Solved Sample Paper 4

Section A

Question 1(i)

If W_A is the work done in path 1 and W_B is the work done in path 2, then

1. W_A < W_B
2. W_A = W_B
3. W_A > W_B
4. All of the above

Answer

W_A = W_B

Reason — Gravitational potential energy (PE) does not depend on the path but on initial and final position of the body. Hence, W_A = W_B

Question 1(ii)

Which of the following wave represents a damped oscillations?

Answer

The below wave represents damped oscillation:

Reason — A damped oscillation is one which fades away with time, hence, its amplitude decreases with time.

Question 1(iii)

A bar magnet is rotated near a coil as shown below

Which of the following statement is correct for given situation?

1. Induced current flows in only one direction.
2. Strength of induced emf increases as the rotation speed is decreased
3. The strength of induced emf is affected due to coil resistance.
4. The direction of induced current changes alternatively.

Answer

The direction of induced current changes alternatively.

Reason — Due to rotation the magnetic polarity of coil keeps on changing. Hence, the direction of induced current changes alternatively. Strength of induced emf decreases as the rotation speed is decreased and the resistance of coil does not affect the strength of induced emf.

Question 1(iv)

Which of the following corresponds to Joule's law of heating?

1. I²R
2. I²Rt
3. V²Rt
4. IR²t

Answer

I²Rt

Reason — According to Joule's Law of heating, the heat produced in a conductor is directly proportional to the square of current, the resistance offered and the duration of time for which the current flows. Hence, H = I²Rt

Question 1(v)

Assertion: Current is the rate of flow of charge.

Reason: Electric current will not flow between two charged bodies when connected, if they are at the same potential

1. Both Assertion and Reason are true.
2. Both Assertion and Reason are false.
3. Assertion is false but Reason is true.
4. Assertion is true but Reason is false

Answer

Both Assertion and Reason are true.

Reason — Current is the rate of flow of charge and it flows form a higher potential to lower potential, hence, electric current will not flow between two charged bodies when connected, if they are at the same potential.

Question 1(vi)

Which of the following are isobars ?

1. $^{12}_6\text{C}$ and $^{14}_6\text{C}$
2. $^{16}_8\text{O}$ and $^{23}_{11}\text{Na}$
3. $^{23}_{11}\text{Na}$ and $^{22}_{11}\text{Mg}$
4. $^{23}_{11}\text{Na}$ and $^{23}_{12}\text{Mg}$

Answer

$^{23}_{11}\text{Na}$ and $^{23}_{12}\text{Mg}$

Reason — Isobars are atoms having the same mass number but different atomic number. Hence, $^{23}_{11}\text{Na}$ and $^{23}_{12}\text{Mg}$ are isobars.

Question 1(vii)

Which of the following is not a type of equilibrium ?

1. Static
2. Dynamic
3. Neutral
4. Rolling

Answer

Rolling

Reason — Rolling is a type of motion and not an equilibrium state.

Question 1(viii)

The correct descending order of the electro magnetic waves according to their wavelengths is

1. UV-rays, γ-rays, microwaves, infrared
2. infrared, microwaves, UV-rays, γ-rays
3. γ-rays, UV rays, Infrared rays, microwaves
4. microwaves, infrared rays, UV rays, γ-rays

Answer

microwaves > infrared rays > UV rays > γ-rays

Reason — The order of electro magnetic waves in terms of decreasing wavelengths is : microwaves > infrared rays > UV rays > γ-rays

Question 1(ix)

Absolute unit of work is

1. joule
2. erg
3. Both (a) and (b)
4. None of these

Answer

Both (a) and (b)

Reason — Both joule and erg are absolute units of work in SI and CGS respectively.

Question 1(x)

The effective diameter of the circular outline of a spherical lens is known as

1. focal length
2. principal focus
3. aperture of lens
4. radius of curvature

Answer

aperture of lens

Reason — The effective diameter of the circular outline of a spherical lens is known as aperture of lens.

Question 1(xi)

The frequency of a source is 20 kHz. The frequency of sound wave produced by it in water and air will be

1. same as that source
2. > 20 kHz
3. < 20 kHz
4. Depend upon velocity

Answer

same as that source

Reason — The frequency does not depend on medium and depends on the frequency of the source. Hence, the frequency of the sound in air and water will be the same.

Question 1(xii)

A ray is made to fall on a prism. Maximum deviation it can suffer is

1. $\dfrac{π}{3}$

2. $\dfrac{π}{4}$

3. $\dfrac{π}{2}$

4. $π$

Answer

$π$

Reason — The maximum deviation produced by a prism is $π$ or 180°, when the ray of light enters the prism at an angle of 90° as shown below.

Question 1(xiii)

Which of the following phenomena is not due to high specific heat of water?

1. Water is used as coolant
2. For fomentation hot water bottles are used.
3. Water is used in cold countries as heat reservoir of wine and juice bottles.
4. None of the above

Answer

None of the above

Reason — All of the above uses of water are due to its high specific heat.

Question 1(xiv)

Lead is not used in calorimeter because

1. it has high specific heat
2. it is not a metal
3. it is a toxic metal
4. it is not a solid metal

Answer

it is a toxic metal

Reason — Lead is not used in calorimeter because it is a toxic metal and can be harmful.

Question 1(xv)

If refractive index of glass w.r.t. air is $\dfrac{3}{2}$. What is the refractive index of air w.r.t. glass?

1. $\dfrac{3}{4}$

2. $\dfrac{4}{3}$

3. $\dfrac{3}{2}$

4. $\dfrac{2}{3}$

Answer

$\dfrac{2}{3}$

Reason — The refractive index of glass w.r.t. air is _aμ_g = $\dfrac{3}{2}$

We know that:

$_{\text{g}}\text{μ}_{\text{a}} = \dfrac{1}{_a\mu_g} \\[1em]$

So, substituting we get,

$_{\text{g}}\text{μ}_{\text{a}} = \dfrac{1}{\dfrac{3}{2}} = \dfrac{2}{3} \\[1em]$

Hence, _gμ_a = $\dfrac{2}{3}$

Question 2(i)

(a) A body of mass 5.5 kg is dropped from the 2nd floor of a building which is at a height of 15 m. What is the force acting on it during its fall? (g = 10 ms^-2)

(b) What are isobars ?

(c) How many pulleys are there in a movable block and tackle system with velocity ratio 5 ?

Answer

(a) Given,

mass (m) = 5.5 kg

height (h) = 15 m

force acting on it during its fall = ?

Force acting on the body during its fall = weight of the body.

And F = mg

Substituting we get,

F = 5.5 x 10 = 55 N

Hence, Force acting on the body during its fall = 55 N

(b) The atoms of different elements which have the same mass number A but different atomic number Z are called isobars.

Example — $^{23}_{11}\text{Na}$ and $^{23}_{12}\text{Mg}$ are isobars.

(c) The number of pulleys required in the movable block of a block and a tackle system with velocity ratio 5 will be 5.

Question 2(ii)

(a) Give one example of pure rotational motion.

(b) Comment on the movement of particles on the axis of rotation in pure rotational motion.

Answer

(a) The motion of the blades of the helicopter is a rotatory motion.

(b) The particles on the axis of rotation are stationary in pure rotational motion.

Question 2(iii)

Point the centre of gravity of the object given below

Answer

Centre of gravity for cylinder is the mid point on the axis of cylinder and for triangle it is the point of intersection of medians:

Question 2(iv)

(a) A brass ball is hanging from a stiff cotton thread. Draw a neat labelled diagram showing the forces acting on the brass ball and cotton thread.

(b) Can the couple acting on a rigid body produce translatory motion?

Answer

(a) In the figure below, the force on the ball is the weight W acting vertically downwards and the force on the thread is the tension T acting upwards.

(b) No, the couple acting on a rigid body cannot produce translatory motion, it can cause only rotatory motion as the resultant force is zero.

Question 2(v)

Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force? Explain it with an example.

Answer

Yes, it is possible.

For example, when an object is moving in a circular path due to centripetal force acting on it.

At any instant of time, a constant acceleration due to the centripetal force acts on the object along the radius towards the centre while the direction of displacement is tangential to the circle i.e., force and displacement are perpendicular to each other and hence, no work is being done.

Question 2(vi)

Can SONAR be used to determine the depth of a sea? Draw a well labelled diagram showing the procedure.

Answer

Yes, SONAR is used for determining the depth of the sea.

To find the depth of the sea, the time interval (t) between the instant when waves are sent and the instant when waves are received, after reflection from the depth is measured.

The depth d of the sea is then —

$\text{d} = \dfrac{Vt}{2} \\[0.5em]$

where, V is the speed of ultrasonic wave in water.

This process is called echo depth sounding.

Question 2(vii)

Of the three connecting wires in a household circuit:

(a) Which two of the three wires are at the same potential?

(b) In which of the three wires should the switch be connected?

Answer

(a) Neutral and earth wires

(b) Live wire

Question 3(i)

A lens forms an erect, magnified and virtual image of an object. Name the lens used and draw a labelled ray diagram to show the image formation.

Answer

Convex lens

Question 3(ii)

(a) The given figure shows three resistors. Find the combined resistance.

(b) Which part of an electrical appliance is earthed?

Answer

(a) In the circuit, there are two parts. In the first part, resistors of R₁ (6 Ω) and R₂ (2 Ω) are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{{R}_P} = \dfrac{1}{6} + \dfrac{1}{2} \\[0.5em] \dfrac{1}{{R}_P} = \dfrac{1 + 3}{6} \\[0.5em] \dfrac{1}{{R}_P} = \dfrac{4}{6} \\[0.5em] \dfrac{1}{{R}_P} = \dfrac{2}{3} \\[0.5em] \Rightarrow {R}_P = \dfrac{3}{2} = 1.5 Ω$

Hence, R_P = 1.5 Ω

In the second part, R_P (1.5 Ω) and R₃ (6 Ω) are connected in series. If the equivalent resistance of this part is R_s then

R_s = (1.5 + 6) Ω = 7.5 Ω

Hence, combined resistance = 7.5 Ω

(b) The earth wire is connected to the metallic body of the appliance.

Question 3(iii)

The diagram below shows a coil connected to the centre zero galvanometer.

What polarity is induced at point X when

(a) the magnet is moved towards the coil

(b) the magnet is moved away from the coil?

Answer

(a) South pole

Reason — A moving south pole of a magnet near a coil induces a current. The induced current creates a magnetic field opposing the change, following Lenz's law. This leads to the coil's side facing the south pole developing a south pole as well.

(b) North pole

Reason — Moving a magnet's south pole away from a coil induces a current. The induced current creates a magnetic field opposing the change, following Lenz's law. This leads to the coil's side facing the south pole developing a north pole.

Question 3(iv)

Some common salt is added to ice. It is found that temperature of mixture drops below 0°C. Explain the fact and where is this used?

Answer

When common salt is added to ice, it takes heat from ice and mixes with the water droplets present on the outer layer of ice. Hence, it absorbs heat from ice and temperature of the mixture falls below 0°C.

Question 3(v)

Suggest one effective process for the safe disposal of nuclear waste.

Answer

The nuclear waste obtained from laboratories, hospitals, scientific establishments or power plants must be first kept in thick casks and then they must be buried in the specially constructed deep underground stores. These underground stores must be far away from the populated areas.

Section B

Question 4(i)

A boy uses monochromatic green light to find the refractive index. When a ray of monochromatic green light enters in a liquid medium from air medium as shown in the figure given below, the angle 1 is 45° and angle 2 is 30°.

(a) Determine the refractive index of the liquid.

(b) Represent in the diagram showing the path of the ray after it strikes the mirror and re-enters air. Mark in the diagram wherever necessary.

(c) Draw the diagram again, if plane mirror becomes normal to the refracted ray inside the liquid. Name the principle used.

Answer

(a) Refractive index of the liquid is given by Snell's law and is shown as below,

_airμ_liquid = $\dfrac{\text{sin i}}{\text{sin r}}$

_airμ_liquid = $\dfrac{\text{sin 45}\degree}{\text{sin 30}\degree}$

$= \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{2}} \\[1em] = \dfrac{2}{\sqrt{2}} \\[1em] = \sqrt{2} \\[1em] = 1.414 \\[1em]$

Hence, the refractive index of the lens is 1.4

(b) Below labelled diagram shows the path of the ray after it strikes the mirror and re-enters in air:

(c) Below labelled diagram shows the path of the ray when it strikes the plane mirror normally inside the liquid:

The principle of reversibility is used.

Question 4(ii)

(a) Mention one difference between reflection of light from a plane mirror and total internal reflection of light from a prism

(b) When an empty test tube is placed in a beaker of water and viewed from above, its surface appears as a mirror. Explain.

Answer

(a) Difference between reflection of light from a plane mirror and total internal reflection of light from a prism are:

(b) Total internal reflection occurs when light traveling through a medium with a higher refractive index (such as water) encounters a boundary with a medium of lower refractive index (such as air) at an angle greater than the critical angle. In this case, the water acts as a more optically dense medium than air.

As a result, the light reflects completely back into the water instead of passing into the air above the test tube. This reflection creates a mirror-like effect on the surface of the water inside the test tube when viewed from above, giving the appearance of a mirror.

Question 4(iii)

If a light of a single colour is passed through a liquid having a piece of glass suspended in it, so on changing the temperature of the liquid, at a particular temperature, the glass piece is not seen.

(a) At what situation, the glass piece will not be seen?

(b) Why the light of a single colour is used?

(c) How can you define the term refractive index of a medium? State whether it can be less than 1 or not.

Answer

(a) The glass piece is not seen when the refractive index of the liquid becomes equal to the refractive index of glass.

(b) The light of a single colour is used as the refractive index of a medium (glass or liquid) is different for the light of different colours.

(c) The refractive index of a medium is defined as the ratio of the speed of light in vacuum (or air) to the speed of light in that medium.

$μ = \dfrac{\text{Speed of light in vacuum or air(c)}}{\text{Speed of light in that medium (V)}}$

The refractive index of a transparent medium is always greater than 1 (it cannot be less than 1) because the speed of light in any medium is always less than that in vacuum ( i.e. V < c ) and μ = 1 for air or vacuum.

Question 5(i)

(a) Define power of a lens.

(b) A child is using a spectacle with power of -2.5D. What is meant by the negative sign?

(c) Find the focal length of the lens used.

Answer

(a) The deviation of the incident light rays produced by a lens on refraction through it is a measure of its power. It can be defined as a reciprocal of focal length:

$\text{Power} = \dfrac{1}{\text{focal length}}$

(b) As the given power is negative hence, we can say that the lens used is concave in nature and the child is suffering from myopia or short sightedness (i.e., unable to see the far objects distinctly).

(c) As we know,

$\text{Power} = \dfrac{1}{\text{focal length}}$

Given,

Power of the lens = -2.5 D

Substituting the value of power in formula we get,

$-\text{2.5} = \dfrac{1}{\text{focal length}} \\[0.5em] \Rightarrow \text{focal length} = -\dfrac{1}{2.5} \\[0.5em] \Rightarrow \text{focal length} = -\text{0.4 m} \\[0.5em] \Rightarrow \text{focal length} = -\text{40 cm} \\[0.5em]$

∴ Focal length = - 40 cm.

Question 5(ii)

The diagram given below shows a lens as combination of glass prisms of different refractive indices. Copy the diagram and answer the following questions.

(a) Name the lens formed by the combination.

(b) Complete the ray diagram when AB is between optical centre and 2F.

Answer

(a) The lens formed by the combination is concave in nature.

(b) The completed ray diagram when AB is between optical centre and 2F is shown below:

Question 5(iii)

A ray is directed towards a lens as shown in the figure below. The lens is being depicted as a combination of a glass slab and two prisms.

Answer the following questions.

(a) Name the lens which is formed by this combination.

(b) What is PQ line called?

(c) Complete the ray diagram.

(d) The incident ray after refraction actually meet or appears to meet line PQ at a point. What is this point called?

Answer

(a) The lens formed by the combination is a convex lens.

(b) The line PQ is called the principal axis.

(c) Below diagram shows the path of the incident ray AB after passing through the convex lens:

(d) The point F is called the focal point.

Question 6(i)

Shyam is trying to balance a meter rod on his finger as shown in the diagram. Two weight 50 gf and 200 gf at 40 cm and 95 cm respectively and his finger at 70 cm.

Find the anti-clock wise moment and clockwise moment.

(b) Find the weight of meter rod.

Answer

(a) Let W be the weight of the rod.

For uniform meter rod, its weight acts at 50 cm.

Clockwise moments = [200 x (95-70)] = 200 x 25 = 5000 Nm

Anticlockwise moments = [50 x (70 - 40)] + [W x (70 - 50)]

= [50 x 30] + [W x 20]

= 1500 + 20W

(b) According to principle of moment :

Anticlockwise moment = Clockwise moment

1500 + 20W = 5000

W = $\dfrac{5000 - 1500}{20} = \dfrac{3500}{20}$ = 175 N

Hence, weight of meter rod = 175 N

As W = 175 N, hence, anticlockwise moment = 1500 + 20W = 1500 + (20 x 175) = 1500 + 3500 = 5000

Hence, anticlockwise moment = 5000 Nm

Question 6(ii)

(a) Draw a diagram of a pulley system of velocity ratio 4. Calculate its mechanical advantage, if its efficiency is 90%.

(b) Why does a rope walker hold a long pole in his hands? Explain.

Answer

(a) Diagram of a pulley system of velocity ratio 4 is shown below:

Given,

Efficiency (η) = 90%

Velocity ratio = 4

MA = ?

We know, $\eta = \dfrac{\text {MA}}{\text {VR}}$

Substituting in the formula above we get,

$\dfrac{90}{100} = \dfrac{\text {MA}}{\text {4}}$

MA = 0.90 x 4 = 3.6

Hence, MA = 3.6 N

(b) The pole increases the walker's moment of inertia or resistance to rotating and prevents him from falling off the wire, allowing the walker time to correct his position and stay balanced. When he feels he is falling towards right, he shifts the pole towards left and vice versa.

In addition, holding the pole low helps to keep his center of gravity low, and this further increases his ability to balance.

Question 6(iii)

(a) A truck and a car are running with the same speed on a highway. If the ratio of the kinetic energies is 25 : 12, then what is the ratio of their masses?

(b) Give two examples in which mechanical energy is conserved.

Answer

(a) Let kinetic energy of truck be KE_t and that of car be KE_c and v be their common speed.

Given,

$\dfrac{\text{KE}_\text{t}}{\text{KE}_\text{c}} = \dfrac{25}{12}$

By formula,

KE = $\dfrac{1}{2}$ mv²

$\phantom{\Rightarrow} \dfrac{\text{KE}_\text{t}}{\text{KE}_\text{c}} = \dfrac{25}{12} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2}\text{m}_\text{t}\text{v}^2}{\dfrac{1}{2}\text{m}_\text{c}\text{v}^2} = \dfrac{25}{12} \\[1em] \Rightarrow \dfrac{\text{m}_\text{t}}{\text{m}_\text{c}} = \dfrac{25}{12}$

Hence, ratio of masses = 25 : 12

(b) Mechanical energy is conserved in:

1. Simple pendulum and
2. Motion of a freely falling body.

Question 7(i)

(a) State two characteristics that a medium should have for propagation of sound.

(b) A pendulum has a frequency of 4 vibrations per second. An observer starts pendulum and fires a gun simultaneously. He hears the echo from a cliff after 8 vibrations of pendulum. If the velocity of sound in air is 340 ms^-1, then what is the distance between cliff and observer?

Answer

(a) Two characteristics that a medium should have for propagation of sound are:

1. The medium must be elastic so that it's particles may come back to their initial position after displacement on either side, i.e., the particles are capable of vibrating about their mean position.
2. The medium must have inertia so that it's particles may store mechanical energy.

(b) Echo is heard after 8 vibrations of pendulum. Since frequency of pendulum is 4 vibrations per sec, so the time taken for the echo to be heard is t = $\dfrac{8}{4}$ = 2s

If d is the distance between the cliff and the observer, then velocity of sound V = $\dfrac{2\text{d}}{\text{t}}$

Substituting the values in the equation above:

340 = $\dfrac{2\text{d}}{\text{2}}$

d = 340 m

Hence, distance between the cliff and the observer = 340 m

Question 7(ii)

(a) Explain in brief the given statement — radioactivity is a nuclear phenomenon.

(b) What is meant by nuclear waste?

Answer

(a) Any physical change (such as change in pressure and temperature) or chemical change (such as excessive heating, freezing, action of strong electric and magnetic fields, chemical treatment, oxidation etc.) does not change the rate of decay and the nature of radiation emitted by the substance.

This clearly shows that the phenomenon of radioactivity cannot be due to the orbital electrons which could easily be affected by such changes. Therefore radioactivity should be the property of the nucleus. Thus, radioactivity is a nuclear phenomenon.

(b) The radioactive material after its use such as nuclear power generation, nuclear weapons production, and nuclear medicine is known as nuclear waste.

Question 7(iii)

A boy tunes a radio channel to a radio station 93.5 MHz.

(a) Name and define the scientific wave phenomenon involved in tuning the radio channel.

(b) Now what is the frequency of the channel? Convert this frequency into SI unit.

Answer

(a) Resonance

Definition — When the frequency of the externally applied periodic force on a body is equal to its natural frequency, the body readily begins to vibrate with an increased amplitude. This phenomenon is known as resonance.

(b) Frequency of the channel is 93.5 MHz.

The SI unit for frequency is the hertz (Hz)

1 MHz = 10⁶ Hz

93.5 MHz = 93.5 x 10⁶ Hz

Question 8(i)

Three resistors are joined together as shown in the figure. The resistors are connected to an ammeter and to a cell of emf 9.0 V

Calculate

(a) the effective resistance of the circuit.

(b) the current drawn from the cell.

Answer

In the circuit, there are two parts. In the first part, resistors of 2.0 and 4.0 Ω are connected in series. If the equivalent resistance of this part is R_s then

R_s = 2 + 4 = 6 Ω

In the second part, R_s = 6.0 Ω and resistor of 6.0 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{6} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1 + 1}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{2}{6} \\[0.5em] R_p = \dfrac{6}{2} \\[0.5em] R_p = 3.0 Ω \\[0.5em]$

Hence, the effective resistance of the circuit = 3 Ω

(b) current drawn = ?

R = 3 Ω

V = 9.0 V

Using Ohm's law,

V = IR

Substituting the values in the formula above we get,

9 = I x 3
⇒ I = $\dfrac{9}{3}$ = 3 A

Hence, the current drawn = 3 A

Question 8(ii)

(a) What are background radiations?

(b) Write an equation of an α-emission from ²³⁸₉₂U

(c) What will be the change in the rate of radioactivity if the temperature of the radioactive substance is raised to four times the initial temperature?

Answer

(a) Background radiations are the radioactive radiations (such as α, β and γ) to which we all are exposed, even in the absence of an actual visible radioactive source. For example, radioactive substances such as potassium (K-40), carbon (C-14) and radium are present inside our body. Cosmic rays also contribute to background radiations.

(b) When a radioactive uranium nucleus $^{238}_{92}\text{U}$ emits an α particle, a new nucleus thorium $^{234}_{90}\text{Th}$ is formed and the change is represented as follows —

$_{92}^{238}\text{U} \longrightarrow \space _{90}^{234}\text{Th} + \space _{2}^{4}\text{He}$

(c) The rate of radioactive decay generally depends on the specific properties of the radioactive substance and is not directly proportional to temperature changes. Hence, increasing the temperature will not affect the rate of radioactivity.

Question 8(iii)

(a) Name the new and old colour codes of the wire which is connected to the switch of an appliance.

(b) Draw the diagrams of a dual control switch when the appliance is switched ON and OFF.

(c) Calculate the electrical energy consumed when a bulb of 40 W is used for 12.5 h everyday for 30 days.

Answer

(a) Live wire is connected to the switch of an appliance.

(b) Diagrams of a dual control switch when the appliance is switched ON and OFF are shown below:

(c) Given,

Power (P) = 40 W

time (t) = 12.5 h for 30 days

As energy consumed (E) = P × t

$E = 40 \times 12.5 \\[0.5em] \Rightarrow E = 500 \text { Wh}$

Hence, electrical energy consumed in one day = 500 Wh

Electrical energy consumed in 30 days = ?

E = 500 x 30
= 15000 Wh
= 15 kWh

Hence, electrical energy consumed in 30 days = 15 kWh

Question 9(i)

(a) Which contains more heat, 1 kg water at 100°C or 1 kg steam at 100°C?

(b) Name a substance that contracts on melting and expands on solidification.

(c) In cold countries, juice bottles are placed underwater, so as to avoid freezing. Why?

Answer

(a) 1 kg of steam at 100°C contains more heat than 1 kg of water at the same temperature. This is because the heat energy in the steam includes the additional latent heat of vaporization, which is the energy required to change water from liquid to vapor at its boiling point.

(b) Ice contracts on melting and expands on solidification.

(c) The reason for keeping juice bottles underwater is that water due to it's high specific heat capacity can impart a large amount of heat before reaching to it's freezing point. Hence, bottles kept in water remain warm and do not freeze when there is considerable fall in surrounding's temperature.

Question 9(ii)

A piece of ice of mass 100 g is dropped into 450 g of water at 45°C. Calculate the final temperature of water after all of the ice has melted. (Specific heat capacity of water 4200 Jkg^-1°C^-1, specific heat of fusion of ice = 336 x 10³ Jkg^-1)

Answer

Given,

m_i = 100 g

Converting g to kg

1000 g = 1 kg

So, 100 g = $\dfrac{100}{1000}$ = 0.1 kg

m_w = 450 g

Converting g to kg

1000 g = 1 kg

So, 450 g = $\dfrac{450}{1000}$ = 0.45 kg

Specific heat capacity of water = 4200 J kg^-1 K^-1

specific latent heat of fusion of ice = 336 x 10³ J kg^-1

final temperature of water = ?

Let final temperature = t

Heat energy given out by water when it cools from 45° C to t° C
= m x c x Δt
= 0.45 × 4200 × (45 – t)
= 85050 – 1890t

Heat energy taken by ice when it converts from ice into water at 0° C
= m x L
= 0.1 × 336 x 10³ J
= 33600 J

Heat energy taken by water when it raises it's temperature from 0° to t° C
= m x c x Δt
= 0.1 × 4200 × (t – 0)
= 0.1 × 4200 × t
= 420t

If there is no loss of energy,

Heat energy gained = heat energy lost

Substituting the values we get,

$33600 + 420t = 85050 - 1890t \\[0.5em] 1890t + 420t = 85050 - 33600 \\[0.5em] 2310t = 51450 \\[0.5em] t = \dfrac{51450}{2310} \\[0.5em] t = 22.27° C$

Hence, final temperature = 22.27°C

Question 9(iii)

(a) Draw a labelled diagram to make an electromagnet from a soft iron bar. Mark the polarity at its ends in your diagram.

(b) What is the nature of magnetic field inside a long straight solenoid carrying current?

(c) Consider a circular loop of wire lying in the plane of a table. Let the current pass through the loop clockwise. Applying the right hand rule, find out the direction of the magnetic field inside and outside the loop.

Answer

(a) Below diagram shows an electromagnet made from a soft iron bar with polarity of the ends labelled:

(b) Magnetic field inside a long straight solenoid carrying current is uniform i.e., same at all points.

(c) Direction of magnetic field:
Inside the loop — Going into the table.
Outside the loop — Emerging out of the table.

Explanation:
On applying the right-hand rule to the downward direction of the current flowing in the circular loop, the direction of magnetic field lines will be as if they are emerging out from the table outside the loop and merging into the table inside the loop.
Similarly, for the upward direction of the current flowing in the circular loop, the direction of magnetic field lines will be as if they are emerging out from the table outside the loop and merging into the table inside the loop.