Question 1

Question 1(i)A certain mass of solid substance is heated for a period of 50 min. Its heating curve is shown below :From the graph, which one of the following statement is correct?The boiling point of the given substance is 40°C.The specific latent heat of fusion of substance is the same as the specific latent heat of vaporisation.The substance has a higher specific heat capacity in solid state than that in liquid state.The substance has a lower specific heat capacity in solid state than that in liquid state.

Accepted Answer

The substance has a higher specific heat capacity in solid state than that in liquid stateReason — From graph, 40°C is the melting point of the substance. It takes 10 mins for melting for temperature change of 20°C (i.e.,40°-20°) and takes 10 mins for boiling for temperature change of 40°C (i.e., 80°-40°).Hence, the latent heat of fusion is less than the latent heat of vapourisationFrom relation,Q = m x c x △TC ∝ 1ΔT\dfrac{1}{\Delta	ext{T}}ΔT1​So, from graph, the specific heat capacity in solid state is greater than that in liquid state.

Question 2

Question 1(ii)The figure given below shows the cooling curve of pure wax material after heating. It cools from A and B and solidifies along BD. If L and c represents the value of latent heat and the specific heat of liquid wax, then the ratio of Lc\dfrac{	ext{L}}{	ext{c}}cL​ is204060100

Accepted Answer

20Reason — Curve AB represents the cooling of wax material, the specific heat of the cooling of wax material is given by C, and temperature change Δt = 90°C − 80°CHeat supplied in 2 min = m x c x △T= m x c x (90-80) = 10mc    .....[1]Curve BD represents the latent heat of fusion during the solidification of wax material for the next 4 minutes. It is represented byQ = mL    .....[2]As process of solidification takes twice as long as the cooling process,∴ 2[10mc] = mL    [From Eq 1 & 2]20mc = mL20c = LHence, Lc\dfrac{	ext{L}}{	ext{c}}cL​ = 20

Question 3

Question 1(viii)A submarine produces an ultrasonic wave of velocity 1500 ms\dfrac{	ext{m}}{	ext{s}}sm​ in water. The officer receives signal after 50 s of emission of ultrasonic waves. Find the distance of object which is present in the bottom of sea.25.7 km37.5 km37.5 m50 km

Accepted Answer

37.5 kmReason — 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,t = 50 sV = 1500 ms-1Substituting the values in the formula above, we get,1500=2d50⇒d=1500×502⇒d=1500×25⇒d=37500 m1500 = \dfrac{2d}{50} \[0.5em] \Rightarrow d = \dfrac{1500 \times 50}{2} \[0.5em] \Rightarrow d = 1500 \times 25 \[0.5em] \Rightarrow d = 37500 \text{ m}1500=502d​⇒d=21500×50​⇒d=1500×25⇒d=37500 mConverting m to km1000 m = 1 km∴ 37500 m = 375001000\dfrac{37500}{1000}100037500​ km = 37.5 kmHence, distance of the submarine from the object = 37.5 km

Question 4

Question 1(xiv)For which position of the object in front of convex lens, the image distance will be negative ?

Accepted Answer

Between optical center and focus of convex lens:Reason — For a convex lens, the image distance is considered negative when the image is formed on the same side as the incident light. This typically occurs when the object is placed within the focal length (between the lens and its focal point). So, if the object is positioned anywhere within the focal length of a convex lens, the image distance will be negative.

Question 5

Question 2(i)(a) Give two reasons as to why the efficiency of a single removable pulley system is always less than 100%.(b)	Why a thick lead container is required to keep a radioactive substance?(c)	What is the principle of moments?

Accepted Answer

(a) The efficiency of a single movable pulley system is not 100% because of the following reasons:There is friction in the pulley bearings or at the axle.The weight of the pulley and string is not zero.(b) The radioactive substances must be kept in a thick lead container with a very narrow opening, so that the walls of the container absorb the radiations which strike on them from the inside and thus radiations only come out through the narrow opening.(c) According to the principle of moments, if the algebraic sum of moments of all the forces, acting on the body, about the axis of rotation is zero, the body is in equilibrium. Therefore, as per the principle of moments,sum of anticlockwise moments = sum of clockwise moments

Question 6

Question 2(iii)Two waves of the same pitch have their amplitudes in the ratio 2:3.(a) What will be the ratio of their loudness?(b) What will be the ratio of their frequencies?

Accepted Answer

(a) Given, amplitude ratio, a1 : a2 = 2 : 3Loudness is directly proportional to square of amplitude∴ L ∝ a2⇒L1L2=a12a22⇒L1L2=2232⇒L1L2=49\phantom{\Rightarrow} \dfrac{\text{L}_1}{\text{L}_2} = \dfrac{\text{a}_1^2}{\text{a}_2^2} \[1em] \Rightarrow \dfrac{\text{L}_1}{\text{L}_2} = \dfrac{2^2}{3^2} \[1em] \Rightarrow \dfrac{\text{L}_1}{\text{L}_2} = \dfrac{4}{9}⇒L2​L1​​=a22​a12​​⇒L2​L1​​=3222​⇒L2​L1​​=94​Hence, the ratio of their loudness will be 4 : 9.(b) Waves with same pitch have the same frequency.Hence, the ratio of their frequencies will be 1 : 1.

Question 7

Question 2(iv)(a)	What is meant by potential energy of body?(b)	State whether the potential energy will be maximum or minimum at the surface of earth.

Accepted Answer

(a) The energy possessed by a body at rest due to its position or size and shape is called the potential energy.(b) Potential energy will be minimum at the surface of earth because potential energy at any point = mgh, and at the surface of earth h = 0, hence, potential energy is minimum.

Question 8

Question 2(v)A uniform hall meter rule balances horizontally on a knife edge at 29 cm mark when a weight of 20 gf is suspended from one end.(a)	Draw a diagram of the arrangement(b)	What is the weight of the half meter rule?

Accepted Answer

(a) Diagram of the arrangement is shown below:(b) If W gf is the weight of half meter rule, by the principle of moments,W×(29−25)=20×(50−29)⇒W×4=20×21⇒W=20×214⇒W=105 gfW \times (29 - 25) = 20 \times (50 - 29) \[0.5em] \Rightarrow W \times 4 = 20 \times 21 \[0.5em] \Rightarrow W = \dfrac{20 \times 21}{4} \[0.5em] \Rightarrow W = 105 \text{ gf}W×(29−25)=20×(50−29)⇒W×4=20×21⇒W=420×21​⇒W=105 gf

Question 9

Question 2(vi)(a) Derive the relation between CGS and SI unit of power.(b) A man is holding a suitcase in his hand at rest. What is the work done by him?

Accepted Answer

(a) SI unit of power = watt = 1 J/s∴ 1 W = 1 J/s = 105 dyne x 102 cm/s [∵ J = N-m]= 107 erg/s [∵ dyne x cm = erg](b) Work done by the man is zero as his displacement is zero.

Question 10

Question 2(vii)A meter scale is balanced in horizontal position as shown in figure below. Find the value of w.

Accepted Answer

From the given figure we can see that,Anticlockwise moment = 6 kg × 60 cmClockwise moment = w × 40 cmAs we know, the principle of moments states thatAnticlockwise moment = Clockwise moment.6×60=w×40⇒w=6×6040⇒w=9 kg6 \times 60 = \text{w} \times 40 \[0.5em] \Rightarrow \text{w} = \dfrac{6 \times 60}{40}\[0.5em] \Rightarrow \text{w} = 9 \text{ kg} \[0.5em]6×60=w×40⇒w=406×60​⇒w=9 kgHence, the value of w = 9 kg

Question 11

Question 3(i)10125 J of heat energy boils off 4.5 g of water at 100°C to steam at 100°C. Find the specific latent heat of steam.

Accepted Answer

Given,Mass (m) = 4.5 g = 4.51000\dfrac{4.5}{1000}10004.5​ = 4.5 x 10-3 kgHeat energy given out (Q) = 10,125 JSpecific latent heat of steam = ?From relation Q = m x LSubstituting the values in the formula we get,10125=4.5×10−3×L⇒L=101254.5×10−3⇒L=2.25×106 J kg−110125 = 4.5 \times 10^{-3} \times L \[0.5em] \Rightarrow L = \dfrac{10125}{4.5 \times 10^{-3}} \[0.5em] \Rightarrow L = 2.25 \times 10^{6} \text{ J kg}^{-1} \[0.5em]10125=4.5×10−3×L⇒L=4.5×10−310125​⇒L=2.25×106 J kg−1Hence, the specific latent heat of steam = 2.25 x 106 J kg-1

Question 12

Question 3(ii)It is established that an electric current through a metallic conductor produces a magnetic field around it. Is there a similar magnetic field produced around a thin beam of moving(a)	α-particles and(b)	neutrons? Justify your answer.

Accepted Answer

When movement of a charged particle takes place, a magnetic field is produced around the path on which the charged particle moves.(a) Yes, a thin beam of α-particles (positively charged) is like a straight current carrying conductor in the direction of motion.(b) No, as neutrons carry no charge, so no magnetic field would be created around its path.

Question 13

Question 3(iii)Write the uses of radio isotopes in(a)	agriculture(b) industries.

Accepted Answer

(a) Agriculture — 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) Industries — Radio isotopes such as   92235U^{235}_{\space \space 92}	ext{U}  92235​U are used as fuel for atomic energy reactors.

Question 14

Question 3(iv)Two lenses have power of(a) +2 D and(b) -4 DWhat is the nature and focal length of each lens?

Accepted Answer

(a) As we know,Power=1focal length\text{Power} = \dfrac{1}{\text{focal length}}Power=focal length1​Given,Power of the lens = +2.0 DAs the given power is positive,Therefore, we can say that the lens used is convex in nature.Substituting the value of power in formula we get,2.0=1focal length⇒focal length=12⇒focal length=0.5 m⇒focal length=50 cm\text{2.0} = \dfrac{1}{\text{focal length}} \[0.5em] \Rightarrow \text{focal length} = \dfrac{1}{2} \[0.5em] \Rightarrow \text{focal length} = \text{0.5 m} \[0.5em] \Rightarrow \text{focal length} = \text{50 cm} \[0.5em]2.0=focal length1​⇒focal length=21​⇒focal length=0.5 m⇒focal length=50 cmTherefore, the focal length is 50 cm and the lens used is convex in nature.(b) As we know,Power=1focal length\text{Power} = \dfrac{1}{\text{focal length}}Power=focal length1​Given,Power of the lens = -4.0 DAs the given power is negative hence, we can say that the lens used is concave in nature.Substituting the value of power in formula we get,−4.0=1focal length⇒focal length=−14⇒focal length=−0.25 m⇒focal length=−25 cm-\text{4.0} = \dfrac{1}{\text{focal length}} \[0.5em] \Rightarrow \text{focal length} = -\dfrac{1}{4} \[0.5em] \Rightarrow \text{focal length} = -\text{0.25 m} \[0.5em] \Rightarrow \text{focal length} = -\text{25 cm} \[0.5em]−4.0=focal length1​⇒focal length=−41​⇒focal length=−0.25 m⇒focal length=−25 cmTherefore, the focal length is 25 cm and the lens used is concave in nature.

Question 15

Question 3(v)Study the following electric circuit. Find the reading of(a)	the ammeter and(b)	the voltmeter

Accepted Answer

(a) In the given circuit, resistances are connected in series, hence, equivalent resistances (Re) = R1 + R2 = 4 + 2 = 6 ΩPotential difference (V) = 3 VCurrent (I) = ?From Ohm's lawV = IRSubstituting the values in the formula above, we get,3=I×6⇒I=36⇒I=0.5A3 = I \times 6 \[0.5em] \Rightarrow I = \dfrac{3}{6} \[0.5em] \Rightarrow I = 0.5 A3=I×6⇒I=63​⇒I=0.5AAs in series combination, current flowing through each component is same and is equal to the total current flowing in the circuit hence, the reading of ammeter = 0.5 A(b) Reading of voltmeter = Potential difference across 2 Ω resistor.Current flowing through bulb = 0.5 AFrom Ohm's lawV = IRSubstituting the values in the formula above, we get,V = 0.5 x 2 = 1 VHence, the reading of voltmeter = 1 V

Question 16

Question 4(iii)(a) A boy uses blue colour of light to find the refractive index of glass, then he repeats the experiment using red colour of light. Will the refractive index be the same or different in the two cases? Give a reason to support your answer.(b) A ray of light incident on a glass slab and follow the path as shown below. If a mirror is placed at S, such that emergent ray falls normally on it, thenI. Draw diagram to show the path of ray reflected from the mirror.II. On the basis of which principle this refraction of ray is done?

Accepted Answer

(a) The speed of blue light in glass is less that of red light.i.e. cb < crWe know that,μ=Speed of light in vacuum or air (c) Speed of light in glass (V)μ = \dfrac{\text{Speed of light in vacuum or air (c) }}{\text {Speed of light in glass (V)}} \[0.5em]μ=Speed of light in glass (V)Speed of light in vacuum or air (c) ​Hence, the refractive index of blue light is greater than that of red light.i.e. μb > μr.Hence, the refractive index of red and blue light in glass will be different.b.I. Below diagram shows the path of the ray reflected from the mirror:II. Refraction of ray is done on the basis of principle of reversibility of light.

Question 17

Question 5(i)A ray of light PQ falls normally on the hypotenuse of a right angled prism ABC as shown below(a) Copy the diagram and complete the path of ray PQ till it emerges from the prism.(b) What is the value of the angle of deviation of the ray?(c) Name an instrument where this action of prism is used.

Accepted Answer

(a) Below diagram shows the path of the ray PQ till it emerges from the prism with all the angles labelled:(b) The angle of deviation of the ray PQ is 180°.(c) This action of the prism is used in Binoculars.

Question 18

Question 5(ii)A linear object is placed on the axis of a lens. An image is formed by refraction in the lens. For all positions of the object on the axis of the lens, the positions of the image are always between the lens and the object.(a)	Name the lens.(b) Draw a ray diagram to show the formation of the image of an object placed infront of the lens at any position of your choice except infinity.

Accepted Answer

(a) The lens is concave as the positions of the image are always between the lens and the object.(b) Ray diagram showing the formation of the image of an object placed infront of the lens at any position except infinity:

Question 19

Question 5(iii)Draw ray diagrams showing the image formation by a convex lens when an object is placed(a) between optical centre and focus of the lens.(b)	between focus and twice the focal length of the lens.(c) at twice the focal length of the lens.(d)	at infinity.

Accepted Answer

(a) between optical centre and focus of the lens:(b)	between focus and twice the focal length of the lens:(c) at twice the focal length of the lens:(d)	at infinity:

Question 20

Question 6(i)(a) A force applied on aI. rigid bodyII. non-rigid bodyHow does the effect of force differ in the two cases?(b)	Give one example of couple force.

Accepted Answer

(a) (i) Rigid body — A force when applied on a rigid object does not change the inter-spacing between it's constituent particles and therefore it does not change the dimensions of the object, but causes only motion in it.(ii) Non-rigid body — A force when applied on a non-rigid object, changes the inter-spacing between it's constituent particles and therefore causes a change in it's dimensions and can also produce motion in it.(b)	Two equal and opposite parallel forces, not acting along the same line form a couple. E.g. tightening the cap of a bottle.

Solved Sample Paper 1

Section A

Section B