Q: Question 3(i)If the refractive index of air with respect to glass is expressed as gμa=sin isin r^g\mu_a = \dfrac{\text{sin i}}{\text{sin r}}gμa=sin rsin i answer the following:(a) Express the similar expression for aμg^a\mu_gaμg in terms of i and r.(b) If angle r = 90°, what is the corresponding angle (i) is called?(c) Write down the physical significance of angle (i) in part (b).

(a) The expression for aμg = sin rsin i\dfrac {\text {sin r}}{\text {sin i}}sin isin r(b) When refractive angle r = 90° then, the corresponding angle of incidence i will be equal to critical angle.(c) The phenomenon of total internal reflection occurs when the angle of incidence exceeds the value of i obtained in part (b).

Question 1

Question 1(iv)What is ratio of force acting on a body while going upwards and downwards under the influence of earth's gravity?2-11-2

Accepted Answer

-1Reason — Acceleration of the body while going down = gAcceleration of the body while going up = - gHence, ratio of acceleration due to gravity upwards and downwards = −gg\dfrac{-g}{g}g−g​ = -1

Question 2

Question 1(xiv)An object is vibrating at 4000 Hz. The time period of sound produced is0.0025 s0.025 s0.00025 s0.005 s

Accepted Answer

0.00025 sReason — Given, frequency (f) = 4000 HzTime period = 1frequency\dfrac{1}{	ext{frequency}}frequency1​ = 14000\dfrac{1}{4000}40001​ = 0.00025 s

Question 3

Question 2(i)(a)	If the moment of force is assigned a negative sign, then will the turning tendency of the force be clockwise or anti-clockwise?(b)	Complete the following nuclear reactions given belowI. 92238P⟶αQ⟶βR⟶βS^{238}_{92}	ext{P} \overset{\alpha} \longrightarrow 	ext{Q} \overset{\beta} \longrightarrow 	ext{R} \overset{\beta} \longrightarrow 	ext{S}92238​P⟶α​Q⟶β​R⟶β​SII. ZAX⟶αX1⟶γX2⟶2βX3^{A}_{Z}	ext{X} \overset{\alpha} \longrightarrow 	ext{X}_{1} \overset{\gamma} \longrightarrow 	ext{X}_{2} \overset{2\beta} \longrightarrow 	ext{X}_{3}ZA​X⟶α​X1​⟶γ​X2​⟶2β​X3​

Accepted Answer

(a) If the moment of force is assigned a negative sign, then the turning tendency of the force will be in clockwise direction.(b)I. When there is an α emission from 92238P^{238}_{92}	ext{P}92238​P , then the atomic number decreases by 2 and mass number decreases by 4. Hence, we get 90234Q^{234}_{90}	ext{Q}90234​Q .Now, when 90234Q^{234}_{90}	ext{Q}90234​Q undergoes β emission, the atomic number increases by 1 and the mass number is unchanged. Hence, we get 91234R^{234}_{91}	ext{R}91234​R.In the final step when 91234R^{234}_{91}	ext{R}91234​R undergoes β emission again the atomic number increases by 1 and the mass number is unchanged. Hence we get 92234S^{234}_{92}	ext{S}92234​S.So the complete nuclear change is as follows —92238P⟶α 90234Q⟶β 91234R⟶β 92234S^{238}_{92}	ext{P} \overset{\alpha} \longrightarrow \space ^{234}_{90}	ext{Q} \overset{\beta} \longrightarrow \space ^{234}_{91}	ext{R} \overset{\beta} \longrightarrow \space ^{234}_{92}	ext{S}92238​P⟶α​ 90234​Q⟶β​ 91234​R⟶β​ 92234​SII. When there is an α emission from ZAX^{A}_{Z}	ext{X}ZA​X , then the atomic number decreases by 2 and mass number decreases by 4. Hence, we get Z−2A−4X1^{A - 4}_{Z - 2}	ext{X}_{1}Z−2A−4​X1​After the γ emission, there is no change in atomic number and mass number. Hence, we get Z−2A−4X2^{A - 4}_{Z - 2}	ext{X}_{2}Z−2A−4​X2​In the final step when there are two β emissions, the atomic number increases by 2 and the mass number is unchanged and we get ZA−4X3^{A - 4}_{Z}	ext{X}_{3}ZA−4​X3​Hence, the nuclear change is as follows —ZAX⟶αZ−2A−4X1⟶γZ−2A−4X2⟶2β     ZA−4X3^{A}_{Z}	ext{X} \overset{\alpha} \longrightarrow ^{A - 4}_{Z - 2}	ext{X}_{1} \overset{\gamma} \longrightarrow ^{A - 4}_{Z - 2}	ext{X}_{2} \overset{2\beta} \longrightarrow _{\space \space \space \space \space Z}^{A - 4}	ext{X}_{3}ZA​X⟶α​Z−2A−4​X1​⟶γ​Z−2A−4​X2​⟶2β​     ZA−4​X3​

Question 4

Question 2(iv)(a)	What is meant by an ideal machine?(b)	Write a relationship between the mechanical advantage (MA) and velocity ratio (VR) of an ideal machine.

Accepted Answer

(a) An ideal machine is that where there is no loss of energy i.e. the work output is equal to the work input. An ideal machine is 100% efficient.(b) For an ideal machine (free from friction etc.), work output is equal to the work input, so the efficiency is equal to 1 ( or 100% ) and the mechanical advantage is numerically equal to the velocity ratio. So,For an ideal machine, M.A. = V.R.

Question 5

Question 2(v)Define work. State the conditions when the work done by a force acting on a body is zero.

Accepted Answer

The definition of work states that when force is applied on a body and the body moves then work is said to be done.The two conditions for the work done to be zero are,(i) When there is no displacement of the body on application of forcei.e. S = 0 and(ii) When displacement is normal to the direction of forcei.e. θ = 90° as cos 90° = 0

Question 6

Question 2(vi)The given figure shows a trace of a sound wave which is produced by a particular tuning fork.(a)	Draw a trace of the sound wave which has a higher pitch than that shown in figure.(b)	Draw a trace of the sound wave which is louder than that shown in figure.

Accepted Answer

Sound waves with higher pitch and which is louder are shown in the figure:

Question 7

Question 2(vii)Why should a fuse with defined rating not be replaced by one with a larger rating?

Accepted Answer

The fuse is rated for a certain maximum current and blows off when a current more than the rated value flows through it. If a fuse is replaced by one with larger ratings, the appliances may get damaged as the protecting fuse will not melt and the circuit will not be broken.

Question 8

Question 3(i)If the refractive index of air with respect to glass is expressed as gμa=sin isin r^g\mu_a = \dfrac{	ext{sin i}}{	ext{sin r}}gμa​=sin rsin i​ answer the following:(a)	Express the similar expression for aμg^a\mu_gaμg​ in terms of i and r.(b) If angle r = 90°, what is the corresponding angle (i) is called?(c) Write down the physical significance of angle (i) in part (b).

Accepted Answer

(a) The expression for aμg = sin rsin i\dfrac {	ext {sin r}}{	ext {sin i}}sin isin r​(b) When refractive angle r = 90° then, the corresponding angle of incidence i will be equal to critical angle.(c) The phenomenon of total internal reflection occurs when the angle of incidence exceeds the value of i obtained in part (b).

Question 9

Question 3(ii)How does increase in the temperature affect the specific resistance of a metal and semiconductor?

Accepted Answer

Specific resistance increases with increase in temperature for metals but it decreases with the increase in temperature for semiconductors.

Question 10

Question 3(iii)The magnetic field produced by a current carrying conductor is shown below. Is there a similar magnetic field produced around a thin beam of moving α-particles and neutrons? Justify your answer.

Accepted Answer

In case of movement of a charged particle, the magnetic field is created around the particle. The field is around the path on which the charged particle moves. So, a thin beam of α-particles (which are positively charged) is like a straight conductor carrying current in the direction of motion of α-particles. But, as neutrons carry no charge, so no magnetic field would be created around their path.

Question 11

Question 3(iv)What do you understand by the following statements?(a) The heat capacity of the body is 60 JK-1.(b) The specific heat capacity of lead is 130 Jkg-1K-1

Accepted Answer

(a) Heat energy required to raise the temperature of a body by 1 K is 60 J.(b) Heat energy required to raise the temperature of 1 Kg of lead by 1 K is 130 J.

Question 12

Question 3(v)(a) What is the origin of β -particles?(b)	Under what condition β -particles produce X rays?

Accepted Answer

(a) β -particles are given out from the nucleus of radioactive atom.(b) β -particles produce X-rays when they are stopped by the metals having high atomic number and high melting point.

Question 13

Question 4(i)An narrow beam of white light is incident on three glass objects as shown below. Comment on the nature of behaviour of the emergent beam in all three cases.

Accepted Answer

(a) The incident beam of light after refraction though glass slab emerges out parallel to the incident beam but laterally shifted. No dispersion takes place in this case.(b) When the beam of light passes through the first prism, it splits into its seven constituent colours, however, these seven colours on passing through the second inverted prisms recombine. The emergent ray is parallel to the incident beam but slightly shifted outwards.(c) When the incident beam passes through the prism, it splits into its constituent colours. These split further on emerging out of the prism and dispersion pattern is observed.

Question 14

Question 4(ii)A diver in water looks obliquely at an object AB in air.(i) Does the object appear taller, shorter or of the same size to the diver?(ii) Show the path of two rays AC & AD starting from the tip of the object as it travels towards the diver in water and hence obtain the image of the object.

Accepted Answer

(i) The object appears taller to the diver.(ii) Below diagram shows the path of two rays AC & AD:

Question 15

Question 4(iii)(a) With the help of a suitable diagram, represent the refraction of light rays fromI. rarer to denser mediumII. denser to rarer medium.(b) Does the depth of a tank of water appear to change or remains the same when viewed normally from the above?

Accepted Answer

(a)I. The below ray diagram shows the refraction of light from rarer (air) to denser (glass) medium:II. The below ray diagram shows the refraction of light from denser (glass) to rarer (air) medium:(b) When a tank is viewed normally from above, the apparent depth of the tank will remain the same as the incident light striking normally on the surface will pass undeviated. Hence, no phenomenon of refraction is observed and depth of pond appears same.

Question 16

Question 5(ii)(a) How many surfaces bound a prism?(b) Give the relationship between wavelength of light and amount by which it deviates when it is passed through a prism.(c) What will be the colour of the emergent light ray when a ray of white light is incident on a thin walled hollow glass prism?

Accepted Answer

(a) A prism is a transparent medium bounded by five plane surfaces i.e., 2 triangular faces and three rectangular lateral surfaces.(b) Wavelength (λ) ∝ 1deviation (δ)\dfrac{1}{	ext{deviation (δ)}}deviation (δ)1​(c) The emergent light will be white because the outer surface of the prism behave like a hollow glass plate.

Question 17

Question 5(iii)(a) Define convex lens and give its nature of refraction.(b) A 6 cm tall object is placed perpendicular to the principal axis of convex lens of focal length 25 cm. The distance of the object from the lens is 40 cm.By calculation determinethe position andthe size of the image formed.

Accepted Answer

(a) A lens which is thicker at the centre and thinner at its ends is called convex lens. It converges the light rays passing through and is called a converging lens.(b) Given,f = 25 cmu = -40 cmv = ?As we know, the lens formula is —1f=1v−1u\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}f1​=v1​−u1​Substituting the values in the formula, we get,125=1v−1−40125=1v+1401v=125−1401v=8−52001v=3200⇒v=2003=66.67 cm\dfrac{1}{25} = \dfrac{1}{v} - \dfrac{1}{-40} \[1em] \dfrac{1}{25} = \dfrac{1}{v} + \dfrac{1}{40} \[1em] \dfrac{1}{v} = \dfrac{1}{25} - \dfrac{1}{40}\[1em] \dfrac{1}{v} = \dfrac{8-5}{200} \[1em] \dfrac{1}{v} = \dfrac{3}{200} \[1em] \Rightarrow v = \dfrac{200}{3} = 66.67 \text{ cm}251​=v1​−−401​251​=v1​+401​v1​=251​−401​v1​=2008−5​v1​=2003​⇒v=3200​=66.67 cm∴ Image distance is 66.67 cmu = -40 cmv = 66.67 cmheight of object (ho) = 6 cmheight of image (hi) = ?From formula:m=vu=hihom = \dfrac{v}{u} = \dfrac{h_i}{h_o}m=uv​=ho​hi​​Substituting the values in the formula, we get,66.67−40=hi6hi=6×−6.6740hi=−10 cm\dfrac{66.67}{-40} = \dfrac{\text{h}_i}{6} \[0.5em] \text{h}_i = 6 \times -\dfrac{6.67}{40} \[0.5em] \text{h}_i = -10 \text{ cm} \[0.5em]−4066.67​=6hi​​hi​=6×−406.67​hi​=−10 cmHence, height of image = -10 cm

Question 18

Question 6(i)A pulley system with V.R. = 4 is used to lift a load of 175 kgf through a vertical height of 15 m. The effort required is 50 kgf in the downward direction. (g = 10 N kg-1).Calculate:(i) Distance moved by the effort.(ii) Work done by the effort.(iii) M.A. of the pulley system.(iv) Efficiency of the pulley system.

Accepted Answer

Given,V.R. = 4L = 175 kgfdL = 15 mE = 50 kgf(i) V.R. = dEdL\dfrac{d_E}{d_L}dL​dE​​Substituting the values we get,4=dE15⇒dE=15×4⇒dE=60 m4 = \dfrac{d_E}{15} \[0.5em] \Rightarrow d_E = 15 \times 4 \[0.5em] \Rightarrow d_E = 60 \text{ m}4=15dE​​⇒dE​=15×4⇒dE​=60 m(ii) Work done by the effort = E x dE     = 50 x 10 x 60     = 3 x 104(iii) M.A. = LE\dfrac{L}{E}EL​Substituting the values we get,M.A.=17550=3.5\text {M.A.} = \dfrac{175}{50} = 3.5M.A.=50175​=3.5(iv) Efficiency η = M.A.V.R.\dfrac{M.A.}{V.R.}V.R.M.A.​Substituting the values we get,η=3.54=0.875∴η=87.5η = \dfrac{3.5}{4} = 0.875 \[0.5em] \therefore η = 87.5 %η=43.5​=0.875∴η=87.5

Question 19

Question 6(ii)(a)	How a beam balance comes to static equilibrium?(b)	How a plane moves at a constant height? On which fact it is based?

Accepted Answer

(a) In beam balance when the beam is balanced in a horizontal position, it is in static equilibrium. The clockwise and anticlockwise moments balance each other hence the beam has no rotational motion and it is in static equilibrium.(b) A plane moves at a constant height when upward lift on it balances its weight acting downwards. It is due to dynamic equilibrium of forces.

Question 20

Question 6(iii)(a)	Calculate the work done whenI. a 20 kg weight is lifted 2 m verticallyII. a bike is moved on a rough road through 35 m against a frictional resistance of 20 N. (g = 9.8 ms-2	)(b) What is the work done by a force when the force is:I. normal to the displacement produced.II. in the same direction as the displacement produced?

Accepted Answer

(a) I. Given:Mass (m) = 20 kgDistance (d) = Height = 2 mForce of gravity on mass of 1 kg = 9.8 ms-2Work done (W) = ?Force (F) = mg = 20 x 9.8 = 196 NWork done = Force x distance = 196 x 2 = 392 JII. Given,Resisting force = 20 NDistance = 35 mWork done = 35 x 20 = 700 J(b) We know that, W = Fs cos θI. when force is normal to the displacement, θ = 90° hence,W = Fs cos 90° = 0 [∵ cos 90° = 0]II. when force is in the same direction as the displacement, θ = 0° hence,W = Fs cos 0° = Fs [∵ cos 0° = 1]

Solved Sample Paper 5

Section A

Section B