# Chapter 2 – Units and Measurements Questions and Answers: NCERT Solutions for Class 11 Physics

Class 11 Physics NCERT book solutions for Chapter 2 - Units and Measurements Questions and Answers.

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Class 11 Physics NCERT book solutions for Chapter 2 - Units and Measurements Questions and Answers.

(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-1covers ………. m in 1 s.

(d) The relative density of lead is 11.3. Its density is …….. g cm-3or ………. kg m-3.

Hence, answer is 10-6

(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 104mm2Hence, answer is 1.5 x 104.

(c) Speed of vehicle = 18 km/h = 18 x 1000/3600 m/s

= 5 m/s ; so the vehicle covers 5 m in 1 s. = 11.3

(d) Density= 11.3 g cm-3

=11.3 x 103kg m-3[1 kg =103g,1m=102cm]

=11.3 x 103kg m-4

(a) 1 kg m2 s-2= …. g cm2s-2

(b) 1 m =………… ly

(c) 3.0 m s-2= …. kmh-2

(d) G = 6.67 x 10-11N m2(kg)-2= …. (cm)3s-2g-1.

“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.

(a) The size of an atom is much smaller than even the sharp tip of a pin.

(c) The mass of Jupiter is very large compared to that of the earth.

(d) The air inside this room contains more number of molecules than in one mole of air.

(e) This is a correct statement.

(f) This is a correct statement.

= Speed of light in vacuum x time taken by light to travel from Sim to Earth = 3 x 108m/ s x 8 min 20 s = 3 x 108m/s x 500 s = 500 x 3 x 108m.

In the new system, the speed of light in vacuum is unity. So, the new unit of length is 3 x 108m.

.•. distance between Sun and Earth = 500 new units.

(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?

(b) Least count of screw gauge =Pitch/No. of divisions on circular scale = 1 x 10-3/100 = 1 x 10-5m

(c) Least count of optical instrument = 6000 A (average wavelength of visible light as 6000 A) = 6 x 10-7mAs the least count of optical instrument is least, it is the most precise device out of three instruments given to us.

and average width of the image of hair as seen by microscope = 3.5 mm

.•. Thickness of hair =3.5 mm/100 = 0.035 mm

(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?

This is equal to the diameter of the thread.

(b) We know that least count = Pitch/number of divisions on circular scale When number of divisions on circular scale is increased, least count is decreased. Hence the accuracy is increased. However, this is only a theoretical idea.Practically speaking, increasing the number of ‘turns would create many difficulties.

As an example, the low resolution of the human eye would make observations difficult. The nearest divisions would not clearly be distinguished as separate. Moreover, it would be technically difficult to maintain uniformity of the pitch of the screw throughout its length.

(c) Due to random errors, a large number of observation will give a more reliable result than smaller number of observations. This is due to the fact that the probability (chance) of making a positive random error of a given magnitude is equal to that of making a negative random error of the same magnitude. Thus in a large number of observations, positive and negative errors are likely to cancel each other. Hence more reliable result can be obtained.

.•. Areal magnification =Area on screen/Area on slide = 1.55 m2/ 1.75 x 10-4m2= 8.857 x 103

.•. Linear magnification

(a) 0.007 m2(b) 2.64 x 104kg

(c) 0.2370 g cm-3(d) 6.320 J

(e) 6.032 N m-2(f) 0.0006032 m2

Volume = (4.234 x 1.005) x (2.01 x 10-2) = 8.55289 x 10-2= 0.0855 m3.

Since the least number of decimal places is 1, therefore, the total mass of the box = 2.3 kg.

(b) Difference of mass = 2.17 – 2.15 = 0.02 g

Since the least number of decimal places is 2 so the difference in masses to the correct significant figures is 0.02 g.

= 4/3 x 3.14 x (0.5 x 10-10) m3= 5.23 x 10-31m3

According to Avagadro’s hypothesis, one mole of hydrogen contains 6.023 x 1023atoms.

Atomic volume of 1 mole of hydrogen atoms

= 6.023 x 1023 x 5.23 x 10-31= 3.15 x 10-7m3.

= 22.4 litre = 22.4 x 10-3m3

Radius of hydrogen molecule = 1A/2

= 0.5 A = 0.5 x 10-10m

Volume of hydrogen molecule = 4/3 πr3

=4/3 x 22/7 (0.5 x 10-10)3m3

= 0.5238 x 10-30m3

One mole contains 6.023 x 1023molecules.

Volume of one mole of hydrogen, VH

= 0.5238 x 10-30x 6.023 x 1023m3= 3.1548 x 10-7m3

Now Vg/VH=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.

θ=b/D,where b=baseline ,D = distance of distant object or star

Since, θ=1″ (s) and b=3 x 1011m

D=b/20=3 x 1011/2 x 4.85 x 10-6m

or D=3 x 1011/9.7 x 10-6m =30 x 1016/9.7 m

= 3.09 x 1016m = 3 x 1016m.

.•. 4.29 light years = 4.29 x 9.46 x 1015= 4.058 x 1016m

Also, 1 parsec = 3.08 x 1016m

.•. 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″.

(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.

Mass of rain-bearing clouds

= area x height x density = 3.3 x 1012x 1 x 1000 kg = 3.3 x 1015kg.

(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.

(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 10-3m3of air has 6.02 x 1023molecules (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

.-. Volume of Sun = 4/3πr3x 3.14 x (7 x 108)3 = 1.437 x 1027m3

As p = M/V, .’. p = 2 x1030/1.437 x 1027= 1391.8 kg m-3= 1.4 x 103kg m-3

Mass density of Sun is in the range of mass densities of solids/liquids and not gases.

= 173.242 x 10-6= 1.73 x 10-4rad

Diameter of Jupiter D = θ x d = 1.73 x 10-4x 824.7 x 109m

=1426.731 x 103= 1.43 x 108m

Dimensions of L.H.S. = Dimensions of R.H.S.

Here, v = tan θ

i. e., [L1T-1] = dimensionless, which is incorrect.

Correcting the L.H.S., we. get

v/u= tan θ, where u is velocity of rain.

Error in 1 second=0.02/100 x 365 x 24 x 60 x 60

=6.34 x 10-12s

.•. Accuracy of 1 part in 1011to 1012.

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 m3-. Are the two densities of the same order of magnitude? If so, why?

Volume of one mole atom of sodium, V = NA .4/3 π R3

V = 6.023 x 1023x –4/3 x 3.14 x (2.5 x 10-10)3m3and mass of one mole atom of sodium, M = 23 g = 23 x 10-3kg

.•. Average mass density of sodium atom, p = M/V

=(23 x 10-3/6.023 x 1023x 4/3 x 3.14 x (2.5 x 10-10))

= 6.96 x 102kg m-3= 0.7 x 10-3kg m-3

The density of sodium in its crystalline phase = 970 kg m-3

= 0.97 x 103kg m-3

Obviously the two densities are of the same order of magnitude (= 103kg 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.

r = r0 A1/3

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.

= 4/3 πr3= 4/3 π [r0A1/3]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)1/3m = 1.2 x 2.844 x 10-15m =3.4128 x 10-15

If distance of enemy submarine be d, then t = 2d/v

.’. d=vt/2 =1450 x 77.0/2 =55825 m=55.8 x 103m or 55.8 km.

t = 3.0 billion years = 3.0 x 109years As 1 ly = 9.46 x 1015m

.•. Distance of quasar from the observer d = 3.0 x 109x 9.46 x 1015 m

= 28.38 x 1024m = 2.8 x 1025m or 2.8 x 1022km.

= 1920″ = 1920 x 4.85 x10-6rad [1″ = 4.85 x 10-6rad]

The earth-moon distance, S = 3.8452 x 108m .’. The diameter of the moon, D = θ x S

= 1920 x 4.85 x 10-6x 3.8452 x 108m = 35806.5024 x 102m = 3581 x 103m 3581 km.

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