Solving ICSE Class 10 Physics Previous Year Question Papers ICSE Class 10 Physics Question Paper 2013 is the best way to boost your preparation for the board exams.
ICSE Class 10 Physics Question Paper 2013 Solved
(Time : 1 ½ hours)
Maximum Marks: 80
General Instructions:
- Answers to this Paper must be written on the paper provided separately.
- You will not be allowed to write during the first 15 minutes.
- This time is to be spent in reading the Question Paper,
- The time given at the head of this Paper is the time allowed for writing the answers.
- Section I is compulsory. Attempt any four questions from Section II.
- The intended marks for questions or parts of questions are given in brackets [ ].
Section I (40 Marks)
Attempt all questions from this Section.
Question 1.
(a) Give any two effects of a force on a non-rigid body. (2)
Answer:
Force when applied on a non-rigid body can :
- change the dimensions (shape/size) of body.
- cause motion in the body.
(b) One end of a spring is kept fixed while the other end is stretched by a force as shown in the diagram.
- Copy the diagram and mark on it the direction of the restoring force.
- Name one instrument which works on the above principle. (2)
Answer:
Restoring force always acts in a direction opposite to the direction of force.
- Spring balance
(c) (i) Where is the centre of gravity of a uniform ring situated ?
(ii) ‘The position of the centre of gravity of a body remains unchanged even when the body is deformed.’ State whether the statement is true or false. (2)
Answer:
(i) At its geometric centre.
(ii) False (Because position of centre of gravity of a body depends on distribution of mass in it so if body is deformed mass distribution varies and so position of centre of gravity also varies).
(d) A force is applied on a body of mass 20 kg moving with a velocity of 40 ms-1. The body attains a velocity of 50 ms-1 in 2 seconds. Calculate the work done by the body. (2)
Answer:
m = 20 kg
u =40 ms-1
v = 50 ms-1
t = 2 s
W.D. = Change in kinetic energy (According to work energy theorem)
= \(\frac{1}{2}\) mv2– \(\frac{1}{2}\) mu2
W.D. = \(\frac{1}{2}\) × 20 x (50)2– \(\frac{1}{2}\) × 20 × (40)2
= \(\frac{1}{2}\) × 20 × 2500 – \(\frac{1}{2}\) × 20 × 1600 .
= 25000 – 16000 = 9000 J or 9 × io3 J
(e) A type of single pulley is very often used as a machine even though it does not give any gain in mechanical advantage.
(i) Name the type of pulley used.
(ii) For what purpose is such a pulley used ? (2)
Answer.
(i) Single fixed pulley.
(ii) It is used only to change the direction of force applied i.e., in a more convenient direction.
Question 2.
(a) (i) In what way does an ‘Ideal machine’ differ from a ‘Practical machine’ ?
(ii) Can a simple machine act as a force multiplier and a speed multiplier at the same time? (2)
Answer:
(i) Ideal machine is free from friction or Output work = Input work or efficiency is 100% whereas for all practical machines due to friction there is always loss in energy and efficiency is less than 100% and V.R. is not equal to M. A. or output work is less than input work.
(ii) No, a simple machine at one time can act as a force multiplier where effort arm is bigger than load arm but at some other time can act as speed multiplier where effort arm is smaller than load arm so it is not possible at one time.
(b) A girl of mass 35 kg climbs up from the first floor of a building at a height 4 m above the ground to the third floor at a height 12 m above the ground. What will be the increase in her gravitational potential energy ? (g = 10 ms-2). (2)
Answer:
m = 35 kg
h1 = 4 m
h2 = 12 m
g = 10 ms-2
Increase in gravitational
P.E. = mg (h2 – h})
= 35 × 10(12 – 4)
= 35 × 10 × 8
= 2800 J
(c) Which class of lever found in the human body is being used by a boy (2)
(i) when he holds a load on the palm of his hand.
(ii) when he raises the weight of his body on his toes ?
Answer:
(i) Class III lever (in holding a load on the palm of his hand).
(ii) Class II lever (in raising the weight of body on toes).
(d) A ray of light is moving from a rarer medium to a denser medium and strikes a plane mirror placed at 90° to the direction of the ray as shown in the diagram.
(i) Copy the diagram and mark arrows to show the path of the ray of light after it is reflected from the mirror.
(ii) Name the principle you have used to mark the arrows to show the direction of the ray.(2)
Answer:
(i) It reverses the path.
(ii) Principle of reversibility of light.
(e) (i) The refractive index of glass with respect to air is 1.5. What is the value of the refractive index of air with respect to glass ?
(ii) A ray of light is incident as a normal ray on the surface of separation of two different mediums. What is the value of the angle of incidence in this case ? (2)
Answer:
(i) airμglass
glassμair = \(\frac{1}{\text { air } \mu_{\text {glass }}}=\frac{1}{1.5}=\frac{2}{3}\) = 0.67
(ii) ∠i = ∠Zr = 0°
Question 3.
(a) A bucket kept under a running tap is getting filled with water. A person sitting at a distance is able to get an idea when the bucket is about to be filled.
(i) What change takes place in the sound to give this idea ?
(ii) What causes the change in the sound ? (2)
Answer:
(i) As the water level in the bucket increases, length of air column decreases, frequency increases so the sound becomes shriller and a person can get an idea of bucket to be filled.
(ii) Due to change in length of air column, frequency changes.
(b) A sound made on the surface of a lake takes 3 s to reach a boatman.
How much time will it take to reach a diver inside the water at the same depth ?
[Velocity of sound in air = 330 ms-1 ; Velocity of sound in water = 1450 ms-1] (2)
Answer:
Given depth is same or d is same.
So d = vt (in air)
= 330 × 3 = 990 m
So time required in water
t = \(\frac{d}{v}=\frac{990}{1450}\) = 0.68 s
(c) Calculate the equivalent resistance between the points A and B for the following combination of resistors: (2)
Answer:
As in the circuit at position C, current goes through three different paths which are parallel to each other :
1st path CDEF
2nd path CGHF
3rd path CF
Resistance of 1st path CDEF
= 4 Ω + 4 Ω + 4 Ω
= 12 Ω
Resistance of 2nd path CGHF
= 2Ω + 2Ω + 2Ω
= 6 Ω
Resistance of 3rd path CF = 4 Ω
So \(\frac{1}{\text { T.R. }}=\frac{1}{12}+\frac{1}{6}+\frac{1}{4}\)
= \(\frac{1+2+3}{12}=\frac{6}{12}=\frac{1}{2}\)
So T.R. in parallel = 2 Ω
Now, AC and FB are in series.
∴ Total resistance = 5 Ω+ 2 Ω + 6 Ω
= 13 Ω
(d) You have been provided with a solenoid AB.
(i) What is the polarity at end A ?
(ii) Give one advantage of an electromagnet over a permanent magnet. (2)
Answer:
(i) North.
(ii) Strength of an electromagnet can be changed – according to its use and it will be a magnet till the time current passes through it whereas strength of permanent magnet cannot be increased and cannot be magnetised and demagnetised in an instance.
(e) (i) Name the device used to protect the electric circuits from overloading and short circuits.
(ii) On what effect of electricity does the above device work ? (2)
Answer:
(i) Fuse.
(ii) Fleating effect of electric current.
Question 4
(a) Define the term ‘Heat capacity’ and state its S.I. unit. (2)
Answer:
Heat capacity is the amount of heat energy required to change the temperature of a given mass of a body by 1°C or by 1 kelvin. S.I. unit is JC_1 or JK-1.
(b) What is meant by Global warming ? (2)
Answer:
Trapping of infrared radiations in atmosphere by certain gases like methane and carbon dioxide causes global warming.
(c) How much heat energy is released when 5 g of water at 20°C changes to ice at 0°C?
[Specific heat capacity of water = 4.2 Jg-1 °C-1; Specific latent heat of fusion of ice = 336 g-1] (2)
Answer:
Heat energy released :
- 5 g of water at 20°C changes to water at 0°C (mcθ)
= 5 × 4.2 × 20 = 420 J. - water at 0°C changes to ice at 0°C (mL)
= 5 × 336= 1680 J
∴ Total heat energy required = 1680 + 420 = 2100 J
(d) Which of the radioactive radiations
(i) can cause severe genetical disorders.
(ii) are deflected by an electric field ?
Answer:
(i) γ
(ii) α as well as β radiations.
(e) A radioactive nucleus undergoes a series of decays according to the sequence
If the mass number and atomic number of X3 are 172 and 69 respectively, what is the mass number and atomic number of X ?
Answer:
If X3 has At. No. 69 and Mass No. 172.
(As emission of α decreases At. No. by 2 and Mass No. by 4)
∴ X has At No. 72, Mass No. 180
(As emission of β increases At. No. by 1 and no change in Mass No.)
Section II (40 Marks)
Attempt any four questions from this Section.
Question 5.
(a) (i) With reference to their direction of action, how does a centripetal force differ from a centrifugal force ?
(ii) State the Principle of conservation of energy.
(iii) Name the form of energy which a body may possess even when it is not in motion. (3)
Answer:
(i) Centripetal force is always directed towards the centre of circle whereas centrifugal force acts on a body away from the centre.
(ii) Energy can neither be created nor can it be destroyed. It only changes from one form to another.
(iii) Potential energy.
(b) A coolie is pushing a box weighing 1500 N up an inclined plane 7.5 m long on to a platform, 2.5 m above the ground.
(i) Calculate the mechanical advantage of the inclined plane.
(ii) Calculate the effort applied by the coolie.
(iii) In actual practice, the coolie needs to apply more effort than what is calculated. Give one reason why you think the coolie needs to apply more effort. (3)
Answer:
(i) M.A. = \(\frac{\text { Length }}{\text { Height }}=\frac{7.5}{2.5}\) = 3
(ii) Effort = ?
M.A. = \(\frac{\mathrm{L}}{\mathrm{E}}\)
3 = \(\frac{1500}{E}\)
E = 500 N
(iii) In actual practice due to friction etc. efficiency is not 100% and V.R. is more than M.A. or output is less than work input. So he needs to apply more effort than calculated.
(c) A block and tackle system of pulley’s a velocity ratio 4.
(i) Draw a labelled diagram of the system indicating clearly the points of application and directions of a load and effort.
(ii) What is the value of the mechanical advantage of the given pulley system if it is an ideal
pulley system ? (4)
Answer:
(i) Load = 4T
E = T
∴ V.R. = 4
Block and tackle system of pulley having V.R. = 4
(ii) M.A. = V.R. if it is ideal pulley
= 4
Question 6
(a) Name the radiations :
(i) that are used for photography at night,
(ii) used for detection of fracture in bones.
(ii) whose wavelength range is from 100 Å to 4000 Å (or 10 nm to 400 nm). (3)
Answer:
(i) Infrared radiations.
(ii) X- rays.
(iii) Ultraviolet
(b) (i) Can the absolute refractive index of a medium be less than one ?
(ii) A coin placed at the bottom of a beaker appears to be raised by 4.0 cm.
If the refractive index of water is 4/3, find the depth of the water in the beaker. (3)
Answer:
(i) No, absolute refractive index of a medium is always greater than 1, as speed of light in any medium is always less than that in vacuum.
(ii) \(\frac{4}{3}=\frac{\text { Real depth }}{\text { Apparent depth }}\)
Let real depth = x
So apparent depth = x – 4
\(\frac{4}{3}=\frac{x}{x-4}\)
4x – 16 = 3x
4x – 3x = 16
x = 16
∴ Real depth = 16 cm
(c) An object AB is placed between 2F1 and F1 on the principal axis of a convex lens as shown in the diagram.
Copy the diagram and using three rays starting from point A, obtain the image of the object formed by the lens. (4)
Answer:
1st ray is ACA’
2nd ray is AOA’
3rd ray is BOB’
Image is formed beyond 2F2 ; real, inverted and enlarged.
Question 7.
(a) (i) What is the principle on which SONAR is based.
(ii) An observer stands at a certain distance away from a cliff and produces a loud sound. He hears the echo of the sound after 1.8 s. Calculate the distance between the cliff and the observer if the velocity of sound in air is 340 ms’. (3)
Answer:
(i) Echo depth sounding.
Ultrasonic waves have the same speed as of audible sound but are not absorbed in the medium. So transmitter sends these waves receiver receives the waves back after striking the rigid obstacle so time taken is recorded. And we can calculate distance d = \(\frac{v t}{2}\)
(ii) t = 1.8 s
d = ?
v = 340 ms-1
As d = \(\frac{v t}{2}\) = \(\frac{340 \times 2}{1.8}\) = 306 m
(b) A vibrating tuning fork is placed over the mouth of a burette filled with water. The tap of the burette is opened and the water level gradually starts falling. It is found that the sound from the tuning fork becomes very loud for a particular length of the water column.
(i) Name the phenomenon taking place when this happens.
(ii) Why does the sound become very loud for this length of the water column ? (3)
Answer:
(i) Resonance
(ii) When the frequency of air column becomes equal to the frequency of tuning fork i.e., vibrations of air column are in resonance with those of fork, loud sound is heard.
(c) (i) What is meant by the terms (1) amplitude (2) frequency of a wave ?
(ii) Explain why stringed musical instruments, like the guitar, are provided with a hollow box. (4)
Answer:
(i) (1) Maximum displacement of wave from its mean position on either side is known as amplitude.
(2) Number of vibrations produced in 1 second is frequency.
(ii) A vibrating string by itself produces a very weak sound which cannot be heard at a distance. Therefore, all stringed instruments are provided with sound boxes having air inside. So when string vibrates the air inside sets into forced vibrations which have same frequency. Due to large air column and resonance loud sound is heard.
Question 8.
(a) (i) It is observed that the temperature of the surrounding starts falling when the ice in a frozen lake starts melting. Give a reason for the observation.
(ii) How is the heat capacity of the body related to its specific heat capacity ? (3)
Answer:
(i) Ice for melting extracts 336 J/g heat energy from surroundings and as surroundings loose a large amount of heat energy, temperature falls.
(ii) Heat capacity is the amount of heat energy required to raise the temperature of given mass of a body by 1 °C or 1 K whereas specific heat capacity is the amount of heat energy required to raise a unit mass of a substance by 1 °C or 1 K.
Heat capacity = mc
Specific heat capacity= c
Units of S.H.C. = J/kg/K whereas of
H.C. = J/K
(b) (i) Why does a bottle of soft drink cool faster when surrounded by ice cubes than by ice cold water, both at 0° C ?
(ii) A certain amount of heat Q will warm 1 g of material X by 3°C and 1 g of material Y by 4°C. Which material has a higher specific heat capacity.
Answer:
(i) Ice cubes extract 336 J/g heat energy extra than ice cold water at 0°C. So it cools soft drink faster.
(ii) As mass of both is same, Q is same.
c = \(\frac{\mathrm{Q}}{m \theta}\), where Q is amount of heat energy given.
So a body having less increase in temperature will have more specific heat capacity.
∴ X has more S.H.C. than Y.
(c) A calorimeter of mass 50 g and specific heat capacity 0.42 J g-1 °C-1 contains some mass of water at 20°C. A metal piece of mass 20 g at 100 °C is dropped into the calorimeter. After stirring, the final temperature of the mixture is found to be 22°C. Find the mass of water used in the calorimeter.
[specific heat capacity of the metal piece = 0.3 Jg 1 °C-1 ; specific heat capacity of water = 4.2 Jg-1 °C-1] (4)
Answer
According to the principle of calorimetry
Heat lost by metal = Heat gained by calorimeter + Heat gained by water
20 × 0.3 × (100 – 22) = 50 × 0.42 × (22 – 20) + m × 4.2 × (22 – 20)
6 × 78 = 42+ 8.4m
468 – 42 = 8.4m
m = \(\frac{426}{8.4}=\frac{4260}{84}\)
= \(\frac{355}{7}\) = 50.71 g
Question 9.
(a) (i) State Ohm’s law.
(ii) A metal wire of resistance 6Ω is stretched so that its length is increased to twice its original length. Calculate its new resistance. (3)
Answer:
(i) Ohm’s law states that physical conditions remaining constant, potential difference is directly proportional . to the current i.e., V ∝ I, if R is kept constant.
So V = IR
(ii) R = 6 Ω,
when length = l, area = a
So R = ρ\(\frac{l}{a}\) ………………….. (i)
When the wire is stretched
New length = 21 d
New area = \(\frac{d}{2}\) , as ρ is same.
So, R’ = ρ \(\frac{2l}{a}\) × 2
= ρ \(\frac{l}{a}\) × 4 ………………. (ii)
Putting the value of ρ \(\frac{l}{a}\) = 6 in equation (ii), we have a
R’ = 6 × 4 = 24 Ω
(b) (i) An electrical gadget can give an electric shock to its user under certain circumstances. Mention any two of these circumstances.
(ii) What preventive measure provided in a gadget can protect a person from an electric shock ?
Answer:
(i) An electric shock may be caused either due to (1) poor insulation of wires (2) when electric appliances are touched with wet hands.
(ii) Insulation of wires must be of good quality and should be checked from time to time so that no wire is left naked and appliances should not be touched with wet hands.
(c) The figure shows a circuit
When the circuit is switched on, the ammeter reads 0.5 A.
(i) Calculate the value of the unknown resistor R.
(ii) Calculate the charge passing through the 3Ω resistor in 1205.
(iii) Calculate the power dissipated in the 3 Ω resistor. (4)
Answer:
(i) As V = IR (Total R)
So Total R = \(\frac{\mathrm{V}}{\mathrm{I}}=\frac{6}{0.5}\) = 12 Ω
∴ R in circuit = 12 – 3 = 9 Ω (as both are in series)
R = 9 Ω
(ii) Charge = Current × Time
= 0.5 × 120 = 60 C
(iii) Power dissipated in 3 × resistor
= I2R = 0.5 × 0.5 × 3 = 0.75 W
Question 10.
(a) Name the three main parts of a Cathode Ray Tube. (3)
Answer:
- Electron gun.
- Deflecting plates.
- Fluorescent screen.
(b) (i) What is meant by Radioactivity ?
(ii) What is meant by nuclear waste ? (3)
(iii) Suggest one effective way for the safe disposal of nuclear waste.
Answer:
(i) Radioactivity is the self spontaneous disintegration of a heavy nucleus into a, (3 and y radiations in order to attain stability.
(ii) The radioactive material after its use is known as nuclear waste.
(iii) The nuclear waste should be kept in thick casks (thick lead containers) and then they must be buried in the specially constructed deep underground stores or in useless mines and these mines should be sealed after storing casks.
(c) (i) Draw a simple labelled diagram of d.c. electric motor.
(ii) What is the function of the split rings in a d.c. motor ?
(iii) State one advantage of a.c. over d.c. (4)
Answer:
(i)
(ii) Function of split ring is to change the direction of current flowing through the coil after each half rotation such that direction of motor is only one.
(iii) Alternating current can be stepped up or stepped down by using step up and step down transformers whereas direct current cannot be. So by doing this, heat losses are reduced e., it prevents huge loss of electrical energy into heat energy.