Posted by Chittoor Live | July 27, 2018
MEASUREMENT OF LENGTH, OUR UNIVERSE GRAVITATION
MEASUREMENT OF LENGTH
SCREW GAUGE
1. PRINCIPLE OF A SCREW GAUGE:-
Screw gauge works on the principle of a screw in a nut or micrometer screw.
2. PITCH OF A SCREW:-
The distance travelled by the tip of the screw for one complete rotation of its head is called the pitch of a screw.
3. LEAST COUNT OF A SCREW GAUGE:-
Least count of a screw gauge is defined as the ratio of pitch of the screw to the number of head scale divisions(H.S.D.).
4. DESCRIPTION OF A SCREW GAUAGE:-
F-‘U’shaped metallic frame, S1-stud, S2-tip of the screw , M-milled head
1. A screw gauge consists of U-shaped metallic frame ‘F’ as shown the
figure. To one end of this frame shaft ‘s1’ is fixed.
2. On the opposite end of ‘F’ a hallow long cylinder ‘C’ is fixed.
3. The hallow cylinder has fine threads cut inside it and it serves as a nut.
4. On the outer surface of the hollow cylinder an index line is present which is divided into some equal divisions(0.5mm or 1mm). This serves as the pitch scale ‘P’.
5. A screw with a flat shaft ‘s2’ having threads moves through the cylinder.
6. To the other end of the screw a milled head ‘M’ is connected.
7. To this milled head, one end of the barrel ‘B’ is connected.
8. The other end of this barrel is tapered and has 100 or 50 divisions on it. This is called head scale ‘H’.
9. When the two flat shafts ‘s1’ and ‘s2’ are exactly opposite to each other and are in contact, the distance between them is zero. In this position the edge of the head scale must coincide with the zeroth division of the pitch scale. Further index line must be coinciding with the zero of the head scale.
Determine the diameter of a wire using a screw gauge
1. The least count of a screw gauge is determined first by using the formula.
2. The given wire is placed between the shafts ‘s1’ and ‘s2’ and the head scale is rotated in clock wise direction such that it holds the wire tightly.
3. The value of the pitch scale division which just precedes the edge of the head scale is noted as pitch scale reading (P.S.R).
4. The value of the head scale division which coincides with the index line is observed as head scale reading (H.S.R.).
5. Now the diameter of the wire is determined using the formula.
d=P.S.R + (H.S.R x L.C.)
PRECAUTIONS:-
1. P.S.R. must be taken with out parallax error.
2. Screw must be rotated by holding the milled head provided with the safety device.
3. Same units must be used for P.S.R and least count.
6. POSITIVE ZERO ERROR OF A SCREW GUAGE:-
Head Scale
Index line of pitch scale
If the zeroth division of the head scale is below the index line of the pitch scale the error is said to be Positive and the correction is Negative.
7. NEGATIVE ZERO ERROR OF A SCREW GUAGE:-
Head scale
Index line of pitch scale
If the zeroth division of the head scale is above the index line of the pitch scale the error is said to be Negative and the correction is Positive.
8. ADVANTAGES OF A SCREW GAUGE:-
Screw gauge is used to measure the thickness of a glass plate, diameter of a wire and the diameter of a small sphere.
9. Draw the diagrams of a) No Zero Error b) Negative Zero Error c) Positive Zero Error
Problems:-
1. Find the least count of a screw gauge whose head scale is divided into 200 divisions, if it moves 5mm distance when the head is rotated through 5 revolutions.
Given
Least count (l.c.)=?
H.S.D=200
Distance moved(x) =5mm;
Number of revolution (n)=5
We know that
2. While measuring the diameter of a lead shot using a screw gauge the reading on the pitch scale is found to be 7.5mm and that on the head scale is 48. If the least count is 0.01mm and zero error is +0.05mm, find the diameter of the lead shot?
Given,
Pitch scale reading (P.S.R.)=7.5mm
Head scale reading (H.S.R.)=48
Least count (L.C)=0.01mm
Zero error=+0.05mm
Zero correction=-0.05mm
d=?
We know that
d=P.S.R. + (C.H.S.RxL.C)
=7.5 + (47.95×0.01)
=7.5 + 0.4795
=7.9795mm
Diameter(d)=7.9795mm
3. The head of a screw gauge is divided into 50 divisions. It advances 1mm when the screw is turned through 2 rotations. Find the pitch of the screw and Least count?
Given
Head scale divisions (H.S.D.)=50
Distance moved(x) = 1mm
No. of rotations (n) = 2
Pitch of the screw (P)=?
L.C. = ?
We know that,
= =0.5mm
=0.01mm
Least Count(L.C.)=0.01mm
4. When the above screw gauge is used to measure the diameter of a nail the P.S.R = 1.5mm and H.S.D=18. Find the diameter of a nail?
Given,
Pitch scale reading (P.S.R)=1.5mm
Head Scale reading (H.S.R.)=18
Least count (L.C.)=0.01mm
We know that,
D=P.S.R. + (H.S.R x L.C)
=1.5 + (18×0.01)
=1.5+0.18mm
=1.68mm
The diameter of the nail d=1.68mm
III Match the following:-
A B
1. Least count of the screw gauge (c) a) Micrometer Screw.
2. Principle of the screw gauge (a) b) Screw gauge.
3. Used for finding the thickness of a (b)
glass plate
4. Distance between two successive (b) a)P.S.R.+(C.H.S.RxL.C)
threads of screw
5. Diameter of a wire as measured (a) b) Pitch of the screw.
by the screw gauge
IV Fill in the blanks:-
1. The smallest length that can be measured accurately using any scale is called least count.
2. Least count of an ordinary scale is 1mm.
3. Smallest length including the thickness of the glass plate can be measured using a screw gauge upto 0.01mm (or) 0.001 cm (or) fraction of 1mm.
4. A mechanical device used to measure the thickness of a glass piece or a glass plate is a screw gauge.
5. A screw gauge works on the principle of a screw in a nut.
6. Distance between two adjacent threads of a screw is called Pitch of the screw.
7. If the zeroth division of the head scale is below the index line of the pitch scale the error is said to be positive and the correction is negative.
8. If the zeroth division of the head scale is below the index line of the pitch scale the error is said to be negative and the correction is positive.
9. Screw gauge consists of pitch scale and head scale.
10. Pitch of the screw
11. For a screw gauge least count=
L.C.
12. For measuring the length of any object we must keep our eyes vertically above the divisions.
13. Small lengths that can be measured as diameter of a wire or thickness a wire or a glass plate can be measured to a fraction of 1mm.
14. The distance traveled by the tip of the screw for one full rotation of the head is called pitch of the screw.
15. The divisions marked on the barrel of a screw gauge is called head scale divisions.
OUR UNIVERSE GRAVITATION
1. Geocentric Theory :-(Ptolemaic Theory)
Ptolemy proposed the geocentric theory. According to this theory, earth is stationary and is at the centre of the universe. The sun, the moon, planets, stars, revolves around the earth.
2. Heliocentric Theory:- (Copernican theory)
2. Copernicus proposed the Helio centric theory. According to this theory
1. The earth and other planets move in perfect circles around the sun.
2. The sun is located at the centre of these circles.
3. The earth and other planets would also rotate about their own axis while orbiting around the sun, in circular orbits of different radii.
3. Differences between Geocentric theory and Heliocentric theory
Geocentric theory | Heliocentric theory |
1. According to this theory the earth is stationary | 1. According to this theory the sun is stationary. |
2. The sun, moon, planets and stars revolve around the earth | 2. The earth and other planets revolve around the sun. |
3. Earth is located at the centre of the universe | 3. The sun is located at the centre of the universe. |
4. It was proposed by Ptolemy. | 4. It was proposed by Copernicus. |
5. It is also called as Ptolemaic theory. | 5. It is also called as Copernican theory. |
4. Derive
Let m1 and m2 be the masses of two bodies. Let ’r’ be the distance between them. According to Newton’s Universal law of gravitation
1. The force of attraction between the two bodies is directly proportional to the product of their masses.
F α m1m2 (1)
2. The gravitational force of attraction between two bodies is inversely proportional to the square of the distance between them.
From equations (1) and (2) we have
Or
Where G is the proportionality constant called Universal gravitational constant.
Note-1:-
S.I unit of G is Newton meter square /kilogram square.
=Nm2/kg2
Note-2:-
The value of G=6.67×10-11 Nm2/kg2.
3. Mass of the earth=6×1024 kg.
4. Radius of the earth(R)=6.4x103km
Or
=6.4×106 m.
5. Acceleration due to gravity:-
The Uniform acceleration produced in a freely falling body due to the gravitational pull of the earth is known as acceleration due to gravity. It is denoted by ‘g’.
6. Relation between acceleration due to gravity(g) and Universal gravitational constant(G):-
Let,
A stone of mass ‘m’ be dropped from a distance ‘r’ from the centre of the earth of mass ‘M’.
The force exerted by the earth on the stone is given by the universal law of gravitation as
(1)
This force produces acceleration ‘g’ in it. According to Newton’s Second law of motion.
Force=massxacceleration
i.e., F=mxa
F=mg (2)
From equation (1) & (2) we have
Note:-
1. Acceleration due to gravity on the earth is 9.8 m/s2.
2. Acceleration due to gravity on the moon is 1.67 m/s2.
3. Acceleration due to gravity on the sun is 27.4 m/s2.
7 What are the factors the influences the value of ‘g’
or
Draw a chart showing variation in the value of ‘g’
Factors influencing ‘g’ | Effect of those factors on the ‘g’ value |
1. Height(altitude) | ‘g’ decreases with altitude |
2. depth | ‘g’ decreases with depth. |
3. local conditions | The value of ‘g’ is slightly affected by geological deposits, massive concrete buildings and topography. |
4. At the poles | As the earth is flattened at the poles ‘g’ is maximum. |
5. At the equator | As the earth is bulged at the equator, ‘g’ value is minimum. |
8. Gravity meters:-
Sensitive instruments used to measure small changes in the value of ‘g’ at a given location are called gravity meters.
Example:-
1. Gulf gravity meter.
2. Boliden gravity meter.
9. Mass:-
The total amount of matter contained in a body is called its mass.
S.I. unit of mass =kg.
C.G.S unit of mass=gm.
10. Weight:-
The force with which a body is attracted towards the center of the earth is called the weight of the body.
W=mxg
W=mg
Where,
‘m’ is the mass and ‘g’ is the acceleration due to gravity.
Note:-
1. S.I.unit of weight =kgwt.
2. C.G.S unit of weight=gm.wt
3. Gravitational unit of force is =kgwt.
4. 1kgwt=9.8 Newtons.
5. 1gmwt=980 dynes.
11. Kilogram Weight (kgwt):-
1 kgwt is that gravitational force which acts on a body of mass 1kg.
12. Differences between:-
Gravitational Constant(G) | Acceleration due to gravity(g) |
1. Gravitational constant is the force of attraction between two unit masses separated by a unit distance. | 1. The uniform acceleration produced in a freely falling body due to gravitational pull of the earth is called acceleration due to gravity. |
2. It is denoted by ‘G’ | 2. It is denoted by ‘g’ |
3. It is a universal constant for all bodies at any place in the universe. | 3. It is the constant on the surface of the earth and it varies from place to place on the earth. |
4. It is a scalar quantity. | 4. It is a vector quantity. |
13. Differences between Mass and Weight:-
Mass | Weight |
1. Mass is the total amount of matter contained in a body. | 1. Weight of a body is the force with which it is attracted towards the center of the earth. |
2. It is denoted by ‘m’ | 2. It is denoted by ‘W’ |
3. Its value is constant at all places. | 3. Its value changes from place to place as the value of ‘g’ changes. |
4. It is a scalar quantity | 4. It is a Vector quantity. |
5. It is measured using a ‘common balance’. | 5. It is measured using a ‘spring balance’. |
14. State and verify Hooke’s law:-
Statement:-
“Within the elastic limit the extension produced in the spring is directly proportional to the force applied”.
Experiment:-
Aim:-
To verify Hooke’s law, and to determine the weight of a body.
Apparatus:-
An extensible spring, pan, pin, weights, retort stand, meter scale, plasticine.
Procedure:-
1. Suspend extensible spring from a stand with a hook.
2. Attach a pan to the lower end of the spring.
3. Fix a pin at the lower end and at right angles to the spring by means of a little plasticine. This pin serves as a pointer.
4. Fix a metre scale vertically so that the pointer can move over the scale.
5. Record the reading of the pointer on the scale when the pan is empty(lo).
6. Add 10 grams of weight to the pan and note the reading of the pointer(l1).
7. Proceed like this by increasing the loads regularly in steps of 10 grams upto 100 grams. Record the values of l1 in each case.
8. Take a second set of readings while the pan is gradually unloaded in the same regular steps. Enter the readings l2 in the tabular column and calculate the mean extension for each load.
Reading of pointer with no load=l0=………………cm
Force F (gm wt) | Balance readings(cm) | Mean l (cm) | Extension of the spring(cm) l- l_{0} | Extension per unit force(cmpergmwt) | |||
Loading (l_{ 1}) | Unloading (l_{ 2}) | ||||||
10
20 30 — — — 100 |
9. The results show that for each reading taken the value of would have nearly the same value.
10. The average value of can be taken as constant. Thus extension is proportional to the force applied. This verifies Hooke’s law.
11. Now plot a graph between extension of the spring (l-l0) against the force F. This gives a straight line passing through the origin. This graph is called Calibration graph.
12. Now place a body on the pan whose weight is to be known. Note the reading of the pointer ‘l1’ and find the extension l1-l0 produced by the weight. Locate the point on the y-axis and read the corresponding value on the x-axis which gives the weight of the body.
Graph:-
15. Two bodies of different masses and sizes when dropped simultaneously from the same height would reach the ground at the same time. What is the reason.
Acceleration due to gravity acts equally on both the bodies and it does not depend upon the mass of the body. So, the two bodies of different sizes reach the ground simultaneously.
16. The value of acceleration due to gravity (g) is maximum at the poles and minimum at the equator. Give reason
Earth is unevenly spherical that is it is bulged at the equator and flattened at the poles.
Due to the flattening of the earth at the poles ‘g’ value is maximum.
Due to the bulging of the earth at the equator, the radius of the earth is maximum
and so ‘g’ value is minimum.
17. Give the value of Universal gravitational constant G
G=6.67×10-11 Nm2/kg2.
18. What is the value of ‘g on the surface of the earth
g= 9.8 m/s2.
19. What is the value of ‘g ‘ on the moon
gM=1.67 m/s2
20. What is the value of ‘g’ on the sun?
gS=27.4 m/s2
21. Write an expression for ‘G’
22. When is the gravitational constant equal to the force of attraction between two unit masses
When the two bodies are separated by ‘Unit distance’.
23. Which law helps us to determine the weight of a body
Hooke’s law.
24. How do you calculate ‘g’ value at a height h
g value at a height can be calculated by using the formula
where,
g=9.8m/s2 and r=Radius of the earth
25. How do you calculate ‘g’ value at a given depth ‘d’
Where,
g=9.8 m/s2
r=Radius of the earth.
Note:-
1. At a height equal to half the radius of the earth ‘g’ value becomes zero
2. At a depth equal to the radius of the earth g value becomes zero
At the center of the earth ‘g’ value is zero.
26. Write the relation between ‘g’ and ‘G’
where,
M= mass of the earth and
R= radius of the earth
27. Who proposed Geocentric theory
Ptolemy.
28. Who proposed Heliocentric theory
Copernicus.
II Match the following:-
A B
2. (c) (b) Force between any objects in the
Universe
(c) Acceleration due to gravity.
III Problems:-
1. Find the gravitational force between sun and Jupiter. Given mass of the sun is equal to 2×1030 kg and mass of Jupiter is 1.89×1027 kg and the radius of Jupiter’s orbit is 7.73x1011m.
Given,
F=?
m1=2×1030 kg
m2=1.89×1027 kg
r=7.73×1011 m
G=6.67×10-11 Nm2/kg2.
We know that,
= 4.22×1023 N
2. Calculate the gravitational force of attraction between a woman weighing 50 kg and her child of 12 kg separated by a distance of 2m
Given,
F=?
m1=50 kg
m2=12kg
r=2m
G=6.67×10-11 Nm2/kg2.
We know that,
F=6.67×10-11×150
F=1.5×10-11 Z
3. Calculate the gravitational force of stone of mass 10kg
Given,
m=10kg
F=?
We know that,
F=mg
F=10×9.8 m/sec2
F=98N
Force=98 Newtons
4. A mango of mass 0.3 kg falls from a tree on to the earth and also acceleration of the earth towards the mango
Given,
M=6x1024kg
m=0.3 kg
g=?
G=6.67×10-11 Nm2/kg2.
r=6.4×106 m.
Ist case:-
Acceleration of the mango towards the earth:-
We know that,
Taking log on both sides
Log g =log 6.67+log 60-2log6.4
Log g=0.8241+1.7782-1.16124
Log g=0.9899
g=Antilog (0.9899)
g=9.770m/s2.
g=9.8 m/sec2
IInd case:-
Acceleration of the earth towards the mango:-
Taking log on both sides
Log g =log 6.67+log 0.3-23log10-2log6.4
Log g=0.8241+log 3-log10-23×1-2×0.8062
Log g=0.8241+1.7782-1-23-1.6124
Log g=24.3112
Or
Taking log on both sides
Log g=log 2.0001-log 40.96
Log g=0.3012-1.612
Log g=
g=0.04885×10-23m/sec2
Thus,
Mango produces an extremely small acceleration a=4.9×10-25 m/sec2 in the earth because of this acceleration we cannot observe the earth moving towards the mango.
5. Acceleration due to gravity on the earth surface is 9.8 m/sec2. What is the value of acceleration due to gravity on a planet whose mass is twice that of the earth and radius is twice that of the earth.
Acceleration due to gravity on earth
9.8 m/s2.
Acceleration due to gravity on Planet.
But,
Given,
Mp=2M
Rp=2R
gp=4.9 m/sec2
6. The force of attraction between two spherical bodies of masses 45 kg and 300kg separated by a distance of 50 cm is 3.59×10-6 N . Find the Universal Gravitational constant
Given,
m1=45 kg
m2=300 kg
F=3.59×10-6 N
r=50cm =50×10-2m.
G=?
We know that,
Taking log on both sides we get
Log G=log 17.95-log 27
Log G=1.2541-1.4314
Log G=
G=0.6648×10-10
G=6.648×10-11 Nm2/kg2
7. Calculate the gravitational force of attraction of two cars 5m apart if the masses of two cars are 1000 kg and 1200 kg respectively.
Given,
m1=1000 kg
m2=1200 kg
F=?
r=5m
G=6.67×10-11 Nm2/kg2
We know that,
F=320160×10-11N
IV Fill in the blanks:-
1. Which of the following theory is no longer valid (b)
a)Copernican theory b)Ptolemaic theory
c)Kepler’s laws d)none
2. The law of gravitation (d)
a)applies only to large bodies such as planets and stars.
b) accounts for all the known forces
c)holds good only in solar system.
d) holds goods every where in the universe.
3. The weight of an object (b)
a)is the quantity of matter in it.
b) force of attraction towards the earth.
c) is basically the same quantity as mass.
d) none.
4. The gravitational force with which the earth attracts the moon is (c)
a)less than the force with which the moon attracts the earth.
b)same as the moon attracts the earth.
c) more than the force with which the moon attracts the earth.
d) varies with the phase of the moon.
5. The weight of 400 gm stone is (c)
a)0.41 N b)0.04N
c)3.9N d)3920N
6. According to Ptolemaic theory earth is stationary.
7. Ptolemaic theory is also called as Geocentric theory.
8. According to heliocentric theory (c,d)
a)earth revolves round the moon
b) earth is stationary
c)earth revolves round the sun.
d)sun is stationary.
9. Heliocentric theory is otherwise known as Copernican theory.
10. The great astronomer who made observations of celestial bodies without the use of telescope is Tyco Brache.
11. The shape of the orbit in which every planet moves is elliptical.
12. The number of foci of an ellipse is two.
13. The time taken by the moon to make one revolution around the earth is 27.3 days.
14. The distance of the moon from the earth 3.85×105 km.
15. The value of acceleration of the moon directed towards the earth 0.0027m/s2.
16. Every object in the universe attracts every other object with a force. This force is called gravitational force.
17. Gravitational force is directly proportional to product of the masses of two bodies.
18. Gravitational force of attraction is inversely proportional to square of the distance between the two bodies.
19. Gravitational force of attraction is inversely proportional to the square of the distance between the two bodies. This law is called (c)
a)law of definite proportion. b) law of multiple proportion.
c)inverse square law d)none.
20. In G is called universal gravitational constant.
21. The expression for finding .
22. The numerical value of G is 6.67×10-11 Nm2/kg2.
23. The relation between g and G is .
24. The relation between gravitational constant G and force of attraction between two unit masses when they are unit distance apart is F=G.
25. The uniform acceleration produced in a freely falling body due to gravitational pull of earth is called acceleration due to gravity.
26. If ‘F’ is a force acting on a body of mass ‘m’ and acceleration due to gravity ‘g’ then F=mg.
27. The value of g on earth is surface is 9.8m/s2.
28. The value of g on the moon is 1.67m/s2.
29. The value of g on the sun is 27.4m/s2.
30. The value of g on the earth is maximum of the poles.
31. The value of g on the earth is minimum at equator.
32. The value of ‘g’ is maximum at the poles because the radius of the earth at poles is minimum.
33. The value of ‘g’ is minimum at the equator because the radius of the earth at the equator is maximum.
34. According to Newtons IInd law of motion F=ma.
35. The instruments used for measuring the variation in ‘g’ value at a given place is gravity meters.
36. The effective mass of the earth is given by .
37. ‘g’ value at the centre of the earth is zero.
38. ‘g’ value at the height equal to half the radius of the earth is zero.
39. The quantity of matter contained in a body is called its mass.
40. The force with which the body is attracted towards the centre of the earth is called weight.
41. S.I. unit of weight is kgwt.
42. C.G.S.unit of weight is gm.wt.
43. 1kgwt=9.8 Newtons.
44. When a spring is fixed at one end and force is applied at the other end, the stretching of the spring is proportional to the force applied. This law is known as Hooke’s law.
45. The law that helps to determine the weight of a body is known Hooke’s law.
46. Kepler modified Copernican theory by showing that the planetary orbits are elliptical.
47. The S.I. unit of acceleration due to gravity is m/s2.
48. Acceleration due to gravity is independent of mass of the body.
49. Acceleration due to gravity increases with decrease in the radius of the earth and decrease with increase in the radius of the earth.
50. ‘g value at the poles is maximum and at the equator is minimum.
51. Mass of a body is constant any where in the universe.
52. The astronomer who gave the summary of the results of Greek Astronomy is Ptolemy.
53. Kepler’s laws of planetary motion supports heliocentric theory.
54. Universal law of gravitation states that every object in the universe attracts every other object.
55. Two objects of different masses when dropped from the same height would reach the ground at the same time.
56. Mass of the earth is 6×1024 kg.
57. Radius of the earth is 6.4×106 m.
58. The value of ‘g’ on the earth varies due to the fact that earth is not a perfect sphere.
59. Due to flattening of the earth at the poles, radius of the earth is minimum.
60. Radius of the earth is not the same at all places on its surface.
61. Due to bulging of the earth at the equator, the radius of the earth is maximum.
62. ‘g’ value at a given height ‘h’ is given by .
63. The value of ‘g’ decreases with the depth .
64. ‘g’ value at a given depth is given by .
65. The value of ‘g’ is slightly affected by geological deposits, massive concrete building and topography.
66. gulf gravity meter and boliden gravity meter are two examples of gravity meters.
67. The mass of a body is inherent and fundamental property of a body.
68. Mass of a body is same at the equator as well as at the poles.
69. Weight of a body is equal to the product of its mass and acceleration due to gravity.
70. The weight of a body on the moon is the weight of the body on the earth.
71. Acceleration due to gravity on moon is of acceleration due to gravity on the earth.
72. The weight of a body decreases as we go upto greater height because the value of ‘g’ decreases with height.
73. The Principle underlying in the working of a spring balance was first investigated by Robert Hooke.
74. The instrument used to measure the weight of a body is called spring balance.
75. The extension produced in the spring is directly proportional to the force applied.
76. The shape of the graph drawn between the extension of an elastic body and the force applied is a straight line passing through the origin.
77. The graph drawn by plotting the force applied on an elastic body and the stretching is called calibration graph.
78. The Calibration graph can be used for measuring the weight of a body.
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