{"id":197405,"date":"2021-03-02T12:34:39","date_gmt":"2021-03-02T12:34:39","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=197405"},"modified":"2023-09-20T08:04:53","modified_gmt":"2023-09-20T08:04:53","slug":"ncert-solutions-for-11th-class-physics-chapter-2-units-and-measurements","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-11th-class-physics-chapter-2-units-and-measurements\/","title":{"rendered":"NCERT Solutions for 11th Class Physics: Chapter 2-Units and Measurements"},"content":{"rendered":"\n
Class 11: Physics Chapter 2 solutions. Complete Class 11 Physics Chapter 2 Notes.<\/p>\n\n\n\n
NCERT Solutions for 11th Class Physics: Chapter 2-Units and Measurements<\/strong><\/h2>\n\n\n\n
2.1. Fill in the blanks<\/strong> (a) The volume of a cube of side 1 cm is equal to…..m3<\/sup> (b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to … (mm)2<\/sup> (c) A vehicle moving with a speed of 18 km h\u20131<\/sup>covers….m in 1 s (d) The relative density of lead is 11.3. Its density is ….g cm\u20133<\/sup>or . …kg m\u20133<\/sup>.<\/strong><\/p>\n\n\n\n
Answer<\/strong><\/p>\n\n\n\n
(a) Length of edge = 1cm = 1\/100 m Volume of the cube = side3<\/sup> Putting the value of side, we get Volume of the cube = (1\/100 m)3<\/sup> <\/p>\n\n\n\n<\/figure>\n\n\n\n
The volume of a cube of side 1 cm is equal to 10<\/p>\n\n\n\n
-6<\/sup> m3<\/sup> (b) Given, Radius, r<\/em> = 2.0 cm = 20 mm (convert cm to mm) Height, h<\/em> = 10.0 cm =100 mm The formula of total surface area of a cylinder S = 2\u03c0r<\/em> (r<\/em> + h<\/em>) Putting the values in this formula, we get Surface area of a cylinder S = 2\u03c0r<\/em> (r<\/em> + h<\/em> = 2 x 3.14 x 20 (20+100) = 15072 = 1.5 \u00d7 104<\/sup> mm2<\/sup> The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to 1.5 \u00d7 10<\/p>\n\n\n\n
4<\/sup> mm2<\/sup><\/p>\n\n\n\n
(c) Using the conversion, Given, Time, t = 1 sec speed = 18 km h-1<\/sup> = 18 km \/ hour 1 km = 1000 m and 1hour = 3600 sec Speed = 18 \u00d7 1000 \/3600 sec = 5 m \/sec Use formula Speed = distance \/ time Cross multiply it, we get Distance = Speed \u00d7 Time = 5 \u00d7 1 = 5 m A vehicle moving with a speed of 18 km h\u20131<\/sup>covers 5<\/p>\n\n\n\n
m in 1 s.<\/p>\n\n\n\n
(d) Density of lead = Relative density of lead \u00d7 Density of water Density of water = 1 g\/cm3<\/sup> Putting the values, we get Density of lead = 11.3 \u00d7 1 g\/ cm3<\/sup> = 11.3 g cm-3<\/sup> 1 cm = (1\/100 m) =10\u20132<\/sup> m3 1 g = 1\/1000 kg = 10-3<\/sup> kg Density of lead = 11.3 g cm-3<\/sup> = 11.3 Putting the value of 1 cm and 1 gram 11.3 g\/cm3<\/sup> = 11.3 \u00d7 10-3<\/sup> kg (10-2<\/sup>m)-3<\/sup> = 11.3 \u00d710\u20133<\/sup> \u00d7 106<\/sup> kg m-3<\/sup> =1.13 \u00d7 103<\/sup> kg m\u20133<\/sup> The relative density of lead is 11.3. Its density is 11.3<\/p>\n\n\n\n
g cm-3 <\/sup>g cm\u20133<\/sup> or 1.13 \u00d7 10<\/p>\n\n\n\n
2.2. Fill in the blanks by suitable conversion of units:<\/strong> (a) 1 kg m2<\/sup>s\u20132<\/sup>= ….g cm2<\/sup> s\u20132<\/sup> (b) 1 m =….. ly (c) 3.0 m s\u20132<\/sup>=…. km h\u20132<\/sup> (d) G= 6.67 \u00d7 10\u201311<\/sup> N m2<\/sup> (kg)\u20132<\/sup>=…. (cm)3<\/sup>s\u20132<\/sup> g\u20131<\/sup>.<\/strong><\/p>\n\n\n\n
Answer<\/strong><\/p>\n\n\n\n
(a) 1 kg = 103<\/sup> g 1 m2<\/sup> = 104<\/sup> cm2<\/sup> 1 kg m2<\/sup> s\u20132<\/sup> = 1 kg \u00d7 1 m2<\/sup> \u00d7 1 s\u20132<\/sup> =103<\/sup> g \u00d7 104<\/sup> cm2 <\/sup>\u00d7 1 s\u20132<\/sup> = 107<\/sup> g cm2<\/sup> s\u20132<\/sup> 1 kg m2<\/sup>s\u20132<\/sup>= 10<\/p>\n\n\n\n
7<\/sup> g cm2<\/sup> s\u20132<\/sup><\/p>\n\n\n\n
(b) Distance = Speed \u00d7 Time Speed of light = 3 \u00d7 108<\/sup> m\/s Time = 1 year = 365 days = 365 \u00d7 24 hours = 365 \u00d7 24 \u00d7 60 \u00d7 60 sec Putting these values in above formula we get 1 light year distance = (3 \u00d7 108<\/sup> m\/s) \u00d7 (365 \u00d7 24 \u00d7 60 \u00d7 60 s) = 9.46 \u00d7 1015<\/sup> m 9.46 \u00d7 1015<\/sup> m = 1 ly So that 1 m = 1\/ 9.46 \u00d7 1015<\/sup> ly = 1.06 \u00d7 10<\/p>\n\n\n\n
-16<\/sup> ly<\/p>\n\n\n\n
(c) 1 hour = 3600 sec so that 1 sec = 1\/3600 hour 1 km = 1000 m so that 1 m = 1\/1000 km 3.0 m s\u20132<\/sup> = 3.0 (1\/1000 km)( 1\/3600 hour)-2<\/sup> = 3.0 \u00d7 10\u20133<\/sup> km \u00d7 ((1\/3600)-2<\/sup> h\u20132) = 3.0 \u00d7 10\u20133<\/sup> km \u00d7 (3600)2<\/sup> h\u20132<\/sup> = 3.88 \u00d7 104<\/sup> km h\u20132<\/sup> 3.0 m s\u20132<\/sup>= 3.88 \u00d7 10<\/p>\n\n\n\n
4<\/sup> km h\u20132<\/sup> (d) Given, G= 6.67 \u00d7 10\u201311<\/sup> N m2<\/sup> (kg)\u20132<\/sup> We know that 1 N = 1 kg m s\u2013<\/sup>2 1 kg = 103<\/sup> g 1 m = 100 cm = 102<\/sup> cm Putting above values, we get 6.67 \u00d7 10\u201311<\/sup> N m2<\/sup> kg\u20132<\/sup> = 6.67 \u00d7 10\u201311<\/sup> \u00d7 (1 kg m s\u20132<\/sup>) (1 m2<\/sup>) (1Kg\u20132<\/sup>) Solve and cancel out the units we get \u21d2 6.67 \u00d7 10\u201311<\/sup> \u00d7 (1 kg\u20131<\/sup> \u00d7 1 m3<\/sup> \u00d7 1 s\u20132<\/sup>) Putting above values to convert Kg to g and m to cm \u21d2 6.67 \u00d7 10\u201311<\/sup> \u00d7 (103<\/sup> g)-1<\/sup> \u00d7 (102<\/sup> cm)3<\/sup> \u00d7 (1 s\u20132<\/sup>) \u21d2 6.67 \u00d7 10\u201311<\/sup> \u00d7 10-3<\/sup> g-1<\/sup> \u00d7 106<\/sup> cm3<\/sup> \u00d7 (1 s\u20132<\/sup>) \u21d2 6.67 \u00d7 10\u20138<\/sup> cm3<\/sup> s\u20132<\/sup> g\u20131<\/sup> G= 6.67 \u00d7 10\u201311<\/sup> N m2<\/sup> (kg)\u20132<\/sup>= 6.67 \u00d7 10<\/p>\n\n\n\n
\u20138<\/sup> (cm)3<\/sup>s\u20132<\/sup> g\u20131<\/sup>. 2.3. A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kg m2<\/sup>s\u20132<\/sup>. Suppose we employ a system of units in which the unit of mass equals \u03b1 kg, the unit of length equals \u03b2 m, the unit of time is \u03b3 s. Show that a calorie has a magnitude 4.2 \u03b1\u20131<\/sup> \u03b2\u20132<\/sup> \u03b32 <\/sup>in terms of the new units.<\/strong><\/p>\n\n\n\n
Answer<\/strong><\/p>\n\n\n\n
Given that, 1 Calorie=4.2 J = 4.2 Kg m2<\/sup> s-2<\/sup> …… (i) As new unit of mass = \u03b1 Kg \u2234 1 Kg = 1\/\u03b1 new unit of mass Similarly, 1 m = \u03b2-1<\/sup> new unit of length 1 s = \u03b3-1<\/sup> new unit of time Putting these values in (i), we get 1 calorie = 4.2 (\u03b1-1<\/sup> new unit of mass) (\u03b2-1<\/sup> new unit of length)2 (\u03b3-1<\/sup> new unit of time)-2 = 4.2 \u03b1-1<\/sup> \u03b2-2<\/sup> \u03b32<\/sup> new unit of energy (Proved)<\/p>\n\n\n\n
2.4. Explain this statement clearly:<\/strong><\/p>\n\n\n\n
\u201cTo call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison\u201d. 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.<\/strong><\/p>\n\n\n\n
Answer<\/strong><\/p>\n\n\n\n
The given statement is true because a dimensionless quantity may be large or small in comparision to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction. (a) An atom is a very small object in comparison to a soccer ball. (b) A jet plane moves with a speed greater than that of a bicycle. (c) Mass of Jupiter is very large as compared to the mass of a cricket ball. (d) The air inside this room contains a large number of molecules as compared to that present in a geometry box. (e) A proton is more massive than an electron. (f) Speed of sound is less than the speed of light.<\/p>\n\n\n\n
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?<\/strong><\/p>\n\n\n\n
Answer<\/strong><\/p>\n\n\n\n
Distance between the Sun and the Earth = Speed of light x Time taken by light to cover the distance Given that in the new unit, speed of light = 1 unit Time taken, t = 8 min 20 s = 500 s \u2234Distance between the Sun and the Earth = 1 x 500 = 500 units<\/p>\n\n\n\n
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?<\/strong><\/p>\n\n\n\n
Answer<\/strong><\/p>\n\n\n\n
(a) Least count of this vernier callipers = 1SD – 1 VD = 1 SD – 19\/20 SD = 1\/20 SD = 1.20 mm = 1\/200 cm = 0.005 cm<\/p>\n\n\n\n
(b) Least count of screw gauge = Pitch\/Number of divisions = 1\/1000 = 0.001 cm.<\/p>\n\n\n\n
(c) Wavelength of light, \u03bb \u2248 10-5<\/sup> cm = 0.00001 cm<\/p>\n\n\n\n
Hence, it can be inferred that an optical instrument is the most suitable device to measure length.<\/p>\n\n\n\n
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?<\/strong><\/p>\n\n\n\n
Magnification of the microscope = 100 Average width of the hair in the field of view of the microscope = 3.5 mm \u2234Actual thickness of the hair is 3.5\/100 = 0.035 mm.<\/p>\n\n\n\n
2.8. Answer the following:<\/strong> <\/p>\n\n\n\n
(a) You are given a thread and a metre scale. How will you estimate the diameter of the thread?<\/strong><\/p>\n\n\n\n
Answer<\/strong><\/p>\n\n\n\n
Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a metre scale. The diameter of the thread is given by the relation, Diameter = Length of thread\/Number of turns<\/p>\n\n\n\n
(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?<\/strong><\/p>\n\n\n\n
Answer<\/strong><\/p>\n\n\n\n
It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number divisions of the circular scale will increase its accuracy to a certain extent only.<\/p>\n\n\n\n
(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?<\/strong><\/p>\n\n\n\n
Answer<\/strong><\/p>\n\n\n\n
A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved in the former are very less as compared to the latter.<\/p>\n\n\n\n
2.9. The photograph of a house occupies an area of 1.75 cm2<\/sup>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<\/sup>. What is the linear magnification of the projector-screen arrangement?<\/strong><\/p>\n\n\n\n
Answer<\/strong><\/p>\n\n\n\n
Area of the house on the slide = 1.75 cm2<\/sup> Area of the image of the house formed on the screen = 1.55 m2 <\/sup>= 1.55 \u00d7 104 cm2<\/sup><\/p>\n\n\n\n<\/figure>\n\n\n\n
2.10. State the number of significant figures in the following:<\/strong> <\/p>\n\n\n\n
(a) 0.007 m2<\/sup><\/strong><\/p>\n\n\n\n
\u25ba 1<\/p>\n\n\n\n
(b) 2.64 x 1024<\/sup> kg <\/strong>\u25ba 3<\/p>\n\n\n\n
(c) 0.2370 g cm-3<\/sup> <\/strong>\u25ba 4<\/p>\n\n\n\n
(d) 6.320 J \u25ba 4<\/p>\n\n\n\n
(e) 6.032 N m-2<\/sup> <\/strong>\u25ba 4<\/p>\n\n\n\n
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.<\/strong><\/p>\n\n\n\n
Answer<\/strong><\/p>\n\n\n\n
Given that, length, l<\/em> = 4.234 m breadth,b = 1<\/em>.005 m thickness, t<\/em> = 2.01 cm = 2.01 \u00d7 10-2<\/sup> m Area of the sheet = 2 (l<\/em> \u00d7 0<\/em> + b<\/em> \u00d7 t<\/em> + t<\/em> \u00d7
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