Model Paper 1

maximum

Reason — The turning effect on the body about an axis is due to the moment of force (or torque) applied on the body which is given by :

Moment of force = Force × Perpendicular distance of the line of action of the force

For producing maximum turning effect on a body by a given force, the force is applied on the body at a point for which the perpendicular distance of the line of action of the force from the axis of rotation is maximum.

Zero

Reason — Work done is equal to the product of force, displacement and cosine of the angle θ between the directions of force F and displacement S i.e., W = F x S cosθ.

When a body moves in a circular path in a horizontal plane, no work is done since the centripetal force on the body at any instant is directed towards the centre of the circular path and the displacement at that instant is along the tangent to the circular path, i.e., normal to the direction of force on the body (i.e. θ = 90°)

Chemical to electrical

Reason — The electric cell maintains a constant difference in potential between the two conductors in the electrolyte by a chemical reaction to obtain a continuous flow of charge between them so a cell stores the chemical energy and it can be used as a source of current and hence chemical energy changes into electrical energy when the cell is in use.

Option (b)

Reason — Increasing order of penetrating power of different radiations is : α - particle < β - particle < γ - rays

violet light

Reason — When light travels from a denser medium to a rarer medium, refraction occurs and the object appears to be raised. The apparent shift depends on the refractive index, which in turn depends on the wavelength of light so shorter wavelengths (like violet) are refracted more because they have higher refractive indices and longer wavelengths (like red) are refracted less.

Reason — When white light passes through a prism, it splits into its constituent colours in such a way so that violet deviates the most (bends the most) because it has the shortest wavelength and highest refractive index while red deviates the least because it has the longest wavelength and lowest refractive index.

Option (d)

Reason — To measure the speed of sound using echo, the student must make a short, sharp sound so that the reflected sound (echo) can be clearly distinguished and the sound travels to the cliff and back, so the total distance travelled by the sound is 2 × the distance to the cliff.

large surface area of the vibrating body

Reason — When the stem of a vibrating tuning fork is pressed against the top of a table, the tuning fork produces forced vibrations in the table top. The table top has a much larger vibrating area than the tuning fork, so the forced vibrations of the table top send forth a greater energy and produces a louder (or more intense) sound than that produced by the tuning fork.

(i) & (iii)

Reason —

(i) Here all resistances are in parallel combination and have same resistance value equal to 3 Ω.

Then,

$\dfrac{1}{R_{eq}}=\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{3}{3}=1 \\[1em] \Rightarrow R_{eq} = 1 \Omega$

(ii) In this case two resistances of value 3 Ω are in parallel combination and this combination is in series with 6 Ω resistance.

For parallel combination,

$R_p = \dfrac{3\times3}{3+3}=\dfrac{9}{6}=\dfrac{3}{2}\ Ω$

For series combination,

$R_{eq} = R_p+6=\dfrac{3}{2}+6=\dfrac{15}{2}\ Ω$

(iii) Here two 1 Ω resistances are in series combination and this whole combination is in parallel combination with itself.

For series combination,

$R_s=1+1=2\ Ω$

For Parallel combination,

$R_{eq} = \dfrac{2\times2}{2+2}=\dfrac{4}{4}= 1\ Ω$

(iv) Here all resistances are in series combination.

$R_{eq}=3+2+3=8\ Ω$

The current at B is greater than that at C.

Reason — All bulbs have equal resistance then the two bulbs in series at C will have a combined resistance greater than the resistance of bulb B therefore, more current will flow through B than through bulb C and the total current from A splits between B and C. So,

(a) Current at B is greater than C — Correct, because B has lower resistance and gets more current.

(b) Current at A equals current at B — Incorrect, because A splits into B and C, so A's current cannot equal B's current.

(c) Current at C is greater than at B — Incorrect, because C has higher equivalent resistance and hence less current flows through C.

(d) Current at A equals current at C — Incorrect, because A's current splits into B and C.

both A and R are true and R is not the correct explanation of A

Explanation

Assertion (A) is true. Lenz’s Law states that the induced current opposes the cause producing it

Reason (R) is true. From Faraday’s law, which says a current is induced in a coil when there is a change in magnetic flux linked with the coil. However, Reason (R) is not the direct explanation of Assertion (A) because R explains the condition for induction, not the direction.

2 : 3

Reason — Let masses of P and Q be $\text{m}_\text{P}$ and $\text{m}_\text{Q}$ and their specific heat capacities $\text{c}_\text{P}$ and $\text{c}_\text{P}$ respectively. Let heat given be Q and temperature change is ΔT.

Given,

$\dfrac{\text{M}_\text{P}}{\text{M}_\text{Q}}=\dfrac{3}{2}$

As, $\text{Q = McΔT}$

According to question,

$\text{M}_\text{P}\text{c}_\text{P}Δ\text T=\text{M}_\text{Q}\text{c}_\text{Q}Δ\text T$

$\text{M}_\text{P}\text{c}_\text{P}=\text{M}_\text{Q}\text{c}_\text{Q}$

$\dfrac{\text{c}_\text{P}}{\text{c}_\text{Q}} = \dfrac{\text{M}_\text{Q}}{\text{M}_\text{P}}=\dfrac{2}{3}$

average kinetic energy of molecules increases

Reason — When a substance is heated, the molecules move faster, and their kinetic energy increases, which results in an increase in temperature.

Option (b)

Reason — Magnetic field is directly proportional to the current which means higher current gives higher magnetic field and inversely proportional to distance so smaller the distance to the wire, higher the magnetic field.

total internal reflection

Reason — Optical fibres work on the principle of Total Internal Reflection. As light entering the fibre is completely reflected back inside at each boundary of the fibre, without escaping, provided the angle of incidence is greater than the critical angle which allows the light to travel long distances through the fibre with minimal loss of energy or information, making it perfect for devices like endoscopes.

(a) Bottle opener is an example of class II lever.

(b) If there are n movable pulley with one fixed pulley, the mechanical advantage is given by 2ⁿ.

(c) According to work energy theorem, the increase in the kinetic energy of a moving body is equal to work done by a force acting in the direction of moving body.

(d) If a machine does a given amount of work in less time, more power is spent by it.

(e) If the length I of wire is increased to 31 by stretching its resistance R increases to 9R.

(f) If a coil is viewed from one end, and the current flows in clockwise direction, then the end is a south pole.

Reason —

(a) In a bottle opener, the fulcrum and the effort are at the two ends of it and the load is somewhere in between the effort and the fulcrum in such a way so that the load and effort are on the same side of the fulcrum but acting in opposite directions same as class II lever so here the effort arm is always longer than the load arm.

(b) As each movable pulley doubles the mechanical advantage by halving the effort required, and these effects multiply as more movable pulleys are added like a single movable pulley which divides the load between 2 rope segments which make the effort needed only half the load and has mechanical advantage equal to 2 so similarly n movable pulleys divide the load between 2n rope segments which make mechanical advantage equal to 2ⁿ.

(c) Work energy theorem states that the increase in kinetic energy of a moving body is equal to the work done by a force acting in the direction of the moving body.

(d) Power spent by a source is given by the amount of work done by the source per unit time taken by the source to do the said work i.e., $\text{Power} = \dfrac{\text {Work done}}{\text {Time taken}}$ So, if a machine does a given amount of work in less time then more power is spent by it.

(e) The resistance of a wire is directly proportional to its length and inversely proportional to the area of cross section of the wire and is given as :

$\text R = \text ρ\dfrac{\text l}{\text a}$

When length of a conductor is increased by 3 times then it's cross sectional area decreases to 1/3 to keep volume constant therefore, resistance increases by 3 × 3 = 9 times.

(f) The polarity at the two faces of a coil depends on the direction of current in the coil so while looking at the face of the coil, if the current in wire around that face is in anticlockwise direction, the face has the north polarity, while if the current at that face is in clockwise direction, the face has the south polarity.

A compressed spring possesses elastic potential energy due to its compressed state. When the spring is released, the potential energy of the spring changes into kinetic energy which does work on the ball placed on it and changes into the kinetic energy of the ball due to which it flies away.

(a) SONAR stands for sound navigation and ranging. It is a technology that uses ultrasonic sound waves to detect and locate objects under water.

(b) Principle of SONAR is when ultrasonic waves are sent in sea water in all directions from the ship then these waves are received after reflection from an obstacle such as an enemy submarine, iceberg, sunken ship, etc. so to find the distance of the obstacle from the ship, the time interval between the instant when waves are sent and the instant when the waves are received, after reflection from the obstacle is measured.

(a) The lens is a convex lens. A convex lens forms an upright and magnified image, when the position of the object is between the lens and F₁.

(b) Below labelled ray diagram shows the image formation:

(a) Miniature circuit breaker (MCB).

(b) Live wire.

(a) As this transformer converts 220 V of inputs into 44 V of output so it is reducing the voltage. Hence, it is a step down transformer.

(b) The core is made from thin rectangular laminated sheets of soft iron of T and U shape, placed alternately one above the other and insulated from each other by a paint (or varnish) coating over them which forms a simple rectangular core.

Metal B should be preferred for calorimeter because it requires a material with low specific heat capacity so the amount of heat energy taken by the calorimeter itself from the contents to acquire its temperature, is very small.

(a) If the current carrying conductor is placed between the poles of magnet with its length in a direction such that the current in it is in the direction parallel to the direction of magnetic field then zero force acts on the wire and it does not move.

(b) If the current carrying conductor is placed between the poles of magnet with its length in a direction such that the current in it is in the direction perpendicular to the direction of magnetic field then maximum force acts on the wire.

(i) Ionizing power → γ < β < α

(ii) Penetrating power → α < β < γ

(a) Nucleus A emits alpha (α) radiation and nucleus B emits beta (β) radiation.
As particles emitted by nucleus A deflect to left hence, they are alpha (α) particles and nucleus B emission deflects to the right hence, they are beta (β) particles.

(b) Fleming's left hand rule.

(c) Nucleus C will have 83 protons and 125 neutrons.

Explanation:

Given,

Before any decay :

Number of protons in A = 84 = Atomic number of A

Number of neutrons in A = 128

Mass number = 84 + 128 = 212

After alpha decay (A → B) :

Loss of protons = 2

Loss of neutrons = 2

New atomic number = 84 – 2 = 82

New mass number = 212 – 4 = 208

After beta decay (B → C) :

As in beta decay, a neutron converts into a proton so that Atomic number increases by 1 and mass number remains the same.

New atomic number = 82 + 1 = 83

Mass number = 208

For C

Number of protons = atomic number of C = 83

Number of neutrons = Mass number - number of proton = 208 - 83 = 125

(a) The pencil appears bent at the water surface; the submerged part looks raised and slightly shortened.

(b) This happens due to refraction of light. Light rays from the submerged pencil bend away from the normal as they pass from water (denser) to air (rarer), making the pencil appear at a higher position than it actually is

(c) Image of candle is a real image as it can be formed on a screen.

Given,

Refractive index of glass with respect to air = _aμ_g = 1.5 = $\dfrac{1}{_\text g\text μ_\text a}$

Angle made by PA with normal = 0°

Angle made by PB with normal = 30°

Angle made by PC with normal = 42°

Angle made by PD with normal = 60°

sin 42° = 2/3

(i) For ray PA

i = 0°

$\text {sin r} = \dfrac{\text{sin i}}{{_g\text{μ}_a}}=\text {sin }0°\times 1.5 =0$

⟹ r = sin^-1(0) = 0°

Ray PA will not deviate and emerge straight.

(ii) For PB

i = 30°

$\text {sin r} = \dfrac{\text{sin i}}{{_\text{g}\text{μ}_a}}=\text {sin}\ 30°\times 1.5 = 0.5\times1.5=0.75$

⟹ r = sin^-1(0.75) ≈ 49°

Ray PB will emerge at an angle of 49° with the normal.

(iii) For PC

i = 42°

$\text {sin r} = \dfrac{\text{sin i}}{{_\text{g}\text{μ}_a}}=\text {sin}\ 42°\times 1.5 = \dfrac{2}{3} \times1.5=1$

⟹ r = sin^-1(1) = 90°

This is the condition for critical angle.

Ray PC will graze along the surface.

(iv) For PD

i = 60°

$\text {sin r} = \dfrac{\text{sin i}}{{_\text {g}\text{μ}_a}}=\text {sin}\ 60°\times 1.5 = \dfrac{\sqrt 3}{2} \times1.5≈1.3$

Here,

sin r > 1 which a condition of total internal reflection.

Ray PD will suffer total internal reflection.

(a) Below ray diagram shows the formation of image I at O itself:

(b) The size of the image is same as that of the object O.

(c) A convex lens of short focal length is used in a microscope to magnify small objects.

(d) Given,

Focal length of convex lens = f = + 25 cm = + 0.25 m

Power of a lens is expressed as reciprocal of its focal length in metre and written as :

$\text P=\dfrac{1}{\text f}=\dfrac{1}{0.25}=+4 D$

Power of the lens is +4 Dioptre

Given,

Focal length of the lens (f) = 15cm

Magnification of the lens (m) = +3 ∵ (image formed is virtual and erect)

(a) This is a convex lens because only convex lens forms a magnified image when the object is between the lens and the focus.

(b) Let,

Distance of object from the lens = u

Distance of image from the lens = v

As for lens,

$\text m = \dfrac{\text v}{\text u} \\[1em] \Rightarrow \dfrac{\text v}{\text u}=3$

⇒ v = 3u .......... (1)

From lens formula,

$\dfrac{1}{\text v}-\dfrac{1}{\text u}=\dfrac{1}{\text f}$

On putting values

$\Rightarrow \dfrac{1}{3\text u}-\dfrac{1}{\text u}=\dfrac{1}{15} \\[1em] \Rightarrow \dfrac{1}{3\text u}-\dfrac{3}{3\text u}=\dfrac{1}{15} \\[1em] \Rightarrow -\dfrac{2}{3\text u}=\dfrac{1}{15} \\[1em] \Rightarrow \text u = \dfrac{-15\times 2}{3}=-10 \text{ cm}$

The object is placed 10 cm in front of the lens.

(c) From (1)

$\text v = 3\text u =3\times (-10)=-\ 30 \text{ cm}$

Distance of the screen from the lens = image distance (v) = -30 cm

The screen is placed 30 cm in front of the lens.

(a) Below diagram shows the effect of the prism on the beam and the spectrum seen on the screen :

(b) When a slit is placed in between the prism and the screen to pass only the light of yellow colour then only yellow light is observed on the screen because the slit blocks other colours (blue and red).

(c) From above experiment we can conclude that the prism does not produce colours, but it only splits the various colours present in the light incident on it.

Below is the completed diagram showing the path of the light ray till it emerges out of the prism with all angles marked:

(a)

Angle of incidence at face AB = 90°
Angle of incidence at face AC = 60°
Angle of incidence at face BC = 90°

(b)

At face AB and AC — The ray of light passes undeviated
At face BC — The ray of light suffers total internal reflection

(a) Given,

Fulcrum (F) is at 30 cm mark.

A 40 gf weight is hanging at 5 cm mark.

Let the weight of the metre scale be W gf, acting at its centre of gravity which is at it's midpoint i.e., at 50 cm mark.

Moment of force due to 40 gf weight:

Distance from F = 30 – 5 = 25 cm (anticlockwise)

So, Anticlockwise moment of force = 40 × 25 = 1000 gf cm

Moment due to the weight of the scale :

Weight W gf acts at 50 cm mark.

Distance from F = 50 – 30 = 20 cm (clockwise)

So, Clockwise moment of force = W × 20 gf cm

Using principle of moments:

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

45 x 25 = W x 20

⇒ 1000 = 20 W

⇒ W = $\dfrac{1000}{20}$ = 50 gf

∴ Weight of the metre scale = 50 gf

(b) Now the 40 gf weight is shifted to the 0 cm mark.

New moment of force due to 40 gf at 0 cm :

Distance from F = 30 – 0 = 30 cm (anticlockwise)

Anticlockwise moment of force = 40 × 30 = 1200 gf cm

Moment of force due to the metre scale (still acting at its centre of gravity i.e., at 50 cm mark) :

Distance from F = 50 – 30 = 20 cm (clockwise)

Clockwise moment of force = 50 × 20 = 1000 gf cm

Here,

Anticlockwise moment of force > Clockwise moment of force, so the scale will tilt anticlockwise unless we correct this.

To balance, the fulcrum F must be shifted towards 0 cm mark to reduce the arm length of the 40 gf weight.

∴ F is shifted towards 0 cm mark.

(a) A man pushes a wall. — No
Reason: There is no displacement of the wall, so no work is done.

(b) A coolie stands with a box on his head for 15 minutes. — No
Reason: The coolie is not moving, so there is no displacement.

(c) A boy climbs up 20 stairs. — Yes
Reason: There is displacement in the direction of force (against gravity), so work is done.

Given,

Mass = 60 kg

(a)

$\text {Loss in potential energy} = \text{mg} (\text h_1 – \text h_2) \\[0.5em] = 60 \times 10\times (75-15) \\[0.5em] = 60 \times10 \times 60 \\[0.5em] = 3.6 \times 10^4\ \text J \\[0.5em]$

Change in gravitational potential energy is 36000 J

(b) When kinetic energy at B is 75% of (3.6 × 10⁴)

$\text{Kinetic energy at B} = \dfrac{75}{100} \times 3.6 \times 10^4 \\[0.5em] = 27000 \\ \text J \\[0.5em] = 2.7 \times 10^4\ \text J \\[0.5em]$

Since,

Kinetic energy = $\dfrac{1}{2}$ mv²

Substituting the values in equation we get,

$27000 = \dfrac{1}{2} \times 60 \times \text v^2 \\[0.5em] \Rightarrow 27000 = 1 \times 30 \times \text v^2 \\[0.5em] \Rightarrow \text v^2 = \dfrac{27000}{30} \\[0.5em] \Rightarrow \text v^2 = 900 \\[0.5em] \Rightarrow \text v = \sqrt{900} \\[0.5em] \Rightarrow \text v = 30 \\[0.5em]$

∴ The speed at which the skier arrives at B = 30 ms^-1.

(c) As there is an interchange between potential energy and kinetic energy and all kinds of frictional forces are absent so the total mechanical energy remains constant, hence no change in mechanical energy occurs.

(a) The arrangement is shown in the figure given below :

Here, the load is being supported by five segments of the string.

∴ L = 5T and E = T

Then,

M.A. = $\dfrac{\text{L}}{\text{E}}$

= $\dfrac{\text{5T}}{\text{T}}$ = 5

∴ Mechanical advantage of the system is 5.

(b) In an ideal block and tackle system, if n segments of rope support the load, then the total length of the rope pulled = n × x.

As, here the load is being supported by five segments of the string then if the load moves up a distance x,

then,

Free end of the string will move up by a distance = 5x

Figure given below shows that two cliffs and the position of the person.

(a) First echo is heard from the nearest cliff.

Let d₁, be the distance of the nearest cliff 1 from the person.

Total distance travelled by the sound in going and then coming back = 2d₁ = 2 x 640 m = 1280 m

Time taken (t) = 4 s

$\text {Speed of sound (v)} = \dfrac{\text {Tot. dist. sound travels} }{\text {Time taken}} \\[1em] {\text {Time taken}}=\dfrac{2d_1}{\text t} \\[1em] =\dfrac{1280}{4}=320\text{ ms}^{-1}$

(b) The second echo is heard from the farther cliff 2. If d₂ is the distance of the farther cliff 2 from the person, then total distance travelled by the sound in going and then coming back = 2d₂.

Time taken (t) = 4 + 3 = 7 s

Now,

$\text {Speed of sound (v)} = \dfrac{\text {Tot. dist. sound travels} }{\text {Time taken}} \\[1em] \Rightarrow \text v = \dfrac{2d_2}{\text t} \\[1em] \Rightarrow \text d_2 = \dfrac{\text{vt}}{2} \\[1em] \Rightarrow \text d_2 = \dfrac{320\times7}{2} \\[1em] \Rightarrow \text d_2 = 1120 \text{ m}$

Distance between the two cliffs = d₁ + d₂ = 640 m + 1120 m = 1760 m

Hence, speed of sound in air is 320 ms^-1 and distance between the cliffs is 1760 m.

(a) C executes free vibrations (damped if medium is present).

(b) D executes forced oscillation.

(c) If the length of D is made equal to C, then the amplitude of vibration of pendulum D increases substantially.

(d) The phenomenon is called resonance.

(a) Both the graphs are correct because when the resistors are joined in series, the equivalent resistance is more than when they are joined in parallel.

Therefore,

(i) On V-I graph, the straight line obtained for a series combination will have higher slope than that for a parallel combination because it's slope gives resistance.

(ii) On I-V graph, the straight line obtained for a series combination will have less slope than that for a parallel combination because it's slope gives conductance which is reciprocal of resistance.

(b) Given,

Length of the wire is increased by 25%.

Let original length = L,

New length = L' = L + 0.25L = 1.25L

If length increases by 25%, to keep the volume constant, area must decrease.

So,

Volume = constant

Let original area be A and new area be A'.

⇒ L ⋅ A = L′ ⋅ A′

$\text A'= \dfrac{\text{L⋅A}}{\text L'}= \dfrac{\text{L⋅A}}{1.25\text L}= \dfrac{\text A}{1.25}=0.8\text A$

So new area = 0.8A.

As,

Original Resistance :

$\text R= \text ρ\dfrac{\text L}{\text A}$

New Resistance :

$\text R'= \text ρ\dfrac{\text L'}{\text A'}=\text ρ\dfrac{1.25\text L}{0.8\text A}=\text ρ\dfrac{\text L}{\text A}\times\dfrac{1.25}{0.8}=1.5625R$

% Increase in Resistance = (1.5625 − 1) × 100 % = 56.25 %

Change in Resistivity:

Resistivity (ρ) is a material property and it does not depend on length, area, or shape.

So,

% change in resistivity = 0%

(c) Although the same current flows through the live wire and the bulb filament, only the filament glows because it has very high resistance. This high resistance converts electrical energy into heat and light, making the filament white-hot and causing it to glow. On the other hand, the live wire has very low resistance, so it does not get heated up significantly and therefore does not glow.

(a) The number of neutrons in a nucleus can be calculated using the formula:

Number of neutrons = Mass number (A) - Atomic number (Z)

Given,

Mass number (A) = 15
Atomic number (Z) = 7
So, number of neutrons = 15 - 7 = 8

(b) The nucleus P can be written as ¹⁵₇P.

(c)

1. After the loss of 1 proton, the mass number and atomic number of the nucleus P will decrease by 1.The new nucleus will be $^{14}_6\text Q$ (say). The change can be written as :
   $^{15}_7\text P \longrightarrow ^{14}_6\text Q + ^{1}_1\text p$

2. After the loss of one β particle, the mass number will remain the same, but the atomic number will increase by 1. The nucleus $^{15}_7\text P$ changes to $^{15}_8\text R$ (say) as follows :
   $^{15}_7\text P \longrightarrow ^{15}_8\text R + ^{0}_{-1}\text e$

3. After the loss of one α particle, the mass number decreases by 4 and the atomic number decreases by 2. The nucleus $^{15}_7\text P$ changes to $^{11}_8\text S$ (say) as follows :
   $^{15}_7\text P \longrightarrow ^{11}_5\text S + ^{4}_2\text {He}$

(a) Given,

Battery Voltage (V) = 6 V

Fixed Resistance (R₁) = 2 Ω

Resistance of wire (R₂) = 4 Ω

R₁ and R₂ are connected in series.

∴ Total resistance of the circuit (R_s) = R₁ + R₂ = 2 + 4 = 6 Ω

By Ohm's law,

$\text {Current (I)}=\dfrac{\text V}{\text R_\text s}=\dfrac{6}{6}=1\text A$

(b) On doubling the length (i.e., 2 x 0.20 = 0.40 m), the resistance of wire (R₂) is doubled, i.e., it becomes 8 Ω.

Total resistance of the circuit (R_s) = R₁ + R₂ = 2 + 8 = 10 Ω

By Ohm's law,

$\text {Current (I)}=\dfrac{\text V}{\text R_\text s}=\dfrac{6}{10}=0.6\text A$

(c) On doubling the area of cross section, the resistance of wire R₂, is reduced to half, i.e., it becomes $\dfrac{1}{2}\times 4$ Ω = 2 Ω.

Total resistance of the circuit (R_s) = R₁ + R₂ = 2 + 2 = 4 Ω

By Ohm's law,

$\text {Current (I)}=\dfrac{\text V}{\text R_\text s}=\dfrac{6}{4}=1.5\text A$

Given,

m = 200 g

Specific heat capacity of water = 4.2 J g⁻¹ °C⁻¹

Specific latent heat of ice = 336 J g⁻¹

total amount of heat energy required = ?

Heat energy taken by ice at 0° C to convert into water at 0° C
= m x L
= 200 × 336
= 67200 J

Heat energy taken by water to raise the temperature from 0° C to 100° C
= m x c x change in temperature
= 200 × 4.2 × (100 - 0)
= 200 × 4.2 × 100
= 84000 J

Total heat energy gained is
= 67200 + 84000
= 151200 J
= 151.2 × 10³ J = 151.2 KJ

Hence, total amount of heat energy required = 151.2 KJ

Temperature-time graph representing the change of ice to steam is shown below:

(a) North pole of needle deflects towards east.

(b) On passing current through the wire AB, a magnetic field is produced around the wire which aligns the magnetic needle in it's direction.

(c) The direction of deflection gets reversed i.e., now it deflects towards west.

(d) An electric bell works on this principle.