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Units And Measurements (Physics) Question 2.1: Fill in the blanks The volume of a cube of side cm is equal to m3 The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to (mm)2 A vehicle moving with a speed of 18 km h–1covers m in s N Ph C ys ER Pa ic T rt s X -1 I The relative density of lead is 11.3 Its density is g cm–3or kg m–3 Answer cm = Volume of the cube = cm3 But, cm3 = cm × cm × cm = ∴1 cm3 = 10–6 m3 Hence, the volume of a cube of side cm is equal to 10–6 m3 The total surface area of a cylinder of radius r and height h is S = 2πr (r + h) Given that, r = cm = × cm = × 10 mm = 20 mm h = 10 cm = 10 × 10 mm = 100 mm = 15072 = 1.5 × 104 mm2 Using the conversion, km/h = Therefore, distance can be obtained using the relation: Distance = Speed × Time = × = m Hence, the vehicle covers m in s N Ph C ys ER Pa ic T rt s X -1 I Relative density of a substance is given by the relation, Relative density = Density of water = g/cm3 Again, 1g = cm3 = 10–6 m3 g/cm3 = ∴ 11.3 g/cm3 = 11.3 × 103 kg/m3 Question 2.2: Fill in the blanks by suitable conversion of units: kg m2s–2= g cm2 s–2 m = ly 3.0 m s–2= km h–2 G= 6.67 × 10–11 N m2 (kg)–2= (cm)3s–2 g–1 Answer kg = 103 g m2 = 104 cm2 kg m2 s–2 = kg × m2 × s–2 N Ph C ys ER Pa ic T rt s X -1 I =103 g × 104 cm2 × s–2 = 107 g cm2 s–2 Light year is the total distance travelled by light in one year ly = Speed of light × One year = (3 × 108 m/s) × (365 × 24 × 60 × 60 s) = 9.46 × 1015 m m = 10–3 km Again, s = s–1 = 3600 h–1 s–2 = (3600)2 h–2 ∴3 m s–2 = (3 × 10–3 km) × ((3600)2 h–2) = 3.88 × 10–4 km h–2 N = kg m s–2 kg = 10–3 g–1 m3 = 106 cm3 ∴ 6.67 × 10–11 N m2 kg–2 = 6.67 × 10–11 × (1 kg m s–2) (1 m2) (1 s–2) = 6.67 × 10–11 × (1 kg × m3 × s–2) = 6.67 × 10–11 × (10–3 g–1) × (106 cm3) × (1 s–2) = 6.67 × 10–8 cm3 s–2 g–1 Question 2.3: A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = kg m2s–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 γ s Show that a calorie has a magnitude 4.2 α–1 β–2 γ2 in terms of the new units N Ph C ys ER Pa ic T rt s X -1 I Answer Given that, calorie = 4.2 (1 kg) (1 m2) (1 s–2) New unit of mass = α kg ew unit, kg = Hence, in terms of the new In terms of the new unit of length, And, in terms of the new unit of time, ∴1 calorie = 4.2 (1 α–1) (1 β–22) (1 γ2) = 4.2 α–1 β–2 γ2 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: atoms are very small objects a jet plane moves with great speed the mass of Jupiter is very large N Ph C ys ER Pa ic T rt s X -1 I the air inside thiss room contains a large number of molecules a proton is much more massive than an electron the speed of sound is much smaller than the speed of light Answer 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 An atom is a very small object in comparison com to a soccer ball A jet plane moves with a speed greater than that of a bicycle Mass of Jupiter is very large as compared to the mass of a cricket ball The air inside this room contains a large number of molecules as compared to that present in a geometry box A proton is more massive than an electron Speed of sound is less than the speed of light Question 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 Ea Earth rth in terms of the new unit if light takes and 20 s to cover this distance? Answer Distance between the Sun and the Earth: = Speed of light × Time taken by light to cover the distance N Ph C ys ER Pa ic T rt s X -1 I Given that in the new unit, speed of light = unit Time taken, t = 20 s = 500 s ∴Distance Distance between the Sun and the Earth = × 500 = 500 units Question 2.6: Which of the following is the most precise device for measuring length: a vernier callipers with 20 divisions on the sliding scale a screw gauge of pitch mm and 100 divisions on the circular scale an optical instrument that can measure length to within a wavelength of light ? Answer A device with minimum count is the most suitable to measure length Least count of vernier callipers = standard division (SD) – vernier division (VD) Least count of screw gauge = Least count of an optical device = Wavelength of light ∼ 10–5 cm = 0.00001 cm Hence, it can be inferred that an optical instrument is the most suitable device to measure length N Ph C ys ER Pa ic T rt s X -1 I Question 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? Answer Magnification of the microscope = 100 Average width of the hair in the field of view of the microscope = 3.5 mm ∴Actual Actual thickness of the hair is = 0.035 mm Question 2.8: Answer the following: You are given a thread and a metre scale How will you estimate the diameter of the thread? 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? 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 tthan a set of measurements only? Answer N Ph C ys ER Pa ic T rt s X -1 I 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, 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 A set of 100 measurements asurements is more reliable than a set of measurements because random errors involved in the former are very less as compared to the latter Question 2.9: The photograph of a house occupies an area of 1.75 cm2on 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 projector arrangement? Answer Area of the house on the slide = 1.75 cm2 Area of the image of the house formed on the screen = 1.55 m2 = 1.55 × 104 cm2 Arial magnification, ma = ∴Linear magnifications, ml = Question 2.10: N Ph C ys ER Pa ic T rt s X -1 I State the number of significant figures in the following: 0.007 m2 2.64 × 1024 kg 0.2370 g cm–3 6.320 J 6.032 N m–2 0.0006032 m2 Answer Answer: The given quantity is 0.007 m2 If the number is less than one, then all zeros on the right of the decimal point (but left to the first non-zero) zero) are insignificant This means that here, two zeros after the decimal are not significant Hence, onlyy is a significant figure in this quantity Answer: The given quantity is 2.64 × 1024 kg Here, the power of 10 is irrelevant for the determination of significant figures Hence, all digits i.e., 2, and are significant figures Answer: The given quantity is 0.2370 g cm–3 For a number with decimals, the trailing zeroes are significant Hence, besides digits 2, and 7, that appears after the decimal point is also a significant figure Answer: The given quantity is 6.320 J For a number with decimals, cimals, the trailing zeroes are significant Hence, all four digits appearing in the given quantity are significant figures N Ph C ys ER Pa ic T rt s X -1 I Answer: The given quantity is 6.032 Nm–2 All zeroes between two non-zero zero digits are always significant Answer: The given quantity is 0.0006032 m2 If the number is less than one, then the zeroes on the right of the decimal point (but left to the first non-zero) zero) are insignificant Hence, all three zeroes appearing before are not significant figures All zeros between two non-zero no zero digits are always significant Hence, the remaining four digits are significant figures Question 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 Length of sheet, l = 4.234 m Orbital radius of the Earth around the Sun, r = 1.5 × 1011 m Time taken by the Earth to complete one revolution around the Sun, T = year = 365.25 days = 365.25 × 24 × 60 × 60 s Universal gravitational constant, G = 6.67 × 10–11 Nm2 kg–2 N Ph C ys ER Pa ic T rt s X -1 I Thus, mass of the Sun can be calculated using the relation, Hence, the mass of the Sun is × 1030 kg Question 8.14: A Saturn year is 29.5 times the earth year How far is the Saturn from the sun if the earth is 1.50 ×108 km away from the sun? Answer Distance of the Earth from the Sun, re = 1.5 × 108 km = 1.5 × 1011 m Time period of the Earth = Te Time period of Saturn, Ts = 29 Te Distance of Saturn from the Sun = rs From Kepler’s third law of planetary motion, we have N Ph C ys ER Pa ic T rt s X -1 I For Saturn and Sun, we can write Hence, the distance between Saturn and the Sun is Question 8.15: A body weighs 63 N on the surface of the earth What is the gravitational force on it due to the earth at a height equal to half the radius of the earth? Answer Weight of the body, W = 63 N Acceleration due to gravity at height h from the Earth’s surface is given by the relation: Where, g = Acceleration due to gravity on the Earth’s surface Re = Radius of the Earth N Ph C ys ER Pa ic T rt s X -1 I Weight of a body of mass m at height h is given as: Question 8.16: Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface? Answer Weight of a body of mass m at the Earth’s surface, W = mg = 250 N Body of mass m is located at depth, Where, = Radius of the Earth Acceleration due to gravity at depth g (d) is given by the relation: N Ph C ys ER Pa ic T rt s X -1 I Weight of the body at depth d, d Question 8.17: A rocket is fired vertically with a speed of km s–1 from the earth’s surface How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G= 6.67 × 10–11 N m2 kg–2 Answer Answer: × 106 m from the centre of the Earth Velocity of the rocket, v = km/s = × 103 m/s Mass of the Earth, Radius of the Earth, Height reached by rocket mass, m = h At the surface of the Earth, Total energy of the rocket = Kinetic energy + Potential ene energy At highest point h, Total energy of the rocket From the law of conservation of energy, we have N Ph C ys ER Pa ic T rt s X -1 I Total energy of the rocket at the Earth’s surface = Total energy at height h Height achieved by the rocket with respect to the centre of the Earth Question 8.18: The escape speed of a projectile on the earth’s surface is 11.2 km s–1 A body is projected out with thrice this speed What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets N Ph C ys ER Pa ic T rt s X -1 I Answer Escape velocity of a projectile from the Earth, vesc = 11.2 km/s Projection velocity of the projectile, vp = 3vesc Mass of the projectile = m Velocity of the projectile far away from the Earth = vf Total energy of the projectile on the Earth Gravitational potential energy of the projectile far away from the Earth is zero Total energy of the projectile far away from the Earth = From the law of conservation of energy, we have Question 8.19: A satellite orbits the earth at a height of 400 km above the surface How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0 ×1024 kg; radius of the earth = 6.4 ×106 m; G = 6.67 × 10–11 N m2 kg–2 Answer N Ph C ys ER Pa ic T rt s X -1 I Mass of the Earth, M = 6.0 × 1024 kg Mass of the satellite, m = 200 kg Radius of the Earth, Re = 6.4 × 106 m Universal gravitational constant, G = 6.67 × 10–11 Nm2kg–2 Height of the satellite, h = 400 km = × 105 m = 0.4 ×106 m Total energy of the satellite at height h Orbital velocity of the satellite, v = Total energy of height, h The negative sign indicates that the satellite is bound to the Earth This is called bound energy of the satellite Energy required to send the satellite out of its orbit = – (Bound energy) Question 8.20: N Ph C ys ER Pa ic T rt s X -1 I Two stars each of one solar mass (= 2× 1030 kg) are approaching each other for a head on collision When they are a distance 109 km, their speeds are negligible What is the speed with which they collide? The radius of each star is 104 km Assume the stars to remain undistorted until they collide (Use the known value of G) Answer Mass of each star, M = × 1030 kg Radius of each star, R = 104 km = 107 m Distance between the stars, r = 109 km = 1012m For negligible speeds, v = total energy of two stars separated at distance r Now, consider the case when the stars are about to collide: Velocity of the stars = v Distance between the centers of the stars = 2R Total kinetic energy of both stars Total potential energy of both stars Total energy of the two stars = N Ph C ys ER Pa ic T rt s X -1 I Using the law of conservation of energy, we can write: Question 8.21: Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table What is the gravitational force and potential at the mid point of the line joining the centers of the spheres? Is an object placed at that point in equilibrium? equilibrium? If so, is the equilibrium stable or unstable? Answer Answer: 0; –2.7 × 10–8 J /kg; Yes; Unstable Explanation: The situation is represented in the given figure: Mass of each sphere, M = 100 kg Separation between the spheres, r = 1m N Ph C ys ER Pa ic T rt s X -1 I X is the mid point between the spheres Gravitational force at point X will be zero This is because gravitational force exerted by each sphere will act in opposite directions Gravitational potential at point X: Any object placed at point X will be in equilibri equilibrium um state, but the equilibrium is unstable This is because any change in the position of the object will change the effective force in that direction Question 8.22: As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth What is the potential due to earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero) Mass of the earth = 6.0 × 1024 kg, radius = 6400 km Answer Mass of the Earth, M = 6.0 × 1024 kg Radius of the Earth, R = 6400 km = 6.4 × 106 m Height of a geostationary satellite from the surface of the Earth, h = 36000 km = 3.6 × 107 m N Ph C ys ER Pa ic T rt s X -1 I Gravitational potential energy due to Earth’s gravity at height h, Question 8.23: A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev per second (Extremely compact stars of this kind are known as neutron stars Certain stellar objects called pulsars belong to this category) category) Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the sun = × 1030 kg) Answer Answer: Yes A body gets stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal rifugal force caused by the rotation of the star Gravitational force, fg Where, M = Mass of the star = 2.5 × × 1030 = × 1030 kg m = Mass of the body R = Radius of the star = 12 km = 1.2 ×104 m Centrifugal force, fc = mrω2 ω = Angular speed = 2πν ν = Angular frequency = 1.2 rev s–1 fc = mR (2πν)2 N Ph C ys ER Pa ic T rt s X -1 I = m × (1.2 ×104) × × (3.14)2 × (1.2)2 = 1.7 ×105m N Since fg > fc, the body will remain stuck to the surface of the star Question 8.24: A spaceship is stationed on Mars How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the Sun = × 1030 kg; mass of mars = 6.4 × 1023 kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 108kg; G= 6.67 × 10–11 m2kg–2 Answer Mass of the spaceship, ms = 1000 kg Mass of the Sun, M = × 1030 kg Mass of Mars, mm = 6.4 × 10 23 kg Orbital radius of Mars, R = 2.28 × 108 kg =2.28 × 1011m Radius of Mars, r = 3395 km = 3.395 × 106 m Universal gravitational constant, G = 6.67 × 10–11 m2kg–2 Potential energy of the spaceship due to the gravitational attraction of the Sun Potential energy of the spaceship due to the gravitational attraction of Mars Since the spaceship is stationed on Mars, its velocity and hence, its kinetic energy will be zero Total energy of the spaceship N Ph C ys ER Pa ic T rt s X -1 I The negative sign indicates that the system is in bound state Energy required for launching the spaceship out of the solar system = – (Total energy of the spaceship) Question 8.25: s–1 If 20% of A rocket is fired ‘vertically’ from the surface of mars with a speed of km s– its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4× 1023 kg; radius of mars = 3395 km; G = 6.67× 10-11 N m2 kg–2 Answer Initial velocity of the rocket, v = km/s = × 103 m/s Mass of Mars, M = 6.4 × 1023 kg Radius of Mars, R = 3395 km = 3.395 × 106 m Universal gravitational constant, G = 6.67× 10–11 N m2 kg–2 Mass of the rocket = m Initial kinetic energy of the rocket = N Ph C ys ER Pa ic T rt s X -1 I Initial potential energy of the rocket Total initial energy If 20 % of initial kinetic energy is lost due to Martian atmospheric resistance, then only 80 % of its kinetic energy helps in reaching a height Total initial energy available Maximum height reached by the rocket = h At this height, the velocity and hence, the kinetic energy of the rocket will become zero Total energy of the rocket at height h Applying the law of conservation of energy for the rocket, we can write: N Ph C ys ER Pa ic T rt s X -1 I