1 X – SCIENCE BOOK BACK PROBLEMS & SOLUTIONS BY T.SAMPATHKUMAR, M.Sc.,M.Ed., S.M.H.HR.SEC.SCHOOL, SIRKALI.
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X – SCIENCE
BOOK BACK PROBLEMS & SOLUTIONS
BY
T.SAMPATHKUMAR, M.Sc.,M.Ed.,
S.M.H.HR.SEC.SCHOOL, SIRKALI.
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PROBLEMS IN BOOK BACK QUESTIONS
CHEMISTRY – LESSON – 10
1. Molecular mass of Nitrogen is 28. Its atomic mass is 14. Find the atomicity of Nitrogen.
Atomicity = = = 2 Atomicity of nitrogen is 2.
2. Gram molecular mass of Oxygen is 32g. Density of Oxygen is 1.429g/cc. Find the gram molecular volume of Oxygen.
Gram molecular volume of Oxygen = = = 22.4 litre.
3. Calculate the gram molecular mass of water from the values of gram atomic mass of Hydrogen and of Oxygen. Gram atomic mass of Hydrogen = 1gGram atomic mass of Oxygen = 16g H2O = 2(H) + 1(O) Gram molecular mass of water = 2(1) + 1(16) = 2 + 16 = 18g. 4. One mole of any substance contains 6.023 x 1023 particles.
If 3.0115 x 1023 particles are present in CO2 find the number of moles.
Number of moles =
=
= 0.5 mole.
5. Calculate the number of moles in: i) 12.046 x 1023 atoms of Copper ii) 27.95g of Iron iii) 1.51 x 1023 molecules of CO2 i) 12.046 x 1023 atoms of Copper
Number of moles = = = 2 moles
ii) 27.95g of Iron
Number of moles = = = 0.5 moles
iii) 1.51 x 1023 molecules of CO2
Number of moles = = = 0.25 moles
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6. Find the gram molecular mass of the following from the datagiven: i) H2O ii) CO2 iii) NaOH iv) NO2 v) H2SO4
i) H2O Gram molecular mass of water H2O = 2(H) + 1(O) = 2(1) + 1(16) = 2 + 16 = 18g. ii) CO2
Gram molecular mass of CO2 = 1(C) + 2 (O) = 1 (12) + 2(16) = 12 + 32 = 44g iii) NaOH
Gram molecular mass of NaOH = 1(Na) + 1(O) + 1(H) = 1(23) + 1(16) + 1(1) = 23 + 16 + 1 = 40 g iv) NO2
Gram molecular mass of NO2 = 1(N) + 2(O)
= 1(14) + 2(16) = 14 + 32 = 46 g. v) H2SO4
Gram molecular mass of H2SO4 = 2(H) + 1(S) + 4(O) = 2(1) + 1(32) + 4( 16) = 2 + 32 + 64 = 98 g. 7.Complete the table given below:
ELEMENT ATOMIC MASS
MOLECULAR MASS
ATOMICITY
Chlorine 35.5 71
Ozone 48 3
Sulphur 32 8
Ans.
ELEMENT ATOMIC MASS
MOLECULAR MASS
ATOMICITY
Chlorine 35.5 71 2
Ozone 16 48 3
Sulphur 32 256 8
ELEMENT SYMBOL ATOMIC NO. MASS NO.
Hydrogen H 1 1
Carbon C 6 12
Oxygen O 8 16
Nitrogen N 7 14
Sodium Na 11 23
Sulphur S 16 32
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Atomicity =
Atomicity of chlorine = = = 2
Atomic mass of ozone = = = 16g.
Molecular mass of sulphur = Atomicity X Atomic mass = 8 x 32 = 256 g.
8. Calculate the number of water molecules present in one drop of water which weighs 0.18 g. Molecular mass of water = 18 gram No. of molecules present in 18g of water = 6.023 x 1023
No. of molecules present in 0.18g of water = x 0.18
No. of molecules present in 0.18g of water = 6.023 x 1023 x 0.01 = 6.023 x 1021 molecules. 9. Fill in the blanks using the given data: The formula of Calcium oxide is CaO. The atomic mass of Ca is 40, Oxygen is 16 and Carbon is 12. i) 1 mole of Ca ( ____g) and 1 mole of Oxygen atom ( ___g) combine to form _____ mole of CaO ( ____g). ii) 1 mole of Ca ( ___g) and 1 mole of C ( ___g) and 3 moles of Oxygen atom ( ___g) combine to form 1 mole of CaCO3 ( ___g) Answer: i) 1 mole of Ca ( _40___g) and 1 mole of Oxygen atom ( __16_g) combine to form _____ mole of CaO ( __56__g). ii) 1 mole of Ca ( _40__g) and 1 mole of C ( _12__g) and 3 moles of Oxygen atom ( _48__g) combine to form 1 mole of CaCO3 ( _100__g) 10. How many grams are there in: i) 5 moles of water ii) 2 moles of Ammonia iii) 2 moles of Glucose
Hint: Number of moles = Mass = Number of moles x Molecular mass i) 5 moles of water
Molecular mass of water = H2O = ( 2 + 16 ) = 18g. 1 mole of water has the molecular mass = 18g 5 moles of water has the molecular mass = 5 x 18 = 90g. ii) 2 moles of Ammonia Molecular mass of ammonia = NH3 = (14 + 3) = 17g 1 mole of ammonia has the molecular mass = 17g 2 moles of ammonia has the molecular mass = 2 x 17 = 34g iii) 2 moles of Glucose Molecular mass of glucose = C6H12O6 = 6(12) + 12(1) + 6(16) = 180g 1 mole of glucose has the molecular mass = 180g 2 moles of glucose has the molecular mass = 2 x 17 = 360g
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11. When ammonia reacts with hydrogen chloride gas, it produces white fumes of ammonium chloride. The volume occupied by NH3 in glass bulb A is three times more than the volume occupied by HCl in glass bulb B at STP. i) How many moles of ammonia are present in glass bulb A? ii) How many grams of NH4Cl will be formed when the stopper is opened? (Atomic mass of N = 14, H = 1, Cl = 35.5) iii) Which gas will remain after completion of the reaction? iv) Write the chemical reaction involved in this process. i) How many moles of ammonia are present in glass bulb A? At STP 22.4 litre of gas contains = 1 mole
At STP 67.2 litre of ammonia gas contains = x 67.2 = 3 moles
ii) How many grams of NH4Cl will be formed when the stopper is opened? When the stopper is opened 1 mole of ammonia reacts with 1 mole of hydrocholoric acid give 1 one mole of ammonium chloride. Molecular mass of NH3Cl = 1(N) + 4(H) + 1(Cl)
= 1 (14) + 4 (1) + 1 (35.5) = 14 + 4 + 35.5 = 53.5g 53.5g grams of NH3Cl will be formed when the stopper is opened. iii) Which gas will remain after completion of the reaction?
NH3 gas iv) Write the chemical reaction involved in this process. NH3 + HCl NH4Cl 12. Nitro glycerine is used as an explosive. The equation for theexplosive reaction is C3H5((NO3))3 12CO2 + 10H2O + 6N2 + O2
(l) (g) (l) (g) (g) (Atomic mass of C = 12, H = 1, N = 14, O=16) i) How many moles does the equation show for i) Nitroglycerine ii) gas moleculesproduced? ii) How many moles of gas molecules are obtained from 1 mole of nitroglycerine? iii) What is the mass of 1 mole of nitroglycerine? i) How many moles does the equation show for i) Nitroglycerine According to the equation given in the text book no. of mole present in Nitroglycerine is 1 mole. But if the equation is balanced 4 moles are present. ii) gas molecules produced? As per the equation given in the book 19moles of gas molecules are produced. 10 moles of water molecules indicated as liquid.If we include this also as a gas molecules 29 moles of gas molecules are produced. ii) How many moles of gas molecules are obtained from 1 mole of nitroglycerine? 4 moles of nitroglycerine gives = 29 moles of gas molecules 1 mole of nitroglycerine gives = 29 / 4 = 7.25 moles iii) What is the mass of 1 mole of nitroglycerine?
Molecular mass of nitroglycerine C3H5(NO3)3 = 3 (C) + 5(H) + 3(N) + 9(O) = 3(12) + 5(1) + 3 (14) + 9 (16) = 227g Mass = No.of moles x molecular mass = 1 x 227 = 227g 13. Sodium bi carbonate breaks down on heating: 2NaHCO3 Na2CO3 + H2O + CO2 (Atomic mass of Na = 23, C = 12, H = 1, O=16)
i) How many moles of sodium bi carbonate are there in the equation? ii) What is the mass of sodium bicarbonate? iii) How many moles of carbon dioxide are there in the equation?
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i) How many moles of sodium bi carbonate are there in the equation? 2 moles of sodium bi carbonate are present in the equation. ii) What is the mass of sodium bicarbonate? Mass of sodium bicarbonate = 1(Na) + 1(H) + 1(C) + 3(O) = 1(23) + 1(1) + 1(12) + 3(16) = 23 + 1 + 12 + 48 = 84g Mass of sodium bicarbonate = no. of moles x mol.mass = 2 x 84 = 168g iii) How many moles of carbon dioxide are there in the equation? 1 mole of carbon dioxide are there in the equation. 14. 100g of calcium was extracted from 174g of calcium oxide(Atomic mass of Ca= 40, O=16) i) What mass of oxygen is there in 174 g of calcium oxide? ii) How many moles of oxygen atoms are there in this ? iii) How many moles of calcium atoms are there in 100g of calcium? iv) What mass of calcium will be obtained from 1000g of calcium oxide? i) What mass of oxygen is there in 174 g of calcium oxide? Molecular mass of CaO = 1(Ca) + 1(O) = 1(40) + 1(16) = 40 + 16 = 56g ii) How many moles of oxygen atoms are there in this ? There are 74g of oxygen atoms are there in this. iii) How many moles of calcium atoms are there in 100g of calcium?
No,of moles of oxygen atoms = = = 4.62 iv) What mass of calcium will be obtained from 1000g of calcium oxide? Mass of calcium obtained from 174g of calcium oxide = 100g
Mass of calcium obtained from 1000g of calcium oxide = x 1000
= 574.713g
15. How many grams are there in the following? i) 1 mole of chlorine molecule, Cl2 ii) 2 moles of sulphur molecules, S8
iii) 4 moles of ozone molecules, O3 iv). 2 moles of nitrogen molecules, N2
i) 1 mole of chlorine molecule, Cl2
Molecular mass of Cl2 = 2(Cl) = 2(35.5) = 71g Mass = No. of moles x molecular mass = 1 x 71 = 71g
ii) 2 moles of sulphur molecules, S8
Molecular mass of S8 = 8(S) = 8 (32) = 256g Mass = No. of moles x molecular mass = 2 x 256 = 512g iii) 4 moles of ozone molecules, O3
Molecular mass of O3 = 3(O) = 3 (16) = 48g Mass = No. of moles x molecular mass = 4 x 48 = 192g Iv). 2 moles of nitrogen molecules, N2
Molecular mass of N2 = 2(N) = 2 (14) = 28g Mass = No. of moles x molecular mass = 2 x 28 = 56g
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16. Find how many moles of atoms are there in: i) 2 g of nitrogen. ii) 23 g of sodium iii) 40 g of calcium. iv) 1.4 g of lithium v) 32 g of sulphur. i) 2 g of nitrogen.
No.of moles of nitrogen atoms = = = 0.142 moles ii) 23 g of sodium
No.of moles of sodium atoms = = = 1 mole iii) 40 g of calcium.
No.of moles of calcium atoms = = = 1 mole iv) 1.4 g of lithium
No.of moles of lithium atoms = = = 0.2 mole v) 32 g of sulphur.
No.of moles of sulphur atoms = = = 1 mole
17. The hydroxide ion concentration of a solution is 1.0 x 10–8M. What is the pH of the
solution? POH = -log 10 (OH-) = -log10 ( 1 x 10-8 ) = -log 10 10-8 = POH = 8 PH = 14 - POH
= 14 – 8 = 6 18. The hydrogen ion concentration of a solution is 1 X 10 -8M
i) What is the PH of the solution? ii) What is the POHof the solution?
iii) Is the given solution, acidic or basic?.
i) What is the PH of the solution?
PH = -log 10 (OH-) = -log10 ( 1 x 10-8 ) = -log 10 10-8 = PH = 8 ii) What is the POHof the solution?
POH = 14 – 8 = 6 iii) Is the given solution, acidic or basic?.
Basic
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PHYSICS PROBLEM- LESSON – 14-PART – B
1. A bullet of mass 20 g moving with a speed of 75 ms-1hits a fixed wooden plank and
comes to rest after penetrating a distance of 5 cm. What is the average resistive force exerted by the wooden plank on the bullet? Mass of the bullet m = 20g or 20 x 10-3kg Initial velocity of the bullet, u = 75ms-1 Final velocity of the bullet, v = 0ms-1
Distance travelled by the bullet, s = 5cm = 5 x 10-2m v2 = u2 + 2as 2as = v2 - u2
a = = = = - 56250ms-2
Force exerted by the wooden plank on the bullet = F1 = ma = 20 x 10-3 x (-56250) = - 1125 N
The resistive force exerted by the wooden plank on bullet = F2 F1 = F2 = -(-1125) = + 1125 N 2. A shopping cart has a mass of 65 kg. In order to accelerate the cart by 0.3ms-2 what force would you exert on it? Mass of cart m = 65 kg Acceleration a = 0.3 ms-2 Force exerted on cart F = ma = 65 x 0.3 = 19.5 N 3.A 10 Kg mass is suspended from a beam 1.2 m long. The beam is fixed to a wall. Findthe magnitude and direction (clockwise or anti-clockwise) of the resulting moment atpoint B. Mass m = 10kg Distance d = 1.2m Moment = F x d = mg x d = 10 x 9.8 x 1.2 = 117.6 Nm Magnitude of the resulting moment at point ‘B’ = 117.6 Nm Direction of the moment = Clockwise 4.If the force experienced by a body of unit mass is gravitational field strength, find the gravitational field strength on the surface of the earth. Mass of the body m = 1kg (unit mass) Mass of the earth M = 5.98 x 1024 kg Gravitational constant G = 6.673 x 10-11 Nm2 kg2 Radius of the earth R = 6.38 x 106 m
F =
mg =
1 x g =
9
G = 9.8 ms-2 5. If the density of the earth is doubled to that of its original value, the radius remaining the same, what will be the change in acceleration due to gravity?
Acceleration due to gravity g =
Density = If the density of the earth is doubled keeping the radius constant, then the mass of earth is also doubled.
So ‘g’ becomes, g =
So,acceleration due to gravity will be doubled to that of its original value. G = 2 x 9.8 = 19.6ms-2
6. Renu is standing in a dining line 6.38 x 103km from the centre of the earth. The mass
of the earth is 6 x 1024kg. i) Find the acceleration due to gravity. ii) Will the value change after she finishes her lunch? i) Mass of the earth M = 6 x 1024kg
Distance between Renu and centre of earth = 6.38 x 103km (or) 6.38 x 106m
Gravitational constant G = 6.673 x 10-11 Nm2 kg2
Acceleration due to gravity g =
= 9.8 ms-2 ii)The value of ‘g’ is the same for all bodies irrespective of their masses. So, the value of ‘g’ does not change after she finishes her lunch. 7. If an angel visits an asteroid called B 612 which has a radius of 20 m and mass of 104 kg, what will be the acceleration due to gravity in B 612 ?. Mass of the asteroid M = 104 kg Radius of the asteroid R = 20 m Gravitational constant G = 6.673 x 10-11 Nm2 kg2
Acceleration due to gravity g =
=
= 1.735 x 10-11 ms -2 8. A man of mass ‘m’ standing on a plank of mass ‘M’ which is placed on a smoothhorizontal surface, is initially at rest. The man suddenly starts running on the plankwith a speed of ‘v’ m/s with respect to the ground. Find the speed of the plank withrespect to the ground. Initial momentum of man = mu Initial momentum of the plank = Mu2 Total initial momentum = mu + Mu2 Final momentum of man = mv Final momentum of the plank = Mv2 Total final momentum = mv + Mv2 According to the law of conservation of momentum Total initial momentum = Total final momentum mu + Mu2 = mv + Mv2 Initial velocities of man (u) and plank (u2) = 0 m(0) + M (0) = mv + Mv2 mv + Mv2 = 0
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Mv2 = - mv
The speed of the plank with respect to the ground v2 = -
9. Two balls of masses in ratio 2:1 are dropped from the same height. Neglecting air resistance, find the ratio of i) the time taken for them to reach the ground. ii) the forces acting on them during motion. iii) their velocities when they strike the ground. iv) their acceleration when they strike the ground. i) the time taken for them to reach the ground. = 1: 1 ii) the forces acting on them during motion. = 1: 1 iii) their velocities when they strike the ground. = 1: 1 iv) their acceleration when they strike the ground. = 1: 1 10. An object of mass 1 kg is dropped from a height of 20 m. It hits the ground and rebounds with the same speed. Find the change in momentum.(Take g=10 m/s2) Mass of the object m = 1kg Height s = 20m Acceleration due to gravity g = 10ms-2 Initial velocity u = 0ms-1 Final velocity v = ? According to the law of equation of motion
v2 = u2 + 2as v2 = (0)2 + 2 x 10 x 20 = 400 v = 20ms-1
Change in momentum = Final momentum – Initial momentum = mv – mu = (1 x 20) – (1 x 0) = 20kgms-1 19. What will be the acceleration due to gravity on the surface of the moon, if its radius is1/4th the radius of the earth and its mass is 1/80 times the mass of the earth.
Mass of the moon = =
Radius of the moon = =
Acceleration due to gravity on the surface of the earth = g =
Acceleration due to gravity on the surface of the moon
G (m) =
=
=
=
11
= ( g)
Acceleration due to gravity on the surface of the moon will be 1/5th the acceleration due to gravity on the surface of the earth.
20. A boy weighing 20 kg is sitting at one end of a see-saw at a distance of 1.2 m from the centre. Where should a man weighing 60 kg sit on the see-saw, so that it stands balanced? Weight of the boy (m1) = 20 kg Weight of the man (m2) = 60 kg Distance between the boy and the centre = 1.2m Distance between the man and the centre = ? Moment of force = F x d
According to principle of moment, Anticlockwise moment = Clockwise moment F1 x d1 = F2 x d2 m1a1 x d1 = m2a2 x d2 20 x 1.2 = 60 x d2
d2 =
d2 = 0.4m A man weighing 60kg should sit a distance of 0.4m from the centre of see-saw.
PART -C .1. i) Newton’s first law of motion gives a qualitative definition of force. Justity. ii) The figure represents two bodies of masses 10 kg and 20 kg, moving with an initialvelocity of 10 ms-1 and 5 ms-1 respectively. They collide with each other. After collision,they move with velocities 12 ms-1 and 4 ms-1 respectively. The time of collision is 2 s.Now calculate F1 and F2.
(ii) Solution:
Data Given: m1 = 10kg. m2 = 20kg.
u1 = 10 ms-1
u2 = 5 ms-1
v1 = 12 ms-1 v2 = 4 ms
-1
Force: (Action) F1 = , = =
Force: (Reaction) F2 = , = =
Action = Reaction F1 = - F2 = 10N = - 10N
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2. A 5 N force acts on a 2.5 kg mass at rest, making it accelerate in a straight line. i) What is the acceleration of the mass? ii) How long will it take to move the mass through 20m? iii) Find its velocity after 3 seconds. Data Given: F = 5N m= 2.5kg.
i) What is the acceleration of the mass?
F = ma; 5 = 2.5 x a
a = ; = =2 ms-2
ii) How long will it take to move the mass through 20m? Distance (d) = 20m U = 0 t = ? a = 2 ms-2 According to newton’s law of motion
S = ut + ½ at2 ; 20 = 0 x t + ½ x 2 x t2 ;
t2= 20 t = ; t = 4.47s
iii) Find its velocity after 3 seconds. u = o a = 2ms-2 t = 3s v = ? v = u + at ; v = o + 2 x 3 v = 2 x 3 = 6ms-1 3. State the law of conservation of momentum. Two billion people jump above the earth’s surface with a speed of 4m/s from the same spot. The mass of the earth is 6x1024 kg. The average mass of one person is 60 kg. i) What is the total momentum of all the people? ii) What will be the effect of this action on the earth? The law of conservation of momentum: In the absence of external unbalanced force, the total momentum of a system of objects remains unchanged. i) What is the total momentum of all the people? Mass of one person = 60 kg. Mass of 2 billion people = 2 x 109 x 60 ( 1 billion = 109 ) = 120 x 109 Velocity (speed) = 4m/s Total momentum of all people,p = mv p = 120 x 109 x 4 = 480 x 109 kgms-1 (OR) = 4.8 x 1011 kgms-1 ii) What will be the effect of this action on the earth? Mass of earth = 6 x 1024 Mass of all the people = 120 x 109 Mass of earth is very high when we compare the mass of all the people. So there will be no effect of this action on the earth. 4. State Newton’s law of gravitation. Write an expression for acceleration due to gravityon the surface of the earth. If the ratio of acceleration due to gravity of two heavenly bodies is 1:4 and the ratio of their radii is 1:3, what will be the ratio of their masses?
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Newton law of gravitation:Every object in the universe attracts every other object with a force which is
directly proportional to the product of their masses and inversely proportional to the square of the distance
between them.
Acceleration due to gravity g =
g1 : g2 = 1 : 4 R1 :R2 = 1 : 3
g1 = ; g2 = ;
M1 = : M2 =
=
= x 2 = x =
Ratio of their masses = 1 : 36
5. A bomb of mass 3 kg, initially at rest, explodes into two parts of 2 kg and 1 kg. The 2 kg mass
travels with a velocity of 3 m/s. At what velocity will the 1 kg mass travel?
Mass of a bomb (m) = 3kg Initial velocity ( u ) = 0 2kg mass = first part 1kg mass = second part Mass of the first part (m1) = 2kg Velocity of the first part (v1) = 3 m/s Mass of the second part (m2) = 1 kg Velocity of the second part (v2) = ? According to the law of conservation of momentum Initial momentum = Final momentum
mu = m1v1 + m2v2 3 x 0 = 2 x 3 + 1 x v2 0 = 6 + v2 v2 = - 6ms-2 Velocity (v2) = - 6ms-2 6. Two ice skaters of weight 60 kg and 50 kg are holding the two ends of a rope. The rope is taut. The 60 kg man pulls the rope with 20 N force. What will be the force exerted by the rope on the other person? What will be their respective acceleration?
Mass of the first ice skater (m1) = 60kg Mass of the second ice skater (m2) = 50kg Force applied by first skater (F1) = 20N According newton’s III law the force applied by the second skater (F2) = -20N Their respective acceleration = F = ma F1 = m1a1 F2 = m2 a2
a1= a2 = = 0.33 ms-2 = = 0.4 ms-2
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LESSON – 16 MODEL EVALUATION
1) The following graph was plotted between V and I values.What would be the values of V / I ratios when the potential difference is 0.5 V and 1 V?
1) If V = 0.5 I = 0.2 then = = 2.5Ω
2) If V = 1.0 I = 0.4 then = = 2.6Ω
2) Observe the circuit given and find the resistance across AB.
R1 and R2 are parallel
= + ;
+ = 1 + 1 = 2;
= 2 ; R12 =
R3 and R4 are parallel
= + ;
+ = 1 + 1 = 2;
= 2 ; R34 =
R12 and R34 are in series
R = R12 + R34
= + = 1
R = 1Ω
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3.How many electrons flow through an electric bulb every second, if the current that passes
through the bulb is 1.6 A.
I = 1.6A t = 1s
Charge Q = I x t ; = 1.6 x 1 =1.6 coulomb
1 coulomb charge contains = 6.25 x 1018 electrons
No. of electrons in 1.6coulomb charge = 1.6 x 6.25 x 1018
= 1 x 1019 electrons.
4. Vani’s hair dryer has a resistance of 50 Ω when it is first turned on.
i) How much current does the hair dryer draw from the 230 V – line in Vani’s house?
ii) What happens to the resistance of the hair dryer when it runs for a long time?
(Hint : As the temperature increases the resistance of the metallic conductor increases.)
i) How much current does the hair dryer draw from the 230 V – line in Vani’s house?
V = 230v R = 50ohm
Current drawn by hair dyer I = = = 4.6A
Current = 4.6A ii) What happens to the resistance of the hair dryer when it runs for a long time?
When it runs for a long time its resistance increases,it is affected and the fuse in the hair dryer
cuts the flow of current.
5. In the given network, find the equivalent resistance between A and B.
R1,R2,R3,R4,R5 = 5Ω R P1,R P2,R P3,R P4 = 10Ω
RA , RB , RC , & RD are in parallel.
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R1 and R2are in series R = 5 + 5 = 10; R + RP1=
RA = 5
RA and R3 are in series R = 5 + 5 = 10; R + RP2=
RB = 5
RB and R4 are in series R = 5 + 5 = 10; R + RP3=
RC = 5
RC and R4 are in series R = 5 + 5 = 10; R + RP4=
RD = 5
The equivalent resistance between A and B = 5Ω 6. Old – fashioned serial lights were connected in a series across a 240V householdline.
i) If a string of these lights consists of 12 bulbs, what is the potential difference across each bulb? Total potential difference = 240V; No. of bulbs = 12
Potential difference across each bulb = =
ii) If the bulbs were connected in parallel, what would be the potential difference across each bulb? If the bulbs were connected in parallel
the potential difference across each bulb is same. i.e. 240V. 7. The figure is a part of a closed circuit. Find the currents i1, i2, i3.
i) i1 + 2A = 3A i1 = 3A – 2A = 1A ii) i2 + 1A = 3A i2 = 3A – 1A = 2A iii) i3 + 1.5A = i2 i3 = i2 – 1.5A ; i3 = 2A – 1.5A = 0.5A i1 = 1A, i2 = 2A, i3 = 0.5A. 8.If the reading of the ideal voltmeter (V) in the given circuit is 6V, then find the reading of the ammeter (A).
Potential difference V = 6v Resistance R = 15Ω
Ammeter reading i.e I = =
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9. A wire of resistance 8Ω is bent into a circle. Find the resistance across the diameter.
The wire is bent into a circle.So the resistance i.e 8Ω is divided into two equal parts.They are in parallel.
= R = 2Ω
10. A wire is bent into a circle. The effective resistance across the diameter is 8Ω. Find the resistance of the wire.
R1+ R2 = R; = R = 16Ω
R1 = 16Ω, R2 = 16Ω
Total resistance = R1+ R2 ; 16 + 16 = 32 Ω
11. Two bulbs of 40 W and 60 W are connected in series to an external potential difference.Which bulb will glow brighter? Why? 40W bulb will glow brighter. Reason: Two bulbs are connected in series.So current through each bulb is the same. 40W has high resistance and dissipate more power.So it glow brighter. 12. Two bulbs of 70 W and 50 W are connected in parallel to an external potential difference.Which bulb will glow brighter? Why? 70W bulb will glow brighter. Reason: Two bulbs are connected in parallel.So potential difference through each bulb is the same.70W has lower the resistance and dissipate more power.So it glow brighter.
PART -C 1. Veena’s car radio will run from a 12 V car battery that produces a current of 0.20 A even when the car engine is turned off. The car battery will no longer operate when it has lost 1.2 x 106 J of energy. If Veena gets out of the car, leaving the radio on by mistake, how long will it take for the car battery to go completely dead, i.e. lose all energy? (1 day =86400 second)
Potential difference of the battery of a car(V) = 12 V Current produced by the battery (I) = 0.2 A Energy at which car battery becomes dead = 1.2 x 106 J Power = V x I; Energy = V x I x t Time taken for the car battery to go completely dead t =
t = ;
= 0.5 x 106 ; 5 x 105s
1 day = 86400 seconds = = 5.78 days
Car battery will lose all of its energy after 5.78 days. 2. Find the total current that passes through the circuit. Find the heat generated across the each resistor.
This system is a series – parallel one.
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Parallel system with R2 and R3
= + ; = + ; = =
Rp = 4Ω R1+ Rp are in series Rs = R1+ Rp = 4 + 4 = 8 Ω
Potential difference (V) = 16V Resistance (R) = 8 Ω Current (I) = = = 2A The total current that passes through the circuit = 2A Heat generated across the each resistor. Heat generated across the resistor R1 = H = I2Rt = (2)2 x 4 x 1 = 16J A similar heat is generated at the other component of the series system R2 and R3 = The total heat generated is 16J + 16J = 32J 3) Find the total current that passes through the circuit given in the diagram. Also find the potential difference across 1 Ω resistor.
R1 , R2 and R3 are in parallel.
= + + ; = + + ;
= = ; Rp = 2Ω
Rp , R4 and R5 are in series.
The resultant resistance R = Rp + R4 + R5 = 2 + 2+ 1= 5Ω
i)Total current that passes through the circuit I = = = 0.3A
ii)Potential difference across 1Ω resistor V = IR = 0.3 x 1 = 0.3V
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4. Raman’s air-conditioner consumes 2160 W of power, when a current of 9.0 A passes through it. i) What is the voltage drop when the air-conditioner is running? Power = 2160 W Current = 9A
Voltage drop V = P = VI )
V = = 240 V
ii) How does this compare to the usual household voltage? Usual household voltage is 220v.So the voltage drop when the air-conditioner is running is 20v higher than household voltage. iii) What would happen if Raman tried connecting his air-conditioner to a 120V line? If Raman tried connecting his air-conditioner to a 120V line , it may cause short circuiting. 5.The effective resistance of three resistors connected in parallel is 60/47 Ω. When one wire breaks, the effective resistance becomes 15/8 ohms. Find the resistance of the wire that is broken.
The effective resistance of three resistors connected in parallel is = Rp = ;
= + + ; = + + ……..(1)
When one wire breaks,i.e. R3 the circuit has two resistors only.So R = ; = …(2)
= + ;
From (1) and (2) = + ;
= - = = =
The resistance of the broken wire is i.e.R3 = 4 Ω 6. Find the resistance across (i) A and D (ii) B and D.
(i) Resistance between A and D. R1 and R2 are connected in series. So RS1 = R1 + R2 = 2 + 2 =4Ω R4 and R5 are connected in series. So RS2 = R4 + R5 = 2 + 2 =4Ω
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RS1, R3 and RS2 are connected in parallel. So = + + =
= 1.33Ω (ii) Resistance between B and D
R1 and R2 are connected in series. So RS1 = R1 + R2 = 2 + 2 =4Ω R4 and R5 are connected in series. So RS2 = R4 + R5 = 2 + 2 =4Ω
RS1, R3 and RS2 are connected in parallel. So = + + =
= 1.33Ω
7. Five resistors of resistance ‘R’ are connected such that they form a letter ‘A’. Find the effective resistance across the free ends.
R1,R2R3,R4 and R5 = R
Resistors R2 and R3 are in series;
RS = R2 + R3 = R + R = 2R
Resistors R5 and RS are in parallel.
= + = + = =
RP =
Resistors RP, R1, andR4 are in series.
Re = RP+ R1+R4= + R + R = = = 2.67R
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LESSON -17
MAGNETIC EFFECT OF ELECTRIC CURRENT AND LIGHT
1. A 3 cm tall bulb is placed at a distance of 20 cm from a diverging lens having a focal length of
10.5 cm. Determine the distance of the image.
Diverging lens means concave lens. u = -20cm f = -10.5cm v = ?
Lens formula: = -
= + ; = + ;
= -( + ; = -( =
v = - 6.89 (or) 6.9 cm
2. A needle placed at 30 cm from the lens forms an image on a screen placed 60cm on the other
side of the lens. Identify the type of lens and determine the focal length.
i) Type of lens = Convex lens
ii) u = -30cm v = 60cm f = ?
Lens formula: = -
= - = + =
f = 20 cm
3. A ray from medium 1 is refracted below while passing to medium 2. Find the refractive index of
the second medium with respect to medium 1.
The refractive index , = ;
= = = x = = 0.707
The refractive index of second medium with respect to medium 1 is 0.707
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4.A real image, 1/5th the size of the object, is formed at a distance of 18cm from a mirror.What is
the nature of the mirror? Calculate its focal length.
i) Nature of the mirror = Concave mirror ( because the image is real)
ii) Focal length =? Magnification m = - ( m is –ve because image is inverted and real)
v = - 18 cm
Magnification, m =
- = - ; = ; = ; u = -90 cm
Mirror formula = = +
= = + ;
= -( + ; = -( = =
Focal length = -15cm
5. Light enters from air to kerosene having refractive index of 1.47. What is the speedof light in
kerosene, if the speed of light in air is 3 x 108m/s.
Refractive index of kerosene, = 1.47
Speed of light in air , c = 3 x 108m/s
= ; 1.47 =
v = = 2.04 x 108m/s 6.
Murugan trims his beard while looking into a concave mirror whose focal length is18 cm. He looks
into it from a distance of 12 cm.
i) How far is Murugan’s image from the mirror?
Concave mirror f = -18cm u = -12cm
Mirror formula = = +
= - ; = - ;
= - + ; = =
v = 36 cm. Murugan’s image is 36cm from the mirror.
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ii) Does it matter whether or not Murugan’s face is closer or farther than the focallength? Explain.
Yes. It matters. Only when the object (Murugan’s face) is placed P and F
an enlarged,virtual and erect image will be formed behind the mirror.
7. Light travels at 1.90 x108m/s in a crystal, what is the crystal’s index of refraction?
Speed of light in air, c = 3 x108 m/s
Speed of light in crystal, v = 1.90 x108m/s
Refractive index of crystal, = ?
= ; = = 1.579 (or) 1.58
8. Ranjini makes arrangements for a candle-light dinner and tops it with a dessertof gelatin filled
blue berries. If a blueberry that appears at an angle of 450 to the normal in air is really located at 300
to the normal in gelatin. What is the index of refraction of the gelatin?
= 450 ; = 300
= ; = = = x = 9.If
9)If the near point of a myopic person is 75cm, what would be the focal length of the lens used to
rectify this defect?
u = -25cm (normal) v = -75cm f = ?
Lens formula: = -
= - ; = +
= = ; f = = 37.5cm
10.Reena and Vani find a discarded plastic lens lying on the beach. The girls discuss what they
learnt in Physics and argue whether the lens is a converging or diverging one.When they look
through the lens, they notice the objects are inverted
i) If an object 25 cm in front of the lens forms an image 20 cm behind the lens. What is the focal
length of the lens? u = -25cm (normal) v = 20cm f = ?
Lens formula: = - ;
= - ; + = =
= , f = = 11.11 cm
ii) Is it a converging or diverging lens? It is a converging lens.(f = positive)
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11. Light which is incident on a flat surface makes an angle of 150 with the surface.
i) What is the angle of incidence? Angle of incidence = 900 – 150 = 750
ii) What is the angle of reflection? = = 750
iii) Find the angle of deviation?. 15 + 15 = 300
PART -C
1. b. The refractive index of diamond is 2.42. What is the meaning of this statement in relation to
the speed of light?
= 2.42. c = 3 x 108 = ;
Speed of light in diamond =
= = 1.24 x108 m/s
The speed of light in diamond is less than the speed of light in air.
3. i) Find the nature, position and magnification of the image formed by a convex lens of focal
length 10cm. If the object is placed at a distance of a) 15cm b) 8 cm
i) If the object is placed at a distance of 15cm
u = -15cm (normal) f = 10cm v = ?
Lens formula: = - ;
= + ; = + ;
= + ; = = - =
v = 30 cm
Magnification = = = -2
Nature of the image: Real and inverted
Position of the image: Beyond 2F
ii) If the object is placed at a distance of 8cm
u = -8cm (normal) f = 10cm v = ?
Lens formula: = - ;
= + ; = + ;
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= + ; = = - =
v = -40 cm
Magnification = = = 5
Nature of the image: virtual and erect
Position of the image: On the same side of the lens.
ii) Which of the above represents the use of convex lens in
a) a film projector = object is at a distance of 15cm.
b) the magnifying glass used by palm reader = object is at a distance of 8cm.
4. An object of 5cm tall is placed at a distance of 10cm from a concave mirror of radius of curvature
30cm.
R = 30cm h = 5cm u = -10cn
Radius of curvature = 2 x focal length; f = R / 2 ; 30 / 2 = 15
Mirror formula = = +
= - ; = - ;
= + ; = =
=
V = 6cm
i) Find the nature, position and size of the image.
Position of the image Nature of the image Size of the image
Behind the mirror (between P and F )
Virtual and erect Enlarged
ii) Draw the ray diagram to represent the above case.
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5. The optical prescription of a pair of spectacle is
Right eye : -3.5 D Left eye: - 4.00 D
i) Name the defect of the eye.
The defect of the eye is myopia. ( Power of concave lens is negative)
ii) Are these lenses thinner at the middle or at the edges?
Thinner at the middle. Because it is a concave lens.
iii) Which lens has a greater focal length?
Focal length of right eye , f = (Power of lens, P = )
f = = - 0.28m
Focal length of left eye , f = (Power of lens, P = )
f = = - 0.2m
So, right eye has a greater focal length.
T.SAMPATHKUMAR, M.Sc.,M.Ed.,
Graduate Teacher,
S.M.H.HR.SEC., SCHOOL,
SIRKALI – 609110.
94438-53723
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