self assessment test unit-i

Self Assessment Test

Unit-I

FORCE

Solutions

Maximum Marks : 30 Timing : 1 Hour

Section-I [20 Marks]

Ans. 1. (a) (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 Nm = 105 dyne × 102cm = 107 dyne cm (b) (i) The given diagram is of class III lever, because Effort(E) is lying between Fulcrum(F) and Load(L). (ii) As Mechanical Advantage = Effort arm / load arm. So, if the distance of the effort from the fulcrum is increased, then the mechanical advantage of the

given lever will also be increased without increasing its length. (c) (i) Two equal and opposite parallel forces, not acting along the same line, form a couple. (ii) Nm (d) (i) Because on a circular path, the direction of motion changes continuously which changes the velocity

of the body continuously. (ii) kilogram or gram (any unit of mass). (e) (i) When a number of forces acting on a body produce no change in its state of rest or motion, the

body is said to be in equilibrium. (ii) Static.Ans. 2. (a) (i) Ring – at the centre of the ring (ii) Rhombus – at the point of intersection of the diagonals (iii) Scalene triangle – At the point of intersection of the medians (iv) Cylinder – At the mid point on the axis (b) (i) Centripetal force – radially inwards towards the centre (ii) Centrifugal force – Radially outward away from the centre (c) Force acting = mg = 1.5 × 9.8 = 14.7 N (d) (i) 1 kgf is the amount of force with which the earth attracts a mass of 1kg towards itself. (ii) 1 kgf = 9.8 N (e) (i) Dimension changes in non-rigid body (ii) No change in dimension of rigid body.

Section-II [10 Marks]

Ans. 3. (a) (i) By principle of moments, 40 × 25 = W × 20 \ W = 50gf (ii) F is shifted towards 0 cm mark.

2 ] Oswaal ICSE Chapter wise & Topic wise Question Bank, PHYSICS, Class – X

(b) (i) Momentum (p) of a body is the quantity of motion contained in a body. It is given by the product of the mass (m) and its velocity (v).

p = mv. It is a vector quantity. (ii) The rate of change of momentum of a body is directly proportional to the force applied on it i.e.

F

pt

∆∝

∆

(iii) When mass is constant, the change in momentum depends only on the change in its velocity. (c) (i) No, the velocity does not remain constant since it is continuously changing its direction. (ii) Yes, the motion is accelerated since the velocity is changing. (iii) Two forces are acting on the stone: Centripetal and centrifugal. (iv) Forces are not similar in nature. Centripetal force is acting radially inward towards the centre and

centrifugal force is acting radially outward away from the centre.

WORK, ENERGY & POWERMaximum Marks : 30 Timing : 1 Hour

Section-I [20 Marks]Ans. 1. (a) One kilowatt hour is defined as the electrical energy which is consumed by the appliance of 1 kW power

in duration of 1 hour. 1 kilowatt hour = 1 kilowatt × 1 hour = 1000 watts × 3600s

= 1000 Js

× 3600 s

= 3600000 J = 3.6 × 106 J (b) (i) In primary cell, chemical energy changes into electrical energy as it is used to convert stored chemical

energy into electrical energy. (ii) In 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. (c) According to question, crane A lifts a heavy load in 5 seconds and crane B does the same work in 2

seconds.

Power = Work doneTime taken

\ For crane A,

PA = W5s

... (i)

{ For crane A, t = 5 s} and for crane B,

PB = W2s

... (ii)

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

PP

A

B =

W

WPP

A

B

5

2

25

⇒ =

or P PA B= 0 4.

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

(d) EE

1

2

1 12

2 22

1

2

1

2

21212

= =

×

m v

m v

mm

vv

SOLUTIONS 3 ]

1259

51

1

2

2

=

×

vv

vv

1

2

21259

15

259

=

×

=

vv

1

2

53

=

(e) (i) A force does work when a displacement is produced in the direction of the force. (ii) Zero work.Ans. 2. (a) (i) q = 0° So, work done = Fs cos q = 0 So, work done is minimum. (ii) q = 90° So, work done = Fs cos q = Fs So, work done is maximum. (b) Work done in 1 beat = 1 J Work done in 60s = 72 J

Power = WorkTime

W= =7260

1 2.

(c) Initial velocity = 3 v Final velocity = v Mass = m

Initial K.E. = 12

m(3v)2

Final K.E. = 12

mv2

Initial K.E.Final K.E.

= =

9212

9 1

2

2

mv

mv:

(d) As the height increases, velocity decreases, so K.E. decreases. Since height increases, P.E. increases. Thus at maximum height when K.E. becomes zero, P.E. becomes maximum. So, K.E. is totally converted into P.E.

(e)


Ans. 3. (a) (i) 1 J = 1 N × 1 m 1 J = 105 dyne × 100 cm 1 J = 107 ergs. (ii) 0°


(b) (i) Mass = m = 50 kg Momentum = P = 3000 kg ms-1

P = mv

v = Pm

=300050

= 60 ms-1

(ii) Kinetic energy = 12

mv2 = 12

× 50 × 602 = 90000 J

(c) (i) Elastic potential energy (ii) The potential energy of the compressed spring is converted into kinetic energy of the ball and so the

ball fly away. (iii) Principle of conservation of energy: Energy can neither be created nor be destroyed. It can only be

transformed from one form to another.

MACHINESMaximum Marks : 30 Timing : 1 Hour


Ans. 1. (a) (i) Second order (or class) lever.

(ii) Mechanical advantage of the lever increases.

(b) (i) Nut cracker

(ii) Hand pump

(c) For a simple physical balance, two arms are of equal length.

Mechanical advantage = Effort armLoad arm

= 1

(d) Mechanical advantage = LoadEffort

= =12050

2 4.

Velocity ratio = 3

\ Efficiency = Mechanical Advantage

Velocity Ratio× = × =100

2 43

100 80.

%

(e) (i) Single fixed pulley.

(ii) To change the direction of effort to a convenient direction.

Ans. 2. (a) Velocity of single fixed pulley is 1

Velocity ratio of single movable pulley is 2.

(b) (i) Class III lever

(ii) Class II lever

(c) In class III lever, effort is in between load and fulcrum.

So, effort arm is always smaller than the load arm.

Since Mechanical advantage = Effort armLoad arm

, it is always less than 1.

Example: Sugar tong

(d) A single movable pulley has efficiency less than 100% because:

(i) The weight of the pulley and its frame is not zero.

(ii) There is a force of friction between pulley wheel and the string.

(e) Load = mg = 6 × 10 = 60 N

Effort = 70 N

Mechanical advantage = LoadEffort

= =6070

0 857.

SOLUTIONS 5 ]


Ans. 3. (a) Fire tong is a class III lever.

m = 250 g = 0.25 kg

Load = mg = 0.25 × 10 = 2.5 N

Load arm = 28 cm = 0.28 m

Effort arm = 7 cm = 0.07 m

According the principle of lever,

Load × load arm = effort × effort arm

2.5 × 0.28 = Effort × 0.07

\ Effort = 10 N

(b) Work input = Effort × displacement of effort

Work output = Load × displacement of load

Efficiency =

Efficiency = LoadEffort

Displacement of loadDispacement of effort

×

Efficiency = Mechanical advantage × Velocity ratio.

(c) (i)

(ii) M.A. = LoadEffort

= =15060

2 5.

(iii) No. It is not an ideal machine.



Unit-IIMaximum Marks : 30 Timing : 1 Hour


Ans. 1. (a) When a ray of light falls normally on a rectangular glass slab, it will be shown as follows :

So, in the given case, Angle of incidence = Angle of refraction = Angle of emergence = 0 i.e., ∠i = ∠r = ∠e = 0 (b)

In the given ray diagram, ∠i = Angle of incidence ∠e = Angle of emergence ∠r1 and ∠r2 are the angle of refractions. AB = Incident ray BC = Refracted ray CD = Emergent ray. 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. (c) (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.

SOLUTIONS 7 ]

(d) Actual depth of pond = ? Apparent depth of pond = 2.7 m.

Refractive index of water = 43

= Actual depth

Apparent depth

43

=

Actual depth2 7.

\ Actual depth = 43

× 2.7 = 3.6 m

(e) (i) Electromagnetic radiations travels with the speed of light in air / vacuum. (ii) Electromagnetic radiations obey the laws of reflection and refraction of light.Ans. 2. (a) (i) Critical angle is the angle of incidence in the denser medium for which the angle of refraction in the

rarer medium is 90°. (ii) Critical angle is affected by : Colour / wavelength of light, temperature and refractive index of the medium. (b) (i) Infra-Red. (ii) They have long wavelength hence less scattered and hence can travel through the fog.

(c) (i) aHw = 1

w aH

(ii) aHw = 53

\ wHa = 35

(d) (i) Dispersion : The phenomenon of splitting of white light, into its constituent colours when it passes through a prism, is known as dispersion of light

(ii) Red colour of light is scattered the least (e) (i) Both are electromagnetic waves and hence have the same velocity in vacuum. So, the ratio of the

velocities is 1 : 1. (ii) Frequency α 1/ wavelength So, 4000Å has higher frequency.


Ans. 3. (a) With increase of angle of incidence, angle of deviation decreases, reaches to minimum value and then increases.

(b)

(i) 90°. (ii) Refracting periscope. (iii) Surface AB.


(c) (i) Real, inverted, diminished.

(ii) 1 1 1v u f− =

1 18

124v

= +−( )

1 18

124v

= −

v = 242

= 12 cm

Image distance = 12 cm

(iii) Magnification = m = vu=−

=−

= −1224

12

0 5.


Unit-IIIMaximum Marks : 30 Timing : 1 Hour


Ans. 1. (a) (i) Amplitudes of A and B are in the ratio 2 : 1. (ii) Wavelengths of A and B are in the ratio 1 : 2. (b) (i) Free vibration : Vibrations of a body in absence of any external periodic force with constant

frequency and amplitude. (ii) The energy is lost to the surrounding due to the friction of the surrounding medium.

(c) (i) Frequency α 1

Length

(ii) Frequency α Tension (d) dB (e) (i) 9 : 16 (ii) 1 : 1Ans. 2. (a) (i) 120 dB (ii) Quality (b) Noise pollution : The disturbance produced in the environment due to undesirable loud and harsh

sound of level above 120 dB. Source of noise pollution: Honking of vehicles in traffic jams.

(c) (i) Number of waves produced in one second = Velocity

Wavelength= =

240 2

120.

(ii) Time required to produce a wave = 1

Number of waves produced in 1 second ms= =

1120

8 3.

(d) Persistence of hearing = t = 0.1s For clear echo, this should be the minimum time the sound should take to travel from source to reflector

and back. So, 2d = 350 × 0.1 = 35 m \ d = 17.5 m (e) Length of air column

SOLUTIONS 9 ]


Ans. 3. (a) (i) Forced vibrations. (ii) Resonance. (iii) Frequency of the tuning fork B matches the natural frequency of the stretched wire and vibrates

with greater amplitude. (b) 1st case : λ1 = 0.332 m T1 = 10–3s

\ f1 = 1

1T = 103 Hz

v = f1λ1 = 103 × 0.332 = 332 m/s 2nd case : T2 = 10–4s

\ f2 = 1

2T = 103 Hz

λ2 = vf2

433210

= = 0.0332 m

(c) (i) The first sound the man hears the sound after 2 s means he hears the sound directly from the gun of the observer.

So, the distance between the man on the shore and the observer is 2 × 320 = 640 m (ii) The second sound the man hears after 3 s means he hears the reflected sound from the cliff. So, the distance between the man on the shore and the cliff is 3 × 320 = 960 m

So, the distance between the observer on the ship and the cliff is 12

( 960 - 640) = 160 m


Unit-IV

CURRENT ELECTRICITY

Maximum Marks : 30 Timing : 1 Hour


Ans. 1. (a) Resistorsaretobeconnectedinseriestohaveatotalresistancemorethan7Ω. (i)

(ii) Equivalent resistance = (2 + 3 +5)Ω=10Ω (b) (i) Live wire (ii) Earth wire (c) 5Ωand4Ωareinseries.So,theequivalentresistance= R1 = 5 + 4 =9Ω

R1and3Ωareinparallel.So,theequivalentresistance= R2 = 3 93 9×+

=2.25Ω

R2and8Ωareinseries.So,theeffectiveresistance= Reff = 8 + 2.25 =10.25Ω


(d) Resistance of a conductor depends upon :

(i) Length of the conductor (R ∝ L) (ii) Area of cross-section of the conductor

1R

A ∝

(iii) Material of the conductor

(e) Specific resistance of a material is the resistance of the material of unit length and unit area of cross-section.

Its S.I. unit is Ohm metre.

Ans. 2. (a) H = i2Rt = (2.5)2 × 20 × (5 × 60) = 37500 J

(b) (i) Colour code: Green or Yellow

(ii) P = I2R, where P = Power, I = current and R = resistance.

(c) E = P × t = 4 × 12 = 48 kWh

Total cost = 48 × 3.25 = ` 156

(d) Resistivity of semiconductor decreases with rise in temperature.

(e)


Ans. 3. (a) Total power, P = 5 × 60 + 2 × 40 = 380 W

Current drawn from mains = I = PV

=380220

= 1.73 A

Excess current available = 5 - 1.73 = 3.27 A

Current drawn by each 100 W bulb at 220 V = 100220

= 0.45 A

So, number of excess bulbs = Excess current

Current drawn by each 100 W bulb=

3 270..445

= 7.3

So, excess 7 bulbs can be lighted.

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

(ii) I = PV

=110220

= 0.5 A

(iii) Semiconductor (example: Silicon, Germanium)

(c) (i) 3Ωand1.5Ωareinparallel.So,theequivalentresistance= R1 = 3 1 53 1 5×+

.

. =1Ω

So, the potential difference across the parallel combination resistors = 0.3 × 1 = 0.3 V

So,thecurrentthrough3Ωresistor= 0.3/3 = 0.1 A

(ii) I = 1 8

41

.R + + r

0.3 = 1 8

1 4.

+ + r 5 + r = 6

\ r =1Ω

SOLUTIONS 11 ]

MAGNETISMMaximum Marks : 30 Timing : 1 Hour


Ans. 1. (a) A current carrying conductor is placed in a magnetic field experiences a mechanical force (F = i B L sinq).

(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°.

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

(ii) The strength of the magnetic field may be increased by:

(1) increasing number of turns of coil.

(2) increasing the current.

(c) (i) x < y

(ii) Induced e.m.f. is directly proportional to the number of turns of the coil.

(d) Speed of rotation may be increased by :

(i) increasing the number of turns of the coil.

(ii) increasing the strength of the magnetic field.

(iii) increasing the area of the coil.

(iv) increasing the current through the coil.

(e) (i) e.m.f. and the number of turns of the coil are directly proportional.

(ii) Alternating current.

Ans. 2. (a) (i) Electric bell

(ii) Electrical energy is converted into mechanical energy.

(b) Compass needle will show a deflection.

This is so because when current passes through a conducting wire, a magnetic field is produced around it.

(c) (i) To convert mechanical energy to electrical energy.

(ii) Faraday’s laws of electromagnetic induction.

(d)

DC motor AC Generator

Converts electrical energy to mechanical energy. Converts mechanical energy to electrical energy.

Based on the principle that a current carrying coil held in a magnetic field experiences a torque.

Based on principle of electromagnetic induction.

(e) The function of a split ring is to reverse the direction of flow of current in armature coil after every half turn, so that it keeps on rotating in the same direction.


Ans. 3. (a) (i) Soft iron core are made of thin laminated sheets which are eight (8) T, E, I 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 :


(b) A.C. Generator :

(c) (i) D.C. Motor :

(ii) In DC motor, the split ring reverses the polarity to the armature coil ends after every half turn, which enables the armature to rotate in the same direction continuously.

(iii) AC can be transported a large distance without much loss.


Unit-VMaximum Marks : 30 Timing : 1 Hour


Ans. 1. (a) (i) SI unit of temperature is Kelvin. °C + 273 = K (ii) X is water. (b) Total heat supplied = 100 × (4 × 60) = 24000 J 150 × L = 24000

\ L = 24000150

= 160 Jg–1

(c) Factors upon which the heat (H) absorbed by a body depends : (i) Mass (H α m) (ii) Change in temperature (H α Dt) (iii) Specific heat capacity (H α c) (d) m × 336 = 300 × 4.2 × (40 - 0)

SOLUTIONS 13 ]

\ m = 300 4 2 40

336× ×.

= 150 g

(e) Heat capacity of a body is defined as the amount of heat energy required to raise the temperature of the body by 1°C.

S.I unit : Joule/KelvinAns. 2. (a) (i) Heat energy (ii) 1 calorie = 4.2 J / 4.186 J / 4.18 J (b) Substance B is a good conductor of heat because specific heat capacity of B is less than that of A and

specific heat capacity. So, to raise the temperature of 1 kg of mass by 1°C, substance B requires less heat energy than the substance A and hence substance B gets heated faster.

(c) To raise the thermal heat capacity it is made thick, so that it can impart sufficient heat to the food. Moreover, it keeps the food warmer for a long time after cooking.

(d)

Heat Temperature

1. It is a form of energy. 1. It is the degree of hotness or coldness of a body

2. S.I. unit is Joule. 2. S.I. unit is Kelvin.

(e)

Heat capacity Specific heat capacity

1. Heat absorbed by a mass of a body to raise its temperature by 1°C.

1. Heat absorbed by unit mass of a body to raise its temperature by 1°C.

2. Depends on mass. 2. Does not depend on mass.


Ans. 3. (a)

(b) From conservation of energy, Heat energy = potential energy mc Dt = mgh

Dt = ghc=

×10 504200

= 0.119°C

(c) Let the final temperature of the mixture be T Heat lost by water at 50°C temperature = (200 × 10–3 ) × 4200 × (50 - T) Heat gained by ice during melting = (40 × 10–3) × (336 × 103) Heat gained by water after melting = (40 × 10–3) × 4200 × (T - 0) Applying the principle of calorimetry, Heat lost = Heat gained (200 × 10–3 ) × 4200 × (50 - T) = (40 × 10–3) × (336 × 103) + (40 × 10–3) × 4200 × (T - 0) 200 × 4.2 × (50 - T) = 40 × 336 + 40 × 4.2 × T 5 × 4.2 × (50 - T) = 336 + 4.2 × T T = 28.33°C



Unit-VIMaximum Marks : 30 Timing : 1 Hour


Ans. 1. (a) According to the question, a nucleus 84X202 of an element emits and alpha partice followed by a beta particle

84202

82198

24

82198

8319

X Z He

X Y

-emision

-emision

α

β

→ +

→ 8810+

∴− e

a b= 83 and =198

(b) No.

Alpha particle is 42He. But the given nucleus does not contain two protons and two neutrons.

(c) (i) Nuclear fusion reaction is called a thermo nuclear reaction because it requires extremely high temperature to take place.

(ii) 3He2 + 2H1 → 4He2 + 1H1 + Energy

(d) (i) Nuclear fusion – The process in which two lighter nuclei of lighter atoms combine to form a heavy and more stable nucleus with the liberation of large amount of heat.

(ii) The product formed is in nuclear fusion not radioactive, hence less harmful to human. Energy produced per nucleon (for same mass of nuclear material) is more than fission.

(e) (i) Elements of same mass number but different atomic numbers are called isobars.

(ii) Example of isobars: 146C, 14

7N

Ans. 2. (a) Properties of alpha particles:

(i) High ionising power and low penetration power.

(ii) Positively charged particles and get deflected by electric and magnetic field.

(b) During radioactivity, changes occur in nucleus of the radioactive element. Hence it is considered to be a nuclear phenomenon.

(c) (i) Gamma radiation

(ii) Alpha and Beta radiations

(d) (i) One should wear special lead-lined apron and hand-gloves.

(ii) Should use long lead tongs while handling radioactive substances.

(e) Mass number change = 238 - 206 = 32

Atomic number change = 92 - 82 = 10

Emission of 8 alpha-particles causes mass number change by 32. It also causes atomic number change by 16.

But here mass number change is 10. So, there must be emission of 6 beta-particles.


Ans. 3. (a) (i) Radioactivity is the spontaneous disintegration of an unstable nucleus to attain stability.

In the process, alpha, beta and gamma radiations are emitted.

(ii) Nuclear waste is the waste of radioactive material which remains after the use of radioactive material.

(iii) Nuclear waste must be put in a thick cask and buried in a specially constructed deep underground store.

SOLUTIONS 15 ]

(d) (i) 92U235

(ii) Fission of 92U238 is possible only by fast neutrons while the fission of 92U235 can be even possible by the slow neutrons.

(iii) 92U238 → 90U234 + 2He4

(c) (i) B (gamma).

(ii) C (Beta) has less mass as compared to A (alpha).

(iii) A.

(iv) C.

self assessment test unit-i

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