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NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements is designed and prepared by the best teachers across India. All the important topics are covered in the exercises and each answer comes with a detailed explanation to help students understand concepts better. These NCERT solutions play a crucial role in your preparation for all exams conducted by the CBSE, including the JEE.
NCERT TEXTBOOK QUESTIONS SOLVED
1. Fill in the 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 2.0 cm and height 10.0 cm is equal to ……..(mm)2.
(c) A vehicle moving with a speed of 18 km h-1 covers ………. m in 1 s.
(d) The relative density of lead is 11.3. Its density is …….. g cm-3 or ………. kg m-3.
Ans. (a) Volume of cube, V = (1 cm)3 = (10-2 m)3 10-6 m3.
Hence, answer is 10-6
(b) Surface area = 2Ï€rh + 2Ï€r2 = 2Ï€r (h + r)
= 2 x 22/7 x 2 x 10 (10 x 10 + 2 x 10) mm2 = 1.5 x 104 mm2 Hence, 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 103 kg m-3 [1 kg =103 g,1m=102 cm]
=11.3 x 103 kg m-4
Ans. Distance between Sun and Earth
= Speed of light in vacuum x time taken by light to travel from Sim to Earth = 3 x 108 m/ s x 8 min 20 s = 3 x 108 m/s x 500 s = 500 x 3 x 108 m.
In the new system, the speed of light in vacuum is unity. So, the new unit of length is 3 x 108 m.
.•. 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?
Ans. (a) Least count of vernier callipers = 1/20 = 0.05 mm = 5 x 10-5 m
(b) Least count of screw gauge =Pitch/No. of divisions on circular scale = 1 x 10-3/100 = 1 x 10-5 m
(c) Least count of optical instrument = 6000 A (average wavelength of visible light as 6000 A) = 6 x 10-7m As the least count of optical instrument is least, it is the most precise device out of three instruments given to us.
Ans. As magnification, m =thickness of image of hair/ real thickness of hair = 100
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?
Ans. (a) Wrap the thread a number of times on a round pencil so as to form a coil having its turns touching each other closely. Measure the length of this coil, mode by the thread, with a metre scale. If n be the number of turns of the coil and l be the length of the coil, then the length occupied by each single turn i.e., the thickness of the thread = 1/n .
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 increasd, 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.
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