Q: Question 2(iii)The distance between two bodies is decreased by a factor of 3. How is the magnitude of gravitational force between them affected ?

As we know,F = G m1m2r2\dfrac{\text{m}_1\text{m}_2}{\text{r}^2}r2m1m2If separation between two bodies is decreased by a factor of 3 then,F = G m1m2(r3)2\dfrac{\text{m}_1\text{m}_2}{{(\dfrac{\text{r}}{3}})^2}(3r)2m1m2 = 9G m1m2r2\dfrac{\text{m}_1\text{m}_2}{\text{r}^2}r2m1m2Therefore, the gravitational force will be increased by a factor of 9.

Question 1

Question 1(ii)If the speed of a moving particle is tripled, then the ratio of its kinetic energy to its momentum will bedoubledtripledquadrupledNone of these

Accepted Answer

tripledReason — Let velocity be v thenKinetic energy (KE) = 12mv2\dfrac{1}{2}mv^221​mv2 andmomentum (ρ) = mvKEρ=12mv2mv=v2\dfrac{KE}{ρ} = \dfrac{\dfrac{1}{2}mv^2}{mv} = \dfrac{v}{2}ρKE​=mv21​mv2​=2v​    [Equation 1]Now, when velocity is tripled thenKE' = 12m(3v)2\dfrac{1}{2}m(3v)^221​m(3v)2 = 92m(v)2\dfrac{9}{2}m(v)^229​m(v)2and ρ' = 3mvKE′ρ′=92mv23mv=3v2\dfrac{KE'}{ρ'} = \dfrac{\dfrac{9}{2}mv^2}{3mv} = \dfrac{3v}{2}ρ′KE′​=3mv29​mv2​=23v​    [Equation 2]On comparing equation 1 and 2, we can see that the ratio gets tripled.

Question 2

Question 1(xii)What is the magnification of the image when a lens of focal length 10 cm and object distance from the lens is 15 cm?2112\dfrac{1}{2}21​4

Accepted Answer

2Reason — As we know, the lens formula is —1f=1v−1u\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \[1em]f1​=v1​−u1​Given,f = 10 cmu = -15 cmv = ?Substituting the values in the formula, we get,110=1v–1−15110=1v+1151v=110−1151v=3−2301v=130⇒v=30 cm\dfrac{1}{10} = \dfrac{1}{v} – \dfrac{1}{-15} \[1em] \dfrac{1}{10} = \dfrac{1}{v} + \dfrac{1}{15} \[1em] \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15} \[1em] \dfrac{1}{v} = \dfrac{3-2}{30} \[1em] \dfrac{1}{v} = \dfrac{1}{30} \[1em] \Rightarrow v = 30 \text{ cm} \[1em]101​=v1​–−151​101​=v1​+151​v1​=101​−151​v1​=303−2​v1​=301​⇒v=30 cm∴ Image distance is 30 cm.As we know, the formula for magnification of a lens is —m=vum = \dfrac{v}{u}m=uv​Given,v = 30 cmu = -15 cmSubstituting the values in the formula, we get,m=30−15m=−2m = \dfrac{30}{- 15} \[1em] m = -2m=−1530​m=−2∴ The magnification is 2, negative sign implies image is inverted.

Question 3

Question 2(ii)A uniform meter scale is in equilibrium as shown in the figure given belowCalculate the weight of the meter scale.

Accepted Answer

Let W gf be the weight of metre rule which acts downwards from 50 cm mark.In equilibrium position, taking moments about fulcrum.Anticlockwise moment = 40 x (30 - 5) = 40 x 25 = 1000 gf cmClockwise moment = W (50 - 30) = 20 W gf cm∴ Using principle of moments:Clockwise moment = anticlockwise momentSo, 20 W = 1000W = 100020\dfrac{1000}{20}201000​ = 50 gf∴ Weight of the metre scale = 50 gf

Question 4

Question 2(iii)The distance between two bodies is decreased by a factor of 3. How is the magnitude of gravitational force between them affected ?

Accepted Answer

As we know,F = G m1m2r2\dfrac{	ext{m}_1	ext{m}_2}{	ext{r}^2}r2m1​m2​​If separation between two bodies is decreased by a factor of 3 then,F = G m1m2(r3)2\dfrac{	ext{m}_1	ext{m}_2}{{(\dfrac{	ext{r}}{3}})^2}(3r​)2m1​m2​​ = 9G m1m2r2\dfrac{	ext{m}_1	ext{m}_2}{	ext{r}^2}r2m1​m2​​Therefore, the gravitational force will be increased by a factor of 9.

Question 5

Question 2(iv)A force of 20 N displaces a body by a distance of 4 m. Find the amount of work done when the displacement is:(a) in the direction of force.(b) at an angle of 60° with the force.

Accepted Answer

Given,Force (F) = 20 NandDisplacement (s) = 4 m(a) When displacement is in the direction of force, θ = 0°W = Fs cos θ°= Fs cos 0°= 20 x 4 x 1 [∵ cos 0° = 1]= 80 J(b) When displacement is at an angle 60° with the direction of force,W = Fs cos θ°= Fs cos 60°= 20 x 4 x 12\dfrac{1}{2}21​ [∵ cos 60° = 12\dfrac{1}{2}21​]= 40 J

Question 6

Question 2(v)(a) Name the two major parts of a SONAR.(b) Name the type of vibrations produced in the given case: A freely suspended pendulum vibrating about its mean position.

Accepted Answer

(a) The two parts of SONAR areTransmitter (for emitting ultrasonic waves)Receiver (for detecting ultrasonic waves)(b) Free vibrations

Question 7

Question 2(vi)Which will be the position of centre of gravity of(a) Parallelogram(b) Hollow cone?

Accepted Answer

(a) Parallelogram — point of intersection of diagonals(b) Hollow cone — at a height h3\dfrac{h}{3}3h​ from the base, on its axis (if h = height of cone).

Question 8

Question 2(vii)(a) Name the colour code of wire which is connected to the metallic body of an appliance according to the new convention.(b) Which part of electrical appliance is earthed?

Accepted Answer

(a) Green(b) Metallic body of an electrical appliance is earthed.

Question 9

Question 3(i)Calculate the mass of ice required to lower the temperature of 600 g of water at 50°C to water at 12°C. (Take, specific latent heat of ice = 366 Jg-1 and specific heat capacity of water = 4.2 Jg-1°C-1)

Accepted Answer

Given,Mass of water (m) = 600 gInitial temperature = 50°CFinal temperature = 12°C∴ fall in temperature = 50 - 12 = 38 °CHeat lost by water = m x c x Δt= 600 x 4.2 x 38 = 95,760 JIf m' g ice is added, heat gained by it to melt to 0 °C = m'L = m' x 336 JBy the principle of method of mixture,heat lost by water = heat gained by ice∴ 95,760 = m' x 366m' = 95,760366\dfrac{95,760}{366}36695,760​ = 261.639 gHence, mass of ice required = 261.64 g

Question 10

Question 3(ii)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.

Accepted Answer

(a) Ray diagram showing the image formation by a convex lens when an object is placed between optical centre and focus of the lens is given below:(b) Ray diagram showing the image formation by a convex lens when an object is placed between focus and twice the focal length of the lens is given below:

Question 11

Question 3(iii)Complete the following nuclear changes.(a) 11Na24 ⟶ X + -1β0(b) 92U238 ⟶ 90Th234 + ............... + energy

Accepted Answer

(a) 11Na24 ⟶ 12Mg24 + -1β0(b) 92U238 ⟶ 90Th234 + 2He4 + energy

Question 12

Question 3(iv)Derive an expression for equivalent resistance in the following caseDecide, which resistances are in series and parallel. Solve for series and then for parallel. Combine both the results to get the equivalent resistance.

Accepted Answer

In the circuit, there are three parts. In the first part, resistors of R1 and R2 Ω are connected in series. If the equivalent resistance of this part is R's thenR's = R1 + R2In the second part, resistors of R3 and R4 are connected in series. If the equivalent resistance of this part is R''s thenR''s = R3 + R4In the third part, the two parts of resistance R's and R''s are connected in parallel. If the equivalent resistance between points A and B is Rp then1Rp=1Rs′+1Rs′′1Rp=1R1+R2+1R3+R41Rp=R3+R4+R1+R2(R1+R2)(R3+R4)Rp=(R1+R2)(R3+R4)R1+R2+R3+R4\dfrac{1}{R_p} = \dfrac{1}{R'_s} + \dfrac{1}{R''_s} \[0.5em] \dfrac{1}{R_p} = \dfrac{1}{R_1 + R_2} + \dfrac{1}{R_3 + R_4} \[0.5em] \dfrac{1}{R_p} = \dfrac{R_3 + R_4 + R_1 + R_2}{(R_1 + R_2)(R_3 + R_4)} \[0.5em] R_p = \dfrac{(R_1 + R_2)(R_3 + R_4)}{R_1 + R_2 + R_3 + R_4} \[0.5em]Rp​1​=Rs′​1​+Rs′′​1​Rp​1​=R1​+R2​1​+R3​+R4​1​Rp​1​=(R1​+R2​)(R3​+R4​)R3​+R4​+R1​+R2​​Rp​=R1​+R2​+R3​+R4​(R1​+R2​)(R3​+R4​)​Hence, equivalent resistance = (R1+R2)(R3+R4)R1+R2+R3+R4\dfrac{(R_1 + R_2)(R_3 + R_4)}{R_1 + R_2 + R_3 + R_4}R1​+R2​+R3​+R4​(R1​+R2​)(R3​+R4​)​

Question 13

Question 3(v)A wire is dropped freely towards earth. Will any emf be induced across the ends of wire, if wire is initially in(a) north-south direction?(b) east-west direction?

Accepted Answer

(a) In the North-South direction, the wire is moving parallel to the Earth's magnetic field, hence, there would be minimal or no induced emf.(b) In the East-West direction, the wire is moving perpendicular to the Earth's magnetic field, hence, there would be an induced emf due to the motion of the wire across the magnetic field lines.

Question 14

Question 4(i)(a) What is the angle of deviation?(b)	Copy the diagram given below and complete the path of the light ray till it emerges out of the prism. The critical angle of the glass is 42°. In your diagram, mark the angles wherever necessary.

Accepted Answer

(a) Angle of deviation is the angle between the direction of incident ray and the emergent ray. It is represented by the greek alphabet δ (delta).(b) The completed ray diagram showing the complete path of the light ray till it emerges out of the prism is given below:

Question 15

Question 4(iii)If a girl puts her pencil into an empty trough and see the pencil from the position as given in the figure below:(a)	What kind of variation will be noticed in the appearance of pencil when water is poured into the trough?(b)	Mention the name of the phenomenon which accounts for the above stated observation.(c)	What is the cause of the observed phenomenon when the pencil is placed in water?

Accepted Answer

(a) When a girl puts her pencil into an empty trough, and later pours water in the same trough, the part of the pencil which is immersed in water will look bent or broken.Below is the diagram showing how the girl's eye sees the pencil through water:(b) The above observation is based on the phenomenon of refraction of light.(c) The cause of above phenomenon is refraction of light. Rays starting from the dipped portion of the pencil suffer refraction at the water-air surface. Air is a rarer medium than water. So, the refracted rays bend away from the normal and appear to come from a point slightly above the actual position of the pencil. Hence, the pencil appears bent or broken.

Question 16

Question 5(i)(a)	Define electromagnetic spectrum.(b)	Write two uses of ultraviolet rays.(c)	Calculate the speed of electromagnetic wave in a glass slab in which its frequency is 4 x 1014 Hz and wavelength is 500 nm.

Accepted Answer

(a) The orderly arrangement of electromagnetic waves in increasing or decreasing order of wavelength or frequency is called electromagnetic spectrum.(b) Two uses of ultraviolet rays are:For sterilising air, surgical equipments, etc.For detecting the purity of gems, eggs, ghee etc.(c) Given,f = 4 x 1014 Hzλ = 500 nm = 500 x 10-9 m = 5 x 10-7 mAs we know,Velocity of wave (c) = frequency (f) x wavelength (λ)Substituting the values in the formula above we get,c = (4 x 1014) x (5 x 10-7) = 2 x 108Hence,Speed of the electromagnetic wave = 2 x 108 ms-1

Question 17

Question 5(ii)A virtual, diminished image is formed when an object is placed between the optical centre and the principal focus of a lens.(a)	Name the type of lens which forms the above image.(b)	Draw a ray diagram to show the formation of the image with the above stated characteristics.

Accepted Answer

(a) Concave lens(b) Ray diagram showing the formation of the image is given below:

Question 18

Question 5(iii)The diagram below shows an object AB kept in front of the lens. The path of one ray coming from the object is shown:(a)	Name the lens L.(b)	Redraw and complete the ray diagram showing the formation of the image.(c)	In which optical instrument, is this kind of image formed?

Accepted Answer

(a) Convex lens(b) Completed ray diagram showing the formation of the image is shown below:(c) Magnifying glass or a simple microscope

Question 19

Question 6(i)A uniform meter scale of mass 60 g, carries masses of 20 g, 30 g and 80 g from points 10 cm, 20 cm and 90 cm marks. Where must be the scale hanged with string to balance the scale?

Accepted Answer

Let, D be the point from where the scale is hanged so that the meter scale and the various masses are balanced.Clockwise moments = 80 x GDAnticlockwise moments = (20 x FD) + (30 x ED) + (60 x CD)As the meter scale is balanced.∴ Sum of clockwise moments = sum of anticlockwise moments80 x GD = (20 x FD) + (30 x ED) + (60 x CD)[80 x {90 - (50 + x)}] = [20 x (50 + x - 10)] + [30 x (50 + x - 20)] + 60x[80 x (90 - 50 - x)] = [20 x (50 + x - 10)] + [30 x (50 + x - 20)] + 60x[80(40 - x)] = [20(40 + x)] + [30(30 + x)] + 60x3200 - 80x = 800 + 20x + 900 + 30x + 60x3200 - 80x = 800 + 20x + 900 + 30x + 60x3200 - 80x = 1700 + 110x3200 - 1700 = 110x + 80x1500 = 190xx = 1500190\dfrac{1500}{190}1901500​ = 7.89 ≈ 7.9 cmHence, the string should be hanged at 50 + 7.9 = 57.9 cm mark.

Question 20

Question 6(ii)(a) What is meant by static equilibrium? Give one example.(b) A pulley system has four pulleys in all and is 95% efficient. CalculateMAEffort required to lift a load of 1000 N

Accepted Answer

(a) When a body remains in a state of rest under the influence of several forces, the body is said to be in static equilibrium.For example, if a book is lying on a table, the weight of the book exerted on the table vertically downwards is balanced by an equal and opposite force of reaction exerted by the table on the book vertically upwards. Thus, the book is in static equilibrium.(b) Given,Efficiency (η) = 95%Number of pulleys = 4Velocity ratio = 4 (as the system has 4 pulleys)MA = ?We know, η=MAVR\eta = \dfrac{	ext {MA}}{	ext {VR}}η=VRMA​Substituting in the formula above we get,95100=MA4\dfrac{95}{100} = \dfrac{	ext {MA}}{	ext {4}}10095​=4MA​MA = 0.95 x 4 = 3.8Hence, MA = 3.8II. From formula : MA = LoadEffort\dfrac{	ext{Load}}{	ext{Effort}}EffortLoad​Substituting in the formula above we get,3.8 = 1000effort\dfrac{1000}{	ext{effort}}effort1000​effort = 10003.8\dfrac{1000}{3.8}3.81000​ = 263.15 NHence, effort required = 263.15 N

Solved Sample Paper 3

Section A

Section B