Q: 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 :√3 : √29 : 42 : 33 : 2

2 : 3Reason — Let masses of P and Q be mP\text{m}_\text{P}mP and mQ\text{m}_\text{Q}mQ and their specific heat capacities cP\text{c}_\text{P}cP and cP\text{c}_\text{P}cP 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}MQMP=23As, Q = McΔT\text{Q = McΔT}Q = McΔTAccording 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 TMPcPΔT=MQcQΔTMPcP=MQcQ\text{M}_\text{P}\text{c}_\text{P}=\text{M}_\text{Q}\text{c}_\text{Q}MPcP=MQcQcPcQ=MQMP=23\dfrac{\text{c}_\text{P}}{\text{c}_\text{Q}} = \dfrac{\text{M}_\text{Q}}{\text{M}_\text{P}}=\dfrac{2}{3}cQcP=MPMQ=32

Question 1

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 :minimumit does not mattermaximumzero

Accepted Answer

maximumReason — 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 forceFor 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 2

Question 1(ix)Which of the following circuits have the same resistance ?iiiiiiiv(i) and (ii)(ii) & (iii)(i) & (iii)(ii) & (iv)

Accepted 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=1⇒Req=1Ω\dfrac{1}{R_{eq}}=\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{3}{3}=1 \[1em] \Rightarrow R_{eq} = 1 \OmegaReq​1​=31​+31​+31​=33​=1⇒Req​=1Ω (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}\ ΩRp​=3+33×3​=69​=23​ ΩFor series combination,Req=Rp+6=32+6=152 ΩR_{eq} = R_p+6=\dfrac{3}{2}+6=\dfrac{15}{2}\ ΩReq​=Rp​+6=23​+6=215​ 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\ ΩRs​=1+1=2 ΩFor Parallel combination,Req=2×22+2=44=1 ΩR_{eq} = \dfrac{2\times2}{2+2}=\dfrac{4}{4}= 1\ ΩReq​=2+22×2​=44​=1 Ω (iv) Here all resistances are in series combination.Req=3+2+3=8 ΩR_{eq}=3+2+3=8\ ΩReq​=3+2+3=8 Ω

Question 3

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 :√3 : √29 : 42 : 33 : 2

Accepted Answer

2 : 3Reason — Let masses of P and Q be mP	ext{m}_	ext{P}mP​ and mQ	ext{m}_	ext{Q}mQ​ and their specific heat capacities cP	ext{c}_	ext{P}cP​ and cP	ext{c}_	ext{P}cP​ respectively. Let heat given be Q and temperature change is ΔT.Given,MPMQ=32\dfrac{	ext{M}_	ext{P}}{	ext{M}_	ext{Q}}=\dfrac{3}{2}MQ​MP​​=23​As, Q = McΔT	ext{Q = McΔT}Q = McΔTAccording to question,MPcPΔT=MQcQΔT	ext{M}_	ext{P}	ext{c}_	ext{P}Δ	ext T=	ext{M}_	ext{Q}	ext{c}_	ext{Q}Δ	ext TMP​cP​ΔT=MQ​cQ​ΔTMPcP=MQcQ	ext{M}_	ext{P}	ext{c}_	ext{P}=	ext{M}_	ext{Q}	ext{c}_	ext{Q}MP​cP​=MQ​cQ​cPcQ=MQMP=23\dfrac{	ext{c}_	ext{P}}{	ext{c}_	ext{Q}} = \dfrac{	ext{M}_	ext{Q}}{	ext{M}_	ext{P}}=\dfrac{2}{3}cQ​cP​​=MP​MQ​​=32​

Question 4

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+1/2n−1][2^n/2^{n+1}/2^{n-1}][2n/2n+1/2n−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]

Accepted 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	ext{Power} = \dfrac{	ext {Work done}}{	ext {Time taken}}Power=Time takenWork done​ 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	ext R = 	ext ρ\dfrac{	ext l}{	ext a}R=ρal​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 5

Question 2(iii)(a) What is 'SONAR'?(b) State the principle on which is based.

Accepted 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 6

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.

Accepted 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:

Question 7

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?

Accepted Answer

(a) Miniature circuit breaker (MCB).(b) Live wire.

Question 8

Question 3(iii)The diagram in the figure given alongside shows the core of a transformer and its input and output connections.(a) Is this transformer a step up or step down?(b) State the material used for the core and describe its structure.

Accepted 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 9

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.

Accepted 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 10

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.

Accepted 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 11

Question 3(vi)Arrange the α, ẞ and γ radiations in ascending order of their (i) ionising power, and (ii) penetrating power.

Accepted Answer

(i) Ionizing power → γ < β < α(ii) Penetrating power → α < β < γ

Question 12

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.(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?

Accepted 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 ANumber of neutrons in A = 128Mass number = 84 + 128 = 212After alpha decay (A → B) :Loss of protons = 2Loss of neutrons = 2New atomic number = 84 – 2 = 82New mass number = 212 – 4 = 208After 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 = 83Mass number = 208For CNumber of protons = atomic number of C = 83Number of neutrons = Mass number - number of proton = 208 - 83 = 125

Question 13

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)

Accepted Answer

Given,Refractive index of glass with respect to air = aμg = 1.5 = 1gμa\dfrac{1}{_	ext g	ext μ_	ext a}g​μa​1​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(i) For ray PAi = 0°sin r=sin igμa=sin 0°×1.5=0	ext {sin r} = \dfrac{	ext{sin i}}{{_g	ext{μ}_a}}=	ext {sin }0°	imes 1.5 =0sin r=g​μa​sin i​=sin 0°×1.5=0⟹ r = sin-1(0) = 0°Ray PA will not deviate and emerge straight.(ii) For PBi = 30°sin r=sin igμa=sin 30°×1.5=0.5×1.5=0.75	ext {sin r} = \dfrac{	ext{sin i}}{{_	ext{g}	ext{μ}_a}}=	ext {sin}\ 30°	imes 1.5 = 0.5	imes1.5=0.75sin r=g​μa​sin i​=sin 30°×1.5=0.5×1.5=0.75⟹ r = sin-1(0.75) ≈ 49°Ray PB will emerge at an angle of 49° with the normal.(iii) For PCi = 42°sin r=sin igμa=sin 42°×1.5=23×1.5=1	ext {sin r} = \dfrac{	ext{sin i}}{{_	ext{g}	ext{μ}_a}}=	ext {sin}\ 42°	imes 1.5 = \dfrac{2}{3} 	imes1.5=1sin r=g​μa​sin i​=sin 42°×1.5=32​×1.5=1⟹ r = sin-1(1) = 90°This is the condition for critical angle.Ray PC will graze along the surface.(iv) For PDi = 60°sin r=sin igμa=sin 60°×1.5=32×1.5≈1.3	ext {sin r} = \dfrac{	ext{sin i}}{{_	ext {g}	ext{μ}_a}}=	ext {sin}\ 60°	imes 1.5 = \dfrac{\sqrt 3}{2} 	imes1.5≈1.3sin r=g​μa​sin i​=sin 60°×1.5=23​​×1.5≈1.3Here,sin r > 1 which a condition of total internal reflection.Ray PD will suffer total internal reflection.

Question 14

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.(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.

Accepted Answer

(a) Below ray diagram shows the formation of image I at O itself:(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 mPower of a lens is expressed as reciprocal of its focal length in metre and written as :P=1f=10.25=+4D	ext P=\dfrac{1}{	ext f}=\dfrac{1}{0.25}=+4 DP=f1​=0.251​=+4DPower of the lens is +4 Dioptre

Question 15

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.

Accepted Answer

Given,Focal length of the lens (f) = 15cmMagnification 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 = uDistance of image from the lens = vAs for lens,m=vu⇒vu=3\text m = \dfrac{\text v}{\text u} \[1em] \Rightarrow \dfrac{\text v}{\text u}=3m=uv​⇒uv​=3⇒ v = 3u .......... (1)From lens formula,1v−1u=1f\dfrac{1}{\text v}-\dfrac{1}{\text u}=\dfrac{1}{\text f}v1​−u1​=f1​On putting values⇒13u−1u=115⇒13u−33u=115⇒−23u=115⇒u=−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}⇒3u1​−u1​=151​⇒3u1​−3u3​=151​⇒−3u2​=151​⇒u=3−15×2​=−10 cmThe 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}v=3u=3×(−10)=− 30 cmDistance of the screen from the lens = image distance (v) = -30 cmThe screen is placed 30 cm in front of the lens.

Question 16

Question 5(ii)The diagram given below shows a ray of blue, yellow and red light incident on one face of a prism.(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 ?

Accepted Answer

(a) Below diagram shows the effect of the prism on the beam and the spectrum seen on the screen :(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 17

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°(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.

Accepted Answer

Below is the completed diagram showing the path of the light ray till it emerges out of the prism with all angles marked:(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 undeviatedAt face BC — The ray of light suffers total internal reflection

Question 18

Question 6(i)A uniform metre scale is in equilibrium as shown in the diagram below :(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.

Accepted 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 cmMoment 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 cmUsing principle of moments:Clockwise moments of force about the fulcrum = Anticlockwise moments of force about the fulcrum45 x 25 = W x 20⇒ 1000 = 20 W⇒ W = 100020\dfrac{1000}{20}201000​ = 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 cmMoment 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 cmHere,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 19

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.

Accepted Answer

(a) A man pushes a wall. — NoReason: 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. — NoReason: The coolie is not moving, so there is no displacement.(c) A boy climbs up 20 stairs. — YesReason: There is displacement in the direction of force (against gravity), so work is done.

Question 20

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?

Accepted Answer

(a) The arrangement is shown in the figure given below :Here, the load is being supported by five segments of the string.∴ L = 5T and E = TThen,M.A. = LE\dfrac{	ext{L}}{	ext{E}}EL​= 5TT\dfrac{	ext{5T}}{	ext{T}}T5T​ = 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

Model Paper 1

Section A

Section B