ICSE Class 10 Physics — 2027 Specimen

Light object close to the pivot; heavy object at the far end on the opposite side.

Reason — Moment = force × perpendicular distance from the pivot. This arrangement maximises the heavy object’s moment and minimises the opposing moment of the light object.

light to chemical

Reason — Photosynthesis converts sunlight into chemical energy stored in food.

Height of A is one-sixth the height of B.

Reason — The gravitational potential energy of a body of mass m raised to a height h is U = mgh.

Given,

m_A = 6m_B and U_A = U_B

So,

$$\begin{aligned}\text m_\text A \text g \text h_\text A = \text m_\text B \text g \text h_\text B \\[1em] \Rightarrow 6\text m_\text B \text g \text h_\text A = \text m_\text B \text g \text h_\text B \\[1em] \Rightarrow \text h_\text A = \frac{\text h_\text B}{6}\end{aligned}$$

Hence, the height of A is one-sixth the height of B.

4

Reason — With the effort directed upwards, two pulleys are in the movable block and one is fixed. Four string segments support the load, so V.R. = 4.

blue

Reason — The refractive index of a given transparent medium decreases with the increase in the wavelength of light. Among red, yellow, green and blue, blue light has the shortest wavelength, so glass has the highest refractive index for blue light.

Since a medium of higher refractive index bends the light more, blue light is refracted (i.e., deviated) the most and red light the least for the same angle of incidence.

Hence, the angle of deviation is greatest for blue light.

49°

Reason — Ray X grazes the liquid surface, so its incidence angle is the critical angle: 90° − 43° = 47°. Ray Y undergoes total internal reflection, requiring an incidence angle greater than 47°; hence, 49°.

inverted and upright prisms respectively

Reason — The upper and lower parts of a concave lens act as prisms with their bases upwards and downwards respectively. Each bends light towards its base, causing the rays to diverge.

Both (A) and (R) are true, and (R) is the correct explanation of (A).

Explanation

(A) is true: blue light has a shorter wavelength and scatters more than red light. (R) is true: scattering intensity varies inversely as the fourth power of wavelength. Therefore, (R) explains (A).

X will be grave, and Y will be shriller.

Reason — The shrillness of a sound wave is directly proportional to its frequency, meaning that higher frequencies result in shriller sounds and lower frequencies result in grave sound.

Mass number = 222, atomic number = 86

Reason — For the given nucleus,

Atomic number Z = number of protons = 90

Mass number A = number of protons + number of neutrons = 90 + 140 = 230

An alpha particle is a helium nucleus ⁴He₂, so each alpha emission decreases the mass number by 4 and the atomic number by 2.

For two successive alpha decays,

New mass number = 230 − (2 × 4) = 230 − 8 = 222

New atomic number = 90 − (2 × 2) = 90 − 4 = 86

Hence, the final nucleus has mass number 222 and atomic number 86.

green

Reason — In the old colour convention, the earth wire is green.

latent heat of fusion

Reason — The heat energy absorbed by a substance in changing its phase from solid to liquid at its melting point, which is not externally manifested by any rise in temperature, is called the latent heat of fusion.

Reason — As the magnet moves towards the coil, the magnetic flux linked with the coil PQ changes, so an e.m.f. is induced in it and a momentary current flows through the galvanometer. This is electromagnetic induction.

By Lenz's law, the direction of the induced current is such that it opposes the cause producing it. Here the south pole of the magnet approaches the end Q, so the end Q of the coil must itself become a south pole in order to repel the approaching south pole. Consequently the other end P of the coil becomes a north pole, and the induced current flows from P to Q through the galvanometer.

Since the end P near the needle now behaves as a north pole, it repels the north pole of the magnetic needle N₁, so the needle is deflected away from the coil.

remains the same

Reason — Specific heat capacity is a property of the material and does not depend on the shape, length or size of the object. When the metallic wire is stretched so that its length becomes twice the original length, only its dimensions change, while the material remains the same.

Therefore, the specific heat capacity remains unchanged.

S1 and S2 only

Reason — The bulb B1 lies in the main circuit, so it can glow only when the circuit is complete. This requires the switch S1 (which lies in the main line) to be ON.

The bulb B2 together with the switch S2 forms one branch, and the switch S3 forms a second branch connected in parallel with it. If S3 is put ON, it provides a path of practically zero resistance across the branch containing B2, so B2 gets short circuited and no current passes through it.

Hence, both the bulbs glow only when S1 and S2 are ON and S3 is OFF.

(a) Friction acts radial to a car undergoing circular motion.

(b) A glass prism is immersed in water. The angle of minimum deviation will decrease.

(c) The CGS unit of specific heat capacity is erg g^-1 K^-1.

(d) The minimum distance for distinct echo formation is greater in water as compared to air.

(e) Two copper wires can have different resistivity if they have different temperatures.

(f) Nuclear fission is possible at ordinary temperature and pressure.

Movement of the body part (Column A)	Class of lever (Column B)
(a) Nodding head	• Class I
(b) Lifting body weight on your toes	• Class II

Reason —

(a) In class I lever, the fulcrum is in between the effort and the load so in the same manner in nodding head, the spine acts as the fulcrum, load is at its front part while effort is at its rear part as shown in below figure.

(b) In class II lever, the fulcrum and the effort are at the two ends of the lever and the load is somewhere in between the effort and the fulcrum so like class II, the weight of the body is raised on toes where the fulcrum is at the toes at one end, the load (i.e., weight of the body) is in the middle and effort by muscles is at the other end as shown in below figure.

(a) Figures b and d.

(b) Figures a and c.

Reason —

(a) Loudness depends on the amplitude and pitch depends on the frequency (i.e., on the wavelength for the same speed), while quality depends on the wave form. Waves b and d have the same amplitude and the same wavelength (50 m), but their wave forms are different. So they represent sounds of the same loudness and pitch but of different quality.

(b) Waves a and c have the same wavelength (125 m), so they have the same pitch, but their amplitudes are different, so they differ in loudness.

(a) The two materials bend the incoming rays by different amounts and produce two focal points. Hence, two images are formed.

(b) Concave lens because irrespective of the position of the object, the image formed by a thin concave (or divergent) lens is always virtual, upright, diminished, and it is situated on the side of the object between the focus and the lens.

Given,

• Power rating of air conditioner (P) = 2000 W
• Initial meter reading = 856 kWh
• Final meter reading = 1567 kWh

Now,

Electrical energy used (E) = 1567 − 856 = 711 kWh

$\text{Time} = \frac{\text{Energy}}{\text{Power}} = \frac{711 \text{ kWh}}{2 \text{ kW}} = 355.5 \text{ h}$

Hence, the time interval is 355.5 h.

(a) X

(b) Z

Reason —

(a) If a piece of clay is attached at point A of the pentagon ABCDE then new centre of gravity will be shifted towards A; hence, the most probable position is X.

(b) If the pentagon is cut along the line PQ then only DCBQPE remains which has equal mass distribution on left and right portions about the point O but has greater mass distributed on the upper portion than the lower portion about the point O. So, new centre of gravity will shift towards upper portion i.e. at a point above O; hence, the most probable position of the new centre of gravity is Z.

(a) True.

(b) The coil on a soft-iron core acts as an electromagnet with either supply.

• With P (AC): an alternating magnetic field repeatedly magnetises the soft iron; a ferromagnetic pin is attracted in both half-cycles (polarity doesn't matter).
• With Q (DC): a steady field magnetises the core continuously, so the pin is attracted.

(a) The heat energy released by a block is Q = mcΔt, where c is the specific heat capacity of the material and Δt = 150 − 50 = 100°C is the same for all the blocks.

Since c and Δt are constant, Q ∝ m. So the graph of heat released against mass is a straight line passing through the origin, as shown below.

(b) The slope of the above straight line is

$\text {slope} = \frac{\text Q}{\text m} = \text c\Delta \text t$

So the specific heat capacity is

$\text c = \frac{\text {slope of the graph}}{\Delta \text t} = \frac{\text {slope of the graph}}{100}$

(a) The nuclear reaction is

⁶Li₃ + ¹n₀ ⟶ ⁴He₂ + ³H₁

Balancing the mass numbers, 6 + 1 = 4 + 3, and balancing the atomic numbers, 3 + 0 = 2 + 1.

Hence, the hydrogen isotope formed is ³H₁ (tritium).

(b) The mass of the reactants in a fission reaction is greater than the mass of the products.

Reason — In a fission reaction a small part of the mass of the reactants disappears. This loss of mass (mass defect) is converted into a large amount of energy in accordance with E = mc², so the total mass of the products is less than the total mass of the reactants.

(a) Total internal reflection.

(b) The two necessary conditions for total internal reflection are :

1. The light must travel from an optically denser medium to an optically rarer medium.
2. The angle of incidence in the denser medium must be greater than the critical angle for that pair of media.

(a) When the light ray travels from Medium 1 to Medium 2, it bends towards the normal, indicating that Medium 2 is optically denser than Medium 1. When it travels from Medium 2 to Medium 3, it bends away from the normal, indicating Medium 2 is also denser than Medium 3. Because light travels slowest in the optically densest medium, its speed is minimum in Medium 2.

(b) The frequency of light does not change on refraction, and speed = frequency × wavelength. Since the speed of light decreases on entering medium 2, the wavelength also decreases.

Hence, the wavelength decreases.

(c) The emergent ray RS is parallel to the incident ray PQ only when the ray suffers equal and opposite bending at the two surfaces, i.e., when the first and the third media are optically identical.

Hence, μ₁ = μ₃.

(a) a > b

Reason — In each graph the lowest point of the curve gives the angle of minimum deviation, so a is the angle of minimum deviation for blue light and b is that for orange light.

The refractive index of the material of a prism decreases with the increase in the wavelength of light. As the wavelength of blue light is smaller than that of orange light, the prism has a greater refractive index for blue light and therefore deviates it more.

Hence, a > b.

(b) The angle of deviation produced by a prism increases with the increase in the angle of the prism.

Hence, on increasing the angle of the prism, the values of both a and b will increase.

(a) A clear image is obtained on a screen, so the image is real and magnified. Only a convex lens can form a real and magnified image.

Hence, Steve used a convex lens.

(b) The candle and the screen lie on opposite sides of the lens, so the sum of the two distances is equal to the separation between them.

u + v = 90 cm

(c) The image is twice the size of the object, so its image distance is twice its object distance.

Hence, v : u = 2 : 1

(d) From part (c), v = 2u

Substituting in u + v = 90, we get

$$\begin{aligned}\text u + 2\text u = 90 \\[1em] \Rightarrow 3\text u = 90 \\[1em] \Rightarrow \text u = 30 \text { cm}\end{aligned}$$

Then,

v = 90 − 30 = 60 cm

Hence, the image distance is 60 cm.

The ray diagram is shown below :

A ray from the top of the object travelling parallel to the principal axis, after refraction, appears to come from the focus F₁ on the same side of the lens, while a ray directed towards the optical centre O passes undeviated. These two refracted rays appear to meet at A'.

The image A'B' so formed is virtual, erect and diminished, and it lies between the optical centre and the focus F₁, on the same side of the lens as the object.

(a) Substance condenses at 150 °C.

(b) 150 °C to 60 °C.

(c) True.

Reason — The amount of heat released when a substance is cooled depends on :

Q = mcΔT

Where:

• Q = heat released
• m = mass of the substance
• c = specific heat capacity
• ΔT = change in temperature

Now, for the same mass and same temperature drop (10°C), the specific heat capacity (c) of a liquid is usually greater than that of its solid form.

So,

Q_liquid > Q_solid

Hence, more heat is released when cooling a substance by 10°C in its liquid state than in its solid state, due to higher specific heat capacity.

Given,

• Mass of copper (m_C) = 0.12 kg = 0.12 x 1000 g = 120 g
• Mass of water (m_W) = 0.84 kg = 0.84 x 1000 = 840 g
• Initial Temperature of water (T₁) = 15 °C
• Final Temperature of water = Final Temperature of copper = (T₂) = 35 °C
• Specific heat capacity of copper (c_C) = 0.4 J g^-1 °C^-1
• Specific heat capacity of water (c_W) = 4.2 J g^-1 °C^-1

Let,

Temperature of the flame = Initial Temperature of copper = T

Now,

Heat lost by copper = m_C x c_C x (T - T₂)

= 120 x 0.4 x (T - 35)

= 48 x (T - 35)

And

Heat gained by water = m_W x c_W x (T₂ - T₁)

= 840 x 4.2 x (35 - 15)

= 840 x 4.2 x 20

= 70560 J

As,

Heat lost by copper = Heat gained by water

⟹ 48 x (T - 35) = 70560

⟹ T - 35 = $\frac{70560}{48}$ = 1470

⟹ T = 1470 + 35 = 1505 °C

Hence, temperature of the flame is 1505 °C.

(a) Given,

• Weight at the end A (W_A) = 100 gf
• Weight at the end B (W_B) = 100 gf
• Length of the rod (L) = 80 cm
• Distance of A from the fulcrum (l_A) = 45 cm
• Distance of B from the fulcrum (l_B) = 80 cm - 45 cm = 35 cm

Let,

Weight of the rod AC is (W_AC) and distance of centre of gravity of the rod AC from the fulcrum is (l_AC).

And

Weight of the rod BC is (W_BC) and distance of centre of gravity of the rod BC from the fulcrum is (l_BC).

Here,

(l_AC) = 45 - 20 = 25 cm,

(l_BC) = 60 - 45 = 15 cm,

Now,

Anticlockwise moment (due to weight of 100 gf at end A and weight of rod AC) = l_A x W_A + l_AC x W_AC = 45 x 100 + 25 x W_AC

Clockwise moment (due to weight of 100 gf at end B and weight of rod BC) = l_B x W_B + l_BC x W_BC = 35 x 100 + 15 x W_BC

As system is balanced then for equilibrium,

Anticlockwise moment of force about the fulcrum = Clockwise moment of force about the fulcrum

45 x 100 + 25 x W_AC = 35 x 100 + 15 x W_BC

⇒ 4500 + 25W_AC = 3500 + 15W_BC

⇒ 15W_BC - 25W_AC = 4500 - 3500 = 1000

⇒ 3W_BC - 5W_AC = 200

⇒ 3W_BC - 5W_AC > 0

⇒ 3W_BC > 5W_AC

$\frac {{\text W}<em>\text {BC}}{\text W</em>\text {AC}} \gt \frac{5}{3} \gt 1$

⇒ W_BC > W_AC

Hence, weight of AC < weight of BC.

(b) Even though the weights present are the same at both ends and the torque arm of B is less than the torque arm of A. This means the moment of the weight of the rod acts from side B and the C.G. lies beyond 45. Thus, more weight is concentrated between C to B.

(a) Zero

(b) Negative

(c) Positive

Reason —

(a) Zero because when a person walks on a levelled road, no work is done against the force of gravity as the displacement of the person is normal to the direction of force of gravity which is vertically downwards.

(b) Negative because when a person climbs a ladder (upward), the displacement covered by the person is opposite to the direction of force of gravity (downwards), so the work done on the person by the force of gravity is negative.

(c) Positive because when a car in neutral gear is coming down the slope, the direction of displacement covered by it has some angle θ with the direction of force of gravity (downwards) such that 0° ≤ θ < 90°, so the work done on the car by the force of gravity is positive.

Given,

• Mass of the empty trolley (m₁) = 10 kg
• Speed of the empty trolley (v₁) = 8 m s^-1
• Mass of the loaded trolley (m₂) = 10 + 10 = 20 kg
• Speed of the loaded trolley (v₂) = 4 m s^-1

(a) Kinetic energy of the empty trolley

$\text K_1 = \frac{1}{2}\text m_1 {\text v_1}^2 = \frac{1}{2} \times 10 \times 8 \times 8 = 320 \text { J}$

Hence, the kinetic energy of the empty trolley is 320 J.

(b) Kinetic energy of the loaded trolley

$\text K_2 = \frac{1}{2}\text m_2 {\text v_2}^2 = \frac{1}{2} \times 20 \times 4 \times 4 = 160 \text { J}$

Hence, the kinetic energy of the loaded trolley is 160 J.

(c) The work done by the braking force in stopping a trolley is equal to its kinetic energy. If the stopping distance is s and the braking force is F, then

F × s = kinetic energy

So,

$\frac{\text F_1}{\text F_2} = \frac{\text K_1 / \text s}{\text K_2 / \text s} = \frac{\text K_1}{\text K_2} = \frac{320}{160} = \frac{2}{1}$

Hence, F₁ : F₂ = 2 : 1, i.e., the braking force needed for the empty trolley is twice that needed for the loaded trolley.

(d) The work-energy theorem, i.e., the work done by the braking force is equal to the change in the kinetic energy of the trolley.

(a) For a lever,

$\text {M.A.} = \frac{\text {effort arm}}{\text {load arm}}$

So the mechanical advantage is greatest when the load arm is as small as possible and the effort arm as large as possible. The load is at P, so the fulcrum must be at the point nearest to P and the effort must be applied at the farthest point.

Hence, the fulcrum should be at Q and the effort should be applied at V.

(b) Load arm = PQ = 20 cm

Effort arm = QV = 120 − 20 = 100 cm

$\text {M.A.} = \frac{\text {effort arm}}{\text {load arm}} = \frac{100}{20} = 5$

Hence, the maximum mechanical advantage is 5.

(c) In this arrangement the fulcrum Q lies in between the load at P and the effort at V.

Hence, it is a class I (first class) lever.

Given,

• Distance of the boy from the right wall = 68 m
• Distance of the boy from the left wall = X
• Time interval between the two echoes = 0.25 s
• Speed of sound in air (v) = 340 m s^-1

Since the boy is nearer to the left wall, the echo from the left wall is heard first and that from the right wall is heard later.

Time taken by the echo from the right wall,

$\text t_2 = \frac{2 \times 68}{340} = \frac{136}{340} = 0.4 \text { s}$

Time taken by the echo from the left wall,

t₁ = t₂ − 0.25 = 0.4 − 0.25 = 0.15 s

Now,

$$\begin{aligned}\text t_1 = \frac{2\text X}{\text v} \\[1em] \Rightarrow 0.15 = \frac{2\text X}{340} \\[1em] \Rightarrow 2\text X = 0.15 \times 340 \\[1em] \Rightarrow 2\text X =51 \\[1em] \Rightarrow \text X = 25.5 \text { m}\end{aligned}$$

Hence, the value of X is 25.5 m.

(a) The phenomenon is resonance.

The vibrating tuning fork applies an external periodic force of frequency 300 Hz on both the wires, so both of them execute forced vibrations. The natural frequency of wire B happens to be exactly equal to 300 Hz, so wire B vibrates with a very large amplitude and sends out a large amount of energy into the surrounding air. Hence a loud sound is heard from wire B only.

(b) f < 300 Hz

Reason — For wires of the same material, same length and under the same tension, a thicker wire has a greater mass per unit length and therefore a lower natural frequency of vibration. Since wire A is thicker than wire B, the natural frequency of A is less than that of B, i.e., less than 300 Hz.

A loud sound will be heard from wire A only when the frequency of the tuning fork is equal to the natural frequency of A, so that resonance occurs.

Hence, f is less than 300 Hz.

Given,

• Current through the 6Ω resistor = ammeter reading = 1 A

(a) The 6Ω and 3Ω resistors are joined in parallel, so the potential difference across both of them is the same.

$$\begin{aligned}1 \times 6 = \text I_1 \times 3 \\[1em] \Rightarrow \text I_1 = \frac{6}{3} = 2 \text { A}\end{aligned}$$

Hence, the current through the 3Ω resistor is 2 A.

(b) In a parallel combination the main current is the sum of the currents in the branches.

I = 1 + I₁ = 1 + 2 = 3 A

Hence, the total current in the circuit is 3 A.

(c) False.

Reason — When the switch S is opened, the 3Ω branch is broken and only the 6Ω resistor is left in the circuit. The resistance of the parallel combination was

$\text R_\text p = \frac{6 \times 3}{6 + 3} = 2\ \Omega$

which now rises to 6 Ω. For the same supply voltage, an increase in resistance decreases the current.

Hence, the ammeter reading will become less than 1 A, and not (1 + I₁) = 3 A.

Given,

• Power rating of the appliance (P) = 2 kW = 2000 W
• Voltage rating (V) = 220 V

(a) The safest circuit is S.

(b) The current drawn by the appliance is

$\text I = \frac{\text P}{\text V} = \frac{2000}{220} = 9.09 \text { A}$

1. Earthing — The appliance has a metal covering. In circuit S the metal body is connected to the earth, so if the live wire accidentally touches the body, a large fault current flows through the earth wire, helping the fuse disconnect the supply and protecting a person touching the appliance. Circuit R has no earth connection, so it is unsafe.

2. Correct fuse rating — The current rating of the fuse must be slightly more than the current drawn by the appliance. In circuit S the fuse is rated 10 A, which is just above 9.09 A, so it carries the normal current safely and blows only when the current becomes excessive. In circuit R the fuse is rated 8 A, which is less than 9.09 A, so it would blow even during the normal working of the appliance.

Hence, circuit S satisfies both the requirements of a safe circuit.

Given,

• Length of nichrome wire X (l_X) = l
• Cross-sectional area of nichrome wire X (A_X) = A
• Potential difference across nichrome wire X (V_X) = 10 V
• Length of nichrome wire Y (l_Y) = 2l
• Cross-sectional area of nichrome wire Y (A_Y) = A/2
• Potential difference across nichrome wire Y (V_Y) = 20 V
• Resistivity of nichrome = ρ

(a) Let 'R_X' and 'R_Y' be the resistance of nichrome wire X and Y respectively.

Now,

$\text R_\text X = \text ρ \frac {\text l}{\text A}$

And

$\text R_\text Y = \text ρ \frac {2\text l}{\frac {\text A}{2}} =\text ρ \frac {4\text l}{\text A} = 4\text R_\text X$

As,

$\frac{\text R_\text Y}{\text R_\text X} = 4$

So, R_X : R_Y = 1 : 4

(b) $\text P_\text X = \frac{{\text V_\text X}^2}{\text R_\text X} = \frac{10^2}{\text R_\text X} = \frac{100}{\text R_\text X}$

And

$\text P_\text Y = \frac{{\text V_\text Y}^2}{\text R_\text Y} = \frac{20^2}{4\text R_\text X} = \frac{400}{4\text R_\text X} =\frac{100}{\text R_\text X} = \text P_\text X$

So, P_X : P_Y = 1 : 1

(c) Given,

Mass density of nichrome = d

Then,

Mass of wire X (m_X) = d x volume = d x Al

And,

Mass of wire Y (m_Y) = d x volume = d x 2l x $\frac{\text A}{2}$ = d x Al = m_X

So, m_X : m_Y = 1 : 1

(d) True.

Reason —

Let, heat gained by the wire X be 'Q_X' and change in temperature be 'ΔT_X'. Similarly, heat gained by the wire Y be 'Q_Y' and change in temperature be 'ΔT_Y'.

Again, let specific heat capacity of nichrome be 'c' and same time be 't'.

Then,

Q_X = P_X x t = m_Xc x ΔT_X

and

Q_Y = P_Y x t = m_Yc x ΔT_Y

As, P_X = P_Y then Q_X = Q_Y for same time t.

⇒ m_Xc x ΔT_X = m_Yc x ΔT_Y

⇒ m_X x ΔT_X = m_Y x ΔT_Y

So,

ΔT_X : ΔT_Y = m_Y : m_X = 1 : 1

Hence, wire X and wire Y both show the same rise in temperature in the same time.

(a) All bulbs are connected in parallel combination because in a parallel connection, voltage across each bulb is the same, and brightness depends directly on power rating. So, the bulb with the highest power rating (P₃) will draw the most current and glow brightest.

(b) In a parallel circuit, all components share the same voltage.

So,

V₁ = V₂ = V₃

(c) Yes because in a parallel circuit, each bulb has an independent path to the power supply and if one bulb fuses, the others continue to receive power and function normally.

(a) When the key is closed, the current in the coil A grows from zero to its steady value, so the magnetic field produced by it changes. Since both the coils are wound on the same soft iron core, the magnetic flux linked with the coil B also changes and an e.m.f. is induced in it by electromagnetic induction. As the circuit of B is closed through the galvanometer, an induced current flows in it and the galvanometer shows a deflection.

The deflection is only momentary because once the current in A becomes steady, the flux linked with B no longer changes and the induced e.m.f. falls to zero.

(b) In the same direction.

(c) On decreasing the variable resistance, the current in the coil A increases. This increases the magnetic flux linked with the coil B, i.e., the flux again increases just as it did at the moment of closing the key.

Since the change in flux is of the same kind as before, by Lenz's law the induced current in B flows in the same direction as earlier, and hence the galvanometer deflects in the same direction.

(a) From the graph,

Number of protons in C = 92

Number of neutrons in C = 146

Mass number of C = Number of protons in C + Number of neutrons in C = 92 + 146 = 238

Number of protons in E = 93

Number of neutrons in E = 145

Mass number of E = Number of protons in E + Number of neutrons in E = 93 + 145 = 238

Since element C has one less proton and one neutron more than the element E, if C decays into E, its proton number increases by one and its neutron number decreases by one, keeping the mass number the same, which happens only in the case of beta radiation.

Thus, the reaction will be as follows :

²³⁸₉₂C → ²³⁸₉₃E + ⁰_-1β

(b) As points D and E are situated on the same vertical line but on different horizontal lines, they have the same number of protons but different numbers of neutrons, i.e., their atomic number is the same but their mass numbers are different, so they are isotopes of each other.

(c) Number of protons in B = 90

Number of neutrons in B = 144

Mass number of B = Number of protons in B + Number of neutrons in B = 90 + 144 = 234

As element B has two protons less than element C and a mass number four less than that of C, during the decay of C into B, two protons and two neutrons are emitted, which is nothing but an alpha particle.

Alpha particles shift towards the negative plate in an electric field.