1st PUC Physics Model Question Paper 1 with Answers

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Karnataka 1st PUC Physics Model Question Paper 1 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.
Name the system of units accepted internationally.
Answer:
S.I unit

Question 2.
Define null vector.
Answer:
A vector whose magnitude is zero and that has any arbitrary direction is known as a null or zero vector.

Question 3.
Give an example for conservative force.
Answer:
Gravitational force.

Question 4.
Write an expression for moment of inertia of a solid sphere of radius R and mass M about its diameter.
Answer:
I = \(\frac{2}{3}\)MR²
M – Mass of the solid sphere
R-Radius of the sphere

Question 5.
Define angular acceleration.
Answer:
The time rate of change of angular velocity is known as angular acceleration.

Question 6.
What is meant by elasticity of a body?
Answer:
Elasticity is the property of a material by virtue of which, it regains its original shape and size when the deforming forces are removed.

Question 7.
How does the fractional change in length of a rod varies with its change in temperature?
Answer:
\(\frac{\Delta \mathrm{L}}{\mathrm{L}}\) ∝ Δ_θ where \(\frac{\Delta \mathrm{L}}{\mathrm{L}}\) is fractional charge in length of rod and A0 is change in the temperature. L E

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_C then T_A = T_B. where T represents the temperatures and A, B, C represent the thermodynamic systems.

Question 9.
Define mean free path of a gas molecule.
Answer:
The average distance a molecule can travel without colliding is called the mean free path.

Question 10.
What is a node in stationary waves?
Answer:
Nodes are points of zero displacement of the particle in a stationary wave.

Part – B

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

Question 11.
Name the strongest and w eakest fundamental forces in nature.
Answer:
Strongest force is strong nuclear force and weekest force is gravitational force.

Question 12.
Write any two limitations of dimensional analysis.
Answer:
Limitations of dimensional analysis:

• The value of dimensionless constants cannot be obtained.
• Formula containing trigonometric, exponential and logarithmic functions cannot be derived.
• If a physical quantity depends on two physical quantities having the same dimensional formula, then equation cannot be derived.
• If a physical quantity depends upon more than three different fundamental physical quantities, then the formula cannot be derived.

Question 13.
Write any two differences between distance and displacement.
Answer:
Distance covered :

1. It is the change in position of a particle without the mention of a direction.
2. It is the actual length of the path covered by the particle.
3. It is not zero at the end of the motion of the particle even if it returns to the point of start.
4. It is a scalar physical quantity.

Displacement :

1. It is change in position of a particle in a specified direction.
2. It is the shortest distance between the initial and the final point described by the particle.
3. It is zero, if the particle returns back to the point of start.
4. It is a vector physical quantity.

Question 14.
Mention any two factors on which the centripetal acceleration depends.
Answer:
Centripetal acceleration, a ∝ v² and a ∝ \(\frac{1}{r}\)

Question 15.
How does the linear momentum of a body vary with its mass and velocity?
Answer:
Linear momentum, p ∝ m and p ∝ v.

Question 16.
State Pascal’s law. Name any one device which works on the principle of Pascal’s law.
Answer:
In a closed system of fluid, if a pressure is applied to it, then it is transmitted equally and in an
undiminished way at all points inside the fluid.

Devices which use the principles of pascal’s law are,

• Hydraulic brakes and
• Sphygmomanometer (used to measure blood pressure).

Question 17.
State and explain Boyle’s law.
Answer:
At constant temperature, the volume of a given mass of a gas is inversely proportional to its pressure.
V ∝ \(\frac{1}{P}\) at constant temperature
So that P₁V₁ = P₂V₂.

Question 18.
What is periodic motion? Give an example.
Answer:
A particle event that repeats itself in regular intervals of time is known as the periodic motion.
Both circular and elliptical motions of electrons around the atomic nucleus are examples of periodic motions.

Part – C

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

Question 19.
Explain triangle method of addition of two vectors.
Answer:
If two vectors acting in the same direction are represented by two sides of a triangle taken in order then their resultant is represented by the closing side of the triangle taken in the opposite order.

Question 20.
Write 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.
Define work done by a force. Under what conditions the work done by a force is maximum and minimum?
Answer:
Work is said to be done when a force applied on the body causes displacement in the direction of force applied.
W_win = 0 when θ = 90° b/w \(\overrightarrow{\mathrm{F}}\) and \(\overrightarrow{\mathrm{S}}\) and W_max = Fs when θ = 0° between \(\overrightarrow{\mathrm{F}}\) and \(\overrightarrow{\mathrm{S}}\)

Question 22.
When does the rigid body said to be in mechanical equilibrium? Write the conditions for the 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 23.
Derive an expression for acceleration due to gravity on the surface of the earth in terms
of mass of the earth 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.
State and explain Hook’s law and hence define modulus of elasticity. Locate yield point and 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 25.
What is meant by viscosity? How does the viscosity change with rise of temperature in case of (i) liquids and (ii) gases?
Answer:
The property by virtue of which it opposes relative motion of layers of liquid is called viscosity.

• Viscosity of liquids decreases with the increase of temperature.
• Viscosity of gases increases with increase of temperature.

Question 26.
What is radiation? Mention any two properties of thermal radiations.
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 and refraction.

Part – D

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

Question 27.
What is meant by velocity – time graph? Derive the equation Jt = v₀t + \(\frac{1}{2}\)at² from 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.
Define elastic potential energy. Derive an expression for potential energy of a spring.
Answer:
The restoring energy in an elongated or compressed spring is called elastic potential enrgy.
Work done by the spring force = W_s = \(-\int_{0}^{x_{m}} \mathrm{F}_{\mathrm{s}} \mathrm{d} \mathrm{x}\)
i.e. W_s = \(-\int_{0}^{x_{\infty}} k x d x\)
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 29.
Derive an expression for kinetic energy of a rolling body. ’
Answer:
K.E of rolling body = Translatory K.E + Rotational K.E

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

Question 30.
Explain the working of Carnot engine with the help of PV diagram.
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 31.
Derive an expression for time period of oscillation of a simple pendulum.
Answer:
The tangential component of force mg is mgsinG.
Torque : -L(mgsinθ) = I∝

Question 32.
Show that only odd harmonics arc present in vibrations 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. Le 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 Cricket ball is thrown at a speed of 28ms^-1 in the direction 30° above the horizontal. Calculate (a) maximum height reached by the projectile and (b) horizontal range of the projectile. ((liven: Acceleration due to gravity = 9.8ms^-2).
Answer:
Given u = 28ms^-1, θ = 30°
H = ? R = ? g = 9.8 ms^-2.
(i) Maximum height

(ii) Horizontal range:

Question 34.
A constant force acting on a body of mass 3.0kg changes its speed from 2.5ms^-1 to 3.5ms^-1 in 25 second. If the direction of the motion of the body remains unchanged, find the magnitude and direction of the force?
Answer:
Given m = 3.0 kg, u = 2.5 ms^-1 to 3.5 ms^-1 t = 25s, F = ?

Force applied is 0.12N in the direction of the acceleration.

Question 35.
An artificial satellite revolves around the earth with an orbital velocity of 5.92kms^-1. Calculate height of the statellite from the earth’s surface, ((liven: Radius of the earth = 6380 km acceleration due to gravity of earth = 9.8 ms^-2)
Answer:
given V₀ = 5.92 × 10³ ms^-1 ; R = 63 80 × 10³ m
g = 9.8 ms^-2

Question 36.
A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block at a temperature of 0°C. Calculate the maximum amount of ice that can melt? (Given: Specific heat of copper = 390 Jkg^-1 K^-1. Latent heat of fusion of ice = 3.35 × 10⁵ kg^-1)
Answer:
Given m = 2.5 kg, θ = 500°C, θ₀ = 0°C.
C_cu = 390 Jkg^-1 K^-1, L = 3.35 × 10⁵Jkg^-1
Since Q = m_cu C_cu Δ_cu = m_ice L_ice
We write m_ice = \(\frac{2.5 \times 390 \times 500}{3.35 \times 10^{5}}\)
∴ m_ice = 1.455 kg
Hence 1.45 kg of ice would melt.

Question 37.
A transverse harmonic wave travelling along positive X-axis on a stretched string is described by Y(x, t) = 0.03 sin (36t – 1.8x + \(\frac{\pi}{4}\) ) Where X and Y are in metre and t is in second. Calculate (a) amplitude (b) initial phase (c) frequency and (d) speed of the wave.
Answer:
Given
y (x, t) = 0.03 sin (36t – 1.8x + \(\frac{\pi}{4}\) ) in comparing this with standard equation,
y(x, t)= A sin (wt – kx + Φ) we write,
(i) amplitude A’ = 0.03m

(ii) Initial phase Φ when t = 0, x = 0 is
Φ = \(\frac{\pi}{4}\) rad

(iii) Angular frequency ω = 36 rad s^-1
i.e., 2πf = 36
∴ f = \(\frac { 36 }{ 2 }\) × 3.142
i.e., f = 5.72Hz

(iv) Speed of the wave V = ω/k, ω = 36 rad s^-1, k = 1.8 rad.m^-1
i.e., v = 36/1.8
i.e., v = 20ms^-1