1st PUC Physics Model Question Paper 3 with Answers

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Karnataka 1st PUC Physics Model Question Paper 3 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.
Define acceleration.
Answer:
The time rate of change of velocity of a particle is known as its acceleration.

Question 2.
What does slope of velocity – time graph’ represent?
Answer:
The slope of the line on a v-t graph gives the acceleration of the particle.

Question 3.
For what angle of projection, does the range of a projectile is maximum?
Answer:
θ = 45°
Angle of projection 0 = 45° so that R_max = \(\frac{u^{2}}{g}\)

Question 4.
State Work-Energy theorem.
Answer:
The difference of RE. or K.E. required by a body at two different positions is equal to the amount of work done.
\(\int_{\mathrm{k}_{\mathrm{i}}}^{\mathrm{k}_{\mathrm{f}}} \mathrm{d}(\mathrm{K} \mathrm{E})=\int_{x_{\mathrm{i}}}^{x_{\mathrm{i}}} \mathrm{Fd}\) = work done
i.e. W = K_f-K_r

Question 5.
Mention an expression for moment of inertia of a thin, circular ring of radius ‘R’ and mass ‘M’ about an axis passing through its diameter.
Answer:
I = \(\frac{M R^{2}}{2}\) for an axis of rotation passing through the diameter of a thin circular ring.

Question 6.
Mention the value of universal gravitational constant.
Answer:
G = 6.67 × 10^-11Nm²kg^-2

Question 7.
Of rubber and steel, which is more elastic?
Answer:
Steel is more elestic then rubber.

Question 8.
What happens to viscosity of a liquid, when it is heated?
Answer:
Viscosity of liquid decreases with the increase of temperature.

Question 9.
State the first law of thermodynamics.
Answer:
∆Q = ∆U + ∆W. Where, ∆Q – energy supplied to the system.
∆U – increase in the internal energy of the system.
∆W – work done on the environment.

Question 10.
Mention the number of degrees of freedom of a monoatomic gas.
Answer:
Number of degrees of freedom of monoatomic gas is 3(translatory).

Part – B

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

Question 11.
Mention any two basic forces in nature.
Answer:

1. Gravitational force
2. Electro – weak force
3. Strong nuclear force

Question 12.
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction of the net force on the pebble (a) During its upward motion (b) During the downward motion.
Answer:

1. Even though the net force is acting upward, the force of gravity acting downward diminises its speed to zero with the passage of time.
2. F = ma, a = -g downward.

Question 13.
Metion the conditions for equilibrium of a rigid body.
Answer:
(1) The vector sum of the forces on the rigid body is zero for translatory equilibrium

(2) The vector sum of the torques on the rigid body is zero for rotatory equilibrium (a=0) i.e.

i.e. the components of X, Y and Z independently vanish to zero for linear equilibrium.

i.e. Sum of X components, Y components and Z components of torque on the particles, vanish for rotational equilibrium.

Question 14.
State and explain Newton’s law of gravitation.
Answer:
Statement: ‘Every body in the universe attracts every other body in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.’
Let M₁ and M₂ be the masses of the two bodies separated by a distance ‘r’. Let \(\overrightarrow{\mathrm{r}_{1}}\) and r \(\overrightarrow{\mathrm{r}_{2}}\) be the position vectors of the two bodies. The force on body (2) due to body (1).

where \(\overrightarrow{\mathrm{F}_{12}}\) is force on (1) due to M₂ (towards M₂ )
\(\overrightarrow{\mathrm{r}_{12}}\) is distance measured from the body (1) to (2)

Question 15.
Distinguish between heat capacity and specific heat capacity.
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 = \(\frac{\mathrm{Q}}{\mathrm{m} \Delta \mathrm{T}}\)
The amount of heat required to raise the given mass of the substance through one kelvin of temperature is known as its thermal capacity.
i.e, Thermal capasity of a substance = ms = \(\left(\frac{\mathrm{Q}}{\Delta \mathrm{T}}\right)\) where ∆T = IK.

Question 16.
Convert 100UC into Fahrenheit temperature.
Answer:
F = \(\frac{9 \mathrm{C}}{5}\) + 32 ; i.e., F = \(\frac{9 \times 100}{5}\) + 32 = 212
Hence 100°C = 212°F

Question 17.
Define period and frequency of oscillation.
Answer:
The smallest interval of time after which the motion of the particle is repeated is called the period of oscillation. The time taken for one complete to and fro motion of a particle is known as the period of oscillation.
The number of times a particle executes a complete to and fro motion in one second is known as its frequency.

Question 18.
Distinguish between longitudinal waves and transverse waves.
Answer:
Longitudinal waves

1. These waves can be set up in all the three material (solids, liquids and gases) media.
2. The particles of the medium vibrate parallel to the direction of propogation of the wave.
3. The density of the medium increases in the region of compressions and decreases in the region of rarefactions
4. These waves can be produced even if the particles are not strongly bounded by cohesive forces, e.g. Sound waves.
5. These waves cannot be polarised

Transverse waves

1. These waves can be setup only in solids and on the surface of liquids
2. The particles of the medium vibrate at right angles to the direction of propogation of the wave.
3. The density of the medium remains the same in the plane of the wave.
4. These waves can be set only in the presence of medium having high cohesiveness among the particles, e.g. Waves on the surface of liquids and solids.
5. These waves can be polarised.

Party – C

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

Question 19.
Check the correctness of an equation v – v₀ = at using dimensional analysis (symbols have usual meaning)
Answer:
Consider v – v₀ = at
LHS has the dimension [LT^-1]
Since [a] = [LT^-2]
We get [LT^-1] = [LT^-2] [T¹]
i.e., [LT^-1] = [LT^-1]
∴ LHS = RHS
Hence the given equation is dimensionally correct.

Question 20.
Mention any three advantages of friction.
Answer:

1. Friction helps to initiate and stop relative motion between the two surfaces in contact.
2. Friction helps in walking, standing, writing, fixing portraits on the wall etc.

Question 21.
Derive an expression for potential energy of a stretched spring.
Answer:
Work done by the spring force W_s= \(-\int_{0}^{\mathrm{x}_{\mathrm{r}}} \mathrm{F}_{\mathrm{s}} \mathrm{d} \mathrm{x}\)
i.e. W_s = \(-\int_{0}^{x_{m}} \mathrm{k} x \mathrm{d} \mathrm{x}\)
W_s = \(-\frac{1}{2} \mathrm{k} x_{\mathrm{m}}^{2}\)
Work done by the external pulling force = W = \(-\frac{1}{2} \mathrm{k} x_{\mathrm{m}}^{2}\)

Question 22.
Explain law of conservation of angular momentum in case of a person sitting in a rotating chair.
Answer:
According to the law of conservation of angular momentum, 1₁ω₁ = I₂ω₂
i.e., ω ∝ \(\frac{1}{1}\)
If the person spreads out his arms and legs then the angular speed of the chair decreases.

Question 23.
Define Young’s modulus. Mention an expression for Young’s modulus of a wire and explain the symbols used.
Answer:
Young’s modulus is defined as the ratio of longitudinal stress to longitudinal strain.
i.e., Y =\(\frac{\text { longitudinal stress }}{\text { longitudinal strain }}\)
For a wire under stretching force, Y = \(\left(\frac{F l}{A \Delta l}\right) N m^{-2}\)
where ‘A’ area of cross section and ;F’ is a stretching force acting normal to the wire.

Question 24.
Derive an expression for liquid pressure at a point inside the 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.
State and explain Dalton’s law of partial pressures.
Answer:
The total pressure of a mixture of ideal gases is the sum of their partial pressures.
The pressure exerted by a gas at the same conditions of volume and temperature if no other gases were present is called partial pressure.

Question 26.
Mention Newton’s formula for speed for sound in gases and explain Laplace’s correction to Newton’s formula.
Answer:
v = \(\sqrt{\frac{p}{\rho}}\) where ‘p’ is pressure on gaseous medium, ρ – density of gaseous medium.
Consider adiabatic variation in the pressure and volume of gaseous medium.
PV^γ = constant.
Since no change of heat takes place between the system and the surrounding.
∆ (PV^γ) = 0
i.e., γPV^{γ – 1} ∆ V + V^γ ∆P = 0 AP
i.e., Bulk modulus = \(-\frac{\Delta \mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}}=\gamma \mathrm{P}\)
Hence in the expression v = \(\sqrt{\frac{E}{\rho}}\), Elasticity ‘E’ is replaced by γρ.

Laplace equation is written as v = \(\sqrt{\frac{\gamma \mathrm{P}}{\rho}}\)
Where γ = 1 + 2/f and ‘f’ is number of degrees of freedom of gaseous molecule.

Part – D

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

Question 27.
Derive
X = v₀6t + \(\frac{1}{2}\) at² using velocity – time graph.
Answer:
A graph of velocity of a particle plotted along the Y-axis and time along the X-axis is known as v-t graph and the curve is known as velocity time curve.
Let ‘ v₀’ be the initial velocity of a particle. Let ‘a’ be the uniform acceleration v-t graph gives a straight line with a constant slope, tanθ = m

i.e. a = \(\frac{v-v_{0}}{t}\) ……. (1)
From the figure OABD is a trapezium.
Area of trapezium OABD = Area of rectangle OACD + Area of triangle ACB
The second term on the right indicates the additional distance covered by the particle due to acceleration

i.e. area of trapezium = (OA)(OD) + 1/2(AC)(BC) = v₀t + 1/2 t(v – v₀) by using (1)
Area of trapezium = v₀t + 1/2 at²
From the dimensional analysis, the right hand terms indicate the distance travelled. From the principle of homogeneity, the left hand side term should indicate the distance covered.
Hence area of the trapezium = x = v₀t + 1/2 at²
For a uniform motion, a = 0 and x = v₀t
For a particle starting from rest v₀ = 0, x = 1/2 at²
In a vector form, \(\vec{x}=\vec{v}_{0} t+\frac{1}{2} \overrightarrow{a t}^{2}\)

Question 28.
Derive an expression for centripetal acceleration.
Answer:
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, \(\vec{r}^{\prime}\) is perpendicular to \(\Delta \vec{r}\). Average acceleration \(\frac{\Delta \vec{v}}{\Delta t}\) is perpendicular to \(a\Delta \vec{r}\) .
The magnitude of \(\vec{a}\) is, by definition, given by \(| \vec{a}_{1}=\lim _{\Delta x \rightarrow 0} \frac{|\Delta \vec{v}|}{\Delta t}\)

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

The centripetal acceleration is always directed towards the centre. The centripetal force = ma.

Question 29.
State and verify law of conservation of linear momentum in case of a system of two bodies.
Answer:
Statement: In an isolated system of collision of bodies, the total linear momentum before impact is equal to the total linear momentum after impact.

Before collision Impact After collision
Let m1 and m, be the masses of two bodies moving along \(\vec{v}_{1 i}\) and \(\vec{v}_{2 i}\)  (. Let \vec{v}_{1 f} and \(\vec{v}_{2 f}\) be the final velocities after the impact.
At the time of impact, the force of action acts on the body B and the force of reaction acts
on A.
Applying Newton’s III law of motion
|ForceofactionortB| = -|ForceofreactiononA|

This shows that the total final linear momentum of the isolated system equals its total initial momentum.

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

Question 30.
What is meant by acceleration due to gravity? Derive an expression for variation of acceleration due to gravity with height from earth’s surface,
Answer:
The constant acceleration of all bodies with which these are accelerated towards the centre of the Earth is known as acceleration due to gravity.
Let M_E be the mass of the Earth,
R_E be the radius of the Earth.
Let’m’ be the mass of the body at ‘h’ above the surface of the Earth.
Acceleration due to gravity on the surface of the Earth (h = 0), g(0) = \(\frac{G M_{F}}{R_{E}^{2}}\)
Acceleration due to gravity at an altitude ‘h’


Question 31.
What is heat engine? With schematic diagram, explain the working of Carnot’s heat engine.
Answer:
Working of Carnot’s engine:
Step 1: The working substance (ideal gas) is enclosed in a non-conducting wall and conducting bottom of a cylinder fitted with air tight non conducting piston. This is placed on the source having an infinite thermal capacity at a steady temperature. The top surface is conducting and the rest non conducting. As a result, the gas expands isothermally. The work done by the system,
Q = W₁ = μRT₁ log \(\frac{V_{2}}{V_{1}}\) (Curve A B)

Step 2 : The working substance is now placed on a non conducting platform, as a result of which no heat exchange takes place between the system and the surroundings. The system expands adiabatically at the expense of its internal energy. The gas cools. The work done by the system,

w₂ = \(\frac{\mu \mathrm{R}}{\gamma-1}\) (T₁ – T₂) (Curve B C)

Step 3 :
The working substance is now placed on the sink maintained at a steady low temperature T₂K. The system undergoes isothermal compression at this temperature. The pressure of gas increases and volume decreases w ithout any change in the internal energy and specific heat of gas remains at infinity .
w₃ = μRT₂log \(\left(\frac{v_{4}}{v_{3}}\right)\) (Curve C D )

Step 4: The working substance is placed on a non-conducting platform, Under thermal isolation, the system undergoes change in its internal energy and its specific heat remains at zero. Adiabatic compression results in increase in the pressure and temperature at the expense of work being done on the system. The system is allowed to reach its initial state. This completes one cycle of operation. The area bounded by the curves gives the amount of heat converted into work.

Question 32.
Derive an expression for total energy in simple harmonic motion.
Answer:
We know that total energy of a particle executing S HM,
TE = KE + PE
Where K.E = \(\frac{1}{2}\) mv²
K.E = \(\frac{1}{2}\) mA² cos² (ωt +Φ ) ω²
and P.E = \(\frac{1}{2}\) ky²
i.e., P.E = \(\frac{1}{2}\) ω². A²sin² (ωt +Φ )
Hence TE = \(\frac{1}{2}\) mω² A²cos² (ωt +Φ ) + sin² (ωt +Φ )
or T.E = \(\frac{1}{2}\) mω². A²
Representing mω² = k
we get TE = \(\frac{1}{2}\) kA²

KE of particle = \(\frac{1}{2}\)mkA²
= \(\frac{1}{2}\) mω²A² – \(\frac{1}{2}\) ω²x²
= \(\frac{1}{2}\) mω²(A² – x²)
where x = A cos(ωt + Φ)
Also K.E. = \(\frac{1}{2}\) mω² (A²)(1 – cos²(ωt + Φ)
i.e., K.E. = \(\frac{1}{2}\) mω²A²sin²(ωt +Φ).

K(t) – K.E. of a particle.

U(t) = P.E. of a particle.
T.E. =K(t) + U(t)= \(\frac{1}{2}\) kA²
where A – amplitude, k = mω², m – mass of the particle.
ω = angular frequency = \(\frac{2 \pi}{\mathrm{T}}\)
T = period of SHM.
Note : T.E = P.E + K.E = \(\frac{1}{2}\) mω² A²cos²(ωt + Φ) + \(\frac{1}{2}\) mω² A²sin²(ωt + Φ) = \(\frac{1}{2}\) mω²A²

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

Question 33.
A cricket ball is thrown at a speed of 50 ms-1 in a direction making an angle 30° with the horizontal. Calculate time of flight and maximum height. (Given g = 9.8 ms^-2)
Answer:
Given u = 50 ms”1, 0 = 30°, T = ?, H = ?
T _ 2u sin 9
(i) w.r.t time of flight T = \(\frac{2 u \sin \theta}{g}\)
i.e., T = \(\frac{2 \times 50 \times \sin 30^{\circ}}{g}\)
i.e., T = 5.1s

(ii) Maximum height reached
H = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
i.e., H = \(\frac{50 \times 50 \times \sin ^{2} 30^{\circ}}{2 \times 9.8}\)
H = 31.9 m

Question 34.
A pump on the ground floor of a building can pump up water to fill a tank of volume 30m3 in 15 minutes. If the tank is 40 m above the ground and the efficiency of the pump is 30%, how much electric power is consumed by the pump?
(Density of water is 10³kgm^-3)
Answer:
Given, V =30m3, t = 15 min = 900s, h = 30m, η = 30%, p_w = 1010³kgm^-3, p = ?

Hence power of the pump required = 32.7kW.

Question 35.
The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds.
(i) What is its angular acceleration?
(ii) How many revolutions does the wheel make during this time?
Answer:
Given ω_i = 1200 rpm; ω_f = 3120 rpm, t = 168, α = ? N = ?
ω_i = \(\frac{1200 \times 2 \pi}{60}\) rad.s^-1
and ω_f = \(\frac{3120 \times 2 \pi}{60}\) rad.s^-1 = 104 πrad.s^-1

(i) angular acceleration oc = \(\frac{\omega_{f}-\omega_{i}}{t}\)
i.e., α = \(\frac{104 \pi-40 \pi}{16}\)
i.e., α = \(\frac{64 \pi}{16}\)
or α = 4π rad. s^-2

(ii) angular distance θ = wt + \(\frac{1}{2}\) αt²
ie θ = 4Oπ × 16 + \(\frac{1}{2}\) × 4π × 16 × 16
i.e., θ = 64π + 512π
i.e., θ = 1 152π rad
Number of rotations = \(\frac{\theta}{2 \pi}=\frac{1152 \pi}{2 \pi}\)
i.e., N = 576

Question 36.
A body cools from 80°C to 50″C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surrounding is 20°C.
Answer:
Given θ₁ = 80°C, θ₂ = 50°C, t₁ = 5 min.
Also θ₁¹ = 60°C , θ₂¹ = 30°C, t₂ = ?,
θ₀= 20° C


Hence the body takes 9 min to cool from 60° to 30°C.

Question 37.
A train, standing at outer signal of a railway station blows a whistle of frequency 400 Hz in still air. What is the frequency of the whistle for a platform observer when the train.
(a) Approaches the platform with a speed of 10 ms^-1.
(b) Recedes from the platform with a speed of 10 ms^-1.
(c) W’hat is the speed of sound in each case? Speed of sound in still air can be taken as 340 ms^-1.
Answer:
Given f₀ = 400 Hz, v_s = 10ms^-1,v = 340 ms^-1., v₀ = 0

(i) The train approaching the observer;

(ii) The train receding the observer: