ICSE Class 10 Physics 2020 Question Paper with Solutions

Ans. (i) The product of magnitude of force and the perpendicular distance of line of action of force from the axis of rotation is known as the moment of force. It is also known as torque.

(ii) Since, the SI unit of moment of force is Newton metre, i.e., Nm and the CGS unit of moment of force is dyne cm.

So, the required relationship between the SI and CGS unit of moment of force will be \(1 \mathrm{Nm}=10^{5}\) dyne \(\times 10^{2} \mathrm{~cm}=10^{7}\) dyne cm

Ans. One kilowatt hour is defined as the electrical energy which is consumed by the appliance of 1 kW power in duration of 1 hour.

So, this is the required relation between kilowatt hour and Joule.

Ans. It is given that a satellite is revolving around the Earth, So, let us represent the given situation through the following diagram, at particular instant.

Note: When the direction of acting force on the body is perpendicular to the direction of displacement, then work done remains zero. So, in this given situation, the work done by the satellite at any instant will be zero. It is because the centripetal force \(\left(\overrightarrow{\mathrm{F}}_{c}\right)\) is perpendicular to the direction of the displacement /motion.

Ans. (i) Consider the following diagram of lever,

The given diagram is of class III lever, because Effort(E) is lying between Fulcrum(F) and Load(L).

(ii) As Mechanical Advantage \(=\frac{\text { Effort arm }}{\text { load arm }}\).

So, if the distance of the effort from the fulcrum is increased, then the mechanical advantage of the above lever will also be increased without increasing its length.

Ans. (i) In case of primary cell, chemical energy changes into electrical energy as it is used to convert stored chemical energy into electrical energy.

(ii) In case of electric bell, the electrical energy changes into sound energy as the sounds produced by the electrical signals is converted into a form of sound energy.

Ans. According to question, crane A lifts a heavy load is 5 seconds and crane B does the same work in 2 seconds. ∵

∴ For crane A,

\(\{\because\) For crane \(\mathrm{A}, t=5 s\}\) and for crane B,

\(\{\because\) For crane \(\mathrm{B}, t=2 s\}\) On dividing eqns. (i) and (ii), we get

or

i.e., the power of crane A is 0.4 times to that of crane B.

Ans. When a ray of light falls normally on a rectangular glass slab, then it will be shown as follows :

So, in the given case, Angle of incidence = Angle of refraction = Angle of emergence = 0 i.e.,

Hence, the above ray diagram is the required ray diagram which shows the path of the ray till it emerges out of the slab.

Ans.

In the given ray diagram,

\(\angle r_{1}\) and \(\angle r_{2}\) are the angle of refractions.

So, this is the required path of the monochromatic light ray AB incident on the surface PQ of equilateral glass prism PQR till it emerges out of the prism due to refraction.

Ans.

(i) To get an enlarged real image, the object should be placed (in front of a convex lens) between F and 2F.

(ii) To get an enlarged virtual image, the object should be placed (in front of a convex lens) between F and O.

Ans. Actual depth of pond = ? Apparent depth of pond \(=2.7 \mathrm{~m}\).

∴ Refractive index of medium

∴

⇒

\(\Rightarrow\) Actual depth \(=3.6 \mathrm{~m}\)

Ans. (i) Since, speed does not change in vacuum, therefore, all colours will move with the same speed, i.e., with the speed of light ( \(3 \times 10^{\wedge 8}\) m/s).

(ii) It is given that wavelength of red light, \(\lambda_{\text {Red }}=\) \(7.8 \times 10^{-7} \mathrm{~m}\) and

wavelength of blue light \(\lambda_{\text {blue }}=4.8 \times 10^{-7} \mathrm{~cm}\) ∵ Velocity = wavelength × frequency

As we know that frequency of light does not change when it enters into a medium, i.e., \(f\) is constant. ∴ eqn. (i) becomes

As given

from eqns (ii) and (iii), we get

It means that the red colour light will have greater speed in glass.

Ans.

This is the required graph between the displacement from mean position and time for a body executing free vibrations in a vacuum.

Ans. It is given that a sound wave is travelling in water has wavelength of 0.4 m, i.e., \(\lambda=0.4 \mathrm{~m}\) Speed of sound in water, \(V=1400 \mathrm{~ms}^{-1}\)

As we know that audible frequency ranges between 20 Hz to 20,000 Hz. And the obtained value of frequency, i.e., 3500 Hz lies between this range, i.e., in between 20 Hz and 20,000 Hz. So, yes this sound wave is audible in water.

Ans. The stone lying in the sun get heated much more than water lying for the same duration of time. It is because the specific heat capacity of stone is less than that of water, due to which the temperature of stone is more.

Ans. It is not advisable to use a piece of copper wire as fuse wire in an electric circuit. It is because copper has low specific resistance and high melting point, due to which it will not break (fuse) during the overloading in the circuit.

Ans. Consider the given diagram,

Let

It can be seen from the figure that \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) are in parallel. So, their equivalent, \(\operatorname{Req}_{1}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}\) ⇒

Now, the above given diagram can be shown as follows

∵ \(\mathrm{Req}_{1}\) and \(\mathrm{R}_{3}\) are in series. ∴ equivalent resistance across AB will be \(R_{\mathrm{AB}}=\operatorname{Req}_{1}+R_{3}=(2+5) \Omega=7 \Omega\).

Ans. Consider a metallic block having mass ‘ \(m\) ‘, specific heat capacity ‘S’, gains heat energy ‘Q’ after rising the temperature ‘ \(\Delta \mathrm{T}\) ‘, then the relation among them is given by, ∵

For block A,

For block B,

As per question,

∴ from eqns. (i) and (ii), we get

It means that specific heat capacity of block P is half of the capacity of block Q.

Ans. It is given that when a current carrying conductor is placed in a magnetic field, it experiences a mechanical force \((F=i B L \sin \theta)\).

(i) So, the force experienced will be zero when the angle between the magnetic field and the length of the conductor is 0°.

(ii) And the force experienced will be maximum when the angle between the magnetic field and the length of conductor is 90°.

Ans. According to the question, a nucleus \({ }_{84} \mathrm{X}^{202}\) of an element emits an alpha particle followed by a beta particle

Ans. (i) The magnetic polarity of the loop, that faces us will be clockwise, i.e., south pole.

(ii) We can increase the strength of the magnetic field produced by the loop in the given diagram, by.

1. increasing number of turns of coil.

2. increasing the current.

Ans. Consider the given diagram,

It is given that Mas of simple pendulum, \(m=200 \mathrm{~g}=0.2 \mathrm{~kg}\). Potential energy at position A is zero, i.e., \(P E_{\mathrm{A}}=0\).

(i) Potential energy at position \(\mathrm{PE}, P E_{\mathrm{B}}=m g h\) \(=0.2 \times 10 \times 5=10 \mathrm{~J}\).

(ii) Whenever the bob of the pendulum rotates, then mechanical energy remains same at all the points according to the law of conservation of mechanical energy.

∴ Mechanical energy at point \(C=10 \mathrm{~J}\).

(iii) As the bob is rotating, so the potential energy of bob at position B will get converted into kinetic energy at point A.

∵ KE at position \(\mathrm{A}, K E_{\mathrm{A}}=\frac{1}{2} m v^{2}\) \(\Rightarrow \quad 10=\frac{1}{2} \times 0.2 \times v^{2}\) \(\Rightarrow \quad 100=v^{2}\) \(\Rightarrow \quad v=10 \mathrm{~ms}^{-1}\)

Ans. (i) With reference to the direction of motion, both centrifugal force and centripetal force acts in opposite direction. Centripetal force acts towards the centre while centrifugal force acts radially outwards.

(ii) No, the centrifugal force is not the force of reaction at centripetal force. It is because the action-reaction pair of forces acts on two bodies,, but here it is acting on a single body.

(iii) The magnitudes of centripetal and centrifugal force are equal.

Ans. (i)

So, this is the required diagram.

(ii) ∴ Velocity ratio is always equals to \(n\).

So, if the weight of movable block is doubled, then its mechanical advantage will get effected but the velocity ratio will remain constant.

Ans. Consider the given diagram,

(i) and (ii)

As we know that when a ray of light enters from air to water, i.e., from rarer to denser medium, it bends toward the normal. So, the above diagram will become as follows : air water 3

Ans.

Ans. (i) As per the information given in the question, we can conclude that the given lens is a convex lens because the formed image is inverted.

(ii) Given,

According to lens formula,

⇒

⇒

⇒

(iii) Magnification, \(m=\frac{v}{u}=\frac{60}{-30}=-2\)

Ans. (i) As clouds scatter all the colours of sunlight (i.e., VIBGYOR scatter) in the sky during the day, therefore, due to this reason, the clouds appear white.

(ii) Sky appears blue because when the light enters from the sun into the sky, the air layers scatter only blue part of light more than any other colour.

Ans. (i) SONAR, i.e., Sound Navigation and Ranging is a system which enables us to locate underwater objects by transmitting ultrasonic waves and detecting the reflecting impulse.

(ii) Frequency and Amplitude are respectively measurable quantities related to pitch and loudness.

Ans. (i) When a tuning fork is held close to ear, it has some sound energy. But when it is held such that its stem is in contact with the table surface, then due to the larger surface area of table top, it produces more vibrations with greater sound energy.

(ii) Let the man is standing infront of cliff at a distance of \(d \mathrm{~cm}\) from the cliff.

So, when he moves closer (after hearing echo) to the cliff by 84 m, then the distance will be \((d-84) \mathrm{cm}\). ∵ we know that

Since, velocity will remain same in both cases, therefore,

Ans. (i) Soft iron core are made of thin laminated sheets which are eight (8) T or U shaped.

(ii) Since the input voltage is greater than the output voltage, therefore, the given transformer is a step-down transformer. So, the given diagram will become as follows :

Ans. (i) The substances which have almost zero resistance at low temperatures, are called superconductors.

(ii) Current drawn by the appliance,

(iii) We know that semiconductor’s resistance decreases with the increase in temperature.

For examples : Silicon, Graphite, etc.

Ans. Consider the given diagram.

(i) Since, \(3 \Omega\) and \(1.5 \Omega\) resistor are in parallel combination. Therefore the potential difference across them will remain same.

Let \(\mathrm{I}_{1}\) be the current passing through \(3 \Omega\) resistor and \(\mathrm{I}_{2}\) be the current passing through \(1.5 \Omega\) resistor.

∵ potential difference across them is same,

From eqns. (i) and (ii), we get

So, the current through \(3 \Omega\) resistor will be, \(I_{1}=0.1 \mathrm{~A}\).

(ii) \(\because 3 \Omega\) and \(1.5 \Omega\) resistors are in parallel combination,

Now, \(\operatorname{Req}_{1}\) and \(4 \Omega\) are in series combination, therefore, their equivalent resistance,

∴ Total resistance, \(R_{\text {tot }}=R e q_{2}+\) Internal resistance

∴ from equn (iii) and (iv), we get

⇒

Ans. (i) The amount of heat which is to be supplied to a given mass of a material to produce a unit change in its temperature, is known as heat capacity.

(ii) The SI unit of heat capacity is joule per kelvin.

(iii) ∵ Heat capacity = mass × specific heat capacity.

\(\Rightarrow \quad C=\mathrm{mS}\) So, this is the required relationship between heat capacity and specific heat capacity of a substance.

Ans. (i) In the given graph, AB represents the specific heat of solid phase and CD represents the specific heat of Liquid phase. Since, the specific heat of solid phase is always greater than that of liquid phase. So, due to this reason, the slope of CD is less than that of AB.

(ii) \(t_{2}\) is the boiling point because, here solid is changing into liquid and \(t_{1}\) is the melting point because, here liquid is changing into gas.

Ans. For Ice,

For water,

Let T be the final temperature of water. Now, apply Principle of Calorimetry,

Ans. (i)

(ii) D.C. motor is used to convert electrical energy into mechanical energy, such as in electric fan etc.

Ans. (i) \begin{tabular}[t]{|l|l|} \hline Nuclear Fusion & Nuclear Fission \\ \hline 1. In this, smaller nuclei combines to form a larger one. & In this, a heavier nuclei breaks into the smaller nuclei. \\ \hline 2. Takes place at high temperature & Takes place at very very high temperature \\ \hline \end{tabular}

(ii) The solid radioactive waste should be sealed in red bag and placed in a cardboard radioactive waste box.

Ans. Nucleus A is composed of 84 protons and 128 neutrons. ∵

∴ The composition of nucleus A will be \({ }_{84}^{212} \mathrm{~A}\)

(i) Since, nucleus A emits an \(\alpha\)-particle, i.e.,

∴ The composition of nucleus B will be \({ }_{82}^{208} \mathrm{~B}\)

(ii) Since, nucleus B emits a \(\beta\)-particle, i.e.,

∴ The composition of nucleus C will be \({ }_{83}^{208} \mathrm{C}\).

(iii) The mass number of the nucleus A is 212.

(iv) If the nucleus C emits gamma radiations, then there will be no change in its composition.