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Units And Measurements

Class 11th Physics Part I CBSE Solution
Exercise
  1. Fill in the Blanks A. The volume of a cube of side 1 cm is equal to.....m^3…
  2. The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to…
  3. A vehicle moving with a speed of 18 km h-1 covers....m in 1 s.
  4. The relative density of lead is 11.3. Its density is ....g cm-3 or ....kg m-3.…
  5. Fill in the blanks by suitable conversion of units. A. 1 kg m^2 s-2 = ....g cm^2 s-2…
  6. 1 m =..... ly
  7. 3.0 m s-2 =.... km h-2
  8. Convert G = 6.67 1011 N m^2 kg=.... (cm)^3 s2 g1.
  9. A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kg m^2…
  10. Explain this statement clearly: “To call a dimensional quantity ‘large’ or ‘small’ is…
  11. A new unit of length is chosen such that the speed of light in a vacuum is unity. What is…
  12. Which of the following is the most precise device for measuring length: (a). A Vernier…
  13. A student measures the thickness of a human hair by looking at it through a microscope of…
  14. Answer the following: A. You are given a thread and a metre scale. How will you estimate…
  15. The photograph of a house occupies an area of 1.75 cm^2 on a 35 mm slide. The slide is…
  16. State the number of significant figures in the following: A. 0.007 m^2 B. 2.64 × 10^24 kg…
  17. The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m,…
  18. The mass of a box measured by a grocer’s balance is 2.300 kg. Two gold pieces of masses…
  19. A physical quantity P is related to four observables a, b, c and d as follows: The…
  20. A book with many printing errors contains four different formulas for the displacement y…
  21. A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle…
  22. The unit of length convenient on the atomic scale is known as an angstrom and is denoted…
  23. One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar…
  24. Explain this common observation clearly: If you look out of the window of a fast moving…
  25. The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of…
  26. The nearest star to our solar system is 4.29 light years away. How much is this distance…
  27. Precise measurements of physical quantities are a need of science. For example, to…
  28. Just as precise measurements are necessary in science, it is equally important to be able…
  29. The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding…
  30. When the planet Jupiter is at a distance of 824.7 million kilometres from the Earth, its…
  31. A man walking briskly in rain with speed v must slant his umbrella forward making an angle…
  32. It is claimed that two cesium clocks, if allowed to run for 100 years, free from any…
  33. Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å.…
  34. The unit of length on the nuclear scale is a Fermi: 1 f = 10-15 m. nuclear sizes obey…
  35. A LASER is a source of very intense, monochromatic, and unidirectional beam of light.…
  36. A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects…
  37. The farthest objects in our Universe discovered by modern astronomers are so distant that…
  38. It is a well-known fact that during a total solar eclipse the disk of the moon almost…
  39. A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of…

Exercise
Question 1.

Fill in the Blanks

A. The volume of a cube of side 1 cm is equal to.....m3


Answer:

The volume of cube of side 1 cm is equal to 10-6 m3

Here we have to find the volume of a cube whose side is given.


We know that in a cube all sides are equal.


Let ‘a’ be the side of a cube.


Therefore ‘a’ = 1 cm


Here we have to find the volume of the cube in m3.


Convert 1 cm into m a = 1 cm = .01 m


Volume of a cube having side ‘a’ = a3


V = (0.01 m) 3


= 10-6 m3



Question 2.

The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...(mm)2


Answer:

Here we have to find the surface area of a solid cylinder whose radius and height is given.


Surface area of a solid cylinder having radius ‘r’ and height ‘h’


S.A = Curved surface area + Area of base and upper portion


In a solid cylinder Base & upper portions are in the form of a circle of radius ‘r’



S.A = 2πrh + 2πr2 . -------------------- (1)


Where r = radius of the upper and lower portions


h = height of the cylinder.


Here values of ‘r’ and ‘h’ are given.


Note: Here we have to convert the values of ‘r & ‘h’ from cm to mm.


r = 2 cm = 20 mm


h = 10 cm = 100 mm


Substituting the values of ‘r’ and ‘h’ in equation (1).


S.A = 2πrh + 2πr2


= 2πr(r +h) π


= 2 × (22/7) × 20(20+100) mm2.


= 1.5 × 104 mm2


The surface area of the solid cylinder = 1.5 × 104 mm2



Question 3.

A vehicle moving with a speed of 18 km h–1 covers....m in 1 s.


Answer:

In this question, we have to find the distance covered by a vehicle in 1 s moving with a particular speed.


We know that distance ‘x’ covered by a vehicle moving with a particular speed ‘v’


x = v × t ------------------- (1)


Where x = distance covered


v = speed of the vehicle


t = time taken to cover distance ‘x’


Here v = 18 km/h = 5 m/s


To convert km/h to m/s multiply the value of km/h with


t = 1 s


Substituting the values of ‘v’ &‘t’ in equation (1)


x = v × t


x = 5 ms-1 × 1 s


x = 5 m


The distance covered by the vehicle in 1s = 5m



Question 4.

The relative density of lead is 11.3. Its density is ....g cm–3 or ....kg m–3.


Answer:

Here the relative density of lead is given. We have to find the density of lead.


Relative density of lead = -------------- (1)


Density of lead = Relative density of lead × Density of water


Relative density of lead = 11.3


Density of water = 1 g/cm3 = 1000 kg/m3


Substituting the above density values in equation (1).


Density of lead = 11.3 × 1 g/cm3


= 11.3 g/cm3


We have to find the density of lead in kg/m3 also.


Substituting the value of density of water in kg/m3 in equation (1)


Density of lead = 11.3 × 1000 kg/m3


= 1.13 × 104 kg/m3


The density of lead = 11.3 g/cm3 or 1.13 × 104 kg/m3



Question 5.

Fill in the blanks by suitable conversion of units.

A. 1 kg m2 s-2 = ....g cm2 s-2


Answer:

Here we have to convert kg to g & m to cm.

1 kg = 1000 g


1 m = 100 cm


Therefore 1 kg m2 s-2 = 1000g × (100 cm)2 s-2


= 103 g× 104 cm2 s-2


= 107 g cm2 s-2


1 kg m2 s-2 = 107 g cm2 s-2



Question 6.

1 m =..... ly


Answer:

Here we have to find 1m is how many light years.


We know that 1 ly = 9.46 × 1015 m


1 m = 1/ (9.46 × 1015) m


= 1.057 × 10-16 ly


1 m = 1.057 × 10-16 ly



Question 7.

3.0 m s-2 =.... km h-2


Answer:

Here we have to convert m/s2 into km/h2


1 km = 1000 m


1 m = 10-3 km


1 hour = 60 min = 60 × 60 sec
or 1 sec = h


Substituting the above values


3 m s-2 = 3 × 10-3 km × h-2


= 38880 km-h-2h

= 3.88 × 104 km h-2


Question 8.

Convert G = 6.67 × 10–11 N m2 kg =.... (cm)3 s–2 g–1.


Answer:

Here the value of a constant ‘G’ is given. We have to express it given unit.


Note: ‘G’ is known as Universal Gravitational Constant. Its value is constant all over the universe.


G = 6.67 × 10-11 N m2 kg


Where, N = Newton


m = meter


Kg = Kilogram


‘N’ can be rewritten as


N = kg × ms-2


Substitute this value in ‘G’


G = 6.67 × 10-11 N m2 kg-2


= 6.67 × 10-11 kg × ms-2 × kg-2


= 6.67 × 10-11 kg-1 m3 s-2


= 6.67 × 10-11 × (102)3 × (10-3)2


= 6.67 × 10-11 cm3 g-1 s-2


G = 6.67 × 10–11 N m2 (kg)–2 = 6.67 × 10-11 (cm)3 s–2 g–1


Question 9.

A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kg m2 s–2.Suppose we employ a system of units in which the unit of mass equals kg, the unit of length equals m, the unit of time is s. Show that a calorie has a magnitude 4.2 in terms of the new units.


Answer:

In this question, we have to apply one of the uses of Dimensional analysis


i.e. Conversion of one system of units into another


Where, n2 = n1 × [M1M2 ]a × [L1/L2]b × [T1/T2]c ----------(1)


M1, L1, T1 = Fundamental units of one system


M2, L2, T2 = Fundamental units of another system


a,b,c are the dimensions of the quantity in mass, length, and time.


n1 = numerical value of quantity in one system


n2 = numerical value of quantity in the other system


The value of 1 calorie in S.I System of units is given. We have to express the value of calorie in another given arbitrary units.


1 calorie = 4.2 J


1 J = 1 kg m2 s–2



Substituting the above values in equation (1).


n2 = 4.2 []1 × []2 × []-2


= 4.2 -1-22


1 calorie = 4.2 α-1 β-2 γ2 in new units.



Question 10.

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:

A. We know that size of an atom is smaller than the tip of a pencil.


B. We know that a jet plane moves faster than a car.


C. The mass of Jupiter is very large when compared with the mass of Earth.


D. The air inside this room contains a large number of molecules when compared with air inside a bottle.


E. There is no need to rephrase the statement. The statement is true.


F. No need to rephrase the above statement. The statement is right.



Question 11.

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?


Answer:

We know that speed of light in vacuum, c = 3 × 108 m/s


Relation between speed & distance travelled


Distance travelled ‘x’ = speed × time taken ------- (1)


Here the value of time taken is given


t = 8 min 20 s = 500 s


So, x = s × t


x = c × t


Where c = speed of light


t = time taken


= 3 × 108 ms-1 × 500 m


= 1.5 × 1011 m


So this is the distance between the sun and the earth.


We have to express the above distance value in terms of new units. In new units, the value of the speed of light is unity.


i.e, c = 1 unit


t = 500 s


Substituting the above values in equation (1)


Distance between sun & Earth x = c × t


= 1 unit × 500 s


= 500 new units


Distance between Sun & Earth = 500 new units



Question 12.

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

(a). A Vernier callipers with 20 divisions on the sliding scale

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

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


Answer:


In this question, three measuring/ calibrating instruments are given. Among these devices, we have to find the most precise device.


To find the most precise device we have to calculate the Least Count of these devices.


(a). Least count of Vernier calliper =


= = 0.05 mm


= 0.05 cm


(b). Least count (L.C) of screw gauge =


Pitch = 1 mm


No. of division on circular scale = 100


L.C = 1 × 10-3 ×


= 0.01 cm


(c). Least count of optical instrument that can measure length to within a wavelength of light is about 0.00001 cm


Here optical instrument has least count among given instruments. So Optical instrument is the most precise device for measuring length.



Question 13.

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 of the thickness of hair?


Answer:

Here given details are


Magnification of Microscope = 100


Average field of view of microscope = 3.5 mm


Thickness of Hair = ------------- (1)


Substituting the above values in equation (1)


Thickness of hair =


= 0.035 mm


Estimate thickness of hair = 0.035 mm



Question 14.

Answer the following:

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

B. A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?

C. The mean diameter of a thin brass rod is to be measured by Vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?


Answer:

A. Take the thread and wrap it around the metre scale. It should be wrapped without giving any space between the coils. Assume that the diameter of the thread is d.


Then number of turns, take n (coils) multiplied with diameter d will give the length up to which you’ve used the thread (t take it l). So, d = l/n


B. It is possible theoretically to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale as the least count is given by pitch/no. of divisions. So, on increasing the no. of divisions, least count will decrease and thus, increasing the accuracy. But practically, it depends on the resolution of eye (to see two different things distinctly and clearly). You can increase no. of divisions upto a certain limit only.


C. The mean diameter of a thin brass rod measured by vernier callipers from a set of 100 measurements is more reliable a more reliable estimate than a set of 5 measurements because the chances of making a negative error and positive error are equally likely and thus they cancel each other giving results with less error.


Question 15.

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:

Given: Area of photograph of house (Aobject) - 1.75 cm2


Area of image formed on screen (Aimage)- 1.55 m2 = 1.55 x 104 cm2


Formula: Areal magnification (m) = Aimage / Aobject ,


where Aimageand Aobject are the areas of image and object respectively.


Since the dimension of areal magnification is L2 and dimension of linear magnification is L,


Linear magnification (ml) = √m


m = (1.55 × 104 cm2)/(1.75 cm2)



⇒ ml = 94.1 (report upto 3 significant figures only)



Question 16.

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


Answer:

A. 1


Explanation: Significant figure- 7. If number is less than one, then 0’s before 7 are insignificant (note that there are only 0’s before 7 and numbers like 0.1007 will have 0 as significant figure)


B. 3


Explanation: Significant figure- 2, 6, 4. Powers of 10 are not taken in counting for significant figures.


C. 4


Explanation: Significant figure- 2,3, 7, 0. Trailing 0’s are significant. These 0’s increase the accuracy of the answer.


D. 4


Explanation: Significant figure- 6, 3, 2, 0. (similar to the problem C)


E. 4


Explanation: Significant figure- 6, 0, 3, 2. 0’s between 2 non-zero digits are significant.


F. 4


Explanation: Significant figure- 6, 0, 3, 2. (similar to the problem A and E)



Question 17.

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


Answer:

Given:

Length (l) = 4.234 m


Breadth (b) = 1.005 m


Thickness (t) = 2.01 cm = 0.0201 m (note that significant figures remains unchanged)


Formulae:


Total surface area of rectangular sheet (S) = 2 (lb + bt + tl)


Volume of rectangular sheet (V) = l × b × h


Calculation:


S = 2×(4.234 ×1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)


S = 8.72 m2 (reported upto 3 significant figures as in multiplication, lowest number of significant figures in all the numbers is chosen for final answer)


V = 4.234 × 1.005× 0.0201


V = 0.0855 m3 (reported upto 3 significant figures as in multiplication, lowest number of significant figures in all the numbers is chosen for final answer)



Question 18.

The mass of a box measured by a grocer’s balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?


Answer:

Given:

Mass of the box (m): 2.300 Kg (2 significant figures, 1 after decimal)


Gold pieces: G1 = 20.15 g = 0.02015 Kg (4 significant figures) and G2 = 20.17 g = 0.02017 Kg (4 significant figures)


(a) total mass of box (M) = m + G1 + G2


M = 2.300 + 0.02015 + 0.02017 = 2.34032 Kg but in addition the final result should have significant digits after decimal equal to the minimum number of significant digits of all the numbers used.


Therefore, M = 2.3 Kg (upto 2 significant digits and 1 after decimal due to 1 significant digit in m)


(b) difference in the masses = G2 – G1 = 0.02 g (1 significant digit). In subtraction the final result should have significant digits after decimal equal to the minimum number of significant digits of all the numbers used.



Question 19.

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:

Given:

The percentage error in a i.e.

The percentage error in b i.e.

The percentage error in c i.e.

The percentage error in d i.e.

(Rule: The relative error in a physical quantity raised to the power k is the k times the relative error in the individual quantity)


Multiplying both sides by 100 will make each quantity % and thus we can directly substitute the given values to find ΔP/P % or percentage error in P.


Since number of significant digits in ΔP/P are 2, P should also have 2 significant digits.

P = 3.8 (rounded off to first decimal place)


Question 20.

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

A. y = a sin 2 t/T

B. y = a sin vt

C. y = (a/T) sin t/a

D.

(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:

The dimensions of the quantities involved are:


y = M0L1T0


a = M0L1T0


T or t=T

A. correct

Explanation: t/T is unitless. Therefore, dimension of y should be the dimension of a which is true.


B. incorrect


Explanation: quantities inside the trigonometric expressions should be unitless but dimension of vt = M0L1T0 which is having unit of length and is therefore wrong.


C. incorrect


Explanation: quantities inside the trigonometric expressions should be unitless but dimension of t/a = M0L-1T1 which is having unit of time/length and is therefore wrong.


D. correct


Explanation: t/T is unitless. Therefore, dimension of y should be the dimension of a which is true.


Question 21.

A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo 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:

The dimension of the quantities involved are:

m = M1L0T0


m0 = M1L0T0


v = M0L1T-1


c = M0L1T-1


For a correct dimensional equation, the dimensions of LHS and RHS should be equal. We can see that dimensions of m and m0 are equal thus the remaining part of the equation should be constant.


should be unitless.


1 is dimensionless so should be dimensionless.


On subtraction, dimension of the 2 quantities should be same. So, dimension of 1 and v2 should be same. Therefore, v2 should be dimensionless and therefore v.


We can see that dimension of v and c are same. So, replacing v by v/c will make it dimensionless and the given equation dimensionally correct.


Hence, correct relation will be,




Question 22.

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?


Answer:

Given

Radius of hydrogen atom, r = 0.5 Å = 0.5 × 10-10 m


Volume of hydrogen atom, V = (4/3)πr3



⇒ V = 5.24 × 10-31 m3


One mole of hydrogen atom contains NA = 6.023 × 1023 atoms.


Where, NA is Avogadro’s number.


So, Volume of one mole of hydrogen atom,


Vt = 6.023 × 1023 × 5.24 × 10-31 m3


⇒ Vt = 3.16 × 10-7 m3



Question 23.

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


Answer:

Given,

Volume of one mole of ideal gas at STP, Vm=22.4L=22.4 × 10-3 m3


Radius of hydrogen atom, r = 1 Å / 2 = 0.5 Å = 0.5 × 10-10 m


Volume of the hydrogen atom, V = (4/3)πr3


⇒ V = (4/3)×3.14×(0.5×10-10 m)3


⇒ V = = 5.24 × 10-31 m3


One mole of hydrogen atom contains NA = 6.023 × 1023 atoms.


Where, NA is Avogadro’s number.


So, Volume of one mole of hydrogen atom,


Vt = 6.023 × 1023 × 5.24 × 10-31 m3


⇒ Vt = 3.16 × 10-7 m3


The ratio of molar volume to atomic volume is



⇒ Vm �/Vt = 7.089 × 104


This ratio is very high due to the fact that the inter-atomic distance is very high as compared to the size of atoms in hydrogen gas.



Question 24.

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:

Due to the fast movement of the train, the line of sight of the passenger for nearby objects changes rapidly. So, the nearby trees, houses, etc. appear to move fast in the opposite direction. But, in case of distant objects, the line of sight changes very slowly. So, the distant objects like hilltops, moon, stars, etc. seem to be stationary.

NOTE: Line of sight is an imaginary line joining an object to the viewer’s eyes.



Question 25.

The principle of ‘parallax’ in section 2.3.1 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 × 1011m. However, even the nearest stars are so distant that with such a long baseline, they show parallax 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:

Given,

Diameter of Earth’s orbit, d = 3 × 1011 m


Radius of earth’s orbit, r = d/2


⇒ r = (3×1011m)/2 = 1.5×1011 m


Parallax angle, θ = 1”


⇒ θ = 1’/60


⇒ θ = 1°/(60×60)





We know,


θ = Arc/Radius


⇒ θ = r/x


⇒ x = r/θ



⇒ x = 3.09 × 1016 m


Hence, 1 parsec = 3.09 × 1016 m


NOTE: The parsec is a unit of length used to measure large distances to astronomical objects outside the Solar System. Parallax is a difference in the apparent position of an object viewed along two different lines of sight, and is measured by the angle or semi-angle of inclination between those two lines of sight.



Question 26.

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:

Given,

Distance of the nearest star, d = 4.29 ly


1 lightyear is the distance travelled by light in one year.


∴ 1 ly = Speed of light × 1 yr


⇒ 1 ly = 3 × 108 m s-1 × (365 × 24 × 60 × 60 s)


⇒ 1 ly = 9.46 × 1015 m


So, d = 4.29 × 9.46 × 1015 m


⇒ d = 4.058 × 1016 m


We know, 1 parsec = 3.09 × 1016 m


So,



⇒ d = 1.313 parsec


We know,



where, θ is the parallax angle


l is the diameter of the Earth’s orbit = 3 × 1011 m


d is the distance of the star from Earth



⇒ θ = 7.39 × 10-6 rad



∴ θ = (7.39 × 10-6 rad)/(π/648000 rad)


⇒ θ = 1.52”


NOTE: It is to be noted that a lightyear is a unit of distance and not time.


Parallax is a difference in the apparent position of an object viewed along two different lines of sight, and is measured by the angle or semi-angle of inclination between those two lines of sight.



Question 27.

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 modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.


Answer:

It is true that the precise measurements of physical quantities is very essential for science. For instance, laser length measurements require a precision in the order of Angstrom (1 Å = 10–10 m). In satellite launching through space rockets, time is an important factor and is measured up to a precision of the order of 1 micro second (1μs = 10-6 s). Mass of sub-atomic particles are measured in the order of 10-30 kg. Inter-atomic spacing is measured in the order of a few Angstroms which needs to be precise to avoid errors in spectroscopy.



Question 28.

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. Height of water column during monsoon is recorded as 215 cm.

H = 200 cm = 2.0 m


Area of the country, A = 3.3 × 1012 m2


Volume of water column, V = AH


⇒ V = 3.3 × 1012 m2 × 2.0 m


⇒ V = 6.6 × 1012 m3


Density of water, ρ = 103 kg m-3


Mass of the rain-water, m = Vρ


⇒ m = 6.6 × 1012 m3 × 103 kg m-3


⇒ m = 6.6 × 1015 kg


The mass of the rain-bearing clouds should be approximately equal to this mass.


B. Consider a boat of known base-area A. Measure the depth upto which it sinks in water when it is empty. Let this height be h.


Volume of water displaced by boat, v = Ah


Measure the depth again when elephant is kept on the boat. Let this depth be h’.


Volume of water displaced, V = Ah’


Volume of water displaced by elephant, V’ = V – v = A(h’-h)


The mass of this volume of water is equal to the mass of the elephant.


Density of water, ρ = 103 kg m-3


Mass of water displaced by elephant, m = V’ρ


⇒ m = A(h’-h)×103 kg


This is approximately equal to the mass of the elephant.


C. Consider a balloon filled with air. When there is no wind, note the height of the balloon from a fixed point. When the wind blows, measure the distance of the balloon from the fixed point after a certain time. The displacement of balloon can be calculated from the angular displacement. Now, the ratio of the displacement to the time of flight is the speed of wind.


Also, an anemometer can be used to measure the speed of the wind. The number of rotations in one second gives the speed of wind.


D. Let A be the area of the head covered with hair.


If r is the radius of hair strand, the base area of hair strand,


a = πr2


So, number of hairs, n = A/a = A/πr2


This gives an approximation of the number of hair strands on a head.


E. If l, b and h are the length, breadth and height of the classroom, then its volume is v = lbh.


If r is the radius of an air molecule, then volume of the air molecule, v’ = (4/3)πr3


So, number of air molecules in the classroom, n = v/v’



⇒ n = 3lbh/4πr3


This gives an approximation of the number of air molecules in the classroom.



Question 29.

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.


Answer:

Given,

Mass of the sun, M = 2.0 × 1030 kg


Radius of the sun, R = 7.0 × 108 m


Volume of the sun, V = (4/3)πR3


⇒ V = (4/3) × 3.14 × (7.0 × 108 m)3


⇒ V = 1.436 × 1027 m3


Density of the sun,



⇒ ρ = 1.39 × 103 kg m-3


This value corresponds to the range of solids and liquids.



Question 30.

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,

Distance of Jupiter from Earth, D = 824.7 million kilometres


⇒ D = 824.7 × 106 × 103 m


⇒ D = 8.247 × 1011 m


Angular diameter, θ = 35.72”


⇒ θ = 35.72 × 1’/60




Let d be the diameter of Jupiter.


We know,


θ = d/D


⇒ d = θD



⇒ d = 1.43 × 108 m



Question 31.

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: tan θ = v and checks that the relation has a correct limit: as v →0, 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:

Dimensions of tan θ = M0L0T0


(∵ All trigonometric functions have no units)


Dimensions for v (velocity) = M0L1T-1


∴ The equation, tan θ = v is Dimensionally incorrect.


To balance the equation,


Let the speed of the rainfall be V, then we can make R.H.S dimensionless by dividing with V.


tan θ =


(∵ velocity dimension = M0L1T-1 always)


Now, the equation is dimensionally balanced and correct.



Question 32.

It is claimed that two cesium clocks, if allowed to run for 100 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:

Time allowed to run (T) = 100 years


= 100×365×24×60×60 sec


∴ T = 3.15×109 s


Difference after 100 years (d) = 0.02 s


In 1 s the time difference shown by clock,


∴ The accuracy of the standard caesium clock in measuring time interval of 1s is given by,


A = s ≈ 1.5×1011 s



Question 33.

Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (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:

Given,


Diameter of the sodium atom, (d)= 2.5 Å


∴ Radius, (r) = 1.25 Å


Density of sodium in crystalline phase, (ρc)= 970 kg m-3


Consider every atom of sodium to be sphere,


Volume (V) = r3 m3


∴ V = ×3.14× (1.25×10-10)3 m3


≈ 8.18×10-30 m3


We know that,


According to Avogadro law, a mole contains 6.023×1023 atoms.


The molecular weight of sodium atom, (M) = 23 g = 23×10-3 kg


Density, (ρ) = kg/m3


∴ ρ = ≈ 4.67×10-7 kg/m3


Thus,


The density of sodium in crystalline phase (ρc) is much higher than the density of general sodium in general configuration (ρ). Because, sodium atoms in crystalline phase have very less inter-atomic separation.



Question 34.

The unit of length on the nuclear scale is a Fermi: 1 f = 10-15 m. nuclear sizes obey roughly the following empirical relation:

r = r0A1/3

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:

The equation for the radius of the nucleus is given by,


r = r0A1/3


∴ Volume on the nucleus, V =


=


=


Mass number is given in the problem as, A.


Thus,


M = Mass number× Mass of single Nucleus


= A×1.67×10-27 kg


(∵ Mass of single Nucleus = 1.67×10-27 kg)


∴ Density, ρ =


= = kg/m3


This equation doesn’t have Mass number A and it has only one variable, Radius of the atom (r). Since, every atom has same radius then the density of sodium atom nuclei remains constant.


∴ Density of sodium atom nucleus, ρ nucleus = kg/m3


2.29×1017 kg/m3



Question 35.

A LASER is a 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:

Given,


Time taken by laser to reflect back from moon to earth = 2.56 s


We know that,


Speed of the light in vacuum = 3×108 m/s


Radius of the lunar orbit (r) = distance between Earth and Moon (d)


Times taken for laser beam to reach moon = 0.5×2.56 = 1.28 s


∴ r = d = speed of light in vacuum × Time taken


= 3×108 × 1.28 m


= 3.84×105 km


Hence, the radius of the lunar orbit is, 3.84×105 km.



Question 36.

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).


Answer:

Given,


Speed of sound in water = 1450 m/s


Total time taken for transmission = 77 s


Let distance between our submarine and enemy submarine be, D.


∴ Time taken to travel the distance D be, t = 0.5×77 = 38.5 s


Thus,


D = Speed of sound in water × Time taken to travel the distance D


= 1450×38.5 m


= 55.8 km



Question 37.

The farthest objects in our Universe discovered by modern 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:

Quasar: The active galactic centre which emit huge amount of energy in the form of light.


Given,


Time taken by Quasar light to reach earth, T = 3 billion years


∴ T = 3×109×365×24×60×60 s


(∵ 1 billion = 109 years)


We know that,


Speed of light in vacuum, V = 3×108 m/s


∴ Distance between the Earth and Quasar is given by,


D= V×T


= 3×108×(3×109×365×24×60×60) m


= 2.8×1022 km



Question 38.

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:

Fig. The Total Solar Eclipse.


Given, from examples 2.3 and 2.4 we have,


Distance between the Sun and the Earth = 1.496×1011 m


Distance between the Moon and the Earth = 3.84×108 m


Diameter of the Sun = 1.39×109 m



Fig. Schematic diagram of positions of Sun, Moon and Earth during Total Solar Eclipse.


From the above figure,


Δ PQT and Δ RST are similar. Thus,


=


Where,


PQ = Diameter of the Sun


RS = Diameter of the Moon


VT = Distance between the Sun and the Earth


UT = Distance between the Moon and the Earth


=


⇒ RS = ×106 m = 3.57×106 m


Hence, The diameter of the Moon = RS = 3.57×103 km.



Question 39.

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:

One relation consists of some fundamental units that give the age of the universe is,


T = × s


Where,


T = Age of the Universe


e = charge of Electron = -1.6×10-19 C


ϵ0 = absolute permittivity


mp = Mass of the Proton = 1.67×10-27 kg


me = Mass of the Electron = 9.1×10-31 kg


c = Speed of light in vacuum = 3×108 m/s


G = Universal Gravitational constant = 6.67×1011 Nm2kg-2


And, = 9×109 Nm2 /C2


∴ T =


≈ 6×109 years


≈ 6 billion years


(∵ 1 billion year = 109 years)


Hence, The age of the universe, T ≈ 6 billion years.