1st PUC Physics Model Question Paper 2 with Answers

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Karnataka 1st PUC Physics Model Question Paper 2 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.
Write the number of significant figures in 287.5 m.
Answer:
The number of significant digits =4.

Question 2.
Give example for a vector multiplied by scalar
Answer:
\(\overrightarrow{\mathrm{V}}_{x}=\left|\overrightarrow{\mathrm{V}}_{x}\right| \hat{i}\) where ‘V_x‘ is scalar and \(\hat{i}\) – unit vector along x- direction .
i.e, n(\(\overrightarrow{\mathrm{V}}_{x}\) ) = \(\left(n \mathbf{V}_{x}\right) \hat{i}\) ‘n’ is a scalar.

Question 3.
When is work said to be maximum?
Answer:
Work is said to be maximum when the angle between \(\overrightarrow{\mathrm{F}} \& \overrightarrow{\mathrm{S}}\) = θ = 0°.

Question 4.
Which law of motion is used to explain rocket propulsion?
Answer:
Newton’s III law.

Question 5.
Write S.I. unit of moment of inertia.
Answer:
S.I unit of moment of inertia is kg m²

Question 6.
Which is more elastic, rubber or steel?
Answer:
Steel is more elartic than rubber.

Question 7.
Define coefficient of volume expansion in soilds.
Answer:
The ratio of increase in volume to its original volume per degree rise in the temperature, is known as the coefficient of volume expansion of solids.

Question 8.
State Zeroth law of thermodynamics.
Answer:
If two thermodynamic systems are in thermal equilibrium with third individually, then the systems are said to be in mutual thermal equilibrium with each other.
If T_A ⇔ T_C, T_B ⇔ T_Cthen T_A ⇔ T_B. where ‘T’ represents the temperatures and A, B, C represent the thermodynamic systems.

Question 9.
How many degrees of freedom are there in diatomic gases?
Answer:
Degrees of freedom f = 5 for diatomic gas.

Question 10.
What is Doppler’s effect?
Answer:
The phenomenon of apparent change in frequency due to relative motion between the source of sound and the listener (observer) is known as Doppler effect.

Part – B

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

Question 11.
Name any two basic forces of nature.
Answer:

1. Gravitational force
2. Electro – Weak force
3. Strong Nuclear force.

Question 12.
Mention any two uses of dimension analysis.
Answer:
Dimensional analysis is used

1. to check the correctness of an equation
2. to express the unit of physical quantity from one system of unit into another.
3. to derive an equation.

Question 13.
Draw velocity time graph of
(a) a body starting from rest and moving with uniform acceleration.
(h) a body moving with uniform negative acceleration.
Answer:

Question 14.
Mention the expression for magnitude of two vectors and explain terms.
Answer:

Where \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are concurrent and coplanar vectors acting at an angle “θ’ between them.

Question 15.
What is the importance of centre of mass of system of particles?
Answer:
The point at which the entire mass of the particles appears to be concentrated is called the centre of mass. Centre of mass lies at the geometric centre..
In the case of a ring, a hollow sphere, a hollow cylinder, the centre of mass of a body lies outside the body.

Question 16.
Define Poisson’s ratio and write the formula for it.
Answer:
The ratio of lateral strain to longitudinal strain is known as poisson’s ratio.

Question 17.
What is meant by 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.

Question 18.
What are damped oscillations? Give one example.
Answer:
A damped oscillation is that in which the amplitude of the oscillating particle / body diminishes gradually due to external damping agencies or forces acting on it.
e.g.: A simple pendulum describes a damped oscillatory motion due to a damping force such as air resistive force and gravitational force of attraction.

Part – C

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

Question 19.
Obtain an expression for range of a projectile.
Answer:
We know that (v_ocosθ) remains constant.
Instantaneous distance along the horizontal is given by ,
x = (v_ocosθ)t
for x = R (range of the projectile), t = T= time of flight.

Question 20.
Mention types of friction.
Answer:

1. Static friction
2. Sliding friction / kinetic friction
3. rolling friction.

Question 21.
Distinguish between elastic and inelastic collision.
Answer:
In an elastic collision between the bodies under isolation, total linear momentum and kinetic energy will be conserved whereas in the case of inelastic collision only linear momentum will be conserved but not the total kinetic energy. In the case of a perfect elastic collision, the coefficient of restitution will be unity but for a perfectly inelastic collision, the coefficient of restitution will be zero. In general, collisions are neither perfectly elastic nor inelastic. The coefficient of restitution for an inelastic collision, lies between 0 and 1. In an elastic collision there will be no loss in the net kinetic energy of the system. The loss in the kinetic energy in an inelastic collision \(\frac{1}{2}\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right) v_{1 i}^{2}\)

Question 22.
State and explain the theorem of perpendicular axes.
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.
Derive the relation between acceleration due to gravity and gravitational constant.
Answer:
Let M be the mass of the Earth and R be its radius.
We know that F = \(-\frac{\mathrm{GMm}}{\mathrm{r}^{2}}\)
where ‘m’ is a mass body very close to the surface of the Earth.
Force of gravity on the mass’m’ is F’ = mg
However F = F’
Therefore mg = \(\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\)
i.e, g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)

Question 24.
Arrive at an expression for pressure inside a liquid.
Answer:
Let ‘h’ be the depth of point ‘p’. Let a liquid cylinder of base area and height ‘IT is resting on the point force exerted by liquid cylinder = mg where m = (Ahρ)
w = (Ahρ)g
By the definition p = \(\frac{w}{A}\)
i,e P = \(\frac{\mathrm{Ahp} g}{\mathrm{A}}\)
or p = hρg

For a given liquid and for low depths (upto 10 km) P∝h
Total pressure at a point inside = P_o + P_t; where P_o = atmospheric pressure, P_t = liquid pressure or gauge pressure.

Question 25.
What is latent heat ? On what factors does it depend? Give its S.I. Unit.
Answer:
The quality of heat required to convert 1 kg of a substance from one physical state into another at a constant temperature is called latent heat.
i.e., Q = mL
SI unit of latent heat (L) is Jkg^-1.

Question 26.
State three assumptions of kinetic theory of gases.
Answer:

1. Collisions of molecules on the walls are perfectly elastic.
2. The size of the molecules are neglected when compared to the volume of the container.
3. The contribution to the net pressure due to intermolecular collisions is neglected.
4. The root mean square value of speed is the average of velocities taken in all the three directions.

Part – D

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

Question 27.
Define centripetal acceleration and obtain expression for it.
Answer:
Definition: The acceleration of a particle directed towards the centre of a circular path is called centripetal acceleration.

Let \(\vec{r}\) and \(\vec{r}^{\prime}\) be the position vectors and \(\vec{v}\) and \(\vec{v}^{\prime}\) velocities of the object when it is at point P and P’. By definition, velocity at a point is along the tangent at that point in the direction of the motion. Since the path is circular, \(\vec{v}\) is perpendicular to \(\vec{r}\) and \(\vec{v}^{\prime}\)  is perpendicular to \(\vec{r}^{\prime}\) .

Therefore, \(\Delta \vec{v}\) is perpendicular to \(\Delta \vec{r}\). Average acceleration \(\frac{\Delta v}{\Delta t}\) is perpendicular to \(\Delta \vec{r}\)

The magnitude of \(\vec{a}\) is, by definition , given by \(|\vec{a}|=\lim _{\Delta \rightarrow 0} \frac{|\Delta \vec{v}|}{\Delta t}\)

The triangle formed by the position vectors is similary to the triangle formed by the velocity vectors.

If \(\Delta t\) is very small, Δ_θ will also small. The arc PP’ is approximately equal to |\Delta \vec{r}|
i.e., \(\lim _{\Delta \rightarrow 0} \frac{|\Delta \vec{r}|}{\Delta t}\) = v Thus, centripetal acceleration \(|\vec{a}|=\frac{v}{r} v=\frac{v^{2}}{r}\) and \(\vec{a}=\frac{v}{r} \frac{d \vec{r}}{d t}\)
The centripetal acceleration is always directed towards the centre. The centripetal force = ma.

Question 28.
State Newton’s second law of motion and 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 torque. Show that torque is equal to the rate of change of angular momentum of particle.
Answer:
The rotating effect of force is called torque and is represented by \(\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}\)
i.e. τ = rFsinθ
where θ is the angle between r and F.
For a system of n particles,

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

Question 30.
State and explain the Newton’s law of cooling.
Answer:
The rate of loss of heat of a body is directly proportional to the difference of temperature of the
body and the surroundings. This law holds good for small differences of temperature.
\(-\frac{\mathrm{dQ}}{\mathrm{dt}}=+\mathrm{K}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)\)
=
Where \(\left(\frac{-\mathrm{dQ}}{\mathrm{dt}}\right)\) is the rate of loss of heat, K is a constant, T₂ – T₁= difference in temperatures.

where dθ – difference between two temperatures, θ₂ – Average of the two temperatures, θ₁ – room temperature.

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

Question 32.
Explain modes of vibration of air column in closed Pipe.
Answer:
Let ‘L’ be the length of the closed pipe. A pipe with one end closed is known as a closed pipe system. Let V be the velocity of sound in air. Length of half segment = \(\frac{\lambda_{0}}{4}\)
i.e., L = \(\frac{\lambda_{0}}{4}\) i.e., λ₀ = 4L.

The least mode of vibration is called fundamental mode i.e.,
fundamental frequency f₀ = \(\frac{\mathbf{v}}{\lambda}=\frac{\mathbf{v}}{4 L}\)
In the second mode of vibration.

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

Question 33.
A player throws a ball upwards with an initial speed of 29.4 ms^-1
(a) What is the velocity at the highest point of its path?
(b) To what height docs the ball rise and after how long does the ball return to the player’s hands?
Take g = 9.8 ms^-2 Neglect air resistance.
Answer:
Given u = 29.4 ms^-1
v = 0 at the highest point
using v² = u² -2gh
ie., 0 = 29.4 × 29.4 – 2 × 9.8 × h
∴ h = \(\frac{29.4 \times 29.4}{19.6}\) m
i.e, h = 44.1m
Using h = ut – 1/2 gt² we get
Net displacement = 0; 0 = 29.4t – \(\frac{9.8 t^{2}}{2}\)
i.e, 4.9t² – 29.4t
i.e, 4.9t = 29.
i.e, t = \(\frac{29.4}{4.9} \mathrm{s}\)
or t = 6s
The ball takes 6s to reach the player’s hands.

Question 34.
A pump on ground floor of building can pump up water to All a tank of volume 30 m3 in 15 minutes. If the tank is 40m above the ground and efficiency of pump is 30%, how much electric power is consumed by the pump.
Take g -9.8 ms^-2, p = 1000 kgm^-3
Answer:
Given V = 30m³, t= 15 min × 60 = 900s
h = 40m, η = 30%, g = 9.8ms^-2, p(Rho) = 1000 kg m^-3

Question 35.
Calculate the acceleration due to gravity.
(a) at height 16 km above the earth’s surface and
(b) at depth 12.8 km below the surface and
Given radius of earth is 6400 km acceleration due to gravity on the surface of earth = 9.8 ms^-2
Answer:
Given R = 6400 km, g = 9.8ms^-2
(i) with altitude, h = 16 km,
g¹ = g \(\left(1-\frac{2 h}{R}\right)\) for h <<R
i.e, g¹ = 9.8 \(\left(1-\frac{2 \times 66}{6400}\right)\)
∴ g¹ = 9.8 × 0.995 or g¹ = 9.75 ms^-2 or g¹ = 9.751ms^-2

(ii) with depth:
i.e., g¹ = g(1-h / R)
i.e., g¹ = 9.8 \(\left(1-\frac{12.8}{6400}\right) \)
i.e., g¹ = 9.8 × 0.998
∴ g¹ = 9.780ms^-2

Question 36.
A Carnot’s engine whose efficiency is 30% takes heat from a source maintained at temperature of 600 K. It is desired to have an engine of efficiency 50%. What should be the intake temperature for same exhaust (sink)?.
Answer:
Given η= 30%, T₁ = 600 K, η’ = 50% T₁¹= ?
w.k.t., η = \(1-\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}\)
ie., 0.30 = \(1-\frac{\mathrm{T}_{2}}{600}\)
T₂ = (1 – 0.30) × 600
or T₂ = 420 K
For η¹ = \(1-\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}^{1}}\)
i.e, \(\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}^{1}}\) = 1-0.50 = 0.5
∴ T₁¹= \(\frac{420}{0.5}\) = 840K

Question 37.
A progressive wave is described by Y(r, t) = 0.005 sin (8πt – 3πx) where x, y are in metre and ‘t’ is in second. Find the (a) amplitude (b) wave length (c) period and frequency of wave.
Answer:
Given y{x, t) = 0.005 sin (80πt – 3πx); A = ?, λ, = ?, T = ?, f = ?
comparing the given equation y [x, t) = A sin (ωt – kx) with the equation
(i) A = 0.005m or A = 5 × 10^-3 m.

(ii) k = 3π rad m^-1
i.e., k = 3 × 3.142 = 12.726 rad m^-1
but k = \(\frac{2 \pi}{\lambda}\)
∴3π = \(\frac{2 \pi}{\lambda}\)
or λ = \(\left(\frac{2}{3}\right)\) = 0.667m

(iii) ω = 80 π
ie., 2πf = 80π
∴ f = 40 Hz

(iv) T = \(\frac{1}{f}\)
∴ T = \(\frac{1}{40}\) = 0.025s
or T = 25 × 10^-3s
or T = 25 ms