Class XI - Physics - 2 - MS - Practice for Everyone

Practice Paper

TERM II (2021 – 2022)

Class – XI

Physics

Time: 2 hours Maximum Marks: 35

General Instructions:

(i) There are 12 questions in all. All questions are compulsory.

(ii) This question paper has three sections: Section A, Section B and Section C.

(iii) Section A contains three questions of two marks each, Section B contains eight questions of three marks each; Section C contains one case study-based question of five marks.

(iv) There is no overall choice. However, an internal choice has been provided in one question of two marks and two questions of three marks. You have to attempt only one of the choices in such questions.

(v) You may use log tables if necessary but use of calculator is not allowed.

SECTION – A

Derive the expression for excess pressure inside a liquid drop.

Let r = radius of a spherical liquid drop of center O.

T – surface tension of the liquid. Let pi and p0 be the values of pressure inside and outside the drop.

∴ Excess of pressure inside the liquid drop = p_i – p₀

Let Δr be the increase in its radius due to excess pressure. It has one free surface outside it.

∴ increase in surface area of the liquid drop.

= 4π (r + Δr)² – 4πr²

= 4π [r² +(Δr)² + 2rΔr – r²]

= 8πr Δr….(i)

(∵ Δr is small Δr² is neglected.)

∴ increase in surface energy of the drop is W = surface tension × increase in area

= T × 8πr Δr ….(ii)

Also W = Force due to excess of pressure × displacement

= Excess of pressure × area of drop × increase in radius

= (p_i – p_o) 4πr² Δr

∴ From (ii) and (iii), we get

(p_i – p_o) 4πr² Δr = T × 8πr Δr

Or (p_i – p_o) = 2T/r

2. The opposite faces of a cubical block of iron of cross-section 4 cm² are kept in contact with steam and melting ice. Calculate the amount of ice melted at the end of 10 minutes if K = 0.2 cal cm^-1 s^-1 °c^-1 for iron. Latent heat of fusion of ice = 80 cal g^-1

L = latent heat of ice= 80 cal g^-1

A = 4 cm²

K = 0.2 cal cm^-1 s^-1 °C^-1

Δθ = θ₂ – θ₁ = 100 – 0 = 100°C

t = time = 10 minutes = 10 × 60 = 600s

Let m = amount of ice melted = ?

Using the formula for heat

we get Q = 0.2×4×100×600

Q = 24000 cal …(i)

Also, Q = mL = m x 80 …(ii)

∴ From (i) and (ii), we get

m × 80 = 24000 ∴ m = 24000/80 = 300 g.

A circular hole of a radius of 1 cm is drilled in a brass sheet kept at 293K. What will be the diameter of the hole when the sheet is heated to 393K? α for brass =18 × 10^-6 K^-1.

Here, α = 18 × 10^-6 K^-1

ΔT = 393 – 293 K = 100 K

r = radius = 1 cm

∴ D = 2r = diameter = 2 cm

and it acts as the original length l (say).

∴ Let D’ be the new diameter = ?

I f Δl be the increase in length, then using the relation,

Δl = αlΔt, we get

Δl = 18 × 10^-6 × 2 × 100 = 36 × 10^-4 cm

∴ increase in diameter, AD = Al = 36 × 10^-4 cm

∴ D’ = D + ΔD = 2 + 0.0036 = 2.0036cm.

3. The temperature of the gas in Kelvin is made 9 times. How does it affect the total K.E., average K.E., r.m.s velocity and pressure?

Total K.E., average K.E. and pressure become 9 times, but the RMS velocity is tripled

SECTION – B

4. Two exactly similar wires of steel and copper are stretched by equal forces. If the total elongation is 1 cm, find by how much each wire is elongated. Given Y for steel = 20 × 10¹¹ dyne cm^-2, Y for copper = 12 × 10¹¹ dyne cm^-2

Let Δl_s and Δl_c be the elongation produced in steel and copper wires respectively.

L_s, L_c be their respective lengths,

L_s = L_c (∵ wires are similar)

Y_s = 20 × 1011 dyne cm^-2

Y_c = 12 × 1011 dyne cm^-2

Δl_s + Δl_c = 1 cm

A = area of cross-section of each wire.

F = equal force applied.

∴ Using the relation,

Y = F/A x L/Δl_c we get

Δl_c = F/A x L_c/Y_c ….. (i)

And Δl_s = F/A x L_s/Y_s= F/A x L_c/Y_c …..(ii)

Dividing (ii) by (i), we get

∴ Δl_s/ Δl_c = Y_c/Y_s= (12 x 10¹¹)/(20 x 10¹¹)

= 3/5 = 0.6 …. (iii)

Also Δl_s + Δl_c = 1 cm

0.6 Δl_c + Δl_c = 1 cm

1.6 Δl_c = 1 cm

Δl_c =1/1.6 = 0.625 cm

From (iii)

∴ Δl_s= 0.6 x 0.625 = 0.325 cm

If B be the bulk modulus of a metal and a pressure P is applied uniformly on all its sides. If D be the density of metal, then find the fractional increase in its density.

Since, Volume of metal will decrease

Hence, change in volume will be negative.

i.e.,

strain = -ΔV/V

We know that

Bulk modules = stress/strain

⇒B = −P/(ΔV/V)

⇒ΔV/V = P/B …..(1)

Now, we know that density (D):

D = m/V

⇒ΔD = −m/V² ΔV

⇒ΔD = −(ΔV/V)

⇒ΔV/V = −ΔD/D ….(2)

equating (1) & (2) we get fraction increase in density at.

ΔD/D = P/B

5. Answer the following:

(a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?

Triple point of water has a unique value i.e., 273.16 K. The melting point and boiling points of ice and water respectively do not have unique values and change with the change in pressure.

(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) Scale?

On Kelvin’s absolute scale, there is only one fixed point, namely, the triple-point of water and there is no other fixed point.

On Celsius scale 0 °C corresponds to the melting point of ice at normal pressure and the value of absolute temperature is 273.15 K. The temperature 273.16 K corresponds to the triple point of water.

6. Derive the expression for the work done during:

(a) Isothermal process

(b) Adiabatic process

Suppose 1 mole of gas is enclosed in isothermal container. Let P₁, V₁, T be initial pressure, volumes and temperature. Let expand to volume V₂ & pressure reduces to P₂ & temperature remain constant. Then, work done is given by

W=∫dW

as PV=RT (n= mole)

=RT[lnV]

=RT[lnV₂ −lnV₁]

W=RTln V₂/V₁

W=2.303RTlog₁₀ V₂/V₁

for constant temperature

P₁/P₂ = V₂/V₁

So, also

W=2.303.RTlog₁₀ P₁/P₂

7. A 50g lead bullet (specific heat 0.02) is initially at 30°C. It is fired vertically upward with a speed of 840 ms^-1. On returning to the starting level, it strikes a cake of ice at 0°C. How much ice is melted? Assume that all energy is spent in melting only. Latent heat of ice = 80 cal g^-1.

Kinetic energy of bullet = mv²/2=1/2 × 0.02 × (840)² = 17640joule=4200cal

heat supplied by lead to ice: ΔH₁=msΔT = 50 × 0.02 × 30 = 30cal

Total heat supplied = 4200 + 30 = 4230cal

let M_i mass of ice melted L is the latent heat of ice

M_iL=4230

M_i × 80=4230

Mi = 52.875 g

8. Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

9. A harmonic oscillation is represented by y = 0.34 cos (3000 t + 0.74) where y and t are in cm and s respectively. Deduce

(i) amplitude,

The given equation is y 0.34 cos (3000 t + 0.74) …. (1)

Comparing eqn. (i) with the standard equation of displacement of harmonic oscillation,

y = r cos (ωt + Φo), we get

r = amplitude = 0.34 cm

(ii) frequency and angular frequency,

1500/π Hz, 3000 rads^-1

(iii) time period,

π/1500 s

(iv) initial phase.

0.74 rad

10. Derive expressions for apparent frequency when

(i) source is moving towards an observer at rest.

(ii) the observer is moving towards the source at rest.

(iii) both source and observer are in motion towards each other.

11. Explain why (or how):

(a) in a sound wave, a displacement node is a pressure antinode and vice versa.

In a sound wave, a decrease in displacement i.e., displacement node causes an increase in the pressure there i.e., a pressure antinode is formed. Also, an increase in displacement is due to the decrease in pressure.

(b) Bats can ascertain distances, directions, nature and sizes of the obstacles without any “eyes”.

Bats emit ultrasonic waves of high frequency from their mouths. These waves after being reflected back from the obstacles on their path are observed by the bats. These waves give them an idea of distance, direction, nature and size of the obstacles.

The quality of a violin note is different from the quality of sitar. Therefore, they emit different harmonics which can be observed by human ear and used to differentiate between the two notes.

What are the characteristics of wave motion?

Wave motion is a form of disturbance that travels in a medium due to repeated periodic motion of the particles of the medium.

The wave velocity is different from the particle velocity.

The vibrating particles of the medium possess both K.E. and P.E.

The particle velocity is different at different positions of its vibrations whereas wave velocity is constant throughout a given medium.

Waves can undergo reflection, refraction, diffraction, dispersion, and interference.

Distinguish between stationary and progressive waves

Progressive waves:

The disturbance travels onward. It is1 handed over from one particle to the next.

Energy is transported in the medium along with the propagation of waves.

Each particle of the medium executes S.H.M. with the same amplitude.

No particle of the medium is permanently at rest.

Changes in pressure and density are the same at all points of the medium.

Stationary waves:

The disturbance is confined to a particular region and there is no onward motion.

No energy is transported in the medium.

All the particles of the medium except at nodes execute S.H.M. with different amplitude.

The particles of the medium at nodes are at rest.

The changes of pressure and density are maximum at nodes and minimum at antinodes.

SECTION C

12. Bernoulli’s Principle:

It states that sum of pressure energy, kinetic energy and potential energy per unit volume of an incompressible, non-viscous fluid in a streamlined, rotational flow remains constant along a streamline.

m = volume x density

m = area of cross-section x length x density

At area of cross section A1 = a₁ v₁ ∆t ρ = a₁ v₁ ∆t ρ

At area of cross section A2 = a₂ v₂ ∆t ρ = a₂ v₂ ∆t ρ

a₁ v₁ = a₂ v₂…… (i)

change in K.E. of fluid = K.E. at B – K.E. at A

= ½ m (v₂²-v₁²)

= ½ a₁ v₁ ∆t ρ (v₂²-v₁²)

change in P.E. of fluid = mg (h₂– h₁)

= P.E. at B – P.E. at A

= a₁ v₁ ∆t ρ g (h₂– h₁)

Net work done (F.S) on the fluid

= work done on fluid at A – Work done on fluid at B

= P₁ a₁ x v₁ ∆t – P₂ a₂ x v₂ ∆t

= a₁ v₁ ∆t (P₁– P₂)

According to law of conservation of energy

Net work done on fluid = Change in KE + change in PE

a₁v₁∆t (P₁– P₂) = ½ a₁v₁∆t ρ(v₂²– v₁²) + a₁v₁∆t ρ g(h₂– h₁)

Dividing both side by a₁ v₁ ∆t, we get

(P₁– P₂) = ½ ρ(v₂²-v₁²) + ρ g (h₂– h₁)

P₁+ ½ ρv₁²+ ρ g h₁= P₂ + ½ ρv₂²+ ρ g h₂

P+ ½ ρv²+ ρ g h= constant

(i) Bernoulli’ s equation is applied to

venturi mrter
Orifice meter
Lifting of aeroplane
All of the above

(ii) Bernoulli’ s principle is based on conservation of

Mass
Energy
Momentum
None of the above

(iii) Bernoulli’s equation holds only for

Low viscosity and incompressible fluids in turbulent flow
high viscosity and incompressible fluids in streamline flow
Low viscosity and compressible fluids in turbulent flow
Low viscosity and incompressible fluids in streamline flow

(iv) If the flow speeds of the upper and lower of the wings of an aeroplane are 260 m/s and 250 m/s, the wings cover an area of 500 m², then what would be the lift generated in kN?

637.5
1275
2550
350

(v) Consider a tank of height 20 m filled with liquid of density 100kg/ m³. The area of tank is 10 m². If the tank has a hole of area 2 m² at the bottom, find the speed of the liquid flowing out through the hole when the height of liquid in the tank is 10 m. Assume speed of liquid descending at top of tank is 5 m/s.

20 m/s
14.14 m/s
15 m/s
20.615 m/s

Class XI – Physics – 2 – MS