Class 11 Physics NCERT Solutions of chapter 2-Units and Measurment - Future Study Point

Class 11 Physics NCERT Solutions of chapter 2-Units and Measurment

Class 11 Physics NCERT Solutions of chapter 2-Units and Measurment

Class 11 Physics NCERT Solutions of chapter 2-Units and Measurement

Class 11 Physics NCERT Solutions of chapter 2-Units and Measurment

Class 11 Physics NCERT Solutions of chapter 2-Units and Measurement are the solutions of physics second chapter of the NCERT class 11 physics which is a basic chapter of physics that is important to study before you go through other chapters in physics because chapter 2 Units and Measurement deals with all types of units used for different physical quantities. The purpose of studying this chapter units and measurements is not only for the preparation of the class 11 CBSE board exam but also to get a basic idea of physics related to units and measurements used in the entire Physics subject from its microscopic to the macroscopic field.

Q2.1-Fill up 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 cm and height 10 cm is equal to….(mm)²

(c)A vehicle moving with a speed of 18 km/h covers ……m in 1 s

(d) The relative density of lead is 11.3.Its density is ….g /cm or….kg/m³

Ans.

(a) The volume of a cube is = side³ =1³= 1 cm³

Since 1 cm3 =10-6 m3

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

(b)The surface area(S) of a solid cylinder of radius r  and height h = 2πr(r +h)

Where r =2 cm,h =10 cm

S = 2×(3.14)×2(2 +10) =150.72 cm² = (150.72 ×10²)mm³ = 15072 mm³

(c)A vehicle moving with a speed of 18 km/h  means it moves 18 km in 1 h

1 h = 60×60 =3600 s and 1 km = 1000 m

18 km/h = [(18×1000)/(3600)] m/s =5 m/s

Therefore the distance covered by the vehicle in 1 second is 5 m

(d) The relative density of lead is 11.3.Its density is ….g /cm or….kg/m³

The relative density of a substance = density of a substance/density of water

The relative density of  lead = density of lead/density of water

The density of lead = the relative density of lead× density of the water =11.3 ×(10³kg/m³) =11.3×10³kg m-3

Since 1 kg = 10³ gm and 1m = 100 cm

11.3×10³kg m-3

= 11.3×10³×10³×(10²)-3

= 11.3 gm cm-3

Hence density of lead is 11.3 gm cm-3 or11.3×10³kg m-3

Q2.2-Fill in the blanks by suitable conversion of units

(a) 1 kg m² s-2 =…..g cm² s-2

(b) 1m = ….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

Ans.

(a) 1 kg m² s-2 = (1000 g)(10²)² cm²s-2 =107 g cm² s-2

(b) Since 1 ly = 9.46×1015 m

Therefore 1 m = 1/( 9.46×1015 )ly ,9.46≈10

Hence 1 m = (1/1016) ly

1m= 10-16ly

(c) 3.0 m s-2 = 3×(10-3 ) km×(60×60)2 h-2

= 3×(3600×3600)×10-3  km s-2

= 3888 ×104 ×10-3  km s-2

= 38880 km s-2= 3.888×10-4 km s-2

(d) Since 1 N =1 Kg  ms-2

G = 6.67 × 10-11 N m2 (kg)-2 = 6.67 x 10-11 ×1 Kg  ms-2m2 (kg)-2 = 6.67 x 10-11  (Kg)-1 m³s-2

=6.67 × 10-11  (10³)-1g-1×(10²)³cm³ s-2

=6.67 ×10-11 × 10-3×106 (cm)3 s-2 g-1

=6.67 ×10-8 (cm)3 s-2 g-1

Q 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 α kg, the unit of length equals β 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.

Ans. The relationship between given unit and new unit is given by

New unit/Given unit = (M1/M2) a(L1/L2) b(T1/T2) c

The dimensional formula for energy is =M1L2T-2

Putting a =1, b = 2 and c = -2

New unit/Given unit=(M1/M2) 1(L1/L2) 2(T1/T2) -2

New unit/4.2 J = (M1/M2) 1(L1/L2) 2(T1/T2) -2

Where M1=1 kg, M2= α kg,L1=1 m,L2=β m,T1=1s,T2=ys

New unit = 4.2(1/α ) (1/β )2(1/y)-2

New unit = 4.2 α-1β-2 y2

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.

Ans.(a) Atoms are very small as compared to the tip of a needle

(b) The speed of a jet plane is very high as compared to the speed of a bus

(c) The mass of the jupitor is very high as compared to the mass of the earth

(d) The air inside this room contains large number of molecules as compared to air molecules inside a inflated baloon.

(e) A proton is much more massive than an electron(it is a complete statement)

(f)The speed of sound is much smaller than the speed of light.(it is a complete statement)

Q 2. 5. A new unit of length is chosen such that the speed of light in a vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

Ans. In the new unit of length, the speed of light is given= 1 unit

The distance between the earth and the sun is = the speed of the light ×time taken by the light to reach the earth

The distance between the earth and and the sun is = 1×(8 min +20 s) =(8×60 +20)  =500 units

Question 2. 6. Which of the following is the most precise device for measuring length:
(a) vernier calipers 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?

Ans. (a) The least count of the vernier calipers is = 1/20 = 0.05 mm =0.05 ×10-3 = 5×10-5m

(b)The least count of the screw gauge is = pitch/no. of division =1/100 = 0.01 mm = 0.01 ×10-3 = 10-5m

(c) The least count of an optical instrument =wavelength of the light =6000Å = 6000×10-10m=6×10-7m

Since the least count of an optical instrument is the least therefore it is the most precise device for measuring length.

Q 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?

Ans. The average width of the hair observed by the microscope is = 3.5 mm

Magnification strength of the microscope = 100

Magnification,m = Image of the thickness of the hair/Real thickness of the hair

100 = 3.5/Real thickness of the hair

Real thickness of the hair = 3.5/100 = 0.035 mm

Q 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 calipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?

Ans.(a) Wrap the thread around a round pencil number of times along its length in such a way that width of the thread are in contact with each other, measuring the length of the portion of the pencil wrapped by the thread with the help of the meter scale and counting the number of turns the thread is wrapped.

Let the number of the turns of the thread around the pencil be = n and the length of the coil is l

The width of the thread = diameter of the thread =l/n

(b) Since least count measurement of the screwguage = pitch/no. of division =1mm/200 =0.005 mm,if we increase the number of division, then least count measurement will be decreased which ensure an increase in the accuracy of the screwguage but it has a limit due to the low resolution of the humane eye.

(c) The number of observations increases yield more accurate measurement because there are more chances that the number of positive random errors cancels the number of negative errors. Therefore a set of 100 measurements of the diameter is expected to yield a more reliable estimate than a set of 5 measurements only.

Q 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?

Ans. Areal magnification = Area on the screen/Area on the slide

Area of the house on the slide =1.75 cm2 =1.75×(10-2)² m² = 1.75 ×10-4

Area of the house on the screen =1.55 m2

Areal magnification =1.55/1.75 ×10-4=0.885×104= 8.885 ×103

Linear magnification =√(areal  magnification) =√(8.885 ×103) =94.1

Q 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

Ans. (a) 1 (b) 3 (c) 4 (d) 4 (e) 4 (f) 4.

Q2 .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.

Ans. Area of the rectangular sheet = 2(Length ×Breadth+Breadth×Height+Height ×Length)

Area of the rectangular sheet = 2(4.234 ×1.005+ 1.005×0.0201+0.0201 ×4.234 )

=2(4.25517+0.0202005+0.0851034) = 2×4.360 =8.72 m²

Area of the rectangle has three significant figures(i.e 8,7 and 2)

Volume of the rectangular sheet = Length×Breadth ×Height = 4.234 × 1.005  ×0.0201 =0.0855m³

The volume of the rectangle has three significant figures(i.e 8,5 and 5)

Q2. 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?

Ans. (a) The mass of the box = 2.3 kg and the mass of the gold pieces are 20.15 g(0.02015 kg) and 20.17 g(0.02017kg)

The total mass of the box = (2.3 + 0.02015+ 0.02017) kg=2.34032 kg

The fig.of least decimal places is 2.3 (i.e only 1 place)

Therefore the total mass of the box is 2,3 kg

Difference between two pieces =20.17 – 20.15 = 0.02 g

Since the least number of decimal places is 2 therefore the required difference in the masses of the pieces is 0.02 g

Q2. 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?

Ans.

The given physical quantity is

Maximum percentage error in the quantity P is given by

The value of P is 3.763

Since the error lies in the first decimal place,therefore let’s round off the value of P upto one decimal place i.e 3.8

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