ICSE Class 10 Physics — All Questions

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.

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.

acceleration should be zero

Reason — A body is said to be in dynamic equilibrium when it is moving with uniform velocity. It means that the body is not at rest, its velocity is constant and hence its acceleration is zero.

Therefore, for dynamic equilibrium, acceleration should be zero.

light to chemical

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

Therefore, the energy transformation in photosynthesis is light energy to chemical energy.

3

Reason — In a block and tackle system, the velocity ratio (V.R.) is equal to the number of strands of the rope supporting the moving block. When the system has two pulleys and the effort is applied in the upward direction, there are 3 supporting strands.

Therefore, the velocity ratio is 3.

$$\begin{aligned}\frac{\text {sin }60^{\circ}}{\text {sin }45^{\circ}} \\[1em]\end{aligned}$$

Reason — The angle given in medium A is 30° with the surface, so the angle with the normal is:

$i = 90^{\circ} - 30^{\circ} = 60^{\circ}$

The angle given in medium B is 45° with the surface, so the angle with the normal is:

$r = 90^{\circ} - 45^{\circ} = 45^{\circ}$

Now, refractive index of medium B with respect to medium A is:

$$\begin{aligned}{}_A\mu_B=\frac{\sin i}{\sin r} \\[1em] \Rightarrow {}_A\mu_B=\frac{\sin 60^\circ}{\sin 45^\circ}\end{aligned}$$

Infrared radiations

Reason — When a blackened bulb thermometer is placed beyond the red end of the visible spectrum, the temperature rises rapidly because this region contains infrared radiations. Infrared rays are invisible but carry significant heat energy, which is absorbed by the blackened bulb, causing the temperature to increase.

Therefore, the correct answer is Infrared radiations.

his total mechanical energy remains constant.

Reason — When the cyclist stops pedalling and continues moving on a hilly track, his kinetic energy gradually converts into potential energy as he moves upward. If there is no loss of energy, the sum of kinetic and potential energy remains unchanged.

Therefore, the total mechanical energy remains constant during the motion.

less than or equal to 15 cm

Reason — A virtual image whose distance never exceeds a fixed finite value is formed by a concave lens. In a concave lens, the image is always virtual and is formed between the optical centre and the principal focus. Hence, the image distance can never be greater than the focal length of the lens.

Therefore, for a lens of focal length 15 cm, the image distance is less than or equal to 15 cm.

Both (A) and (R) are true and (R) is not the correct explanation of (A).

Explanation:
Assertion is true because blue light has a shorter wavelength than red light, so it is scattered more by tiny air molecules. Reason is also true because the refractive index of a medium is greater for blue light than for red light. However, scattering depends mainly on wavelength and not on refractive index.

Therefore, both (A) and (R) are true, but (R) is not the correct explanation of (A).

L₂

Reason — In the circuit, L₁ and L₃ are placed in the main vertical branch, so if either of them fails, the circuit path breaks and current cannot reach the other lamps.

L₄ is also part of the lower main path; its failure would interrupt the current in that branch.

However, L₂ is connected in a separate parallel branch between the two sides of the circuit so if L₂ fails, only that branch becomes open while current can still flow through the rest of the circuit containing L₁, L₃, and L₄.

Therefore, the failure of L₂ would not interrupt the power supply to the other lamps.

B

Reason — When air is blown across the mouth of a jar, the pitch of the sound depends on the length of the air column above the water. A shorter air column produces a higher-pitched (shriller) sound, while a longer air column produces a lower pitch.

As, volume of a cylindrical jar is given by,

Volume = πr²h

Where r is the radius and h is the height of the jar and since equal volumes of water are poured into jars with radii r_B < r_A < r_C, the jar with the smallest radius (B) will have the greatest height of water, leaving the shortest air column.

Therefore, jar B produces the shrillest note.

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.

$$\begin{aligned}\frac{\text R_1\text R_2}{\text R_1 + \text R_2} \\[1em]\end{aligned}$$

Reason — When two resistors are connected in parallel, the potential difference across each resistor is the same, while the current divides between the branches. For two resistors R₁ and R₂, the reciprocal of the equivalent resistance equals the sum of the reciprocals of the individual resistances i.e.,

$$\begin{aligned}\frac{1}{\text R} = \frac{1}{\text R_1} + \frac{1}{\text R_2} \\[1em] \Rightarrow \frac{1}{\text R} = \frac{\text R_1 + \text R_2}{\text R_1\text R_2} \\[1em] \Rightarrow \text R = \frac{\text R_1\text R_2}{\text R_1 + \text R_2} \\[1em]\end{aligned}$$

Only 1 is aligned

Reason — A straight current-carrying wire produces circular magnetic field lines around it so using the right-hand thumb rule, if the current is into the plane of the paper (shown by X), the magnetic field lines circulate clockwise.

A compass needle aligns tangentially to these field lines and at the position of compass 1, the field direction is towards the right, so compass 1 is correctly aligned.

At the position of compass 2, the magnetic field should point downward, but the compass is shown pointing upward, so it is not correctly aligned.

Therefore, only compass 1 is aligned correctly.

Substance B

Reason — When substances are at their melting point, the heat supplied is used for melting and is given by Q = mL, where L is the specific latent heat.

Since the masses are the same and heat is absorbed at the same rate, the substance that takes the longest time to melt requires more heat and therefore has a higher specific latent heat.

Among A (5 min), B (7 min), and C (3 min), substance B takes the longest time.

Hence, substance B has the highest specific latent heat.

7

Reason — The mass number of an atom is the total number of protons and neutrons present in its nucleus. In a lithium atom, there are 3 protons and 4 neutrons, so the mass number is 3 + 4 = 7.

Therefore, the mass number of lithium is 7.

(a) A car is moving in uniform circular motion. The direction of friction between the tyres and the path is tangential to the path.

(b) When a ray of light passes from a denser to a rarer medium, its wavelength increases.

(c) The lid of a calorimeter minimises heat loss by convection.

(d) Quality of sound depends on its waveform.

(e) A substance whose resistance becomes almost negligible at a temperature near absolute zero is called a superconductor.

(f) Alpha radiation deviates minimum in a magnetic field.

The position of the centre of gravity of a body depends mainly on two factors:

1. Shape of the body
2. Distribution of mass in the body

Explanation:

The position of the centre of gravity depends on the shape of the body and on how its mass is distributed.
For example, two rods may have the same shape and size, but if one rod is uniform and the other is heavier at one end, their centres of gravity will be at different positions.

(a) The potato was closer to her hand in Case 1.

(b) The cutter acts as a Class II lever, in which the load lies between the fulcrum and the effort. When the potato is placed closer to the hand applying effort, it is farther from the fulcrum, so the load arm becomes longer. For a fixed effort arm, a longer load arm requires greater effort to cut. Since E₁ > E₂, the potato must have been closer to her hand in the Case 1.

(a) The lens used is a convex lens.

(b) If the object is placed between F and 2F, the magnification will be more than 1.

Explanation:

(a) The graph shows that for an object placed in front of the lens, a real image is formed on the other side and the image distance changes according to the position of the object. This behaviour is shown by a convex lens.

(b) For a convex lens, when the object is placed between F and 2F, the image is formed beyond 2F. The image is real, inverted and magnified. Therefore, the magnification is more than 1.

(a) The voltmeter reading decreases.

(b) As, voltmeter reading is given by,

V = ε - Ir

where,

• I is the current flowing in the circuit,
• ε is the emf of the cell and
• r is the internal resistance of the cell.

So, when the rheostat is adjusted to increase the current in the circuit, the current drawn from the cell becomes larger and due to the internal resistance of the cell, a greater potential drop occurs inside the cell.

As a result, the terminal voltage of the cell decreases and since the voltmeter measures this terminal voltage across the cell, its reading decreases.

(a) The periodic vibrations of a body in the absence of any external force on it, are called natural (or free) vibrations.

(b) In natural vibrations, the amplitude remains constant because there is no loss of energy while in damped vibrations, the amplitude gradually decreases with time due to energy loss caused by the presence of resistive force.

Given,

• Thermal capacity of the metal piece = 40 JK^-1
• Heat absorbed by the substance = 800 J

Let, rise in temperature be T.

As, thermal capacity of a substance is given by,

$$\begin{aligned}\text {Thermal capacity} = \frac{\text {Heat absorbed}}{\text {Temperature rise}} \\[1em] \Rightarrow \text {Temperature rise} = \frac{\text {Heat absorbed}}{\text {Thermal capacity}} \\[1em] \Rightarrow \text T = \frac{800}{40} \\[1em] \Rightarrow \text T = 20\ \text K\end{aligned}$$

Since a rise of 1 K = 1°C, the rise in temperature is also 20°C.

Hence, the rise in the temperature of the metal piece is 20°C.

(a) Soft iron core

(b) Carbon brushes

(a) The angle of incidence is maximum for prism B.

(b) The angle of incidence is minimum for prism A.

(c) The angle of incidence is equal to the angle of emergence for prism C.

Explanation:

For identical prisms, a larger angle of incidence produces a refracted ray inside the prism making a larger angle with the normal. Hence, diagram B corresponds to maximum incidence, while diagram A corresponds to minimum incidence.
When the angle of incidence is equal to the angle of emergence, the path of the ray through the prism is symmetrical, and the refracted ray inside the prism becomes parallel to the base. This is shown in diagram C.

(a) At the first surface, from air to glass:

$\frac{\text {sin i}_1}{\text {sin r}_1} = \text μ$

where μ is the refractive index of glass with respect to air.

At the second surface, from glass to air:

$\frac{\text {sin i}_2}{\text {sin r}_2} = \frac{1}{\text μ}$ .

Therefore, the product of the two ratios is given by,

$\frac{\text {sin i}_1}{\text {sin r}_1}\times \frac{\text {sin i}_2}{\text {sin r}_2} = \text μ \times \frac{1}{\text μ} = 1$

Hence, the product of the ratio is 1.

(b) The opposite surfaces are not parallel.

(c) If the opposite surfaces were parallel, the angle of refraction at the first surface would be equal to the angle of incidence at the second surface and since these angles are not equal, the surfaces cannot be parallel.

(a) Given,

• Refractive index of the glass block = 1.8
• Speed of light in air = 3 x 10⁸ m s^-1

As, refractive index (μ) of the glass block is given by,

$$\begin{aligned}\text μ = \frac{\text {Speed of light in vacuum or air}}{\text {Speed of light in the glass block}} \\[1em] \Rightarrow \text {Speed of light in the glass block} = \frac{\text {Speed of light in vacuum or air}}{\text μ} \\[1em] = \frac{3\times 10^8}{1.8} \\[1em] = \frac{30\times 10^8}{18} \\[1em] = \frac{5\times 10^8}{3} \\[1em] \approx 1.67 \times 10^8 \text { m s}^{-1}\end{aligned}$$

Hence, speed of light in the glass block is approximately 1.67 x 10⁸ m s^-1.

(b) If the width of the glass block is doubled, the speed of light does not change because the speed of light in a medium depends only on the refractive index of the material, not on its thickness or width.

Therefore, the speed remains 1.67 x 10⁸ m s^-1.

(a) Ultraviolet (UV) radiation

(b) The completed diagram is shown below :

Explanation:

• The ray AB enters the prism normally at point B, so it passes straight into the prism without bending.
• It then travels vertically upward and strikes the slant face of the prism.
• The angle between this face and the vertical side is 30°, so the angle of incidence at the slant face becomes 60°.
• Since this angle is greater than the critical angle (42°), the ray undergoes total internal reflection at the slant surface.
• After reflection, the ray travels at an angle of 60° with the normal of slant face and move towards opposite face.
• The ray incident on this face at an angle of 30° which is smaller than the critical angle (42°) so the ray refract at this face.

(a) Initially,

Let, initial object distance be u, image distance be v and focal length be f.

According to question,

Since the image is formed on a screen so it is real and inverted and it is also of the same size as the object which means magnification produced by the lens is -1.

Magnification (m) of a lens is given by,

$$\begin{aligned}\text m = \frac{\text {v}}{\text {u}} \\[1em] \Rightarrow -1 = \frac{\text {v}}{\text {u}} \\[1em] \Rightarrow \text {u} = -\text {v}\end{aligned}$$

From lens formula,

$$\begin{aligned}\frac{1}{\text f} = \frac{1}{\text v} - \frac{1}{\text u} \\[1em] \frac{1}{\text f} = \frac{1}{\text v} - \frac{1}{(-\text v)} \\[1em] \frac{1}{\text f} = \frac{1}{\text v} + \frac{1}{\text v} \\[1em] \frac{1}{\text f} = \frac{2}{\text v} \\[1em] \Rightarrow \text f = \frac{\text v}{2}\end{aligned}$$

Let, final object distance be u' and final image distance be v'.

According to question,

u' = u + 12 and m' = -3

Magnification (m') of a lens is given by,

$$\begin{aligned}\text m' = \frac{\text {v'}}{\text {u'}} \\[1em] \Rightarrow -3 = \frac{\text {v'}}{\text {u'}} \\[1em] \Rightarrow \text {u'} = -\frac{\text {v'}}{3} \\[1em] \Rightarrow \text {u} + 12 = -\frac{\text {v'}}{3} \\[1em] \Rightarrow -\text {v} + 12 = -\frac{\text {v'}}{3} \\[1em] \Rightarrow -3(\text {v} - 12) = -\text {v'} \\[1em] \Rightarrow 3(\text {v} - 12) = \text {v'} \\[1em] \Rightarrow 3\text {v} - 36 = \text {v'} \\[1em]\end{aligned}$$

From lens formula,

$$\begin{aligned}\frac{1}{\text f} = \frac{1}{\text v'} - \frac{1}{\text u'} \\[1em] \frac{1}{\text f} = \frac{1}{\text v'} - \frac{3}{-\text v'} \\[1em] \frac{1}{\text f} = \frac{1}{\text v'} + \frac{3}{\text v'} \\[1em] \frac{1}{\text f} = \frac{4}{\text v'} \\[1em] \Rightarrow \text f = \frac{\text v'}{4} \\[1em] \Rightarrow \text f = \frac{3\text {v} - 36}{4} \\[1em] \Rightarrow \text f = \frac{3\text {v}}{4} - \frac{36}{4} \\[1em] \Rightarrow \text f = \frac{3\text {f}}{2} - 9 \\[1em] \Rightarrow \frac{3\text {f}}{2} - \text f = 9 \\[1em] \Rightarrow \frac{3\text {f}}{2} - \frac {2\text f}{2} = 9 \\[1em] \Rightarrow \frac {\text f}{2} = 9 \\[1em] \Rightarrow \text f = 2\times 9 \\[1em] \Rightarrow \text f = 18 \text { cm}\end{aligned}$$

Hence, the focal length of the lens is 18 cm.

(a) False
Reason — After the hail-strom, ice starts absorbing the heat energy required for its melting from the surroundings, so the temperature of the surroundings falls further down.

(b) Making the base of a frying pan heavier increases its heat capacity because heat capacity depends on the mass of the object so that a heavier base requires more heat to raise its temperature.

(c) The principle of calorimetry states that when a hot body is brought in contact with a cold body, then heat lost by the hot body is equal to the heat gained by the cold body, provided there is no heat loss in the environment.
Heat energy lost by the hot body = Heat energy gained by the cold body

(a) The freezing temperature of the liquid is 20 °C because at this temperature, the graph becomes horizontal, indicating that the temperature remains constant during the phase change.

(b) In the region QR, the phase change taking place is freezing (liquid changing into solid).

(c) The substance liberates heat at a faster rate in the solid state which is shown by the steeper slope of the graph after point R, indicating a faster decrease in temperature compared to the liquid state before point Q, where the slope is less steep.

(a) The force F₁ produces the negative moment because it tends to rotate the wheel in the clockwise direction, which is taken as negative.

(b) No, the wheel is not in equilibrium.

(c) For equilibrium, the sum of clockwise moments must be equal to the sum of anticlockwise moments about the centre O. Although F₁ and F₂ have equal magnitudes, their perpendicular distances from the centre (moment arms) are different. Hence, the moments produced by them are not equal and opposite, so the net moment is not zero and the wheel is not in equilibrium.

(a) electron volt (eV)

(b) Class I lever

(c) Given,

• AJB is a semi-circle.
• θ < 90°
• AB = 20 m

Work done by a force is given by

W = Force x Displacement x cos θ,

where θ is the angle between force and displacement.

• Case 1 : The stone moves along a semicircular path, so the force is always perpendicular to the direction of displacement which makes θ = 90°.

Hence, work done is zero.

• Case 2 : The force acts along the direction of displacement from A to B, so θ = 0°.

Hence, work done is maximum.

• Case 3: The force acts at an angle θ < 90° to the displacement, so only the component Fcosθ does work.

Hence, the work done is less than Case 2 but more than Case 1.

Therefore, the ascending order of work done is Case 1 < Case 3 < Case 2

(a) Given,

• Mass of the ball (m) = 20 g = 0.02 kg
• Initial height (h₁) = 45 m
• Final height (h₂) = 40 m
• g = 10 m s^-2

(a) Initial potential energy of the ball is given by,

PE = mgh₁
= 0.02 x 10 x 45
= 9 J

So, the initial potential energy is 9 J.

(b) Due to law of conservation of mechanical energy, just before hitting the ground, the initial potential energy converts into kinetic energy.

Let speed of the ball be v.

So,

$$\begin{aligned}\text {Kinetic energy} = 9 \\[1em] \Rightarrow \frac{1}{2} \text {mv}^2 = 9 \\[1em] \Rightarrow \frac{1}{2} \times 0.02\times \text {v}^2 = 9 \\[1em] \Rightarrow 0.01\times \text {v}^2 = 9 \\[1em] \Rightarrow \text {v}^2 = \frac{9}{0.01} \\[1em] \Rightarrow \text {v}^2 = 900 \\[1em] \Rightarrow \text {v} = \sqrt {900} \\[1em] \Rightarrow \text {v} = 30 \text { m s}^{-1}\end{aligned}$$

So, the speed of the ball on hitting the ground = 30 m s^-1.

(c) Final kinetic energy after rebound is equal to potential energy at 40 m which is given by,

Final kinetic energy = mgh₂
= 0.02 x 10 x 40
= 8 J

Loss in KE = Initial kinetic energy - Final kinetic energy = 9 − 8 = 1 J

So, the loss in kinetic energy = 1 J.

Given,

• Load = 30 kgf

(a) As, mechanical advantage (MA) of a machine is given by,

$\text {MA} = \frac{\text {Load}}{\text {Effort}}$

For a single fixed pulley, the mechanical advantage (MA) is 1.

Then,

$$\begin{aligned}\text {MA} = \frac{\text {Load}}{\text {Effort} (\text E_1)} \\[1em] \Rightarrow \text E_1 = \frac{\text {Load}}{\text {MA}} \\[1em] \Rightarrow \text E_1 = \frac{30}{1} \\[1em] \Rightarrow \text E_1 = 30 \text { kgf}\end{aligned}$$

For a single movable pulley, the mechanical advantage (MA) is 2.

Then,

$$\begin{aligned}\text {MA} = \frac{\text {Load}}{\text {Effort} (\text E_2)} \\[1em] \Rightarrow \text E_2 = \frac{\text {Load}}{\text {MA}} \\[1em] \Rightarrow \text E_2 = \frac{30}{2} \\[1em] \Rightarrow \text E_2 = 15 \text { kgf}\end{aligned}$$

Ratio of efforts = $\text E_1 : \text E_2$ = 30 : 15 = 2 : 1

Hence, the ratio of the efforts in the two cases is 2 : 1.

(b) Potential energy is given by,

Potential energy gained = Load × Height raised

If the effort displacement is the same:

• In a fixed pulley, load rises by the same distance as the effort.
• In a movable pulley, load rises by half the distance of the effort.

Let, for the fixed pulley, height raised by the load be h.

Therefore,

Ratio of the potential energies in two cases is given by,

$$\begin{aligned}\text {Ratio of PE} = \frac{\text {PE gained by the load in fixed pulley}}{\text {PE gained by the load in movable pulley}} \\[1em] = \frac {30 \times \text h}{30 \times \frac{\text h}{2}} \\[1em] = \frac {30\text h \times 2}{30 \text h} \\[1em] = \frac {2}{1} \\[1em] = 2 : 1\end{aligned}$$

Hence, the ratio of the potential energies in two cases is 2 : 1.

(c) In an ideal machine, efficiency = 100 % for both pulleys.

Then,

Efficiency ratio = 1 : 1

Hence, the ratio of the efficiencies in the two cases is 1 : 1.

(a) When one end of the plastic ruler is held firmly at the edge of the table and the other end is plucked, the ruler vibrates on its own after being displaced and these vibrations are called natural vibrations.

(b) The frequency of vibration increases because when the ruler is pushed further inside, the length of the free vibrating part decreases and since a shorter vibrating length oscillates more rapidly, so the frequency increases.

Given,

• Distance between A and B = 170 m
• Time taken by the person B to hear the echo = 3 s
• Speed of sound in air = 340 ms^-1

(a)

1. Let, the distance of the person B from the cliff be d.

So,

$$\begin{aligned}\text d = \frac {\text {Speed of sound in air} \times \text {Time taken}}{2} \\[1em] = \frac{340 \times 3}{2} \\[1em] = 170 \times 3 \\[1em] = 510\ \text m\end{aligned}$$

Hence, the distance of the person B from the cliff is 510 m.

2. Person B first hears the direct sound of the gunshot and only afterwards hears its echo from the cliff. Therefore, the minimum time in which B hears the gunshot is the time taken by the sound to travel directly from person A to person B, who is 170 m away.

Then, minimum time (t) is given by,

$$\begin{aligned}\text t = \frac{\text {Distance between A and B}}{\text {Speed of sound in air}} \\[1em] = \frac{170}{340} \\[1em] = \frac{1}{2} \\[1em] = 0.5\ \text s\end{aligned}$$

Hence, the minimum time in which B hears the gunshot fired by A is 0.5 second.

(b) The echo is softer (less loud) than the original sound due to the decrease in amplitude of the wave.

Given,

• Power rating of the bulb A ($\text P_\text A$) = 160 W
• Voltage rating of the bulb A ($\text V_\text A$) = 40 V
• Power rating of the bulb B ($\text P_\text B$) = 40 W
• Voltage rating of the bulb B ($\text V_\text B$) = 40 V
• Supply voltage ($\text V$) = 40 V

(a) Resistance of bulb A is given by,

$$\begin{aligned}\text R_\text A = \frac{\text V_\text A^2}{\text P_\text A} \\[1em] = \frac{40^2}{160} \\[1em] = \frac{1600}{160} \\[1em] = 10\ \text Ω\end{aligned}$$

Similarly,

Resistance of bulb B is given by,

$$\begin{aligned}\text R_\text B = \frac{\text V_\text B^2}{\text P_\text B} \\[1em] = \frac{40^2}{40} \\[1em] = \frac{1600}{40} \\[1em] = 40\ \text Ω\end{aligned}$$

From Ohm's law,

Potential drop (V) = Current (I) x Resistance (R)

Since the bulbs are in series, so the same current flows through both bulbs and the potential difference divides in the ratio of their resistances i.e.,

V₁ : V₂ = $\text R_\text A : \text R_\text B$ = 10 : 40 = 1 : 4

Hence, the ratio V₁ : V₂ = 1 : 4.

(b) False because if bulb A fuses, the circuit becomes open, so no current can flow through the circuit. Hence the current becomes zero, not the same.

(a) Fault in the connection : The live (red) wire is incorrectly connected to the earth terminal (top thick pin) instead of the live terminal and the earth (green) wire is also wrongly connected to the live terminal. Thus, the live and earth wires should be interchanged.

(b) If the appliance is used in this condition, the outer metallic body of the appliance may become live, creating a risk of electric shock to the user.

(c) No, the appliance will not function because the wiring connections are incorrect.

(a) When the switch is open, the middle connection does not join the two branches. So the circuit has two separate branches in parallel i.e., top and bottom branches where 12 Ω and 6 Ω resistances are in series.

Then net resistance of the top branch is given by,

$\text R_1 = 12 + 6 = 18 Ω$

Similarly, net resistance of the bottom branch is given by,

$\text R_2 = 12 + 6 = 18 Ω$

Now, $\text R_1$ and $\text R_2$ are in parallel connection and the net resistance is given by,

$$\begin{aligned}\frac{1}{\text R} = \frac{1}{\text R_1} + \frac{1}{\text R_2} \\[1em] = \frac {1}{18} + \frac {1}{18} \\[1em] = \frac {2}{18} \\[1em] \Rightarrow \frac{1}{\text R} = \frac {1}{9} \\[1em] \Rightarrow \text R = 9\ \text Ω\end{aligned}$$

Hence, the resistance across AB when the switch S is open is 9 Ω.

(b) When switch S is closed, the midpoints are connected which rearranges the circuit into two branches of parallel resistors that are connected in series as shown below.

Then net resistance of the left branch (containing 12 Ω and 6 Ω resistances) is given by,

$$\begin{aligned}\frac{1}{\text R_1} = \frac{1}{12} + \frac{1}{6} \\[1em] = \frac {1}{12} + \frac {2}{12} \\[1em] = \frac {3}{12} \\[1em] \Rightarrow \frac{1}{\text R_1} = \frac {1}{4} \\[1em] \Rightarrow \text R_1 = 4\ \text Ω\end{aligned}$$

Similarly, net resistance of the right branch (containing 12 Ω and 6 Ω resistances) is given by,

$$\begin{aligned}\frac{1}{\text R_2} = \frac{1}{12} + \frac{1}{6} \\[1em] = \frac {1}{12} + \frac {2}{12} \\[1em] = \frac {3}{12} \\[1em] \Rightarrow \frac{1}{\text R_2} = \frac {1}{4} \\[1em] \Rightarrow \text R_2 = 4\ \text Ω\end{aligned}$$

Now, $\text R_1$ and $\text R_2$ are in series connection and the net resistance is given by,

$$\begin{aligned}\text R = \text R_1 + \text R_2 \\[1em] = 4 + 4 \\[1em] = 8\ \text Ω\end{aligned}$$

Hence, the resistance across AB when the switch S is closed is 8 Ω.

Given,

• Power rating of the iron ($\text P$) = 1100 W
• Voltage rating of the iron ($\text V$) = 220 V
• Operating time = 5 hours

(a) Power consumed by an appliance is given by,

$$\begin{aligned}\text P = \text {VI} \\[1em] \Rightarrow \text I = \frac{\text P}{\text V} \\[1em] = \frac{1100}{220} \\[1em] = 5\ \text A\end{aligned}$$

Here, $\text I$ is the operating current flowing through the appliance.

Since the fuse rating should be slightly higher than the operating current, so the minimum suitable fuse rating is 6 A.

Hence, 6 A is the minimum rating of the fuse required.

(b) Energy consumed by the iron = Power consumed x Operating time

= 1100 x 5

= 5500 Wh

= 5.5 x 10³ Wh

= 5.5 kWh

Hence, the energy consumed by the iron is 5.5 kWh.

(c) Given,

• Electricity rate = ₹ 10 per unit

As, 1 unit = 1 kWh

Then,

The cost of energy consumed by the iron = Energy consumed x Electricity rate

= 5.5 x 10

= ₹ 55

Hence, the cost of the energy consumed is ₹ 55.

The deflection of the galvanometer depends on the rate of change of magnetic flux, which depends on the relative motion between the magnet and the coil.

(a) Both the magnet and the coil move in the same direction with the same speed, so there is no relative motion between them and hence, the magnetic flux through the coil does not change.

Therefore, no current is induced and the galvanometer shows no deflection.

(b) Since the magnet and the coil are moving opposite to each other so now the relative speed between them is given by the sum of individual speeds and becomes 10 m s^-1 (5 + 5) which increases the rate of change of magnetic flux.

Therefore, the direction of deflection remains the same (to the right), but the magnitude of deflection increases.

(a)

1. Lead is used in the lining of the special aprons worn by workers in nuclear power plants.
2. Lead is preferred because it is a dense metal and absorbs harmful nuclear radiations effectively.

(b)

1. The radiation that does not change the mass number but is deflected by a magnetic field is beta (β) radiation.
   Hence, beta (β) radiation is emitted by $_{\textbf {11}}^{\textbf {24}}\textbf {Na}$.
2. Radioactive decay equation:
   $_{11}^{24}\text {Na} \longrightarrow _{12}^{24}\text {Na} + _{-1}^{0}\text {β}$
   In this decay, a beta particle (electron) is emitted, the atomic number increases by 1, while the mass number remains unchanged.

Heat energy to electrical energy

Reason — A thermocouple is made by joining two wires of different metals so as to form two junctions. When one junction is kept hot and the other is kept cold, a current begins to flow in the circuit. Thus the heat energy supplied at the hot junction is converted into electrical energy.

joule

Reason — As momentum p = mv, so

$\frac{p^2}{2m} = \frac{(mv)^2}{2m} = \frac{m^2v^2}{2m} = \frac{1}{2}mv^2$

which is the expression for kinetic energy. Since energy is measured in joule, the unit of $\frac{p^2}{2m}$ is joule.

3

Reason — In a block and tackle system in which the effort is applied in the downward direction, the tension in the n segments of the string supports the load, so

L = nT

where n is the total number of pulleys used in both the blocks.

Here, L = 5T, so n = 5.

Then,

Number of pulleys in the fixed block = Total number of pulleys - Number of pulleys in the movable block = 5 - 2 = 3

least for F₃ and maximum for F₁

Reason — The moment of a force is the product of the force and the perpendicular distance of its line of action from the pivot. If a force F acts at A at an angle θ with the rod and OA = r, then

$\tau = \text{F r}\sin \theta$

All the three forces are equal and act at the same point A, so the moment of each depends only on the angle θ.

F₃ acts along the rod, so θ = 0° and its line of action passes through O, giving τ₃ = 0, the least.

F₁ acts perpendicular to the rod, so θ = 90° and τ₁ = F r, the maximum.

F₂ acts at an angle between 0° and 90°, so its moment lies between the two.

0°

Reason — From the diagram, the light ray enters the prism through one refracting face, suffers total internal reflection at the base of the prism and finally emerges out through the other refracting face. The emergent ray is parallel to the incident ray and it travels in the same direction as the incident ray.

Since the emergent ray has not turned from the direction of the incident ray, the net angle of deviation produced by the prism is 0°.

total internal reflection

Reason — A crack in a glass vessel contains a thin layer of air. When light travelling inside the glass (denser medium) strikes this air layer (rarer medium) at an angle of incidence greater than the critical angle for the glass-air interface, the entire light is reflected back into the glass. Because 100% of the light is returned, the crack appears bright and shines like a mirror.

virtual and magnified

Reason — The sign of magnification tells the nature of the image and its numerical value tells the size of the image.

Here, magnification is positive, which means the image is erect and hence virtual, and its numerical value is 2 (i.e., greater than 1), which means the image is twice the size of the object and hence magnified.

Radio waves

Reason — The scattering of a radiation decreases as its wavelength increases. Among the given radiations, radio waves have the longest wavelength, so they are scattered the least. Gamma rays, having the shortest wavelength, are scattered the most.

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

Explanation

Assertion (A) is correct: The buzzing of a mosquito is a shrill, that is, a high pitched sound, whereas the roar of a lion is a low pitched sound. The roar is much louder than the buzz, but loudness depends on the amplitude of the sound wave, while shrillness depends on its pitch.

Reason (R) is also correct: A mosquito vibrates its wings very rapidly and therefore produces a sound of high frequency, while the roar of a lion is produced by the slower vibration of its vocal cords and is of low frequency.

Since the shrillness, or pitch, of a sound is decided by its frequency — the greater the frequency, the shriller the sound — the higher frequency of the mosquito is exactly the cause of its sound being shriller.

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

R

Reason — The resistance of a wire is given by

$\text{R} = \rho\frac{l}{a}$

i.e., it depends on the resistivity of the material, the length of the wire and its area of cross-section.

On bending the wire into a U shape, neither its length, nor its area of cross-section, nor its material changes. Hence its resistance remains R.

It has high conductivity.

Reason — A fuse wire must get heated up quickly and melt when the current in the circuit exceeds the safe value. For this it is made of an alloy of lead and tin which has a high resistance and a low melting point. High conductivity means low resistance, so the wire would not get heated enough to melt. Hence high conductivity is not a characteristic of a fuse wire.

Kinetic energy of the molecules remains the same.

Reason — During a change of phase, the temperature of the substance does not change. Since the kinetic energy of the molecules depends on the temperature, the kinetic energy of the molecules remains the same. The heat energy supplied (i.e., the latent heat) is used up in increasing the potential energy of the molecules by increasing the separation between molecules against their attractive forces.

1 g of water through 1 K

Reason — One calorie is defined as the amount of heat energy required to raise the temperature of 1 gram of water through 1 K (or 1 °C). Note that a change of 1 K is numerically equal to a change of 1 °C, so both express the same rise in temperature.

mass of the conductor

Reason — For a conductor normal to the field, the force depends on the current, field strength and length of wire in the field; its mass does not determine the magnetic force.

Flemming's Left Hand Rule

Reason — The α-particles and β-particles are charged particles, so a beam of these radiations is equivalent to a current flowing in a conductor. When such a beam passes through a magnetic field, it experiences a force whose direction is obtained by Fleming's left hand rule, which gives the direction of the force acting on a current carrying conductor placed in a magnetic field.

(a) If the displacement of a load is greater than the displacement of the effort, then the velocity ratio of the machine is less than one.

(b) Out of multiple images of an object, formed by a thick plane glass mirror, the second image is the brightest.

(c) No negative temperature is possible on the Kelvin scale.

(d) 250 dB is louder than 10 dB.

(e) In a series combination, across each resistor, the current remains the same.

(f) ₁₈Ar⁴⁰ and ₂₀Ca⁴⁰ are the examples of isobars.

(a) The knife edge should be kept on the steel part.

Reason — Steel is much denser than wood, so for the same length the steel half is heavier than the wooden half. Hence the centre of gravity of the ruler is shifted from the 50 cm mark towards the steel half, and the ruler will balance only when the knife edge is placed at the centre of gravity, i.e., somewhere on the steel part.

(b) Principle of moments — according to which, in the equilibrium position, the sum of the anticlockwise moments about the knife edge is equal to the sum of the clockwise moments about it.

(a) In a second-class lever, the ratio of the effort arm to the load arm is greater than 1.

Reason — In a class II lever the load lies in between the fulcrum and the effort, so the effort arm is always longer than the load arm. Hence the ratio $\frac{\text{effort arm}}{\text{load arm}}$, which is the mechanical advantage, is always greater than 1.

(b) In a single movable pulley the effort has to be applied in the upward direction, which is inconvenient. By adding a fixed pulley to it, the direction of the effort is changed, so that the effort can now be applied in the downward direction, which is much more convenient since the weight of the person applying the effort also helps. The mechanical advantage remains 2 in both cases.

(a) It is a concave (diverging) lens.

(b) The lowest part of the lens behaves like a small prism. From the diagram, this part is thicker at its lower edge and thinner at its upper edge, i.e., the base of this prism points away from the principal axis of the lens.

This means that as we move from the lower edge of the lens towards its middle, the lens becomes thinner. A lens which is thin at the middle and thick at the edges is a concave lens. Further, since each prism-like part bends the light towards its base, i.e., away from the principal axis, the lens diverges the incident light. Hence it is a concave lens.

(a) High power appliances such as heaters and radiators draw a large current. Since the resistance of a wire is inversely proportional to its area of cross-section, a thick wire has a large area of cross-section and hence a low resistance. Therefore the heat produced (H = I²Rt) in a thick wire is small and it can carry a heavy current without getting over-heated, so there is no danger of the insulation burning or of a fire.

(b) A fuse (or a miniature circuit breaker, MCB) is used to protect the electric circuits from overloading and short circuits.

In a three pin socket, the upper bigger hole is for the earth wire, the hole on the right side is for the live wire and the hole on the left side is for the neutral wire.

So here, C is earth, B is live and A is neutral.

(a) Yes, the bulb will glow.

Reason — The live wire B is at a high potential (220 V) while the earth wire C is at zero potential, so there is a potential difference of 220 V across the bulb and a current flows through it.

(b) No, the bulb will not glow.

Reason — Both the neutral wire A and the earth wire C are at zero potential, so there is no potential difference across the bulb and hence no current flows through it.

Given,

• Temperature of hot liquid = 80 °C
• Temperature of cold liquid = 20 °C
• Temperature of mixture = 60 °C
• Ratio of masses m₁ : m₂ = 8 : 5

Let the masses be m₁ = 8m and m₂ = 5m, and their specific heat capacities be c₁ and c₂ respectively.

Heat energy lost by the hot liquid = m₁ x c₁ x (80 - 60) = 8m x c₁ x 20 = 160 m c₁

Heat energy gained by the cold liquid = m₂ x c₂ x (60 - 20) = 5m x c₂ x 40 = 200 m c₂

By the principle of calorimetry,

Heat energy lost by the hot liquid = Heat energy gained by the cold liquid

$$\begin{aligned}160\ \text m\ \text c_1 = 200\ \text m\ \text c_2 \\[1em] \Rightarrow \frac{\text c_1}{\text c_2} = \frac{200}{160} = \frac{5}{4}\end{aligned}$$

Hence, the ratio of their specific heat capacities is 5 : 4.

(a) When the key 'K' is closed, then the bulb glows only for a moment.

(b) On closing the key, the current in coil A grows from zero to its maximum value. During this short interval the magnetic flux linked with coil B changes, so an e.m.f. is induced in coil B and an induced current flows through the bulb, making it glow.

Once the current in coil A becomes steady, the magnetic flux linked with coil B no longer changes. Hence no e.m.f. is induced, no current flows in coil B and the bulb stops glowing.

Given,

• Atomic number of uranium (Z) = 92
• Mass number of uranium (A) = 235

On emission of one α-particle — the atomic number decreases by 2 and the mass number decreases by 4.

Atomic number = 92 - 2 = 90

Mass number = 235 - 4 = 231

On emission of two β-particles — the atomic number increases by 1 for each β-particle and the mass number remains unchanged.

So, for emission of two β-particles:

Atomic number = 90 + 2 = 92

Mass number = 231

Hence, the daughter nucleus has atomic number 92 and mass number 231.

(a) The angle of refraction will be more in water (beaker B).

Reason — The refractive index of a medium is:

$\text{μ} = \frac{\text c}{\text v}$

Since the speed of light in water is greater than that in kerosene oil,

μ _water $\lt$ μ _kerosene

For the same angle of incidence, light bends less towards the normal in the optically rarer medium. Therefore, the angle of refraction is greater in water (beaker B).

(b) Given,

• Speed of light in water (v_w) = 2.25 x 10⁸ m s^-1
• Speed of light in kerosene oil (v_k) = 2.15 x 10⁸ m s^-1

As, refractive index of a medium = $\frac{\text{speed of light in air}}{\text{speed of light in the medium}} = \frac{\text c}{\text v}$

Then,

$$\begin{aligned}\frac{\text{μ}_\text w}{\text{μ}_\text k} = \frac{\frac{\text c}{\text v_\text w}}{\frac{\text c}{\text v_\text k}} \\[1em] = \frac{\text v_\text k}{\text v_\text w} \\[1em] = \frac{2.15 \times 10^8}{2.25 \times 10^8} \\[1em] = \frac{215}{225} \\[1em] = \frac{43}{45}\end{aligned}$$

Hence, the ratio of the refractive indices of water to kerosene oil is 43 : 45.

(c) Liquid A is kerosene oil. On heating, the refractive index of kerosene oil decreases. Therefore the light ray bends less towards the normal, i.e., the angle of refraction increases and the refracted ray shifts away from the normal.

(a) The phenomenon is Total internal reflection.

Total internal reflection is the phenomenon in which a ray of light travelling from an optically denser medium to an optically rarer medium is completely reflected back into the denser medium when the angle of incidence is greater than the critical angle.

On a hot afternoon the air near the road is hot and rarer, while the air above it is cooler and denser. Light coming from the sky travels from the denser upper layers to the rarer lower layers, bending away from the normal at each layer, until it strikes a layer at an angle greater than the critical angle. It is then totally internally reflected and reaches the driver's eye from below, so the driver sees the image of the sky on the road and the road appears wet.

(b) A diamond sparkles because of total internal reflection.

Given,

• Focal length of the concave lens L1 = 20 cm
• Focal length of the convex lens L2 = 12 cm

(a) Length x :

From the diagram, the ray travelling between the two lenses is parallel to the principal axis. A ray of light incident parallel to the principal axis of a convex lens passes, after refraction, through its second focal point.

Hence P is the second focal point of the convex lens L2, so

x = focal length of L2 = 12 cm

Length y :

The dotted line in the diagram shows that the ray incident on the concave lens L1 is directed towards the point P. After refraction through L1 this ray becomes parallel to the principal axis. This happens only when the incident ray is directed towards the first focal point of the concave lens.

Hence P is the first focal point of L1, so

Distance of P from L1 = focal length of L1 = 20 cm

Then,

y + x = 20

⇒ y = 20 - x = 20 - 12 = 8 cm

Hence, x = 12 cm and y = 8 cm.

(b) With respect to the concave lens L1, the point P is its first focal point (first focus) i.e., the point on the principal axis of the lens such that the incident rays of light appearing to meet at it become parallel to the principal axis after refraction through the lens.

(a) The refractive index of the glass will increase.

Reason — The refractive index of a medium increases as the wavelength of light decreases. Green light has a shorter wavelength than yellow light, so the refractive index of glass for green light is greater than that for yellow light.

(b) The angle of deviation is 0° (zero), because a ray falling normally on the surface is incident along the normal, so it passes into the second medium undeviated.

(c) NO

Reason — A critical angle exists only when light travels from a denser medium to a rarer medium. Here the light would travel from water to glass, i.e., from a rarer medium to a denser medium, so total internal reflection is not possible and there is no critical angle.

(a) The two phenomena taking place at the surface AB are :

1. Refraction of light, and
2. Dispersion of light.

(b) No, the separate colours cannot be seen when viewed from the side DC.

Reason — The two refracting surfaces AB and DC of a rectangular glass block are parallel to each other. At the first surface AB the colours get separated because each colour is deviated by a different amount. At the second surface DC, each colour is refracted in such a way that its emergent ray becomes parallel to the incident ray, so no further separation of colours takes place as it does in a prism. Hence the emergent coloured rays are parallel to one another and are so close together that they overlap, and the emergent light appears almost white.

(a) The temperature remains constant at 150 °C along the portion BC of the graph, during which the substance condenses from vapour to liquid.

Hence, the boiling point of the substance is 150 °C.

(b) Along DE the temperature remains constant at 100 °C while the substance is still giving out heat. Hence in the region DE the substance changes from the liquid state to the solid state, i.e., it is freezing (solidification).

(c) The substance is completely a liquid from the end of condensation (point C, at 150 °C) up to the beginning of freezing (point D, at 100 °C).

Hence, the substance is in the liquid state in the temperature range 150 °C to 100 °C.

Given,

• Mass of ice (m₁) = 280 g at 0 °C
• Mass of water (m₂) = 600 g at 50 °C
• Mass of the metal container (m₃) = 300 g at 50 °C
• Final temperature of the mixture = 10 °C
• Specific latent heat of ice (L) = 336 J g^-1
• Specific heat capacity of water (c_w) = 4.2 J g^-1 K^-1

Let the specific heat capacity of the metal container be c.

Heat energy gained by the ice :

Heat used in melting the ice = m₁ x L = 280 x 336 = 94080 J

Heat used in raising the temperature of the melted water from 0 °C to 10 °C

= m₁ x c_w x ΔT
= 280 x 4.2 x 10
= 11760 J

Total heat energy gained = 94080 + 11760 = 105840 J

Heat energy lost by the water and the container :

Heat lost by water = m₂ x c_w x ΔT
= 600 x 4.2 x (50 - 10)
= 600 x 4.2 x 40
= 100800 J

Heat lost by the container = m₃ x c x ΔT
= 300 x c x 40
= 12000 c

By the principle of calorimetry,

Heat energy gained = Heat energy lost

$$\begin{aligned}105840 = 100800 + 12000\ \text c \\[1em] \Rightarrow 12000\ \text c = 105840 - 100800 \\[1em] \Rightarrow 12000\ \text c = 5040 \\[1em] \Rightarrow \text c = \frac{5040}{12000} = 0.42\end{aligned}$$

Hence, the specific heat capacity of the metal container is 0.42 J g^-1 K^-1.

Given,

• Weight suspended at the 0 cm mark = 100 gf
• Weight of the uniform metre scale = 200 gf, acting at its centre of gravity i.e., at the 50 cm mark
• Weight suspended at the 100 cm mark = 200 gf

Let the ruler be balanced when the knife edge is placed at the x cm mark.

Taking moments about the knife edge,

Anticlockwise moments (due to the forces to the left of the knife edge)

= 100 x (x - 0) + 200 x (x - 50)

Clockwise moment (due to the force to the right of the knife edge)

= 200 x (100 - x)

By the principle of moments,

Anticlockwise moments = Clockwise moment

$$\begin{aligned}100(\text x - 0) + 200(\text x - 50) = 200(100 - \text x) \\[1em] \Rightarrow 100\text x + 200\text x - 10000 = 20000 - 200\text x \\[1em] \Rightarrow 300\text x - 10000 = 20000 - 200\text x \\[1em] \Rightarrow 300\text x + 200\text x = 20000 + 10000 \\[1em] \Rightarrow 500\text x = 30000 \\[1em] \Rightarrow \text x = \frac{30000}{500} = 60\end{aligned}$$

Hence, the ruler will be balanced at the 60 cm mark.

(a) The moment of force is greater in Case 1.

(b) Reason — The moment of a force is the product of the force and the perpendicular distance of its line of action from the axis of rotation.

In Case 1, the ruler turns about the pivot at the end B, so the perpendicular distance of the force F at A from the axis is the whole length AB.

In Case 2, the ruler is free, so it turns about its centre of gravity which lies at the middle of the uniform ruler. Hence the perpendicular distance is only $\frac{\text{AB}}{2}$.

Since the force F is the same in both cases and the perpendicular distance in Case 1 is twice that in Case 2, the moment of force is greater in Case 1.

(c) The ruler exhibits both translational and rotational motion simultaneously in Case 2, because the ruler is free to move, so the force makes its centre of gravity move forward (translational motion) and at the same time turns the ruler about the centre of gravity (rotational motion).

(a) Let a machine overcome a load L by the application of an effort E. In the process, let the displacement of the effort be d_E and the displacement of the load be d_L.

Then,

Work input = effort x displacement of effort = E x d_E ...(i)

Work output = load x displacement of load = L x d_L ...(ii)

By definition,

$\text{Efficiency }\text{η} = \frac{\text{work output}}{\text{work input}}$

From eqns. (i) and (ii),

$$\begin{aligned}\text{η} = \frac{\text L \times \text d_\text L}{\text E \times \text d_\text E} = \frac{\text L}{\text E} \times \frac{\text d_\text L}{\text d_\text E} \\[1em] = \frac{\text L}{\text E} \times \frac{1}{\frac{\text d_\text E}{\text d_\text L}}\end{aligned}$$

But,

$\frac{\text L}{\text E}$ = M.A. and $\frac{\text d_\text E}{\text d_\text L}$ = V.R.

$$\begin{aligned}\therefore \text{η} = \frac{\text{M.A.}}{\text{V.R.}} \\[1em] \Rightarrow \text{M.A.} = \text{V.R.} \times \text{η}\end{aligned}$$

Thus, the mechanical advantage of a machine is equal to the product of its efficiency and velocity ratio.

(b)

1. It is a speed multiplier.

2. Reason — In a class III lever the effort is applied in between the fulcrum and the load, so the effort arm is always shorter than the load arm. Hence its mechanical advantage and velocity ratio are always less than 1. Such a lever does not give a gain in force, but it gives a gain in speed, i.e., a larger displacement of the load is obtained by a smaller displacement of the effort.

As, in a block and tackle system,

V.R. = total number of pulleys used in both the blocks

So, for V.R. = 5, a total of five pulleys are needed. For the convenience of applying the effort in the downward direction, the upper (fixed) block is given three pulleys and the lower (movable) block is given two pulleys, and one end of the string is tied to the hook of the lower (movable) block.

The labelled diagram of the system is shown below :

Here, the load L is attached to the hook of the lower movable block and the effort E is applied at the free end of the string in the downward direction. The tension T is the same all along the string, and the load is supported by the tension in the five segments of the string, so

L = 5T and E = T

(a) Resonance.

(b) A high rise building is tall, so its natural time period of vibration is large (i.e., its natural frequency is low). When long period seismic waves reach such a building, the time period of these waves becomes equal to the natural time period of the building. The building therefore vibrates under resonance and the amplitude of its vibration becomes very large, so the building gets damaged.

Given,

• Time after which the first echo is heard (t₁) = 3 s
• Distance between the two cliffs = 1760 m
• Speed of sound (v) = 320 m s^-1

The first echo is heard from the nearer cliff.

Let the distance of the person from the nearer cliff be d₁.

$$\begin{aligned}\text{Speed of sound} = \frac{2\text d_1}{\text t_1} \\[1em] \Rightarrow 2\text d_1 = \text v \times \text t_1 = 320 \times 3 = 960\ \text m \\[1em] \Rightarrow \text d_1 = \frac{960}{2} = 480\ \text m\end{aligned}$$

Then,

Distance of the person from the farther cliff (d₂) = 1760 - 480 = 1280 m

Let the second echo be heard after time t₂.

$$\begin{aligned}\text t_2 = \frac{2\text d_2}{\text v} = \frac{2 \times 1280}{320} \\[1em] = \frac{2560}{320} = 8\ \text s\end{aligned}$$

Hence, he will hear the second echo 8 s after producing the sound, i.e., 5 s after the first echo.

Given,

• E.m.f. of the cell (ε) = 6 V
• Internal resistance of the cell (r) = 0.6 Ω
• Resistances in parallel = 6 Ω and 4 Ω
• Resistance in series = 3 Ω

Let the resultant of the two parallel resistances be R_P.

$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{6} + \frac{1}{4} \\[1em] \Rightarrow \frac{1}{\text R_\text P} = \frac{2 + 3}{12} = \frac{5}{12} \\[1em] \Rightarrow \text R_\text P = \frac{12}{5} = 2.4\ \text{Ω}\end{aligned}$$

(a) Now R_P is in series with the 3 Ω resistor and the internal resistance of the cell.

Total resistance of the circuit (R_S) = 2.4 + 3 + 0.6 = 6 Ω

Current drawn from the cell = $\frac{\text ε}{\text R_\text S} = \frac{6}{6}$ = 1 A

Since the 3 Ω resistor is in series with the cell, the whole of this current passes through it.

Hence, the current through the 3 Ω resistor is 1 A.

(b) When the 3 Ω resistance is replaced by a connecting wire of negligible resistance,

Total resistance of the circuit = 2.4 + 0.6 = 3 Ω

Current drawn from the cell = $\frac{\text ε}{\text R} = \frac{6}{3}$ = 2 A

Hence, the current drawn from the cell is 2 A.

(a) Graph C shows the correct relation.

Reason — An ohmic resistor obeys Ohm's law, according to which the potential difference across the resistor is directly proportional to the current flowing through it (V ∝ I), provided the physical conditions remain unchanged. A directly proportional relation is represented by a straight line passing through the origin, which is the case only in graph C.

In graph A the line is straight but it does not pass through the origin, and in graph B the potential difference decreases with increase in current. Hence neither A nor B represents an ohmic resistor.

(b) All metallic conductors are ohmic resistors, for example a copper wire (or a resistor made of an alloy such as manganin or constantan).

(a) 1. B should be the live wire.

Reason — From the diagram, the two dual control (two-way) switches used to operate the oven from two different places are connected in the wire that leads to the terminal B. A switch must always be connected in the live wire, so that when the oven is switched off, it gets disconnected from the high potential point of the supply and it is then safe to touch the appliance. If the switches were in the neutral wire, the oven would remain connected to the live wire even when switched off, which is dangerous.

2. No, the fuse will not serve its purpose.

Reason — In the given circuit the fuse has been connected in the wire joining the metal body of the oven to the earth terminal C, and not in the live wire. During an overload, the excessive current flows through the live wire and the oven coil, while the earth wire normally carries no current. Hence this fuse will not melt and the circuit will not be broken. A fuse is always connected in the live wire before the appliance, so that when the current exceeds the rated value, the fuse melts and prevents the heavy current from reaching the appliance.

(b) The statement 'A bulb is rated 600 W, 220 V' means that if the bulb is connected across a supply of 220 V, it will consume 600 W of electrical power, i.e., it will consume 600 joule of electrical energy in one second.

Given,

• Power rating of each bulb (P) = 40 W
• Voltage rating of each bulb (V) = 12 V
• E.m.f. of the battery = 12 V

(a) As,

$$\begin{aligned}\text P = \frac{\text V^2}{\text R} \\[1em] \Rightarrow \text R = \frac{\text V^2}{\text P} = \frac{12^2}{40} = \frac{144}{40} = 3.6\thinspace\text{Ω}\end{aligned}$$

Hence, the resistance of each bulb is 3.6 Ω.

(b) 1. When S₁ is closed and S₂ and S₃ are open :

The current flows from the battery through the first bulb, then through the closed switch S₁ and then through the second bulb. Hence the two bulbs are connected in series across the 12 V battery.

Total resistance (R_s) = 3.6 + 3.6 = 7.2 Ω

Total power consumed = $\frac{\text V^2}{\text R_s} = \frac{12^2}{7.2} = \frac{144}{7.2}$ = 20 W

Hence, the total power consumed by the circuit is 20 W.

2. When S₁ is open and S₂ and S₃ are closed :

The closed switch S₃ provides a path from one terminal of the battery directly to the second bulb, and the closed switch S₂ provides a path from the first bulb directly to the other terminal of the battery. Hence each bulb now gets connected directly across the 12 V battery, i.e., the two bulbs are connected in parallel.

So, total resistance:

$$\begin{aligned}\frac{1}{\text R_p}=\frac{1}{3.6}+ \frac{1}{3.6}\\[1em] \frac{1}{\text R_p} = \frac{2}{3.6}\\[1em] \text R_p= \frac{3.6}{2}\\[1em] \text R_p=1.8\thinspace\text{Ω}\end{aligned}$$

Total power consumed = $\frac{\text V^2}{\text R_p} = \frac{12^2}{1.8} = \frac{144}{1.8}$ = 80 W

Hence, the total power consumed by the circuit is 80 W.

(a) The magnetic lines of force around a straight wire carrying current are concentric circles, lying in a plane perpendicular to the wire, with their centres on the wire, as shown below :

(b) On increasing the current through the wire, the strength of the magnetic field increases. The magnetic field lines become denser (i.e., more crowded) near the wire, and the field remains effective up to a larger distance from the wire.

(a) The process is called carbon dating.

(b) ₆C¹⁴ and ₆C¹² are collectively known as isotopes (of carbon).

(c) ₆C¹⁴ is radioactive and it goes on decaying into nitrogen by the emission of β-particles, while ₆C¹² is stable and its amount does not change.

While the organism is alive, it keeps taking in carbon from its surroundings, so the ratio ₆C¹⁴ : ₆C¹² in it remains the same as in a fresh sample. After its death this intake stops, so the ₆C¹⁴ present in it goes on decreasing due to decay while the ₆C¹² remains unchanged. Hence in the fossil the ratio ₆C¹⁴ : ₆C¹² is less than that in a fresh sample.

(d) In a nuclear reaction, the mass number and the atomic number are conserved on both sides.

Let the atomic number of X be Z and its mass number be A.

Equating the mass numbers :

A + 1 = 14 + 1

⇒ A = 14

Equating the atomic numbers :

Z + 0 = 6 + 1

⇒ Z = 7

The element with atomic number 7 is nitrogen (N).

Hence, the complete equation is ₇N¹⁴ + ₀n¹ ⟶ ₆C¹⁴ + ₁P¹.

positive

Reason — Conventionally, if the effect on the body is to turn it anticlockwise, the moment of couple is called anticlockwise moment and it is taken positive, while if the effect on the body is to turn it clockwise, the moment of couple is called clockwise moment and it is taken negative.

momentum

Reason — As,

Kinetic Energy = $\frac{1}{2}$ mv²

And,

Momentum (p) = mv

Then,

$$\begin{aligned}\text {Kinetic Energy} = \frac{1}{2\text m}\text m^2 \text v^2 \\[1em] = \frac{1}{2\text m}\text p^2\end{aligned}$$

So, the kinetic energy of a given body depends on its momentum.

heat → mechanical → electrical

Reason — Coal's chemical energy is first released as heat in the boiler (steam). During power production, that heat drives the turbine (mechanical energy), and the generator converts it to electrical energy. Hence the sequence shown is heat → mechanical → electrical.

Velocity Ratio remains unchanged and efficiency increases.

Reason — For an actual single movable pulley, due to (i) friction in the pulley bearings or at the axle, and (ii) the weight of the pulley and string, the effort needed to lift up a load L will be more than $\frac {\text L}{2}$, so the mechanical advantage will be less than 2. Since the velocity ratio remains 2, hence the efficiency will always be less than 1 or 100%.

As velocity ratio for a single movable pulley always remains constant (equal to 2) so on lubricating the pulley the frictional forces reduces which increases the mechanical advantage but velocity ratio remains the same and since efficiency is given by

$\text {Efficiency} = \frac {\text {Mechanical Advantage}}{\text {Velocity Ratio}}$

so it will also increases.

So, on lubricating the pulley, mechanical advantage and efficiency increases but velocity ratio remains constant.

slows down less and refracts less

Reason — In a prism, the refractive index of a given transparent medium decreases with the increase in the wavelength of light. Thus, for visible light the refractive index of the material of a prism is greater for blue colour and minimum for red colour i.e., speed of red colour is greater than speed of blue colour. Consequently, a given prism deviates blue light more as compared to red light.

refraction

Reason — The shimmering effect around a campfire is primarily caused by the refraction of light through varying temperature and density layers in the air.

away from the lens.

Reason —

Given,

Focal length (f) = +12 cm

Object distance (u) = -20 cm

Blurred image distance (v') = +23 cm

From the given information we can conclude that the object is placed between F₁ and 2F₁ of the convex lens. So, the image will be formed beyond 2F₂ i.e., the image distance will be greater than 24 cm.

Hence, in order to obtain the clear image, the screen should be moved away from the lens.

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

Explanation

Assertion (A) is true because poor visibility can cause misjudgment of target location, increasing the chance of accidental damage or friendly fire.

Reason (R) is true because fog contains tiny water droplets that cause scattering of light which reduces visibility.

Since, the reduced visibility caused by scattering is precisely why soldiers avoid firing as they can’t see the target clearly so here reason clearly explains the assertion.

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.

current through 2Ω = current through 5Ω

Reason —

Given,

Emf of the cell (ε) = 12 V

As, same current flows through 3 Ω and 2 Ω resistances so they are connected in series and their net resistance will be :

R_S = 2 + 3 = 5 Ω

Since, R_S and 5 Ω have same potential difference so they are in parallel combination.

As resistance in both the arms is the same (5 Ω) so equal current flows through (3 Ω + 2 Ω) combination and 5 Ω.

Hence, current through 2Ω = current through 5Ω.

green

Reason — The colour coding is given below :

$\frac{\text {power}_\text{primary}}{\text {power}_\text{secondary}} = 1$

Reason — The step up transformer is used to change a low alternating voltage to a high alternating voltage (of same frequency) i.e.,

V_Secondary > V_Primary

⟹ $\frac{\text {Voltage}_\text{secondary}}{\text {Voltage}_\text{primary}} \gt 1$

Also,

In a step up transformer, the number of turns in the secondary coil is more than the number of turns in the primary coil i.e.,

number of turns_Secondary > number of turns_Primary

⟹ $\frac{\text {number of turns}_\text{secondary}}{\text {number of turns}_\text{primary}} \gt 1$

And

The current in primary coil is more than in the secondary coil i.e.,

Current_primary > Current_secondary

⟹ $\frac{\text {Current}_\text{primary}}{\text {Current}_\text{secondary}} \gt 1$

As, for an ideal transformer, when there is no energy loss, the output power will be equal to the input power. i.e.,

power_primary = power_secondary

⟹ $\frac{\text {power}_\text{primary}}{\text {power}_\text{secondary}} = 1$

energy needed to raise the temperature of a body by 1°C

Reason — The heat capacity of a body is the amount of heat energy required to raise its temperature by 1°C (or 1 K).

latent heat of fusion

Reason — Heat energy absorbed in change of phase that is not externally manifested by any rise in temperature is called the latent heat of fusion.

protons = 93, neutrons = 144

Reason — An alpha particle consists of two protons and two neutrons. Emission of alpha particle reduces the number of protons by 2 and also the number of neutrons by 2. Hence, number of protons = 95 - 2 = 93 and number of neutrons = 146 - 2 = 144.

(a) A class II lever will always have M.A. > 1.

(b) A boy uses a GPS device to locate his missing friend in a crowded area; the system primarily uses microwaves to track the location.

(c) Unit of specific heat capacity is m²s^-2K^-1.

(d) The threshold of hearing is 0 dB.

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

(f) The reaction responsible for the production of energy in the sun is Nuclear Fusion.

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) Ultrasonic waves.

(b) Ultrasonic waves are used for echo depth sounding because they travel long distance without deviation and is not easily absorbed by the medium or can be confined to a narrow beam.

(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.

(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.

Given,

Mass of solid (m_s) = 60 g

Mass of water (m_w) = 150 g

Fall in temperature of solid (△t_s) = (100 - 25) = 75°C

Rise in temperature of water (△t_w) = (25 - 20) = 5°C

Specific heat capacity of water (c_w) = 4.2 Jg^-1K^-1

Let specific heat capacity of solid be c_s.

Then,

Heat energy given by solid = m_sc_s△t_s = 60 x c_s x 75 = 4500 x c_s ......(1)

Heat energy taken by water = m_wc_w△t_w = 150 × 4.2 × 5 = 3150 ......(2)

Assuming that there is no loss of heat energy,

Heat energy given by solid = Heat energy taken by water.

Equating equations 1 & 2, we get,

4500 × c_s = 3150

$⇒ \text c_\text s = \frac{3150}{4500}$

⇒ c_s = 0.7 J g^-1 K^-1

Now,

Heat capacity of solid = mass x specific heat capacity = 60 x 0.7 = 42 J K^-1

Hence, Heat capacity of solid = 42 J K^-1.

(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) Gamma (γ) radiation.

(b) No, the emission of gamma radiation does not result in a change in the mass number as they are a form of electromagnetic radiation, essentially high-energy photons, and they have no mass or charge. Since they are massless, their emission does not affect the number of protons or neutrons in the nucleus, and therefore the mass number remains unchanged.

(a) Total internal reflection.

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

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

(a) Given,

• Actual depth of the object = 25 cm
• Refractive index of the glycerin = μ = 1.25
• The ruler is seen to be touching the object visually, so it is lowered up to the apparent position of the object (i.e., the image due to refraction)

As,

$\text {Apparent depth} = \frac{\text {Real depth}}{\text μ} = \frac{25}{1.25}$ = 20 cm

So, the ruler will get wet up to the 20 cm mark.

(b) As,

$\text {Apparent depth} = \frac{\text {Real depth}}{\text μ}$

Here, apparent depth depends inversely on refractive index (μ) of the medium so if another liquid of μ > 1.25 is used then apparent depth decreases.

Hence, the wet length of the ruler will be less than 20 cm.

(a) The ray will pass through different points on principal axis after refraction through the convex lens as the refractive index of a medium decreases with the increase in wavelength so red colour will deviate the least, followed by green and then blue will deviate the most.

(b) Ultraviolet radiations.

(c) Ultraviolet radiations are used for sterilizing air, surgical equipment, etc.

(d) Infrared radiation.

(a) As, the object distance (= 24 cm) is greater than the focal length (= 8 cm) of the convex lens which means the object placed beyond it's focus so the lens will form a real and inverted image.

(b)

Given,

Object distance (u) = -24 cm

Focal length (f) = +8 cm

From lens formula,

$$\begin{aligned}\frac{1}{\text f} = \frac{1}{\text v} - \frac{1}{\text u} \\[1 em] ⇒ \frac{1}{\text v} = \frac{1}{\text f} + \frac{1}{\text u} \\[1 em] ⇒ \frac{1}{\text v} = \frac{1}{8}+\frac{1}{(-24)} \\[1 em] ⇒\frac{1}{8}-\frac{1}{24}= \frac{3-1}{24}\\[1 em] ⇒\frac{1}{\text v} = \frac{2}{24} \\[1 em] ⇒ \text v = \frac{24}{2} = 12 \text { cm}\end{aligned}$$

∴ The image is formed at a distance of 12 cm behind the lens.

(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

$\frac {\text W_\text {BC}}{\text W_\text {AC}} \gt \frac{5}{3} \gt 1$

⇒ 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, 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 such that 0° ≤ θ < 90°, so the work done on the car by the force of gravity is positive.

Given,

• Diameter (2R) = 360 cm = 3.6 m
• Radius (R) = 1.8 m
• Mass of the sphere (m) = 500 g = 0.5 kg
• AE = $\frac {\text R}{3} = \frac {1.8}{3}$ = 0.6 m
• Acceleration due to gravity (g) = 10 m s^-2

Condition : No energy loss so energy is conserved.

(a) As, point B is on the ground then

Height of point A relative to point B = radius of hemisphere (R) = 1.8 m

Potential energy U_A at A relative to B is given by

U_A = m x g x R

Substituting the values we get,

U_A = 0.5 x 10 x 1.8

⇒ U_A = 9 J

Hence, the potential energy at point A relative to point B is 9 J.

(b) As there are no energy losses so total mechanical energy is same at all points of the path due to conservation of mechanical energy and all the potential energy at point A will be converted to kinetic energy at point B.

Let, speed at point B be 'v_B'.

Then,

Kinetic energy at B = U_A

$$\begin{aligned}\Rightarrow \frac{1}{2} \text m {\text v_ \text B}^2 = 9 \\[1em] \Rightarrow \frac{1}{2} \times 0.5 \times {\text v_ \text B}^2 = 9 \\[1em] \Rightarrow {\text v_ \text B}^2 = \frac{2\times 9}{0.5} = 36 \\[1em] \Rightarrow \text v_ \text B = \sqrt{36} = 6 \text { m s}^{-1} \\[1em]\end{aligned}$$

So, speed at point B is 6 m s^-1.

(c) As points D and E are on the same height so at these points their potential and kinetic energies are equal i.e.,

Potential energy at point D = Potential energy at point E

And

Kinetic energy at point D = Kinetic energy at point E

Now,

Distance of point D from the ground = R - AE = 1.8 - 0.6 = 1.2 m

Thus, potential energy at point D = mg x 1.2 = 0.5 x 10 x 1.2 = 6 J

Mechanical energy at point D = Potential energy at point D + Kinetic energy at point D

As, mechanical energy is conserved,

Mechanical energy at point D = Potential energy at point A

⇒ Potential energy at point D + Kinetic energy at point D = 9

⇒ 6 + Kinetic energy at point D = 9

⇒ Kinetic energy at point D = 9 - 6 = 3 J

So, kinetic energy at point D is 3 J.

(a) The labelled diagram of the system indicating clearly, the direction of the load and effort is shown below :

(b) Given,

m = 100 kg

h = 5 m

g = 10 m s^-2

Then,

Potential Energy (U) = mgh = 100 × 10 × 5 = 5000 J

(a) Given,

Time taken to hear the echo = 3s

Speed of sound = 336 ms^-1

So,

$$\begin{aligned}\text {Distance of the person from the cliff} = \frac {\text {Speed of sound} \times \text {Time taken}}{2} \\[1 em] =\frac{336 \times 3}{2} = \frac{1008}{2} \\[1 em] = 504\ \text m\end{aligned}$$

So, distance of the person from the cliff is 504 m.

(b) Given,

Time taken to hear the new echo = 3 s + 1.5 s = 4.5 s

So,

$$\begin{aligned}\text {New distance of the person from the cliff} = \frac {\text {Speed of sound} \times \text {Time taken}}{2} \\[1 em] =\frac{336 \times 4.5}{2} = \frac{1512}{2} \\[1 em] = 756\ \text m\end{aligned}$$

Distance moved by the person = 756 - 504 = 252 m

So, distance moved by the person is 252 m.

(a) At position A, the mother should give a constant periodic push to the swing every time in the forward direction to increase the amplitude of the swing because it is an extreme position where speed of the swing is zero and direction of motion changes from backward direction to forward so at this position direction of motion and direction of push are in parallel while for C it is opposite and B speed is maximum on either direction which can collide with mother and hurt her.

(b) Resonance phenomenon.

(c) The natural frequency of the swing will match the frequency of the force applied by mother this will increase the amplitude of the swing.

(a) As the resistance depends directly on length and inversely on area of the resistor and since here tube 2 is thicker and shorter than tube 1 thus, resistance of the tube 2 < resistance of the tube 1.

(b) The voltmeter reading for tube 1 is the same as the voltmeter reading for tube 2 because according to Ohm's law potential difference across a resistor depends only on the current applied not on the resistance of the resistor.

(c) As the specific resistance of a resistor is independent of its dimensions but depends only on the contained material and since here both tubes contain only mercury thus, the specific resistance in both the cases is the same.

Given,

• Power rating (P) = 2 KW = 2000 W
• Voltage rating (V) = 220 V

(a) The safest circuit is S.

(b) There are two criteria to choose the circuit S :

1. It is earthed so that in the case short circuit all the excess current can flow to earth without harming the person.
2. The current in the supply should be less than or equal to the current rating of the fuse.

So, current in the circuit = $\frac {\text P}{\text V} = \frac {2000}{220}$ = 9.09 A

Circuit P

As the circuit is not earthed so it is not safe.

Circuit Q

• Circuit is earthed properly.
• Current in the circuit is 9.09 A but it has fuse which has current rating of 8 A so it will get damaged and hence someone can be hurt.

Circuit R

• Circuit is not earthed properly so it is not safe.
• Current in the circuit is 9.09 A and it has fuse which has current rating of 10 A.

Circuit S

• Circuit is earthed properly.
• Current in the circuit is 9.09 A and it has fuse which has current rating of 10 A.

So circuit S satisfy all the necessary criteria for 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) As, the physical quantity indicated by 'a' is induced emf or output voltage of the a.c. generator so on doubling the speed of the generator, the output voltage also gets doubled.

(b) As, the physical quantity indicated by 'b' is time period of the generator so on doubling the speed, the period becomes half.

(c) When in an a.c generator the speed at which the coil rotates is doubled the frequency of output voltage is also doubled and since time period is reciprocal of the frequency so it gets halved.

(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.

have only rotational motion

Reason — Forces are equal and opposite, but not along same straight line.

So, the forces will form a couple and the body will have rotational motion only.

moment of couple

Reason — A couple has both magnitude and direction, making it a vector quantity. Its direction is perpendicular to the plane in which the two forces lie. The magnitude is calculated by multiplying one of the forces with the perpendicular distance between their lines of action.

chemical → heat + light

Reason — When a candle burns, chemical energy stored in the wax is transformed into heat and light energy through a combustion reaction.

Frequency of the wave

Reason — Frequency of wave depends on the source and not on the medium through which it propagates.

Gamma-rays, visible light, microwaves

Reason — Wavelength of gamma rays is less than 0.01 nm, wavelength of visible light ranges from 400 to 800 nm and of microwave ranges from 1 mm to 1 m.

Mechanical advantage

Reason — Load vs effort graph is a straight line and slope of the graph = Load/effort = Mechanical advantage.

beyond 2F

Reason — Here, real image size < object size and this happens when the object is placed beyond 2F.

60°

Reason — As, the ray is normally incident on face AB so it passes undeviated. Angle of incidence on face AC is 60˚. The reflected ray is incident normally on face BC and passes undeviated.

Now, from the above diagram,

∠NRS = 60˚

∴ ∠SRC = 30˚

∠ARQ = 30˚

∴ ∠CRT = 30˚

RT is the original path of the incident ray.

So, total deviation = ∠SRT = 30˚ + 30˚ = 60˚

V = ε – Ir

Reason — : The circuit is :

When current flows,

ε = V + Ir

∴ V = ε – Ir

0.5 J g⁻¹ K⁻¹

Reason — As,

$\text {Specific heat capacity} = \frac {\text {Thermal capacity}}{\text {mass}} = \frac{2.5\ \text J \text K⁻¹}{5\ \text g} = 0.5\ \text J\ \text g⁻¹\ \text K⁻¹$

(A) is true but (R) is false.

Explanation:

Assertion (A) is true because as the water column increases, air column decreases, frequency increases.

Reason (R) is false because the frequency is directly proportional to the length of water column.

parallel in both the circuits

Reason — Circuit Y may be redrawn as :

Both the resistors are connected across points A and B. Hence, those are connected in parallel.

Circuit Z may be redrawn as :

Both the resistors are connected across points A and B. Hence, those are connected in parallel.

Two groups to the left of P

Reason — The reaction is :

$$\begin{aligned}_{\text Z}^{\text A}\text{P}\ \longrightarrow \\ _{\text Z - 2}^{\text A - 4}\text{Q}\ + \\ _{2}^{4}\text{He}\end{aligned}$$

In periodic table elements are arranged in ascending order of atomic number and the atomic number of Q is 2 less than that of P. Hence, position of Q will 2 groups left of P.

Lead

Reason — The substance whose specific heat capacity is the lowest will attain the highest temperature.

anticlockwise when magnet is above the ring and clockwise when the magnet is below the ring

Reason

Magnet above the ring (approaching it)

• Flux change The north pole of the magnet faces the ring, so magnetic field lines enter the ring downwards. As the magnet falls closer, the downward flux through the ring increases.

• Opposition required To oppose this increase in downward flux, the ring must create an upward field (its own north pole must face the magnet).

• Current direction Using the right-hand rule, a coil whose top acts as a north pole carries anticlockwise current when viewed from above.

Result: Anticlockwise current while the magnet is above the ring.

Magnet inside and leaving the ring (moving away below it)

• Flux change Once the magnet has passed through, the same downward field lines are now moving away, so the downward flux decreases.

• Opposition required The ring tries to maintain the original flux by producing more downward field—i.e., its top must now behave like a south pole.

• Current direction A coil whose top is a south pole carries clockwise current when viewed from above.

Result: Clockwise current while the magnet is below the ring.

(a) In uniform circular motion the centrifugal force acts away from the centre.

(b) Refractive index of a medium is independent of angle of incidence.

(c) Heat absorbed during change of phase depends on mass of the substance.

(d) Emf of a cell is equal to the terminal voltage when the cell is in open circuit.

(e) In a step-up transformer the turns ratio is more than 1.

(f) The nuclear radiation with lowest ionising power is γ.

(a) R is the most probable position of its centre of gravity.

(b) When a body is freely suspended from a point, it comes to rest (i.e. stops oscillating) in such a position that its centre of gravity lies vertically below the point of suspension.
So, in the figure, the string from the branch passes vertically through point R, hence R is the most probable position of the kite’s centre of gravity.

(a) As PR is the time period of the vibration and reduction of time period means the increase of frequency. Also, Frequency is proportional to square root of the tension.

So, tension is to be increased to increase the frequency i.e., to reduce the time period.

(b) Since, pitch depends on frequency, pitch will be affected by reduction of PR.

This ray will not undergo total internal reflection because for total internal reflection, the light ray should travel from denser to rarer medium but in the given situation light ray travels from rarer to denser medium since refractive index of glass is more than that of water. So, there will be no total internal reflection.

(a) Earth wire is connected to the metal body of electrical appliances and according to new convention the colour of earth wire is green or yellow.

(b) Live wire is connected to the switch of an appliance and according to new convention the colour of live wire is brown.

(a) Alternating current produces a varying magnetic field when it flows through a conductor because alternating current changes its magnitude as well as direction periodically and thus, it provides a varying current and hence varying magnetic field as magnetic field is directly proportional to the applied current.

(b) 50 Hz.

Given,

Mass of paraffin (m) = 200 g

Specific latent heat of fusion of paraffin (L) = 146 J g⁻¹

Let, required heat be Q.

Then,

$\text Q = \text {mL} = 200 \times 146 = 29200\ \text J = 29.2\ \text {kJ}$

(a) Since, looking from the F side current is flowing in anti-clockwise direction and hence, north pole developed at that end and south pole at the T end.

The labelled diagram is given below :

(b)

(a) As current flows for a long period of time, heat is generated and temperature increases. So, the resistance of metallic wire increases as resistance is directly proportional to the temperature of a conductor.

(b) As temperature increases, resistivity of metallic wire increases.

(a) As, Curium has 96 protons and 247 – 96 = 151 neutrons and point P represents 96 protons and 151 neutrons. Hence, P point indicates element Curium.

(b) After the radioactive decay of 1 α and 2 β particles, the atomic number and mass number of daughter nucleus will be 243 and 96 respectively. So, it will have 243 – 96 = 147 neutrons.

As, point R represents 96 protons and 147 neutrons. Hence R point indicates the daughter nucleus.

(c) Mass number of daughter nucleus = Mass number of Curium – Mass number of α particle = 247 – 4 = 243

(a)

From the above diagram, it is visible that inside the prism, the angles of incidence of rays I, J and H are respectively 30˚, 60˚ and 30˚. Hence only the ray J will suffer total internal reflection.

(b)

(a) Given,

Refractive index of glass = 1.5

Apparent depth of air bubble = 4 cm

As,

$$\begin{aligned}\text {Refractive index of glass} = \frac{\text {Real depth of air bubble}}{\text {Apparent depth of air bubble}} \\[1em] ⇒ \text {Real depth of air bubble} = \text {Refractive index of glass}\times \text {Apparent depth of air bubble} \\[1em] = 1.5 \times 4 = 6 \text { cm}\end{aligned}$$

So, actual depth of the air bubble from the surface AB is 6 cm.

(b) As,

$$\begin{aligned}\text {Refractive index of glass} = \frac{\text {Real depth of air bubble}}{\text {Apparent depth of air bubble}} \\[1em] ⇒ \text {Real depth of air bubble} = \text {Refractive index of glass}\times \text {Apparent depth of air bubble} \\[1em]\end{aligned}$$

Now, apparent depth will increase, if refractive index decreases and as wavelength increases, refractive index decreases. As wavelength of yellow light is greater than wavelength of blue light so, refractive index for yellow light is smaller than refractive index of blue light.

Hence, for yellow light apparent depth will be greater.

(c) When the block is turned upside down, then the real depth of the bubble is 15 – 6 = 9 cm

And,

$$\begin{aligned}\text {Apparent depth of air bubble} = \frac{\text {Real depth of air bubble}}{\text {Refractive index of glass}}=\frac{9}{1.5} =6 \text { cm}\\[1 em]\end{aligned}$$

So, the statement is false.

(a) Ray diagram showing the formation of the image when the object is placed at 2F position of a convex lens is shown below:

(b) If we ONLY replace the original lens with another lens of greater focal length then object distance (u) which was at 2F₁ earlier, now comes between F₁ and 2F₁.

And

When an object is placed between F₁ and 2F₁ of the convex lens then it produces real, inverted and magnified image beyond 2F₂ behind the lens.

Hence, if we, ONLY replace the original lens with another lens of a greater focal length then the size of the image increases.

(a) Given,

Object distance (u) = -45 cm

Image distance (v) = -30 cm

From lens formula,

$$\begin{aligned}\frac{1}{\text f} = \frac{1}{\text v} - \frac{1}{\text u} \\[1em] \frac{1}{\text f} = \frac{1}{-30}-\frac{1}{(-45)}\\[1em] \frac{1}{\text f}=-\frac{1}{30}+\frac{1}{45}\\[1em] \frac{1}{\text f} = \frac{-3+2}{90}\\[1em] \frac{1}{\text f} = -\frac{1}{90} \\[1 em] \text f = -\frac{90}{1}\\[1 em] \text f= - 90 \text { cm}\end{aligned}$$

(a) Angle of incidence for both the rays being same, the directions of both the rays remain same. As the refracted ray corresponding to RS bends towards the normal is more than that corresponding to ray PQ hence, the lateral displacement of the emergent ray corresponding to incident ray RS will also be more.

Diagram shows, x₁ and x₂ are the lateral shifts of the emergent rays corresponding to incident rays PQ and RS.

So, here x₂ > x₁

(b) Since, the refracted ray corresponding to incident ray PQ bends less towards the normal, hence this travels with greater speed.

(c) Speed corresponding to incident ray PQ is more than the speed corresponding to incident ray RS. So, wavelength corresponding to incident ray PQ is more than the wavelength corresponding to incident ray RS. So, RS will scatter more in atmosphere.

(a)

1. For ultraviolet radiation, a quartz prism is used since, ordinary glass absorbs UV radiation.
2. Infra red radiation.
3. Visible light radiation (Red, Yellow, Green)

(b) All electromagnetic radiations travels with the same speed ie., 3 × 10⁸ ms^-1.

The diagram of the arrangement is shown below :

Given,

Weight at the end X (W₁) = 2 N

Weight at the end Y (W₂) = 5 N

Distance of X from the knife (R₁) = 70 cm

Distance of Y from the knife (R₂) = 100 cm - 70 cm = 30 cm

Let,

W = weight of the scale

Now,

Anticlockwise moment (due to 2N) = R₁ x W₁ = 2 × 70 = 140 Nm

Clockwise moment (due to 5N) = R₂ x W₂ = 5 × 30 = 150 Nm

Anticlockwise moment (due to weight of the scale) = W × 20 = 20 W

For equilibrium,

Anticlockwise moment of force about knife = Clockwise moment of force about knife

$$\begin{aligned}150 = 140 + 20\text { W} \\[1em] 20 \text W = 150 - 140 = 10 \\[1em] \text W = \frac{10}{20} = 0.5\ \text N\end{aligned}$$

So, weight of the scale is 0.5 N.

(a) All are class I levers since, the fulcrum is in between the load and effort.

(b) As,