# Oswal Model Specimen Papers ICSE Class 10 Physics Solutions (Specimen Paper - 9)

## Section-A

1. (i) (d) All of the above

(ii) (c) it is travelling from optically rarer medium to a optically denser medium.

Explanation :

Speed of light decreases as it travels from rarer to denser medium.

(iii) (c) Kinetic energy and potential energy

Explanation :

Basic forms of mechanical energy are kinetic and potential energy.

(iv) (c) Total number of pulleys

Explanation :

Cranes and hoist are examples of block and tackle system of pulleys.

(v) (d) both (a) and (b)

Explanation :

Sparkling is because of entrapping of light due to large cut surfaces resulting in low critical angle.

(vi) (a) Total energy of the oscillator decreases with time

Explanation :

Vibrations are considered to be damped when the energy of a vibrating system is progressively dissipated by various resisting forces. The vibrations gradually decrease in frequency and after sometime the system returns to its equilibrium state. Thus in damped vibration, total energy of the oscillator decreases with time.

(vii) (a) 52300 Joule

Explanation :

We know, 1 cal = 4.184 Joule

Or 1 kcal = 4184 Joule

Thus 12.5 kcal = 4184 × 12.5 = 52300 Joule.

(viii) (c) 8 Ω-m

Explanation :

We know that resistivity, ρ = RA/l. where R is the resistance, A is the cross sectional area of the wire and l is the length of the wire.

Given: R = 4 Ω; A = 4 m2; l = 2 m.

Thus, ρ = (4 × 4)/2 = 8 Ω-m.

(ix) (d) 600 Hz

Explanation :

The topmost wave has frequency of 100 Hz, i.e., fundamental frequency. Now, it can be seen that the bottom most wave is 5th overtone or 6th harmonic. Thus, the frequency of the bottom most wave will be

= 6 × 100 = 600 Hz.

(x) (a) Class III lever

Explanation :

The Mechanical Advantage and Velocity Ratio of Class III lever is always less than one because the load arm is greater than the effort arm. Hence it can be used as a speed gainer.

(xi) (d) energy

Explanation :

The law of energy conservation is the foundation of Lenz’s law. According to this rule, the induced current always tends to oppose the source that creates it.

(xii) (a) Melting point increases.

Explanation :

The melting point of materials that expand when melted, such as wax, lead, etc., increases on rising in pressure. In contrast, the melting point of substances that contract when melted, such as ice, decreases as the pressure rises.

(xiii) (d) Both (a) and (c)

Explanation :

The material absorbs heat energy during the melting and boiling processes. During freezing, however, the material releases heat energy.

(xiv) (c) Circuit Breaker

Explanation :

The instrument presented in the figure is called ‘Circuit Breaker’.

(xv) (a) Ammeter

Explanation :

Ammeter is used to measure current in a circuit while voltmeter measures voltage difference.

2. (i) (a) Refractive index of a medium is the ratio of velocity of light in vacuum to the velocity of light in that medium.

(b) Ray 2.

(ii) The position of the object in front of a converging lens if :

(a) It produces a real and same size image as that of an object is when the object is placed at 2F.

(b) The type of lens is a convex lens as it is converging in nature.

(ii)

(iv) (a) The light must travel from a denser to a rarer medium.

(b) The angle of incidence must be greater than the critical angle for the pair of media.

(v) (a) Class II lever.

(b) Given : FA = 40 cm, AB = 60 cm

∴ FB = (40 + 60) cm = 100 cm

and,

$$\text{M.A.} =\frac{\text{Load}}{\text{Effort}}\\=\frac{\text{Effort arm (FB)}}{\text{Load arm (FA)}}\\=\frac{100}{40} = 2.5.$$

(vi) (a)

(b) Angles are marked in the diagram.

(vii) Given : For solid (Hot body) : m1 = 50 g, T1 = 150°C, T = 20°C

Fall in temperature of solid, ∆T1 = (150 – 20)°C = 130°C

C1 = ?

For water (Cold body) : m2 = 100 g, T2 = 11°C, T = 20°C

Rise in temperature of water, ∆T2 = (20 – 11)°C = 9°C

C2 = 4.2 Jg–1 °C–1

Heat lost by hot body = m1C1 ∆T1

= 50 × C1 × 130 joule

Heat gained by cold body = m2C2 ∆T2

= 100 × 4.2 × 9 joule

From the principle of calorimetry, if the system is fully insulated :

Heat lost by hot body = Heat gained by cold body

or 50 × C1 × 130 = 100 × 4.2 × 9

$$\text{or}\\\space{\space C}_{1} =\frac{100×4.2×9}{50×130}\text{Jg}^{\normalsize-1}\degree\text{C}^{\normalsize-1}$$

= 0.58 Jg–1°C–1

3. (i) The diagram showing the use of a total reflecting prism to turn a ray of light through 90° is given in the figure.

This action of prism is used in periscope.

(ii) (a) The angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90° is called critical angle.

(b) The relation between the refractive index (µ) of the medium and critical angle (c) is

$$\mu = \frac{1}{\text{sin c}}$$

(iii) (a) The statement is Correct. Specific heat capacity of water is 4184 J·kg−1·K−1 and for ice it is 2108 J·kg−1·K−1. So, Specific heat capacity of water is more compared to ice. Ice has a covalent chemical structure, which makes it easier to break intermolecular bonds than water. As a result, ice has a lower specific heat capacity than water.

(b) According to the principle of Calorimetry, when a hot body is kept in contact with a cold body, heat energy is transferred from the hot body to the cold body until both bodies reach the same temperature. If no heat energy is lost to the surroundings, then Heat energy lost by the hot body is equal to the Heat energy gained by the cold body.

(iv) (a) Differences:

1. A chemical change occurs when orbital electrons change, whereas a nuclear change occurs when nucleons inside the nucleus change.
2. A nuclear change requires a significantly larger amount of energy than a chemical change.

(b) Beneficial effects: Radiations emitted during the decay process are used to treat such as cancer, tumor etc.

Harmful effect: Radiations has the potential to harm and cause permanent damage to living tissues.

$$\text{(v) (a) The given isotope of}\space \text{carbon is}\space^{14}_{6}\text{C}.$$

Number of protons = 6

Number of neutrons = mass number – atomic number = 14 – 6 = 8.

(b) After decaying a particle, the daughter element became

$$^{14}_{7}\text{X}\space\text{Means,}\space\text{the mass number}$$

remain the same but the atomic number is increased by 1. Thus it is a β-decay.

$$\text{The equation is,}\space^{14}_{6}\text{C}\xrightarrow{-\beta}\space ^{14}_{7}\text{X.}$$

## Section-B

4. (i) (a) In equilibrium position of lever, sum of the clockwise moments about the fulcrum is equal to the sum of the anticlockwise moments about the fulcrum.

(b) Lever of class III, example: forceps or a pair of sugar tongs.

(c) By the principle of moments, E × 100 = 190 × 20

∴ E = 38 N.

(ii) (a) For single fixed pulley,

Since,

$$\text{M.A.}=\frac{\text{Load}}{\text{Effort}}\\\therefore\space\text{Effort (E}_1) \text{applied =}\frac{\text{Load}}{\text{M.A.}}\\=\frac{50}{1}= 50\space\text{kgf}$$

For a single movable pulley,

L = 50 kgf

M.A. = 2

$$\therefore\space\text{Effort (E}_2) \text{applied} =\frac{\text{Load}}{\text{M.A.}}\\=\frac{50}{2} = 25\space\text{kgf}$$

Ratio of effort applied, E1 : E2 = 50 : 25 = 2 : 1

(b) In uniform linear motion, the speed and velocity are constant and acceleration is zero, whereas in a uniform circular motion the velocity is variable (even though speed is uniform), so, it is an accelerated motion.

(iii) Given : Velocity Ratio (V.R.) = 4

= 175 × 10 N = 1750 N

Displacement of load (dL) = 15 m

∴ Effort (E) = 50 kgf

= 50 × 10 N = 500 N

g = 10 N kg–1

(a)

$$\text{V.R.} =\frac{\text{Distance moved by effort (d}_\text{E})}{\text{Distance moved by load(d}_{\text{L}})}\\\text{or} \space 4 =\frac{d_{\text{E}}}{15}$$

∴ Distance moved by effort (dE) = 4 × 15 m = 60 m

(b) Work done by the effort = E × dE

= 500 × 60 J = 30,000 J

$$\text{(c)\space M.A}=\frac{\text{L}}{\text{E}}=\frac{1750}{500}= 3.5\\\text{(d)\space}\text{Efficiency(η)} =\frac{\text{M.A.}}{\text{V.R.}}×100\%\\=\frac{3.5}{4}×100\% = 87.5\%$$

5. (i) (a) Angle of minimum deviation δm = 37°

Angle of incidence corresponding to minimum deviation im = 46°

(b) For deviation δ1 = 51°, we get

δ1 = i1 + i2 – A

51 = 40 + (X) – 62

X = 51 + 22 = 73°

(ii) Concave lens.

(iii) The radiations used for :

(a) Photography at night : Infra-red radiations.

(b) Detecting fractures in bones : X rays.

(c) Whose wavelength ranges from 100°A to 4000°A : UV radiations.

6. (i) (a) Amplitude

(b) Waveform

(c) Frequency

(ii) (a) The periodic vibrations of decreasing amplitude in the presence of resistive force are called damped vibrations.

(b) A tuning fork when stroked on a rubber pad, executes damped vibrations in air.

(c) Forced vibrations.

(iii) (a) Given : f1 = 256 Hz, l1 = 80 cm, f2 = 1024 Hz, l2 = ?

$$\text{Since, f}∝\frac{1}{l}$$

fl = constant

or f1l1 = f2l2

or 256 × 80 = 1024 × l2

$$\therefore\space l_{2}=\frac{256×80}{1024} = 20\space\text{cm.}$$

The length of wire which will have frequency 1024 Hz under similar conditions is 20 cm.

(b) Given : f = 256 hertz, λ = 1.3 m

(1) Speed of sound (v) = f λ

= 256 × 1.3 ms–1

= 332.8 ms–1

(2) Given : v = 332.8 ms–1, λ = 2.6 m

$$\therefore\space \text{Frequency, f =}\frac{v}{\lambda}\\=\frac{332.8}{2.6} = 128\space\text{hertz.}$$

The second sound of wavelength 2.6 m will have low pitch and sound will be flat compared to the fixed sound of wavelength 1.3 m.

7. (i) (a) According to Lenz’s law, the direction of induced e.m.f is such that it opposes the cause which produces it. Thus, the polarity at the rightmost point of the solenoid would be South pole.

The magnet will be repelled. Because the nearest point of the solenoid would be North pole and North-North repelled.

(b) Bar shaped electromagnets are commonly used in relay or switching devices. Also it is used for magnetic experiments, in medical procedures, in electronic devices such as telephones, radios, televisions etc.

(ii) (a) Given, V = 4 V, I = 2 A, l = 10 m.

Thus the resistance of 10 m length of wire is,

$$\text{R}=\frac{\text{V}}{\text{I}} =\frac{4}{2} = 2\space\text{ohm.}$$

(b) The resistance per unit length of the wire is

$$=\frac{\text{R}}{\text{l}}=\frac{2}{10} = 0.2\space\text{ohm\space m}^{\normalsize-1}.$$

(c) The resistance of 2 m length of wire = resistance per unit length × length = 0.2 × 2 = 0.4 ohm.

(iii) (a) The vibrations of a body that occur under the influence of a periodic external force operating on it, is called forced vibrations.

(b) For instance, while a guitar is being played, the artist uses the guitar’s strings to create forced vibrations.

(c) We know that loudness can be determined by amplitude, but frequency can not be measured by amplitude. Hence, pitch cannot be determined as it is independent of its amplitude.

8. (i) (a) Using the Clock rule, looking at the face of the loop it can be said that end A will develop south pole and end B will develop north pole.

(b) The magnetic field may be increased by,

1. increasing the current strength in the coil,
2. increasing the number of turns in the coil,

(ii) (a) The resistances are connected in parallel fashion.

Thus the equivalent resistance,

$$\frac{1}{\text{R}_{eq}} =\frac{1}{4} + \frac{1}{6},\\\text{or}\space\text{R}_{eq} =\frac{12}{5} = 2.4\space Ω$$

(b) The current flowing through the battery is

$$\text{I} =\frac{\text{V}}{\text{R}_{eq}} =\frac{6}{2.4} = 2.5\space\text{A}.$$

As the resistances are connected in parallel, the voltage across each resistor will be same.

So the current through 4 Ω resistor is =

$$\frac{6}{4} = 1.5 \text{A}.$$

(iii) (a) The heat energy required to raise the temperature by 5 °C is,

= heat capacity × temperature change

= 200 J K–1 × 5 = 1000 J = 103 J.

(b) Specific heat capacity =

$$\frac{\text{Heat capacity}}{\text{Mass of the metal}}\\=\frac{200}{0.4}$$

= 500 J kg–1 K–1.

(c) The thermal energy necessary to melt a unit amount of ice at 0°C to water at 0°C without changing the temperature is known as the specific heat of melting of ice. The specific latent heat of fusion of ice, on the other hand, is the heat energy generated when a unit amount of water at 0°C freezes to ice at 0°C without a temperature change. Thus, for a pure substance latent heat of fusion is the same as the latent heat of melting.

9. (i) (a) From graph B it can be said that the direction of the current is not changing, thus it is direct current. On the other hand graph A shows the periodical change in direction of the current with time, thus it is alternating current.

(b) Differences between AC and DC:

1. In direct current (DC), the electric current only flows in one direction. In alternating current (AC), on the other hand, direction changes periodically.
2. Power consumption in AC is significantly less compared to DC.

(c) Alternating current is five times as harmful as direct current. The major cause of this harmful effect on the human body is the alternating current’s frequency. Even a little voltage of around 25 volts may kill a human if the frequency of alternating current is approximately 60 Hz.

(ii) (a) The amount of sound energy travelling per second, across unit area, at a point of the medium is the intensity of a sound wave at that location. Its unit is watt per metre2 (W/m2).

(b) The intensity of a sound wave depends upon the amplitude of vibrations, the frequency of vibrations, and the density of medium. It is proportional to the square of the amplitude of vibrations, the square of the frequency of vibrations, and the density of air.

(iii) (a) Polarity of Q is North pole and So polarity of R is south pole.

(a) Electromagnetic induction

(c) Lenz’s law

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