# Oswal Model Specimen Papers ICSE Class 10 Physics Solutions (Specimen Paper - 8)

## Section-A

**1.** (i) (c) The prism produce colours

(ii) (c) Effort × distance moved by effort

(iii) (a) 2 J/sec

** Explanation : **

$$\text{P} =\frac{\text{Work done}}{\text{sec}}$$

= Work done per beat × no. of beat in 1 sec

$$ = 2\text{J}×\frac{70}{60} = 2.33\text{J/sec}$$

(iv) (a) Rainbow formation

** Explanation : **

When sunlight enters the atmosphere of the earth, the atoms and molecules of different gasses present in the air absorb the light. Then these atoms re-emit light in all directions. This process is known as Scattering of light.

Some of the applications of scattering of light are Blue colour of the sky,reddish colour of the sky in the evening and colour of water in deep seas and oceans.

But Rainbow formation is due to dispersion of light and not by scattering of light.

(v) (c) Centrifugal force

(vi) (d) 0 to 80 dB

** Explanation : **

Safe limit of level of sound for having is 0 to 80 dB. Beyond the sound level of 120 dB can cause permanent damage to ears.

(vii) (d) Conductance

** Explanation : **

According to Ohm’s law,

$$\frac{\text{V}}{\text{I}} =\text{R}\\\therefore\space \text{ Slope of given graph is =}\frac{1}{\text{V}} =\frac{1}{\text{R}}$$

(viii) (c) 0.5 Ω

** Explanation : **

$$\text{I} =\frac{\epsilon}{\text{R + r}}$$

∴ IR + Ir = ε

or 0.2 × 10 + 0.2 × r = 2.1

or 0.2r = 0.1

or r = 0.5 Ω

(ix) (d) upward

** Explanation : **

The direction of motion of proton is the direction of current I. The direction of force F on the proton is towards north (as the proton is deflected towards north). Applying Fleming’s left hand rule, the direction of magnetic field (B) is upwards.

(x) (b) Loudness decreases

** Explanation : **

Amplitude and Loudness are directly proportional to each other, when the amplitude of sound decreases, the Loudness also decreases.

(xi) (c) Water

** Explanation : **

Water has the largest specific heat i.e., 1 cal g^{–1} °C^{–1}.

(xii) (a) Specific heat capacity

** Explanation : **

Weight W = mg, which is mass dependent. Specific heat capacity is a property of the body’s material that is independent of its mass, whereas heat capacity is dependent on both the substance and the mass of the body.

(xiii) (a) Current in the coil is in direction DCBA

** Explanation : **

From figure, the current in the coil is flowing in direction ABCD. Using Fleming’s left hand rule, in the arm AB, the force is inward at right angle to the plane of the coil. On the arm BC, no force acts. On the arm CD, the force is outward perpendicular to the plane of the coil. On the arm DA, no force acts.

(xiv) (c) 30°C

** Explanation : **

Let the temperature of water after mixing is T.

Heat lost = Heat gained

100(90 – T) = 600 (T – 20)

9000 – 100 T = 600 T – 12000

700 T = 21000

∴ T = 30°C

(xv) (d) All of these

** Explanation : **

The amount of heat energy contained in a body varies material substance to substance. Heat energy contained in a body depends upon mass, temperature, material properties of the body.

**2.** (i) The energy transformation taking place in the following appliances are :

(a) Microphone : Electrical energy to sound energy.

(b) Electric bulb : Electrical energy to light energy.

(c) Steam engine : Combustion of fuel energy to heat energy to mechanical kinetic energy.

(ii) (a) The complete diagram is shown below :

(b) Incident angle at surface AB of the prism = 0°.

Emergent angle at AC :

∠A = 60° as the prism is equilateral.

As, ∠ARM = 90° and so, ∠ARP = 180° – (90° + 60°) = 30°.

∠PRN = 90° – 30° = 60° = ∠SRN.

(iii) Two ways by which efficiency of the block and tackle system of pulleys can be increased are :

(a) reducing the weight of the lower movable block

(b) reducing the friction between bearings of pulleys by using lubricants.

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

(a) 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) Yes, it is possible to burn a piece of paper by using a convex lens as illustrated in the figure.

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

**Example :** (a) Vibrations produced by a guitar string in air.

(b) Vibrations produced by a tuning fork in air.

(vii) Copper wire has very low resistance, and high melting point than that of mercury. Fuse wire must have high resistance.

**3.** (i) Given : Mass (m) = 60 kg, Momentum (p) = 3000 kg ms^{–1}

$$\text{(a)\space}\text{Kinetic energy (K) =}\frac{p^{2}}{\text{2m}}\\=\frac{3000×3000}{2×60}\text{J}\\= 75000 J = 7.5 × 10^{4}\text{J}\\\text{(b)\space}\text{Velocity (v) =}\frac{p}{m}\\=\frac{3000}{60} = 50\space ms^{\normalsize-1}.$$

(ii) (a) The diagram is shown alongside.

(b) The fixed pulley is used to change the direction of effort.

(c) This pulley system is superior than a single fixed pulley because its mechanical advantage is almost double than that of a single fixed pulley.

(iii) (a) Magnetic field produced at A is into the plane of the paper whereas at B, it is outward to the plane of the paper.

(b) Magnetic field at any point due to a straight current carrying conductor is inversely proportional to the distance of the point from the conductor. As r_{1} > r_{2}, thus magnetic field at B will be larger as compared to that at point A.

(iv) (a) Sound of higher pitch is called shriller sound. The pitch of the sound depends upon frequency of sound. The sound which have higher frequency of vibrations have higher pitch.

(b) Loudness of sound is proportional to the square of amplitude of the vibrations producing the sound. When the amplitude of vibration is large, the sound produced is loud when the amplitude is small, the sound produced is feeble.

(v) (a) For the region AB, there is no change in temperature till the whole ice melts. In this process, the heat is being continuously supplied but the temperature of ice and water does not change. Here, the heat supplied is used in changing the state from solid (ice) to liquid (water). The process of change of state from solid to liquid is called melting.

(b) After the whole ice is converted into water in the region AB, if we continue heating, the temperature of the water begins to rise till it reaches at 100°C, where it again becomes steady. For the portion of the graph BC, the heat supplied is used to change water from liquid state to vapour or gaseous state. The change of state from liquid to vapour or gas, is called vaporisation.

## Section-B

**4.** (i) (a) The weight of rule 100 gf will act at the mid-point of rule i.e., at the mark 50 cm since the rule is uniform.

Taking moment about the knife edge,

w × (40 – 20) = 100 × (50 – 40)

$$\therefore\space w =\frac{100×10}{20} = 50 gf$$

(b) Resultant moment = 50 × (40 – 10) – 100 × (50 – 40)

= 1500 – 1000

= 500 gf cm. (anticlockwise)

It will rotate the meter rule anticlockwise.

(c) To balance the meter rule in part (b), a clockwise moment is required for which the weight 20 gf is to be kept on right side of knife edge at distance x cm (say). Then

20 × x = 500 or x = 25 cm

Thus, the weight 20 gf is to be suspended at the mark 40 + 25 = 65 cm.

(ii) (a) The arrangement of pulley system is shown alongside.

(b) Velocity ratio = Number of pulleys = 4

$$\text{Mechanical advantage =}\frac{\text{Load}}{\text{Effort}}\\=\frac{300 kgf}{100 kgf} = 3\\\text{Efficiency =}\frac{\text{Mechanical advantage}}{\text{Velocity ratio}}\\=\frac{3}{4}=0.75\space(\text{or} 75\%)\\\text{Efficiency =}\\\frac{\text{Output work (Work done by load)}}{\text{Input work (Work done by effort)}}\\\therefore\space\frac{3}{4}=\frac{300 kgf× 10 m}{\text{W}_{\text{effort}}}$$

∴ W_{effort} = 400 kgf × 10 m = (400 × 10) N × 10 m = 4 × 10^{4} J.

(iii) (a) (1) Energy is the capacity of doing work, while power is the rate of doing work.

(2) Energy does not depend on the time in which the work is done, while the power depends on the time in which the work is done.

(b) Given : W = 60 kgf = 60 × 10 N = 600 N, h = 10 m, t = 15 min = 15 × 60 s = 900 s

Increase in potential energy = mgh = W × h = 600 × 10 = 6000 J

$$\text{Power spent =}\frac{\text{Work done}}{\text{Time}}\\=\frac{6000\space\text{J}}{900\space\text{s}} = 6.67\space\text{W}$$

**5.** (i) (a) Since the object just disappears in the liquid, the refractive index of the liquid equals that of the object, i.e., 1.8.

$$\therefore\space 1.8 =\frac{\text{Speed of light in air}}{\text{Speed of light in liquid}}$$

∴ Speed of light in liquid =

$$\frac{3×10^{8}}{1.8}\text{ms}^{\normalsize-1}$$

= 1.67 × 10^{8} ms^{–1}

(b) (1) The image is formed at the 2F point on the other side of the lens.

(2) The image has the same size as the object.

(ii) (a) The bird appears to be higher than its actual position.

(b) The relevant ray diagram is shown here.

(c) We have refractive index of water w.r.t. air

$$=\frac{\text{Apparent height of the bird}}{\text{Real height of the bird}}$$

(iii) (a) Yes. A ray, incident along the principal axis of a lens falls normally on both its surfaces. This is so because the principal axis, by definition, is the line joining the centres of curvature of the two surfaces of the lens.

(b) The four common characteristics of the images formed by a concave lens are :

(1) It is always virtual.

(2) It is always erect.

(3) It is always diminished.

(4) It is always formed between the optical centre and the focus of the lens.

**6.** (i) (a) The vibration in Fig. (c) is of largest amplitude.

(b) The vibration in Fig. (c) is of least frequency.

(c) If the frequency of the principal note in Fig. (c) is f, the frequency of note in Fig. (a) is 3f.

Thus, f_{1} = 3f and f_{c} = f

Ratio of frequency between (a) and (c) is

f_{a} : f_{c}

= 3 : 1

(ii) (a) Potential Energy = mgh

= 0.450 × 10 × 80 = 360 J

(b) When coconut strikes the ground, its potential energy is converted to kinetic energy. This principle is known as conservation of energy. According to conservation of energy “Energy is neither be created nor be destroyed, It can be transform from one type to another.”

Hence, kinetic energy with which it strikes the ground is 360J.

(iii) (a) For first sound, f = 256 Hz, λ = 1.3 m

∴ Speed, v = fλ = 256 × 1.3 = 332.8 ms^{–1}

(b) For second sound, v = 332.8 ms^{–1}, λ = 2.6 m

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

The frequency (i.e., pitch) of first sound is 256 Hz which is twice that of the second sound (which is 128 Hz). Hence, to the listener, the first sound will be shriller than the second sound.

**7.** (i) (a) The current appears anti-clockwise when viewed from end A because end A will form a north pole.

(b) No deflection is observed as there is no relative motion between the magnet and coil.

(ii) (a) ΔQ = SmDT = 0.5 × 10 × 5 = 25 cal

(b) ΔQ = mL = 10 × 80 = 800 cal

(c) ΔQ = SmDT = 1 × 10 × 100 = 1000 cal

(iii) (a)

(b) The radioactive substance S is kept in thick lead container with a very narrow opening to stop the radiations coming out from other directions because they may cause biological damage.

(c) β radiations deflected the most.

(d) γ radiations are unaffected by the electrostatic field and shows no deflection.

**8.** (i) (a) The direction of magnetic field for A will be downwards and perpendicular to the plane of the paper.

The direction of magnetic field for B will be upwards and perpendicular to the plane of the paper.

(b) For A, it is South pole.

For B, it is North pole.

(ii) (a) The oscillations of pendulum D are free oscillations.

(b) The pendulums A and C having different lengths are made to oscillate with the frequency of D which is different from their natural frequency, hence the oscillations of A and C are forced oscillations.

(iii) (a) Mass number of Thorium = 235 – 4 = 231

(b) Atomic number of Thorium = 92 – 2 = 90

(c) _{92}U^{235} — _{2}He^{4} + _{90}Th^{231}

**9.** (i) (a) For R_{2} and R_{3},

$$\frac{1}{\text{R}} = \frac{1}{\text{R}_{2}} +\frac{1}{\text{R}_{3}}\\\text{Or\space}\text{R'}=\frac{\text{R}_{2}\text{R}_{3}}{\text{R}_{2} + \text{R}_{3}}\\=\frac{4×4}{4+4} = 2Ω$$

Total resistance of the circuit, R = R_{1} + R’ = 2 + 2 = 4 Ω

(b) Total current flowing through the circuit,

$$\space \text{I} =\frac{\text{V}}{\text{R}}\\\therefore\text{I}=\frac{20}{4} = 5\text{\text{A}}$$

(c) Potential difference across R_{1},

V = IR_{1} = 5 × 2 = 10 V

(ii) (a) Given that, mass of the lead bullet, m = 1 kg

Heat required to raise its temperature from 27°C to 327°C

ΔQ_{1} = SmΔT

= 125 × 1 × (327 – 27) = 3.75 × 10^{4} J

(b) Heat required to melt the bullet,

ΔQ_{2} = mL = 1 × 2.5 × 10^{4} J = 2.5 × 10^{4} J

(c) Given that, 50% of kinetic energy of the bullet is used to heat it.

$$\therefore\space\text{Heat developed =}\frac{1}{2}×\text{K.E}\\\text{or}\space\Delta Q_{1} + \Delta Q_{2}=\frac{1}{2}×\frac{1}{2}mv^{2}$$

where v is the initial speed of the bullet.

∴ (3.75 × 10^{4} + 2.5 × 10^{4}) =

$$\frac{1}{4}×1×v^{2}\\\text{or}\space 6.25×10^{4} =\frac{v^{2}}{4}\\\text{or\space} v =\sqrt{4×6.25×10^{4}}\\= 5×10^{2}\text{m/s}$$

= 500 m/s

(iii) (a) Power of the electric oven, P = 2 kW = 2 × 1000 W = 2000 W

(b) Given, potential difference or voltage, V = 220V

We know that, Power P = V × I

$$\text{I} =\frac{\text{P}}{\text{V}} =\frac{2000}{220} = 9\text{A}$$

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