dimensions of diffusion constant D are:
Options:
(a) M⁰ L T²
(b) M⁰ L² T⁻¹
(c) M⁰ L T³
(d) M⁰ L² T³

Answer: (b) M⁰ L² T⁻¹
Explanation:

• N has dimension [L⁻²T⁻¹] (particles per unit area per unit time)
• (n₂-n₁)/(x₂-x₁) has dimension [L⁻⁴] / [L] = L⁻⁴ / L = L⁻³? Actually n = particles per unit volume = L⁻³, so (n₂-n₁)/(x₂-x₁) = L⁻³ / L = L⁻⁴
• N = D * (n₂-n₁)/(x₂-x₁) → [D] = [N]/[gradient n] = (L⁻² T⁻¹)/(L⁻⁴) = L² T⁻¹

Question 5: A physical quantity is X = M^α L^β T^γ. If percentage errors in M, L, T are a%, b%, y%, then total percentage error is:
Options:
(a) (αa – βb + γc)%
(b) (αa + βb + γc)%
(c) (αa – βb – γc)%
(d) none of the above

Answer: (b) (αa + βb + γc)%
Explanation: For quantities raised to powers, percentage errors add linearly with powers.

Question 6: ‘Parsec’ is the unit of:
Options:
(a) Time
(b) Distance
(c) Frequency
(d) Angular acceleration

Answer: (b) Distance
Explanation: 1 parsec = distance at which 1 AU subtends angle 1 arcsecond.

Question 7: The density of a cube is measured by mass and side length. Maximum errors in mass and length = 3% and 2%. Maximum error in density = ?
Options:
(a) 9%
(b) 7%
(c) 5%
(d) 1%

Answer: (b) 7%
Explanation: Density ρ = m/L³ → Δρ/ρ = Δm/m + 3 ΔL/L = 3% + 3×2% = 3% + 6% = 9% ? Wait let’s check:

• Δρ/ρ = Δm/m + 3 (ΔL/L) = 3% + 3×2% = 3% + 6% = 9%

Correct answer: (a) 9%

Question 8: A wire has mass 0.3 ± 0.003 g, radius 0.5 ± 0.005 mm, length 6 ± 0.06 cm. Maximum percentage error in density:

• Density ρ = m / (π r² L) → %error: Δρ/ρ = Δm/m + 2 Δr/r + ΔL/L
• Δm/m = 0.003 / 0.3 ×100 = 1%
• Δr/r = 0.005 / 0.5 ×100 = 1%
• ΔL/L = 0.06 / 6 ×100 = 1%
• Total %error = 1 + 2×1 + 1 = 1 + 2 + 1 = 4%

Closest option: (d) 5%

V. Question On High Order Thinking Skills (HOTS)
Question 1. A laser light beam sent to the moon takes 2.56 s to return after reflection at the Moon’s surface. Calculate the radius of the lunar orbit around the earth.

Solution:
The laser travels from Earth to Moon and back, so total distance = 2 × radius of lunar orbit (R).

Time = Distance / Speed

t = 2R / c

R = (c × t) / 2

Substitute values:

R = (3 × 10^8 × 2.56) / 2
R = 7.68 × 10^8 / 2
R = 3.84 × 10^8 m

Question 2. The parallactic angle subtended by a distant star is 0.76 on the earth’s orbital diameter (1.5 x 10¹¹ m). Calculate the distance of the star from the earth.
Answer:

Question 3. The heat dissipated in a resistance can be obtained by the measurement of resistance, the current and time. If the maximum error in the measurement of these quantities is 1 %, 2 % and 1 % respectively, what is the maximum error in determination of the dissipated heat?
Answer:

Question 4. E, m, 1 and G denote energy, mass, angular momentum and gravitational constant respectively.Determine the dimensions of El²/m⁵G².
Answer:

Question 5. The Reynold’s number n_R for a liquid flowing through a pipe depends upon:

• (i) the density of the liquid ρ,
• (ii) the coefficient of viscosity η,
• (iii) the speed of flow of the liquid v, and
• (iv) the f radius of the tube r.

Obtain dimensionally an expression for n_R. Given, n_R is directly proportional to r.
Answer:

Note in Eqn. (1) we have used the information that n_R is directly proportional to r. If this information was not available there will be four unknowns. By equating powers of M, L and ‘ T only three independent equations will be obtained and they cannot give values of the four
unknowns. Now

Question 6.

Answer:

Question 7. The speed of light in air is 3.00 x 108 ms 1. The distance travelled by light in one year (i.e., 365 days = 3.154 x 10⁷ s) is known as light year. A student calculates one light year = 9.462 x 10¹⁵ m. Do you agree with the student? If not, write the correct value of one light year.

Answer: One light year = speed x time = 9.462 x 10¹⁵ m.

When two physical quantities are multiplied, the significant figures retained in the final result should not be greater than the least number of significant figures in any of the two quantities.

Since, in this case significant figures in one quantity (3.00 x 10⁸ ms^–1) are 3 and the significant figures in the other quantity (3.154 x 10⁷ s) are 4,

therefore, the final result should have 3 significant figures. Thus, the correct value of one light year = 9.46 x 10¹⁵ m.

Question 8.

Answer:

Question 9. If velocity of sound in a gas depends on its elasticity and density, derive the relation for the velocity of sound in a medium by the method of dimensions.
Answer: If v be the velocity of sound, E the elasticity of the medium and p the density of the medium, then

where k is a dimensionless constant of proportionality. Writing down the dimensions of both sides of equation (i), we get

Question 10. Reynold’s number N_R (a dimensionless quantity) determines the condition of laminar flow of a viscous liquid through a pipe. N_R is a function of the density of the liquid r, its average speed is v and the coefficient of viscosity of the liquid is h. If N_R is given directly proportional to d (the diameter of the pipe),

Answer:  As the Reynold’s number N_R depends on density p, average speed v and coefficient of viscosity η, then let us say

Question 11. ft is required to find the volume of a rectangular Mock. A Vernier Caliper is used to measure the length, width and height of the Mock. The measured values are found to be 1.37 cm, 4.11 cm and 2.56 cm respectively.
Answer: The measured (nominal) volume of the block is,
V = l x w x h
= (1.37 x 4.11 x 2.56) cm³ = 14.41 cm³
The least count of Vernier Caliper is ± 0.01 cm Uncertain values can be written as
l = (1.37 ± 0.01) cm w = (4.11 ± 0.01) cm h = (2.56 ± 0.01) cm Lower limit of the volume of the block is,
V ) = (1.37 – 0.01) x (4.11 – 0.01) x (2.56 – 0.01) cm³ = (1.36 x 4.10 x 2.55) cm³ = 14.22 cm³
This is 0.19 cm³ lower than the nominal measured value.
Similarly the upper limit can also be calculated as follows.
V(max) = (1.37 + 0.01) x (4.11 + 0.01) x (2.56 + 0.01) cm³ = (1.38 x 4.12 x 2.57) cm³ = 14.61 cm³
This is 0.20 cm³ higher than the measured value.
But we choose the higher of these two values as the uncertainty i.e. (14.41 ± 0.20) cm³

Question 12. In an experiment on determining the density of a ectangular Mock, the dimensions of the Mock are measured with a Venier Caliper (with a least count of 0.01 cm) and its mass is measured with a beam balance of least count of 0.1 gm. How do we report our result for the density of the block?
Answer:  Let the measured values be :
Mass of the block (m) = 39.3 g
length (1) = 5.12 cm
breadth (b) = 2.56 cm
thickness (f) = 0.37 cm
The density of the block is given by

You can access the official NCERT Solutions for Class 10 Mathematics on the NCERT website at the following link:

NCERT Class 10 Mathematics Solutions