SCERT Maharashtra Question Bank Pdf Solution Q.3.rd Answer the Following subquestions. There in Pdf total 94 Answer the Following subquestions. We have given full Solve in this page.
Maharashtra Question Bank Class 10 Answer the Following Solution
Q.3.rd:Answer the Following. (3m.each)
1) Observe the given figure and answer these following questions.
a) What is the conclusion about the orbit of a planet.
Answer: From the above diagram it is concluded that the orbit of planet in which planet revolves is an elliptical with the sun at its one of the foci.
2) Write the Kepler’s law.
Answer: To describe the planetary motion of the planets kepler’s had given three laws which are given below.
Kepler’s First Law: Kepler’s First law states that the orbit in which the planet revolves is an ellipse with the sun at its one of the foci. The location of sun is denoted by the S.
Kepler’s Second Law: Kepler’s Second law states that the line joining the sun and the planet occupies equal areas in equal interval of time.
Kepler’s Third Law: Kepler’s Third law states that the square of period of revolution of the planet around the sun is equal to or directly proportional to the cube of the mean distance of the planet from the sun. i.e. T2 α r3 Where r is the distance of planet from the sun and T is the period of revolution of the planet around the sun.
3) State Newton’s universal law of gravitation. Express it with the mathematical form of force of gravitation?
Answer: Newton’s Universal law of gravitation states that every object in the universe attracts other objects of the universe with definite force. This force is known as gravitation force which is directly proportional to the product of the masses if the two objects and is inversely proportional to the square of the distance between them.
Mathematic form of force of gravitation: Let us consider two masses m1 and m2 and the surface of these objects are uniformly distributed. Let r be the distance between the centres of these two masses.
As we know that the force of gravitation is directly proportional to the product of masses and inversely proportional to the square of distance between them.
Therefore, F = G m1m2/r^2,
here G is the proportionality constant known as universal gravitational constant.
4) An object takes 5 s to reach the ground from a height of 5 m on a planet. What is the value of g on the planet?
Solution,
Given: u= 0m/s, s = 5m, t = 5s,
To find: The value of g.
Calculations:
S= 1/2 gt^2, by putting given values
Therefore, g = 2/5 m/s^2
Therefore the value of g on the planet is 0.4 m/s2.
5) The radius of planet A is half the radius of planet B. If the mass of A is MA, what must be the mass of B so that the value of g on B is half that of its value on A?
Solution,
Given: The radius of planet A is half the radius of planet B. i.e. RA = RB/2. The mass of A is MA. it is also given that the value of gravity on B is half the value of gravity on A. i.e.
gB = 1/2 g_A .
To find: MB Mass of planet B.
g = GM/R^2 , therefore gA = (GM_A)/R^2 and gB = (GM_B)/R^2
Therefore, g_B/g_A = M_B/M_A .R_A/R_B ,
1/2=M_B/M_A .1/4,
Therefore, M_B/M_A = 4/2
Therefore MB = 2 MA.
6) The mass and weight of an object on earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is 1/6th of that on the earth.
Solution,
Given: m = 5kg, W = 49 N,
Assume that the Acceleration due to gravity on the moon is 1/6 of that on the earth i.e.
gM = g_E/6
To find: a. Mass on the moon
- Mass on the earth
Calculations: a. it is known that the mass of the object on the moon is equal to the mass of object on the earth which is equal to 5kg.
- We know that W = mg.
Therefore, W_M/W_E = (mg_M)/(mg_E ) = g_M/g_E = 1/6
Therefore, W_M= W_E/6= (49 N)/6
= 8.17 N
Therefore the weight of the object on the moon is 8.17 N.
7) An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity? How long will the object take to come back to the earth? Assume g = 10
Solution,
Given,
S = 500m , V= 0, A = -19 m/s2
According to newton’s laws of equation,
V2 = U2 + 2as
By putting the given values,
0 = U2 – 10000
U2 = 10000
U = 100
Therefore the initial velocity is found to be 100 m/s.
According to Newton’s first law of equation,
V = U + at
0 = 100 – 10t
t = 100/10
t = 10s
Therefore, the time required for object to go up is 10s.
Total time = Time of ascent + time of descent
= 10 + 10
= 20s
Hence the total time taken to come back at the earth is 20 sec.
8) A ball falls off a table and reaches the ground in 1 s. Assuming g = 10 m/s 2 , calculate its speed on reaching the ground and the height of the table?
Solution,
Given,
Time taken by the ball to reach ground is 1s.
Acceleration due to gravity g = 10 m/s.
According to newton’s laws of equation,
S = ut + ½ gt2
By putting values in equation we get,
Here initial velocity is zero.
S = 0 x 1 + ½ x 10 x1
S = 0 + 5
S = 5 m
Therefore, the height of table is 5m.
9) The masses of the earth and moon are 6 x 10 24 kg and 7.4 x 10 22 kg, respectively. The distance between them is 3.84 x 10 5 km. Calculate the gravitational force of attraction between the two? (Use G = 6.7 x 10 -11 N m 2/kg 2)
Solution,
The mass of earth M = 6 x10 24 kg
Mass of moon m= 7.4 x 1022kg
The distance between earth and moon (r) is 3.84 x 108m
G = 6.7 x 10-11 N m2/kg 2
Therefore, the force exerted by earth on moon is given as,
F= G
By putting values in formula we get,
F = 6.7 x 10-11 N m2/kg 2 x 7.4 x 1022kg x 6 x 10 24 kg/ (3.84 x 108m)2
= 2 x 1020 N
Therefore the force exerted by the earth on moon is 2 x 1020 N.
10) The mass of the earth is 6 x 10 24 kg. The distance between the earth and the Sun is 1.5 x 10 11 m. If the gravitational force between the two is 3.5 x 10 22 N, what is the mass of the Sun? Use G = 6.7 x 10 -11 N m 2kg -2 .
Solution,
Given: the mass of earth m1 is given by 6 × Kg. The distance between the earth and the sun (r) is 1.5 × m, F = 3.5 x 10 22 N and G = 6.7 x 10 -11 N m 2kg -2.
To find: mass of the sun (m2).
Calculations: We know that, F : G m1m2/r^2, m2 = Fr2/Gm1
Therefore by putting values in above formula,
m2 = (3.5 × 10^22 N×(1.5 ×10^11)m)/(6.7×10^(-11) N.m^2 kg^(-2)×6×10^24 )
Therefore, mass of the sun = 1.96 × 10^30 kg.
11) A tennis ball is thrown up and reaches a height of 4.05 m before coming down. What was its initial velocity? How much total time will it take to come down? Assume g = 10 m/s
Solution,
Given,
Distance travelled by tennis ball (s) = 4.05m
Accelaration = a = 10 m/s2
To Find: initial velocity (u) and Time taken (t)
According to newton’s laws of motion
V2 = U2 + 2as
For upward motion of ball V = 0,
Therefore, 0 = u2 + 2 (-10) × 4.05
U2 = 81
U= 9 m/s
For downward motion u = o
According to Newton’s law of motion,
S = ut + ½ at2
By putting values in above formula we get,
4.05 = 0 + ½ × 10 t2
T = 0.9 s
Time taken by the ball for upward motion is same as the downward motion and hence total time taken by ball is given as,
Total time taken
a) State mendeleev’s periodic law. On which basis mendeleev organised periodic law ?
b) State limitations of Dobereiner’s law of triads ?
c) Describe the merits of the mendeleev’s periodic table?
d) What are the demerits of mendeleev’s periodic table?
e) Write name of elements, symbol, atomic number, electronic configuration of second period in modern periodic tables?
f) 10, 20, 7 write inormation about the given atomic number in the table.
Atomic Number |
Electronic Configuration | Groups | Periods |
Elements |
10 | ||||
20 | ||||
7 |
12) An x element with atomic number 11 and Y element with atomic number 13 belong s to the third period in the modern periodic table with this information given in the answers of the below question.
a) Which elements are more metallic characters from these two elements?
b) what is the valency of X and Y elements?
c) atoms size of Y element is smaller than atoms size of X element? explain?
Answer:
a) X element is more metallic characters from the above two elements.
b) Valency of X is 1.
Valency of Y is 3.
c) While going from left to right in period atomic number increases. Positive charge on nucleus increases and electron added in outermost shell. This increases nuclear charge and results in the decrease of size of atom. Hence, the atomic radius decreases while going from left to right In period. Hence, Y element has small atom size than X element.
13) Identify periods of elements and block of elements from given electronic configuration.?
a) 2, 8, 2 b) 2, 8, 7 c) 2, 1
Answer:
a) 2,8,2 : The electronic configuration of 3 Period and s-block elements is 2,8,2.
b) 2,8,7 : The electronic configuration of 3rd period and p-block elements is 2,8,7.
c) 2,1 : The electronic configuration of 2in period and s-block elements is 2,1
14) Position of A, B & C three elements is given in the table from the modern periodic table. Answer the following questions?
Periods | Group 2 | Group 17 |
2 | ———– | A |
3 | B | ——— |
4 | ———– | C |
a) what is A element metal or nonmetals?
b) which is the outermost orbit of element B ?
C) identify C element and it’s physical state?
Answer:
a) Element A is Non-metal.
b) Outermost orbit of element B is M.
c) Element is is Bromine and the physical state of Bromine is Liquid state.
15) 3,1,2 electrons are in valence shell of X , Y , Z elements from this in which groups it belongs and write it’s valency.
Answer:
a) Element with the valence shell electron 3 belongs to the group 13 and valency of the element belonging to this group is 3.
b) Element with the valence shell electron 1 belongs to the 1st group and valency of these elements is 1.
c) Elements with the valence shell electron 2 belongs to the 2nd group and valency of the elements is
16) Match the columns.
Reactants |
Product |
Types of chemical reaction |
MgH2 | Mg + H2 | Endothermic |
2H2S + SO2 → | 3S + 2H2O | Oxidation |
CaO + H2O → | Ca(OH)2 + heat | Exothermic |
redox |
Answer:
Reactants | Product | Types of chemical reaction |
MgH2 | Mg + H2 | Oxidation |
2H2S + SO2 → | 3S + 2H2O | Redox |
CaO + H2O → | Ca(OH)2 + heat | Exothermic |
17) Write three steps of writing chemical equations with example?
Answer: Step I: Write the chemical equation,
SO2(g )+ H2S(aq) → S(s) + H2O(l)
Step II: Write the number of atoms of the various elements present on both side of equation.
Elements | No of atoms in reactant side | No. of atoms in product side |
S | 2 | 1 |
O | 2 | 1 |
H | 2 | 2 |
Step III: The number of hydrogen atoms in both the sides is equal and therefore there is no need to equalize the number of atoms of hydrogen, therefore equalize the number of atoms of sulphur and oxygen atom. To balance the number of sulphur atoms we have to use 2 as a factor in product.
Number of atoms of sulphur | In reactant side | In product side | |
SO2 | H2S | ||
Initially | 1 1 | 1 | |
To balance | 1 1 | 1×2 | |
Hence the equation becomes
SO2 + H2S → 2S + H2O
To balance the number of atoms of oxygen on both side, we have to use 2 as the factor in the product side.
Number of atoms of oxygen |
In reactant side |
In product side |
Initially | 2 | 1 |
To balance | 2 | 1×2 |
Hence the equation becomes
SO2 + H2S → 2S + 2H2O
Now to equalize the number of atoms of hydrogen, we have to use 2 as a factor in the reactant side.
Number of atoms of oxygen | In reactant side | In product side |
Initially | 2 | 4 |
To balance | 2×2 | 4 |
Hence the equation becomes
SO2 +2 H2S → 2S + 2H2O
Now, the number of sulphur atoms found unequal on both side and hence equal the number of sulphur atoms on product side and rewrite the balanced chemical equation.
SO2n(g) + 2H2S(aq) →3S(s) + 2H2O(l).
18) Identify the following reactions the reactants undergo oxidation and reduction and write it?
a) 2Ag2O → 4Ag + O2
b) 2Mg + O2 → 2MgO
c) NiO + H2 → Ni +H2O
Answer:
a) 2Ag2O → 4Ag + O2
In this reaction, reduction of 2Ag2O takes place and the products 4Ag and O2 are formed.
b) 2Mg + O2 → 2MgO
In this reaction, oxidation of Mg takes place and the product 2MgO is formed.
c) NiO + H2 → Ni +H2O
In this reaction, oxidation of H2 and reduction of NiO takes place and the products Ni and H2o are formed.
19) Answer the following questions
a) What is corrosion?
Answer: The process in which the metals get vanished or destroyed slowly by the reaction of air and or other chemicals on the surface of metals. It is known as corrosion. Mostly corrosion occurs with the iron. This iron gets reddish by the action of atmospheric air. Corrosion can be prevented by cutting of the contact of metal with other metals. Corrosion is also prevented by coating a metal with other oil paints or non reactive metals. To prevent the corrosion of any metal grease is applied to the metal surface. The rusting is prevented by this method.
b) What is Electrolysis?
Answer: When the electric current passes through the water, water get decomposed into hydrogen and oxygen. This decomposition takes place in presence of electrical energy and therefore it is called as electrolysis.
c) which changes occur during chemical changes?
Answer: When chemical reaction occurs, the chemical composition of substance used in chemical reaction changes and hence new product is formed.
d) what is called The reaction which involves oxidation and reductions simultaneously ? explain with one example?
Answer: The reaction which involves oxidation and reduction simultaneously is called as redox reaction. In this reaction one reactant get oxidized and other get reduced in during chemical reaction.
In redox reaction, One reactant get oxidized and other get reduced. The reductant get oxidized by oxidant and the oxidant is reduced by reductant. For example, When CuO in solid form reacts with H2 which is in gaseous form Cu in solid form and Water in gaseous form. In this reaction Reactant is oxidized and reduced and new product is formed.
CuO(s)+ H2(g) ——–> Cu(s) + H2O(g)
20) Explain the reaction given in figure?
The above diagram shows the Decomposition of calcium carbonate. The apparatus used in the decomposition of calcium carbonate are two test tubes, bent tube and rubber cork and burner, etc. One test tube is filled with the calcium carbonate and other is filled with the lime water. The two test tubes are fixed and connected with the one another by the rubber cork with the help of bent tube. The powdered calcium carbonate is heated. The calcium carbonate present in the test tube decomposed due to the heating and the carbon dioxide formed turns lime water milky. The calcium oxide powder remains in the first test tube.
CaCO3(s) → CaO(s) + CO2↑….
21) Name any three appliances based on the heating effect of electric current.
Answer: Electric heater, electric cooker and electric bulb are the three appliances based on the heating effect of electric current.
22) Name any three appliances based on the magnetic effect of electric current.
Answer: Telephone receiver, Stereo speaker and electric generator are the appliances based on the magnetic effect of electric current.
Write laws:-
a) Fleming right hand rule : Place your thumb, index finger and the middle finger in such a way that they will be perpendicular to each other, in which middle finger shows the direction of induced current, the thumb shows the direction of motion of the conductor and index finger gives the direction of the magnetic field. This is known as Fleming’s left hand rule.
b) Fleming left hand rule : Fleming’s left hand rule tells us that the direction of the force exerted is perpendicular to the direction of the magnetic field and electric field. In which left hand thumb, index finger and the middle finger are stretched in such a way that the index finger gives the direction of magnetic field, the middle finger gives the direction of current and thumb represents the direction of force on the conductor. This rule is known as Fleming’s left hand rule.
c) Right hand thumb rule : Fleming’s right hand thumb rule gives direction of magnetic field generated by a current flowing through an electrical conductor. Imagine that you have held in right hand a conductor in such a way that the thumb indicated the direction of current. Now turn your fingers around conductor which gives the direction of the magnetic lines of force. This is known as Fleming’s right hand thumb rule.
23) An electrical iron 1100 wt is operated for 2hours daily what will be the electrical expenses for that in the month of April(the electrical charges 5 rs. per unit of energy)
Solution,
Given: P = 1100 W, t = 2 × 30 = 60h, Rs 5 per unit of energy
N = Pt/(1000W.h/unit) = (1100 W ×60h)/(1000W.h/unit)= 66 units
Therefore the electricity consumption expenses is given as,
= 66 units × rs 5 per unit
= Rs 330
24) What is overloading?when does it occur?What does it cause?How can overloading be avoided?
Answer: When a large amount of current suddenly flows through the circuit, the suddenly short circuit occurs. And due to this large flow of current overloading happens. When a number of appliances are connected to the single socket which causes short circuit.
To avoid the overloading we can use the high resistive wires or we can connect single appliance to the single socket.
25) Explain the construction and working of electric motors in short.
Answer: Electric motor is an device which converts electrical energy into mechanical energy.
Construction: The electric motor made up of copper wired rectangular loop with the resistive coating. This loop is placed between the north pole and south pole of the magnet. The loop is placed in such a way that the branches AB and CD are perpendicular to the direction of magnetic field. The ends of the loop are connected to the split ring. The two parts of the rings are coated with the resistive coating. These tow part of the split ring has conducting surfaces in connection with the two stable carbon brushes (E and F) respectively.
Working: When the current flows in the branch AB through the carbon brushes. As the direction of magnetic field is from north pole to south pole. According to the Fleming right hand rule branch AB experiences force which is downward direction. As the current flowing in the branch CD is in opposite direction to the Branch AB, the force exerted by it is in upward direction. Hence, loop and the axle rotates in anticlockwise direction. After a half rotation, the current starts flowing from DCBA. The force exerted in branch AB in upward direction and the force exerted in CD branch in downward direction. Therefore the current starts flowing from DCBA. And loop starts rotating in the anticlockwise direction. Hence the current repeats its motion after each half rotation. The loop and axle continues its motion in anticlockwise direction.
26) Write a short note on the galvanometer.
Answer: Galvanometer is an device used to detect the presence of current in electric circuit. It is also used for some electrical measurements like current, voltage. The deflection in the galvanometer always points the current value. A galvanometer is now upgraded so that it can suitably measure both current in ammeter and voltage in voltmeter. It is consist of coil and the pole piece of magnets. A coil is fixed between the two pole pieces of magnets. When the small amount of current flows through the coil, the coil rotates and this rotation leads to the deflection of pointer. The deflection moves in both sides depending on the direction of current. The galvanometer works on the principle of faraday’s electromagnetic induction.
27) What is the use of earthing wire?
Answer: To avoid the electric short circuits earthing wire is used in houses.
28) Explain the application of heating effect of electric current in an electric bulb with diagram.
Answer: Electric bulb is based on the heating effect of electric current. In this device, the conductors with the high resistance are used. The tungsten wire is used in electric bulb. When the current flows through this wire, wire get heated at high temperature 3400 Degree C and it emits light. The high temperature or hot wire also emits heat.
29) Draw a neat labelled diagram to show the magnetic effect of electric current.
30) Name the following diagrams and explain the concept behind them.
Fig. a. shows the Fleming right hand rule
Place your thumb, index finger and the middle finger in such a way that they will be perpendicular to each other, in which middle finger shows the direction of induced current, the thumb shows the direction of motion of the conductor and index finger gives the direction of the magnetic field. This is known as Fleming’s left hand rule.
Fig. b shows the Fleming left hand rule
Fleming’s left hand rule tells us that the direction of the force exerted is perpendicular to the direction of the magnetic field and electric field. In which left hand thumb, index finger and the middle finger are stretched in such a way that the index finger gives the direction of magnetic field, the middle finger gives the direction of current and thumb represents the direction of force on the conductor. This rule is known as Fleming’s left hand rule.
31) Identify the given figure, write the labels of it.
Answer:
32) Who will spend more electrical energy? 500 W TV Set in 30 mins, or 600 W heater in 20 mins?
Answer:
The electrical energy is found out by the formula,
Energy = Power × time
For T.V. set, Power is given 500W, Time period is 30 min = 30 × 60 s
Therefore the electrical energy is given as
Energy = 500 × 30 × 60 = 900 × J……………….. (1)
For Heater, Power is given 600 W, time period is 20 min = 20 × 60 s
Therefore the electrical energy is given as,
Energy = 600 × 20 × 60 = 720 × J ………………. (2)
From equation (1) and (2) it is found that TV set consumes more energy that heater.
33) Which principle is used to measure the specific heat capacity of a substance?
Answer: The principle of heat of exchange is used to measure the specific heat capacity of a substance by using calorimetry method.
34) Decide the unit for specific heat capacity.
Answer: The unit for specific heat capacity is J⋅kg −1 ⋅K −1
35) In cold regions in winter, the rocks crack due to anomalous expansion of water Explain term.
Answer: The water get contracts instead of expanding when it is heated from 0 degree to 4 degree C. the volume of water becomes minimum at 4 degree C. If we further provide heat to the water, water get expand in between temperature range 0 degree to 4 degree. The volume of water increases and hence the density of water is maximum at 4 degree C. This behaviour of water is called as anomalous behaviour of water. In cold region, due to the decrease in temperature the volume of water inside the fissures of rock increases and at the same time rocks filled with water undergoes contraction and rocks cracks.
36) Explain how heat capacity of a solid can be determined by the method of mixture.
Answer: In the method of mixture, the solid substance is heated whose specific heat to be determined. After heating the substance is mixed with water with the calorimeter. The solid substance loses its heat energy in water and calorimeter. This process continues unless the temperature of all becomes equal. The principle of heat exchange states that the heat lost by solid substance (Q) is equal to the sum of heat gained by calorimeter (Q1) and the heat gained by water in the calorimeter (Q2). We know that Q is given by the product of mass of solid substance, its specific heat and its temperature. By using this formula all other quantities are measured for calorimeter and water too.
37) What is meant by latent heat? How will the state of matter transform if latent heat is given off?
Answer: When the substance changes its state from solid to liquid or liquid to gas or solid to gas, some heat energy Is released or absorbed by the substance at constant temperature. This heat energy is known as latent heat. When the latent heat given off from the substance, the substance changes its state from gas to liquid and from liquid to solid.
38) What is the role of anomalous behaviour of water in preserving aquatic life in regions of cold climate?
Answer: The water get contracts instead of expanding when it is heated from 0 degree to 4 degree C. the volume of water becomes minimum at 4 degree C. If we further provide heat to the water, water get expand in between temperature range 0 degree to 4 degree. The volume of water increases and hence the density of water is maximum at 4 degree C. This behaviour of water is called as anomalous behaviour of water.
39) How can you relate the formation of water droplets on the outer surface of a bottle taken out of the refrigerator with formation of dew?
Answer: When we take cold water bottle from refrigerator, drop of water vapours observed on the outer surface of water bottle. The temperature of water bottle is less as compare to the outer surrounding atmosphere. The bottle cools the temperature of surrounding air and the temperature of surrounding air reached to the temperature of water bottle temperature. As a result the water vapours in the atmosphere converted into tiny droplets. This is due to the dew point temperature which is decided by the amount of vapour in air.
40) ‘Geeta observed white trail at the back of aeroplane in a clear sky to answer the question from this incident given below.
i)what will be the effect of relative humidity of the air surrounding the plane?
ii)i)what will be the effect of relative humidity of the air surrounding the plane is low?
iii) when the air is dry and humid?
Answer:
i. White trail is formed on if the relative humidity of surrounding air is high. It takes longer time to disappear this white trail.
If the relative humidity of the air surrounding the plane is low then no white trail formed.
iii. if the relative humidity is below 60%, air is said to be dry and if the relative humidity is above the 60% the air is said to be dry.
41) Observe the given picture & answer the following questions.
a) which property do you understand in this picture?
b) what is the temperature of the water at the surface?
c) what is the temperature below the layer of ice on the surface?
Answer:
a) The property shown in picture is anomalous behaviour of water.
b) At the surface of water the temperature is 0 degree.
c) The temperature below the layer of ice on the surface is approximately o degree and it increases with increase in depth.
42) read this activity and Answer the following questions.
1) Take three spheres of iron, copper and lead. lead of equal mass.
2)Put all the three spheres in boiling water in the beaker for some time.
3) Take the three spheres out of the water.
4)All the spheres will be at temperature 100°c C. Put them immediately on the thick slab of wax 5)Note, the depth that each of the sphere goes into the wax
Que: a)which property is determined from this activity?
b) give name to that property.
c) explain the term Principal of heat exchange with the help of this activity.
Answer:
a) The property shows that the amount of heat absorbed by a sphere is different for the different spheres.
b) property of specific heat capacity is shown in the activity.
c) The spheres which absorbs more heat goes deeper in wax which causes melting of wax more. This is due to the principle of heat exchange. According to this principle, heat gained by cold object = heat lost by hot object in isolated system.
43) The cold object the hot object enclosed in a one box of heat resistant material
a) what changes will occur in the two objects when temperature flows from those objects ?
b) which principle can show that the energy exchange takes place between two objects only ?
Answer:
a) The process of heat exchange will continue until the cold object and hot object aquire the same temperature. This process includes gain and loss of heat energy.
b) The principle of heat exchange shows that the energy exchange takes place between two objects only.
44) Rainbow is a beautiful natural phenomenon. It is the combined effect of a natural three processes together produced by light. write it into the circle.
Answer: 1.) Refraction
2.) Dispersion
3.) Total internal reflection
46) Observe the given figure and write appropriate phenomenon of light in the box.
Answer:
Figure 1. shows the phenomenon of total internal refraction of light.
Figure no. 2 shows the phenomenon of refraction of light.
Figure no. 3 shows the dispersion of light.
47) Observe the given figure and answer the following questions.
a) Which colour light rays bends most?
b) Which colour light rays bends least.
c) What is the wavelength of violet light rays
Answer:
a) The violet colour light ray bends the most.
b) The red colour light ray bends the least.
c) The wavelength of violet light is approximately 400 nm.
48) Find the power of a convex lenses of focal length of + 25 c.m.
Solution,
Given,
The focal length of the convex lens is given +25 cm.
Power of lens (P) = 1/ f(m)
Therefore, P = 1/ 0.25 = 4D
The power of the convex lens is found to be 4 D.
49) If a convex lense focal length is 20 c.m at what is the power of the lens?
Solution,
Given, The focal length of convex lens is 20 cm = 0.2 m
Power of lens (P) is given by formula, P = 1/ f(m)
Therefore, P = 1/0.2 = 5D
The power of the convex lens is found to be 5 D.
50) If each two concave lense of focal length 30 c.m are kept in contact with each other what will be the power of combination.
Solution,
Given, Let f1 be the focal length of first lens and f2 be the focal length of second lens. Given that f1 = f2 = -30 cm.
Therefore the power of combination (P) is given by formula,
P =
P = -20/3 = 6.7 D
Therefore the power of combination is found to be – 6.7 D.
51) An object is placed at a distance of 10 c.m a convex lens of focal length 12 c.m found at what distance object placed from the lens . find position and nature of image
Solution,
Given,
u = -10 cm and f = 12 cm
According to lens formula
= 5.45 cm
Magnification = v/u = – 0.54
Hence the image formed is inverted.
52) 5 cm high object is placed at a distance of 20 cm from a converging lens of focal length of 10 cm. Determine the position, size and type of the image.
Solution,
Given,
Object is placed at a distance of -20 cm i.e. u = -20cm, object placed at a height of 5 cm i.e. (h) = 5cm
Radius of curvature (R) = 30 cm
We know that radius of curvature is 2 times the focal length.
R = 2f
f = 15 cm
We know that,
Therefore, v = 8.57 cm.
The positive value of v shows that the image formed is behind the mirror.
Now magnification of the object is given by formula, -image distance/ object distance = -8.57 / -20 = 0.428.
The value of magnification is positive and this shows that the image formed is virtual.
Now Magnification = height of image/ height of the object = hi/ho
Therefore, ho = m × hi = 5 × 0.428 = 21.4 cm
This shows that the image formed is erected.
Hence image formed is erect, virtual and smaller.
53) An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? how much bigger as compared to the object?
Solution,
Given,
An object is placed vertically at a distance of 20 cm hence the distance of the object (u) = -20 cm, the height of the object is (h1) = 5cm and the focal length (f) = 10 cm.
Magnification is given as –
1/v−1/u=1/f
Calculation :
1/v = 1/u + 1/f
1/v = 1/-20 + 1/10
1/v = (-1 +2)/ 20
1/v = 1/20
V = 20 cm
And now from the relation, h2/h1 = v/u
Therefore, h2 = v/u x h1
By putting respective values we get h2 = – 5 cm
Therefore, M = u/v = 20/(- 20)= -1.
The negative sign of height shows that the image is inverted and real and it is below the principal axis.
54) Two convex lenses of focal length 30 c.m and 10 c.m each are kept in contact with each other .Find the power of their combination.
Solution:
Given, Two convex lenses of focal length 30 cm and 10 cm.
Let f1 be the focal length of first lens and f2 be the focal length of second lens and their respective length are given as 30 cm and 10 cm.
Therefore the power of combination (P) is given as P = 1/f = 1/f1 + 1/f2
By putting given values,
P = 1/f= 1/0.30+ 1/0.10
= 1/0.30+ 3/0.30 = 4/0.30 = 13.33 D.
Therefore the power of combination is 13.33 D.
56) Write the function of the human eye and label parts of the figure given below.
Answer:
- Double convex transparent crystalline lens: This lens is just behind the pupil and creates inverted and real image of object inside the eye.
- Retina: Retina is the screen made up of light sensitive cells. When light falls on these cells it generates electric signals. These signals are sent to brain via optic nerve and brain analyses signals and converts it into the image.
- Optic Nerve: Optic nerve are the sensitive cells on the retina. These cells get excited when light fall on it and produces electrical signals which are then sent to brain for the analyses.
- Pupil: Depending on the intensity of light pupil contracts or widens. It controls the light entering in the eye.
59) Write laws in given figure.
1.) The refracted ray passes through the principle focus when the incident ray parallel to the principal axis.
2.) If the incident ray passes through the principal focus, the refracted ray get paralleled to the principle axis.
3.) Refracted ray passes through the optical centre of lens without changing its direction as the incident ray passes through the optical centre.
60) Observe the given figure and answer the following questions.
a)Where are the above types of lens construction used ?
b)What type of image is formed by an objective lens?
e)What happens instead of placing at Fo if the object is placed in between O and FO ?
Answer:
A) the lens constructions given in figure is used in compound microscope.
B) Real and inverted image if formed at the a distance within the focal length of eyepiece.
C) Virtual image is formed when the object is placed in between O and Fo.
61) Identify and Explain concepts given in this Diagram
Answer:
The given figure shows the concept of persistence of vision. When the object removed from the front of the eyes after 0.16 sec the image on the retina remain imprinted. This phenomenon of persistence is called as persistence of vision. When sheet of paper with cage drawn on either side of cardboard are tied to thread and is set into rotation by twisting. As a result human being gets the picture of the monkey being inside the cage. It is due to the persistence of vision. Before the vanishing of image of cage from the retina, an image of monkey created on the retina and hence the monkey in cage is seen inside the cage.
62) Complete paragraph by choosing the right options given below. (minimum, near point, 25 cm, farthest, farthest distance)
The …… distance of an object from a normal eye, at which it is clearly visible without stress on the eye, is called the minimum distance of distinct vision. The position of the object at this distance is called the …… of the eye, for a normal human eye, the near point is at ……. The ……. distance of an object from a human eye, at which it is clearly visible without stress on the eye is called ……… of distinct vision. The position of the object at this distance is called the ……… of the eye.
Answer:
The minimum distance of an object from a normal eye, at which it is clearly visible without stress on the eye, is called the minimum distance of distinct vision. The position of the object at this distance is called the near point of the eye, for a normal human eye, the near point is at 25 cm. The farthest distance of an object from a human eye, at which it is clearly visible without stress on the eye is called farthest distance of distinct vision. The position of the object at this distance is called the far point of the eye.
63) Choose the correct option from the bracket and complete the stanza. (colour blind, actual, conical , light sensitive, rodlike, colours )
The retina in our eyes is made up of many ——— cells. These cells are shaped like a rod and like a cone.The ——- cells respond to the intensity of light and give information about the brightness or dimness of the object to the brain.The ———- cells respond to the colour and give information about the colour of the object to the brain.Brain processes all the information received and we see the —— image of the object.Rod like cells respond to the faint light also but —— cells do not.Some people lack conical cells responding to certain colours. These persons can not recognize those colours or can not distinguish between different ———.These persons are said to be ——–.
Answer: The retina in our eyes is made up of many Light sensitive cells. These cells are shaped like a rod and like a cone. The rodlike cells respond to the intensity of light and give information about the brightness or dimness of the object to the brain. The conical cells respond to the colour and give information about the colour of the object to the brain. Brain processes all the information received and we see the actual image of the object. Rod like cells respond to the faint light also but conical cells do not. Some people lack conical cells responding to certain colours. These persons can not recognize those colours or can not distinguish between different colours. These persons are said to be colour blind.
A) What is corrosion ?
Answer: The process in which the metals get vanished or destroyed slowly by the reaction of air and or other chemicals on the surface of metals. It is known as corrosion.
B) Write the chemical name of Corrosion.
Answer: Rusting is the chemical name of corrosion.
C) Write a molecular formula for corrosion.
Answer: The molecular formula for corrosion is Fe2O3× H2O.
64) Observe the following diagram and identify the type of reaction and write observation .
Answer: Above diagram shows the displacement reaction. In this reaction copper is displaced from copper sulphate solution by iron. It is found that iron is more reactive than metallic copper. No reaction takes place with the copper wire in ferrous sulphate solution. In this reaction blur copper sulphate changes converts to colourless. This reaction is takes placed in test tube containing iron nail in one test tube and copper wire in another test tube.
65) Observe the following diagram and give answers.
A) Name the method of prevention of corrosion.
B) For prevention of which metal this method is used?
C) What is used as Anode in this method?
Answer:
A) the above diagram indicates the prevention of corrosion method called anodization.
B) Copper and aluminium metals are prevented from corrosion in this method.
C) Copper and aluminium are used as anode in this method.
66) Explain the Hydraulic separation method with a neat labelled diagram.
The hydraulic separation method is based on working of mill. It has tapering vessel like a grinding mill. It is opened in container. The container has water outlet at upper side and water inlet at lower side. When the powdered ore is added into the container. A fast stream of water flows from the lower of tank. The gangue particle removes out with the flow of water on the upper side and the heavy particles are collected at the lower side of the tank. This method is follows the law of gravitation in which particles are separated by its weight.
67) Observe the following diagram and write answers.
A) Name the method :———–
B) Explain the method .
C) Give two examples of this method.
Answer:
A) Electroplating
B) In this method less reactive metal is coated with the high reactive metal by the process of electrolysis is called electroplating.
C) Gold plated Jewellery, mobile phones and silver plated utensils.
68) Observe the following diagram and write answers.
A) Name the method :———–
B) Write Anode reaction and Cathode reaction .
C) Why fluorspar and cryolite are added in the mixture?
Answer:
A) Electrolysis reduction of Alumina
B) Anode Reaction: 2O-2→ O2 + 4 e–
Cathode Reaction: Al3+ + 3 e– → Al(l)
C) In order to carry out electrolytic reduction of alumina at very low temperature Cryolite and Fluorspar are mixed with alumina. Because fluorspar and cryolite reduces melting point of alumina.
69) Identify the following method of concentration of ores and explain briefly.
Answer: The above mentioned diagram shows the Wilfley table method.
In this the method of separation based on gravitation which separates heavy particle of ores from the light particles of gangue by gravitational method. In this method one table is used made up of narrow thin wooden riffles on inclined surface called Wilfley table. The table keeps vibrating. By using ball mill lumps of ore is converted into powdered ore. This powdered ore is spread on table and water is also released from the upper side of the table. This results in the separation of gangue particles with the flow of water. Heavier particles are blocked by the wooden riffles and collected on the slits.
70) Observe the following diagram and write answers.
A) Write the name of two metals which react with water.
B) Write the name of two moderately reactive metals .
C) Which is highly reactive and less reactive metal?
Answer:
A) Potassium and sodium are the metals which reacts with water.
B) Magnesium and Aluminium are the two moderately reactive metals.
C) Sodium is highly reactive metal while mercury is a less reactive metal.
71) Explain the Froth floatation method with a neat labelled diagram.
Answer:
The froth floatation method depends on the opposite properties. The two opposite properties of particles are hydrophilic and hydrophobic. In this method the particles of metal sulphides get wetted with oil due to its hydrophobic property and gauge particle get wetted with water due to hydrophilic property. In this method ore is dropped into big tank. The big tank contains the ample amount of water. Some vegetable oil is added in water for the formation of froth. Pressurised air is blown through the water when the pine oil is added in it. There is one agitator is fitted at the centre of tank. Some bubbles are formed due to the pressurised air. Due to agitation, foam is formed due to the oil, water and air bubbles. This foams floats on the surface of water. Hence this method is known as froth floatation process. Sulphide minerals get floated with the water and the gangue particles are wetted by water which settles down at the bottom of tank.
72) Read the following passage and answer the questions.
According to the reactivity series Zinc is more reactive than Iron, Iron is more reactive than silver. During study of this a student deeped the iron nails in silver nitrate solution.
A.) What is reactivity series?
B.) What will happen when iron nails are dipped in silver nitrate solution ?
C.) Which type of reaction happens when iron metal reacts with silver nitrate solution ?
D.) What will happen if a Zinc rod is used other than Iron nail ?
Answer:
A.) The arrangement of metals in the form of series in the decreasing order of their reactivity.
B.) When iron nails are dipped in silver nitrate solution ferrous nitrate is formed and silver. This turns the solution brown.
C.) Displacement reaction occurs when iron metal reacts with silver nitrate solution.
D.) When zinc rod is used instead of iron nail, zinc nitrate and silver is formed on reacting with silver nitrate.
73) Complete the following flowchart.
Answer:-
74) Complete the following flowchart.
75) Complete the following flowchart.
76) Homologous series of Alkanes.
Name | Molecular formula | Condensed structural formula | Number of carbon atoms | Number of
-CH2- units |
Boiling point |
Methane | CH4 | CH4 | 1 | 1 | -162 |
Ethane | C2H6 | CH3 -CH3 | 2 | 2 | -88.5 |
Propane | C3H8 | CH3 -CH2 -CH3 | 3 | 3 | -42 |
Butane | C4H10 | CH3 -CH2 -CH2 -CH | 0 | ||
Pentane | C5H12 | CH3 -CH2 -CH2 -CH2 -CH3 | 36 | ||
Hexane | C6H14 | CH3 -CH2 -CH2 -CH2 -CH2 -CH3 | 69 |
Answer:
Name | Molecular formula | Condensed structural formula | Number of carbon atoms | Number of
-CH2- units |
Boiling point |
Methane | CH4 | CH4 | 1 | 1 | -162 |
Ethane | C2H6 | CH3 -CH3 | 2 | 2 | -88.5 |
Propane | C3H8 | CH3 -CH2 -CH3 | 3 | 3 | -42 |
Butane | C4H10 | CH3 -CH2 -CH2 -CH | 4 | 4 | 0 |
Pentane | C5H12 | CH3 -CH2 -CH2 -CH2 -CH3 | 5 | 5 | 36 |
Hexane | C6H14 | CH3 -CH2 -CH2 -CH2 -CH2 -CH3 | 6 | 6 | 69 |
Name | Molecular Formula | Condensed Structural Formula | Number of carbon atoms | Number of
–CH2- units |
Boiling point |
Methanol | CH4 O | CH3 –OH | 1 | 1 | 63 |
Ethanol | C2 H6 O | CH3 -CH2 –OH | 2 | 2 | 78 |
Propanol | C3 H8 O | CH3 -CH2 -CH2 -OH | 3 | 3 | 97 |
Butanol | C4 H10O | CH3 -CH2 -CH2 -CH2 -OH | 4 | 4 | 118 |
77) Homologous series of Alcohols
Name | Molecular Formula | Condensed Structural Formula | Number of carbon atoms | Number of
–CH2- units |
Boiling point |
Methanol | CH4 O | CH3 –OH | 1 | 1 | 63 |
Ethanol | C2 H6 O | CH3 -CH2 -OH | 2 | 2 | 78 |
Propanol | C3 H8 O | CH3 -CH2 -CH2 -OH | 97 | ||
Butanol | C4 H10O | CH3 -CH2 -CH2 -CH2 –OH | 118 |
78) Homologous series of Alkenes
Name | Molecular Formula | Condensed Structural formula | Number of carbon atoms | Number of –CH2- units | Boiling point
o C |
Ethene | C2H4 | CH2 = CH2 | 2 | 0 | -102 |
Propene | C3H6 | CH3 -CH=CH2 | 3 | 1 | -48 |
1- Butene | C4H8 | CH3 -CH2 -CH=CH2 | -6.5 | ||
1-Pentene | C5H10 | CH3 -CH2 -CH2 -CH=CH2 | 30 |
Answer:-
Name | Molecular Formula | Condensed Structural formula | Number of carbon atoms | Number of –CH2- units | Boiling point
o C |
Ethene | C2H4 | CH2 = CH2 | 2 | 0 | -102 |
Propene | C3H6 | CH3 -CH=CH2 | 3 | 1 | -48 |
1- Butene | C4H8 | CH3 -CH2 -CH=CH2 | 4 | 2 | -6.5 |
1-Pentene | C5H10 | CH3 -CH2 -CH2 -CH=CH2 | 5 | 3 | 30 |
79) Complete the given chart with writing the correct functional group of carbon compounds. (Ester, Aldehyde, Ketone, Carboxylic acid, Alcohol, Ether)
Answer:
80) Complete the following table with writing correct structural formula and molecular formula.
Straight chain of carbon atoms |
Structural formula | Molecular formula |
Name |
C – C |
|
C2H6 | Ethane |
C – C – C – C | C4H10 | Butane | |
C – C – C – C- C- C- C | C7H16 | Heptane | |
C – C – C – C- C- C- C – C | C8H18 | Octane |
81) Complete the following table with writing IUPAC name of carbon compound.
Answer:
Sr. No. |
Common Name | Structural Formula |
IUPAC Name |
1 | Ethylene | CH2 =CH2 | Ethene |
2 | Acetylene | HC =CH | Ethyne |
3 | Acetic acid | CH3 -COOH | Ethanoic acid |
4 | Methyl Alcohol | CH3 –OH | Methanol |
5 | Ethyl Alcohol | CH3 -CH2 -OH | Ethanol |
6 | Acetaldehyde | CH3 –CHO | Ethanal |
7 | Acetone | CH3 -CO-CH3 | Propanone |
8 | Ethyl methyl Ketone | CH3 -CO-CH2 – CH3 | Butanone |
9 | Ethyl amine | CH3 -CH2 -NH2 | Ethanamine |
10 | n-Propyl chloride | CH3 – CH-CH2 -Cl | 1- Chloropropane |
82) Complete the following activity
Answer-
83) complete the following activity.
Write the names of the hydrocarbons for the following structures.
(isobutylene, cyclohexane, propene, cyclohexene, cyclopentane, benzene, propyne, isobutene, propene)
84) Complete the following acitivity
Answer:
85) Observe the structural formula and answer the following questions.
1.) Write the name of the given hydrocarbon.
2.) The given hydrocarbon included in which type of hydrocarbon?
3.) What kind of compounds with the above characteristic structure are called?
Answer:
1.) The given hydrocarbon is Benzene.
2.) The given hydrocarbon includes cyclic unsaturated hydrocarbon.
3.) In this hydrocarbons three alternate double bonds in six membered ring structure is shown. The compounds having this characteristic structure are called aromatic compounds.
86) Complete the following chart by using examples given in brackets. (isobutylene, cyclohexane, propene, cyclohexene, cyclopentane, benzenei, propyne, isobutane, propene)
Answer:
Straight Chain Hydrocarbons |
Branched Chain Hydrocarbons |
Cyclic Hydrocarbons |
Propane | Isobutylene | Cyclohexane |
Propyne | Isobutane | Cyclohexene |
Propene | Cyclopentane | |
Benzene |
87) Write the properties of Ionic compounds.
Answer: 1.) Ionic compounds are crystalline in nature.
2.) Ionic compounds exists in solid state because of the strong force of attraction between positively and negatively charged ion.
3.) Ionic compounds can be broken into pieces because of its property of brittleness.
4.) Ionic compounds are insoluble in solvents (kerosene or petrol)
5.) Ionic compounds have high melting and boiling points because the intermolecular force of attraction is high and hence large energy is required.
6.) When the ionic compounds are in solid state they cannot conduct electricity.
88) Observe the figure and write the answers.
A) Name the outer orbit.
B) Which satellites revolve in low earth orbits ?
C) Which various orbits are given in the figure?
D) Give an example of a launch vehicle based on Newton’s third law.
Answer:
A) High earth orbit.
B) International space station and Hubble telescope revolves in low earth orbits.
C) High Earth orbit, Medium earth orbit and low earth orbit are shown in figure.
D) Polar Satellite Launch Vehicle is the launch vehicle works on Newton’s Third Law.
89) Explain why spacecraft take longer to reach the moon than light?
Answer: Moon is the nearer to earth. As we know that light travels at a speed of 3 . Hence light takes 1 second to reach moon. Therefore spacecraft takes longer to reach the moon than light.
90) Write functions of Military satellite and Navigational satellite.
Answer: Military satellite keeps our nation safe from the other nations. Military satellite helps to monitor all movements of neighbouring nations or enemy countries. It gathers the information from the enemy country.
Navigation satellite gives the location of any place on the earth surface in terms of its precise latitude and longitude.
91) What is meant by Artificial satellite ? How are they classified depending on their functions?
Answer: A satellite manufactured by human being revolving around earth and any other planet is called Artificial satellite.
On the basis of their function, artificial satellites are classified into 6 types.
1.) Weather satellite: Weather satellite gives the prediction and study of weather.
2.) Communication satellite: communication satellite communicates with the people or satellites from different locations in the world.
3.) Broadcast satellite: Broadcast satellite is used for the telecast of television programs.
4.) Navigational satellite: Navigational satellite is used to fix the location in terms of latitude and longitude.
5.) Military satellite: Military satellite are used keep nation safe from the sovereign.
6.) Earth observation satellites: This satellite provides the information about the earth. This satellite collects the information about the resources and about the natural phenomenon.
92) If the mass of a planet is eight times the mass of the earth and its radius is twice the radius of the earth, What will be the escape velocity for that planet?
Solution,
Given, The mass of the planet is eight times the mass of the earth. Let ME be the mass of the earth and Mp be the mass of planet.
Mp = 8ME…………….. 1
Let Rp be the radius of planet and RE be the radius of earth.
Rp = 2 RE ………………. 2
Escape velocity for the earth is 11.2 m/s
To find out: The escape velocity of the moon.
Therefore, √4 ×11.2 = 22.4km/s
Hence the value of escape velocity for planet is 22.4 km/s.
93) Explain : Escape velocity on the moon is less than escape velocity on the earth.
Answer: Escape velocity is the velocity with which the spacecraft escapes earth gravitational force to travel into the outer space.
The escape velocity on the earth is different from the moon. It is given below as,
Therefore the escape velocity of moon is less than that of earth. i.e. Vesc (moon)< Vesc(earth).
94) Complete the following chart.
Answer: