Model Paper 1

Solutions for Physics, Class 10, ICSE

Section A

25 questions

Question 1(i)

For producing the maximum turning effect on a body by a given force, the perpendicular distance of the line of action of force from the axis of rotation should be :

  1. minimum
  2. it does not matter
  3. maximum
  4. zero
Section A

Answer:

maximum

Reason — The turning effect on the body about an axis is due to the moment of force (or torque) applied on the body which is given by :

Moment of force = Force × Perpendicular distance of the line of action of the force

For producing maximum turning effect on a body by a given force, the force is applied on the body at a point for which the perpendicular distance of the line of action of the force from the axis of rotation is maximum.

Question 1(ii)

A man moves in a circular path in a horizontal plane, the work done is :

  1. Positive
  2. Negative
  3. Zero
  4. Cannot say
Section A

Answer:

Zero

Reason — Work done is equal to the product of force, displacement and cosine of the angle θ between the directions of force F and displacement S i.e., W = F x S cosθ.

When a body moves in a circular path in a horizontal plane, no work is done since the centripetal force on the body at any instant is directed towards the centre of the circular path and the displacement at that instant is along the tangent to the circular path, i.e., normal to the direction of force on the body (i.e. θ = 90°)

Question 1(iii)

In an electrical cell while in use, the change in energy is from :

  1. Electrical to magnetic
  2. Chemical to electrical
  3. Mechanical to electrical
  4. Chemical to heat
Section A

Answer:

Chemical to electrical

Reason — The electric cell maintains a constant difference in potential between the two conductors in the electrolyte by a chemical reaction to obtain a continuous flow of charge between them so a cell stores the chemical energy and it can be used as a source of current and hence chemical energy changes into electrical energy when the cell is in use.

Question 1(iv)

A radioactive source emits three types of radiations R, S and T. The diagram given below shows an experimental set up to study the penetrating properties of R, S and T. Which type of radiation are R, S and T ?

A radioactive source emits three types of radiations R, S and T. The diagram given below shows an experimental set up to study the penetrating properties of R, S and T. Which type of radiation are R, S and T. Concise Physics Solutions ICSE Class 10.
 
(a)R - α - particleS - β - particleT - γ - rays
(b)R - α - particleS - γ - raysT - β - particle
(c)R - β - particleS - α - particleT - γ - rays
(d)R - γ - raysS - β - particleT - α - particle
  1. Option (a)
  2. Option (b)
  3. Option (c)
  4. Option (d)
Section A

Answer:

Option (b)

Reason — Increasing order of penetrating power of different radiations is : α - particle < β - particle < γ - rays

Question 1(v)

An object in a denser medium when viewed from a rarer medium appears to be raised. The shift is maximum for :

  1. red light
  2. violet light
  3. yellow light
  4. green light
Section A

Answer:

violet light

Reason — When light travels from a denser medium to a rarer medium, refraction occurs and the object appears to be raised. The apparent shift depends on the refractive index, which in turn depends on the wavelength of light so shorter wavelengths (like violet) are refracted more because they have higher refractive indices and longer wavelengths (like red) are refracted less.

Question 1(vi)

Identify the figure which depicts the correct dispersion and deviation of light.

Identify the figure which depicts the correct dispersion and deviation of light. Concise Physics Solutions ICSE Class 10.
Identify the figure which depicts the correct dispersion and deviation of light. Concise Physics Solutions ICSE Class 10.
Identify the figure which depicts the correct dispersion and deviation of light. Concise Physics Solutions ICSE Class 10.
Identify the figure which depicts the correct dispersion and deviation of light. Concise Physics Solutions ICSE Class 10.
Section A

Answer:

Identify the figure which depicts the correct dispersion and deviation of light. Concise Physics Solutions ICSE Class 10.

Reason — When white light passes through a prism, it splits into its constituent colours in such a way so that violet deviates the most (bends the most) because it has the shortest wavelength and highest refractive index while red deviates the least because it has the longest wavelength and lowest refractive index.

Question 1(vii)

A student wishes to measure the speed of sound in air. He plans to measure the time between making the sound and hearing its echo from a cliff. What type of sound should he make and which distance should he use in his calculations ?

A student wishes to measure the speed of sound in air. He plans to measure the time between making the sound and hearing its echo from a cliff. What type of sound should he make and which distance should he use in his calculations. Concise Physics Solutions ICSE Class 10.
 SoundDistance
(a)Continuous soundDistance to cliff2\dfrac{\text{Distance to cliff}}{2}
(b)Continuous soundDistance to cliff x 2
(c)Short, sharp soundDistance to cliff2\dfrac{\text{Distance to cliff}}{2}
(d)Short, sharp soundDistance to cliff x 2
  1. Option (a)
  2. Option (b)
  3. Option (c)
  4. Option (d)
Section A

Answer:

Option (d)

Reason — To measure the speed of sound using echo, the student must make a short, sharp sound so that the reflected sound (echo) can be clearly distinguished and the sound travels to the cliff and back, so the total distance travelled by the sound is 2 × the distance to the cliff.

Question 1(viii)

When a tuning fork is sounded in air, the sound given by it is feeble. But when it is placed on a table top, the sound becomes much louder. The reason is :

  1. amplitude of the wave
  2. its prongs start vibrating rapidly
  3. presence of a surrounding wave
  4. large surface area of the vibrating body
Section A

Answer:

large surface area of the vibrating body

Reason — When the stem of a vibrating tuning fork is pressed against the top of a table, the tuning fork produces forced vibrations in the table top. The table top has a much larger vibrating area than the tuning fork, so the forced vibrations of the table top send forth a greater energy and produces a louder (or more intense) sound than that produced by the tuning fork.

Question 1(ix)

Which of the following circuits have the same resistance ?

i

Which of the following circuits have the same resistance. Concise Physics Solutions ICSE Class 10.

ii

Which of the following circuits have the same resistance. Concise Physics Solutions ICSE Class 10.

iii

Which of the following circuits have the same resistance. Concise Physics Solutions ICSE Class 10.

iv

Which of the following circuits have the same resistance. Concise Physics Solutions ICSE Class 10.
  1. (i) and (ii)
  2. (ii) & (iii)
  3. (i) & (iii)
  4. (ii) & (iv)
Section A

Answer:

(i) & (iii)

Reason

(i) Here all resistances are in parallel combination and have same resistance value equal to 3 Ω.

Then,

1Req=13+13+13=33=1Req=1Ω\dfrac{1}{R_{eq}}=\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{3}{3}=1 \\[1em] \Rightarrow R_{eq} = 1 \Omega

 

(ii) In this case two resistances of value 3 Ω are in parallel combination and this combination is in series with 6 Ω resistance.

For parallel combination,

Rp=3×33+3=96=32 ΩR_p = \dfrac{3\times3}{3+3}=\dfrac{9}{6}=\dfrac{3}{2}\ Ω

For series combination,

Req=Rp+6=32+6=152 ΩR_{eq} = R_p+6=\dfrac{3}{2}+6=\dfrac{15}{2}\ Ω

 

(iii) Here two 1 Ω resistances are in series combination and this whole combination is in parallel combination with itself.

For series combination,

Rs=1+1=2 ΩR_s=1+1=2\ Ω

For Parallel combination,

Req=2×22+2=44=1 ΩR_{eq} = \dfrac{2\times2}{2+2}=\dfrac{4}{4}= 1\ Ω

 

(iv) Here all resistances are in series combination.

Req=3+2+3=8 ΩR_{eq}=3+2+3=8\ Ω

Question 1(x)

Three bulbs of equal resistance are connected as shown in figure. Out of the following statements which one is correct ?

Three bulbs of equal resistance are connected as shown in figure. Out of the following statements which one is correct. Concise Physics Solutions ICSE Class 10.
  1. The current at B is greater than that at C.
  2. The current at A equals the current at B.
  3. The current at C is greater than that at B.
  4. The current at A equals the current at C.
Section A

Answer:

The current at B is greater than that at C.

Reason — All bulbs have equal resistance then the two bulbs in series at C will have a combined resistance greater than the resistance of bulb B therefore, more current will flow through B than through bulb C and the total current from A splits between B and C. So,

(a) Current at B is greater than C — Correct, because B has lower resistance and gets more current.

(b) Current at A equals current at B — Incorrect, because A splits into B and C, so A's current cannot equal B's current.

(c) Current at C is greater than at B — Incorrect, because C has higher equivalent resistance and hence less current flows through C.

(d) Current at A equals current at C — Incorrect, because A's current splits into B and C.

Question 1(xi)

Assertion (A) : The direction of induced current is such that it opposes the cause which produces it.

Reason (R) : A current is induced in the coil if there is a change in the magnetic flux linked with the coil.

  1. both A and R are true and R is the correct explanation of A
  2. both A and R are true and R is not the correct explanation of A
  3. assertion is false but reason is true
  4. assertion is true but reason is false
Section A

Answer:

both A and R are true and R is not the correct explanation of A

Explanation

Assertion (A) is true. Lenz’s Law states that the induced current opposes the cause producing it

Reason (R) is true. From Faraday’s law, which says a current is induced in a coil when there is a change in magnetic flux linked with the coil. However, Reason (R) is not the direct explanation of Assertion (A) because R explains the condition for induction, not the direction.

Question 1(xii)

Two blocks P and Q of different metals having their masses in the ratio 3 : 2 are given the same amount of heat. Their temperature rise by the same amount. The ratio of their specific heat capacities is :

  1. √3 : √2
  2. 9 : 4
  3. 2 : 3
  4. 3 : 2
Section A

Answer:

2 : 3

Reason — Let masses of P and Q be mP\text{m}_\text{P} and mQ\text{m}_\text{Q} and their specific heat capacities cP\text{c}_\text{P} and cP\text{c}_\text{P} respectively. Let heat given be Q and temperature change is ΔT.

Given,

MPMQ=32\dfrac{\text{M}_\text{P}}{\text{M}_\text{Q}}=\dfrac{3}{2}

As, Q = McΔT\text{Q = McΔT}

According to question,

MPcPΔT=MQcQΔT\text{M}_\text{P}\text{c}_\text{P}Δ\text T=\text{M}_\text{Q}\text{c}_\text{Q}Δ\text T

MPcP=MQcQ\text{M}_\text{P}\text{c}_\text{P}=\text{M}_\text{Q}\text{c}_\text{Q}

cPcQ=MQMP=23\dfrac{\text{c}_\text{P}}{\text{c}_\text{Q}} = \dfrac{\text{M}_\text{Q}}{\text{M}_\text{P}}=\dfrac{2}{3}

Question 1(xiii)

The temperature of a substance on heating rises because :

  1. average potential energy of molecules increases
  2. average kinetic energy of molecules increases
  3. average kinetic energy of molecules decreases
  4. average potential energy of molecules remains the same
Section A

Answer:

average kinetic energy of molecules increases

Reason — When a substance is heated, the molecules move faster, and their kinetic energy increases, which results in an increase in temperature.

Question 1(xiv)

The magnetic field lines produced by passing current in a long wire is shown alongside. Identify the correct condition for the intensity of the magnetic field to be maximum.

The magnetic field lines produced by passing current in a long wire is shown alongside. Identify the correct condition for the intensity of the magnetic field to be maximum. Concise Physics Solutions ICSE Class 10.
 Distance from the wireThe amount of current
(a)smallsmall
(b)smalllarge
(c)largesmall
(d)largelarge
  1. Option (a)
  2. Option (b)
  3. Option (c)
  4. Option (d)
Section A

Answer:

Option (b)

Reason — Magnetic field is directly proportional to the current which means higher current gives higher magnetic field and inversely proportional to distance so smaller the distance to the wire, higher the magnetic field.

Question 1(xv)

An endoscope uses optical fibre to transmit high resolution images of internal organs without the loss of information. The phenomenon of light that governs the functioning of the optical fibres is :

  1. refraction
  2. reflection
  3. scattering
  4. total internal reflection
Section A

Answer:

total internal reflection

Reason — Optical fibres work on the principle of Total Internal Reflection. As light entering the fibre is completely reflected back inside at each boundary of the fibre, without escaping, provided the angle of incidence is greater than the critical angle which allows the light to travel long distances through the fibre with minimal loss of energy or information, making it perfect for devices like endoscopes.

Question 2(i)

Complete the following by choosing the correct answer from the bracket:

(a) Bottle opener is an example of ............... [class I/class II/class III] lever.

(b) If there are n movable pulley with one fixed pulley, the mechanical advantage is given by ............... [2n/2n+1/2n1][2^n/2^{n+1}/2^{n-1}]

(c) According to work energy theorem, the increase in the kinetic energy of a moving body is equal to ............... by a force acting in the direction of moving body [velocity/acceleration/work done].

(d) If a machine does a given amount of work in less time, ............... power is spent by it. [less/more/same]

(e) If the length l of wire is increased to 3l by stretching its resistance R increases to ............... [3R/6R/9R]

(f) If a coil is viewed from one end, and the current flows in clockwise direction, then the end is a .............. pole. [north/south]

Section A

Answer:

(a) Bottle opener is an example of class II lever.

(b) If there are n movable pulley with one fixed pulley, the mechanical advantage is given by 2n.

(c) According to work energy theorem, the increase in the kinetic energy of a moving body is equal to work done by a force acting in the direction of moving body.

(d) If a machine does a given amount of work in less time, more power is spent by it.

(e) If the length I of wire is increased to 31 by stretching its resistance R increases to 9R.

(f) If a coil is viewed from one end, and the current flows in clockwise direction, then the end is a south pole.

Reason

(a) In a bottle opener, the fulcrum and the effort are at the two ends of it and the load is somewhere in between the effort and the fulcrum in such a way so that the load and effort are on the same side of the fulcrum but acting in opposite directions same as class II lever so here the effort arm is always longer than the load arm.

(b) As each movable pulley doubles the mechanical advantage by halving the effort required, and these effects multiply as more movable pulleys are added like a single movable pulley which divides the load between 2 rope segments which make the effort needed only half the load and has mechanical advantage equal to 2 so similarly n movable pulleys divide the load between 2n rope segments which make mechanical advantage equal to 2n.

(c) Work energy theorem states that the increase in kinetic energy of a moving body is equal to the work done by a force acting in the direction of the moving body.

(d) Power spent by a source is given by the amount of work done by the source per unit time taken by the source to do the said work i.e., Power=Work doneTime taken\text{Power} = \dfrac{\text {Work done}}{\text {Time taken}} So, if a machine does a given amount of work in less time then more power is spent by it.

(e) The resistance of a wire is directly proportional to its length and inversely proportional to the area of cross section of the wire and is given as :

R=ρla\text R = \text ρ\dfrac{\text l}{\text a}

When length of a conductor is increased by 3 times then it's cross sectional area decreases to 1/3 to keep volume constant therefore, resistance increases by 3 × 3 = 9 times.

(f) The polarity at the two faces of a coil depends on the direction of current in the coil so while looking at the face of the coil, if the current in wire around that face is in anticlockwise direction, the face has the north polarity, while if the current at that face is in clockwise direction, the face has the south polarity.

Question 2(ii)

A ball is placed on a compressed spring, which form of energy does the spring possess? On release of the spring, the ball flies away. Give reason.

Section A

Answer:

A compressed spring possesses elastic potential energy due to its compressed state. When the spring is released, the potential energy of the spring changes into kinetic energy which does work on the ball placed on it and changes into the kinetic energy of the ball due to which it flies away.

Question 2(iii)

(a) What is 'SONAR'?

(b) State the principle on which is based.

Section A

Answer:

(a) SONAR stands for sound navigation and ranging. It is a technology that uses ultrasonic sound waves to detect and locate objects under water.

(b) Principle of SONAR is when ultrasonic waves are sent in sea water in all directions from the ship then these waves are received after reflection from an obstacle such as an enemy submarine, iceberg, sunken ship, etc. so to find the distance of the obstacle from the ship, the time interval between the instant when waves are sent and the instant when the waves are received, after reflection from the obstacle is measured.

Question 3(i)

A lens forms an upright and magnified image of an object. Identify the lens and draw a labelled ray diagram to show the image formation.

Section A

Answer:

(a) The lens is a convex lens. A convex lens forms an upright and magnified image, when the position of the object is between the lens and F1.

(b) Below labelled ray diagram shows the image formation:

Draw a ray diagram to show how a converging lens is used as a magnifying glass to observe a small object. Mark on your diagram the foci of the lens and the position of the eye. Refraction through a lens, Concise Physics Class 10 Solutions.

Question 3(ii)

(a) Which electrical component protects the electric circuit in case of excess current and can also be used a switch ?

(b) Name the wire to which this electrical component is connected in an electric circuit?

Section A

Answer:

(a) Miniature circuit breaker (MCB).

(b) Live wire.

Question 3(iii)

The diagram in the figure given alongside shows the core of a transformer and its input and output connections.

The diagram in figure shows the core of a transformer and it's input and output connections. Electromagnetism, Concise Physics Solutions ICSE Class 10.

(a) Is this transformer a step up or step down?

(b) State the material used for the core and describe its structure.

Section A

Answer:

(a) As this transformer converts 220 V of inputs into 44 V of output so it is reducing the voltage. Hence, it is a step down transformer.

(b) The core is made from thin rectangular laminated sheets of soft iron of T and U shape, placed alternately one above the other and insulated from each other by a paint (or varnish) coating over them which forms a simple rectangular core.

Question 3(iv)

Out of the following metals A, B and C of specific heat capacities 900 J kg⁻¹ °C⁻¹, 380 J kg⁻¹ °C⁻¹ and 460 J kg⁻¹ °C⁻¹ respectively, which one will you prefer for calorimeter? Give reason.

Section A

Answer:

Metal B should be preferred for calorimeter because it requires a material with low specific heat capacity so the amount of heat energy taken by the calorimeter itself from the contents to acquire its temperature, is very small.

Question 3(v)

State the condition in each case for the magnitude of force on a current carrying conductor placed in a magnetic field to be (a) zero and (b) maximum.

Section A

Answer:

(a) If the current carrying conductor is placed between the poles of magnet with its length in a direction such that the current in it is in the direction parallel to the direction of magnetic field then zero force acts on the wire and it does not move.

(b) If the current carrying conductor is placed between the poles of magnet with its length in a direction such that the current in it is in the direction perpendicular to the direction of magnetic field then maximum force acts on the wire.

Question 3(vi)

Arrange the α, ẞ and γ radiations in ascending order of their (i) ionising power, and (ii) penetrating power.

Section A

Answer:

(i) Ionizing power → γ < β < α

(ii) Penetrating power → α < β < γ

Question 3(vii)

The figure given below shows a radioactive nucleus A composed of 84 protons and 128 neutrons kept in a thick lead walled container. The emitted particles pass through a magnetic field in a direction perpendicular to the plane of paper inwards as shown by X. The nucleus A emits a particle which deflects to the left and is transformed into nucleus B. Nucleus B further emits a particle which deflects towards right and transforms into nucleus C.

The figure given below shows a radioactive nucleus A composed of 84 protons and 128 neutrons kept in a thick lead walled container. Concise Physics Solutions ICSE Class 10.

(a) Name the radiations emitted by nucleus A and B.

(b) Name the law used to identify the radiations.

(c) What is the composition of nucleus C?

Section A

Answer:

(a) Nucleus A emits alpha (α) radiation and nucleus B emits beta (β) radiation.
As particles emitted by nucleus A deflect to left hence, they are alpha (α) particles and nucleus B emission deflects to the right hence, they are beta (β) particles.

(b) Fleming's left hand rule.

(c) Nucleus C will have 83 protons and 125 neutrons.

Explanation:

Given,

Before any decay :

Number of protons in A = 84 = Atomic number of A

Number of neutrons in A = 128

Mass number = 84 + 128 = 212

After alpha decay (A → B) :

Loss of protons = 2

Loss of neutrons = 2

New atomic number = 84 – 2 = 82

New mass number = 212 – 4 = 208

After beta decay (B → C) :

As in beta decay, a neutron converts into a proton so that Atomic number increases by 1 and mass number remains the same.

New atomic number = 82 + 1 = 83

Mass number = 208

For C

Number of protons = atomic number of C = 83

Number of neutrons = Mass number - number of proton = 208 - 83 = 125

Section B

18 questions

Question 4(i)

Jatin puts a pencil into a glass container having water and is surprised to see the pencil in a different state.

Jatin puts a pencil into a glass container having water and is surprised to see the pencil in a different state. Concise Physics Solutions ICSE Class 10.

(a) What change is observed in the appearance of the pencil?

(b) What is the reason for this change?

(c) Classify the following as a real or virtual image:
The image of a candle flame formed on a screen by a convex lens.

Section B

Answer:

(a) The pencil appears bent at the water surface; the submerged part looks raised and slightly shortened.

(b) This happens due to refraction of light. Light rays from the submerged pencil bend away from the normal as they pass from water (denser) to air (rarer), making the pencil appear at a higher position than it actually is

(c) Image of candle is a real image as it can be formed on a screen.

Question 4(ii)

The adjacent diagram represents a glass slab of refractive index 1.5. If PA, PB, PC and PD represent the rays of light from a point P at the bottom of the block, draw the approximate directions of these rays as they emerge out of the glass slab. (sin 42° = 2/3)

The adjacent diagram represents a glass slab of refractive index 1.5. If PA, PB, PC and PD represent the rays of light from a point P at the bottom of the block, draw the approximate directions of these rays as they emerge out of the glass slab. Concise Physics Solutions ICSE Class 10.
Section B

Answer:

Given,

Refractive index of glass with respect to air = aμg = 1.5 = 1gμa\dfrac{1}{_\text g\text μ_\text a}

Angle made by PA with normal = 0°

Angle made by PB with normal = 30°

Angle made by PC with normal = 42°

Angle made by PD with normal = 60°

sin 42° = 2/3

The adjacent diagram represents a glass slab of refractive index 1.5. If PA, PB, PC and PD represent the rays of light from a point P at the bottom of the block, draw the approximate directions of these rays as they emerge out of the glass slab. Concise Physics Solutions ICSE Class 10.

(i) For ray PA

i = 0°

sin r=sin igμa=sin 0°×1.5=0\text {sin r} = \dfrac{\text{sin i}}{{_g\text{μ}_a}}=\text {sin }0°\times 1.5 =0

⟹ r = sin-1(0) = 0°

Ray PA will not deviate and emerge straight.

(ii) For PB

i = 30°

sin r=sin igμa=sin 30°×1.5=0.5×1.5=0.75\text {sin r} = \dfrac{\text{sin i}}{{_\text{g}\text{μ}_a}}=\text {sin}\ 30°\times 1.5 = 0.5\times1.5=0.75

⟹ r = sin-1(0.75) ≈ 49°

Ray PB will emerge at an angle of 49° with the normal.

(iii) For PC

i = 42°

sin r=sin igμa=sin 42°×1.5=23×1.5=1\text {sin r} = \dfrac{\text{sin i}}{{_\text{g}\text{μ}_a}}=\text {sin}\ 42°\times 1.5 = \dfrac{2}{3} \times1.5=1

⟹ r = sin-1(1) = 90°

This is the condition for critical angle.

Ray PC will graze along the surface.

(iv) For PD

i = 60°

sin r=sin igμa=sin 60°×1.5=32×1.51.3\text {sin r} = \dfrac{\text{sin i}}{{_\text {g}\text{μ}_a}}=\text {sin}\ 60°\times 1.5 = \dfrac{\sqrt 3}{2} \times1.5≈1.3

Here,

sin r > 1 which a condition of total internal reflection.

Ray PD will suffer total internal reflection.

Question 4(iii)

The figure given below shows the experimental set up for the determination of focal length of a lens using a plane mirror.

The diagram in figure shows experimental setup for determination of the focal length of a lens using a plane mirror. Draw two rays from the point O of the object to show the formation of image I at O itself. What is the size of the image I? State two more characteristics of the image I. Name the distance of the object O from the optical centre of the lens. To what point will the rays return if the mirror is moved away from the lens by a distance equal to the focal length of the lens? Refraction through a lens, Concise Physics Class 10 Solutions.

(a) Draw two rays from the point O of the object to show the formation of image I at O itself.

(b) What is the size of the image I?

(c) A convex lens of short focal length is used in which optical instrument?

(d) The focal length of a convex lens is 25 cm. Express its power with sign.

Section B

Answer:

(a) Below ray diagram shows the formation of image I at O itself:

The diagram in figure shows experimental setup for determination of the focal length of a lens using a plane mirror. Draw two rays from the point O of the object to show the formation of image I at O itself. What is the size of the image I? A convex lens of short focal length is used in which optical instrument? The focal length of a convex lens is 25 cm. Express its power with sign. Model Paper 1, Concise Physics Class 10 Solutions.

(b) The size of the image is same as that of the object O.

(c) A convex lens of short focal length is used in a microscope to magnify small objects.

(d) Given,

Focal length of convex lens = f = + 25 cm = + 0.25 m

Power of a lens is expressed as reciprocal of its focal length in metre and written as :

P=1f=10.25=+4D\text P=\dfrac{1}{\text f}=\dfrac{1}{0.25}=+4 D

Power of the lens is +4 Dioptre

Question 5(i)

A lens of focal length 15 cm forms an erect image three times the size of the object on a screen.

(a) What kind of lens is this?

(b) Find the position of the object from the lens.

(c) Find the position of the screen from the lens.

Section B

Answer:

Given,

Focal length of the lens (f) = 15cm

Magnification of the lens (m) = +3 ∵ (image formed is virtual and erect)

(a) This is a convex lens because only convex lens forms a magnified image when the object is between the lens and the focus.

(b) Let,

Distance of object from the lens = u

Distance of image from the lens = v

As for lens,

m=vuvu=3\text m = \dfrac{\text v}{\text u} \\[1em] \Rightarrow \dfrac{\text v}{\text u}=3

⇒ v = 3u .......... (1)

From lens formula,

1v1u=1f\dfrac{1}{\text v}-\dfrac{1}{\text u}=\dfrac{1}{\text f}

On putting values

13u1u=11513u33u=11523u=115u=15×23=10 cm\Rightarrow \dfrac{1}{3\text u}-\dfrac{1}{\text u}=\dfrac{1}{15} \\[1em] \Rightarrow \dfrac{1}{3\text u}-\dfrac{3}{3\text u}=\dfrac{1}{15} \\[1em] \Rightarrow -\dfrac{2}{3\text u}=\dfrac{1}{15} \\[1em] \Rightarrow \text u = \dfrac{-15\times 2}{3}=-10 \text{ cm}

The object is placed 10 cm in front of the lens.

(c) From (1)

v=3u=3×(10)= 30 cm\text v = 3\text u =3\times (-10)=-\ 30 \text{ cm}

Distance of the screen from the lens = image distance (v) = -30 cm

The screen is placed 30 cm in front of the lens.

Question 5(ii)

The diagram given below shows a ray of blue, yellow and red light incident on one face of a prism.

The diagram given below shows a ray of blue, yellow and red light incident on one face of a prism. Concise Physics Solutions ICSE Class 10.

(a) Complete the diagram to show the effect of the prism on the ray of light and to show what is seen on the screen.

(b) A slit is placed in between the prism and the screen to pass yellow light only. What will be your observation on the screen ?

(c) What conclusion do you draw from the observation in part (ii) above ?

Section B

Answer:

(a) Below diagram shows the effect of the prism on the beam and the spectrum seen on the screen :

Below diagram shows the effect of the prism on the beam and the spectrum seen on the screen Concise Physics Solutions ICSE Class 10.

(b) When a slit is placed in between the prism and the screen to pass only the light of yellow colour then only yellow light is observed on the screen because the slit blocks other colours (blue and red).

(c) From above experiment we can conclude that the prism does not produce colours, but it only splits the various colours present in the light incident on it.

Question 5(iii)

In the figure given below, a ray of light PQ is incident normally on the face AB of an equilateral glass prism. Complete the ray diagram showing its emergence into air after passing through the prism. Take critical angle for glass = 42°

Copy the diagram given below and complete the path of the light ray till it emerges out of the prism. The critical angle of glass is 42°. In your diagram mark the angles wherever necessary. Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

(a) Write the angles of incidence at the faces AB and AC of the prism.

(b) Name the phenomenon suffered by the ray of light at the face AB, AC and BC of the prism respectively.

Section B

Answer:

Below is the completed diagram showing the path of the light ray till it emerges out of the prism with all angles marked:

Completed diagram showing the path of the light ray till it emerges out of the prism with all angles marked. The critical angle of glass is 42°. Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

(a)

Angle of incidence at face AB = 90°
Angle of incidence at face AC = 60°
Angle of incidence at face BC = 90°

(b)

At face AB and AC — The ray of light passes undeviated
At face BC — The ray of light suffers total internal reflection

Question 6(i)

A uniform metre scale is in equilibrium as shown in the diagram below :

A uniform metre scale is in equilibrium as shown in the diagram. Calculate the weight of the metre scale. Which of the following options is correct to keep the ruler in equilibrium when 40 gf wt is shifted to 0 cm mark? F is shifted towards 0 cm or F is shifted towards 100 cm. Model Paper 1, Concise Physics Class 10 Solutions.

(a) Calculate the weight of the metre scale.

(b) Which of the following options is correct to keep the ruler in equilibrium when 40 gf wt is shifted 0 cm mark ? F is shifted towards 0 cm or F is shifted towards 100 cm.

Section B

Answer:

(a) Given,

Fulcrum (F) is at 30 cm mark.

A 40 gf weight is hanging at 5 cm mark.

Let the weight of the metre scale be W gf, acting at its centre of gravity which is at it's midpoint i.e., at 50 cm mark.

Moment of force due to 40 gf weight:

Distance from F = 30 – 5 = 25 cm (anticlockwise)

So, Anticlockwise moment of force = 40 × 25 = 1000 gf cm

Moment due to the weight of the scale :

Weight W gf acts at 50 cm mark.

Distance from F = 50 – 30 = 20 cm (clockwise)

So, Clockwise moment of force = W × 20 gf cm

Using principle of moments:

Clockwise moments of force about the fulcrum = Anticlockwise moments of force about the fulcrum

45 x 25 = W x 20

⇒ 1000 = 20 W

⇒ W = 100020\dfrac{1000}{20} = 50 gf

∴ Weight of the metre scale = 50 gf

(b) Now the 40 gf weight is shifted to the 0 cm mark.

New moment of force due to 40 gf at 0 cm :

Distance from F = 30 – 0 = 30 cm (anticlockwise)

Anticlockwise moment of force = 40 × 30 = 1200 gf cm

Moment of force due to the metre scale (still acting at its centre of gravity i.e., at 50 cm mark) :

Distance from F = 50 – 30 = 20 cm (clockwise)

Clockwise moment of force = 50 × 20 = 1000 gf cm

Here,

Anticlockwise moment of force > Clockwise moment of force, so the scale will tilt anticlockwise unless we correct this.

To balance, the fulcrum F must be shifted towards 0 cm mark to reduce the arm length of the 40 gf weight.

∴ F is shifted towards 0 cm mark.

Question 6(ii)

State whether work is done or not in the following cases, by writing Yes or No :

(a) A man pushes a wall.

(b) A coolie stands with a box on his head for 15 minutes.

(c) A boy climbs up 20 stairs.

Section B

Answer:

(a) A man pushes a wall. — No
Reason: There is no displacement of the wall, so no work is done.

(b) A coolie stands with a box on his head for 15 minutes. — No
Reason: The coolie is not moving, so there is no displacement.

(c) A boy climbs up 20 stairs. — Yes
Reason: There is displacement in the direction of force (against gravity), so work is done.

Question 6(iii)

The diagram given alongside shows a ski jump. A skier weighing 60 kgf stands at A which is the top of the ski jump. He moves from A and takes off for his jump at B.

The diagram shows a ski jump. A skier weighing 60kgf stands at A at the top of ski jump. He moves from A and takes off for his jump at B. Calculate the change in the gravitational potential energy of the skier between A and B. Work, Energy, Power Concise Physics Class 10 Solutions.

(a) Calculate the change in the gravitational potential energy of the skier between A and B.

(b) If 75% of the energy in part (i) becomes kinetic energy at B, calculate the speed at which the skier arrives at B. [Take g = 10 ms-2]

(c) Does total mechanical energy change during the ski jump?

Section B

Answer:

Given,

Mass = 60 kg

(a)

Loss in potential energy=mg(h1h2)=60×10×(7515)=60×10×60=3.6×104 J\text {Loss in potential energy} = \text{mg} (\text h_1 – \text h_2) \\[0.5em] = 60 \times 10\times (75-15) \\[0.5em] = 60 \times10 \times 60 \\[0.5em] = 3.6 \times 10^4\ \text J \\[0.5em]

Change in gravitational potential energy is 36000 J

(b) When kinetic energy at B is 75% of (3.6 × 104)

Kinetic energy at B=75100×3.6×104=27000J=2.7×104 J\text{Kinetic energy at B} = \dfrac{75}{100} \times 3.6 \times 10^4 \\[0.5em] = 27000 \\ \text J \\[0.5em] = 2.7 \times 10^4\ \text J \\[0.5em]

Since,

Kinetic energy = 12\dfrac{1}{2} mv2

Substituting the values in equation we get,

27000=12×60×v227000=1×30×v2v2=2700030v2=900v=900v=3027000 = \dfrac{1}{2} \times 60 \times \text v^2 \\[0.5em] \Rightarrow 27000 = 1 \times 30 \times \text v^2 \\[0.5em] \Rightarrow \text v^2 = \dfrac{27000}{30} \\[0.5em] \Rightarrow \text v^2 = 900 \\[0.5em] \Rightarrow \text v = \sqrt{900} \\[0.5em] \Rightarrow \text v = 30 \\[0.5em]

∴ The speed at which the skier arrives at B = 30 ms-1.

(c) As there is an interchange between potential energy and kinetic energy and all kinds of frictional forces are absent so the total mechanical energy remains constant, hence no change in mechanical energy occurs.

Question 7(i)

A block and tackle has two pulleys in each block with the tackle tied to the hook of the lower block and the effort being applied upwards.

(a) Draw a neat diagram to show this arrangement and calculate its mechanical advantage.

(b) If the load moves up by a distance x, by what distance will the free end of the string move up?

Section B

Answer:

(a) The arrangement is shown in the figure given below :

The arrangement is shown in the figure given below. Concise Physics Solutions ICSE Class 10.

Here, the load is being supported by five segments of the string.

∴ L = 5T and E = T

Then,

M.A. = LE\dfrac{\text{L}}{\text{E}}

= 5TT\dfrac{\text{5T}}{\text{T}} = 5

∴ Mechanical advantage of the system is 5.

(b) In an ideal block and tackle system, if n segments of rope support the load, then the total length of the rope pulled = n × x.

As, here the load is being supported by five segments of the string then if the load moves up a distance x,

then,

Free end of the string will move up by a distance = 5x

Question 7(ii)

A person standing between two vertical cliffs and 640 m away from the nearest cliff produces a sound. He hears the first echo after 4 s and the second echo 3 s later. Calculate :

(a) The speed of sound in air and

(b) The distance between the cliffs

Section B

Answer:

Figure given below shows that two cliffs and the position of the person.

Figure given below shows that two cliffs and the position of the person. Concise Physics Solutions ICSE Class 10.

(a) First echo is heard from the nearest cliff.

Let d1, be the distance of the nearest cliff 1 from the person.

Total distance travelled by the sound in going and then coming back = 2d₁ = 2 x 640 m = 1280 m

Time taken (t) = 4 s

Speed of sound (v)=Tot. dist. sound travelsTime takenTime taken=2d1t=12804=320 ms1\text {Speed of sound (v)} = \dfrac{\text {Tot. dist. sound travels} }{\text {Time taken}} \\[1em] {\text {Time taken}}=\dfrac{2d_1}{\text t} \\[1em] =\dfrac{1280}{4}=320\text{ ms}^{-1}

(b) The second echo is heard from the farther cliff 2. If d2 is the distance of the farther cliff 2 from the person, then total distance travelled by the sound in going and then coming back = 2d2.

Time taken (t) = 4 + 3 = 7 s

Now,

Speed of sound (v)=Tot. dist. sound travelsTime takenv=2d2td2=vt2d2=320×72d2=1120 m\text {Speed of sound (v)} = \dfrac{\text {Tot. dist. sound travels} }{\text {Time taken}} \\[1em] \Rightarrow \text v = \dfrac{2d_2}{\text t} \\[1em] \Rightarrow \text d_2 = \dfrac{\text{vt}}{2} \\[1em] \Rightarrow \text d_2 = \dfrac{320\times7}{2} \\[1em] \Rightarrow \text d_2 = 1120 \text{ m}

Distance between the two cliffs = d1 + d2 = 640 m + 1120 m = 1760 m

Hence, speed of sound in air is 320 ms-1 and distance between the cliffs is 1760 m.

Question 7(iii)

Two pendulums C and D are suspended from a wire as shown in the figure given below. Pendulum C is made to oscillate by displacing it from its mean position. It is seen that D also starts oscillating.

Two pendulums C and D are suspended from a wire as shown in the given figure. Pendulum C is made to oscillate by displacing it from it's mean position. It is seen that D also starts oscillating. Name the type of oscillation, C will execute. Name the type of oscillation, D will execute. If the length of D is made equal to C, then what difference will you notice in the oscillations of D? What is the name of the phenomenon when the length of D is made equal to C? ICSE 2019 Physics Solved Question Paper.

(a) Name the type of oscillations that C will execute.

(b) Name the type of oscillations that D will execute.

(c) If the length of D is made equal to C, then what difference will you notice in the oscillations of D?

(d) What is the name of the phenomenon when the length of D is made equal to С?

Section B

Answer:

(a) C executes free vibrations (damped if medium is present).

(b) D executes forced oscillation.

(c) If the length of D is made equal to C, then the amplitude of vibration of pendulum D increases substantially.

(d) The phenomenon is called resonance.

Question 8(i)

(a) Students of a class perform an experiment on series and parallel combinations of two resistances R1 and R2. Some of the students plotted graph (1) and remaining plotted graph (2). Identify the correct graph and give reason.

A

Identify the correct graph and give reason. Concise Physics Solutions ICSE Class 10.

B

Identify the correct graph and give reason. Concise Physics Solutions ICSE Class 10.

(b) The length of an electric wire is increased by 25%. What is the percentage increase in the resistance and resistivity?

(c) Same current flows through an electric live wire and a bulb filament, but only the filament glows. Give reason.

Section B

Answer:

(a) Both the graphs are correct because when the resistors are joined in series, the equivalent resistance is more than when they are joined in parallel.

Therefore,

(i) On V-I graph, the straight line obtained for a series combination will have higher slope than that for a parallel combination because it's slope gives resistance.

(ii) On I-V graph, the straight line obtained for a series combination will have less slope than that for a parallel combination because it's slope gives conductance which is reciprocal of resistance.

(b) Given,

Length of the wire is increased by 25%.

Let original length = L,

New length = L' = L + 0.25L = 1.25L

If length increases by 25%, to keep the volume constant, area must decrease.

So,

Volume = constant

Let original area be A and new area be A'.

⇒ L ⋅ A = L′ ⋅ A′

A=L⋅AL=L⋅A1.25L=A1.25=0.8A\text A'= \dfrac{\text{L⋅A}}{\text L'}= \dfrac{\text{L⋅A}}{1.25\text L}= \dfrac{\text A}{1.25}=0.8\text A

So new area = 0.8A.

As,

Original Resistance :

R=ρLA\text R= \text ρ\dfrac{\text L}{\text A}

New Resistance :

R=ρLA=ρ1.25L0.8A=ρLA×1.250.8=1.5625R\text R'= \text ρ\dfrac{\text L'}{\text A'}=\text ρ\dfrac{1.25\text L}{0.8\text A}=\text ρ\dfrac{\text L}{\text A}\times\dfrac{1.25}{0.8}=1.5625R

% Increase in Resistance = (1.5625 − 1) × 100 % = 56.25 %

Change in Resistivity:

Resistivity (ρ) is a material property and it does not depend on length, area, or shape.

So,

% change in resistivity = 0%

(c) Although the same current flows through the live wire and the bulb filament, only the filament glows because it has very high resistance. This high resistance converts electrical energy into heat and light, making the filament white-hot and causing it to glow. On the other hand, the live wire has very low resistance, so it does not get heated up significantly and therefore does not glow.

Question 8(ii)

A certain nucleus P has a mass number 15 and atomic number 7.

(a) Find the number of neutrons.

(b) Write the symbol for the nucleus P.

(c) Write the symbols of the resulting nuclei in the following cases if :
the nucleus P loses :

  1. one proton
  2. one B-particle
  3. one α-particle

Express each change by a reaction.

Section B

Answer:

(a) The number of neutrons in a nucleus can be calculated using the formula:

Number of neutrons = Mass number (A) - Atomic number (Z)

Given,

Mass number (A) = 15
Atomic number (Z) = 7
So, number of neutrons = 15 - 7 = 8

(b) The nucleus P can be written as 157P.

(c)

  1. After the loss of 1 proton, the mass number and atomic number of the nucleus P will decrease by 1.The new nucleus will be 614Q^{14}_6\text Q (say). The change can be written as :
    715P614Q+11p^{15}_7\text P \longrightarrow ^{14}_6\text Q + ^{1}_1\text p

  2. After the loss of one β particle, the mass number will remain the same, but the atomic number will increase by 1. The nucleus 715P^{15}_7\text P changes to 815R^{15}_8\text R (say) as follows :
    715P815R+10e^{15}_7\text P \longrightarrow ^{15}_8\text R + ^{0}_{-1}\text e

  3. After the loss of one α particle, the mass number decreases by 4 and the atomic number decreases by 2. The nucleus 715P^{15}_7\text P changes to 811S^{11}_8\text S (say) as follows :
    715P511S+24He^{15}_7\text P \longrightarrow ^{11}_5\text S + ^{4}_2\text {He}

Question 8(iii)

The circuit diagram shown alongside includes a 6 V battery, an ammeter A, a fixed resistor R1, of 2 Ω and resistance wire R2, connected between the terminals A and B. The resistance of the battery and ammeter may be neglected. Calculate the ammeter readings when the wire R2, is of :

Identify the correct graph and give reason. Concise Physics Solutions ICSE Class 10.

(a) 0.20 m length and of resistance 4 Ω.

(b) 0.40 m length and of the same thickness and material as in case (i).

(c) 0.20 m length and having an area of cross section double than that in case (i).

Section B

Answer:

(a) Given,

Battery Voltage (V) = 6 V

Fixed Resistance (R1) = 2 Ω

Resistance of wire (R2) = 4 Ω

R1 and R2 are connected in series.

∴ Total resistance of the circuit (Rs) = R1 + R2 = 2 + 4 = 6 Ω

By Ohm's law,

Current (I)=VRs=66=1A\text {Current (I)}=\dfrac{\text V}{\text R_\text s}=\dfrac{6}{6}=1\text A

(b) On doubling the length (i.e., 2 x 0.20 = 0.40 m), the resistance of wire (R2) is doubled, i.e., it becomes 8 Ω.

Total resistance of the circuit (Rs) = R1 + R2 = 2 + 8 = 10 Ω

By Ohm's law,

Current (I)=VRs=610=0.6A\text {Current (I)}=\dfrac{\text V}{\text R_\text s}=\dfrac{6}{10}=0.6\text A

(c) On doubling the area of cross section, the resistance of wire R2, is reduced to half, i.e., it becomes 12×4\dfrac{1}{2}\times 4 Ω = 2 Ω.

Total resistance of the circuit (Rs) = R1 + R2 = 2 + 2 = 4 Ω

By Ohm's law,

Current (I)=VRs=64=1.5A\text {Current (I)}=\dfrac{\text V}{\text R_\text s}=\dfrac{6}{4}=1.5\text A

Question 9(i)

Calculate the total amount of heat required to melt 200 g of ice at 0°C to water at 100°C. (Specific latent heat of ice = 336 J g⁻¹, specific heat capacity of water = 4.2 J g⁻¹ °C⁻¹)

Section B

Answer:

Given,

m = 200 g

Specific heat capacity of water = 4.2 J g⁻¹ °C⁻¹

Specific latent heat of ice = 336 J g⁻¹

total amount of heat energy required = ?

Heat energy taken by ice at 0° C to convert into water at 0° C
= m x L
= 200 × 336
= 67200 J

Heat energy taken by water to raise the temperature from 0° C to 100° C
= m x c x change in temperature
= 200 × 4.2 × (100 - 0)
= 200 × 4.2 × 100
= 84000 J

Total heat energy gained is
= 67200 + 84000
= 151200 J
= 151.2 × 103 J = 151.2 KJ

Hence, total amount of heat energy required = 151.2 KJ

Question 9(ii)

Some ice is heated at a constant rate and its temperature is recorded after every few seconds till steam is formed at 100°C. Draw a temperature-time graph to represent the change. Label the two phase changes in your graph.

Section B

Answer:

Temperature-time graph representing the change of ice to steam is shown below:

Some ice is heated at a constant rate, and it's temperature is recorded after every few seconds, till steam is formed at 100°C. Draw a temperature time graph to represent the change. Label the two phase changes in your graph. ICSE 2016 Physics Solved Question Paper.

Question 9(iii)

The diagram below shows a magnetic needle kept just below the conductor AB which is kept in North-South direction.

The diagram below shows a magnetic needle kept just below the conductor AB which is kept in North South direction. In which direction does the needle deflect when the key is closed? Why is the deflection produced? What will be the change in the deflection if the magnetic needle is taken just above the conductor AB? Name one device which works on this principle. ICSE 2019 Physics Solved Question Paper.

(a) In which direction will the needle deflect when the key is closed ?

(b) Why is deflection produced ?

(c) What will be the change in deflection if the magnetic needle is taken just above the conductor AВ ?

(d) Name one device which works on this principle.

Section B

Answer:

(a) North pole of needle deflects towards east.

(b) On passing current through the wire AB, a magnetic field is produced around the wire which aligns the magnetic needle in it's direction.

(c) The direction of deflection gets reversed i.e., now it deflects towards west.

(d) An electric bell works on this principle.