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Tuesday, 19 November 2013

F.Sc Notes: Physics XI: Chapter 07 Oscillations Exercise Short Questions:

 FSc Notes: Physics XI: Chapter 07 Oscillations Exercise SQ:

Question 7.1 Name two characteristics of simple harmonic notion (SHM)?
Answers 7.1
Characteristics of simple harmonic motion are as follow
  1. Simple harmonic motion is a vibrating motion.
  2. Acceleration is directly proportional to displacement.
  3. Acceleration is always directed towards mean position.

Question 7.2 Does frequency depends on amplitude for harmonic oscillators?
Answer 7.2
No, frequency is independent of amplitude it depends on Time period
                                          Time Period = T = 1/f.

Question 7.3 Can we realize an ideal simple pendulum?
Answer 7.3
No, because a friction less system cannot be made. We need mass less bob, in-extensible string and suspend it from a friction less support.

Question 7.4 What is the total distance traveled by an object moving with SHM in a time equal to its period, if its amplitude is A?
Answer 7.4
As T  is the time period for one complete vibration. Its maximum displacement,
                    xo = r = A. so total distance traveled will be 4A.

Question 7.5 What happens to the period of a simple pendulum if its length is doubled? What happens if the suspended mass is doubled?
Answer 7.5
    For simple pendulum,                
                                                      T = 2π√( l / g)
As the length is doubled so  l = 2l
                                   T = 2π√ (2 l / g) = √2 x 2π√ l / g = √2 T
So the time period increases by √2 (=1.414) times, as length is doubled. There will be no change, when suspended mass is doubled. As time period, T Time Period is independent of mass m.

Question 7.6 Does the acceleration of a simple harmonic oscillator remain constant during its motion? Is the acceleration ever zero? Explain.
Answer 7.6 
No acceleration depends upon displacement x. As the relation between acceleration and displacement is a = -w(2)x.
The acceleration is zero at mean position  i.e. (x=0), and maximum at extreme position (x=xo) .So,
acceleration does not remain constant during motion.

Question 7.7 What is meant by phase angle? Does it define angle between maximum displacement and the driving force?
Answer 7.7
Phase angle (or phase): “The angle θ = wt which specifies the displacement as well as the direction of motion of the point executing SHM”.
It indicates the:
  1. State of motion of a vibrating particle
  2. Direction of motion of a vibrating particle.
No. It does not define angle between maximum displacement and the driving force.

Question 7.8 Under what conditions the addition of two simple harmonic motions does produces a resultant. Which is also simple harmonic?
Answer 7.8
When there is a constant phase difference in amplitude and frequency are same. Exp are interference and beats.

Question 7.9 Show that in SHM the acceleration is zero when the velocity is greatest and the velocity is zero when the acceleration is greatest?
Answer 7.9
For SHM. v = ω √ ( xo2 – x2) & a = - ω2 x .
  1. At mean position, from the above equations, x = 0 then a = 0 & v = ω xo i.e. acceleration is zero and velocity is maximum.
  2. At extreme positions x = xo then v = 0 & a = -ω xo. i. e. velocity is zero when acceleration is maximum.

Question 7.10 In relation to SHM, explain the equations:
  1. y = A sin (ω t + ϕ )
  2. a = - ω2 x
Answers 7.10
  1. y = A sin (ω t + ϕ). ϕ initial phase, y Instantaneous displacement, A Amplitude time t,  (ω t + ϕ)  State of motion. This equation shows that displacement of SHM as a function of amplitude and phase angle depending upon time.
  2. a = - ω2 x where a = acceleration of a particle executing SHM ω = constant angular frequency  x=instantaneous displacement from the mean position. This equation shows that acceleration is directly proportional to displacement and is directed towards mean position.

Question 7.11  Explain the relation between total energy, potential energy and kinetic energy for a body oscillating with SHM.
Answer 7.11 
  For a body executing SHM. Etotal=P.E+K.E. Since Total energy of SHM remains constant in the  absence of frictional forces, the K.E and P.E interchange continuously from one form to another. At  mean position, the energy is totally K.E. P.E = 0. In between it is partially P.E and K.E.

Question 7.12 Describe some common phenomena in which resonance plays an important role.
Answer 7.12
Important role of resonance:
  1.     Microwave oven Microwaves (of frequency 2450 MHz) with λ = 12 cm, are absorbed due to resonance by water and fat molecules in the food, heating them up and so cooking the food.
  2.     Children’s swing In order to raise the swing to a great height, we must give it a push at the right moment and in the right direction.
  3.     Musical instruments in some instruments (e.g. drums) air columns resonate in the wooden box. In string instruments (e.g. sitar) strings resonate with their frequencies and loud music is heard.
  4.     Tuning radio/TV we change the frequency with knob. When it becomes equal to a particular transmitted station, resonance occurs. Then we receive amplified audio/video signals.

Question 7.13 If a mass spring system is hung vertically and set into oscillations, why does the motion eventually stop?
Answer 7.13
Mass spring system is hung vertically and set into oscillations, the motion eventually stop due to energy dissipation, friction and damping.

Written By: Asad Hussain

Thursday, 14 November 2013

F.Sc Notes: Physics XI: Chapter 08 Waves Exercise Short Questions:

FSc Notes: Physics XI: Chapter 08 Waves Exercise Short Answer:

Question 8.1 What features do longitudinal waves have in common with transverse waves?
Answer 8.1
Common features between longitudinal and transverse waves.
  1. They produce disturbance in the medium in which they are traveling.
  2. Transport energy from one place to another.
  3. The equation v = f λ is valid for both waves.

Question 8.2 The five possible waveform obtained, when the output from a microphone is fed into the Y-input of the cathode ray oscilloscope, with the time base on, are shown in fig8.23 (on book). These waveform are obtained under the same adjustment of the cathode ray oscilloscope controls. Indicate the waveform.
a. which trace represents the loudest note?
b. which trace represents the highest frequency?
Answer 8.2
Loudness of sound depends on the intensity of sound which is directly related with the square of amplitude of the waves. Hence fig b represents the loudest note. Trace b has the highest frequency as the wavelength is small than other waves.

Question 8.3 Is it possible for two identical waves traveling in the same direction along a string to give rise to a stationary wave?
Answer 8.3 No
, there is no possibility for two waves traveling in the same direction along a string to give rise to a stationary wave. As stationary wave is produced by the superposition of two identical waves traveling in opposite direction.

Question 8.4 A wave is produced along a stretched string but some of its particles permanently show zero displacement. What type of wave is it?
Answer 8.4
The wave in which some of its particles permanently show zero displacement is known as stationary wave and these zero displacement maximum tension points are known as nodes.

Question 8.5 Explain the terms crest, trough, node and anti nodes.
Answer 8.5 Crest:
The portion of wave above mean position in a traverse wave is called crest.
Trough: The portion of wave below mean position in a traverse wave is called trough.
Node: Points having zero amplitude in a stationary wave is called node.
Anti nodes: Points having maximum amplitude in a stationary wave is called anti nodes.

Question 8.6 Why does a sound travel faster in solids than in gases?
Answer 8.6
In the relation v = √ (E / ρ) ‘E’ Elasticity of the medium is directly proportional to velocity of sound    its value is greater for solids than in gases. The effect of density, ρ is very less as compared to E. The ratio of  (E / ρ) is greater for solid than gases, so sound travel faster in solids then in gases.

Question 8.7 How are the beats useful in tuning musical instruments?
Answer 8.7
A new instrument is tuned as it is played with the standard musical instruments. They are sounded together thus beats are produced. The frequency of the new instrument is changed until the resonance occurs.

Question 8.8 When two notes of frequency f1 and f2 are sounded together, beats are formed if f1>f2. What will be the frequency of beats?
  1. f1+f2
  2. 1/2(f1+f2)
  3. f1-f2
  4. ½(f1-f2)
Answer 8.8 Since we know that no of beats per second is equal to the difference between the frequencies of the tuning forks. Hence option iii) f1-f2 is correct.

Question 8.9 As a result of distant explosion, an observer senses a ground tremor and then hears the explosion. Explain the time difference.
Answer 8.9
Sound waves travel faster in solids than in air. The sound waves produced by the explosion travel from two paths. One through ground reaches faster than traveling through air. This is the reason for time difference.

Question 8.10 Explain why sound travels faster in warm air than in cold air.
Answer 8.10
                        v ∝ √T  and  v = √ (γ P / ρ)
The speed of sound varies directly as the square root of absolute temperature and inversely to the square root of density of medium. As the temperature of air increases, the pressure increases and density decreases. So, speed of sound increases in warm air.

Question 8.11 How should a sound source move with respect to an observer so that the frequency of its sound does not change?
Answer 8.11
Both the source and the observer should not move so their relative speed be zero. From Doppler Effect, Then there will be no change in the frequency of its sound.

Written By: Asad Hussain

F.Sc Notes: Physics XI: Chapter 06 Fluid Dynamics Exercise Short Questions:

FSc Notes: Physics XI: Chapter 06 Fluid Dynamics Exercise Short Question:

Question 6.1 Explain what do you understand by the term viscosity?
Answer 6.1
Viscosity means how much force is required to slide one layer over another in a fluid. Substances that do flow easily like honey or tar have large co-efficient of viscosity. On the other hand fluid those  flows easily like water have low value of co-efficient of viscosities.

Question 6.2 What is meant by drag force? What are the factors upon which drag force acting upon a small sphere of radius r moving down through a liquid, depend?
Answer 6.2
An object moving through a fluid experience a retarding force which opposes its motion is known as drag force.
By Stoke’ Law its eq is: F = 6πηrv
Velocity: Drag force is directly proportional to the velocity of the body in the fluid.
Radius: Drag force also depends upon the radius of the body.
Nature of Fluid: It also depends on the co-efficient of viscosity of the fluid. i.e. nature of fluid.

Question 6.3 Why fog droplets appear to be suspended in air?
Answer 6.3
As we know that: vt=mg/6πηr.
As mass of the fog droplet is very small so its terminal velocity will be very small. For this reason fog droplet appears to be suspended in air.

Question 6.4 Explain the difference between laminar flow and turbulent flow?
Answer 6.4
The flow is said to be laminar if every particle that passes a particular point moves along exactly the same path as followed by particles which passed that points earlier. i.e. the traffic on motorway. The irregular or unsteady flow of the fluid is called turbulent flow. i.e. as a driver changes line randomly on a road.

Question 6.5 State Bernoulli’s relation for a liquid in motion and describe some of its applications?
Answer 6.5
Bernoulli’s relation P + ½ ρ v2 + ρgh = constant
Torricelli’s Theorem: The speed of efflux is equal to the velocity gained by the fluid in falling through the distance (h1-h2) under the action of gravity.          v2 = √2g (h1 – h2)                            
Venturi Relation: The effect of the decrease in pressure with the increase in speed of the fluids in a horizontal pipe is known a venturi relation.   P1 - P2 = ½ ρ v2(2).

Question 6.6 A person is standing near a fast moving train. Is there any danger that he will fall towards it?
Answer 6.6
From Bernoulli's Theorem. Where speed is faster streamlines are forced together and pressure will be low there. Between train and a person streamlines are close due to high speed but pressure exerted by fluid will be low. On the other hand streamlines are further and pressure will be greater. Therefore there will be a danger that a person will fall towards a fast moving train.

Question 6.7 Identify the correct answer. What do you infer from Bernoulli's theorem?
  1. Where the speed of the fluid is high the pressure will be low.
  2. Where the speed of the fluid is high the pressure is also high.
  3. The theorem is valid only for turbulent flow of the liquids.
Answer 6.7 Correct answer is (i) Where the speed of the fluid is high the pressure will be low.

Question 6.8 Two row boats moving parallel in the same direction are pulled towards each other. Explain?
Answer 6.8
According to Torricelli Theorem Where the speed is high, the pressure will be low. Between row boats velocity is high and pressure is low outside row boats velocity is low but pressure is high so boats are pulled towards each other.

Question 6.9 Explain, how the swing is produced in a fast moving cricket ball?
Answer 6.9
When a bowler bowls a cricket ball it spins and moves forward. The velocity of air on the shiny Surface is greater than on the dull surface and pressure is low on the shiny surface and greater on the dull surface. This gives an extra curvature to the ball which is known as swing which may deceive the batsman.

Question 6.10 Explain the working of a carburetor of a motorcar using by Bernoulli's principle.
Answer 6.10
The carburetor of the car uses venturi duct to feed the correct mix of air and petrol to the cylinders. Air is drawn through the duct and along a pipe to the cylinders. A tiny inlet at the side of the duct is fed with petrol. The air through the duct moves very fast, creating low pressure in the duct, which draws petrol vapors into the air stream.

Question 6.11 For which position will the maximum blood pressure in the body have the smallest values?
  1. Standing up right.
  2. Sitting.
  3. Lying horizontally.
  4. Standing on one’s head.
Answer 6.11 The maximum blood pressure in the body has the smallest value when body is lying horizontally

Question 6.12 In an orbiting space station, would the blood pressure in major arteries in the leg ever be greater than the blood pressure in major arteries in the neck?
Answer 6.12 No
. Due to lack of force of gravity, the blood pressure in major arteries in the leg will be equal to that in arteries of the neck, due to weightlessness.

Written By: Asad Hussain

Wednesday, 13 November 2013

F.Sc Notes: Physics XI: Chapter 02 (11-21) Vectors and Equilibrium Exercise Short Questions:

FSc Notes: Physics XI: Chapter 02 (11-21) Vectors and Equilibrium Exercise SQ:

Majority of these question need diagram to explain but i have not drawn the diagrams.

Question 2.11 Two vectors have unequal magnitudes. Can their sum be zero? Explain.
Answer 2.11 No
. The sum of two unequal vectors cannot be zero. For the sum of vectors to be zero, the  vectors must have equal magnitude with opposite directions.

Question 2.12 Show that the sum and differences of two perpendicular vectors of equal lengths are also perpendicular and of the same length.
Answer 2.12
Vectors A and B are perpendicular to each other having equal lengths we have:
(A + B) is ⊥ to (A - B) i.e. sum and difference of the vectors are perpendicular to each other.

Question 2.13 How would the two vectors of the same magnitude have to be oriented, if they were to be combined to give a resultant equal to a vector of the same magnitude?
Answer 2.13
When the angle between two vectors of same magnitude is 120 degree, the magnitudes of the resultant will be same.

Question 2.14  The two vectors to be combined have magnitude 60N and 35N. Pick the correct answer from those given and tell why it is only one of the three that is correct. (i) 100N (ii) 70N (iii) 20N.
Answer 2.14
  A1 = 60N and A2 = 35N
Answer (ii) 70N is correct.
For maximum value, both vectors in same direction,
A1 + A2 = 60 + 35 = 95 → cannot be (i) 100N.
For minimum value, both vectors having opposite direction,
A1 - A2 = 60 + 35 = 25 → cannot be (iii) 20N.

Question 2.15 Suppose the sides of a closed polygon represent vector arranged head to tail .What is the sum of these vectors?
Answer 2.15
The sum will be zero. Because the tail of first vector meets the head of last vector. Whenever vector representation makes a closed shape its resultant vector is a null vector.

Question 2.16  Identify the correct answer.
  1. Two ships X and Y are traveling in different directions at equal speeds. The actual direction of motion of X is due north but to an observer on Y, the apparent direction of motion of X is north-east. The actual direction of motion of Y as observed from shore will be (A) East (B) West (C) south-east (D) south-west.
  2. A horizontal force F is applied to a small object P of mass m at rest on a smooth plane inclined at an angle θ to the horizontal as shown in fig The magnitude of the resultant force acting up and along the surface of the plane, on the object is (a) F cos θ -mg sin θ  (b) F sin θ -mg cos θ (c) F cos θ +mg cos θ (d) Fsin θ+mg sin θ (e) mgtan θ.
Answer 2.16 (1) The correct answer is (B) West.
(2)  The correct answer is (a) F cos θ - mg sin θ the resultant force acting Up and along the surface of the plane is = F cos θ - mg sin θ.

Question 2.17   If all the components of the vectors, A1 and A2 were reversed, how would this alter A1 x A2?
Answer 2.17
    For A1 = - A1& A2= - A2
 -A1 x -A2 = A1 x A2  
There will be no effect, if all the components of the vectors A1 & A2 are reversed.

Question 2.18   Name the three different conditions that could make A1 x A2 = 0.
Answer 2.18
   A1 x A2 could be zero, if
  1. A1 is null vector; 0 x A2 = 0
  2. A2 is null vector; A1 x 0 = 0      
  3. A1 x A2 are parallel or anti-parallel, i.e. A1 x A2 = A1 A2 sin θ = A1 A2 sin 180 degree = 0

Question 2.19   Identify true or false statement and explain the reason.
  1. A body in equilibrium implies that it is neither moving nor rotating.
  2. If co planar forces acting on a body from a closed polygon, then the body is said to be in equilibrium.
Answer 2.19 a) It is false. A body moving with constant velocity can also be in equilibrium.
b) It is true. The vector sum will be zero, for the coplanar forces forming a closed polygon, fulfills the 1st condition of equilibrium.

Question 2.20   A picture is suspended from a wall by two strings. Show by diagram the configuration of the strings  for which the tension in the strings will be minimum.
Answer 2.20
    For T minimum, θ = 90 degree
Σ Fy = 0                 Ty + Ty - w = 0                    2Ty - w = 0
2 T sin θ = w                                           T = w / 2 sinθ
For minimum T, θ = 90 degree  i.e. T = w / 2 sin 90 degree 
T = w / 2.

Question 2.21 Can a body rotate about its center of gravity under the action of its weight?
Answer 2.21 No
. A body cannot rotate about its center of gravity under the action of its weight. Because moment arm will be zero, so torque or turning effect will be zero.  

Written By: Asad Hussain

F.Sc Notes: Physics XI: Chapter 02 (1-10) Vectors and Equilibrium Exercise Short Questions:

                 FSc Notes: Physics XI: Chapter 02 (1-10) 

                  Vectors and Equilibrium Exercise SQ:

Majority of these question need diagram to explain but i have not drawn the diagrams.

Question 2.1 Define the terms (i) unit vector (ii) Position vector (iii) Components of a vector?
Answer 2.1
Definition of these terms is given below
  1. Unit Vector: A unit vector in a given direction is a vector with magnitude one in that direction.
  2. Position Vector: The position vector r is a vector that describes the location of a particle with respect to the origin.
  3. Components of a vector: A component of a vector is its effective value in a given direction.

Question 2.2 The vector sum of three vectors gives a zero resultant. What can be the orientation of the vectors?
Answer 2.2
If three vectors are drawn, to make a closed triangle, then their vector sum will be zero.

Question 2.3 Vector A lies in the xy plane. For what orientation will both of its rectangular components be negative? For what orientation will its components have opposite signs?
Answer 2.3
Orientation of a vector having components negative and opposite is given below:
  1. Vector A lies in 3rd quadrant, its rectangular components will be negative
  2. When the vector will lie in 2nd or 4th quadrant, its components will have opposite signs.

Question 2.4 If one of the rectangular components of a vector is not zero, can its magnitude be zero? Explain.
Answer 2.4 No
. Its magnitude cannot be zero, when one of the components of the vector is not zero. E.g. if      Ax ≠0 & Ay = 0 then A = √ ((Ax) 2 + (0)2) = √ (Ax) 2 = Ax ≠ 0

Question 2.5 Can a vector have a component greater than the vector’s magnitude?
Answer 2.5
No. A vector cannot have a component greater than the vector’s magnitude. 
                                              As A = √ ((Ax) 2 + (Ay) 2)

Question 2.6 Can the magnitude of a vector have a negative value?
Answer 2.6 No
. The magnitude of a vector has always positive values; As A = √ ((Ax) 2 + (Ay) 2).

Question 2.7 If A+B=0, what can you say about the components of the two vectors?
Answer 2.7
A + B = 0 Two Vectors sum will be zero vector if their components are equal and opposite.

Question 2.8 Under what circumstances would a vector have components that are equal in magnitude?
Answer 2.8
When θ = 45 degree , the components will have equal magnitude for a vector making angle 45 degree  with X-axis.

Question 2.9 Is it possible to add a vector quantity to a scalar quantity? Explain.
Answer 2.9 No. It is not possible to add a vector quantity to a scalar quantity. Different physical quantities cannot be added according to the rules of algebra.

Question 2.10 Can you add zero to a null vector?
Answer 2.10 No
. We cannot add zero to a null vector. Because zero, a scalar quantity cannot be added with a vector quantity the null vector.

Written By: Asad Hussain

F.Sc Notes: Physics XI: Chapter 05 Circular Motion Exercise Short Questions:

 FSc Notes: Physics XI: Chapter 05 Circular Motion Exercise SQ:

Question 5.1 Explain the difference between tangential velocity and the angular velocity. If one of these is given for a wheel of known radius, how will you find the other?
Answer 5.1
Linear velocity of a particle moving along a curve or circle directed along the tangent at any point on the curve is known as tangential velocity .
Rate of angular displacement to unit time is known as angular velocity .                                      

Question 5.2 Explain what is meant by centripetal force and why it must be furnished to an object if the object is to follow a circular path?
Answer 5.2
The force which compels the body to move in a circle is known as centripetal force.When a body follows a circular path its direction changes every moment. To change the direction of motion continuously a continuous perpendicular force is required which is known as centripetal force it is directed towards the center of the circle.

Question 5.3 What is meant by moment of inertia? Explain its significance.
Answer 5.3
A property of a body to resist its uniform circular motion. Product of mass and perpendicular distance of the body from the axis of rotation is known as moment of inertia.

Question 5.4 What is meant by angular momentum? Explain the law of conservation of angular momentum?
Answer 5.4
Cross product of linear momentum P and position vector r is known as angular momentum.
Law of conservation of Angular Momentum. The angular momentum of an isolated system remains the same if no external torque is applied on it.
If I decrease its w increases so product remains the same.

Question 5.5 Show that orbital angular momentum Lo = mvr?
Answer 5.5
We know that  and its magnitude is given by
L=r*p = rpsinθ = mvrsinθ
where θ  is the angle between r and p .
In case of circular orbital motion θ =90 degree between r and p .
Hence                       Lo = mvrsin90 = mvr
Lo = mvr                               ( sin90=1)

Question 5.6 Describe what should be the minimum velocity, for a satellite, to orbit close to the Earth around it?
Answer 5.6
We that minimum to put a satellite in orbit vo=√gR .
As low flying satellite has g = 9.8m\s(2)
And    R= 6400m.
putting the values vo=√9.8*6400  =7.9km\s .
This is the minimum velocity for a satellite to orbit close to earth.

Question 5.7 State the direction of the following vectors in the simple situations angular momentum and angular velocity?
Answer 5.7
In simple situations angular momentum and angular velocity both are parallel to the axis of rotation.

Question 5.8 Explain why an object, orbiting the Earth, is said to be free falling. Use your explanation to point out why object appear weightless under certain circumstances.
Answer 5.8
If the object is thrown fast enough parallel to the earth, the curvature of its path match the curvature of the earth the object in this case will simply circle around the earth this is known as free fall.
Weightlessness is a sensation experienced by any individual when there are no external objects touching one's body and exerting a push or pull upon it. In free fall, the only force acting upon your body is the force of gravity  a non-contact force. So the object appears weightless.

Question 5.9 When mud flies off the tire of a moving bicycle, in what direction does it fly explain?
Answer 5.9
When mud flies off the tire of a moving bicycle it flies along the tangent to the tire. When tire is moving every part of it has tangential velocity, angular momentum and centripetal force. The adhesive forces of mud provides the required centripetal force when speed of tire increases. When required centripetal force increases the adhesive force the mud flies off the tire.

Question 5.10 A disc and a hoop start moving from the top of an inclined plane at the same time. Which one will be  moving faster on reaching the bottom?
Answer 5.10
The velocity of disc on reaching the bottom of the inclined plane,
v=√4gh/3=√4/3√gh  and for hoop  velocity at the bottom of the inclined plane v1 =√gh
So  v=√4/3v1=1.15v1
Hence Disc will reach the bottom earlier.

Question 5.11 Why does a diver change his body positions before and after diving in the pool?
Answer 5.11
To follow law of conservation of momentum.

Question 5.12 A student holds two dumb-bells with stretched arms while sitting on a turn table. He is given a push until he is rotating at certain angular velocity. The student then pulls the dumb-bells towards his chest. What will be the effect on rate of rotation?
Answer 5.12
Using Law of conservation of momentum: I1wi=I2wf .
With stretched arms moment of inertia I1and angular velocity wi .The product I1wi is the angular momentum remains the same when he pulls the dumb bells towards his chest his moment of inertia I2 decrease and angular velocity wf increases to keep angular momentum I2wf  constant.

Question 5.13 Explain how many minimum number of Geo-stationary satellites are required for global coverage of TV transmission?
Answer 5.13
A Geo-stationary satellite covers an angle of 120 degree so whole of the earth’s surface can be covered by three correctly poisoned Geo-stationary satellites for global coverage of TV transmission.

Written By: Asad Hussain

Tuesday, 12 November 2013

F.Sc Notes: Physics XI: Chapter 04 Work and Energy Exercise Short Questions:

F.Sc Notes: Physics XI: Chapter 04 Work and Energy Exercise Short Questions:

Question 4.1 A person holds a bag of groceries while standing still, talking to a friend. A car is stationary with its engine running. From the stand point of work, how are these two situations similar?
Answer 4.1
Since Work = Force* displacement. In both cases displacement is zero so work done in both cases is zero.

Question 4.2 Calculate the work done in kilo joules in lifting a mass of 10 kg (at a steady velocity) through a vertical height of 10 m.
Answer 4.2
In this case work is stored in the body in the form of potential energy
So Work = P.E = mgh = 10*9.8*10 = 980 J = 0.98 KJ.

Question 4.3 A force F acts through a distance L. The force is then increased to 3F, and then acts through a further distance of 2L. Draw the work diagram to scale.
Answer 4.3
  A force F acts through a distance L, then work done =FL.
If force is increased to 3F and distance is increased to 2L.
Then work done = 3F.2L = 6FL. Total work done = 6FL+FL=7FL.

Question 4.4 In which case is more work done? When a 50 kg bag of books is lifted through 50 cm, when a 50 kg crate is pushed through 2m across the floor with a force of 50N?
Answer 4.4
(a) m= 50kg h= 50cm = 0.5m.
So Work Done = W1= mgh= 50*9.8*.5.
Work done = W1 = 245J.     

(b) m= 50kg  d= 2m F=50N.
So Work Done =  W2= F.d= 50*2.
Work done = W2 = 100J.

Question 4.5 An object has 1J of potential energy. Explain what does it mean?
Answer 4.5
An object has potential energy when it is in a force field such as gravitational field or it is in a constrained state such as stretched spring.
  1. When an object is raised by doing 1J work. P.E = mgh = W.h =1J (1N*1m). When 1N force is applied on a body to lift it through 1m. The work appears as gravitational P.E.
  2. When a spring is stretched or extended by doing 1J of work. The work is conserved in it as elastic P.E.

Question 4.6 A ball of mass m is held at a height h1 above a table. The table top is at a height h2 above the floor. One student says that the ball has potential energy mgh1 but another says that it is mg(h1+h2). Who is correct?
Answer 4.6
Both statements are correct. It is a matter of relative position for specifying the P.E of the ball. The P.E relative to the table top is mgh1 and P.E relative to ground is mg (h1+h2).

Question 4.7 When a rocket re-enters the atmosphere, its nose cone becomes very hot. Where this heat does comes from?
Answer 4.7
When a rocket re-enters the atmosphere, its nose cone becomes very hot due to air friction and dust particles.

Question 4.8 What sort of energy is in the following:
  1. Compressed spring
  2. Water in a high dam
  3. A moving car.
Answer 4.8 Energy stored in stated above is given below:
  1. Elastic P.E.
  2. Gravitational P.E.
  3. Kinetic Energy.

Question 4.9 A girl drops a cup from a certain height, which breaks into pieces. What energy changes are involved?
Answer 4.9
The cup has gravitational P.E due to its height. When it is dropped its P.E changes to K.E just before hitting the ground the P.E completely changes to K.E. On hitting the ground K.E changes to heat, sound, and work done in breaking the cup.

Question 4.10 A boy uses a catapult to throw a stone which accidentally smashes a green house window. List the possible energy changes?
Answer 4.10
The elastic P.E of the catapult is given to stone as its K.E. When the stone smashes the green house window the K.E of the stone is changed into sound, heat and work done in breaking the glass.

Written By: Asad Hussain

F.Sc Notes: Physics XI: Chapter 03 Motion and Force Exercise Short Questions:

FSc Notes: Physics XI: Chapter 03 Motion and Force Exercise Short Questions:

Question 3.1 What is the difference between uniform and variable velocity? From the explanation of variable velocity, define acceleration. Give SI units of velocity and acceleration.
Answer 3.1
Uniform Velocity:  The velocity of a body is said to be uniform if its direction and magnitude does not change with time. e.g. i) velocity of earth ii) velocity of satellites.
Variable Velocity: The velocity of a body is said to be variable if its direction or magnitude or both changes with time. A motion with variable velocity is called accelerated motion. In this case velocity may be increasing or decreasing. For e.g. motion of a car on road.
Acceleration:  Rate of change of velocity is known as acceleration. When velocity is increasing acceleration is positive and when velocity is decreasing acceleration is negative.Units: The SI units of velocity is m/s and that of acceleration is m/s-2

Question 3.2 An object is thrown vertically upward. Discuss the sign of acceleration due to gravity, relative to velocity, while the object is in air.
Answer 3.2
When an object is thrown upward its velocity is +ve and its acceleration due to gravity is –ve, as the object is moving against the direction of gravitational force. But at maximum height its velocity becomes zero and then it starts moving downward. Now its acceleration and velocity is +ve.

Question 3.3 Can the velocity of an object reverse the direction when acceleration is constant? If so, give an example.
Answer 3.3
Yes, velocity of an object can reverse direction when initially acceleration and velocity are opposite in direction. For e.g. when a body is thrown vertically upward its velocity goes on decreasing due to gravity, becomes zero at maximum height, and then the direction of the body is reversed.

Question 3.4 Specify the correct statements:
  1. An object can have a constant velocity even its speed is changing.
  2. An object can have a constant speed even its velocity is changing.
  3. An object can have a zero velocity even its acceleration is not zero.
  4. An object subjected to a constant acceleration can reverse its velocity.
Answer 3.4 All statement are true expect: 1). 
2) Circular motion.
3) When body is thrown upward, at maximum height.
4) Vertical motion.

Question 3.5 A man standing on the top of a tower throws a ball straight up with initial velocity vi and at the same time throws a second ball straight downward with the same speed. Which ball will have larger speed when it strikes the ground? Ignore air friction.
Answer 3.5
Both the balls will have same speed when striking the ground. The ball thrown upward will pass from the same path with the same velocity while moving down and gains the same velocity as that of the ball thrown vertically downward, until it reaches the surface of the ground.

Question 3.6 Explain the circumstances in which the velocity v and the acceleration a of a car are:
  1. Parallel
  2. Anti-Parallel
  3. Perpendicular to one another
  4. v is zero but a is not zero
  5. a is zero but v is not zero.
Answer 3.6 Circumstances in velocity v and acceleration a of a car are given below:
  1. When v and a are parallel the velocity of the car is increasing.
  2. When v and a are anti-parallel the velocity of the car is decreasing.
  3. When v and a are perpendicular, the car is moving in a circular path. Here velocity is along the tangent and acceleration is along the radius.
  4. v is zero when car comes to rest but acceleration is –ve but not zero or when there is a hurdle in front of the car it cannot move in the passage of car.
  5. When car move on a straight road with uniform speed (neglecting friction).

Question 3.7 Motion with constant velocity is a special case of motion with constant acceleration, Is this statement true? Discuss.
Answer 3.7
Yes, motion with constant velocity is a special case of motion with constant acceleration. In this case the acceleration of the object is zero and velocity is uniform.

Question 3.8 Find the change in momentum for an object subjected to a given force for a given time and state law of motion in terms of momentum. 
Answer 3.8 Consider a body mass m moving with an initial velocity vi. Suppose an external force F acts upon it time t after which velocity becomes vf the acceleration a produced by the force is given by:
                                                               a= vf – vi /t
by second law on                                 Newton a=F/m
comparing two equations               F /m = vf – vi /t   or   F x t= mvf – mvi
where mvf is the final momentum and mvi is the initial momentum. F = (mvf – mvi )/t
Time rate of change of momentum of a body equals the applied force

Question 3.9 Define impulse and show that how it is related to linear momentum?
Answer 3.9
Impulse is defined as the product of force and time i.e.
                                                    Impulse = Force x Time
                                                    I=F x t= mvf - mvi
                                                    I = pf - pi= p change in momentum
                                                    I= p 

Question 3.10 State the law of conservation of linear momentum, pointing out the importance of isolated system. Explain, why under certain conditions, the law is useful even through the system is not completely isolated?
Answer 3.10
Law of conservation of linear momentum: Total linear momentum of an isolated system remains constant. An isolated system is a system of bodies free from external force. This law holds good only for an isolated system. But under certain circumstances where external force is very small as compared to mutually interacting forces, the law can be applied to good approximation.

Question 3.11 Explain the difference between elastic and inelastic collisions. Explain how would a bouncing ball behave in each case? Give plausible reasons for the fact that K.E is not conserved in most cases?
Answer 3.11

A collision in which K.E remains constant after and before collision is called elastic collision.
In a collision where K.E does not remains constant is called inelastic collision.
Ideally the bouncing ball will rebound to some height in case of an elastic collision, but if it rebounds to nearly to the initial height, the collision is considered elastic one. In case of inelastic collision the ball will rebound to a small height as compared to dropping point.

Question 3.12 Explain what is meant by projectile motion. Derive expressions for
  1. the time of flight
  2. the range of projectile.                               
Show that the range of projectile is maximum when projectile is thrown at an angle of 45 degree with the horizontal.
Answer 3.12 Projectile Motion:
It is a two dimensional motion under constant acceleration due to gravity

1) The time of flight:

 The time taken by the body to cover the distances from the place of its projection to the place where it hits the ground at the same level is called the time of flight. This can be obtained by taking S=h=0, because the body goes up and comes back to the same level, thus covering no vertical distance. If the body is projecting with velocity v making angle  with a horizontal, then its vertical components will be visinθ
We have S = 0 ; vi = vi sinθ ; a = -g
Using S = vi t + ½ a t2
⇒ 0 = vi sinθ t – ½ g t2
or t = 2 vi sinθ/g
t= 2vi sin/g   
where t is the time of flight of the projectile when it is projected from the ground.

2) Range of the Projectile: 

Maximum distance which a projectile covers in the horizontal direction is called the range of the projectile.To determine the range R of the projectile, we multiply the horizontal components of the velocity of projection with the time taken by the body after leaving the point of projection thus. We have
S = R , vx = vi cosθ, t = 2vi sinθ
Using S = v t g
⇒ R = (vi cosθ x 2 vi sinθ)/g = (vi2 2 sinθ cosθ)/g
or R = (v sin 2θ)/g
for maximum range, sin 2θ should have maximum value ( i.e.) = 1
sin 2θ = 1 ⇒ 2θ = 90 ⇒ θ = 45

Question 3.13 At what point or points in its path does a projectile have its minimum speed, its maximum speed?
Answer 3.13
A projectile has its minimum speed at the maximum height where vertical component of the velocity becomes zero. A projectile has its maximum speed at its minimum height if air friction is neglected.

Question 3.14 Each of the following questions is followed by four answers, one of what correct answer.
Identify that answer.

1)What is meant by a ballistic trajectory?
  1. The paths followed by an un-powered and unguided projectile.
  2. The paths followed by the powered and unguided projectile.
  3. The paths followed by an un-powered but guided projectile.
  4. The paths followed by an powered and guided projectile.

2)What happens when a system of two bodies undergoes an elastic collision?

  1. The momentum of the system changes.
  2. The momentum of the system does not change.
  3. The bodies come to rest after collision.
  4. The energy conservation law is violated.

Answer 3.14
i) The statement (a) is correct as ballistic trajectory follows an un-powered and unguided projectile.
ii) The statement (b) is correct as momentum of the system does not change.                                                                                                                                                                                

Written By: Asad Hussain


F.Sc Notes: Physics XI: Chapter 01 Measurements Exercise Short Questions:

FSc Notes: Physics XI: Chapter 01 Measurements Exercise Short Questions:

Question 1.1 Name several repetitive phenomenons’ occurring in nature which could serve as reasonable time standards?
Answer 1.1
Every type of natural phenomenon which is repetitive after same intervals of time acts as reasonable time standards for e.g.
  1. Rotation of earth around the sun in 365 days,
  2. Change of shadows of objects position of sun position of star,
  3. Moon revolve around earth in 28 to 29 days.

Question 1.2 Give the draw backs to use time period of a pendulum as a time standards?
Answer 1.2
Drawbacks to use time period of a pendulum as a time standard
  1. Air friction can change the time period.
  2. Due to increase in temp length of pendulum
  3. Increase and hence the time period increases.
  4. Due to change in position from sea level gravitational acceleration changes hence time period changes.

Question 1.3 Why do we find it useful to have two units for the amount of substance, the kilogram and the mole?
Answer 1.3
It is very useful to have two units for the amount of substance i.e. kg mole.
  1. Kilogram is a macro level unit it only gives the amount of substance.
  2. Mole is a micro level unit one mole of substance contains the same no of molecules, no of radicals  no of atoms etc.

Question 1.4 Three students measured the length of a needle with a scale on which minimum
division is 1mm and recorded as    (1)0.2145m    (2).21m    (3).214m
which is correct and why?
Answer 1.4
In these readings option (3).214m is correct because the least count is 1m=.001mm it measures accurately up to three decimal places.

Question 1.5 An old saying is that “A chain is as strong as its weakest link” what analogous statement can you make regarding experimental data used in computation?
Answer 1.5
The analogous statement is: A given data is as much accurate as its least accurate values.

Question 1.6 The period of a simple pendulum is measured by a stop watch what type of errors are possible in the time period?
Answer 1.6
Following errors is possible in the time period:
  1. Systematic error
  2. Random error
  3. Personal  error
  4. Air friction.

Question 1.7 Does a dimensional analysis give any information on constant of proportionality that may appear in an algebraic expression? Explain.
Answer 1.7
Two answers are possible here: Yes, we can find the units of constants by using dimensional analysis (e.g.  G). If the constant is a number such as 2π and 6π, then dimensional analysis gives no information about the constant.

Question 1.8    Write the dimensions of (1) pressure and (2) density?
Answer 1.8 
    (1) Pressure:      P = F/A= ma/A
So [P] = [m] [a]/ [A] = [M] [LT-2]/ [L2]

[P]= [M1] [L1] [T-2] [L-2] = [M1] [L-1] [T-2]

[P] = [ML-1T-2]

(2) Density:         =m/v

[d] = [M] / [L3] = [ML-3]

[d] = [ML-3]

Question 1.9   The wavelength ƛ of a wave depends on the speed v of the wave and its
            frequency f. Knowing that               [ƛ] = [L]                [v] = [LT-1]   [f] = [T-1].
            Decide which of the following is correct.      f=v ƛ      or            f=v/ ƛ.
Answer 1.9
    Given    [ƛ] = [L]                [v] = [LT-1]   [f] = [ T-1]

(i)                  f=v ƛ

[T -1]= [LT-1] [L]

Or [T -1] = [L2T-1]

As the dimension on both are not same so ( I )  is incorrect.

(ii)           f=v/ ƛ

[T -1] = [LT-1]/ [L]

[T -1] = [T-1]

As the dimension on both sides are same so formula (ii) is correct.

Written By: Asad Hussain