Chapter – 1
Unit And Measurement 1
2.1 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.
Answer:
(a) Volume of cube, V = (1 cm)3 = (10-2 m)3 = 10-6 m3
(b) Surface area = curved area + area on top /base = 2πrh + 2πr2 = 2πr (h + r)r = 2 cm = 20 mmh = 10 cm = 100 mmSurface area = 2πr (h + r) = 2 x 3.14 x 20 (100 + 20) = 15072 mm2Hence, the answer is 15072 mm2
(c) Speed of vehicle = 18 km/h1 km = 1000 m1 hr = 60 x 60 = 3600 s1 km/hr = 1000 m/3600 s = 5/18 m/s18 km/h = = (18 x 1000)/3600= 5 m/sDistance travelled by the vehicle in 1 s = 5 m
(d) The relative density of lead is 11.3 g cm-3=> 11.3 x 103 kg m-3 [1 kilogram = 103g, 1 meter = 102 cm]=> 11.3 x 103 kg m-4
2.2 Fill in the blanks by suitable conversion of units.
(a) 1 kg m2 s–2 = ….g cm2 s–2
(b) 1 m = ….. ly
(c) 3.0 m s–2 = …. km h–2
(d) G = 6.67 × 10–11 N m2 (kg)–2 = …. (cm)3s–2 g–1
Answer:
(a) 1 kg m2 s–2 = ….g cm2 s–21 kg m2 s-2 = 1kg x 1m2 x 1s -2We know that,1kg = 1031m = 100cm = 102cmWhen the values are put together, we get1kg x 1m2 x 1s-2 = 103g x (102cm)2 x 1s-2 = 103g x 104 cm2 x 1s-2 = 107 gcm2s-2=>1kg m2 s-2 = 107 gcm2s-2
(b) 1 m = ….. lyUsing the formula,Distance = speed x timeSpeed of light = 3 x 108 m/sTime = 1 yr = 365 days = 365 x 24 hr = 365 x 24 x 60 x 60 secPut these values in the formula mentioned above, and we getOne light year distance = (3 x 108 m/s) x (365 x 24 x 60 x 60) = 9.46×1015m9.46 x 1015 m = 1lySo that, 1m = 1/9.46 x 1015ly=> 1.06 x 10-16ly=>1 meter = 1.06 x 10-16ly
(c) 3.0 m s–2 = …. km h–21 km = 1000m so that 1m = 1/1000 km3.0 m s-2 = 3.0 (1/1000 km) (1/3600 hour) -2 = 3.0 x 10-3 km x ((1/3600)-2h-2)= 3 x 10-3km x (3600)2 hr-2 = 3.88 x 104 km h-2
=> 3.0 m s -2 = 3.88 x 104 km h-2
(d) G = 6.67 × 10–11 N m2 (kg)–2 = …. (cm)3s–2 g–1G = 6.67 x 10-11 N m2 (kg)-2We know that,1N = 1kg m s-21 kg = 103 g1m = 100cm= 102 cmPut the values together, we get
=> 6.67 x 10-11 Nm 2 kg -2 = 6.67 x 10-11 x (1 kg m s -2) (1 m2) (1 kg -2)Solve the following, and cancelling out the units, we get=> 6.67 x 10-11 x (1 kg -1 x 1 m3 x 1 s -2)Put the above values together to convert kg to g and m to cm.
=> 6.67 x 10-11 x (103g)-1 x (102 cm)3 x (1s-2)
=> 6.67 x 10-8 cm3 s-2 g -1
=>G = 6.67 x 10-11 Nm 2(kg)-2
= 6.67 x 10-8 (cm)3 s -2 g -1
2.3 A calorie is a unit of heat (energy in transit), which 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 α kg, the unit of length equals β m, and the unit of time is γ s. Show that a calorie has a magnitude of 4.2 α–1 β–2 γ2 in terms of the new units.
Answer
1 calorie = 4.2 J = 4.2 kg m2 s–2The standard formula for the conversion is Dimensional formula for energy = Here, x = 1, y = 2 and z =- 2M1 = 1 kg, L1 = 1m, T1 = 1sand M2 = α kg, L2 = β m, T2 = γ s Calorie = 4.2 α–1 β–2 γ2
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.(a) Atoms are small object
Answer:
(a) In comparison with a soccer ball, atoms are very small
(b) When compared with a bicycle, a jet plane travels at high speed.
(c) When compared with the mass of a cricket ball, the mass of Jupiter is very large.
(d) As compared with the air inside a lunch box, the air inside the room has a large number of molecules.
(e) A proton is massive when compared with an electron.
(f) Like comparing the speed of a bicycle and a jet plane, the speed of light is more than the speed of sound.
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?
Answer:
Distance between them = Speed of light x Time taken by light to cover the distance Speed of light = 1 unit Time taken = 8 x 60 + 20 = 480 + 20 = 500sThe distance between Sun and Earth = 1 x 500 = 500 units
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
Answer:
(a) Least count = 1- = 0.01cm
(b) Least count = = = 0.001 cm
(c) least count = wavelength of light = 10-5 cm= 0.00001 cm We can come to the conclusion that the optical instrument is the most precise device used to measure length.
2.7. A student measures the thickness of a human hair by looking at it through am icroscope 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 the hair?
Answer
Magnification of the microscope = 100The average width of the hair in the field of view of the microscope = 3.5 mm Actual thickness of hair =3.5 mm/100 = 0.035 mm
2. 8. Answer the following :
(a) You are given a thread and a metre scale. How will you estimate the diameter ofthe 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 arbitrarily increase the accuracy of the screw gauge 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. Whyis a set of 100 measurements of the diameter expected to yield a more reliableestimate than a set of 5 measurements only?
Answer
(a) The thread should be wrapped around a pencil a number of times to form a coil having its turns touching each other closely. Measure the length of this coil with a metre scale. If L be the length of the coil and n be the number of turns of the coil, then the diameter of the thread is given by the relation Diameter = L/n.
(b) Least count of the screw gauge = Pitch/number of divisions on the circular scaleSo, theoretically, when the number of divisions on the circular scale is increased, the least count of the screw gauge will decrease. Hence, the accuracy of the screw gauge will increase. However, this is only a theoretical idea. Practically, there will be many other difficulties when the number of turns is increased.
(c) The probability of making random errors can be reduced to a larger extent in 100 observations than in the case of 5 observations.
2.9 . The photograph of a house occupies an area of 1.75 cm 2 on a 35 mm slide. The slide is projected onto 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
Arial Magnification = Area of the image/Area of the object= 1.55/1.75 x 104= 8.857x 103Linear Magnification = √Arial magnification= √8.857x 103 = 94. 1
2.10 State the number of significant figures in the following:
(a) 0.007 m2
(b) 2.64 × 1024 kg
(c) 0.2370 g cm–
(d) 6.320 J
(e) 6.032 N m–2
(f) 0.0006032 m 2
Answer:
(a) 0.007 m 2 The given value is 0.007 m2.Only one significant digit. It is 7.
(b) 2.64 × 1024 kgAnswer:The value is 2.64 × 1024 kgFor the determination of significant values, the power of 10 is irrelevant. The digits 2, 6, and 4 are significant figures. The number of significant digits is 3.
(c) 0.2370 g cm–3Answer:The value is 0.2370 g cm–3For the given value with decimals, all the numbers 2, 3, 7, and 0 are significant. The 0 before the decimal point is not significant
(d) All the numbers are significant. The number of significant figures here is 4.
(e) 6, 0, 3, and 2 are significant figures. Therefore, the number of significant figures is 4.
(f) 6, 0, 3, and 2 are significant figures. The number of significant figures is 4.
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.
Answer
Area of the rectangular sheet = length x breadth= 4.234 x 1.005 = 4.255 m2= 4.3 m2The volume of the rectangular sheet = length x breadth x thickness = 4.234 x 1.005 x 2.01 x 10-2 = 8.55 x 10-2 m3.
2.12 The mass of a box measured by a grocer’s balance is 2.30 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:
The mass of the box = 2.30 k gand the mass of the first gold piece = 20.15 gThe mass of the second gold piece = 20.17 gThe total mass = 2.300 + 0.2015 + 0.2017 = 2.7032 kg Since 1 is the least number of decimal places, the total mass = 2.7 kg.The mass difference = 20.17 – 20.15 = 0.02 g Since 2 is the least number of decimal places, the total mass = 0.02 g.