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Solutions for Motion in a plane conceptual questions
2. a
3. d
4.a,b
5.a,b
6.d
7.b,c
8. a,b,c
9.a,b,d
10.a,b
Solution for Gravitation conceptual question
2.b
this can derived using kepler law T^{2} =kR^{3}
3.b
4.c
5.b
6.a,c
7.d
8.a
9.c
10.a
Conceptual question for Newton's law
(a) Direction of motion
(b) Momentum
(c) Kinetic energy
(d) All the above
2. there are two statements
A Rate of change of momentum correspondes to force
B Rate of change of momentum corresponds to Kinetic Energy
Which of the following is correct
(a) A only
(b) B only
(c) Both A and B are correct
(d) Both A and B are wrong
3. A truck and car are moving on a plane road with same kinetic Energy. They are brought to rest by application of brakes which provide equal retarding forces. Which of the following statements is true.
(a) Distance travelled by truck is shorter then car before comming to rest
(b) Distance travelled by car is shorter then truck before comming to rest
(c) Distance travelled depends on individual velocity of both the vehicles
(d) Both will travel same distance before comming to rest
4. A block A of mass m_{1} is released from top of smooth inclined plane and it slides down the plane. Another block of mass m_{2} such that m_{2} > m_{1} is dropped from the same point and falls vertically downwards. Which one of the following statements will be true if the friction offered by air is negligible?
(a) Both blocks will reach ground at same time
(b) Both blocks will reach ground with the same speed
(c) speed of both the blocks when they reach ground will depend on their masses
(d) Block A reaches ground brfore block B
5. Which of the following observer is/are non interial
a.A child revolving in the merry ground
b. A driver in a train moving with constant velocity v
c. A passenger in a bus which is slowing down to a stop
d None of these
6. there are two statements
A Newtons first law in valid from the pilot in an aircraft which is taking off
B Newtons first law in valid from the observer in a train moving with constant velocity
Which of the following is correct
(a) A only
(b) B only
(c) Both A and B are correct
(d) Both A and B are wrong
7.A IITJEE text book of mass M rests flat on a horizontal table of mass m placed on the ground.Let R_{X>Y} be the constant force exerted by the body x on body Y.According to Newton third law,which of the following is an actionreaction pair of forces?
a. (M+m)g and R_{table>book}
b. R_{ground>table} and mg+R_{book>table}
c. R_{ground>table} and R_{table>ground}
d Mg and R_{table>book}
8. Choose the correct alternative
a. The acceleration of a moving particle is always in the direction of its velocity
b. Velocity and acceleraion vectors of a moving particle may have any angle between 0 and 360 between them
c.Velocity and acceleraion vectors of a moving particle may have any angle between 0 and 180 between them
d. If the acceleration vector is always perpendicular to the velocity vector of a moving particle,the velocity vector does not change
Solutions
Detailed Solutions
Ans 1:
A body acted uopn by a certain force produces acceleration i.e. it undergoes change in it's velocity. hence choice (d) is correct
Ans 4.
since velocity of block when it reaches the ground is given by v=(2gh)^{1/2} and it is independent of the mass the correct choice will be (b).
Ans 6
We know that Newtons law is invalid in Non interial frame of refrence, now since acceleration is present in Ist case..So answer is B
Ans 7.
Action reaction pair acts on diffrent body and always are in oppsite direction.We will have to draw free body diagram for each part in this case.
Newtons law of Motion
(1) Newton's first law of motion:
'A body continues to be in state of rest or uniform motion unless it is acted upon by some external force to act otherwise'
(2) Newton's second law of motion:
'Rate of change of momentum of a body is proportional to the applied force and takes place in the direction of action of force applied'
Mathematically,
F=dp/dt
=ma
where, F=external force applied
p=mv,momentum of the body
a=acceleration
SI unit of force is Newton (N).
(3) Impulse is the product of force and time which is equal to the change in momentum
Impulse=FΔt
=Δp
(4) Newton's third law of motion:
'To every action there is always an equal and opposite reaction'
F_{AB}=F_{BA}
(5) According to the law of conservation of linear momentum
Initial momentum = final momentum
m_{1}v1+m_{2}v2=m_{1}v1^{'}+m_{2}v2^{'}
For equilibrium of a body
F_{1}+F_{2}+F_{3}=0
Some points to note
1. An accelerated frame is called non interial frame while am non accelerated frame is called interial frame
2. Newton Ist law are valid in intertial frame only
3 Apparent weight of a body in the lift
Going Upward with acceleration a
W=m(g+a)
Going Down with acceleration a
W=m(ga)
4.Always draw free body diagram to solve the force related problems
Thermodynamics objective solutions
1. b
Hint:Workdone is area enclosed
2. b
Hint:Workdone is area enclosed
3.d
Hint:Workdone is area enclosed
4.a
5.d
6.c
7.d
8.b
9.c
hint for PV^{n}=constant,Molar Specific Heat=R/y1 + R/1n
10 a
hint for TV^{y1}=constant
Feb 4 Solutions
1.a
dU=dQdW
2.d
3.d
4.a
hint V_{rms}=√3RT/M
5.b
Graphical Question for Kinematics
2. The displacement time graph of a moving particle is shown below.The instantanous velocity of the particle is negative at the point
a. C
b. D
c. B
d. A
3.The velocity time graph of a moving particle is shown below.Total displacement of the particle during the time interval when there is nonzero acceleration and retardation is
a. 60m
b. 40m
c. 50m
d. 30m
4. Figure below shows the displacement time graph of two particles.Mark the correct statement about their relative velocity
a. It first increases and then decreases
b. It is a non zero constant
c. it is zero
d. none of the above
5.Four position time graph are shown below.What all graph shows motion with positive velocities
a. a and c only
b all the four
c. b and D only
d. b only
6. What all graph shows motion with negative velocities
a. b and d only
b all the four
c. a and c only
d. c only
7.which of the folliwing graph correctly represents velocitytime relationships for a particle released from rest to fall under gravity
8.The vx graph(fig1) of a particle moving along a straight line is shown below.Which of the following below graph(A,B,C,D) shows ax graph
Solutions
Relative velocity Conceptual questions
1. Bullet one is fired in the north direction with the muzzle velocity u.Find the velocity of the bullet as seen from the observer on the earth
a u+v
b uv
c u
d v
2. find the velocity of the bullet as seen from the observer on the moving car
a u+vw
b uvw
c u
d v
3.Bullet one is fired in the south direction with the muzzle velocity u.Find the velocity of the bullet as seen from the observer on the earth
a u+v
b vu
c u
d v
4.find the velocity of the second bullet as seen from the observer on the moving car
a u+vw
b vuw
c u
d v
B.A train is moving in the west direction with a velocity 15m/s.A monkey runs on the roof of the train against its motion with a velocity 5m/s with respect to train .Take the motion along west as positive
5.Velocity of train relative to its driver
a. 0
b. 15 m/s
c. 15 m/s
d. 20 m/s
6. What is the velocty of train with respect to monkey
a. 5m/s
b 5 m/s
c. 15 m/s
d 15 m/s
7. find the velocty of ground with respect to monkey
a. 5 m/s
b. 5 m/s
c. 10 m/s
d. 10 m/s
C.Two particles start from the origin of the horizontal xy plane.Particle A moves along +x axis with uniform velocity u.Particle B moves along y axis with velocity u. i and j are the unit vectors along x and y direction
8. find the relative velocity of particle B wrt to A at time t
a u(i+j)
b u(i+j)
c 0
d none of the above
9. find the relative velocity of particle A wrt to B at time t
a u(i+j)
b u(i+j)
c 0
d none of the above
10. find the relative position vector of particle A wrt to B at time t
a ut(i+j)
b ut(i+j)
c. 0
d. None of the above
Solutions
Detailed Solutions
Ans 1
Velocity of bullet w.rt to train=u
Velcity of train=v
Velocity of bullet w.rt to train=Velocity of bullet w.r.t earth Velocity of train w.r.t earth
So Velocity of bullet w.r.t earth =Velocity of bullet w.rt to train+Velocity of train w.r.t earth
=u+v
Ans 4.
Velocity of bullet w.rt to train=u
Velcity of train=v
velocity of car
Velocity of bullet w.rt to car=Velocity of bullet w.r.t earth Velocity of car w.r.t earth
= vuw
Ans 5
Velocity of train w.r.t to driver=Velocity of train w.r.t earth Velocity of driver w.r.t earth
so =0
Ans 6
Given
velocity of train=15m/s
velocity of monkey w.r.t train=5 m/s
Now velocity of monkey w.r.t train=velocity of train w.r.t monkey
So velocity of train w.r.t monkey=5 m/s
Ans 8.
Velocity of A=ui
Velocity of B =uj
Velocity of B w.r.t to A=Velocity of BVelocity of A
=u(i+j)
Ans 9
Velocity of A=ui
Velocity of B =uj
Velocity of A w.r.t to B=Velocity of AVelocity of B
=u(i+j)
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Relative velocity
Relative velocity is defined as
V_{BA}=V_{B}V_{A}
where V_{BA} is relative velocity of B relative to A
Similiary relative velocity of A relative to B
V_{AB}=V_{A}V_{B}
Special cases.
For straight line motion,we can just add or subtract to get the relative velocity.
If the objects are moving in the same direction,relative velocity can be get by substracting other.If they are moving in opposite direction ,relative velocity will be get by adding the velocities example like train problems
For two dimensions motion,we will need to add or subtracting components along x & y direction to get the relative velocity
Suppose
v_{A}=v_{xa}i + v_{ya}j
v_{B}=v_{xb}i + v_{yb}j
Relative velocity of B relative to A
=v_{xb}i + v_{yb}j (v_{xa}i + v_{ya}j)
=i(v_{xb}v_{xa}) + j(v_{yb}v_{ya})
For three dimensions motion
v_{A}=v_{xa}i + v_{ya}j +v_{za}z
v_{B}=v_{xb}i + v_{yb}j + v_{zb}z
Relative velocity of B relative to A
=v_{xb}i + v_{yb}j + v_{zb}z (v_{xa}i + v_{ya}j +v_{za}z)
=i(v_{xb}v_{xa}) + j(v_{yb}v_{ya})+z(v_{yb}v_{ya})
Kinematics Numerical
2. A ball is thrown upward from a point on the side of a hill which slopes upward uniformly at angle 45°.Intial velocty of the ball is u_{0} m/s and at an angle 65°(with respect to the horizotal).At what distance up the slope the ball strike and in what time?
3.A cannon on a level plain is aimed at an angle θ above the horizontal.A shell is fired with a muzzle velocity v_{0} towards a pole which is distance R away.It hits the pole at height H.
a find the timetaken to reach the pole
b. find the value of H in terms of θ,R and v_{0}
4. The displacement of the body x(in meters) varies with time t (in sec) as
x=(2/3)t^{2} +16t+2
find following
a. what is the velocty at t=0,t=1
b. what is the acceleration at t=0
c. what is the displacement at t=0
d .what will the displacement when it comes to rest
e .How much time it take to come to rest.
5. A man runs at a speed at 4 m/s to overtake a standing bus.When he is 6 m behind the door at t=0,the bus moves forward and continues with constant acceleration of 1.2 m/s^{2}
find the following
a. how long does it take for the man to gain the door
b if in the beginning he is 10m behind the door ,will he running at the same speed ever catch up bus?
Detailed Solutions:
3. The time taken to reach the pole is given
R=v_{0}cosθ t
or t=R/v_{0}cosθ
Now equation of motion of vertical motion
h=v_{0}sinθt1/2gt^{2}
at t=R/v_{0}cosθ h=H
so
H=(v_{0}sinθ)X(R/v_{0}cosθ ) (1/2)g( R/v_{0}cosθ)^{2}
or H=RtanθgR^{2}/2v_{0}^{2}cos^{2}θ
4.
Given x=(2/3)t^{2} +16t+2
Velocity=dx/dt=(4/3)t+16
so velocity at t=0 =16
and velocity at t=1 =(4/3)+16=44/3
Acceleration is given as =d^{2}x/dt^{2}=4/3
so acceleration is time independent and it is constant
Displacement at t=0 can be found by simply substituting the values of t=0 in equation (1)
=2
Now v=(4/3)t+16
when v=0
then t=4/3 sec
Displacement can be found by substituing the value of t=4/3 in equation (1)
=32/27+64/3+2
=(32+576+54)/27=598/27 m
Solution for thermo conceptual question
1. (a)
There was no restriction for it expansion.So no tensile or compressive force developed.Longitudinal strains happens only when tensile or compressive force developed in the rod.So answer a
2.(c)
3.(d)
4.(c)
5. (a),(b)
6.(b)
7.(a),(c)
Due to thermal expansion,the diameter of the disc as well of the hole will increase.therefore the moment of inertia will increase resulting in a increase in the angular speed.
Solution of PART 2
1 (c)
2. (b)
3. (b)
4.(c)
5. (a),(c)
Solution of PART 3
1. (a)
dU=dQdW or dU=dQ as dQ is negative & dW=o
so dU is negative .dU <0 now dU =nC_{v}dT
so nC_{v}dT <0 or dT<0
2.(d)
3.(c) as dW=pdV so if dW> 0 the dV >0
4. (b),(c) as temperature is constant.
5. (a)
6. (a),(b)
7. (a),(b),(d)
8. (c)
9.(a),(c)
10. (a),(b)
Solution of PART 4
1. (d)
2. (c)
3. (b)
4. (a)
5. (d)
6. (d)
7. (a),(b)
8.(d)
More detailed solution for each question will be given later on
Objective question for Kinematics
x=2t
y=2t^{2}
where x and y are horizontal and vertical displacement respectivley at time t.Which one of the following is true
a.The trajectory of the body is a parabola
b The trajectory of the body is a straight line
c.the velocity vector at point t is 2i+4tj
d the acceleration vector at time t is 4j
2.when the projectile is at the highest point of its trajectory,the direction of its velocity and acceleration are
a. parallel to each other
b.antiparallel to each other
c. Inclined to each other at 45
d. Perpendicular to each other
3.The horizontal and vertical displacement of the projectile at time t are
x=36t
y=48t4.9t^{2}
where x and y are in meters and t in second.Intial velocity of the projectile in m/s
a. 15
c. 30
b. 45
d. 60
4.Displacement(y) of the particle is given by
y=2t+t^{2}2t^{2}
The velocity of the particle when acceleration is zero is given by
a. 5/2
b. 9/4
c. 13/6
d. 17/8
5. A car ,starting from rest is accelerated at constant rate a until it attains speed V.It is then retarted at a constant rate b until it comes to rest.which of the following is true
a. the average speed for the whole motion is av/2b
b. the average speed for the whole motion is v/2
c. Total time taken for the journey is v(1/a+1/b)
d. none of the above
6.The intial velocty of the particle is u=4i+3j m/s.It is moving with uniform acceleration a=.4i+.3j m/s^{2}.which of the following is true
a. the magnitude of the velocity after 10 sec is 10m/s
b. The velocity vector at time t is given by (4+.4t)i +(3+.3t)j
c. the displacement at time t is (4t+.2^{2})i +(3t+.15^{2})j
d . none of the above
7.It is possible to project an body with a given speed in two possible ways so that it has a same horizontal range.The product of time taken by it in tow possible ways is
a. R/g
b. 2R/g
c. 3R/g
d. 4R/g
8.A projectile has a range R and time of flight T.If the range is tripled by the increasing the speed of the projection ,without changing the angle of projection.The time of the flight will become
a T/√3
b T√3
c. T/3
d. 3T
9.A large number of stones are fired in all the direction with same speed V.The maximum area of the ground on which this stone will spread is
a. πV^{4}/g^{2}
b. π^{2}V^{4}/g^{2}
c. πV^{4}/g
d. πV/g^{2}
10.A body is projected horizontally from the top of the building 39.2 m high.How long will it take to hit the ground
a. 2 sec
b. 4 sec
c. 1 sec
d 2√2 sec
Detailed Solutions
Ans 4:
given
y=2t+t^{2}2t^{3}
Velocity is given
v=dy/dt=2+2t6t^{2}
a=dv/dt=212t
Now acceleratio is zero
212t=0
t=1/6
Putting this value velocity equation
v=2+2/66(1/6)^{2}
=2+1/31/6
=13/6
Ans 5:
The distance s_{1} covered by the car during the time it is accelerated is given by
2as_{1}=v^{2}
Which gives
s_{1}=v^{2}/2a
Similary in the decelerated motion,distance covered is
2bs_{2}=v^{2}
Which gives
s_{2}=v^{2}/2b
SO Total distance travelled during the whole journey
s=s_{1}+s_{2}=v^{2}/2(1/a+1/b)
Let t_{1} be the time taken during accelerated motion then
v=at_{1} or t_{1}=v/a
and Let t_{2} be the time taken during deccelerated motion then
v=bt_{2} or t_{2}=v/b
Total timetaken
t=t_{1}+t_{2}
=v(1/a+1/b)
So average speed =total distance/total time taken
=s/t=v/2
Ans 7:
If a body is projected with a given velocity u at angle θ and (90θ) to the horizontal,it will have same range R given by
R=u^{2}sin2θ/g
The corresponding times if flight are
t_{1}=2usinθ/g
t_{2}=2usin(90θ)/g=2ucosθ/g
t_{1}t_{2}=2u^{2}(2sinθcosθ)/g^{2}
=2u^{2}sin2θ/g^{2}=2R/g
Ans 8
Range is given by
R=u^{2}sin2θ/g(1)
and time taken by
T=2usinθ/g(2)
Now
3R=u_{c}^{2}sin2θ/g(3)
T_{c}=2u_{c}sinθ/g(4)
Dividing 1 by 3
1/3=(u/u_{c})^{2} 5
Dividing 2 by 4
T/T_{c}=u/u_{c} 6
From 5 and 6
T/T_{c}=1/√3
T_{c}=T/√3
Ans 10:
Intial vertical component of velocity is zero
so h=(1/2)gt^{2}
39.2=(1/2)*9.8*t^{2}
t=2√2 sec
Biography of Newton
Sir Isaac Newton was an English physicist and mathematician and the greatest scientist of his era.
Isaac Newton was born on 4 January 1643 in Woolsthorpe, Lincolnshire. His father was a prosperous farmer, who died three months before Newton was born. His mother remarried and Newton was left in the care of his grandparents. In 1661, he went to Cambridge University where he became interested in mathematics, optics, physics and astronomy. In October 1665, a plague epidemic forced the university to close and Newton returned to Woolsthorpe. The two years he spent there were an extremely fruitful time during which he began to think about gravity, and also devoted time to optics and mathematics, working out his ideas about 'fluxions' (calculus).
In 1667, Newton returned to Cambridge, where he became a fellow of Trinity College. Two years later he was appointed second Lucasian Professor of Mathematics. It was Newton's reflecting telescope, made in 1668, that finally brought him to the attention of the scientific community and in 1672 he was made a Fellow of the Royal Society. From the mid1660s, Newton conducted a series of experiments on the composition of light, discovering that white light is composed of the same system of colours that can be seen in a rainbow and establishing the modern study of optics (or the behaviour of light). In 1704 Newton published 'The Opticks' which dealt with light and colour. He also studied and published works on history, theology and alchemy.
However, in 1687, with the support of his friend the astronomer Edmond Halley, Newton published his single greatest work, the 'Philosophiae Naturalis Principia Mathematica' ('Mathematical Principles of Natural Philosophy'). This showed how a universal force, gravity, applied to all objects in all parts of the universe.
In 1689, Newton was elected MP for Cambridge University (1689  1690 and 1701  1702). In 1696 Newton was appointed warden of the Royal Mint, settling in London. He took his duties at the Mint very seriously and campaigned against corruption and inefficiency within the organisation. In 1703, he was elected president of the Royal Society, an office he held until his death. He was knighted in 1705.
Newton was a difficult man, prone to depression and often involved in bitter arguments with other scientists, but by the early 1700s he was the dominant figure in British and European science. He died on 31 March 1727 and was buried in Westminster Abbey.
Short notes in Motion in a plane
i,j,z are the unit vector along x,y and z axis
Notes: Bold letter are used to denote vector quantity
r=xi
v=(dx/dt)i
a=dv/dt=(d^{2}x/dt^{2})i
a=vdv/dr
v=u+at
s=ut+1/2at^{2}
v^{2}=u^{2}+2as
r=∫vdt
v=∫adt
Short notes for motion in two dimension
r=xi +yj
v=dr/dt=(dx/dt)i + (dy/dt)j
a=dv/dt=(d^{2}x/dt^{2})i+(d^{2}x/dt^{2})j
a=vdv/dr
v=u+at
s=ut+1/2at^{2}
v^{2}=u^{2}+2as
r=∫vdt
v=∫adt
Short notes for motion in three dimension
r=xi +yj+zk
v=dr/dt=(dx/dt)i + (dy/dt)j+(dz/dt)k
a=dv/dt=(d^{2}x/dt^{2})i+(d^{2}x/dt^{2})j+(d^{2}x/dt^{2})k
a=vdv/dr
v=u+at
s=ut+1/2at^{2}
v^{2}=u^{2}+2as
r=∫vdt
v=∫adt
Gravitation
Kepler's law
Acceleration due to grawing of earth
Acceleration due to grawing below and above earth surface.
(1) Introduction :
 In our dairy life we have noticed things falling freely downwords towards earth when thrown upwards or dispped from some hight.
 Fact that all bodies irrespective of their masses are accelerated towards the earth mith a constant aceeleation was first recognized by galive (15641642)
 The motion of celesh'al bodies such as moon. earth plametes etc. and attrachieve of moon towards earth and arth towards sun is an interasting subject of study since long time.
 Now the question's what is the force that produces such acceleration is which earth attract all bodies towards the centre and what is the law governing this force.
 Is this law is same for both earthly and weshal bodies.
 Answer to this question was given by Newton as he declared that "laws of nature are same for earthly and weshal Bodies".
 The force between any object falling freely towards earth and that between earth and moon are gowerned by the same laws.
 Johnaase kepler (15711631) Studied the planetary motion in detail and formulated his three laws of planetary motion, which were available Universal law of grawitation.
(2) Kepler's Law :
Kepler's law of planetary motion are :
(i) Law of orbits :
Each planet revolues around the sun in an elliptical orbit with sun at one of the foci of the ellipse .
Fig (a) An ellipse traced by planet sevolary round the sun.
AO = a  Sewi major axis
BO = b  Sewi minor axis
P  hearest point between planet and sun k/as perihetion
A  farthest point between planet and sun apheiton.
(ii) Law of areas :
"The line joining planet and the sun sweeps
equal area in equal intervals of time"
This law follows from th e observation that when planet is nearer to the sun
its velocity increases and It appears to be slower when it is farther from the sun.
(iii) Law of periods :
"The squre of time period of any planet about the sun is propotional to the cube of the semimajor axis."
 If T is the time period of semi major axis a of elliptical orbit then.
T^{2} x a^{3} (1)
 If T_{1} and T_{2} are time periods of any two planets and a_{1} and a_{2} being their semi major axis resp. then
T_{1}^{2} =ka_{1}^{3}
T_{2}^{2} =ka_{2}^{3} (2)
This question (2) can be used to find the time period of a planet, when the time period of the other planet and the semimajor axis of orbits of two planets.
(3) Universal law of gravition :
 Everybody in the universe attracts every other body with a force which is directly proportional to the product of their masses and invessly proportional to the square of distance between them.  Mathematically Newton's gravitation law is if F is the force acting between two bodies of masses M_{1} and M_{2} and the distance between them is R then majnitade of force is given as
F =  Gm_{1}m_{2} 
r^{2} 
In vector notation
F =  Gm_{1}m_{2}(r^) 
r^{2} 
F =  Gm_{1}m_{2}r^ 
r^{2} 
where G universal gravitational constant
r^  unit vector from m_{1} to m_{2} and r^ = r_{1}^  r_{2}^
Gravitational force is attractive constant is
SI  G= 6.67 x 10^{11} nm^{2} kg^{2}
CGS  G= 6.67 x 10^{8} dyn cm^{2} g^{2}
Deminsional formula of C1 is [M^{1}L^{3}T^{2}]
(4) The gravitational constant : (Measurement)
(5) Accelesation due to gravity of earth :
 Earth attracts every object lying an its surface towards its centre with a force known as gravitational towards its centre with a force known as gravitational pull or gravity.
 Whenever force acts on any body it produces acceleration and in case of gravitation this acceleration produced under effect of gravity is known as accelesation due to gravity (g)
 Value of accelesation due to gravity is independent of mass of the body and its value near surface of earth is 9.8 ms^{2}
 Expression for acceleration due to gravity
Consider mass of earth to be as M_{E} and its redius be R_{E} Suppose a body of mass M (much smaller then that fo earth) is kept at the earth surface. Force eseerted by earth on the body of mass m is
F =  GMM_{E} 
R_{E}^{2} 
Let the above equation is (4)
The force for the body due to earth produces acceleration due to gravity
(g) in the motion of the body. From Newton's Second law of motion
f=mg (5)
from (4) and (5)
g =  GM_{E} 
R_{E}^{2} 
which is acceleration due to gravity at earth's surface.
(6) Acceleration due to gravity below and above the earth surface :
(i) Above earth's surface
 An object of mass m is placed at hight h above the earth's surface
F =  GMM_{E} 
(R_{E}+h)^{2} 
from this it can be concluded that value of g decreases as distance above surface of earth measures now,
g =  GM_{E} 
R_{E}(1+h/R_{E})^{2} 
g =  g_{o} 
(1+h/R_{E})^{2} 
which is equation no (7)
Where
g_{o} =  GM_{E} 
R_{e}^{2} 
* eqn (7) tells us that for small hight h above surface of earth. value of g decreases by factor (12h/R_{E})
 for h is smaller compared to R
g=g_{o}(1+h/R_{E})^{2}
g=g_{o}(12h/R_{E}) Expanding by Binomial theriom
(ii) Below the earth's surface
 If one goes inside the earth surface the value of g again decreases
 P = density of material of earth them
m=4/3R_{E}^{3}P
from this accelesation due to gravity at earth’s Surface is
g =  G4/33R_{E}^{3}P 
R_{E}^{2} 
g=4/3GR_{E}P (8)
g acceleration due to gravity at depth d below earth's surface
 Body at depth d will experience force only due to portion of reduce (R_{E}d) of earth's
 outer spnerical shell of thickness d will not exeperience any force
 M is mas of the portion of earth with radius (R_{E}d) then
g =  GM 
R_{E}^{2} 
M=4/3(R_{E}d)^{3} P
g =  G4/3(R_{E}d)^{3} P 
(R_{E}^{2}d)^{2} 
g= 4/3G (R_{E}d)P (9)
Dividing epn (9) by (8)
g/go= (1d/R_{E})
or = g=go (1d/R_{E})  (10)
from eqn (10) it is clearer that acceeleration due to gravity also decreases with depth.
Solution for Jan 28 Problems
a. Let P be the Pressure of the gas.
Then
Pressure of gas=weight of the Piston/A + Atmospheric pressure
=Mg/A + P N/m^{2}
b.Pressure remain constant during the heating process
Intial state of Gas
P_{1}=Mg/A + P
T_{1}=T
Now we know that
PV=nRT
V_{1} =  nRT_{1} 
P_{1} 
V_{1} =  nRT 
Mg/A + P 
Final State
P_{2}=Mg/A + P
T_{2}=2T
Now we know that
PV=nRT
V_{2} =  nRT_{2} 
P_{2} 
V_{2} =  2nRT 
Mg/A + P 
Workdone by gas=P(V_{2}V_{1})
Substituting all the values we get
Workdone=nRT
Change in Internal Energy=nC^{V}Î”T
=n*3R/2*T
=3nRT/2
Now from first law of thermodynamics
Heat supplied=Change in Internal Energy + Workdone by gas
=5nRT/2
Solution 2:
Given that
Pressure at the ice point P_{ice}= 80 cm of Hg
Pressure at the steam point P_{steam}= 90 cm of Hg
Pressure at the wax bath P_{wax}= 100 cm of Hg
The temperature of the wax bath measured by the thermometer is
T =  (P_{wax}P_{ice})*100 
P_{steam}P_{ice} 
T = (100  80)X100/(9080)
= 20X100/10
=200° C
Solution 3:
dQ=mCdT
Now integrating with upper and lower limit as T_{1} and T_{2}
Q=∫mC_{0}(1+aT)dT
Q=mC_{0}[T+aT^{2}/2]
Q =  mC_{0}[2T_{2}2T_{1}a(T_{1}^{2}T_{2}^{2}] 
2 
Q =  mC_{0}(T_{2}T_{1})(2+a(T_{2}+T_{1})) 
2 
Solution 4:
Workdone in Process AB=P(3VV)
=2PV
Workdone in Process BC=0 as volume is constant
Workdone in Process CD=3P(V3V)
=6PV
Workdone in Process DA=0 as volume is constant
So net workdone=4PV
Net heat =Net workdone=4V
Change in Internal energy =0
Solution of Jan 25 Problems
Total forces on the piston
Weight of the piston acting downward=200N
Force due to Atmospheric pressure downward =100*10^{3}*20*10^{4} N=200N
Let P be the Pressure of the gas
Then Force acting due to pressure of gas on the piston upward=PA
Force with the piston is being pulled upward=100N
Total Upward force=Total Downward force
PA + 100=200 +200
20*10^{4}*P=300
P=150kPa
Solution 2:
Since O_{2} is a diaatomic gas
C_{P}=7R/2,C_{V}=5R/2
Molecular mass=32 gm
Intial state
Intial volume of gas
V=nRT/P
n=2*10^{3}/32
T=303K
R=8.3
P=500*10^{3} N/m^{2}
so V=.314 m^{3}
Second state when At top volume becomes double
Volume=2*.314 =.628m^{3}
Pressure=500*10^{3} N/m^{2} remains same
So as per ideal gas equation
T=606K
Workdone by the gas=PV=500*10^{3}*.314=157*10^{3} J
Heat transfered=nC_{P}(T_{2} T_{1})
=2*10^{3} *7*8.3*303/32*2
=550*10^{3} J
Third state when Pressure becomes double
P=1000*10^{3} N/m^{2}
V=.628m^{3}
As per ideal gas equation
Temperature=1212K
Heat transfered=nC_{V}(T_{3} T_{2})
=2*10^{3} *5*8.3*606/32*2
=785 kJ
So total heat tranfered=550+785=1335 kJ
Final Temperature=1216 K
Solution 3:
Since O_{2} is a diaatomic gas
C_{P}=7R/2,C_{V}=5R/2
Molecular mass=32 gm
Intial state
Intial volume of gas
V=nRT/P
n=1*10^{3}/32
T=303K
R=8.3
P_{0}=500*10^{3} N/m^{2}
so V=.157 m^{3}
Second state ( when piston reaches the spring)
v=.2 m^{3}
P=500*10^{3} N/m^{2}
As per ideal gas equation
Temperature becomes=385.5 K
So heat tranfer till that point=nC_{P}(T_{2} T_{1})
=1*10^{3} *7*8.3*82.5/32*2
=74.8 KJ
Third state ( when it compresses the spring)V=.2+0.1*0.25=0.225 m^{3}
P=500*10^{3} + kx/A
=500*10^{3} + 120*10^{3}*25*10^{2}/.1
=800*10^{3} N/m^{2}
As per ideal equation
Temperature becomes=694K
Change in Internal energy=nC_{V}(T_{3} T_{2})=5nR(T_{3} T_{2})/2
=5*1*10^{3}*8.3*308.5/32*2
=200 KJ
Workdone by the gas =P_{0}Ax + kx^{2}/2
=500*10^{3}*.1*.25 + 120*10^{3}*.25*.25/2
=16.25 KJ
Total heat supplied in this Process=200+16.25=216.25KJ
So net Heat transfer=291.05 KJ
Solution 4:
Doubling the system should double the energy, so U is an extensive variable.
Solution:5
By repeated application of the Zeroth Law, we can state that all M+N systems are in thermal equilibrium with each other.
Solution:6
Total heat supplied =Workdone + Change in internal energy
So work done=21401580=560 J
Let s be the distance moved then
the workdone is given by =Fs
Fs=560
s=560/F
=560/102*10
s=.54 m
Solution:7
As per dimension analysis Unit on both sides should be equal
Now since R & R_{0} both unit are same
Quantity 1+aT+bT^{5} should be dimension less
so at should be dimension less
so a unit is C^{1}
similarly bt^{5} should be dimensionless
so b unit is C^{5}
Solutions for Jan 18 problems
With increase in temperature from 0° C to 100° C diameter of ring increases
using
L=L_{0}(1+Î±Î”T)
where L_{0}=2.0 cm
Î± = 2.3 * 10^{3} / ° C
Î”T=(100 0)=100 ° C
we can find diameter at 100° C
L=2(1+2.3*10^{3}*100)
=2.46 cm
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