Q: Question 1(vi)A ray of light incident at an angle of 48° on a prism of refracting angle 60°, suffers minimum deviation. The angle of minimum deviation would be12°24°36°48°

36°Reason — As we know,δmin=2i−Aδ_ {min} = 2i - Aδmin=2i−AGiven,i = 48°A = 60°So, substituting the values in the formula above we get,δmin=(2×48°)−60°δmin=96°−60°⇒δmin=36°δ_ {min} = (2 \times 48°) - 60° \\[0.5em] δ_ {min} = 96° - 60° \\[0.5em] \Rightarrow δ_ {min} = 36° \\[0.5em]δmin=(2×48°)−60°δmin=96°−60°⇒δmin=36°Hence, the angle of minimum deviation is equal to 36°

Question 1

Question 1(iv)The number of beta particles emitted by radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter nuclei is asisobar of parentisomer of parentisotone of parentisotope of parent

Accepted Answer

isotope of parentReason — Given, number of beta particles emitted by radioactive substance is twice the number of alpha particles emitted by it.Hence,After alpha decay : ZAX⟶Z−2A−4Y+24He_{Z}^{A}	ext{X} \longrightarrow _{Z-2}^{A-4}	ext{Y} + _{2}^{4}	ext{He}ZA​X⟶Z−2A−4​Y+24​HeAfter beta decay : Z−2A−4Y⟶Z−1A−4Z⟶ZA−4W_{Z-2}^{A-4}	ext{Y} \longrightarrow _{Z-1}^{A-4}	ext{Z} \longrightarrow _{Z}^{A-4}	ext{W}Z−2A−4​Y⟶Z−1A−4​Z⟶ZA−4​WThe daughter nucleus after alpha decay has its mass number reduced by 4.There will be no change in the reduced mass number after beta decay.The obtained daughter nucleus has a different mass number = (A-4) and the same atomic number = Z as the parent nucleus. So, the resulting daughter nuclei is an isotope of the parent.

Question 2

Question 1(vi)A ray of light incident at an angle of 48° on a prism of refracting angle 60°, suffers minimum deviation. The angle of minimum deviation would be12°24°36°48°

Accepted Answer

36°Reason — As we know,δmin=2i−Aδ_ {min} = 2i - Aδmin​=2i−AGiven,i = 48°A = 60°So, substituting the values in the formula above we get,δmin=(2×48°)−60°δmin=96°−60°⇒δmin=36°δ_ {min} = (2 \times 48°) - 60° \[0.5em] δ_ {min} = 96° - 60° \[0.5em] \Rightarrow δ_ {min} = 36° \[0.5em]δmin​=(2×48°)−60°δmin​=96°−60°⇒δmin​=36°Hence, the angle of minimum deviation is equal to 36°

Question 3

Question 1(xi)Observe the heating curve of waterWhich of the following represent boiling?OAABBCCD

Accepted Answer

BCReason — In part BC of the graph, the temperature of water does not increase even though more heat is supplied with time. Hence, BC represents boiling.

Question 4

Question 1(xv)Which of the following is not a representation of electric power?I2RV2R\dfrac{	ext{V}^2}{	ext{R}}RV2​VII2V

Accepted Answer

I2RReason — One watt is the electric power consumed when a current of 1 ampere flows through a circuit having potential difference of 1 volt. Hence, P = VI = V2R=I2R\dfrac{	ext{V}^2}{	ext{R}} = 	ext{I}^2	ext{R}RV2​=I2R

Question 5

Question 2(i)(a)	Write the two uses of radioactivity.(b)	A crow-bar of length 80 cm has its fulcrum situated at a distance of 10 cm from the load. Calculate the mechanical advantage of the crow-bar.(c)	Why fission nuclei undergo β-decay?

Accepted Answer

(a) Uses of radioactivity :Many diseases such as leukemia, cancer, etc., are cured by radiation therapy.Radioisotopes are used to trace and study the uptake of nutrients by plants. For example, the use of radioactive phosphorous 32P in fertilizers has revealed how phosphorous is absorbed by plants.(b) Given,Total length of a crowbar = 80 cmEffort arm = 80 - 10 = 70 cmLoad arm = 10 cmWe know that,M.A.=Effort armLoad armM.A..=7010⇒M.A.=7\text{M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \[0.5em] \text{M.A.}. = \dfrac{70}{10} \[0.5em] \Rightarrow\text{M.A.} = 7 \[0.5em]M.A.=Load armEffort arm​M.A..=1070​⇒M.A.=7Hence, mechanical advantage = 7(c) Fission reactions primarily release energy in the form of neutrons. Some of the neutrons produced during fission are initially in an excited state. Beta decay occurs when a neutron in this excited state transforms into a proton, emitting a beta particle (an electron) and an antineutrino. This process helps stabilize the nucleus by converting a neutron into a proton, maintaining a balance between protons and neutrons in the nucleus.

Question 6

Question 2(ii)A person standing 22 m away from a wall produces a sound and receives the reflected sound.(a)	In what time interval, he receives the reflected sound, if speed of sound in air is 330 ms-1 ?(b)	At which distance should he stand to receive the sound in 0.12 s?

Accepted Answer

(a) As we know,Speed of sound (V)=Total distance travelled (2d)time interval (t)\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \[0.5em]Speed of sound (V)=time interval (t)Total distance travelled (2d)​Total distance travelled by the sound in going and then coming back = 2dGiven,d = 22 mV = 330 ms-1Substituting the values in the formula above, we get,330=2×22t⇒t=44330⇒t=430⇒t=0.13 s330 = \dfrac{2 \times 22}{t} \[0.5em] \Rightarrow t = \dfrac{44}{330} \[0.5em] \Rightarrow t = \dfrac{4}{30} \[0.5em] \Rightarrow t = 0.13 \text{ s} \[0.5em]330=t2×22​⇒t=33044​⇒t=304​⇒t=0.13 sHence, the time after which he receives the reflected sound = 0.13 seconds(b) As we know,Speed of sound (V)=Total distance travelled (2d)time interval (t)\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \[0.5em]Speed of sound (V)=time interval (t)Total distance travelled (2d)​t = 0.12 sV = 330 ms-1d = ?Substituting the values in the formula above, we get,330=2d0.12⇒d=330×0.122⇒d=19.8 m330 = \dfrac{2d}{0.12} \[0.5em] \Rightarrow d = \dfrac{330 \times 0.12}{2} \[0.5em] \Rightarrow d = 19.8 \text{ m} \[0.5em]330=0.122d​⇒d=2330×0.12​⇒d=19.8 mHence, distance to receive the sound in 0.12 s = 19.8 m

Question 7

Question 2(iii)The moment of a force of 20 N about a point O is 18 N-m. Find the distance of point of application of the force from the point O.

Accepted Answer

Given,moment of a force = 18 N-mforce = 20 Ndistance = ?Moment of force = force × distanceSubstituting the values in the formula above, we get,18=20×d⇒d=1820⇒d=0.9 m18 = 20 \times \text{d} \[0.5em] \Rightarrow d = \dfrac{18}{20} \[0.5em] \Rightarrow d = 0.9 \text{ m} \[0.5em]18=20×d⇒d=2018​⇒d=0.9 mHence, distance of point of application of the force from the point O = 0.9 m

Question 8

Question 2(iv)Define the term 'centre of gravity of a body'.

Accepted Answer

The centre of gravity of a body is defined as the point about which the algebraic sum of moments of weights of particles constituting the body is zero and the entire weight of the body is to act at this point.

Question 9

Question 2(v)The diagram below shows a lever in use:(a)	To which class of levers does it belong?(b) Without changing the dimensions of the lever, if the load is shifted towards the fulcrum, what happens to the mechanical advantage of the lever?

Accepted Answer

(a) It belongs to second class lever as load is situated between fulcrum and effort.(b) The mechanical advantage of the lever will increase when load is shifted towards the fulcrum.

Question 10

Question 2(vi)A beam of metal is acted upon by two forces (parallel) of 15 N and 20 N at a distance of 35 cm from each other. At what point the beam will be balanced ?

Accepted Answer

Let the balance be at x cm from 20 N force.Anticlockwise moment = 20(35-x)Clockwise moment = 15xAccording to principle of moment :Anticlockwise moment = Clockwise momentSubstituting the values in the formula above, we get,20(35−x)=15x⇒(20×35)−20x=15x⇒(20×35)=(15+20)x⇒(20×35)=35x⇒x=(20×35)35=20 cm20(35-\text{x}) = 15\text{x} \[0.5em] \Rightarrow (20 \times 35) - 20 \text{x} = 15\text{x} \[0.5em] \Rightarrow (20 \times 35) = (15 + 20)\text{x} \[0.5em] \Rightarrow (20 \times 35) = 35 \text{x} \[0.5em] \Rightarrow \text{x} = \dfrac{(20 \times 35)}{35} = 20 \text{ cm} \[0.5em]20(35−x)=15x⇒(20×35)−20x=15x⇒(20×35)=(15+20)x⇒(20×35)=35x⇒x=35(20×35)​=20 cmHence, the beam will be balanced at 20 cm

Question 11

Question 2(vii)Calculate the equivalent resistance between P and Q from the following diagram:

Accepted Answer

In the circuit, there are two parts. In the first part, two resistors of 10 Ω each are connected in series. If the equivalent resistance of this part is R's thenR's = (10 + 10) Ω = 20 ΩIn the second part, resistance R's = 20 Ω and 5 Ω are connected in parallel. If the equivalent resistance of this part is Rp then1Rp=120+151Rp=1+4201Rp=5201Rp=14⇒Rp=4Ω\dfrac{1}{R_p} = \dfrac{1}{20} + \dfrac{1}{5} \[0.5em] \dfrac{1}{R_p} = \dfrac{1 + 4}{20} \[0.5em] \dfrac{1}{R_p} = \dfrac{5}{20} \[0.5em] \dfrac{1}{R_p} = \dfrac{1}{4} \[0.5em] \Rightarrow R_p = 4 Ω \[0.5em]Rp​1​=201​+51​Rp​1​=201+4​Rp​1​=205​Rp​1​=41​⇒Rp​=4Ω∴ Equivalent resistance of second part = 4 ΩIn the last part, 3 Ω, Rp = 4 Ω and 2 Ω are connected in series. If the equivalent resistance of this part is Rs thenRs = (3 + 4 + 2) Ω = 9 Ω∴ Equivalent resistance between the points P and Q = 9 Ω

Question 12

Question 3(i)(a) The diagram shows a bar magnet surrounded by four plotting compasses. Copy the diagram and mark the direction of the compass needle for each of the cases B, C and D.(b) Which is the North pole, X or Y?

Accepted Answer

(a) Below diagram shows the direction of the compass needle for cases B, C and D:(b) X is the north pole as the magnetic field lines are coming out of the bar magnet at point X.

Question 13

Question 3(ii)An object is placed in front of a converging lens. Then, draw the ray diagrams for following cases(a)	When object distance (u) > 2f.(b)	When image formed has the same size as that of object.

Accepted Answer

(a) Ray diagram for the formation of image when object distance (u) > 2f for a convex lens is shown below:(b) Ray diagram for the formation of image when image formed has the same size as that of object for a convex lens is shown below:

Question 14

Question 3(iii)(a)	Ring system is a better wiring system, why?(b)	State the relation between electrical power, resistance and potential difference in an electrical circuit.

Accepted Answer

(a) Advantages of a ring system are —Each appliance has a separate fuse. Therefore, if due to some fault, the fuse of one appliance burns, it does not affect the operation of the other appliances.In this system all the plugs and sockets used can be of same size, but each socket should have it's own fuse of rating suitable for the appliances to be connected with it.(b) The relation between electrical power, resistance and potential difference in an electrical circuit is given by:P = V2R\dfrac{	ext{V}^2}{	ext{R}}RV2​

Question 15

Question 3(iv)Explain any two factors on which the internal resistance of a cell depends.

Accepted Answer

Two factors on which the internal resistance of a cell depends are:The surface area of the electrodes — Larger the surface area of the electrodes, less is the internal resistance.The distance between the electrodes — More the distance between the electrodes, greater is the internal resistance.

Question 16

Question 3(v)When 2240 J of heat is supplied to a metal piece of mass 450 g, its temperature rises from 25°C to 65°C. Then, find(a)	heat capacity of metal piece and(b) specific heat capacity of metal.

Accepted Answer

Given,Amount of heat supplied (Q)= 2240 JRise in temperature (△t) = (65 – 25)°C = 40°CHeat capacity = ?(a) Heat capacity (C') = Q△t\dfrac{	ext{	ext{Q}}}{	ext{△t}}△tQ​Substituting the values in the formula above we get,=224040=56 J°C−1= \dfrac{2240}{40} = 56 	ext{ J°C}^{-1}=402240​=56 J°C−1Hence, heat capacity of metal piece = 56 J per °C(b) Specific heat capacity (c) = ?mass (m) = 450 gWe know that,c=C′m	ext{c} = \dfrac{	ext{C}'}{	ext{m}}c=mC′​Substituting the values in the formula above we get,c=56450=0.12 J g−1°C−1	ext{c} = \dfrac{56}{450} = 0.12	ext{ J g}^{-1} 	ext{°C}^{-1}c=45056​=0.12 J g−1°C−1Hence, Specific heat capacity = 0.12 J g-1 °C-1

Question 17

Question 4(i)An object of height 5 cm is placed perpendicular to the principal axis of a concave lens of focal length 10 cm. If the distance of the object from the optical centre of the lens is 20 cm, determine the position, nature and size of the image formed using the lens formula.

Accepted Answer

Given,f = -10 cmu = -20 cmv = ?Lens formula is —1v−1u=1f\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \[0.5em]v1​−u1​=f1​Substituting the values in the formula, we get,1v−1−20=1−101v+120=−1101v=−110−1201v=−2−1201v=−320⇒v=−203⇒v=−6.67\dfrac{1}{v} - \dfrac{1}{-20} = \dfrac{1}{-10} \[0.5em] \dfrac{1}{v} + \dfrac{1}{20} = - \dfrac{1}{10} \[0.5em] \dfrac{1}{v} = -\dfrac{1}{10} - \dfrac{1}{20} \[0.5em] \dfrac{1}{v} = \dfrac{-2-1}{20} \[0.5em] \dfrac{1}{v} = -\dfrac{3}{20} \[0.5em] \Rightarrow v = -\dfrac{20}{3} \[0.5em] \Rightarrow v = -6.67v1​−−201​=−101​v1​+201​=−101​v1​=−101​−201​v1​=20−2−1​v1​=−203​⇒v=−320​⇒v=−6.67Therefore, the image in formed 6.67 cm in front of the lens.As we know,the formula for magnification of a lens is —m=vu=hihom = \dfrac{v}{u} = \dfrac{h_i}{h_o} \[0.5em]m=uv​=ho​hi​​ Given,u = -20 cmv = -6.67 cmheight of object (ho) = 5 cmheight of image (hi) = ?Substituting the values in the formula, we get,−6.67−20=hi5hi=5×−6.67−20hi=1.67 cm\dfrac{-6.67}{-20} = \dfrac{\text{h}_i}{5} \[0.5em] \text{h}_i = 5 \times \dfrac{-6.67}{-20} \[0.5em] \text{h}_i = 1.67 \text{ cm} \[0.5em]−20−6.67​=5hi​​hi​=5×−20−6.67​hi​=1.67 cmHence, the image is formed 6.67 cm in front of the lens, it is of size 1.67 cm and it is virtual (as the magnification is positive), erect and diminished.

Question 18

Question 4(ii)(a) Figs. I and II show the incident and transmitted rays through lens kept in a box in each case. Draw the lens and complete the path of rays through it.(b)	What is the difference between double convex and bi-convex lens?(c)	Does convex lens always form real images?

Accepted Answer

(a) In Fig I, the rays are diverging, hence a concave lens is placed inside the box:In Fig II, the rays are converging, hence a convex lens is placed inside the box:(b) In case of double convex lens the two surfaces may have any values for the two radii of curvature of the two surfaces while in bi-convex lens the radii of curvature of the two surfaces, have the same value. So, all double convex lenses may not be bi-convex but every bi-convex lens is a double convex lens.(c) No, when the distance of the object from the lens is less than the focal length of the lens, then convex lens will form virtual image.

Question 19

Question 4(iii)(a) The velocity of light in diamond is 121000 km/s. What is its refractive index?(b)	With the help of a labelled diagram, show that the apparent depth of an object such as a coin in water is less than its real depth.

Accepted Answer

(a) Given,velocity of light in diamond = 121000 kms-1To convert kms-1 into ms-1, we multiply by 1000121000 x 1000 ms-1 = 121 x 106 ms-1Speed of light in air, c = 3 × 108 ms-1As we know,μ=Speed of light in vacuum or air (c) Speed of light in diamond (V)μ = \dfrac{\text{Speed of light in vacuum or air (c) }}{\text {Speed of light in diamond (V)}} \[0.5em]μ=Speed of light in diamond (V)Speed of light in vacuum or air (c) ​Substituting the values in the formula above we get,μ=3×108121×106μ=3×1081.21×108μ=2.48μ = \dfrac{ 3 \times 10^8 } {121 \times 10^6} \[0.5em] μ = \dfrac{ 3 \times 10^8 } {1.21 \times 10^8} \[0.5em] μ = 2.48 \[0.5em]μ=121×1063×108​μ=1.21×1083×108​μ=2.48Hence, refractive index = 2.48(b) Below labelled diagram, shows that the apparent depth of a coin in water is less than its real depth:

Question 20

Question 5(ii)(a)	What are the necessary conditions for TIR?(b)	Draw a diagram to show how rays of light are totally reflected when fall normally on the face of a right angled prism. Where is this action of prism used?

Accepted Answer

(a) The two necessary conditions for total internal reflection are :The light must travel from a denser medium to a rarer medium.The angle of incidence must be greater than the critical angle for the pair of media.(b) Below ray diagram shows the total internal reflection of light rays falling normally on the face of a right angled prism:This action of prism is used in a periscope where a total reflecting prism is preferred over a plane mirror.

Solved Sample Paper 2

Section A

Section B