Oswal 61 Sample Question Papers ICSE Class 9 Physics Solutions

Section-A

Answer 1.

(i) (b) for measuring the inner diameter

Explanation :

Inside jaws are used to measure the internal diameter of a hollow cylinder or pipe.

(ii) (c) x / n

Explanation :

1 MSD = x cm

n VSD = (n – 1) MSD

∴ 1 VSD = ((n-1)/2) MSD

∴ L.C. = 1 MSD – 1 VSD

= 1 MSD –((n-1)/2) MSD

= n-(n-1)/n MSD

=1/n MSD

1 MSD = x cm

L. C. = x/n cm

(iii) (b) 2 R

Explanation :

Total time of motion on circular track
= 2 × 60 + 20 = 140 s.
Time period of revolution = 40 s. Therefore displacement in time = (3 × 40 = 120 s) = zero as athlete will
be reaching at the starting point.

(iv) (d) 1 m/s²

Explanation :

Here, u = 72 km/h = 20 m/s a = (v²-u²)/2s =(0-(20)2)/2 x 200 = – 1 m/s Retardation = - a = 1 m/s²

(v) (d) mv

Explanation :

Use momentum = mass × linear velocity shift to numerical section.

(vi) (c) impulse of force

Explanation :

Since, impulse is calculated as the product of force and time the unit Ns is defining the quantity impulse.

(vii) (c) Hydraulic turbine

Explanation :

Hydraulic turbine does not follow the principle of a hydraulic machine. It works on the speed of flow of
water.

(viii) (a) Vertically above its centre of gravity

Explanation :

For stable equilibrium the centre of buoyancy should be vertically above its centre of gravity so that it does not sink.

(ix) (d) Photosynthetic plants

Explanation :

The producers (photosynthetic plants) synthesize organic substances by the process of photosynthesis (i.e., they bind the simple compounds with the help of solar energy into the complex organic substances).

(x) (c) Hope’s experiment

Explanation :

In 1805, the scientist T.C. Hope derived a simple arrangement, known as Hope’s apparatus for
demonstrating the anomalous expansion of water.

(xi) (c) 60°

Explanation :

The situation is shown in figure.

(xii) (b) 30 cm

Explanation :

Here, f = – 20 cm, m = – 2 ( image is real)

m = − v /u , v = – mu = 2u

Using, 1/u+1/v=1/f

we get, =1/u+1/2u=-1/20

or 3/2u=-1/20 or u = – 30 cm

(xiii) (b) 3.4 km

Explanation :

Distance of thunder cloud
= Speed of sound × time
= 340 × 10
= 3400 m = 3.4 km

(xiv) (b) 4950 C

Explanation :

Current =Electric charge/Time

0.55=Electric charge/[150×60] sec

Electric charge = 0.55 × 9000

= 4950 C

(xv) (b) Steel

Explanation :

Preferred material for electromagnet that can be demagnetised easily alloy of aluminium, nickel and
cobalt give a substance called AlNiCo which has easy tempering properties and hence gets easily
demagnetised.

Answer 2.

(i) (a) Given : Size of particle = 5.2 μ Since 1 μ = 10–6 m ∴ Size of particle = 5.2 × 10^–6 m.
(b) The frequency of oscillation of a second’s pendulum is 0.5 s–1. No, it does not depend on the amplitude of oscillation.
(c) T = 1/f

(ii) (a) Fundamental quantities are those quantities which do not depend upon other quantities. The examples are : Mass, length, time etc.
(b) Given, Oscillations in 20 s = 80
∴ Oscillations in 1 s = 80/20=4
∴ Frequency = 4 Hz
Also, Time period = 1/Frequency=1/4=0.25s.

(iii) Distance = 3 × Circumference of field
Given : r = 7 m
∴ Distance = 3 × 2πr
= 3 × 2 ×22 x 7/7=132m.

Displacement = 0, because initial and final position of the athlete are the same.

(iv) When a carpet is hit with a stick, it sets the carpet in motion, whereas the dust particles embedded in it remain in the initial position of rest due to inertia. Thus, the dust particles are removed from the carpet.

(v) Given : Weight of body in air, W = 75.0 g
Weight of body in water, W1 = 70.5 g
Weight of body in liquid, W2 = 66.0 g
R.D. = (W-W2)/(W-W1)
∴ Relative density of the liquid =(75 66) /(75-70.5)=9/4.5=2.

(vi) Both, the stone and the pencil will reach the ground simultaneously in vacuum since acceleration due to gravity is same on both.

(vii) Iron nail has a less area and volume, so it readily sinks into water, but in case of a ship, it has a large amount of void space filled with air which reduces the average density of the ship, making it much lighter. So it floats in water.

Answer 3.

(i) (a) When the object is placed between pole and principal focus of the concave mirror.
(b) When the object is placed at the centre of curvature of the concave mirror.

(ii) (a) A convex mirror always produces an erect and virtual image. The image formed is shorter than the object.
(b) The maximum distance at which an image in a convex mirror can be obtained is the focal length. The location of the object will be at infinity.

(iii) Given : v = 332 ms–1, f = 480 Hz, λ = ?
We know, v = f λ
or 332 = 480 × λ
(a) ∴ λ =332/480=0.691 m
(b) Also, T = 1/f=1/480=0.00208 s.

(iv) (a) The potential difference between two points is said to be 1 volt (1V) if 1 joule of work is done in moving 1 coulomb of electric charge from one point to the other.
(b) A compass needle gets deflected when brought near a bar magnet because the bar magnet exerts a magnetic force on the compass needle which is itself a tiny pivoted magnet capable of moving in a horizontal plane.

(v) Given : Current, I = 0.25 A, Charge, Q = ?, Time, t = 10 minute = 10 × 60 s = 600 s
I = Q/l
0.25= Q/600
⇒ Q = 0.25 × 600
or Q = 150 C
Thus, the amount of electric charge that flows through the circuit is 150 C.

Section-B

Answer 4.

(i)

	Contact force	Non-contact force
1.	The force caused between two bodies due to actual physical contact is called contact force.	The force caused between two bodies separated by a distance is called as non-contact force.
2.	It is independent of distance of separation between the bodies.	It is inversely proportional to square of distance between the two bodies.

(ii) Given : Initial velocity, u = 0, Acceleration, a = 5 ms^–2
(a) Distance travelled in t = 4 s,
s₁ = ut + 1 / 2 at²
= 0 × 4 +1/2 x 5 x (4)²=40m
(b) Distance travelled in t = 5 s,
s₂ = ut + 1 / 2 at²
= 0 × 5 +1/2 x 5 x (5)²=62.5m
(c) Distance travelled in 5th second = Distance travelled in 5 s – Distance travelled in 4 s
= s₂ – s₁
= (62.5 – 40) m = 22.5 m

(iii) Given : u = 50 m/s, g = – 10 m/s2,v = 0
(a) v² = u² + 2gh
0 = (50)² + 2 × (–10) × h
∴ h = 50 x 50 /20 = 125 m.
(b) v = u + gt
or t = (v-u)/g=(0-50)/-10=5s

Answer 5.

(i) (a) Sound of frequency greater than 20 kHz is known as ultrasound.
(b) The two properties of ultrasound that make it useful to us are :
1. High energy content
2. High directivity.

(ii) (a) The range of frequency within which the sound can be heard by a human being is called the audible range of frequency.
(b) The audible range of frequency for humans is 20 Hz to 20 kHz.
(c) Human ears are most sensitive to frequencies between 2000 Hz to 3000 Hz.

(iii) (a) No effect
(b) Speed of sound increases with increase in temperature.
(c) No effect
(d) Speed of sound increases with increase of moisture in air.

Answer 6.

(i) (a) A concave mirror is used in the headlights of a car.
(b) A convex mirror is used as the side/rear view mirror of a vehicle.
(c) A concave mirror is used in a solar furnace.

(ii) Rules for drawing geometric images in concave mirror are :
1. Any ray of light travelling parallel to the principal axis of a concave mirror, after reflection passes through the principal focus of the mirror.
2. Any ray of light which first passes through the principal focus, after reflection, will travel parallel to the principal axis of a concave mirror.
3. A ray of light which first passes through the centre of curvature of a concave mirror after reflection will retrace its path.

(iii) (a) It shows that fluid pressure is directly proportional to depth.
(b) It shows that fluid pressure at a given depth is same in all direction.

Answer 7.

(i) (a) The expansion of a substance on heating is called the thermal expansion of that substance.
(b) Two substances that contract on heating are : water from 0°C to 4°C, and silver iodide from 80°C to 141°C.

(ii)

Temporary Magnet	Permanent Magnet
1. They are made of soft iron.	They are made from steel.
2. They behave like a magnet only as long as there is an inducing magnet.	They act as a magnet even on the removal of the inducing magnet.
3. They have very poor magnetic retention.	They have a very high magnetic retention.

(iii) (a) Pressure on piston P =Force/Area=F/a
(b) Pressure exerted by fluid =F/a
(c) Pressure exerted on piston Q = Pressure in fluid = F / a
(d) Thrust [force] acting on piston Q in upward direction = Force × area=F x A /a

Answer 8.

(i) (a) A real image is one which can be obtained on a screen.
(b) A concave mirror can be used to obtain a real image of an object.
(c) No, it forms a virtual image if the object is placed between the focus and pole of the mirror.

(ii) (a) The density of water is maximum at a temperature of 4°C and its value is 1000 kg m^–3.
(b) The graph showing variation in density of water with temperature in range 0°C to 10°C is shown below.

(iii) Three ways to use energy efficiently:
1. The use of compact fluorescent lights (CFL) saves 67% energy and may last 6 to 10 times longer than the incandescent lamps.
2. The use of advanced boilers and furnaces in industry can save sufficient amount of energy in attaining high temperatures while burning less fuel. Such technologies are more efficient and less polluting.
3. The fuel efficiency in the vehicles can be increased by reducing the weight of the vehicle, using the advanced tyres and computer controlled engines.

Answer 9.

(i) It will rest in the geographic north-south direction with north pole towards the geographic north, making some angle with the horizontal as shown in the figure.

$$\text{(ii) } \because \text{\space g =}\frac{\text{GM}}{\text{R}^2}\\\text{}\\ \space\space\space\space\space\space\space\space \therefore \text{\space\space\space\space M = }\frac{\text{gR}^2}{\text{G}}\\\text{}\\ \text{\space\space\space\space\space\space\space\space Here, g = 9.8 m/s}^2,\text{R = 6370 km = 6370× 10}^3\text{m and G = 6.67}×10^{-11} \text{ Nm}^2\text{kg}^{-2}\\ \space\space\space\space\space\space\space\space \text{Putting the values},\\\\\space\space M = \frac{9.80 \space m/s^2 × (6370×10^3)^2m^2}{6.67×10^{-11}\space Nm^2kg^{-2}}\\ = \frac{9.80 \space m/s^2 × (6370×10^3)^2m^2}{6.67×10^{-11}\space kg\space ms^{-2}\space mkg^{-2}} [\because \space 1\space N = 1\space kg\space ms^{-2}]\\ \\ \text{\space\space\space = }\frac{\text{9.80} ×\text{ (6370}× \text{10}^3\text{)}^2}{\text{6.67 }× \text{10}^{-11}\text{ kg}^{-1}}\text{ = 5.96 × 10}^\text{24}\text{kg.} \\ $$

(iii) (a) The completed diagram is as follows :

(b) Magnet 2 is the weaker magnet.

ICSE 61 Sample Question Papers

All Subjects Combined for Class 9 Exam 2024

Buy Now

The dot mark ◉ field are mandatory, So please fill them in carefully
To download the Sample Paper (PDF File), Please fill & submit the form below.