Physics Model Paper NEW 3

8/17/2019 Physics Model Paper NEW 3

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Board of Intermediate EducationJunior Inter - Physics

Model Paper (English Version)

TIME : 3 Hours MAX. MARKS : 60

SECTION - A

Note: 1) Answer All the questions. 10 × 2 = 20 2) Each question carries TWO marks.

1. Which of the following has symmetry?

a) Acceleration due to gravity.

b) Law of gravitation.

2. Express unified atomic mass unit in kg?

3. Give an example where the velocity of an object is zero, but its acceleration is

not zero?

4. Which physical quantity remains constant

(i) in an elastic collision (ii) in an inelastic collision?

5. Is it necessary that a mass should be present at the centre of mass of any system?

6. The displacement in S.H.M. is given by y = a sin (20 t + 4). What is the

displacement when it is increased by 2π / ω?

7. What are water proofing agents and water wetting agents? What do they do?

8. State the examples of nearly perfect elastic and plastic bodies.

9. Find the increase in temperature of Aluminium rod if its length is to be increased

by 1%.

(α −for Aluminium = 25 × 10−6

/ °C)10. What is the expression between pressure and KE of gas molecule?

SECTION - B

Note: 1) Answer any SIX questions. 6 × 4 = 242) Each question carries FOUR Marks.

u2 sin2θ11. Show that the maximum height and range of projectile are and

2g

u2sin2θ

respectively, where terms have their regular meaning.g

www.eenadupratibha.net



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12. Mention the methods to minimise the friction?

13. Show that in case of one dimensional elastic collision, the relative velocity of

approach of two colliding bodies before collision is equal to relative velocity of

separation after collision.

14. Find the centre of mass of three particles at the vertices of an equilateral triangle.

The masses of particles are 100 gm, 150 gm, 200 gm respectively. Each side of

equilateral triangle is 0.5 m long.

15. Obtain on equation for the frequency of oscillation of a spring of force constant

'k' to which mass 'm' is attached.

16. How does acceleration due to gravity change for the same values of height (h)

and depth (d).

17. What is Toricellis Law? Explain how the speed of efflux is determined with anexperiment.

18. (a) State Newton's law of cooling.

(b) A body cools down from 60° C to 50° C in 5 minutes and to 40° C inanother 8 minutes. Find the temperature of surroundings.

SECTION - C

Note: 1) Answer any TWO questions. 2 × 8 = 16

2) Each question carries EIGHT marks.19. Develop the notations of Work and Kinetic energy and show that it leads to

Work - Energy theorem.

20. Describe simple harmonic motion. Show that the motion of projection of a

particle performing uniform circular motion on any diameter is simple

harmonic.

21. Explain reversible and irreversible processes. Describe the working of Carnot's

engine. Obtain an expression for the efficiency.




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ANSWERSSECTION - A

1. Which of the following has symmetry?

a) Acceleration due to gravityb) Law of gravitation.

A: Acceleration due to gravity varies from place to place. Hence Law of gravitation

has symmetry.

2. Express unified atomic mass unit in kg?

A: Unified atomic mass unit (a.m.u) = 1.67 × 10−27 kg1

By the definition 1 amu = × mass of an atom of C1212

1 12= × 12 6.023 × 1023

= 1.67 × 10−24 gm.3. Give an example, where the velocity of an object is zero; but its acceleration is

not zero.

A: A body thrown up vertically, at maximum height its velocity v = 0. But

acceleration due to gravity acts on it so, even though v = 0, g ≠ 0.

4. Which physical quantity remains constant(i) in an elastic collision (ii) in an inelastic collision?

A: (i) Both momentum and K.E. remains constant.

(ii) Momentum remains constant.

5. Is it necessary that mass should present at centre of mass of any system.

A: It is not necessary for mass to be there at the centre of mass.

e.g.: ring, horse shoe magnet.

6. The displacement in S.H.M. is given by y = a sin (20t + 4). What is the

displacement when it is increased by 2π / ω?

A: There is no change in displacement because 2π / ω is time period (T).

7. What are water proofing agents, water wetting agents? What do they do?

A: Water proofing agents are the substances which are used to increase the angle of

contact.

e.g.: Wax.

Wetting agents are the substance used to decrease the angle of contact.

e.g.: Soaps, Detergents.




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8. State the examples of nearly perfect elastic and plastic bodies.

A: Nearly perfect elastic bodies are Quartz, Fibre.

Nearly perfect plastic bodies are Clay, Dough and Plaster of paris.

9. Find the increase in temperature of Aluminium rod if its length is to be

increased by 1%.

αAl = 25 × 10−6 / ° C.l2 − l1 1

A: = 1% = l1 100

l2 − l1α =

l1 ∆t

l2 − l1∆t =

l1(α)1 1

= × 100 25 × 10−6

= 400°C

10. What is the expression between pressure and KE of gas molecule?

2A: P = E

3

Where P − Pressure exerted by ideal gas.

E − Mean KE of translation per unit volume of gas.

SECTION - Bu2 sin2θ

11. Show that the maximum height and range of projectile are and

2gu2sin2θ respectively, where terms have their regular meaning.

g

A: Maximum height: The maximum vertical displacement of projectile during its

journey is called maximum height. At maximum height vertical velocity.

uy = u sin θ = 0, a = −g, s = Hmax∴ v2 − u2 = 2as

0 − (u sin θ)2 = 2 (−g) Hmax




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u2 sin2θ⇒ Hmax =

2g

Range: The maximum horizontal distance travelled by a projectile during time of

flight is called range.

ux = u cos θ, ax = 0

2u sin θT =

g

1s = ut + at2

21

R = ux T + (0) t2

2

R = ux T

2u sin θ⇒ R = u cos θ ×

g

u2 2sin θ cos θ∴ R =

g

u2 sin 2θ∴ R =

g

12. Mention the methods to minimise the friction.

A: Polishing: The roughness of surface is reduced by polishing (or) rubbing the

surface with sand paper.

Ball bearing: Since rolling friction is less than sliding friction. So vehicles like

bicycles, two wheelers, motor cars provided with ball bearings to reduce the friction.

Lubrication: Friction in machines is reduced by lubrication. A thin layer of an

oil (or) a fluid is used between surfaces in contact to reduce the friction. In heavy

and fast moving machine thick oil of high viscosity is used.

Streamlining: Automobiles and aeroplanes are specially designed with curved

surfaces, so that air layers may get streamlined during motion, friction will be

reduced.



u cos θu

Hmax

← Range →


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13. Show that incase of one dimensional elastic collision, the relative velocity of

approach of two colliding bodies before collision is equal to relative velocity

of separation after collision.

A: Let us consider two bodies of masses m1, m2 are moving with velocities u1, u2

along same line in same direction collided elastically; v1, v2 are final velocities.

By using Law of conservation of linear momentum

m1u1 + m2u2 = m1v1 + m2v2

m1 (u1 − v1) = m2 (v2 − u2) ..........................(1)from Law of conservation of KE1 1 1 1 m1u1

2 + m2u22 = m1v1

2 + m2v

22

2 2 2 21 1 m

1(u

12 − v

12) = m2(v2

2 − u22) .........................(2)

2 2

(2) ÷ (1)1 1 m

1(u

1

2 − v1

2) m2(v

2

2 − u2

2)2 2

=

m1(u1 − v1) m2(v2 − u2)

(u1 + v1)(u1 − v1) (v2 + u2)(v2 − u2) =

(u1 − v1) (v2 − u2)⇒ u1 + v1 = v2 + u2 ⇒ u1 − u2 = v2 − v1

∴ Relative velocity of approach before collision = Relative velocity of separationafter collision.

14. Find centre of mass three particles at vertices of an equilateral triangle. Themasses of particles are 100 gm, 150 gm, 200 gm respectively. Each side of

equilateral triangle is 0.5 m long.

A:



→ → → ← → > →

m1

u1 u2 v1 v2

m2 m1 m2

0 . 5 m

0.5 m150 gm

200 gm

0.5/20.5/2A(0, 0)

100 gm

B (0.5, 0)

0.5 0.5C ( , √

3)2 2


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Co-ordinates of A are (0, 0) m1x1 + m2x2 + m3x3Co-ordinates of B are (0.5, 0) Xcm =

0.5 0.5√

3m1 + m2 + m3

Co-ordinates of C are ( , ) m1y1 + m2y2 + m3y32 2 Ycm = m1

+ m2

+ m3m1 = 100 gm, m2 = 150 gm, m3 = 200 gm

(x1, y1) = (0, 0)

(x2, y2) = (0.5, 0)

0.5 0.5√

3(x3, y3) = ( , )2 2

m1x1 + m2x2 + m3x3Xcm = m1 + m2 + m3

0.5100(0) + 150(0.5) + 200( )2

= 100 + 150 + 200

75 + 50 125 5= = =

350 + 100 450 18

m1y1 + m2y2 + m3y3Ycm = m1 + m2 + m3

0.5√

3100(0) + 150(0) + 200(

)2= 350 + 100

50√

3 √

3= =

450 9

5 √

3∴ (Xcm, Ycm) = ( , ) m18 9

15. Obtain an equation for frequency of oscillation of a spring of force constant

k to which mass 'm' is attached.

A: Consider a spring of force constant 'k' loaded with mass 'm'. When it is pulled

down and released, the body makes vertical oscillations which are simple

hormonic. Let 'y' be the displacement of the mass. Restoring force

F ∝ − y⇒ F = −ky .................... (1)Where 'k' is called force constant. '-ve' sign indicates that restoring force and

displacement are opposite in direction. But F = ma ........... (2)from (1) & (2)




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ma = −kyk

a = − y = −ω2y ...................(3)m

k

[...ω2 =

]m

y yIn SHM time period, T = 2Π √

= 2Π √

a ω2y1 2Π

T = 2Π √

= ω2 ω

m∴ T = 2Π √

k

1But frequency of oscillation of spring (n) = T

1 k ∴ n = √

2Π m

16. How does acceleration due to gravity change for the same values of height

(h) and depth (d).

A: Variation of 'g' with height:

Consider a body of mass 'm' on the surface of earth. Let 'M' be the mass of earth,

'R' be its radius, if 'g' is acceleration due to gravity on surface of earth,

GMg = ...........................(1)

R2

gh is acceleration due to gravity at height 'h' from the surface of earth, then

GMgh = .................. (2)(R + h)2

(2) ÷ (1)

GM

gh (R + h)2 R2

= = g GM (R + h)2

R2

R 2∴ gh = g ( )R + h

1 h −2g

h= g

[

]= g

[1 +

](1 + h R)2 R



mm

x↓↑....

.....


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2hgh= g [1 − ] ...................(3)R∴ gh < gVariation of 'g' with depth:

When we go deep into the ground 'g' decreases. Assume that earth to be a

homogeneous uniform sphere of radius R and mass M, uniform density ρ.GM

Acceleration due to gravity on the surface of earth g = R2

But M = density × volume4

= ρ × ΠR33

4

G( ΠR3 ρ)3∴ g =

R2

4= ΠRGρ ..................(4)

3gd is acceleration due to gravity at depth 'd'.

4gd = ΠG (R − d) ρ .....................(5)

3

(2) ÷ (1)4 Π G(R − d)ρ

gd 3 R − d = =

g 4 R ΠGRρ3

dgd = g[1 − ] ............ (6)Rfrom equations (3) and (6) for same values of 'h' and 'd', gd > gh

17. What is Toricellis Law? Explain how the speed of efflux is determined with

an experiment.

A: Statement: The speed of efflux from oriffice (small hole) is numerically equal to

the speed of a freely falling body falls through a height 'h' in free

space.

∴ v = √

2gh

Proof: Consider a tank containing liquid of density 'ρ' with a small hole at adepth of 'h' from the surface. Let 'v' be the velocity of efflux. Area of hole



RM

d

>

>

m

(R−d)

m

h↑

↓

MR


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dθ = θ1 - θ2 = 60 - 50 = 10°Cθ1 + θ2 60 + 50

θ = = = 55° C2 2

10∴

= K[55-

θ0) ....................... (1)5 × 60Now the body cools from 50° C to 40° C in 8 minutes.θ1 - θ2 = 50° - 40° = 10° Cθ1 + θ2 50 + 40 = = 45°C

2 2

10 = K[45 - θ0] ........................ (2)8 × 60➁ K[45

-

θ0] = ① K[55 - θ0]

10

5 × 60 8= =

10 5

8 × 60⇒ 3θ0 - 85 ⇒ θ0 = 28.34°C

SECTION − C19. Develop notations of work and KE and show that it leads to Work-Energy

theorem.

A: Work: Work is said to be done when a force applied on the body displaces the

body through a distance in the direction of applied force

(or)

Workdone by a force is the product of component of force in the direction ofdisplacement and magnitude of displacement.

W = (F cos θ) SW = F

−.S−

If θ = 0° W = FSUnits: N-m (or) Joule

D-cm (or) erg.

It is a scalar, D.F. = [ML2T−2

]

Kinetic Energy (KE): The energy possessed by a body by virtue of its motion is

called Kinetic Energy.




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KE = 1

2mv2

m = mass of the body

v = velocity

D.F. = [ML2T

−2]

It is a scalar.

Units: N - m (or) Joule

D - cm (or) erg

Work-Energy theorem:

Work done by the resultant force on a body is equal to change in its Kinetic

Energy.W = 1

2mv2 - 1

2mu2

Proof: Consider a body of mass 'm' moving with an initial velocity u. Under action of

constant force F, it gains uniform acceleration 'a' and final velocity v, after

displacement of 's'.

W = Fs = mas

(v2 − u2)= m s

... v2 − u2 = 2as2s [ v2 − u2 ]a =

2s

v2 − u2∴ W = m ( )2

1 1W = mv2 − mu2

2 2

W = KEf − KEi

KEf , KEi are initial and final Kinetic energies. Hence theorem is proved.

★ This theorem can be applicable for a single particle as well as for a system.

★ It can also be applicable to a system under action of variable forces, conservative

and non-conservative forces.



m m

s

u- v-


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20. Describe simple harmonic motion. Show that the motion of projection of a

particle performing uniform circular motion on any diameter is simple

harmonic.

A: A body is said to be in simple harmonic motion if it moves to and fro along a

straight line, about its mean position and its acceleration is directly proportional

to its displacement in magnitude and opposite in direction, always directed

towards its mean position.

a ∝ - y ⇒ a = -kyk is constant of proportionality.

Consider a particle is moving with constant angular velocity 'ω' along thecircumference of a circle of radius 'A' in anticlockwise direction. This circle is called

"reference circle" and particle is called "reference particle".

XOX', YOY' are two mutually perpendicular diameters of the circle with

centre "O". Let "P" be the position of particle at an instant of time "t" and N is the foot

of perpendicular dropped on to YOY'. As "P" moves along the circumference of the

circle, foot of perpendicular makes simple harmonic motion on diameter along X-axis

(or) Y-axis. At t = 0, particle is at point X and reached point P after time "t". The foot

of perpendicular moves from O to N and its displacement is 'y' at that instant.

Angular displacement of particle is

XOP = θON

sin θ = ⇒ ON = OP sin θOP

y = A sin ωt [... θ = ωt]

If θ = ωt + φ, then y = A sin (ωt + φ)

where φ is phase angle.

Centripetal acceleration of P is aP

= A

ω2 along 'PO'

resolve aP into components. They are

aN = aP sin θ = Aω2 sin ωt

= ω2 (A sin ωt)

aN = -ω2 y ⇒ aN ∝ - y

-ve sign indicates that aN and y are in opposite direction.

Thus projecton of particle in uniform circular motion on to a diameter is simple

harmonic.




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t21. Explain reversible and irreversible processes. Describe the working of Carnot's engine. Obtain an expression for the efficiency.A: Reversible Process: A process that can be retraced back in the opposite

direction, so that it passes through the same states as in direct process (or)

forward process. Finally the system must be in thermal equilibrium with

surroundings, return to the original state is called reversible process.

Conditions: A reversible process must obey the following conditions. (i) The process

should be quasi-static. (ii) There is no loss of energy in any form. (iii) No amount

of heat to be converted into electric and magnetic forms.

Example: Peltier effect and Seebeck effect.

Fusion of ice and vapourisation of water.

Irreversible Process: A thermodynamical process that cannot be taken back inopposite direction is called an irreversible process.

e.g.: Work done against friction.

Magnetization of materials.

Diffusion of gases.

Carnot's Engine:

Working: A heat engine is device which convert heat into work.

Carnot's heat engine consists four parts, they are:

1) Working substance - An ideal gas taken in cylinder with non conducting walls

and conducting base.




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2) Source of heat - A hot body of high thermal capacity at temperature T1 K.

3) Sink - A cold body at low temperature T2 K (surrounding temperature).

4) Insulating stand - To make whole system perfectly a non-conducting,

cylinder is placed on this stand.

The working substance undergo a cyclic operation which has four steps.

Step-I: When cylinder is placed on source (T1), heat absorbed by the gas (Q1), from

source at temperature (T1), the states of gas changes from (P1V1T1) to (P2V2T1)

under isothermal process (expansion).

V2Workdone by the gas W1 = Q1 = nRT1 loge ( ) ..................(1)V1

= Area of ABB'A'A

Step-II: Cylinder is shifted from source to non - conducting base, states of gaschanges from (P2 V2 T1) to (P3 V3 T2). Workdone by the gas due to adiabatic

expansion.

nRWorkdone W2 = (T1 − T2) .....................(2)γ − 1

= Area of BCC'B'B

Step-III: Cylinder is placed on sink maintained at T2 K. 'Q2' be the heat released by

the gas to the sink at temperature T2 K due to Isothermal compression.

V3W3 = Q2 = -nRT2 log ( ) .....................(3)V4= Area of CC'DD'C

-ve sign represents compression & work is done on the system.

Step -IV: Cylinder is placed on non-conducting base. Due to adiabatic compression

workdone on the gas is W4

−nR (T1 −T2)W

4

= ..............................(4)

γ − 1= Area of DD'A'AD

Total workdone by the gas in complete cycle is

W = W1 + W2 + W3 + W4

V2 nR (T1 − T2) V3 nR (T1 −T2)= nRT1 log ( ) + − nRT2 log ( ) − V1 γ - 1 V4 γ − 1

V2 V3W = nRT

1 log

(

)− nRT

2 log

(

)...............(5)

V1 V4

W = Q1 − Q2 = Area of ABCD




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W Q1 − Q2Efficiency η = =

Q1 Q1

Q2= 1 −

Q1 V3

( )nRT2 log V4 V3 V2= 1 − [... = ]V2

V4 V1nRT1 log ( )V1T2∴ η = 1 −

T1



R. SUDHAR. SUDHA RANIRANI

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Physics Model Paper NEW 3