Q: Question 3(iii)State two factors affecting the speed of rotation of the coil in a D.C. motor.

The speed of rotation of coil can be increased by the following two ways :by increasing the strength of current in the coil,by increasing the number of turns in the coil.

Question 1

Question 1(ii)A door lock is opened by turning the lever (handle) of length 0.2 m. If the moment of force produced is 1 Nm, then the minimum force required is:5 N10 N20 N0.2 N

Accepted Answer

5 NReason —Given,Length of lever = Perpendicular distance of the line of action of the force = 0.2 mMoment of force = 1 NmAs, Moment of force = Force × Perpendicular distance of the line of action of the force1 = force x 0.2force = 10.2=102\dfrac{1}{0.2}=\dfrac{10}{2}0.21​=210​ = 5 N

Question 2

Question 1(xiv)The graph given below shows heat energy supplied against change in temperature when no energy is lost to the surrounding. The slope of this graph will give :Specific heat capacityLatent heat of fusionLatent heat of vaporizationHeat capacity

Accepted Answer

Specific heat capacityReason — When no energy is lost to the surrounding, the amount of heat energy supplied is directly proportional to the change in temperature △t i.e.Q∝△torQ=mc△t\text Q ∝ △\text t\[1em] \text {or} \[1em] \text Q = \text {mc} △\text tQ∝△torQ=mc△tWhere,m : mass of a substancec : specific heat capacity of the substanceSo,Q△t=mc\dfrac {\text Q}{△\text t} = \text {mc}△tQ​=mcFor a given mass,Slope of Q vs △t graph=Q△t∝c=constant\text {Slope of Q vs △\text t graph} = \dfrac {\text Q}{△\text t} ∝ \text c = \text {constant}Slope of Q vs △t graph=△tQ​∝c=constantHence, slope of Q vs △t graph gives the value of the specific heat capacity of the substance.

Question 3

Question 2(iii)A non uniform beam of weight 120 N pivoted at one end is shown in the diagram below. Calculate the value of F to keep the beam in equilibrium.

Accepted Answer

Given,Weight (W) = 120 NFrom the figure,Distance of C.G. from the pivot point O = r = 0.20 mDistance of F from the pivot point O = R = 0.80 mNow,Moment of force about pivot point O due to C.G. = r x W = 0.20 x 120 = 24 Nm (clocKwise direction)andMoment of force about pivot point O due to F = R x F = 0.80 x F (anticlocKwise direction)For equilibrium,Anticlockwise moment of force due to F about pivot point O = Clockwise moment of force due to C.G. about pivot point O0.80×F=24⇒F=240.80=2408=30 N0.80 \times \text F = 24 \[1 em] \Rightarrow \text F = \dfrac{24}{0.80}= \dfrac{240}{8} =30\ \text N0.80×F=24⇒F=0.8024​=8240​=30 NSo, value of F = 30 N.

Question 4

Question 2(iv)Meera chose to use a block and tackle system of ‘9’ pulleys instead of a single movable pulley to lift a heavy load.(a) What is the advantage of using a block and tackle system over a single movable pulley?(b) Why should she connect more number of pulleys in the upper fixed block?

Accepted Answer

(a) A block and tackle system with multiple pulleys provides a higher mechanical advantage, which means less effort is required to lift a heavy load compared to a single movable pulley since in a single movable pulley, mechanical advantage (M.A.) is 2 but in a block and tackle system with 9 pulleys, the M.A. can go up to 9 (ideally), making it much easier to lift heavier loads with the same effort.(b) To ensure smooth and balanced operation, more pulleys should be in the upper fixed block because :It reduces the weight of the movable block, making it easier to lift.It helps in maintaining stability and reduces oscillation during lifting.It also makes arrangement and threading of ropes more efficient and manageable.

Question 5

Question 2(v)Sumit does 600 J of work in 10 min and Amit does 300 J of work in 20 min. Calculate the ratio of the powers delivered by them.

Accepted Answer

Given,Work done by Sumit (WS) = 600 JTime taken by Sumit (tS) = 10 min = 10 x 60 = 600 secondsWork done by Amit (WA) = 300 JTime taken by Amit (tA) = 20 min = 20 x 60 = 1200 secondsThen,Power delivered by Sumit (PS) = Work done by Sumit (WS)Time taken by Sumit (tS)=600600=1 W\dfrac{	ext {Work done by Sumit}\ (	ext W_	ext S)}{	ext {Time taken by Sumit}\ (	ext t_	ext S)} =\dfrac{600}{600} = 1\ 	ext WTime taken by Sumit (tS​)Work done by Sumit (WS​)​=600600​=1 WandPower delivered by Amit (PA) = Work done by Amit (WA)Time taken by Amit (tA)=3001200=0.25 W\dfrac{	ext {Work done by Amit}\ (	ext W_	ext A)}{	ext {Time taken by Amit}\ (	ext t_	ext A)} =\dfrac{300}{1200} = 0.25\ 	ext WTime taken by Amit (tA​)Work done by Amit (WA​)​=1200300​=0.25 WSo,PSPA=10.25=4\dfrac{	ext P_	ext S}{	ext P_	ext A}=\dfrac{1}{0.25} =4PA​PS​​=0.251​=4

Question 6

Question 2(vi)5 bulbs are connected in series in a room. One bulb is fused. It is removed and remaining 4 bulbs are again connected in series to the same circuit. What will be the effect on the following physical quantities? (Increases, Decreases, Remain Same)(a) Resistance(b) Intensity of light

Accepted Answer

(a) In a series connection, total resistance is the sum of individual resistances.Let resistance of each bulb be R.Initially :Rtotal = 5RAfter removing one bulb:Rtotal = 4RSo, resistance will decrease.(b) When resistance decreases, total current in the circuit (assuming the same voltage supply) will increase :I=VR	ext I = \dfrac{	ext V}{	ext R}I=RV​So, more current → more power delivered to each bulb.Since,Power (P) = I2R,Hence, intensity of light increases for each remaining bulb.

Question 7

Question 2(vii)Rohan conducted experiments on echo in different media. He observed that a minimum distance of ‘x’ meters is required for the echo to be heard in oxygen and ‘y’ meters in benzene. Compare ‘x’ and ‘y’. Justify your answer.Speed of sound in oxygen: 340 ms-1Speed of sound in benzene: 200 ms-1

Accepted Answer

Given,Speed of sound in oxygen = 340 ms-1Speed of sound in benzene = 200 ms-1Minimum distance required to hear an echo = distance sound travels in 0.1 s (since the human ear can distinguish sound if the reflected sound reaches after 0.1 s)So,Minimum distance for echo=Speed of sound×Minimum time to hear an echo2=v×tmin2	ext {Minimum distance for echo} = \dfrac{	ext {Speed of sound}	imes 	ext {Minimum time to hear an echo}}{2}=\dfrac{	ext v 	imes 	ext t_	ext{min}}{2}Minimum distance for echo=2Speed of sound×Minimum time to hear an echo​=2v×tmin​​andtmin = 0.1 secFor medium containing oxygen :x=340×0.12=342=17 m	ext x = \dfrac{340	imes 0.1}{2}=\dfrac{34}{2} = 17\ 	ext mx=2340×0.1​=234​=17 mFor medium containing benzene :y=200×0.12=202=10 m	ext y = \dfrac{200	imes 0.1}{2}=\dfrac{20}{2} = 10\ 	ext my=2200×0.1​=220​=10 mSo, from above calculations x > y.

Question 8

Question 3(i)(a) In a reading glass what is the position of the object with respect to the convex lens used?(b) Why can we not use concave lens for the same purpose?

Accepted Answer

(a) In a reading glass, the object is placed between the optical center and the principal focus of the convex lens to produce a virtual and enlarged image of the object so that a magnified view of the object can be seen.(b) A concave lens always forms a virtual, erect, but diminished (smaller) image of the object so it cannot magnify the object.

Question 9

Question 3(ii)A fuse is rated 5 A. Can it be used with a geyser rated 1540 W, 220 V. Write Yes or No. Give supporting calculations to justify your answer.

Accepted Answer

Given,Maximum current passing through the fuse (Imax) = 5 AInput voltage of the geyser (v) = 220 VPower rating of the geyser (P) = 1540 WLet, current passing through geyser be I.Then,P=IVI=PV=1540220=7 A\text P = \text {IV} \[1em] \text I = \dfrac{\text P}{\text V}=\dfrac{1540}{220}=7\ \text AP=IVI=VP​=2201540​=7 AAs,Imax = 5 A < I = 7 ASo if the fuse is used with the geyser then it will draw more current greater than the maximum current and in turn fuse will melt.No, the fuse can not be used with the geyser

Question 10

Question 3(iii)State two factors affecting the speed of rotation of the coil in a D.C. motor.

Accepted Answer

The speed of rotation of coil can be increased by the following two ways :by increasing the strength of current in the coil,by increasing the number of turns in the coil.

Question 11

Question 3(iv)How much heat is required to convert 500 g of ice at 0°C to water at 0°C? The latent heat of fusion of ice is 330 J g⁻¹.

Accepted Answer

Given,m = 500 glatent heat of fusion of ice (L) = 330 J g⁻¹Let, required heat be Q.Then,Q=mL=500×330=165000 J=165 kJ	ext Q = 	ext {mL} = 500 	imes 330 = 165000\ 	ext J = 165\ 	ext {kJ}Q=mL=500×330=165000 J=165 kJ

Question 12

Question 3(v)Copy and complete the nuclear reaction by filling in the blanks.92U235+0n1⟶56Ba⋯+ ⋯Kr92+3 0n1_{92}	ext{U}^{235} + _0	ext n^1 \longrightarrow _{56}	ext {Ba}^{\cdots} + \space _{\cdots}	ext {Kr}^{92} + 3 \space _0	ext n^192​U235+0​n1⟶56​Ba⋯+ ⋯​Kr92+3 0​n1

Accepted Answer

As only neutrons are involved in the reaction which number of protons and total nucleon number (i.e., number of protons + neutrons) remain constant.Then,Initially,No. of protons (np) = Atomic number of 92U235_{92}	ext{U}^{235}92​U235 = 92No. of nucleons (n) = Mass number of 92U235_{92}	ext{U}^{235}92​U235 + mass number of 0n1_0	ext n^10​n1 = 235 + 1 = 236Now,Finally,Let atomic number of Kr is Z and mass number of Ba is A.No. of protons (np) = Atomic number of 56BaA_{56}	ext{Ba}^{	ext A}56​BaA + Atomic number of ZKr92_{	ext Z}	ext{Kr}^{92}Z​Kr92 = 56 + ZandNo. of nucleons (n) = mass number of 56BaA_{56}	ext{Ba}^{	ext A}56​BaA + mass number of ZKr92_{	ext Z}	ext{Kr}^{92}Z​Kr92 + 3 x mass number of 0n1_0	ext n^10​n1 = A + 92 + 3 = A + 95As number of protons are constant :⇒ 92 = 56 + Z⇒ Z = 92 - 56 = 36Also, number of nucleons are constant :⇒ 236 = A + 95⇒ A = 236 - 95 = 141So reaction will be :92U235+0n1⟶56Ba141+ 36Kr92+3 0n1_{92}	ext{U}^{235} + _0	ext{n}^1 \longrightarrow _{56}	ext {Ba}^{141} + \space _{36}	ext {Kr}^{92} + 3 \space _0	ext n^192​U235+0​n1⟶56​Ba141+ 36​Kr92+3 0​n1

Question 13

Question 4(i)The image of a candle flame placed at a distance of 36 cm from a spherical lens, is formed on a screen placed at a distance of 72 cm from the lens. Calculate the focal length of the lens and its power.

Accepted Answer

Given,Object distance (u) = - 36 cmAs, image is formed on a screen then,Image distance (v) = + 72 cmLet, focal length be f.From lens formula,1f=1v−1u1f=172−1(−36)=172+136=1+2721f=372f=723=+24 cm\dfrac{1}{\text f} = \dfrac{1}{\text v} - \dfrac{1}{\text u} \[1 em] \dfrac{1}{\text f} = \dfrac{1}{72}-\dfrac{1}{(-36)}=\dfrac{1}{72}+\dfrac{1}{36}= \dfrac{1+2}{72} \[1em] \dfrac{1}{\text f} = \dfrac{3}{72} \[1 em] \text f = \dfrac{72}{3} = + 24 \text { cm}f1​=v1​−u1​f1​=721​−(−36)1​=721​+361​=721+2​f1​=723​f=372​=+24 cmNow,Power=1f (in metre)=124×10−2=10024≈4.17 D\text {Power} = \dfrac{1}{\text {f (in metre)}} = \dfrac{1}{24\times 10^{-2}}=\dfrac{100}{24} ≈ 4.17\ \text DPower=f (in metre)1​=24×10−21​=24100​≈4.17 D

Question 14

Question 4(iii)(a) Why do we use red colour as a danger signal on the top of a skyscraper?(b) The diagram below shows the path of a blue ray through the prism :Calculate the critical angle of the material of the prism for blue colour.What is the measure of the angle of this prism (A)?Which colour should replace the blue ray, for the ray to undergo Total Internal Reflection?

Accepted Answer

(a) Red colour is used as a danger signal on the top of a skyscraper because red light has the longest wavelength among visible light, so it is scattered the least by air molecules and dust and remains visible from a long distance and in low visibility, making it ideal for danger signals.(b)Here, the refracted ray PQ is normal to the surface AB of the prism (i.e., ∠r = 0°), so the incident ray at the point P should also be normal to the surface AB so that ∠i = 0°.At point Q, the angle of incidence for the ray PQ = 133° - 90° = 43°Now, this angle should be equal to the critical angle because ray PQ is refracted at 90°, i.e., it is refracted along QC. Hence, the critical angle of the prism is 43°.In △AРQ∠APQ + ∠PQA + ∠PAQ = 180°⇒ 90° + 47° + ∠PAQ = 180°⇒ ∠PAQ = 180° - 90° - 47°⇒ ∠ PAQ = 43°Hence, ∠A= 43°As the light used and its wavelength determine the angle of deviation since with a decrease in the light wavelength, the angle of deviation increases so the beam will experience total internal reflection if we swap out the blue light with lower wavelength i.e., indigo.

Question 15

Question 5(i)(a) Refractive index of glass with respect to water is 98\dfrac{9}{8}89​. Find the refractive index of water with respect to glass.(b) Name the principle used to find the value in part (a).(c) If we change the temperature of water,then will the ratio 98\dfrac{9}{8}89​ remain the same? Write yes or no.

Accepted Answer

(a) Given,Refractive index of glass with respect to water = 98\dfrac{9}{8}89​Then, according to the principle of reversibility of lightRefractive index of water with respect to glass =1Refractive index of glass w.r.t water=198=89\dfrac{1}{	ext {Refractive index of glass w.r.t water}} = \dfrac{1}{\dfrac{9}{8}}=\dfrac{8}{9}Refractive index of glass w.r.t water1​=89​1​=98​(b) Principle of reversibility of light(c) If temperature is changed of water then the ratio 98\dfrac{9}{8}89​ will also change as refractive index of medium is inversely proportional to the temperature i.e., on increasing temperature refractive index of a medium decreases and vice versa.So, If we change the temperature of water,then the ratio 98\dfrac{9}{8}89​ will not remain the same.

Question 16

Question 5(ii)Light travels a distance of ‘10x’ units in time ‘t1’ in vacuum and it travels a distance of ‘x’ units in time ‘t2’ in a denser medium. Using this information answer the question that follows :(a) ‘Light covers a distance of ‘20x’ units in time ‘t1’ in diamond.’ State true or false.(b) Calculate the refractive index of the medium in terms of ‘t1’ and ‘t2’.

Accepted Answer

(a) False because speed of light in vacuum is highest and diamond is a denser medium so speed of light will be lower and hence, it can not travel a distance 20x in diamond which is two times of distance travelled by light in vacuum.(b) Given,Distance travelled by light in vacuum = 10xTime taken by light in vacuum = t1Distance travelled by light in the medium = xTime taken by light in diamond = t2As,Speed of light in vacuum (c)=Distance travelled by light in vacuumTime taken by light in vacuum=10xt1\text {Speed of light in vacuum (c)} = \dfrac {\text {Distance travelled by light in vacuum}}{\text {Time taken by light in vacuum}}= \dfrac {10\text x}{\text t_1}Speed of light in vacuum (c)=Time taken by light in vacuumDistance travelled by light in vacuum​=t1​10x​andSpeed of light in diamond (v)=Distance travelled by light in diamondTime taken by light in diamond=xt2\text {Speed of light in diamond (v)} = \dfrac {\text {Distance travelled by light in diamond}}{\text {Time taken by light in diamond}}= \dfrac {\text x}{\text t_2}Speed of light in diamond (v)=Time taken by light in diamondDistance travelled by light in diamond​=t2​x​then,The refractive index of diamond=Speed of light in vacuum (c)speed of light in diamond (v)=10xt1xt2=10t2t1\text {The refractive index of diamond} = \dfrac{\text{Speed of light in vacuum (c)}}{\text {speed of light in diamond (v)}}= \dfrac{\dfrac {10\text x}{\text t_1}}{\dfrac {\text x}{\text t_2}} \[1em] =\dfrac{10 \text t_2}{\text t_1}The refractive index of diamond=speed of light in diamond (v)Speed of light in vacuum (c)​=t2​x​t1​10x​​=t1​10t2​​

Question 17

Question 5(iii)A monochromatic ray of light is incident on an equilateral prism placed at minimum deviation position with an angle of incidence 45° as shown in the diagram.(a) Copy the diagram and complete the path of the ray PQ.(b) State two factors on which the angle of deviation depends.

Accepted Answer

(a) In below figure, I is the position of source O when seen through the surface RS.(b) The angle of deviation depends on the following factors :the angle of incidence (i),the material of prism (i.e., on refractive index μ).

Question 18

Question 6(i)(a) Define Centre of Gravity.(b) A hollow ice cream cone has height 6 cm.Where is the position of its centre of gravity from the broad base?Will its position change when it is filled completely with honey? Write Yes or No.

Accepted Answer

(a) The centre of gravity (C.G.) of a body is the point about which the algebraic sum of moments of weights of all the particles constituting the body is zero and the entire weight of the body can be considered to act at this point, howsoever the body is placed.(b)As the centre of gravity of a hollow cone is at h/3 from its base on its axis where h is the height of the cone.Given,Height of the cone (h) = 6 cmCentre of gravity of cone = h3=63=2 cm\dfrac{	ext h}{3} = \dfrac{6}{3} = 2 	ext{ cm}3h​=36​=2 cmYes, now cone is a solid cone. So, the position of the new center of gravity from the base will be h/4 сm i.e. =64=1.5  cm=\dfrac{6}{4} = 1.5\ 	ext{ cm}=46​=1.5  cm from the base of the cone.

Question 19

Question 6(ii)Two identical marbles A and B are rolled down along Path 1 and Path 2 respectively. Path 1 is frictionless and Path 2 is rough.(a) Which marble will surely reach the next peak?(b) Along which path/s the mechanical energy will be conserved?(c) Along which path/s is the law of conservation of energy obeyed?

Accepted Answer

(a) As Path 2 is rough and Path 1 is frictionless then marble A will surely reach the top because there will be no energy lost on the frictionless road while some energy will be degraded om rough path.(b) As total mechanical energy remains constant when there is no frictional force, hence, mechanical energy will be preserved along path 1.(c) Since energy cannot be created or destroyed, the law of conservation of energy applies to both scenarios.

Question 20

Question 6(iii)Given are two pulleys.(a) Copy and complete the labelled diagram connecting the two pulleys with a tackle to obtain Velocity Ratio = 2.(b) If Load = 48 kgf and efficiency is 80% then calculate :Mechanical Advantage.Effort needed to lift the load

Accepted Answer

(a) The diagram is given below :(b)Given,Velocity Ratio = 2Load = 48 kgfEfficiency = 80% = 0.8As,Efficiency=Mechanical AdvantageVelocity Ratio⇒Mechanical Advantage=Efficiency×Velocity Ratio=0.8×2=1.6\text {Efficiency} = \dfrac {\text {Mechanical Advantage}}{\text {Velocity Ratio}}\[1 em] ⇒ \text {Mechanical Advantage} = \text {Efficiency} \times \text {Velocity Ratio}=0.8 \times 2 = 1.6Efficiency=Velocity RatioMechanical Advantage​⇒Mechanical Advantage=Efficiency×Velocity Ratio=0.8×2=1.6As,Mechanical Advantage=LoadEffort⇒Effort=LoadMechanical Advantage=481.6=30 kgf\text {Mechanical Advantage} = \dfrac{\text {Load}}{\text {Effort}}\[1 em] ⇒ \text {Effort} = \dfrac{\text {Load}}{\text {Mechanical Advantage}}= \dfrac{48}{1.6}=30 \text { kgf}Mechanical Advantage=EffortLoad​⇒Effort=Mechanical AdvantageLoad​=1.648​=30 kgf

Solved 2024 Question Paper ICSE Class 10 Physics

Section A

Section B