1st PUC Physics Model Question Paper 4 with Answers

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Karnataka 1st PUC Physics Model Question Paper 4 with Answers

Time: 3.15 Hours
Max Marks: 100

General instructions:

1. All parts are compulsory.
2. Answers without relevant diagram/figures wherever necessary will not carry any marks.
3. Numerical problems solved without writing the relevant formulae carry no marks.

Part – A

I. Answer all the following questions : ( 10 × 1 = 1 )

Question 1.
Which component of velocity of a projectile .remains constant throughout the time of flight?
Answer:
‘u cos θ’ component remain contant.

Question 2.
State work-energy theorem.
Answer:
The difference ofP.lv. or K..E. acquired by a body at two different positions is equal to the amount of work done.
\(\int_{\mathrm{k}_{1}}^{\mathrm{k}_{\mathrm{f}}} \mathrm{d}(\mathrm{K} \mathrm{E})=\int_{x_{1}}^{x_{\mathrm{f}}} \mathrm{Fd} x=\text { work done }\)
i.e. W = K_f -K_i

Question 3.
In which type of collision, the two bodies stick together after collision?
Answer:
Inelastic collision

Question 4.
What is the total external force on a system of particles when its total momentum remains constant?
Answer:
External force = 0, thereby dp / dt= 0 and p = constant.

Question 5.
Write the relation between escape velocity and orbital velocity of a satellite revovling close to Earth surface.
Answer:
V_esc = √2V_orb

Question 6.
W hat is a Venturimeter?
Answer:
Venturimeter is a device to measure pressure difference between any two different areas of cross section of tubes.

Question 7.
What is the Mode of heat transfer through a solid?
Answer:
Conduction.

Question 8.
Mention the S.I unit of Young’s Modulus.
Answer:
Nm^-2

Question 9.
Define specific heat capacity of the substance.
Answer:
The amount of heat required to raise one kilogram of a substance through one degree kelvin of temperature is known as the specific heat of that substance.
C = \(\frac{\mathrm{Q}}{\mathrm{m} \Delta \mathrm{T}}\)

Question 10.
How docs average kinetic energy of a gas molecule depend on the absolute temperature?
Answer:
E_K ∝ T

Part – B

II. Answer any FIVE of the following questions: ( 5 × 2 = 10 )

Question 11.
Mention the strongest and weakest force among the fundamental forces in nature.
Answer:
The strongest force of nature is strong nuclear force and the weakest force of nature is gravitation force.

Question 12.
Draw position time graph for a particle moving with:
(i) negative unifom uniform velocity.
(ii) positive acceleration.
Answer:

Question 13.
What is a Projectile? Give an example.
Answer:
Aparticle thrown into the space at an angle and whose motion guided by the action of Earth’s gravity is called projectile motion.
eg: a bullet fired from a gun at an angle into air.

Question 14.
Show that power is equal to the dot product of force and velocity?
Answer:
The instantaneous power is defined as the limiting value of average power, as the time interval approaches zero.
p = \(\frac{\mathrm{d} \mathrm{W}}{\mathrm{dt}}\)
P = \(\overrightarrow{\mathrm{F}} \cdot \frac{\mathrm{d} \overrightarrow{\mathrm{r}}}{\mathrm{dt}}\) or P = \(\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{v}}\)
where \(\overrightarrow{\mathrm{V}}\) is velocity vector, \(\overrightarrow{\mathrm{F}}\) is force vector.

Question 15.
Define radius of Gyration. Write the expression for Moment of Inertia in terms of Radius of Gyration.
Answer:
The radius of gyration of a body about an axis may be defined as the distance from the axis of a mass point whose mass is equal to the entire body and whose moment of inertia is equal to the Moment of inertia depends on the mass of the rigid body and square of the radius of gyration.
I = MK².

Question 16.
State Kepler’s laws of Planetary motion.
Answer:
Kepler’s I law (Law of orbit): All planets revolve in elliptical orbits with Sun as one of its foci.
2a-Major axis length
2b – Minor axis length
S – Sun at one focus
S’ – The other focus of the ellipse
P – Perihelion position of the planet.
A-Apehelion position of the planet.

Kepler’s II law (Law of areas) : The line joining the planet and the Sun sweeps out equal areas in equal intervals of time.
\(\frac{\Delta \overrightarrow{\mathrm{A}}}{\Delta \mathrm{t}}=\frac{1}{2 \mathrm{m}} \overrightarrow{\mathrm{L}}\)

For a conservation force \(|\overrightarrow{\mathrm{L}}|\) is constant and hence \(\left|\frac{\Delta \vec{A}}{\Delta t}\right|\) is also constant

Kepler’s III law (Law of periods): The square of the period of revolution of a planet around the

For a conservative force T is constant and hence sun is directly proportional to the cube of the semi major axis of the ellipse.
i.e., T² ∝ a³ so that for two planets
\(\left(\frac{T_{1}}{T_{2}}\right)^{2}=\left(\frac{a_{1}}{a_{2}}\right)^{3}\)

Question 17.
Suggest two methods of reducing friction.
Answer:

1. Wear and tear can be minimised by polishing the surfaces and lubricating machinery parts with oil and grease etc.
2. Streamlining the bodies reduce fluid (air or water) resistance.

Question 18.
At which point is kinetic energy of a particle executing SHM:
(i) Maximum
(ii) Minimum?
Answer:
K.E is maximum at mean position and minimum at extreme position.

Part – C

III. Answer any FIVE of the following questions : ( 5 × 3 = 15 )

Question 19.
Using the principle of homogeneity of dimensions, check the dimensional consistency of the equation T² = \(\frac{4 \pi^{2} \mathbf{r}^{3}}{\mathbf{G} \mathbf{M}}\) Where, T is time period, G is gravitational constant, M is mass and r is orbital radius.
Answer:

∴ the given equation is dimenstionally correct.

Question 20.
Obtain an expression for time of fight of a projectile. Ans. Expression for time of flight:
Consider v = (v₀ sin θ) – gt

Put v = 0, for t = time of ascent = t vosin0
t_a = \(\frac{v_{0} \sin \theta}{g}\)
but time of ascent = time of descent and time of flight T = t_a + t_d
i.e. T = \(\frac{2 v_{0} \sin \theta}{g}\) and T∝ v₀ for a given angle of projection.

Question 21.
What is static friction? State the laws of static friction.
Answer:
The force of opposition that arises between the two surfaces in contact which tend to move relative to each other is known as static force.

1. Limiting friction depends on the nature of the surfaces in contact but not on the surface area as long as normal reaction remains the same.
2. Force of friction always acts tangential to the surface and opposite to the direction of motion.
3. The limiting friction is directly proportional to the normal reaction between two surfaces in contact.

Question 22.
State and explain theorem of perpendicular axes applicable to a planar body.
Answer:
Statement: The moment of inertia about an axis perpendicular to two other axes acting in the same plane with their point of intersection being a point on it and the (third) axis passing through the common point, is equal to the sum of moments of inertia about the two axes e.g . I_z = I_x + I_y
Let M be the mass of the disk of radius R.
M.I. about a point passing through the centre and perpendicular to the plane containing X and Y
is I_z = \(\frac{M R^{2}}{2}\)

Since X and Y are in the same plane, I_x = I_y.
I_z = I_x+ I_y becomes I_z = 2I_x
hence I_x = \(\frac{I_{Z}}{2}=\frac{M R^{2}}{4}\)
i. e. moment of inertia of a circular disc about the diameter = \(\frac{M R^{2}}{4}\)

Question 23.
Draw a typical stress-strain curve for a metal. Hence, locate:
(i) Yield point (ii) fracture point ‘
Answer:
Statement: The ratio of stress to strain is a constant for a material within the elastic limit.
Modulus of elasticity = \(\frac{\text { Stress }}{\text { Strain }}\)

Within the elastic limit, stress v/s strain is a straight line ‘A’ is the elastic limit upto which Hooke’s law is applicable. Beyond ‘B’ the yielding point, the wire extends but does not return to the initial state when the deforming force is removed. ‘F’ is the breaking point. ‘EF’ allows the material to be malleable and ‘DE’, ductile.

Question 24.
Distinguish between streamline flow and turbulent flow.
Answer:
Streamline flow :

1. This is characterized by a steady flow.
2. The velocity will be less then the critical value.
3. The speed and direction of a particle of the liquid at a point, remains the same.
4. Reynold’s No. < 1000.

Turbulent flow

1. This is characterized by an unsteady flow.
2. The velocity of flow exceeds the critical value.
3. The speed and direction of the liquid at a point varies.
4. Reynold’s No. > 2000.
5. 

Question 25.
Deduce α_v = \(\frac{1}{r}\) for an ideal gas with usual notation.
Answer:
At constant pressure, V ∝ T
i.e., V = KT. For an increase in the temperature AT, the corresponding increase in the volume of the gas will be ∆V.
i.e., (V + ∆V) = K(T + ∆T)
i.e., V + ∆V = KT + K∆T

i.e., \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{T}}=\mathrm{K}=\frac{\mathrm{V}}{\mathrm{T}}\) or \(\frac{\Delta \mathrm{V}}{\mathrm{V} \Delta \mathrm{T}}=\frac{1}{\mathrm{T}}=\gamma\) where Y is volume coefficient ot an ideal gas.

Question 26.
What is a degree of freedom? Mention the number of degrees of freedom of a monoatomic and diatomic gas molecule.
Answer:
The number of independent co-ordinates required to specify the position of a molecule is called degrees of freedom of a molecule.
Number of degrees freedom = 3 translatory degrees of freedom + 2 rotatory degrees of freedom =5.
For monootomic gas f= 3

Pard – D

IV Answer any TWO of the following questions : ( 2 × 5=10 )

Question 27.
Using velocity time graph, derive v² = v₀² + 2ax where the symbols have their usual significance.
Answer:
Let ‘v₀’ be the initial velocity of a particle describing uniform accelerated motion ‘a’.
Slope of the line AB = a = tanθ = \(\frac{\mathrm{BC}}{\mathrm{AC}}\) Where BC = v – v₀  and AC = t
i.e., a = \(\frac{v-v_{0}}{t}\)
x = 1/2 x OD x (OA + BD)

i.e x = 1/2 t(v₀ + v )
From (1), t = \(\frac{v-v_{0}}{a}\)

Hence (2) may be written as 2s = \(\left(\frac{v-v_{0}}{a}\right)\left(v+v_{0}\right)\)
i.e,. v² – v₀² = 2ax
∴ v² = v₀² = 2ax

Question 28.
State Newton’s second law of motion. Hence, derive F = ma
Answer:
Newton’s II law of motion :
Statement: “The rate of change in the linear momentum of a body is directly proportional to i the impressed force and takes place in the direction of force applied.

To show that \(\vec{F}=m \vec{a}\)
Let m be the mass of the body. Let \(\vec{p}_{i}\) be the initial linear momentum. Let \(\vec{p}_{f}\) be the final linear momentum as a result of the impressed force.
By definition \(\lim _{\Delta y \rightarrow 0} \frac{\Delta \vec{p}}{\Delta t}=\frac{d \vec{p}}{d t}\) where \(\Delta \vec{p}=\vec{p}_{f}-\vec{p}_{i} \), and \(\frac{d \vec{p}}{d t}\) is instantaneous acceleration.
From Newton’s II law of motion, dp –

Question 29.
Define angular momentum of a particle. Prove that the time rate of change of angular momentum is equal to torque acting on the particle.
Answer:
Moment of linear momentum is known as angular momentum L = Iω = m v r.
where, I = moment of inertia, ω – angular velocity.
For a system of n particles,

Question 30.
What is thermal radiation? List four properties of thermal radiation.
Answer:
Radiation is one of the modes of heat transfer which requires no material medium for its transmission.
Thermal radiation exhibits the properties like rectilinear propagation, reflection, refraction, interference, diffraction and polarisation.

Question 31.
Derive an expression for time period and hence frequency of an oscillating simple pendulum.
Answer:
The tangential component of force mg is mgsinG.
Torque : -L(mgsinθ) = I∝

Question 32.
What is an open pipe? Discuss the modes of vibration of the air column in an open pipe.
Answer:
A pipe with both ends open is called open pipe system.

I mode of vibration :
Let L be the length of the open pipe system.
Let v be the velocity of sound in air
and f₀ be the fundamental frequency.
from the fig L = \(\frac{\lambda_{0}}{2}\) i.e. λ₀ = 2L
But  f₀ = \(\mathrm{f}_{0}=\frac{\mathrm{v}}{\lambda_{0}}=\frac{\mathrm{v}}{2 \mathrm{L}}\)

II mode of vibration :
Here L =2 \(\left(\frac{\lambda_{i}}{2}\right)\) λ₁
Hence I overtone f₁ = \(\frac{\mathrm{v}}{\lambda_{1}}=\frac{2 \mathrm{v}}{2 \lambda_{1}}=2 \mathrm{f}_{0}\) ……..(1)

III mode of vibration :
Here L = \(3\left(\frac{\lambda_{2}}{2}\right)\)
so that II overtone f₂ = \(\frac{\mathrm{v}}{\lambda_{2}}\)
or f₂ = \(\frac{3 \mathrm{V}}{2 \mathrm{L}}\)
Hence, the fundamental, I overtone, II overtone are in the ratio f₀: f₁ : f₂ :: 1:2:3. The nth harmonic = (n – l)^th overtone = nf₀.

VI. Answer any THREE of the following questions : ( 3 × 5 = 15 )

Question 33.
A football player kicks a ball at a distance 10m from a vertical pole at au angle of 45°. The ball just clears the tip of the pole and falls at a distance of 10 m on the other side. Determine the height of the vertical pole.
Answer:
Given x = 10 m θ = 45° x¹ = 10m y = ?
Since R = x + x¹ = 20km w.k.t
Y = xtanθ – \(\frac{x^{2} \tan \theta}{\mathrm{R}}\)
i.e., y = 10 × tan 45° – \(\frac{100 \times \tan 45^{\circ}}{2 \theta}\)
i.e., y = 10 – 5 = 5m
∴ height of the pole = 5m.

Question 34.
A car weighing 900 kg moving with a velocity of 20ms-1 is uniformly accelerated with an acceleration of 2ms^-2 for 10 second. Calculate the work done and the force required to accelerate the car.
Answer:
Given
m = 900 kg, u = 20ms^-1, a = 2ms^-2, t = 10s, W = ? F = ?
Since F = ma,
F = 900 × 2 = 1800N
i.e., F = 1.80 × 10³N
F = 1.80x 103N
Applying s = ut + 1/2 at²
We gets = 20 + 10 + \(\frac{1}{2}\) × 2 × 100
∴ s = 300m
By definition W = Fs
hence work done = 1.80 × 10³ × 300
ie., W = 5.40 × 10⁵ J

Question 35.
An artificial satellite revolves round the Earth at a height 700 km from the surface. Calculate its (i) kinetic energy (ii) potential energy (iii) total energy. Given : Mass of satellite = 150 kg, Mass of earth = 6 x 10²⁴ kg, Radius of earth = 6400 km.
Answer:

Question 36.
A Carnot heat engine absorbs 600J of heat from source of temperature 800K and rejects 300J of heat to the sink. Calculate efficiency and temperature of the sink.
Answer:
Given, Q₁ = 600J, T₁ = 800K, Q₂ = 300J, η = ?
Using \(\frac{Q_{1}}{Q_{2}}=\frac{T_{1}}{T_{2}}\)
We get T₂ = \(\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}} \cdot \mathrm{T}_{1}=\frac{300}{600} \times 800 \mathrm{K}\)
T₂ = 400k
By using = 1 – \(\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}\) we get
= 1 – \(\frac{400}{800}\)
i.e. efficiency of Carnot’s heat engine is 50%.

Question 37.
A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
Answer:
Spring constant = \(\frac{50 \mathrm{kg}}{0.2 \mathrm{m}}\) = 250kgm^-1
We know that T – 2π \(\sqrt{\frac{m}{k}}\) ; T = 0.6s (given)
i.e. m = \(\frac{\mathrm{kT}^{2}}{4 \pi^{2}}=\frac{250 \times(0.6)^{2}}{4 \times(3.142)^{2}}\)
m =2.279 kg.
Mass of the body = 2.279 kg and its weight is 2.279 kgwt.