SOLUTION MODEL SPECIMEN PAPER–3 - Oswal Publishers · 2018. 7. 18. · (iii) Electric toaster Alloy of nickel and chromium, nichrome High resistance (iv) Fuse wire Alloy of lead

SOLUTION

MODEL SPECIMEN PAPER–3SECTION–I

Answer 1.(a) (i) The weight of the truck = 1050 kgf = 1050 × 10 = 10500 N.

The acceleration of the truck = (72 – 36) ×

518

2 × 60 = 0·083 ms– 2.

Hence the displacement,

s = ⎝⎜⎛

⎠⎟⎞

36 × 518 × 120 +

12 × 0·0833 ×1202

= 1800 mTherefore, the work done can be calculated as

W = 10500 × 1800 = 18900000 J.

(ii) Power, P =Wt =

189000002 × 60 = 15750 W.

(b) Building won’t have any momentum as there is no net force acting on the building andthus the building won’t have any motion.

(c) (i) Class III Lever(ii) Class II Lever

(d) Potential energy is the energy possessed by an object even when the body is not in motion.Example : Mass attached to a spring.

(e) (i) The lever whose mechanical advantage is greater than one are used as a forcemultiplier.

(ii) The lever whose mechanical advantage is equal to one can be used as a physicalbalance.

Answer 2.

(a) The index of refraction is a function of the wavelength of the light. The wavelength of redlight is longer than the wavelength of blue light. Therefore, blue light bends more when itpasses from air to glass. As the angle of deviation in both cases will be different, therefractive index will also be different. It will be more in case of blue light than red light.

(b) The brilliance of the diamond is due to the total internal reflection of light. The criticalangle for diamond air interface is 24·4°. The diamond is cut suitably so that light enteringthe diamond from any face falls at an angle greater than 24·4°. Hence, the light suffersmultiple total internal reflections. Finally, the light emerges out in certain directions. Thismake the diamond to shine.

(c) The quartz prism is required for obtaining spectrum of ultraviolet light as these radiationare not absorbed by quartz, while other ordinary glass absorbs the ultraviolet light.

(d) (i) We know that, Loudness ∝ (Amplitude)2

Hence, the ratio of loudness = ⎝⎜⎛

⎠⎟⎞2

3

2 =

⎝⎜⎛

⎠⎟⎞4

9 = 4 : 9.

(ii) Frequency remains same if the pitch is unchanged. Hence, the ratio of frequencies= 1 : 1.

(e) For a resonance to occur the object must have its natural frequency and an external forcemust be applied on the object to set it into vibration.

2 | ICSE Model Specimen Papers, X

At resonance, a compression of one wave falls on the compression of another wave and ararefaction on the rarefaction. Because of this, the amplitude of vibrating particlesincreases. Since the intensity of sound is directly proportional to the square of theamplitude of the vibrating particles (I ∝ a2), maximum sound is heard at resonance pipefor same fundamental frequency.

Answer 3.

(a) The time taken to reach the hill, t = 12 s, Speed of sound in air, v = 350 ms– 1.

So, the distance between the boy and the hill,

s = vt = 350 × 12 = 175 m.

(b) The strength of the electromagnet can be increased by :(i) increasing the current in the coil, and(ii) increasing the number of turns in the coil.

(c) Soft iron core is used as a core of the electromagnet in electric bell because it creates strongmagnetic field and loses magnetism as soon as the current is switched off.

(d) A split-ring commutator (sometimes just called a commutator) is used to reverse thedirection of current through an armature after every half rotation.

(e) Lumps of ice at 0°C are more effective in cooling a drink than water at 0°C because ice willconvert into the liquid state by absorbing the latent heat of fusion. On the other hand, nosuch heat will be absorbed by water as it is already in the liquid state. Thus, the drinkliberates an additional amount of heat energy to ice than to water at 0°C. Therefore,cooling produced by lumps of ice is much more than that by water at 0°C.

Answer 4.(a) (i) The melting point is lowered with an increase in pressure. So, ice melts at a

temperature less than 0°C.(ii) The boiling point of water increases with an increase in pressure. So the water will

boil at temperature more than 100°C.(b) Water might have some impurities (e.g., salt dissolved in it), as the boiling point of water

increases by the addition of impurities in it.(c) (i) The β radiations are a stream of electrons which are emitted from a nucleus.

(ii) Alpha particles can be blocked by a few pieces of paper. Beta particles pass throughpaper, but are stopped by aluminium foil. Gamma rays are the most difficult to stopand require concrete, lead or other heavy shielding to block them and thus have themaximum penetrating power.

(d) The two harmful effects of radioactivity are given below :(i) Exposure to γ-rays destroy the cells of the body and cause cancer.(ii) Long exposure to Iodine-131 can affect the functioning of thyroid gland.

(e) A radioactive substance should not be touched by hand because they emit nuclearradiations continuously which can produce harmful effects on the body.

SECTION–IIAnswer 5.(a) Couple is the set of two equal and opposite forces having a different line of action.

Moment of a couple is the product of either of the forces of a couple and the perpendiculardistance between them. The S.I. unit of moment of couple is newton-metre (Nm).Let us consider two equal and opposite forces acting at points A and B. In the given figure,the two forces rotate the bar in anticlockwise direction.

Physics | 3

Moment of force F at the end A = F × OA (Anticlockwise)Moment of force F at the end B = F × OB (Anticlockwise)Total moment of couple (i.e., moment of both the forces,

= F × OA + F × OB= F × (OA + OB) = F × AB= F × d (Anticlockwise)

Moment of couple = Either force × Perpendicular distance betweenthe two forces (or couple arm)

(b) Let the man sit at a distance of x m from the centre.From the principle of moments,

2 × 30 + 2·5 × 50 = x × 74

⇒ x =2 × 30 + 2·5 × 50

74 = 2·5 m

(c) (i) The effort is marked as ‘E’.

(ii) As there are 5 strings in the system, velocity ratio (V.R.) = 5

Also, V.R. =dEdL

or dE = V.R. × dL

= 5 × 1

= 5 m

(iii) There are 5 strands of string to tackle the load.

(iv) The Mechanical Advantage is given as, M.A. = LoadEffort =

5TT = 5

Answer 6.

(a) (i) We know that,

v = λf

But, the frequency of light does not change when it travels in any medium.

∴ v ∝ λ

So, on increasing the wavelength of light, speed of light also increases.


(ii) The electromagnetic spectrum is the range of frequencies of electromagneticradiation and their respective wavelengths. The electromagnetic spectrum coverselectromagnetic waves with frequencies ranging from below one hertz to above1025 hertz, corresponding to wavelengths from thousands of kilometres down to afraction of the size of an atomic nucleus. This frequency range is divided intoseparate bands and the electromagnetic waves within each frequency band arecalled by different names like radiowaves, microwaves, infrared, visible light,ultraviolet, X-rays and gamma rays.

(b) The phenomenon of splitting of white light into its constituent colours on passing througha transparent medium is called dispersion of white light.Different components of white light have different refractive indices due to which theybend through different angles on passing through a prism. The red component of whitelight deviates the least while the violet component deviates the most. Thus, we get a bandof seven colours, i.e., VIBGYOR.

(c) (i) The special name is the Critical Angle.(ii) The angle of refraction is equal to 90°.(iii) The name of the phenomenon is Total Internal Reflection.(iv) The image of the object P is at P′ in the figure :

Answer 7.(a) We know that v = f × λ

Since, the medium is same,∴ v1 = v2⇒ f1λ1 = f2λ2⇒ 40 × 825 = 32 × f2

⇒ f2 =40 × 825

32 = 412532

= 1031·25 Hz(b) For cliff A : t1 = 3 s, v = 330 ms– 1

d1 =vt12

=330 × 3

2 = 495 m

For cliff B : t2 = 5 s, v = 330 ms– 1

d2 =vt22

=330 × 5

2 = 825 m

Distance between cliff A and B = d1 + d2 = 495 + 825 = 1320 m.

Physics | 5

(c) (i) The conditions to hear an echo are :(1) The minimum distance between the source of sound and obstacle should be 17

m.(2) The size of reflector must be large enough as compared to the wavelength of

sound wave.(3) The intensity of sound should be such that the reflected sound is sufficiently

intense to be audible.(ii) Let ‘t’ be the time after which an echo is heard, ‘d’ is the distance between the source

and the reflecting body and v is the speed of sound.Then in time t, distance travelled by sound = 2d

or, in one second distance travelled by sound = 2dt

But distance travelled by sound in one second = Speed of sound (v)

∴ v =2dt

⇒ d =vt2

Answer 8.

(a) (i) Net resistance between points A and B,

RAB = (R1 + R2) || (R3 + R4)

=(R1 + R2) (R3 + R4)

(R1 + R2) + (R3 + R4)

RAB =(2 + 2) (2 + 2)

(2 + 2) + (2 + 2) = 4 × 44 + 4 = 2 Ω

(ii) Net resistance between points A and B,RAB = (R1 + R2) || R3 || (R4 + R5)

1RAB

=1

(R1 + R2) + 1

R3 +

1R4 + R5

1RAB

=1

(2 + 3) + 11 +

1(2 + 3)

=1 + 5 + 1

5 = 75

⇒ RAB =57 Ω

(b) (i) The direction of magnetic field would be towards east.(ii) The rule that is used to determine the direction is the Right Hand Thumb Rule.

(c) S. No. Device Material Reason(i) Connecting wires Copper High conductivity(ii) Electric bulb filament Tungsten High melting point(iii) Electric toaster Alloy of nickel and

chromium, nichromeHigh resistance

(iv) Fuse wire Alloy of lead and tin, solder High resistance and lowmelting point.

Answer 9.(a) (i) Cold water should be preferred as specific heat capacity of cold water is more as

compared to that of hot water. So, it would absorb more heat from the burn andthus, soothes the burns faster.

(ii) Specific heat capacity is the amount of heat required to increase the temperature ofthe substance by 1°C.Since with the same amount of heat, temperature of material X increases by 3°C,while that of Y increases by 5°C, therefore, material X has greater specific heatcapacity.


(b) Let the final temperature of water be t°C.Fall in temperature of hot water = (80 – t)°CRise in temperature of cold water = (t – 25)°CSpecific heat capacity of water, c = 4200 J kg– 1 K– 1

If the heat absorbed by the bucket is neglected, then, by the principle of calorimetry,Heat lost by hot water = Heat gained by cold water

m1c (t1 – t) = m2c (t – t2)2 × 4200 × (80 – t) = 8 × 4200 × (t – 25)

80 – t = 4 (t –25)

t =1805 = 36°C

(c) Given : t = 2·5 min. = 2·5 × 60 = 150 s, P = 104 W, c = 0·91 Jg– 1 °C– 1 = 0·91 × 103 J kg– 1 °C– 1

The energy used in drilling, Q = P × t = 1·5 × 106 JSince, 50% of energy is lost to the surroundings,

∴ Useful energy, Q′ =Q2 =

1·5 × 106

2 = 7·5 × 105J

Now, we know thatQ′ = mc ΔT

⇒ 7·5 × 105 = 8 × 0·91 × 103 × ΔT⇒ 7·5 × 105 = 7·28 × 103 × ΔT

⇒ ΔT =7·5 × 105

7·28 × 103 ≈ 103°C

Answer 10.

(a) (i) 92U238 8 α

⎯→ xXy 6 β

⎯→ PbFrom the above decay,

x = 92 – (8 × 2) ; y = 238 – (8 × 4) [˙.˙ α particle = 2He4]⇒ x = 76 ; y = 206∴ Atomic number of Pb = 76 – (6 × – 1) = 82and, Atomic mass = 206 – 0 = 206.Thus, the atomic number of lead is 82 and its atomic weight is 206.

(ii) (1) Nuclear reactions involve a change in an atom’s nucleus, usually producing adifferent element. Chemical reactions, on the other hand, involve only arearrangement of electrons and do not involve changes in the nuclei.

(2) Different isotopes of an element normally behave similarly in chemical reactions.The nuclear chemistry of different isotopes vary greatly from each other.

(3) The rate of chemical reaction is influenced by temperature and catalysts whilethe rate of nuclear reaction is unaffected by such factors.

(b) (i) Infrared rays are heat waves while γ-rays are neutral in nature.(1) The wavelength of the infrared rays is much more than that of the gamma rays.(2) Penetrating power of gamma rays is much more than that of infrared rays.

(ii) Ionizing power of alpha particles is maximum followed by the beta particles and theleast ionizing power is of the gamma rays.

(c) (i) The number of electrons in the neutral atom is equal to its atomic number i.e., 92.(ii) The number of protons is equal to the number of electrons i.e., 92 and the number of

neutrons is equal to the difference between the mass number and atomic number ofthe atom i.e., (235 – 92) = 143.

(iii) Isotopes don’t have the same number of neutrons rather they have same number ofprotons.

(iv) The number of protons is equal to the atomic number of the atom i.e., 92 and thenumber of neutrons in it is equal to (238 – 92) = 146.

❏❏

SOLUTION


Answer 1.(a) (i) When the force is applied on the rigid body, the body changes its position due to the

application of force.(ii) When the force is applied on a non-rigid body, the shape of the body gets affected.

(b) (i) Scissors belong to class I lever with the fulcrum being in between the load and theeffort.

(ii) Pliers also belong to class I lever with the fulcrum being in between the load and theeffort.

(c) The principle of moments state that, under equilibrium position, the moment of loadabout the fulcrum is equal to the moment of effort about the fulcrum. These two momentsacts in opposite directions.Example : See-saw

(d) From the principle of moments,FL × dL = FE × dE

⇒ 200 × 1 = 20 × dE⇒ dE = 10 m

Now, M.A. =LoadEffort =

20020 = 10

(e) Let the mass of the meter scale be m.The centre of gravity of meter scale is at 50 cm.The scale balances horizontally on a knife edge at 55 cm, when a weight of 10 g issuspended from one end.So, the difference in centre of gravity and the point of balance,

x = 55 – 50 = 5 cmSince the weight is suspended at one of the extremes,∴ Distance from the point of knife edge to the weight,

y = 100 – 55 = 45 cmNow, by the principle of moments,

m × x = 10 × y⇒ m × 5 = 10 × 45⇒ m = 90 g

Answer 2.(a) (i) The lens having both the focal lengths equal is a biconvex or biconcave lens having

the radii of curvature equal.(ii) The ray must have passed through the optical centre.

(b) We know that the refractive index of a medium is given by :

aμm =Real depth

Apparent depthFor a given medium of water,

aμw =43

Real depth = 15 cm

⇒43 =

15Apparent depth

⇒ Apparent depth =15 × 3

4 = 454 cm


Thus, the coin seems to be raised by :

⎝⎜⎛

⎠⎟⎞

45 – 454 =

1354 cm

(c)

(d) Natural frequency of vibration of a stretched string is inversely proportional to the radius(thickness) of the string so notes of different frequencies can be obtained by producingvibrations in the different strings.

(e) Two applications of resonance :(i) To produce loud sound of musical instruments.(ii) In the tuning of radio and television sets.

Answer 3.(a) RADAR is an abbreviation of Radio Detection and Ranging and is used to detect the

position or movements of objects. Echo is heard distinctly only when the distance betweensource of sound and obstacle is large enough to allow the reflected sound to reach thesource at least 0·1 s after the original sound is heard.

(b) Given : For bulb R :

PR = 60 W

VR = 220 V

For bulb S :

PS = 60 W

VS = 110 V

We know, P =V2

R

For same power, V2 ∝ R

or ⎝⎜⎛

⎠⎟⎞VR

VS

2=

RRRS

⇒RRRS

= ⎝⎜⎛

⎠⎟⎞220

110

2

= 4 : 1

(c) (i) An alloy of lead and tin i.e., solder wire is used as a fuse wire.

(ii) Fuse is always made up of materials which have very high resistance and low meltingpoint so that the material can withstand the high power at instant and break-down onpower failure.

(d) (i) The resistivity increases with increase in temperature for metals.

(ii) The resistivity decreases with increase in temperature for semiconductors.

(e) Water has a high specific heat capacity and thus it does not cool down fast. It thereforeprotects the plants from frost.

Physics | 3

Answer 4.

(a) Mass of ice melted = (1000 – 600) g = 400 g = 0·4 kg

Mass of steam condensed = (450 – 400) g = 50 g = 0·05 kg

Heat taken in by 0·4 kg of ice during melting = 0·4 × 336000 J = 134400 J

Heat given out by 0·05 kg of steam during condensation = 0·05 × L

where, L = specific latent heat of vaporisation of steam.

Again, heat given out by 0·05 kg of water at 100°C in cooling down to 0°C

= 0·05 × 4200 × 100 J = 21000 J

Now, Heat lost = Heat gained

or 21000 + 0·05 × L = 134400

or 0·05 L = (134400 – 21000) J = 113400 J

∴ L =113400

0·05 = 2268000 Jkg– 1

(b) Given : m = 30 g, latent heat of fusion (L) = 80 cal g– 1

Heat energy required = mL

= 30 × 80

= 2400 cal.

(c) Radioactivity is the spontaneous emission of energy from unstable atoms. There are stableatoms, which remain the same always and unstable atoms, which break down or decayinto new atoms. These unstable atoms are said to be ‘radioactive’, because they emitradioactivity from the nucleus as they decay. The phenomenon of radioactivity isobserved in heavy elements such as Uranium. So radioactivity refers to the emissionparticles from nuclei as a result of nuclear instability.

(d) (i) Use : Radioactivity tracers are used in the medical field and also in the study of plantsand animals.

Harmful effect : Radioactivity rays ionize the molecules in the human body and killthe cells.

(ii) The biggest source of background radiation is Radon (Rn-222), a radioactive gaswhich arises from the decay of Ra-226, present in all rocks and soils.

(e) Gamma rays are the most dangerous as they have the maximum penetration power.SECTION–II

Answer 5.

(a) Given : m = 2 kg, h = 5 m, g = 10 ms– 2

(i) The potential energy possessed by the ball when initially at rest = mgh = 2 × 10 × 5= 100 J.

(ii) The kinetic energy possessed by the ball just before it hits the ground = The potentialenergy possessed by the ball when initially at rest = 100 J.

(iii) The mechanical energy is lost in the form of heat and sound when the ball hits theground.

(b) Class II lever. A common example of class II lever is the wheel barrow, where the wheel isthe fulcrum. In this, the load rests within the box, and the force is the lift that is suppliedby the user.

To increase the mechanical advantage of class II lever, length of effort arm must beincreased and distance between load and fulcrum must be decreased.


(c) (i) In a single fixed pulley, the mechanical advantage is always less than the velocityratio because mechanical advantage decreases due to the friction and weight ofmoving parts of the machine whereas the velocity ratio remains constant.

(ii) Movable pulley has efficiency less than 100% because some energy is lost inovercoming friction of the moving parts.

(iii) With an increase in the number of pulleys in the arrangement of block and tackle,the number of strands of strings that support the load increases and therefore, themechanical advantage increases.

(iv) The lower block of the pulley system must have a negligible weight. The reason canbe given as follows :

The actual Mechanical Advantage (M.A.) = n – W′P , where W′ is the weight of the

lower block. Hence smaller is the value of W′, more will be the mechanical advantage.Answer 6.(a) The rules for the construction of ray diagrams is as follows :

(i) Pick a point on the top of the object and draw at least two incident rays travellingtowards the lens.

(ii) Once these incident rays strike the lens, then refract them according to the rules ofrefraction.

(iii) After this, mark the image of the top of the object.(iv) The same needs to be repeated for the bottom of the object and the complete image

of the object can be obtained.(b) Given : h2 = 2h1 …(i)

and, the distance between the object and the screen i.e., u + v = 45 cm …(ii)We know,

Magnification (m) =h2h1

= vu

⇒ v = 2u [from (i)] …(iii)Solving equations (ii) and (iii), we get

2u + u = 45⇒ 3u = 45

⇒ u =453 = 15 cm

∴ v = 2u = 2 × 15 = 30 cmNow, by using lens formula,

1f =

1v –

1u =

130 –

1(– 15) =

330

⇒ f =303 = 10 cm

(c)

Physics | 5

(i) The rays should fall normal to the first face of the prism.(ii) Total Internal Reflection is responsible for this action of the prism as the glass has the

critical angle equal to 42°.(iii) The image is inverted with respect to the object.

Answer 7.(a) (i) When acoustic resonance takes place, the natural frequency of vibration of body

becomes equal to the frequency of external force and the amplitude of vibration of thebody becomes very large. Since the loudness is proportional to the square of theamplitude, therefore a loud sound is produced.

(ii) (1) The pitch is highest for the highest frequency, hence the highest pitch is forTrumpet.

(2) Tuning fork should resonate with Guitar.(b) (i) Let’s assume the speed of sound as 320 ms– 1 and we need 0·1 s to differentiate the

two sounds, hence the minimum distance required for producing the echo will be

given as, d = v × t

2 = 320 × 0·1

2 = 16 m, i.e., the minimum distance between source of

sound and reflecting body should be 16 m, therefore no echo would be heard at adistance of 11 m.

(ii) Given : Distance between man and wall (d) = 30 m, Speed of sound (v) = 330 ms– 1

We know, t =2dv =

2 × 30330

v = 0·182 sec.

(c) (i) The vibrations who amplitude decreases with time due to the presence of resistiveforce are called damped vibrations.

(ii) Given : The frequency of pendulum = 5 vibrations per second

∴ Time taken for 1 vibration =15 sec

Hence for 8 vibrations, the time taken = 85 sec.

Also, speed of sound, v = 340 ms– 1

So, we can find the distance between the cliff as :

d =vt2 =

340 × 82 × 5

= 272 mAnswer 8.(a) (i) (1)


(2) The equivalent resistance can be found as :

Req =6 × 46 + 4 + 4·8 =

2410 + 4·8 = 2·4 + 4·8 = 7·2 Ω

(ii) An A.C. generator works on the principle of electromagnetic induction i.e., when themagnetic flux linked with the coil changes, an e.m.f. is induced in it.

(b) Given : Power of motor, P1 = 1 H.P. = 746 W

Power of 2 fans, P2 = 2 × 80 W = 160 W

Current through each bulb, i = 0·2 A

Voltage supply, V = 220 V

We know, P = VI

∴ Power of one bulb, P3 = 220 × 0·2 = 44 W

Power of 5 bulbs = 5 × 44 = 220 WTotal power of all appliances = 746 + 160 + 220 = 1126 W

Energy consumed, E = P × tAnd each appliance works for 5 hours daily∴ E = 1126 × 5 = 5630 Wh = 5·63 kWh

Cost of electrical energy consumed = 20 paise per unit

∴ Total cost =20

100 × 5·63 = 1·126.

(c) (i) Given : V = 4 V, I = 0·4 A, R1 = 40 Ω, R2 = R

(1) The total resistance, Rtotal =VI =

40·4 = 10 Ω

(2) Since the two resistances are connected in parallel,

∴1

Rtotal=

1R1

+ 1

R2

⇒110 =

140 +

1R

⇒1R =

340

R =403 Ω

(ii) Resistance of each wire, R = 1·5 × 3= 4·5 Ω

And, effective resistance of the given arrangement,

R′ =R3 =

4·53 = 1·5 Ω

Answer 9.(a) Pure water has a lower boiling point as compared to impure water and thus it would start

boiling first. This is because impurities increase the boiling point.The boiling point of a liquid increases with the increase in pressure and decreases with thedecrease in pressure.

(b) Given : Mass, m = 200 g, Initial temperature, t = 0°CEnergy absorbed, E = 67200 J

We know,E = mL

Where, L = Specific latent heat of fusion

or L =Em =

67200200 = 336 Jg– 1

Now, Final temperature, T = 80°C

Physics | 7

Specific heat capacity, c = 4200 Jkg– 1 K– 1

Energy required, E = mc Δ T= 0·2 × 4200 × (80 – 0)= 0·2 × 4200 × 80= 67200 J

(c) Rise in temperature = (t2 – t1) °C

∴ Heat needed by calorimeter + Stirrer = x × s1 × (t2 – t1)

Heat needed by water initially present = z × s2 × (t2 – t1)Also,Heat given out by steam in condensing into water at 100°C = y × LAnd heat given out by this condensed water, at 100°C, in cooling into water at t2°C

= y × s2 × (100 – t2)Hence, x × s1 × (t2 – t1) + z × s2 × (t2 – t1)= y × L + y × s2 × (100 – t2)∴ yL = (xs1 + zs2) (t2 – t1) – ys2 (100 – t2)

∴ L = ⎣⎢⎢⎡

⎦⎥⎥⎤(xs1 + zs2) (t2 – t1)

y – [s2 (100 – t2)]

Answer 10.(a) (i) According to the question,

AZ X

– β⎯→ AZ + 1S

α⎯→ A – 4

Z + 1 – 2Y

Since the final resulting nucleus is PQY

From the above equation, P = A – 4; Q = Z + 1 – 2⇒ P = A – 4; Q = Z – 1

(ii) Isotones are elements having the same number of neutrons. Examples of isotones areChlorine-37 and Potassium-39.Isobars are atoms of different elements having the same atomic mass but differentatomic number. Examples are Carbon -12 and Boron-12.

(b) (i) The ‘enrichment’ of uranium means increasing the percent composition ofUranium -235 by the process of isotope separation. This is done because Uraniumtaken directly from the Earth is not suitable as fuel for most nuclear reactors andrequires enrichment to make it usable.

(ii) Control rods are used in nuclear reactors so that the fission rate of uranium andplutonium can be controlled. They are composed of chemical elements such asboron, silver, indium and cadmium which are capable of absorbing many neutronswithout undergoing fission.

(iii) Given : Energy, E = 1 × 106 kWh = 1 × 106 × 3·6 × 106 J = 3·6 × 1012 JWe know, E = Δmc2

∴ Δm =Ec2 =

3·6 × 1012

(3 × 108)2 = 4 × 10– 5 kg

(c) (i) Einstein’s mass energy relation is given by E = mc2, where c is the velocity of light.Given : m = 1 amu = 1 × 1·66 × 10– 27 = 1·66 × 10– 27 kgThus the equivalent energy will be,

E = mc2 = 1·66 × 10– 27 × (3 × 108)2

= 14·94 × 10– 11 J

=14·94 × 10– 11

1·6 × 10– 13 = 933·75 MeV

(ii) Co-60 is a radioactive isotope which is used in the treatment of Cancer.❏❏

SOLUTION


Answer 1.(a) We lean forward so as to keep the balance and that makes the feet in a straight line

between the centre of gravity and the centre of the earth.

(b) (i) Nut Cracker can be used to multiply forces.

(ii) Ideal pulley can be used to change the point of application of effort.

(c) If two bodies of equal masses are dropped from the same height then at any instant oftime, all the given physical quantities will remain same for both of them as the externalconditions are same for both of them.

(d) Given : F = 300 N, v = 18 km h– 1 = 18 × 5

18 ms– 1 = 5 ms–1

We know, P = Fv = 300 × 5 = 1500 W

=1500736 H.P. = 2·03 H.P.

(e) The situations are :(i) The vector sum of all the forces is zero.(ii) The sum of torques around the axis of the plane is zero.

Answer 2.(a) (i) Angle of deviation is equal to 0° as the ray of light would pass undeviated.

(ii) Angle of emergence is equal to angle of incidence i.e., 0°, with respect to the normal.(b) Given : λ = 0·01 Å = 10– 12 m

We know, c = λf

or f =cλ =

3 × 108

10– 12 = 3 × 1020 Hz

The wave is γ-rays.(c) (i) The wavelength of light is directly proportional to the speed of light in a medium.

Thus on increasing the wavelength, the speed of light increases.(ii) Minimum possible value for absolute refractive index is the refractive index of air

i.e., 1.(d) (i) Deviate the path of a ray of light by 90°.

(ii) Turn a ray of light through 180°.(e) (i) Decibel (dB).

(ii) Watt per square metre (Wm– 2).Answer 3.(a) No sound will not be audible on the surface of the Moon. This is because, sound needs a

medium to propagate and since there is no atmosphere on the surface of the Moon,therefore, the sound will not be heard.

(b) Given : v = 1·7 km s– 1 = 1700 ms–1, f = 1·2 MHz = 1·2 × 106 Hz

We know, v = λf

or λ =vf =

17001·2 × 106 = 0·00141 m

= 1·41 mm(c) Superconductivity is a phenomenon where the materials show zero electrical resistance at

very low absolute temperature. If a superconductor could work at room temperature, the


advantage would be that the current can be conducted over long distances without anypower dissipation as superconductors have zero resistance.

(d) Given : Power consumed by bulb, P1 = 100 WPower consumed by fan, P2 = 60 W

Time, t = 5 min = 5 × 60 = 300 sTotal power consumed, P = P1 + P2

= 100 + 60= 160 W

We know, E = P × t= 160 × 300= 4·8 × 104 J

=4·8 × 104

3·6 × 106 kWh

= 0·013 kWh(e) (i) The number of turns of primary coil in step-up transformer is more than that of the

secondary coil.(ii) The primary coil is made of thicker wire in step-up transformer.

Answer 4.(a) Given :

For hot body :mH = 60 gcH = ?tH = 100° C

For cold body :mC = 100 gcC = 4·2 J g– 1 °C– 1

tC = 18° CFinal temperature = 20° CFrom the principle of method of mixtures,

mH cH Δt = mC cC Δt′60 × cH × (100 – 20) = 100 × 4·2 × (20 – 18)

60 × cH × 80 = 420 × 2

cH =8404800 = 0·175

cH = 0·175 J g– 1 °C– 1

(b) All plants and animals have high content of water in their bodies because water with highspecific heat capacity helps them in regulating their body temperatures.

(c) At higher altitudes, the atmospheric pressure is low and since boiling point of waterdecreases with decrease in pressure so water starts boiling at temperature lower than100°C. Due to this, heat provided by water is not sufficient to cook vegetables. Therefore,it is difficult to cook vegetables on hills and mountains.

(d) Neutron is the most unstable particle as it decays to produce electrons and protons.(e) β-particles are fast moving electrons with a very low mass and thus have high charge to

mass density so they are deflected much more than the heavier alpha particles.SECTION–II

Answer 5.(a) (i) Bottle opener and nut cracker are the examples of force multiplier as their effort arm

is always longer than the load arm, so less effort is needed to overcome a large load.(ii) Knife is an example of speed multiplier in which effort arm is shorter than load arm

so larger displacement is obtained by a smaller displacement of effort.

Physics | 3

(b) (i) Yes, in uniform circular motion, the velocity of particle is variable i.e. has acceleration,while the speed remains constant.

(ii) Kinetic energy of a body of mass m moving with velocity v is given as,

K.E. =12 mv2

=12

m2v2

m

=12

(mv)2

mBut, mv = p (linear momentum)

∴ K.E. =p2

2m(c) (i) Given : E = 220 kJ = 220 × 103 J = 2·2 × 105 J, t = 55 s

We know, E = P × t

or P =Et =

2·2 × 105

55 = 0·04 × 105 W

= 4 × 103 W = 4 kW

(ii) Given : m = 500 kg, h = 80 m, t = 10 s, η = 40%

Here, work done by the pump is equal to potential energy at a height h.∴ W = mgh

= 500 × 10 × 80= 4 × 105 J

Power, P =Wt =

4 × 105

10 = 4 × 104 W

But, η = 40%∴ Power rating of the pump,

P′ =40

100 × 4 × 104

= 16000 WAnswer 6.(a) The ray diagram can be drawn as follows :

(b) (i) Given : μ2 < 1

⇒μ2μ1

< 1

⇒ μ2 < μ1

i.e., refractive index of medium 2 is less than that of medium 1.

∴ Medium 2 is rarer and medium 1 is denser.(ii) The refractive index of medium 1 with respect to medium 2 would be given by

2μ1 =1

1μ2 > 1


(c) (i) (1) Concave lens.(2) The image can be drawn as follows :

(ii) (1) In the collimator of a spectroscope, convex lens is used for obtaining parallelbeam of light.

(2) Convex lens is used as a objective lens in telescope, camera etc.Answer 7.(a) (i) Music is a pleasant, continuous and uniform sound produced by the regular and

periodic vibrations. For example, the sounds produced by a violin, piano, flute etc.are musical sounds.

(ii) Echolocation is a process of using reflected sound or echoes to determine the positionof objects in space. Bats are able to use echolocation process to determine the size,shape, distance and also the type of prey.To echolocate, bats send out sound waves from their mouth or nose in all directions.When the sound waves hit the prey, they produce echoes. The echo bounces off theprey and returns to the bat’s ears. Bats listen to the echoes to figure out where theprey is ? how big it is ? and its shape.

(b) (i) The phenomenon behind this is resonance. When the frequency of the enginebecomes equal to the body of the vehicle, some parts of the vehicle start vibratingviolently and thus a rattling sound is heard.

(ii) The sound heard after reflection from a distant obstacle, after the original sound hasceased, is called an echo. To hear an echo distinctly, the distance between thereflecting surface and the listener should be large enough to allow the reflected soundto reach the listener at least 0·1 second after the original sound is heard.

(c) (i) Given :A1A2

=45

We know that, the intensity of sound is directly proportional to the square of theamplitude i.e., I ∝ A2.Therefore, the ratio of intensities,

I1I2

=A2

1

A22 =

42

52 = 1625

(ii) The frequency of the transverse vibration is given by

f =12l

Tπr2d =

12l

Tm

Here, l is the length of the vibrating string, m is the mass per unit length and T is thetension in the string.

Therefore, the frequency of the transverse vibration of the stretched string can bedecreased by :

(1) increasing the length of the string.

(2) increasing the radius of the string.(3) decreasing the tension T in the string.

Physics | 5

Answer 8.(a) (i) The instrument X is Ammeter and Y is Voltmeter.

(ii) When switch S is open :No current flows through the arm containing 2 Ω resistance.∴ Net resistance of the circuit, R = 3 + 2 = 5 Ω

Current, I =VR =

105 = 2 A

Voltage across 3 Ω resistance = I × 3= 2 × 3 = 6 V

∴ Reading of ammeter (X) = 2 A∴ Reading of voltmeter (Y) = 6 V

(iii) When switch S is closed :

Net resistance of the circuit, R′ = 3 + ⎝⎜⎛

⎠⎟⎞2 × 2

2 + 2

= 3 + 1 = 4 Ω

Current, I′ =VR′ =

104 = 2·5 A

Voltage across 3 Ω resistance = 3 × 2·5 = 7·5 V∴ Reading of ammeter (X) = 2·5 A∴ Reading of voltmeter (Y) = 7·5 V

(b) (i) A, A’ are brushes and B, B’ are commutators.(ii) The arm MN would move upwards and the arm OP move downwards i.e., clockwise

direction.(iii) By increasing the pitch of the propeller.

(c) (i) A transformer is an electrical device that transfers electrical energy between two ormore circuits. As the current is varied in one coil of the transformer it produces atime varying magnetic field, which in turn induces a time varying e.m.f. in thesecond coil.

(ii) The pins are splitted at the ends of the three way pin plug to provide a spring actionso that they fit in the socket holes tightly.

(iii) The arms AC and CB are in series and thecombination of these two is parallel to arm AB.

∴ Net resistance, R = 60 + 60 × 6060 + 60

= 60 + 30 = 90 Ω

Current, I =VR =

490 = 0·04 A

Answer 9.(a) Heat capacity of a body is defined as the amount of heat required to raise its temperature

by 1°C.If on imparting Q amount of heat to a body, its temperature rises through ΔT, then its heatcapacity is given by

C =QΔT

Specific heat capacity is the amount of heat energy required to raise the temperature of aunit mass of substance by 1°C.

Specific heat capacity, c =Q

m × Δt


(b) (i) Energy required to melt 10 kg of ice at 0°C,

Q1 = mL = 10 × 3·36 × 105

= 3·36 × 106 J

And, energy required to raise the temperature of water from 0°C to 100°C.

Q2 = mcΔT = 10 × 4·2 × 103 × (100 – 0)

= 4·2 × 106 J

Hence,Q1Q2

=3·364·2 =

45

(ii) A large amount of heat energy is required for melting the snow which is absorbedfrom the surrounding atmosphere. As a result, the temperature of the surroundingfalls and the weather becomes cool.

(c) (i) Given : Mass of brass rod, mb = 0·2 kg, Its temperature, tb = 100°C

Mass of water, mw = 0·5 kg, Its temperatures, tw = 20°C

Final temperature, T = 23°C

By the principle of the method of mixtures,

Heat lost by rod = Heat gained by water

⇒ mbcb Δt = mwcwΔ t′

⇒ 0·2 × cb × (100 – 23) = 0·5 × 4200 × (23 – 20)

⇒ 0·2 × 77 × cb = 0·5 × 4200 × 3

⇒ cb = 409·09 J kg– 1 K– 1

(ii) S. No. Heat Temperature1. Heat is a form of internal energy

obtained due to random andattractive force of molecules in asubstance.

Temperature is a quantity whichdetermines the direction of flow ofheat on keeping the two bodies atdifferent temperatures in contact.

2. The S.I. unit of heat is joule. The S.I. unit of temperature iskelvin.

3. Heat is measured by principle ofcalorimetry.

Temperature is measured bythermometer.

Physics | 7

Answer 10.(a) The diagram can be drawn as follows :

(b) Medical uses of α-rays :

(i) α-rays are used in the treatment of various forms of cancer.

(ii) α-rays are used as energy source to power heart pacemakers.

Industrial uses of β-rays :

(i) To check thickness of an item.

(ii) To locate leaks in oil or gas lines.

(c) (i) Since, the atomic number of the product has decreased by 2 and out of alpha, betaand gamma radiations, atoms of alpha radiations have atomic number 2, therefore,alpha rays are emitted in the given decay.

84X ⎯→ 82Y + 2He4

(ii) Given : E = 3 × 106 kWh = 3 × 106 × 3·6 × 106J

We know, E = Δmc2

or Δm =Ec2

=3 × 106 × 3·6 × 106

(3 × 108)2

= 1·2 × 10– 4 kg.❏❏

SOLUTION


Answer 1.(a) Given : Weight, mg = 750 N, h = 16 m, t = 5 s

We know, Power (P) =Work done

time

=mgh

t

=750 × 16

5 = 2400 W.

(b) Given : Force = 2 kgf = 2 kg × 9·8 ms–2 = 19·6 NBy Newton’s second law,

F = ma

or a =Fm

=19·610

= 1·96 ms–2

(c) The efficiency of a machine is the ratio of its ouput to its input, if both input and outputare expressed in the same units of energy or power.

(d)

Moment of force F at the end A = F × OA (Anticlockwise)Moment of force F at the end B = F × OB (Anticlockwise)Total moment of couple (i.e., moment of both the forces)

= F × OA + F × OB= F × (OA + OB) = F × AB= F × d (Anticlockwise)

Moment of couple = Either force × Couple arm(e) We have, E = Pt

= 3 × 103

= 30 kWh= 30 × 3·6 × 106 = 1·08 × 108 J

Answer 2.(a) (i) A prism refracting light by 90° can be drawn as :


(ii) A prism refracting light by 180° can be drawn as :

(b) Advantages of using right angled prism as a reflector :

(i) There is 100% total internal reflection.

(ii) There is no absorption of light in a prism, so sharp and bright image is produced.

(c) The primary reason of why the colour red is used for danger signals is that red light isscattered the least by air molecules as scattering is inversely related to the fourth power ofthe wavelength of a colour.

(d) What colour our eyes register is determined by the frequency of the light wave hitting oureyes. Frequency (f) and wavelength (λ) are inversely proportional, the relationshipbetween them is v = f λ, where v is the speed of the wave.

(e) Audible range of sound is 20 Hz to 20,000 Hz but the frequency of vibrating pendulum isless than 20 Hz which can not be heard by human ears. Therefore, we does not hear thesound of the vibrating pendulum.

Answer 3.(a) Let d be the distance between the man and the hill in the beginning.

v =2dt

v =2d5 …(i)

He moves 310 m towards the hill. Therefore, distance will be (d – 310) m.

v = 2 (d – 310)

2 …(ii)

Since velocity of sound is same, equating (i) and (ii),

2d5 = 2

(d – 310)3

3d = 5d – 1550

2d = 1550

d = 775 m

Velocity of sound, v = 2 × 7755

= 310 ms–1

(b) Resonance occurs when the frequency of the forced vibration is the same as the naturalfrequency of the vibrating body.

(c) The filament in the electric bulb is coiled to get appropriate resistance so that the filamentreaches the proper temperature to produce sufficient amount of heat and light.

Physics | 3

(d) The resistivity of a material depends on its nature and the temperature of the conductor.

(e) A copper wire cannot be used as a fuse wire because it has a high melting point. It will notmelt easily when a high electric current passes through it and may damage the electricalappliances.

Answer 4.(a) (i) Calorie can be defined as the energy needed to raise the temperature of 1 gram of

water through 1°C (now usually defined as 4·1868 joules).

(ii) Calorimetry is the process of measuring the heat evolved or absorbed in a chemicalreaction or physical change.

(b) Boiling is the process by which a liquid turns into a vapour when it is heated to its boilingpoint. The heat energy supplied to the liquid is absorbed by the liquid.

It is called latent heat because there is no change in temperature of the liquid and there isonly a change in state.

(c) (i) Latent heat of fusion is absorbed by the ice.(ii) Hot water bottles used for fermentation.

(d) (i) The reaction is possible.(ii) The reaction is possible.

(e) (i) The position of the daughter element is two positions to the left of the parentelement, as in alpha decay, the atomic number decreases by a factor of 2.

(ii) The position of the daughter element is one position to the right of the parentelement, as in beta decay, the atomic number increases by a factor of 1.

(iii) There is no change in the position of the element, as gamma radiation does notinvolve any change in the atomic number.

SECTION–IIAnswer 5.(a) (i) Work is said to be done when the force applied on a body displaces it in the same

direction. It is equal to the product of force and displacement.Given : L = 72 kgf = 72 × 10 N = 720 N, E = 2 × 107 dyne = 2 × 107 × 10–5 N = 200 N

(1) Mechanical Advantage, (M.A.) =LE

=720200 = 3·6

(2) Here, d = 20 m, t = 1·5 min. = 1·5 × 60 sec = 90 sec

Useful power output, P =Wt =

L × dt

=720 × 20

90

= 160 W(ii) Given : h = 5 m + 10 m = 15 m, m = 50 kg

Kinetic energy just before hitting the ground = Potential energy of the falling body⇒ K.E. = mgh

= 50 × 10 × 15= 7500 J

(b) Given : m = 800 kg, h = 60 m, t = 10 s.(i) Work done, W = F × d

= mg × h= 800 × 10 × 60= 480000 J = 4·8 × 105 J


(ii) Power, P =Work done

Time = 4·8 × 105

10 = 4·8 × 104 W

(iii) Here, η = 50% = 0·5

Efficiency of the pump =Power at which pump works

Power rating of the pump

⇒ Power rating =Pη =

4·8 × 104

0·5 = 9·6 × 104 W

(c) (i) Given : L = 500 N, E = 780 N, dL = 5 m, t = 10 s

Mechanical Advantage, M.A. =LoadEffort =

500780 =

2539

(ii) We know, M.A. =Effort ArmLoad Arm

⇒2539 =

Effort ArmLoad Arm

⇒ Efort Arm =2539 × Load Arm =

2539 × 5 =

12539

Hence Velocity ratio =Effort ArmLoad Arm =

2539

(iii) Efficiency =M. A.

Velocity Ratio =

25392539

= 1 = 100%

(iv) Energy gained = Potential energy at height 5 m

= mgh

= 500 × 10 = 5000 J.

Answer 6.

(a) (i) No, the parallel rays will not remain parallel inside the glass as the refractive indexfor different wavelengths of light is different so they suffer different angles ofbending.

(ii) The wavelength of red light is more, hence its refractive index is less as compared toviolet. That results in speed of red coloured light being more than that of the violetcoloured light, inside the glass.

(b) The glass piece will not be visible in the liquid having refractive index 1·61 as thereis nodifference in refractive index of the liquid and the glass piece, so glass piece will not beseen in it.

(c) (i) (1) The graph can be drawn as :

Physics | 5

(2) The ray diagram can be drawn as :

(ii) (1) If a blackened bulb thermometer is moved from violet end towards the redend, there is a steady increase in the temperature but as the thermometer goesbeyond the red end, there is a rapid increase in the temperature. This showsthe existence of some kind of radiation producing heating effect beyond the redend of the spectrum.If the radiations from the red end are made to fall on the silver chloridesolution, it almost remains unaffected upto violet end of the spectrum, butbeyond the violet end it first turns violet and finally turn dark brown.

(2) Given : λ = 446 nm = 446 × 10– 9 m, c = 3 × 108 ms– 1

We know,c = λν

or ν =cλ =

3 × 108

446 × 10–9 = 6·72 × 1014 Hz

Answer 7.(a) (i) The sound heard after reflection from a distant obstacle, after the original sound has

ceased is called echo.(ii) A mechanical wave is a wave that is an oscillation of matter that transfers energy

through a medium from one place to another.(iii) Ultrasonic waves.

(b) Given : λ1 = 40 cm; f1 = 82 Hz

λ2 = 32 cm; f2 = ?

Since while travelling in the same medium, speed of waves does not change.∴ f1λ1 = f2λ2

⇒ 40 × 825 = 32 × f2

⇒ f2 =40 × 825

32 = 4125

4 = 1031·25 Hz

(c) (i) The frequency of sound produced by a stretched string depends on the followingfactors :

(1) The length of the string.

(2) The linear mass density of the string.

(3) The tension in the string.

(ii) Given : t = 2·5 s, v = 1800 ms– 1.Let d be the depth of the sea.

Here, v =2dt

1400 =2d2·5

⇒ d =vt2 =

1400 × 2·52 = 1750 m


Answer 8.(a) The three 6 Ω resistances in both the branches are connected in parallel.

∴1

R1 =

1R2

=16 +

16 +

16 =

36 =

12

⇒ R1 = R2 = 2 ΩNow, these two resistances are connected in series.∴ Net resistance between A and B,

R = R1 + R2 = 2 + 2 = 4 Ω(b) (i) Given : P = 100 W, V = 230 V

We know P = VI

⇒ I =PV =

100230 = 0·43 A

Also, P =V2

R

⇒ R =V2

P = (230)2

100 = 529 Ω.

(ii) The circuit diagram for dual control switch is :

(c) (i) Commutator is a split part of a copper slip ring whose function is to keep the torqueon a D.C. motor from reversing every time the coil moves through the planeperpendicular to the magnetic field.

(ii) (1) The transformer works on the principle of mutual induction, for which currentin one coil must change continuously. If D.C. supply is given, the current willnot change due to constant supply and thus, transformer will not work.

(2) 1. Using thick wires of low resistance.2. Using soft iron core that is easily magnetized and de-magnetized.

Answer 9.(a) Let the specific latent heat of ice be L J kg– 1.

Here, Total heat energy, E = mcice Δt + mLice + mcwater ΔTE = 2 × (0 – (–10)) × 2100 + 2L + 2 × (100 – 0) × 4200

⇒ E = 42000 + 2L + 840000⇒ E = 2L + 882000⇒ 280100 = 2L + 882000⇒ 2L = 280100 – 882000 = – 559900

⇒ L =–559900

2 = – 279950 Jkg– 1

(b) (i) the temperature at which water changes from liquid state to gaseous state.(ii) decrease(iii) increases

Physics | 7

(c) (i) The heating curve for water can be drawn as :

(ii) Given : m = 8 kg = 8000 g, t = 2·5 min. = 2·5 × 60 = 150 sec and P = 104 WTotal energy used in drilling,

Q = P × t = 1·5 × 106 J

But, Useful energy, Q′ = Q × 50

100 = 1·5 × 106 × 50

100 = 7·5 × 105 J

Now, we know Q = mcΔT⇒ 7·5 × 105 = 8000 × 0·91 × ΔT

⇒ ΔT =7·5 × 105

7·28 × 103 ≈ 103°C

Answer 10.(a) (i) Let the number of alpha particles emitted be ‘n’ and the number of beta particles

emitted be ‘m’.Thus, we get :

234 90Th ⎯→ 206

82Pb + n 42He + m 0 –1e

Hence we have :234 = 206 + 4n

⇒234 – 206

4 = n

⇒ n = 7and 90 = 82 + 2n – m⇒ 90 – 82 – 14 = – m⇒ m = 6Hence 7 alpha and 6 beta particles are emitted.

(ii) Isotopes are elements with same atomic number and different mass number.(b) From a radioactive substance, α, β or γ rays are emitted. These particles destroy the living

cells and cause biological damage. Therefore, radioactive substances should not betouched by hands.

(c) Alpha rays are positively charged particles, beta rays are negatively charged and gammarays are neutral in nature.Properties of alpha particles :(i) An alpha particle strongly ionizes the gas through which it passes.(ii) Alpha particles affect a photographic plate.Properties of beta particles :(i) Beta particles are the fast moving electrons emitted from the nucleus of an atom.(ii) Beta particles cause fluorescence on striking a fluorescent material.Properties of gamma rays :(i) The speed of gamma radiations is same as the speed of light i.e., 3 × 108 ms–1.(ii) Gamma radiations are not deflected by the electric and magnetic fields.

❏❏

SOLUTION


Answer 1.(a) (i) The S.I. unit of force is newton which is defined as the force required to accelerate an

object of mass 1 kg with 1 ms– 2.

(ii) In C.G.S. system, the unit of force is dyne where 1 dyne = 105 N.

(b) The arrow pointing downwards represents the weight of the 3 kg block and the arrowpointing upwards represents the tension in the string.

(c) (i) The centre of gravity of an object of given mass depends on its shape i.e., on thedistribution of mass in it.

(ii) Ring, hollow cylinder, toroid, etc.

(d) The torque required is the same in both cases, hence equating the torques, we get :

100 × 32 = F × 3

Thus, the force applied is equal to F = 50 N at the end of the door.

(e) We know that ,

Torque = Force × Displacement

20 × 10 = F × 2·5100

Hence, F = 8000 NAnswer 2.(a) (i) White light is considered to be polychromatic in nature as it is a combination of seven

colours of light.

(ii) It would have been black (no colour) like the sky above the surface of the Moon asthere would be no atoms to scatter light coming from the sun.

(b) The refractive index of a medium is defined as the ratio of the speed of light in air to thespeed of light in that medium.

Given that the refractive index of glass is equal to 1·5 means that speed of light is 1·5 timesslower in glass than the speed of light in air.

(c) (i) A ray of light would speed up if it goes from water to air, as the density of air is lessthan that of water.

(ii) A ray of light would slow down when it goes from water to glass, as the density ofglass is more than that of water.

(d) (i) The angle of incidence would be 0° as it is incident normally.(ii) The angle of refraction from the face of incidence would be 0° as it is incident

normally and would go straight in that face.(e) While moving at constant speed, the train applies a periodic force on the rails and if the

frequency of this vibration becomes equal to the natural frequency of vibration of thebridge, there is resonance and the bridge begins to vibrate with large amplitude. Thus,there is danger of breaking of the bridge.

Answer 3.(a) (i) Changing the length of the string.

(ii) Changing the tension of the wire.

(b) (i) Decibel (dB) is the unit for expressing level of loudness.

(ii) The waveform of a musical sound is regular while that of noise is irregular.


(c) Given : I = 4 A, V = 12 V, t = 10 min. = 10 × 60 = 600 sThe power consumed is given by,

P = VI = 12 × 4 = 48 WThus, the energy consumed,

E = P × t = 48 × 600 = 28800 J(d) The two 4 Ω resistances are connected in series.

∴ R = 4 + 4 = 8 ΩNow the 8 Ω resistance and 5 Ω resistance are connected in parallel.∴ Equivalent resistance between P and Q,

1RPQ

=1R +

15 =

18 +

15 =

1340

⇒ RPQ =4013 Ω = 3·07 Ω

(e) (i) Efficiency can be increased by :(1) Using thick wires of low resistance.(2) Using soft iron core that is easily magnetized and demagnetized.

(ii) The total power consumed by bulbs is given asP = 40 × 100 + 50 × 60 = 7000 W = 7 kW

The number of units consumed in one day,E = 7 × 10 = 70 kWh

So, the units consumed in 30 days is given as,

E′ = 70 × 30 = 2100 kWh

Hence the cost incurred = 2100 × 2·5 = 5,250.

Answer 4.

(a) The wet soil contains water and thus when it is heated, first the water evaporates and afterthat it starts gaining temperature, while the sandy soil has no water in it, so it heats upquickly.

(b) The specific heat capacity of aluminium is more than that of copper, so, if aluminium isused, it would absorb more heat from the contents of calorimeter and thus, alter theresults. So, we will not get the desired results.

(c) Let the specific heat capacity of the solid be c.

We know, Q = mc Δt

⇒ 8000 =2001000 × c × (120 – 20)

⇒ c = 40 J kg– 1 K– 1

(d) (i) Alpha rays. (ii) Gamma rays.

(e) Two uses of radioisotopes are :

(i) Cobalt-60 is used to treat cancer.

(ii) Uranium-235 is used as a fuel for atomic energy reactors.SECTION–II

Answer 5.(a) (i) (1) Crow bar is a class I lever in which fulcrum (F) is in between the effort (E) and

the load (L).

(2) Nail cutter is a class I lever in which fulcrum (F) is in between effort (E) and theload (L).

Physics | 3

(ii) Given : m = 50 kg, p = 3000 kg ms–1

We know, Kinetic energy, E =12 mv2 =

12

m2v2

m = p2

2m = (3000)2

2 × 50 = 90,000 J

(b) (i) Kinetic energy before striking the ground,

K.E.i =1001000 × 10 × 10 = 10 J

(ii) From third equation of motion, we havev2 = u2 – 2gh

When it strikes the ground the final velocity would be equal to 0, hence02 = u2 – 2 × 10 × 4

⇒ u2 = 80⇒ u = √⎯⎯80 ms–1

Therefore, kinetic energy after striking the ground,

K.E.f =12 ×

1001000 × 80 = 4 J

(iii) The loss in energy is given as,ΔK.E. = K.E.f – K.E.i = 4 – 10 = – 6 J

The negative sign indicates loss in energy.(c) (i) Solar energy to electrical energy.

(ii) Kinetic energy to mechanical energy and further into electrical energy.(iii) Gravitational potential energy to electrical energy.(iv) Nuclear energy to electrical energy.

Answer 6.(a) (i) LL’ is a concave lens.

(ii) O and O’ are the principal foci of the concave lens.(iii) The image would be formed between the focus (O′) and the optical centre (X).(iv) The nature of the image is virtual and diminished.

(b) (i) S. No. Real image Virtual image1. A real image is formed due to actual

intersection of the rays refracted bythe lens.

A virtual image is formed when therays refracted by the lens appear tomeet if they are producedbackwards.

2. A real image can be obtained on ascreen.

A virtual image cannot be obtainedon a screen.

3. A real image is inverted with respectto the object.

A virtual image is erect with respectto the object.

(ii) The factors affecting the critical angle for the pair of media are :(1) Colour of light : On increasing the wavelength of light, the refractive index for a

given pair of media decreases, so critical angle increases.(2) Temperature : On increasing the temperature, the refractive index of a medium

decreases, so the critical angle increases.(c) (i) The scattering of light is the splitting of white light into its constituent colours due to

the refractive index of the surface and the wavelength of the light.(ii)


Answer 7.(a) Examples of natural vibrations are :

(i) Weight suspended from a spring, when stretched and then released, starts vibratingwith its natural frequency.

(ii) When a tuning fork is struck against a hard rubber pad, it vibrates with its naturalfrequency.

(iii) When the string in the instruments like sitar, violen, guitar, etc. is plucked, thetransverse vibrations of a definite natural frequency are produced in the string.

(b) When sound travels in a given medium strikes the surface of another medium, it bouncesback in some other direction. This phenomenon is called the reflection of sound. Thereflection of sound waves does not require a smooth and shining surface. Sound wavesget reflected from any surface whether smooth or hard. The only requirement for thereflection of sound wave is that the size of reflecting surface must be bigger than thewavelength of the sound wave.

(c) S. No. Forced vibrations Resonant vibrations1. The vibrations of a body under an

external periodic force of frequencydifferent from the natural frequency ofthe body, are called the forcedvibrations.

The vibrations of a body under anexternal periodic force of frequencyexactly equal to the natural frequencyof the body, are called the resonantvibrations.

2. The amplitude of vibration is usuallysmall.

The amplitude of vibration is verylarge.

3. The vibrations of the body are not inphase with the external periodic force.

The vibrations of the body are in phasewith the external periodic force.

4. These vibrations last for a very smalltime after the periodic force has ceasedto act.

These vibrations last for a long timeafter the periodic force has ceased toact.

Answer 8.(a) (i) Resistivity of a conductor does not depend on the dimensions of the conductor, so

on doubling the length of the wire, resistivity does not change.(ii) Adding more resistors in parallel is equivalent to providing more branches through

which current can flow easily and that decreases the resistance.(iii) Series arangement is not used for domestic circuits due to the following reasons :

(1) In series arrangement, same current will flow through all the appliances, whichis not required.

(2) Total resistance of domestic circuit will be sum of the resistance of all appliancesand hence current drawn by the circuit will be less.

(3) We cannot use independent on/off switches with individual appliances.(4) All appliances are to be used simultaneously even if we do not need them.

(b) (i) Given : P = 80 W, V = 220 V, t = 20 hour.We know, Energy consumed, E = P × t

= 80 × 20= 1600 Wh= 1·6 kWh

(ii) Given : lP = 30 m, lQ = 10 m, rP = 2 mm = 2 × 10– 3 m, rQ = 1 mm = 10– 3 m.

We know, R = ρ l

A

or R ∝l

r2 (˙.˙ A = πr2)

Physics | 5

⇒RPRQ

=lPr2P ×

r2Q

lQ =

30(2 × 10– 3)2

× (10– 3)2

10 = 34

⇒RPRQ

=34

So, resistance of wire P is less.(c) (i) Fleming’s right hand rule : Stretch the thumb, central finger and forefinger of right

hand mutually perpendicular to each other. If the forefinger points in the direction ofmagnetic field and the thumb points in the direction of motion of the conductor, thenthe direction in which the central finger points, gives the direction of induced current.

(ii) (1)

(2) Right hand thumb rule.Answer 9.(a) Let θ° C be the initial temperature of cold water.

∴ Final temperature of the mixture = (θ + 10) °C

Hence fall in temperature of hot water = [50 – (θ + 10)] °C = (40 – θ) °C

Now, Heat gained by cold water = Heat lost by hot water

or 300 × s × 10 = 150 × s × (40 – θ)

or 40 – θ = 20

∴ θ = 20°C

(b) (i) The slope represents the change in temperature of the steam with the amount of heatsupplied.

(ii) It means that boiling of cube needs 2·5 times more heat energy as compared to heatenergy required for melting of solid cube.

(iii) The part AB represents melting of solid cube.

(c) (i) Given : P = 2000 W, m = 50 kg, T1 = 30°C. T2 = 36°C , t = 2 min = 2 × 60 s = 120 s

Energy supplied, Q = P × t

= 2000 × 120

= 2,40,000 Wh

= 240 kWh

= 240 × 3·6 × 106 J

= 8·64 × 108 J

(1) Heat capacity, C′ =QΔT =

8·64 × 108

(36 – 30)= 1·44 × 108J °C– 1

(2) Specific heat capacity, c =C′m


=1·44 × 108

50

= 2·88 × 106 J kg– 1 °C– 1

(ii) Heat capacity =Energy supplied

Rise in temperature = 800020 = 400 J °C– 1

Answer 10.(a) (i) Hydrogen has one proton and no neutron.

(ii) Deuterium has one proton and one neutron.

(iii) Tritium has one proton and two neutrons.

(b) We have, 5B11, i.e.,

Atomic number = 5

Atomic mass = 11

We know, Atomic number = No. of protons = No. of electrons

∴ No. of electrons = No. of protons = 5

and Atomic mass = No. of protons + No. of neutrons

⇒ 11 = 5 + No. of neutrons

⇒ No. of neutrons = 11 – 5 = 6

Thus, boron atom would have 5 electrons, 5 protons and 6 neutrons.

(c) (i) Mass number of Aluminium would be 27 and its atomic number would be 13.

(ii) The resulting nucleus would be of Aluminium itself as gamma radiations arepackets of energies.

(iii) Isotopes.

(iv) A nucleus is radioactive when it is unstable due to the presence of large number ofneutrons in it.

❏❏

SOLUTION


Answer 1.(a) The two conditions are :

(i) When there is no displacement.

(ii) When the angle between the force and displacement is 90°.

(b) (i) When the displacement of the body is along the direction of the force applied, workdone is positive.

(ii) When the direction of displacement of the body is opposite to the direction of forceapplied, work done is said to be negative.

(c) When a body moves in a circular path, no work is done by the body. This is because, whilemoving in a circular path, centripetal force acts on the body which is along the radius ofthe circle and perpendicular to the motion of the body, i.e.,

θ = 90°

∴ W = 0

(d) Given : m = 0·2 kg and v = 20 ms– 1

Here, the maximum potential energy at the highest point = Initial kinetic energy at thebottom most point.

∴ The maximum potential energy = 12 mv2 =

12 × 0·2 × 20 × 20 = 40 J

(e) Given : P = 2 H.P. = 2 × 746 W = 1492 W, V = 5 m3, h = 15 mWe know, density of water, ρ = 1000 kg m– 3

∴ Mass of water = V × ρ

= 5 × 1000 = 5000 kg

Work done in raising the water, W = mgh

= 5000 × 10 × 15

= 7,50,000 JAlso, Work done, W′ = P × t

= 1492 × t˙.˙ W = W′∴ 1492 t = 7,50,000

⇒ t =7‚50‚000

1492= 502·68 s

Answer 2.(a)


(b) Total reflecting prisms are right angled isosceles prism. They are based on the principle oftotal internal reflection. These prisms are used to rotate light through 180° or 90°.

(c) (i) The angle which an incident ray makes with the normal at the point of incidence.(ii) The angle between the two faces of a prism is known as the angle of prism.

(d) (i) Here, 20 waves passes through a point in 2 seconds i.e., in 1 second, 10 waves passes.∴ f = 10 Hz

(ii) Wavelength = Double the distance between crest and adjacent trough = 2 × 1·5 = 3 m(e) Sound waves are called mechanical waves because they need a material medium for their

propagation, like air or liquid like water or metals like silver, etc.Answer 3.(a) A SONAR works on the principle of echoes. A strong and short sound signal is sent

towards the bottom of the ocean. The reflected signal (echo) is then detected by it. Fromthe knowledge of the time after which the reflected signal (echo) reaches it, the depth ofthe ocean can be calculated.

(b) (i) Given : Q = 80 C, t = 2 min = 2 × 60 = 120 s

I =Qt =

80120 =

23 A

(ii) Q = It = 4 × 8 = 32 C.(c) The earth pin is longer so that the appliance is first earthed. This ensures the safety of the

user, because if the appliance is defective, then as soon as the live pin gets connected, thecurrent passes to the earth and the fuse blows off. The earth pin is thicker so, that even bymistake it cannot be inserted into the hole of socket for the live or neutral connection.

(d) (i) Step-down transformer.(ii) A transformer works on the principle of electromagnetic induction, i.e., whenever the

magnetic flux linked with a coil changes, an e.m.f. is induced in it.(e) Heat energy needed = mc ΔT

Here, m = 2 kg, c = 480 J kg– 1 °C–1, ΔT = 15° C∴ H = 2 × 480 ×15

= 14400 JAnswer 4.(a) (i) Heating curve :

Physics | 3

(ii) Cooling curve :

(b) Water is preferred to any other liquid in the hot water bags as it has high specific heatcapacity due to which it does not cool fast and can be used for longer period.

(c) Loss in mass = 0·0246 a.m.u.∴ Equivalent energy released = 0·0246 × 931

= 22·9 MeV(d) The nuclear reactions take place when an unstable nucleus decays to form a stable

nucleus. In the process, the nucleons get transformed. This results in a loss of energy andis related to the difference in the actual mass and the theoretically calculated mass. Thisdifference in mass is called the mass defect and the energy corresponding to it iscalculated as E = Δmc2.

(e) Mass of reactants = Mass of (A + B)= 5 × 10– 27 + 2 × 10– 27

= 7 × 10– 27 kgMass of products = Mass of (C + D)

= 3 × 10– 27 + 1·9 × 10– 27

= 4·9 × 10–27 kgMass defect, Δm = Mass of reactants – Mass of product

= 7 × 10– 27 – 4·9 × 10– 27

= 2·1 × 10– 27 kgThus, the energy released will be calculated as,

E = Δmc2

= (2·1 × 10– 27) × (3 × 108)2

= 1·89 × 10– 10 JSECTION–II

Answer 5.(a) The velocity ratio is defined as

Velocity ratio =Distance moved by effortDistance moved by load

⇒ Distance moved by effort = Velocity ratio × Distance moved by Load= 10 × 5 = 50 m

(i) Work obtained from the machine, W = Fl × dl = 2000 × 5 = 10000 J

(ii) Mechanical Advantage, M.A. =LoadEffort =

2000250 = 8

(iii) Efficiency, η = Work outputWork input × 100 =

Fl × dlFe × de

× 100

= 2000 × 5250 × 50 × 100 = 80%


(b) (i) The work done by the force F is equal to W = 150 × 5 = 750 J.(ii) Potential energy at the ground,

P1 = mg × h1= 200 × 0= 0

Potential energy when it reaches the top,P2 = mg × h2

= 200 × 3 = 600 JChange in potential energy,

ΔP = P2 – P1= 600 – 0 = 600 J.

(iii) The difference in the gained potential energy and the work done is equal to thekinetic energy of the block as the block accelerates under the action of force.

(c) (i) The capacity of a body to do work is called energy. The S. I. unit is joule (J).(ii) Power, P = F × v = 500 × 15 = 7500 J

Answer 6.(a) When a ray of white light is passed through a prism, it splits into its seven colours. These

colours form a spectrum of white light. This process of splitting of light into its constituentcolours is called dispersion.

(b) Given : h1 = 4 cm, h2 = 16 cm, u = 6 cm(i) We know,

Magnification, m =Image heightObject height =

Image distanceObject distance

⇒h2h1

=vu

⇒164 =

v(–6)

⇒ v = – 24 cm

The image is virtual, erect and magnified.

(ii) From lens formula,1f =

1v –

1u =

1(– 24) –

1(– 6) =

– 1 + 424 =

324

⇒ f =243 = 8 cm

(iii) Now, u′ = 24 cm, f = 8 cmAgain, from lens formula,

1f =

1v′ –

1u′

18 =

1v′ –

1(– 24)

⇒1v =

18 –

124 =

3 – 124 =

224 =

112

Physics | 5

⇒ v′ = 12 cmSo, the image will be formed at a distance of 12 cm on the other side of the lens.

(c) The ray diagrams are :(i) Object between F and 2F.

The image formed will be real, inverted and magnified.(ii) Beyond 2F.

The image formed will be real, inverted and diminished.

(iii) At 2F.

The image formed will be real, inverted and of same size as the object.

Answer 7.

(a) (i) (1) The largest amplitude is the one which has the maximum displacement fromthe mean position i.e., option (b).

(2) The fundamental note is the one which has only oscillation, which isrepresented in option (c).

(ii) The ratio between frequencies is 3 : 2.

(b) (i) The note is absolute pitch of a sound which corresponds to a particular frequency,while the tone is the characteristic quality of a sound wave.

(ii) The sound is produced by the vibration of the vessel. More the amplitude andfrequency of the vibration, more is the noise.

(iii) In stringed instruments, the frequency of vibrating string is given as,


f =12l

Tm

i.e., f ∝1l

Tm

Thus, by changing the length of wire, tension of the wire or by using differentthickness wire, desired pitch can be obtained.

(c) (i) The repetition of sound resulting from multiple reflection of sound waves is knownas reverberation.

(ii) A traveller locates shoal of fish in deep water by sending sound waves down thewater. The time taken by the sound waves to come back after reflection from shoal offish gives the idea about the depth of the fish from the traveller’s position.

(iii) (1) Loudness is proportional to the square of the amplitude.Hence, the ratio of loudness will be 22 : 32, i.e., 4 : 9

(2) Amplitude does not vary with frequency.Answer 8.(a) (i) For electromagnetic induction to occur, magnetic field in a coil of wire must change.

(ii) The total number of magnetic lines of force crossing any surface in the magnetic fieldis termed as magnetic flux.The S.I. unit of magnetic flux is weber.

(b) Given : r = 0·2 Ω, R = 2·4 Ω, V = 2 V, Number of e.m.f. cells = 3Total e.m.f. = 3 × 2 V = 6 V

Total resistance, R = R + 3r = 2·4 + 3 (0·2) = 3 Ω(i) From Ohm’s law,

V = IR′

or I =VR′ =

63 = 2 A

(ii) Potential drop = 3 × Ir= 3 × (2 × 0·2)= 3 × 0·4 = 1·2 V

(iii) Terminal voltage = IR= 2 × 2·4= 4·8 V

(c) (i) Given : V = 12 V, P = 24 W(1) Here, t = 1 hr

We know, E = P × t= 24 × 1 = 24 Wh = 24 × 3600 J = 86400 J

(2) We know, P = VI

or I =PV =

2412 = 2 A

(ii) Let the e.m.f. of the cell be ε and the internal resistance of the cell be r, then

0·6 = ε

2 + r …(i)

0·2 =ε

7 + r …(ii)

On dividing equations (i) by (ii), we get0·60·2 =

7 + r2 + r

⇒ (0·6) (2 + r) = (0·2) (7 + r)

Physics | 7

⇒ 1·2 + 0·6r = 1·4 + 0·2r

⇒ 0·4r = 0·2

⇒ r =0·20·4 =

12 Ω

Substituting the value of r in equation (i), we get

ε = 0·6 (2 + r) = 0·6 (2 + 0·5) = 1·5 VAnswer 9.

(a) S. No. Heat Temperature

1. Heat is a form of internal energyobtained due to random motion andattractive force of molecules in asubstance.

Temperature is a quantity whichdetermines the direction of flow of heaton keeping the two bodies at differenttemperatures in contact.

2. The amount of heat contained in abody depends on mass, temperatureand substance of body.

The temperature of a body depends onthe average kinetic energy of itsmolecules due to their random motion.

3. Two bodies having same quantity ofheat may differ in their temperature.

Two bodies at same temperature maydiffer in the quantities of heatcontained in them.

4. When two bodies are placed incontact, the total amount of heat isequal to the sum of heat of individualbody.

When two bodies at differenttemperatures are placed in contact, theresultant temperature is a temperaturtein between the two temperatures.

(b) (i) Calorie : The amount of heat energy required to raise the temperature of 1g of waterby 1°C.Relation between joule and caloriel : 1 calorie = 4 ·184 joule

(ii) The term background radiation refers to the ionizing radiation that we are constantlyexposed to even in the absence of an actual visible radioactive source.

No, it is not possible for us to keep ourselves away from the background radiations.

(c) (i) Let the final temperature of mixture be T.

Rise in temperature, (θR) = T – T2 and the fall in temperature (θf) = T1 – T.Now, by principle of method of mixtures,Heat gained by substance at lower temperature = Heat lost by substance at highertemperature.Hence, we get :

m2c2 (T – T2) = m1c1 (T1 – T)⇒ m2c2T – m2c2T2 = m1cT1 – m1c1T

⇒ (m2c2 + m1c1) T = m1c1T1 + m2c2T2

⇒ T =m1c1T1 + m2c2T2

(m2c2 + m1c1)

(ii) Given : P = 1000 W, m = 5 kg, T1 = 25°C, T2 = 31°C, t = 2 min. = 2 × 60 = 120 s

Change in temperature, ΔT = T2 – T1

= 31 – 25 = 6°CNow, we know, E = P × t

= 1000 × 120 = 1·2 × 105 J

(1) Also, Heat capacity, C′ =Energy supplied

Change in temperature


=1·2 × 105

6 = 2 × 104 J °C– 1

(2) Specific heat capacity, c =Heat capacity

Mass

=2 × 104

5

= 4000 J kg– 1 °C– 1

Answer 10.(a) (i)

(ii) Flemings left hand rule.

(b) (i) We have,235 92U + 10 n ⎯→ y

57 La + 85x Br + 3 10 n + 186 MeV

Since, number of nucleons remain conserved in nuclear reactions.

92 + 0 = 57 + x + 0

⇒ x = 92 – 57

= 35

And, 235 + 1 = y + 85 + (3 × 1)

⇒ 236 = y + 88

⇒ y = 236 – 88 = 148

(ii) Both the nuclear fusion and fission result so that a stable nuclei is formed.

(c) (i) (1) α < β < γ.

(2) γ < β < α

(ii) The resultant can be found as :293 90Th ⎯→ 0–1 e + 293

91Np

293 91Np ⎯→ 0–1 e + 293

92Pu

Hence, the atomic number of the resultant nucleus is 92 and the mass number is 293.❏❏

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