Question 1

Question 1(ii)When the speed of a moving object is doubled, then its kinetic energy:remains the samedecreasesis doubledbecomes four times

Accepted Answer

becomes four timesReason — According to the formula :Kinetic energy = 12\dfrac{1}{2}21​mv2Hence, when speed (v) is doubled the kinetic energy increases by 4 times.

Question 2

Question 1(xiv)When a ray of light travels normal to the given surface, then the angle of refraction is:180°90°0°45°

Accepted Answer

0°Reason — The ray of light incident normally on a plane glass slab passes undeviated, ( i.e. such a ray suffers no bending at the surface). Thus, if angle of incidence ∠i = 0°, then the angle of refraction ∠r = 0°.

Question 3

Question 2(ii)(a) What is the position of centre of gravity of a triangular lamina?(b) When this triangular lamina is suspended freely from any one vertex, what is the moment of force produced by its own weight in its rest position?

Accepted Answer

(a) The centre of gravity for a triangular lamina is at the point of intersection of the medians.(b) When the triangular lamina is suspended freely from any one vertex, the moment of force produced by its own weight in its rest position is zero. This is because the center of gravity of the triangular lamina is at the point where its medians intersect and the triangle is in rotational equilibrium. Hence, we get zero moment of force.

Question 4

Question 2(iii)The diagram shows wheel O pivoted at point A. Three equal forces F1, F2 and F3 act at point B on the wheel.(a) Which force will produce maximum moment about A?(b) Give a reason for your answer in (a).

Accepted Answer

(a) Force F1 produces the maximum moment.(b) Moment of force = force x perpendicular distance from the point of application of the force to the pivot point.As the three forces are equal in magnitude, the magnitude of moment of force will depend on the perpendicular distance from the point of application of the force to the pivot point.The perpendicular distance from the line of action of F1 to the pivot point A is equal to the diameter of the wheel, while the perpendicular distance from line of action of F2 and F3 is less than the diameter, therefore, the moment produced by F1 will be greater than the moments produced by F2 and F3.

Question 5

Question 2(iv)(a) What should be the angle between the direction of force and the direction of displacement, for work to be negative?(b) Name the physical quantity obtained using the formula Uh\dfrac{	ext{U}}{	ext{h}}hU​, where U is the potential energy and h is the height.

Accepted Answer

(a) 180°(b) As gravitational potential energy [U] = mghHence, Uh\dfrac{U}{h}hU​ = mgTherefore, the physical quantity obtained is weight of the object.

Question 6

Question 2(v)Calculate the power spent by a crane while lifting a load of mass 2000 kg, at velocity of 1.5 ms-1. (g = 10 ms-2).

Accepted Answer

Given,mass = 2000 kgvelocity = 1.5 ms-1weight or downward force = mg = 2000 x 10 = 20,000 NFrom formula, P = F x vP = 20,000 x 1.5 = 30,000 W = 30 kWHence, power spent by the crane = 30 kW.

Question 7

Question 2(vi)A metal foot ruler is held at the edge of a table. It is pressed at its free end and then released. It vibrates.(a) Name the vibrations produced.(b) State one way to increase the frequency of these vibrations.

Accepted Answer

(a) The vibrations produced are known as damped vibrations.(b) If we place the ruler partially off the table then a shorter length of the ruler is free to vibrate, hence the frequency of vibration increases. The shorter the length of the ruler, the higher the frequency of its natural vibrations.

Question 8

Question 2(vii)'A geyser is rated 240 W – 220 V'. Explain the meaning of this statement.

Accepted Answer

It means that if the geyser is operated on a 220 V supply, the electric power consumed by it is 240 W i.e., 240 J of electric energy is consumed by the geyser in 1 s.

Question 9

Question 3(i)(a) Is it possible for a concave lens to form an image of size two times that of the object? Write Yes or No.(b) What will happen to the focal length of the lens if a part of the lens is covered with an opaque paper?

Accepted Answer

(a) No.(b) The focal length of the lens will not change.Reason — The focal length of a lens is a characteristic property of the lens itself and is determined by its shape and refractive index. Covering a portion of the lens with opaque paper will not alter the lens's inherent optical properties.

Question 10

Question 3(ii)(a) Which electrical component protects the electric circuit in case of excess current and which can also be used as a switch?(b) Name the wire to which this electrical component is connected in an electric circuit.

Accepted Answer

(a) Miniature circuit breaker [MCB] can be used to protect the electric circuit in case of excess current and can also be used as a switch.(b) The MCB is always connected in series with the live wire in the household circuit.

Question 11

Question 3(iii)A copper conductor is placed over two stretched copper wires whose ends are connected to a D.C. supply as shown in the diagram.(a) What should be the magnetic poles at the points A and B lying on either side of the conductor to experience the force in the upward direction?(b) Name the law used to find these polarities.

Accepted Answer

(a) The magnetic poles at the points A and B should be of opposite polarity for the conductor to experience the force in the upward direction. So, if point A is North pole then point B should be South pole and vice versa.(b) Fleming's left hand rule can be used to find these polarities.

Question 12

Question 3(iv)Thermal capacities of substances A and B are same. If mass of A is more than mass of B then:(a) Which substance will have more specific heat capacity?(b) Which substance will show greater rise in temperature if the same amount of heat is supplied to both?

Accepted Answer

(a) B will have more specific heat capacity.Reason —Specific heat capacity (c)=Heat Capacity of body (C’)mass of body (m)∴Specific heat capacity (c)∝1mass\text{Specific heat capacity (c)} = \dfrac{\text{Heat Capacity of body (C')}}{\text{mass of body (m)}} \[1em] \therefore \text{Specific heat capacity (c)} ∝ \dfrac{1}{\text{mass}}Specific heat capacity (c)=mass of body (m)Heat Capacity of body (C’)​∴Specific heat capacity (c)∝mass1​∴ Specific heat capacity of B will be more as its mass is less.(b) Both substances will show equal rise in temperature as their thermal capacities are the same.

Question 13

Question 3(v)How is the radioactivity of a radio isotope affected if it undergoes a chemical change? Give a reason for your answer.

Accepted Answer

The radioactivity of a radioactive isotope is not affected by chemical changes. This is because radioactivity is a nuclear phenomenon specifically related to the stability of the isotope's nucleus, while chemical changes involve the rearrangement of electrons and the formation or breaking of chemical bonds.

Question 14

Question 4(i)The diagram below shows the ray OP travelling through an equilateral prism of a certain material.(a) Calculate the value of i2, if the angle of deviation is 43°.(b) What is the ray QS called?

Accepted Answer

(a) Given,angle of prism (A) = 60° for an equilateral prism.angle of incidence (i1) = 65°Angle of deviation (δ) = 43°𝑒 = i2 = ?From formula,Angle of deviation (δ) = 𝑖 + 𝑒 − 𝐴Substituting the values we get:43 = 65 + i2 - 60°i2 = 43 - 5 = 38°Hence, i2 = 38°(b) Emergent ray

Question 15

Question 4(iii)The diagram below shows two parallel rays A (Orange) & B (Blue) incident from air, on air-glass boundary.(a) Copy and complete the path of the rays A and B.(b) How do the speeds of these rays differ in glass?(c) Are the two refracted rays in glass parallel? Give a reason.

Accepted Answer

(a) Ray B (Blue) will bend more than ray A (Orange) as shown in the diagram below:(b) In glass, speed of orange (A) light is more than that of blue light (B).(c) The two refracted rays are not parallel in glass as orange light with more speed bends less while blue light with slow speed bends more.

Question 16

Question 5(i)A convex lens of focal length 10 cm is placed at a distance of 60 cm from a screen. How far from the lens should an object be placed so as to obtain a real image on the screen?

Accepted Answer

Given, focal length = 10 cmImage distance (v) = 60 cmObject distance (u) = ?From lens formula —1v−1u=1f\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \[1em]v1​−u1​=f1​Substituting the values in the formula we get,160−1u=110⇒160−110=1u⇒1−660=1u⇒−560=1u⇒1u=−112⇒u=−12\dfrac{1}{60} - \dfrac{1}{\text{u}} = \dfrac{1}{10} \[1em] \Rightarrow \dfrac{1}{60} - \dfrac{1}{10} = \dfrac{1}{\text{u}} \[1em] \Rightarrow \dfrac{1 - 6}{60} = \dfrac{1}{\text{u}} \[1em] \Rightarrow \dfrac{-5}{60} = \dfrac{1}{\text{u}} \[1em] \Rightarrow \dfrac{1}{\text{u}} = \dfrac{-1}{12} \[1em] \Rightarrow u = -12601​−u1​=101​⇒601​−101​=u1​⇒601−6​=u1​⇒60−5​=u1​⇒u1​=12−1​⇒u=−12∴ The object should be placed at a distance of 12 cm in front of the lens.

Question 17

Question 5(ii)(a) A coin kept inside water [µ=43\dfrac{4}{3}34​] when viewed from air in a vertical direction appears to be raised by 3.0 mm. Find the depth of the coin in water.(b) How is the critical angle related to the refractive index of a medium?

Accepted Answer

(a) Given, µ=43\dfrac{4}{3}34​Shift = 3.0 mmWe know,refractive index = real depthapparent depth\dfrac{\text{real depth}}{\text{apparent depth}}apparent depthreal depth​43\dfrac{4}{3}34​ = xapparent depth\dfrac{\text{x}}{\text{apparent depth}}apparent depthx​apparent depth = 3x4\dfrac{\text{3x}}{4}43x​Shift = real depth - apparent depth3=x−3x4⇒3=4x - 3x4⇒3=x4⇒x=4×3=123 = \text{x} - \dfrac{\text{3x}}{4} \[1em] \Rightarrow 3 = \dfrac{\text{4x - 3x}}{4} \[1em] \Rightarrow 3 = \dfrac{\text{x}}{4} \[1em] \Rightarrow \text{x} = 4 \times 3 = 123=x−43x​⇒3=44x - 3x​⇒3=4x​⇒x=4×3=12∴ Depth of coin in water = 12 mm(b) Critical angle [C] = sin-1 1μ\dfrac{1}{μ}μ1​

Question 18

Question 5(iii)(a) Infrared radiations are used in warfare. Explain with reason, why.(b) A ray of light is incident at 45° on an equilateral prism in the diagram below.Name the phenomenon exhibited by the ray of light when it enters and emerges out of the prism.State the cause of the above phenomenon mentioned by you.

Accepted Answer

(a) Infrared radiations are used for signals during war because they are not visible and they are not absorbed much in the medium.(b)Dispersion.The cause of dispersion of white light is the change in speed of light with wavelength.When white light passes through a glass prism, its constituent colours (red, orange, yellow, green, blue, indigo, violet) travel with different speeds in the prism because each colour has different wavelength. This causes the dispersion of light.

Question 19

Question 6(i)A block and tackle system of pulleys has velocity ratio 4.(a) Draw a labelled diagram of the system indicating clearly, the direction of the load and the effort.(b) What is the value of the mechanical advantage of the given pulley system if it is an ideal pulley system?

Accepted Answer

(a) Labelled diagram of the block and tackle system of pulleys with velocity ratio 4 is shown below:(b) The load is being supported by five segments of the string. Therefore, L = 5T and E = T∴ M.A. = loadeffort\dfrac{	ext{load}}{	ext{effort}}effortload​ = 5TT\dfrac{5	ext{T}}{	ext{T}}T5T​ = 5

Question 20

Question 6(ii)A metre scale of weight 50 gf can be balanced at 40 cm mark without any weight suspended on it.(a) If this ruler is cut at its centre then state which part [0 to 50 cm or 50 to 100 cm] of the ruler will weigh more than 25 gf.(b) What minimum weight placed on this metre ruler can balance this ruler when it is pivoted at its centre?

Accepted Answer

(a) Given, wt of scale = 50 gfLet wt. from [0 to 40] cm be W1 and [40 to 100] cm be W2W1 + W2 = 50⇒ W1 = 50 - W2 .....(1)In equilibrium,Anti-clockwise moment = Clockwise moment(40 - 0) × W1 = (100 - 40)W240W1 = 60W240 [50 - W2] = 60W2 [.....From (1)]40 x 50 - 40W2 = 60W22000 = 60W2 + 40W22000 = 100W2W2 = 2000100\dfrac{2000}{100}1002000​W2 = 20 gfNow, W1 = 50 - 20 = 30 gf∴ [0 to 40 cm] part of ruler weighs 30 gf.Hence, the part [ 0 to 50 cm] of the ruler will weigh more 25 gf.(b) Given the metre scale balances at 40 cm mark, so its weight of 50 gf acts at 40 cm mark.Its weight of 50 gf produces anticlockwise moment about the centre.Let the minimum weight needed be m.In equilibrium,Anti-clockwise moment = Clockwise moment50 x (50 - 40) = m x (100 - 50)⇒ 50 x 10 = m x 50⇒ m = 50×1050\dfrac{50 	imes 10}{50}5050×10​⇒ m = 10 gfHence, minimum weight needed to balance the metre ruler when it is pivoted at its centre = 10 gf.

Solved 2023 Question Paper ICSE Class 10 Physics

Section A 40 Marks

Section B 40 Marks