UNIT II :2019-20

CLASS: XI

TIME:3 Hrs SUBJECT:PHYSICS M.M.:70

Q.No VALUE POINT/EXPECTED ANSWERS MARKS TOT.

MARK

SECTION-A-

1. (c) 1 1

2. (a) 1 1

3. (b) 1 1

4. (b) 1 1

5. (c) 1 1

6. (d) 1 1

7. (c) 1 1

8. (d) OR (d) 1 1

9. (a) 1 1

10. (a) OR (d) 1 1

11. The ratio Stress/Strain decreases with reason. ½+½ 1

12. It's a scalar quantity. 1 1

13. The bulb of a thermometer is made up of diathermic wall . 1

14. Yes, this is possible when the entire heat supplied to the

system is utilised in expansion.

As ∆Q = ∆U + ∆W and ∆U = nCv∆T

∆Q = nCv∆T+ ∆W

If temperature remains constant, then ∆T = 0, this implies

∆Q = ∆W. This implies that heat supplied should perform

work against the surroundings.

½+½ 1

15. V2=200 cc 1 1

16. Longitudinal strain and shearing strain

1

17. (i)Conservation of mass

(ii) Bernoulli’s theorem

½+½ 1

18. =2𝐾1𝐾2

𝐾1+𝐾2 1 1

19. Molar heat capacity (Cp) 1 1

20. Work to Heat

½+½ 1

SECTION-B-

21. Hydraulic pressure exerted on the glass slab, p = 10 atm =

Ans.

OR

Formula for l and l’

So, the ratio of elongation in the two wires is 1:1.

½

½

1

½+½

1

2

22. since scale pass is adjusted to zero without putting block in

water, so the final reading on the scale when block is

immersed in water will be equal to upthrust on the block

due to water :

=V ρw g = 𝑀

ρ ρw g

1

1

2

23. According to wien’s displacement law, λmt = Constant

λ1 < λ3 < λ2

T1>T3>T2

½

½

1

2

24.

1

1

2

25. ∆U = 100W = 100J/s

∆W = 75J/s

increase in internal energy, ∆U = ?

∆Q = ∆U + ∆W

•°• ∆U = ∆Q - ∆W = 100 - 75

= 25J/s

½

½

1

OR

Efficiency of the engine = Output energy / Input energy

∴ η = W / H = 5.4 × 108 / (3.6 × 10

9) = 0.15

Hence, the percentage efficiency of the engine is 15 %.

Amount of heat wasted = 3.6 × 109 – 5.4 × 10

8

= 30.6 × 108 = 3.06 × 10

9 J

1

1

2

26.

½

½

1

2

27.

½

½

1

2

28. Diagram

Hooke's Law, F = -kx, where F is the force and x is the

elongation.

½

½

The work done = energy stored in stretched string = F.dx

The energy stored can be found from integrating by

substituting for force,

and we find,

The energy stored =1

2 kx

2, where x is the final elongation.

½

½

1

3

29. 1) Since young’s Modulus is given by the

slope of stress – strain graph, Since slop of

A is more than that of B, hence it

has greater young’s Modulus.

2) Ductility is the extent of plastic

deformation and it is greater for A.

3) Tensile strength is the direct measure

of stress required, from by graph, it is

greater for A.

1

1

1

3

30.

(i)(a)Defination

(b) Diagram

(c)Derivation

OR

𝟏𝟏

𝟐

½

½

½

½

½

3

1

1

31.

𝟏𝟏

𝟐

𝟏𝟏

𝟐

3

32.

1

2

3

33. { RELATION BETWEEN Cp and Cv }

To establish the relation, we need to begin from the first law

of thermodynamics for 1 mole of gas.

ΔQ = ΔU + pΔV

If heat ΔQ is absorbed at constant volume,

∴ pΔV = 0 and ΔQ = CvΔT for one mole of a gas

Now, ΔV = 0

Then, Cv = (ΔQ/ΔT)v = (ΔU/ΔT)v = (ΔU/ΔT) ----------> (1)

where the V is dropped in the last step, since U of an ideal

gas depends only on the temperature, not on the volume.

Now, heat ΔQ is absorbed at constant pressure, then

ΔQ = CpΔT

Cp = (ΔQ/ΔT)p = (ΔU/ΔT)p = (ΔU/ΔT)p

Now, p can be the dropped from the first term since U of an

ideal gas depends only on T, not on pressure.

Now, by using Eq. (2)

or Cp = Cv +p(ΔV/Δp)p --------------------> (2)

Now, for 1 mole of an ideal gas, PV = RT

If the pressure is kept constant

p(ΔV/ΔT)p = R -----------------------> (3)

From Eq. (1) , (2) and (3)

{ Cp - Cv = R }

Here, Cp and Cv are molar specific heat capacities of an ideal

gas at constant pressure and volume and R is the universal

gas constant.

1

1

1

3

34. Degree of freedom

Discussion of degrees of freedom of diatomic along with

diagrams

OR

Statement and explaination of equipartition of energy:

1

2

According to the equipartition theorem an ideal gas

distributes its energy equally in all degrees of freedom. The

average energy of a molecule in a gas associated with each

degree of freedom is kT/2 where k is the Boltzmann constant

and T the temperature.

1

2

3

35. According to Bernoulli’s theorem the sum of pressure

energy, potential energy and kinetic energy per unit mass is

constant at all cross-section in the streamline flow of an ideal

liquid. i.e.

1

(a) Suppose that a liquid is taken in a

container having a large cross-

sectional area A1. There is also a small

hole of cross-sectional area A2 on the

wall of the container.

Suppose that the velocity of efflux of

the liquid out of the container is v2. The velocity of a liquid

particle at the surface, at that instant, is v1 . The height of the

liquid column at any instant from the position of the hole to

the upper end of the tank is h, and the atmospheric pressure is

Po.From the principle of continuity, we get

v1A1 = v2A2 ….(1) Applying Bernoulli’s principle at the

free end outside the hole and also at the surface, we get

….(2)

Therefore, from equations (1) and (2), we get,

…(3)

If A2 << A1 then,

……(4)

(b) V1 = 70m/s, V2 = 63 m/s,

Area of wing A=2.5m2, Density of air ρ=1.3Kgm

−3

According to Bernoulli's theorem :

Where P1 & P2 is the pressure

on the upper & lower surface of the ring respectively.

Lift on the wing =(P2−P1)A

2

Therefore lift on the wing of the aeroplane =1.51×10

3N

OR

(a)

(b)

2

3

5

2

36. Ans.

(i) Graph

The construction of a heat engine following Carnot

cycle is :

1) Source of heat :- It is maintained at higher

temperature T1

2) Sink of heat – It is maintained at lower temperature

T2

3) Working base :- A perfect ideal gas is the working

substance.

Theory :- Carnot cycle consist of four stages:

1) Isothermal expansion

2) Adiabatic expansion

3) Isothermal compression

4) Adiabatic compression.

1

3

5

NO, The efficiency of a reversible heat engine is also

independent of the working fluid and depends only on the

temperatures of the reservoirs between which it operates.

NO.

OR

½

½

1

2

2

37. Diagram

Derivation :

Statement of equipartition Energy

Average kinetic energy depends only upon the

(i)absolute temperature and

(ii)is directly proportional to it.

OR

½

3

½

½

½

1

1

1

2

5