Update ICSE - X Physics Board Paper 2024.txt

ICSE - X Physics Board Paper 2024.txt

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-ICSE EXAMINATION PAPER - 2024
-PHYSICS
-Class-10th
-(Solved)
-
-Maximum Marks: 80
-Time allowed: Two hours
-Answers to this Paper must be written on the paper provided separately.
-You will not be allowed to write during first 15 minutes.
-This time is to be spent in reading the question paper.
-The time given at the head of this Paper is the time allowed for writing the answers.
-Section A is compulsory. Attempt any four questions from Section B.
-The intended marks for questions or parts of questions are given in brackets [ ].
-
-SECTION A - 40 MARKS
-
-Oswaal ICSE PHYSICS, Class-X
-
-ANSWERS
-SECTION A
-1. (i) Option (b) is correct.
-		
-Explanation: Bell fixed on a cycle rings when
-pressure by finger is applied. So, mechanical energy
-is converted into sound energy.
-(ii) Option (a) is correct
-		
-Explanation: Moment of a force = Force x
-perpendicular distance from the line of action of the
-force
-		 Or,
-1 = F × 0.2
-		
-∴
-F =5N
-(iii) Option (c) is correct.
-		
-Explanation: Displacement should be in the direction
-of force.
-(iv) Option (a) is correct.
-		
-Explanation: Number of nucleons i.e., mass no. does
-not change due to β emission Only atomic number
-will increase by 1.
-(v) Option (b) is correct.
-		
-Explanation: As wavelength decreases, scattering
-increases.
-		 Wavelength of ultraviolet is less than the wavelength
-of microwave.
-		 Hence, scattering of ultraviolet is more than that of
-microwave.
-		 Hence, the assertion is true but reason is false.
-(vi) Option (c) is correct.
-		
-Explanation: When stem of a vibrating tuning fork is
-pressed on a table, the table top starts vibrating due
-to the periodic force applied by the tuning fork. This
-is an example of forced vibration.
-(vii) Option (d) is correct.
-		
-Explanation: For class III lever force is applied in
-between the fulcrum and the load.
-		 In human forearm, elbow is the pivot, bicep muscle
-applies the force and load is placed on the palm.
-Hence, it is a class III lever.
-(viii) Option (b) is correct.
-		
-Explanation: Specific resistance is independent of
-the dimension of the conductor. It depends on the
-material of the conductor.
-(ix) Option (a) is correct.
-		
-Explanation: From mains, blue wire (NEUTRAL)
-should directly reach the appliance and the brown
-wire (LIVE) should reach the appliance through a
-switch.
-(x) Option (a) is correct.
-		
-Explanation: Potential difference between terminals
-of a cell, in open circuited condition, is termed as
-electromotive force. Potential difference between
-terminals of a cell, in closed circuited condition, is
-termed as terminal voltage.
-(xi) Option (b) is correct.
-		
-Explanation: During melting, ice requires latent
-heat (a form of energy) and temperature remains
-constant at its melting point.
-
-(xii) Option (a) is correct.
-		
-Explanation: Concave lens always produces virtual,
-erect image, smaller than the object. Hence, the
-magnification is less than 1.
-(xiii) Option (c) is correct.
-		
-Explanation: Rate of radioactive elements is
-determined by the properties of the nucleus of
-an atom. so, it does not change by environment
-(external factors).
-(xiv) Option (d) is correct.
-		
-Explanation: Slope the graph gives the amount of
-heat required for unit rise of temperature. This is
-termed as the heat capacity.
-(xv) Option (a) is correct.
-		
-Explanation: The light passing through glass slab
-will bend towards the normal and after coming out
-in the air, the light will converge out at a point away
-from the glass slab.
-
-2. (i) (a) C14 is a radioisotope because of the instability
-of the nucleus. Used for carbon dating.
-		 (b) Class II lever.
-(ii) (a) Centripetal
-		(b) zero
-(iii)		The weight of the beam will produce a clockwise
-moment and the applied force F will produce a
-clockwise moment.
-		 Equating the moments of the force,
-			
+ICSE CLASS 10 PHYSICS BOARD PAPER 2024 - COMPLETE ANSWER KEY
+
+SECTION A (40 MARKS)
+
+Question 1 (Multiple Choice - 15 marks)
+(i) (b) kinetic energy to sound energy
+(ii) (a) 5 N
+(iii) (c) 5 m
+(iv) (a) 128
+(v) (b) A is true but R is false.
+(vi) (c) forced vibrations
+(vii) (d) Human forearm
+(viii) (b) material
+(ix) (a) APPLIANCE Mains Blue wire
+(x) (a) terminal voltage
+(xi) (b) energy is absorbed and temperature remains constant.
+(xii) (a) m < 1
+(xiii) (c) Remain unchanged
+(xiv) (d) Heat capacity
+(xv) (a) Move away from the slab
+
+Question 2 (15 marks)
+(i)(a) C¹⁴ is a radioisotope because of the instability of the nucleus. Used for carbon dating.
+(i)(b) Class II lever.
+
+(ii)(a) Centripetal
+(ii)(b) zero
+
+(iii) Solution:
 120 × 0.2 = F × 0.8
-		
-∴
 F = 30 N
-(iv) (a) Mechanical advantage using a block and tackle
-system of 9 pulleys is 9 whereas the mechanical
-advantage of a single moveable pulley is 2.
-		(b) Increase of number of
-pulley increases
-mechanical advantage.
-WSumit
-600
-tSumit
-(v)		PSumit / PAmit =
-= 10 = 4:1
-300
-WAmit
-20
-tAmit
-(vi) (a) Resistance decreases
-		(b) Current (Intensity of light) increases as resistance
-decreases
-(vii) Applying the formula
-0.1
-			
-d =v×
-2
-		 For Oxygen,
-
-x = 340 ×
-
-0.1
-= 17 m
-2
-
-Solved Paper - 2024
-		 For Benzene,
-
-y = 200 ×
-
-0.1
-= 10 m
-2
-
-		 So,
-x : y = 17 : 10
-3. (i) (a) In between optical centre and focus
-		(b) Because concave lens produces reduced size
-image.
-(ii) No.
-P
-1540
-		 The geyser draws I =
-=
-= 7A.
-V
-220
-		 Hence, 5A rated fuse will blow off for the geyser.
-(iii) Speed of rotation may be increased by
-		 Increasing the number of turns of the armature.
-		 Increasing the applied battery e.m.f.
-(iv) Required heat Q = mL= 500 × 330 = 165000 J
-(v) 92U235 + 0n1 → 46Ba141 + 36Kr92 + 30n1
-
-SECTION B
-4. (i) Using lens formula,
-1 1
-1
-− =
-f
-v
-u
-			
-		 Or,
-
-1
-1
-1
-+
-=
-72 36
-f
-
-		Or,
-
-3
-1
-=
-72
-f
-
-		
-∴
-
-f = 24 cm
-1
-100
-		 Power of the lens =
-=
-= 4.17 D
-24
-f
-(ii) (a) Microwave
-		(b) X-rays
-		(c) Both have same speed in vacuum
-(iii) (a) Red colour is used as danger signal because it is
-scattered least due to its longest wavelength in
-visible range. Hence, it is visible from a very long
-distance.
-		(b) 1. Critical angle is 43°
-A
-43°
-90°
-
-Blue
-ray
-
-47°
-43°
-
-		(b) The principle of reversibility
-		(c) No.
-(ii) (a) The statement is false.
-			 Diamond is a denser medium compared to
-vacuum. Hence, in it light cannot cover longer
-distance than that in vacuum.
-10 x
-		(b) Speed of light in vacuum =
-t1
-Speed of light in the medium =
-
-			
-
-10 x
-t
-So, refractive index of the medium = 1
-x
-t2
-=
-
-			
-(iii) (a)
-
-			
-
-Hence, the Refractive index of water with respect
-8
-to glass =
-9
-
-10t2
-t1
-
-		(b) Angle of deviation depends on
-			 l Angle of incidence
-			 l Refractive index
-6. (i) (a) Centre of gravity: Centre of gravity is the
-point where the mass of the whole body appears
-to be concentrated.
-		(b) 1. The centre of gravity from the broad base is at
-h
-6
-a height
-=
-= 2 cm.
-3
-3
-			 2. Yes.
-(ii) (a) Marble A
-		(b) Along both the paths.
-		(c) Along both the paths.
-(iii) (a)
-
-B
-
-C
-2. From the above figure, Angle of prism = 43°
-3. If the blue ray is replaced by indigo or violet,
-there will be total internal reflection.
-			
-Critical angles for indigo and violets are
-less than that of blue. Hence, the angle of
-incidence 43° will be greater than the critical
-angle and total internal reflection will occur.
-5. (i) (a) Refractive index of glass with respect to water
-9
-=
-8
-
-x
-t2
-
-			
-
-133°
-
-			
-			
-			
-
-7
 
-		(b) 1.
+(iv)(a) Mechanical advantage using a block and tackle system of 9 pulleys is 9 whereas the mechanical advantage of a single moveable pulley is 2.
+(iv)(b) Increase of number of pulley increases mechanical advantage.
 
-Efficiency =
+(v) Solution:
+Pₛᵤₘᵢₜ = 600 J / 10 min = 60 J/min
+Pₐₘᵢₜ = 300 J / 20 min = 15 J/min
+Ratio = 60:15 = 4:1
 
-M.A.
-V.R.
-M.A.
-2
+(vi)(a) Resistance decreases
+(vi)(b) Current (Intensity of light) increases as resistance decreases
 
-			
+(vii) Solution:
+For Oxygen: x = 340 × (0.1/2) = 17 m
+For Benzene: y = 200 × (0.1/2) = 10 m
+So, x:y = 17:10
 
-Or,
+Question 3 (10 marks)
+(i)(a) In between optical centre and focus
+(i)(b) Because concave lens produces reduced size image.
 
-0.8 =
-
-			
-
-∴
-
-M.A. = 1.6
-
-8
-
-Oswaal ICSE PHYSICS, Class-X
-
-			
-
-2.
-
-Effort =
-
-Load
-M.A.
-
-			
-
-Or,
-
-Effort =
-
-48
-1.6
-
-			
-
-∴
-
-Effort = 30 kgf
-
-7. (i) (a) Ultrasonic wave
-			
-
-(b) Lata cannot clearly distinguish the sound
-reflected from cliff A. Since the distance is less
-than the minimum distance required to hear
-a distinct echo.
-
-			
-
-So, she will hear the sound reflected from
-cliff B as 1st echo.
-
-			
-
-So, she will hear the 1st echo after time t =
-=
-
-(ii) (a)
-
-2d
-v
+(ii) No.
+Justification: The geyser draws I = P/V = 1540/220 = 7A.
+Hence, 5A rated fuse will blow off for the geyser.
 
-2 × 160
-= 1 s.
-320
+(iii) Two factors:
+1. Increasing the number of turns of the armature
+2. Increasing the applied battery e.m.f.
 
-234
-230
-4
-234
-0
-90 X → 88 Y + 2 He → 91 Z+ −1 e
+(iv) Solution:
+Q = mL = 500 × 330 = 165,000 J
 
-		(b) U-235
-(iii) (a) Resonance
-		(b) Natural frequency of the air column matched
-with the frequency of the tuning fork.
-		(c) Length of air column should be decrease. Since
-frequency is inversely proportional to the air
-column.
-8. (i) (a) Non-ohmic
-		(b) The resistance is not constant. It increases as
-voltage and current increases.
-		 (c) Ohm’s law:
-			
+(v) Complete reaction:
+₉₂U²³⁵ + ₀n¹ → ₅₆Ba¹⁴¹ + ₃₆Kr⁹² + 3₀n¹
 
-The flow of current through the conductor is
-directly proportional to the potential difference
-established across the conductor provide its
-physical conditions (like, temperature, pressure,
-density, volume etc.) are constant.
+SECTION B (40 MARKS)
 
-(ii) (a) Step down transformer
-		(b) Wire of secondary coil is thicker because
-secondary current is higher than primary
-current.
-(iii) (a) 10 Ω and 6 Ω are in series. Hence, equivalent
-resistance R1 = 16 Ω.
-			 12 Ω and 4 Ω are in series. Hence, equivalent
-resistance R2 = 16 Ω.
-			 R1 and R2 are in parallel. Hence, the equivalent
-resistance = Req = 8 Ω.
-		(b) Current I =
-
-V
-4V
-=
-= 0.5A
-R eq
-8W
-
-		(c) Current in both the resistors are equal because
-both the arms have same resistance value.
-9. (i) Let the amount of ice = m gram
-		 Heat lost by water = 85 × 4.2 × (30 – 5)
-		 Heat lost by ice = m × 336 + m × 4.2 × (5 – 0)
-		 Applying principle of calorimetry,
-			 85 × 4.2 × (30 – 5) = m × 336 + m × 4.2 × (5 – 0)
-		 Or,
-8925 = 357 m
-		
-∴
-m = 25 g
-(ii) (a) When lake freezes, it released latent heat to the
-atmosphere which warms the atmosphere. So,
-the atmosphere becomes pleasantly warm.
-		(b) Since there is no change of temperature there is
-no change in kinetic energy.
-		(c) 1 g of ice (solid) at 1°C has extra heat in the form
-of its Latent Heat while 1 g of water (liquid) at
-1oC has no Latent heat.
-			 Hence, heat energy in 1 g of ice at 0°C has more
-heat energy than the 1 g water at 0°C.
-(iii) (a) Number of turns in armature may be increased.
-		(b) 1. Maxwell right hand thumb rule.
-			 2. Deflection being clockwise, the direction of
-current is from X to Y.
-			 3. As R increases, current decreases. So, strength
-of magnetic field decreases and hence, the
-density of magnetic lines of force decreases.
-
-
-
-
+Question 4 (10 marks)
+(i) Solution:
+1/v - 1/u = 1/f
+1/72 + 1/36 = 1/f
+3/72 = 1/f
+f = 24 cm
+Power = 100/24 = 4.17 D
+
+(ii)(a) Microwave
+(ii)(b) X-rays
+(ii)(c) Both have same speed in vacuum
+
+(iii)(a) Red colour is used as danger signal because it is scattered least due to its longest wavelength in visible range. Hence, it is visible from a very long distance.
+(iii)(b)1. Critical angle is 43°
+   2. Angle of prism = 43°
+   3. If the blue ray is replaced by indigo or violet, there will be total internal reflection.
+
+Question 5 (10 marks)
+(i)(a) 8/9
+(i)(b) The principle of reversibility
+(i)(c) No
+
+(ii)(a) False
+(ii)(b) Solution:
+Refractive index = (10x/t₁)/(x/t₂) = 10t₂/t₁
+
+(iii)(a) [Diagram showing completed ray path]
+(iii)(b) Two factors:
+1. Angle of incidence
+2. Refractive index
+
+Question 6 (10 marks)
+(i)(a) Centre of gravity is the point where the mass of the whole body appears to be concentrated.
+(i)(b)1. 2 cm from base (h/3 = 6/3)
+    2. Yes
+
+(ii)(a) Marble A
+(ii)(b) Along both the paths
+(ii)(c) Along both the paths
+
+(iii)(a) [Diagram showing tackle system with VR=2]
+(iii)(b)1. Solution:
+    Efficiency = M.A./V.R.
+    0.8 = M.A./2
+    M.A. = 1.6
+    2. Effort = Load/M.A. = 48/1.6 = 30 kgf
+
+Question 7 (10 marks)
+(i)(a) Ultrasonic wave
+(i)(b) Solution:
+t = 2d/v = (2×160)/320 = 1 s
+
+(ii)(a) ²³⁴₉₀Y → ²³⁴₉₁Z + ⁰₋₁e
+(ii)(b) U-235
+
+(iii)(a) Resonance
+(iii)(b) Natural frequency of the air column matched with the frequency of the tuning fork.
+(iii)(c) Length of air column should be decrease. Since frequency is inversely proportional to the air column.
+
+Question 8 (10 marks)
+(i)(a) Non-ohmic
+(i)(b) The resistance is not constant. It increases as voltage and current increases.
+(i)(c) The flow of current through the conductor is directly proportional to the potential difference established across the conductor provide its physical conditions are constant.
+
+(ii)(a) Step down transformer
+(ii)(b) Wire of secondary coil is thicker because secondary current is higher than primary current.
+
+(iii)(a) Solution:
+R₁ = 10 + 6 = 16Ω
+R₂ = 12 + 4 = 16Ω
+Rₑq = (16×16)/(16+16) = 8Ω
+(iii)(b) I = V/R = 4/8 = 0.5A
+(iii)(c) Current in both the resistors are equal because both the arms have same resistance value.
+
+Question 9 (10 marks)
+(i) Solution:
+85 × 4.2 × 25 = m × 336 + m × 4.2 × 5
+8925 = 357m
+m = 25g
+
+(ii)(a) When lake freezes, it released latent heat to the atmosphere which warms the atmosphere.
+(ii)(b) Since there is no change of temperature there is no change in kinetic energy.
+(ii)(c) 1 g of ice at 0°C has more heat energy than 1 g water at 0°C.
+
+(iii)(a) Number of turns in armature may be increased.
+(iii)(b)1. Maxwell right hand thumb rule.
+    2. Direction of current is from X to Y.
+    3. As R increases, current decreases. So, strength of magnetic field decreases.