NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements.

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements

Single Correct Answer Type

Q1. The number of significant figures in 0.06900 is
(a) 5 (b) 4 (c) 2 (d) 3
Sol: (b)
Key concept: Significant figures in the measured value of a physical quantity tell the number of digits in which we have confidence. Larger the number of significant figures obtained in a measurement, greater is the accuracy of the measurement. The reverse is also true.
The following rules are observed in counting the number of significant figures in a given measured quantity.
1. All non-zero digits are significant.
2. A zero becomes significant figure if it appears between two non¬zero digits.
3. Leading zeros or the zeros placed to the left of the number are never significant.
4. Trailing zeros or the zeros placed to the right of the number are significant.
5. In exponential notation, the numerical portion gives the number of significant figures.
Leading zeros or the zeros placed to the left of the number are never

Hence, number of significant figures are four.

Q2. The sum of the numbers 436.32, 227.2 and 0.301 inappropriate significant figures is
(a) 663.821 (b) 664 (c) 663.8 (d) 663.82
Sol: (b) The result of an addition or subtraction in the number having different precisions should be reported to the same number of decimal places as present in the number having the least number of decimal places.

The final result should, therefore, be rounded off to one decimal place, i.e. 664.

Q3. The mass and volume of a body are 4.237 g and 2.5 cm³, respectively. The density of the material of the body in correct significant figures is
(a) 1. 6048 g cm^-3
(b) 1.69 g cm^-3
(c) 1.7 g cm ³                                           
(d) 1.695 g cm^-3

Sol: (c) The answer to a multiplication or division is rounded off to the same number of significant figures as possessed by the least precise term used in the calculation. The final result should retain as many significant figures as are there in the original number with the least significant figures. In the given question, density should be reported to two significant figures

After rounding off the number, we get density =1.7

Q4. The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give
(a) 2.75 and 2.74
(b) 2.74 and 2.73
(c) 2.75 and 2.73
(d) 2.74 and 2.74
Sol: (d)
Key concept: While rounding off measurements, we use the following rules by convention:
1. If the digit to be dropped is less than 5, then the preceding digit is left unchanged.
2. If the digit to be dropped is more than 5, then the preceding digit is raised by one.
3. If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit is raised by one.
4. If digit to be dropped is 5 or 5 followed by zeros, then preceding digit is left unchanged, if it is even.
5. If digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one, if it is odd.
Units and Measurements
Let us round off 2.745 to 3 significant figures.
Here the digit to be dropped is 5, then preceding digit is left unchanged, if it is even.
Hence on rounding off 2.745, it would be 2.74.
Now consider 2.737, here also the digit to be dropped is 5, then the preceding digit is raised by one, if it is odd. Hence on rounding off 2.735 to 3 significant figures, it would be 2.74.

Q5. The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm, re­spectively. The area of the sheet in appropriate significant figures and error is
(a) 164 ±3 cm²
(b) 163.62 ± 2.6 cm²
(c) 163.6 ±2.6 cm²                                  
(d) 163.62 ±3 cm²
Sol: (a)

Q6. Which of the following pairs of physical quantities does not have same dimensional formula?
(a) Work and torque
(b) Angular momentum and Planck’s constant
(c) Tension and surface tension
(d) Impulse and linear momentum

Q7. Measure of two quantities along with the precision of respective measuring instrument is
A = 2.5 ms^-1 ± 0.5 ms^-1, B = 0.10 s ± 0.01 s. The value of AB will be
(a) (0.25 ± 0.08) m
(b)  (0.25       ± 0.5) m
(c) (0.25 ± 0.05) m
(d)  (0.25    ± 0.135) m

Q8. You measure two quantities as A = 1.0 m ± 0.2 m, B = 2.0 m ± 0.2 m. We should report correct value for √AB  as
(a) 1.4 m± 0.4 m
(b) 1.41 m± 0.15 m
(c) 1.4 m + 0.3 m
(d) 1.4 m± 0.2 m

Q9. Which of the following measurements is most precise?
(a) 5.00 mm
(b) 5.00 cm
(c) 5.00 m
(d) 5.00 km
Sol:(a)
Key concept: Precision is the degree to which several measurements provide answers very close to each other. It is an indicator of the scatter in the data. The lesser the scatter, higher the precision.
Let us first check the units. In all the options magnitude is same but units of measurement are different. As here 5.00 mm has the smallest unit. All given measurements are correct upto two decimal places. However, the absolute error in (a) is 0.01 mm which is least of all the four. So it is most precise.

Q10. The mean length of an object is 5 cm. Which of the following measurements is most accurate?
(a) 4.9 cm
(b) 4.805 cm
(c) 5.25 cm
(d) 5.4 cm
Sol: (a)
Key concept: Accuracy describes the nearness of a measurement to the standard or true value, i.e. a highly accurate measuring device will provide measurements very close to the standard, true or known values.
Example: In target shooting, a high score indicates the nearness to the bull’s eye and is a measure of the shooter’s accuracy.

Q11. Young’s modulus of steel is 1.9 x 10¹¹ N/m². When expressed in CGS units of dyne/cm², it will be equal to (1 N = 10⁵ dyne, 1 m² = 10⁴ cm²)                     .
(a) 1.9 xlO¹⁰                                            
(b) 1.9×10¹²
(c) 1.9 xlO¹²                                           
(d) 1.9 xlO¹³

Q12. If momentum (p), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula

More Than One Correct Answer Type
Q13. On the basis of dimensions, decide which of the following relations for the displacement of a particle undergoing simple harmonic motion is not correct?

Hence, (c) is not the correct option.
=> LHS ≠ RHS.
So, option (b) is also not correct.

Q14. If P, Q, R are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity?
(a) (P-Q)/R          
(b) PQ-R
(c) PQ/R                                  
(d) (PR-Q²)/R
(e)(R + Q)/P
Sol: (a, e)
Key concept: Principle of Homogeneity of dimensions: It states that in a correct equation, the dimensions of each term added or subtracted must be same. Every correct equation must have same dimensions on both sides of the equation.
According to the problem P, Q and R are having different dimensions, since, sum and difference of physical dimensions, are meaningless, i.e., (P – Q) and (R + Q) are not meaningful.
So in option (b) and (c), PQ may have the same dimensions as those of R and in option (d) PR and Q² may have same dimensions as those of R.
Hence, they cannot be added or subtracted, so we can say that (a) and (e) are not meaningful.

Q15. Photon is a quantum of radiation with energy E = hv, where v is frequency and h is Planck’s constant. The dimensions of h are the same as that of
(a) Linear impulse
(b) Angular impulse
(c) Linear momentum                          
(d) Angular momentum

Q16. If Planck’s constant (h) and speed of light in vacuum (c) are taken as two fundamental quantities, which one of the following can, in addition, be taken to express length, mass and time in terms of the three chosen fundamental quantities?
(a) Mass of electron (m_e)             
(b) Universal gravitational constant (G)
(c) Charge of electron (e)             
(d) Mass of proton (m_p)

Q17. Which of the following ratios express pressure?
(a) Force/Area                                       
(b) Energy/Volume
(c) Energy/Area                                     
(d) Force/Volume

Sol: (a, b) Let us first express the relation of pressure with other physical quantities one by one with the help of dimensional analysis.

Q18. Which of the following are not a unit of time?
(a). Second
(b) Parsec
(c) Year
(d) Lightyear
Sol: (b, d) Parsec and light year are those practical units which are used to measure large distances. For example, the distance between sun and earth or other celestial bodies. So they are the units of length not time. Here, second and year represent time.
Important point: 1 light year (distance that light travels in 1 year with speed = 3 x 10⁸ m/s.) = 9.46 x 10¹¹ m And 1 par see = 3.08 x 10¹⁶ m

Very Short Answer Type Questions

Q19. Why do we have different units for the same physical quantity?
Sol: Magnitude of any given physical quantity may vary over a wide range, therefore, different units of same physical quantity are required.
For example:
1.Mass ranges from 10^-30 kg (for an electron) to 10⁵³ kg (for the known universe). We need different units to measure them like miligram, gram, kilogram etc.
2.The length of a pen can be easily measured in cm, the height of a tree can be measured in metres, the distance between two cities can be measured in kilometres and distance between two heavenly bodies can be measured in light year.

Q20. The radius of atom is of the order of 1 A and radius of nucleus is of the order of fermi. How many magnitudes higher is the volume of atom as compared to the volume of nucleus?

Q21. Name the device used for measuring the mass of atoms and molecules.
Sol: A mass spectrograph is a device which is used for measuring the mass of atoms and molecules.

Q22. Express unified atomic mass unit in kg.
Sol: The unified atomic mass unit is the standard unit that is used for indicating mass on an atomic or molecular scale (atomic mass). One unified atomic mass unit is approximately the mass of one nucleon (either a single proton or neutron) and is numerically equivalent to 1 g/mol. It is defined as one- twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state.

Q24. Why length, mass and time are chosen as base quantities in mechanics?
Sol: Normally each physical quantity requires a unit or standard for its specification, so it appears that there must be as many units as there are physical quantities. However, it is not so. It has been found that if in mechanics we choose arbitrarily units of any three physical quantities we can express the units of all other physical quantities in mechanics in terms of these. So, length, mass and time are chosen as base quantities in mechanics because
(i) Length, mass and time cannot be derived from one another, that is these quantities are independent.
(ii) All other quantities in mechanics can be expressed in terms of length, mass and time.

Short Answer Type Questions
25. (a) The earth-moon distance is about 60 earth radius. What will be the . diameter of the earth (approximately in degrees) as seen from the moon?
(b) Moon is seen to be of (1/2)° diameter from the earth. What must be the relative size compared to the earth?
(c) From parallax measurement, the sun is found to be at a distance of about 400 times the earth-moon distance. Estimate the ratio of sun-earth diameters.

Q26. Which of the following time measuring devices is most precise?
(a) A wallclock                                        
(b) A stop watch
(c) A digital watch                                  
(d) An atomic clock
Given reason for your answer.
Sol: Option (d) is correct because a clock can measure time correctly up to one second. A stop watch can measure time correctly up to a fraction of a second. A digital watch can measure time up to a fraction of second whereas an atomic clock is the most accurate timekeeper and is based on characteristic frequencies of radiation emitted by certain atoms having precision of about 1 second in 300,000 years. So an atomic clock can measure time most precisely as precision of this clock is about 1 s in 10¹³ s.

Q27. The distance of a galaxy is of the order of 10²⁵ Calculate the order of magnitude of time taken by light to reach us from the galaxy.
Sol: According to the problem, distance of the galaxy = 10²⁵m.
Speed of light = 3 x 10⁸ m/s
Hence, time taken by light to reach us from galaxy is

Q28. The Vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, calculate the minimum inaccuracy in the measurement of distance.

Q29. During a total solar eclipse the moon almost entirely covers the sphere of the sun. Write the relation between the distances and sizes of the sun and moon.
Sol: Key point: In geometry, a solid angle (symbol: Ω or w) is the two­dimensional angle in three-dimensional space that an object subtends at a point. It is a measure of how large the object appears to an observer looking from that point. In the International System of Units (SI), a solid angle is expressed in a dimensionless unit called a steradian (symbol: sr).

Q30. If the unit of force is 100 N, unit of length is 10 m and unit of time is 100 s, what is the unit of mass in this system of units?

Q31. Give an example of
(a) a physical quantity which has a unit but no dimensions
(b) a physical quantity which has neither unit nor dimensions
(c) a constant which has a unit
(d) a constant which has no unit

Q32. Calculate the length of the arc of a circle of radius 31.0 cm which subtends an angle of π/6 at the centre.

Q33. Calculate the solid angle subtended by the periphery of an area of 1 cm² at a point situated symmetrically at a distance of 5 cm from the area.

Important point: Please keep in mind that solid angle is for 3-D figure like sphere, cone etc and plane angle is for plane objects or 2-D figures like circle, arc etc.

Q34. The displacement of a progressive wave is represented by y =A sin(wt– kx), where x is distance and / is time. Write the dimensional formula of (i) w and (ii) k 
Sol: We have to apply principle of homogeneity to solve this problem. Principle of homogeneity states that in a correct equation, the dimensions of each term added or subtracted must be same, i.e., dimensions of LHS and RHS should be equal.
According to the problem

Q35. Time for 20 oscillations of a pendulum is measured as t₁ =39.6 s; t₂ = 39.9 s and t₃ = 39.5 s. What is the precision in the measurements? What is the accuracy of the measurement?
Sol: According to the problem, time for 20 oscillations of a pendulum,
t₁ = 39.6 s, t₂ = 39.9 s and t₃ = 39.5 s
It is quite obvious from these observations that the least count of the watch is 0.1 s. As measurements have only one decimal place. Precision in the measurement = Least count of the measuring instrument= 0.1 s
Precision in 20 oscillations = 0.1

Long Answer Type Questions

Q36. A new system of units is proposed in which unit of mass is α kg, unit of length β m and unit of time γ s). How much will 5J measure in this new system?
Sol: For solving this problem, dimensions of physical quantity will remain same whatever be the system of units of its measurement.

Q40. If velocity of light c, Planck’s constant h and gravitational constant G are taken as fundamental quantities, then express mass, length and time in terms of dimensions of these quantities.
Sol: We have to apply principle of homogeneity to solve this problem. Principle of homogeneity states that in a correct equation, the dimensions of each term added or subtracted must be same, i.e., dimensions of LHS and RHS should be equal,
We know that, dimensions of

Q41. An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional

Q42. In an experiment to estimate ‘the size of a molecule of oleic acid, 1 mL of oleic acid is dissolved in 19 mL of alcohol. Then 1 mL of this solution is diluted to 20 mL by adding alcohol. Now, 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule.

Read the passage carefully and answer the following questions.

• Why do we dissolve oleic acid in alcohol?
• What is the role of lycopodium powder?
• What would be the volume of oleic acid in each mL of solution prepared?
• How will you calculate the volume of n drops of this solution of oleic
• What will be the volume of oleic acid in one drop of this solution?

Sol: (a) Since Oleic acid does not dissolve in water, hence it is dissolved in alcohol.

(b)Lycopodium powder spreads on the entire surface of water when it is sprinkled evenly. When a drop of prepared solution of oleic acid and alcohol is dropped on water, oleic acid does not dissolve in water. Instead it spreads on the water surface pushing the lycopodium powder away to clear a circular area where the drop falls. We can thus be able to measure the area over which oleic acid spreads.

(c)Since 20 mL (1 mL oleic acid + 19 mL alcohol) contains 1 mL of oleic acid, oleic acid in each mL of the solution =1/20 mL. Further, as this 1 mL is diluted to 20 mL by adding alcohol. In each mL of solution prepared, volume of oleic acid = 1/20 mL x 1/20  = 1/400 mL

(d) Volume of n drops of this solution of oleic acid can be calculated by means of a burette (used to make solution in the form of countable drops) and measuring cylinder and measuring the number of drops.

(e) As 1 mL of solution contains n number of drops, then the volume of oleic acid in one drop will be = 1/(400)n mL

Q43. (a) How many astronomical units (AU) make 1 parsec?
(b) Consider the sun like a star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be (1/2)° from the earth. Due to atmospheric fluctuations, eye cannot resolve objects smaller than 1 arc minute.
(c) Mars has approximately half of the earth’s diameter. When it is closest to the earth it is at about 1/2 AU from the earth. Calculate at what size it will appear when seen through the same telescope.

Q44. Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass (m) to energy (E) as E= mc², where c is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV, where 1 MeV = 1.6 x 10^-13 J; the masses are measured in unified atomic mass unit (u) where, 1 u = 67 x 10^-27 kg.
(a) Show that the energy equivalent of 1 u is 931.5 MeV.
(b) A student writes the relation as 1 u = 931.5 MeV. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.
Sol: (a) We can apply Einstein’s mass-energy relation in this problem, E = mc², to calculate the energy equivalent of the given mass.
Here

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