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GRAVITATION:--
EXERCISES- Q21_to_Q30
21. The gravitational field in a region is given by E = (2i+3j) N/kg. Show that no work is done by the gravitational field when a particle is moved on the line 3y+2x=5.
ANSWER: Let the angle between the x-axis an E be α. Then,
cosα = E.i/|E|.|i|
Now,
E.i = (2i+3j).i =2
|E| = √(2²+3²) =√13 and |i| = 1
∴ cosα = 2/√13
The equation of line 3y+2x = 5 can be written as,
y = -2x/3+5/3
It is in the form of y=mx+c where m = tanß
(ß is the angle between the line and the x-axis)
Here tanß = -2/3
Since cosα=2/√13
→secα=√13/2
→tan²α=sec²α-1 =13/4 - 1 =9/4
→tanα = 3/2
∴tan(ß-α)=tanß-tanα/(1+tanß.tanα)
={(-3/2) - (2/3)}/{1+(3/2)(-2/3)}
→tan(ß-α)={3/2+2/3}/0 = ∞
→ß-α = 90°
So the angle between the direction of the field and the given line is 90°. The component of the field along this line will be zero hence no work will be done by the field if a particle is moved along this line.
22. Find the height above the earth's surface at which the weight of the body becomes half of its value at the surface.
ANSWER: The weight of the body at the surface = GMm/R²
Let the required height above the earth's surface = h.
The weight at this height = GMm/(R+h)²
By the condition, GMm/(R+h)² = ½*GMm/R²
→(R+h)² = 2R²
→R+h = √2R
→ h = (√2-1)R
Hence the required height is (√2-1) times the radius of the earth.
23. What is the acceleration due to gravity on the top of Mount Everest? Mount Everest is the highest mountain peak of the world at the height of 8848 m. The value at sea level is 9.80 m/s².
ANSWER: The height of the Mount Everest above the sea level
h = 8.848 km
The acceleration due to gravity at this place
g' = g{1-2h/R}
=g{1-2*8.848/6400} m/s²
=9.80*{0.997} m/s²
=9.77 m/s²
24. Find the acceleration due to gravity in a mine of depth 640 m if the value at the surface is 9.800 m/s². The radius of the earth is 6400 km.
ANSWER: The depth of the mine h = 640 m = 0.640 km
At surface g = 9.800 m/s²
The acceleration due to gravity in a mine is given by
g' = g{1-h/R} Where R =6400 km
→g' = 9.800*{1-0.640/6400} m/s²
=9.800*0.999 m/s²
=9.799 m/s²
25. A body is weighed by a spring balance to be 1.000 kg at the north pole. How much will it weigh at the equator? Account for the earth's rotation only.
ANSWER: The angular speed of the earth
⍵ = 2π/(24*3600) rad/s
Assume R =6400 km = 6400x1000 m = 6400000 m
The net acceleration due to gravity at the equator due to earth's rotation g' = g-⍵²R
=9.800 - ⍵²R =9.80 -4π²*6400000/(24*24*3600*3600)
=9.800 - 4*9.869*640/(24*24*36*36)
=9.800 - 9.869*10/(3*3*9*36)
=9.800 - 98.69/2916
=9.800 - 0.033 m/s²
=9.767 m/s²
So the weight at equator = mg' =1.000*9.767 = 9.767 N =9.767/9.80 kg =0.997 kg
26. A body stretches a spring by a particular length at the earth's surface at the equator. At what height above the south pole will it stretch the same spring by the same length? Assume the earth to be spherical.
ANSWER: The required height will have the acceleration due to gravity same as on the equator so that the weight mg' of the body is the same and it stretches the spring by the same length.
Let the acceleration due to gravity at the South pole = g
At equator g' = g-⍵²R
and at the required height h above the South pole
g' = g[1-2h/R]
Equating the two
g-⍵²R = g[1-2h/R] =g-2gh/R
→2gh/R = ⍵²R
→h = ⍵²R²/2g
=[4π²/24*3600*24*3600]*[64x10⁵*64x10⁵]/{2*9.80}
=4*9.869*64*64x10⁶/{2*9.8*24*36*24*36}
=2*1.007*64*64*10⁶/24*36*24*36 m
=2.014*64*10⁶/9*36*36 m
= 0.011*10⁶ m
=0.011*1000 km
=11 km
≈10 km approx
27. At what rate the earth should rotate so that the apparent g at the equator becomes zero? What will be the length of the day in this situation?
ANSWER: The apparent g at the equator due to the earth's rotation = g' =g-⍵²R, for g' to be zero
⍵²R = g
→⍵ =√(g/R)
=√{9.80/64x10⁵} rad/s
=√{1.531x10⁻⁶} rad/s
=1.237x10⁻³ rad/s
Let the length of the day for this rotation = T hours.
Angular speed for this time = 2π/(T*3600) rad/s
Equating the two angular speeds
2π/(T*3600) = 1.237x10⁻³
→ T = 2π/(3600*1.237x10⁻³) hours
= 2*3.14/3.6*1.237
=1.41 hours
28. A pendulum having a bob of mass m is hanging in a ship sailing along the equator from east to west. When the ship is stationary with respect to water the tension in the string is T₀. (a) Find the speed of the ship due to rotation of the earth about its axis. (b) Find the difference between T₀ and the earth's attraction on the bob. (c) If the ship sails at speed v, what is the tension in the string? The angular speed of earth's rotation is ⍵ and radius of the earth is R.
ANSWER: (a) When the ship is stationary with respect to water its angular speed about the center of the earth will be the same as the earth. So its speed due to rotation of the earth =⍵R
(b) The earth's attraction on the bob = mg
The tension in the string T₀ = mg' {where g' is the apparent acceleration due to gravity at the equator and g' = g-⍵²R}
→T₀ = mg-m⍵²R
→mg-T₀ = m⍵²R
So the difference between T₀ and the earth's attraction on the bob =m⍵²R
(c) If the ship sails from east to west with a speed v, its angular speed about the center of the earth =v/R but it is opposite to ⍵. So net angular speed ⍵' = ⍵-v/R
Now the tension in the string = T = m(g-⍵'²R)
=m{g-(⍵-v/R)²R}
=m{g-(⍵²+v²/R²-2⍵v/R)R}
=m{g-⍵²R-v²/R+2⍵v}
=m{(g-⍵²R)+2⍵v-v²/R}
=m{g'+2⍵v-v²/R}
=mg'+2m⍵v-mv²/R {Since mv²/R is a very small quantity}
=T₀+2m⍵v (Approx.)
29. The time taken by Mars to revolve around the sun is 1.88 years. Find the ratio of the average distance between Mars and the sun to that between the earth and the sun.
ANSWER: Let the average distance between the sun and the earth =R and the sun and the Mars = R'.
The time period of earth's revolution = T = 1 year
The time period of Mars T' = 1.88 year
According to the third law of Kepler, the square of the time period of a planet is proportional to the cube of the radius for a circular orbit.
T² ∝ R³
Hence T'²/T² = R'³/R³
→(1.88/1.0)² = (R'/R)³
→R'/R = ∛1.88² =∛3.534 =1.52
30. The moon takes about 27.3 days to revolve around the earth in a nearly circular orbit of radius 3.84x10⁵ km. Calculate the mass of the earth from these data.
ANSWER: The time period of a satellite T is given by,
T² = 4π²a³/GM {where a is the radius of the orbit and M is the mass of the earth}
→M = 4π²a³/GT²
Given that a = 3.84x10⁵ km =3.84x10⁸ m
T =27.3x24x3600 s
Hence M = 4π²(3.84x10⁸)³/{6.67x10⁻¹¹*(27.3x24x3600)²}
=4*9.869*3.84*3.84*3.84x10²⁴⁺¹¹/6.67*5.564x10¹²
=2235.26x10³⁵⁻¹²/37.11
=60.23x10²³
=6.023x10²⁴ kg
ANSWER: Let the angle between the x-axis an E be α. Then,
cosα = E.i/|E|.|i|
Now,
E.i = (2i+3j).i =2
|E| = √(2²+3²) =√13 and |i| = 1
∴ cosα = 2/√13
The equation of line 3y+2x = 5 can be written as,
y = -2x/3+5/3
It is in the form of y=mx+c where m = tanß
(ß is the angle between the line and the x-axis)
Here tanß = -2/3
 |
| Diagram for Q-21 |
→secα=√13/2
→tan²α=sec²α-1 =13/4 - 1 =9/4
→tanα = 3/2
∴tan(ß-α)=tanß-tanα/(1+tanß.tanα)
={(-3/2) - (2/3)}/{1+(3/2)(-2/3)}
→tan(ß-α)={3/2+2/3}/0 = ∞
→ß-α = 90°
So the angle between the direction of the field and the given line is 90°. The component of the field along this line will be zero hence no work will be done by the field if a particle is moved along this line.
22. Find the height above the earth's surface at which the weight of the body becomes half of its value at the surface.
ANSWER: The weight of the body at the surface = GMm/R²
Let the required height above the earth's surface = h.
The weight at this height = GMm/(R+h)²
By the condition, GMm/(R+h)² = ½*GMm/R²
→(R+h)² = 2R²
→R+h = √2R
→ h = (√2-1)R
Hence the required height is (√2-1) times the radius of the earth.
23. What is the acceleration due to gravity on the top of Mount Everest? Mount Everest is the highest mountain peak of the world at the height of 8848 m. The value at sea level is 9.80 m/s².
ANSWER: The height of the Mount Everest above the sea level
h = 8.848 km
The acceleration due to gravity at this place
g' = g{1-2h/R}
=g{1-2*8.848/6400} m/s²
=9.80*{0.997} m/s²
=9.77 m/s²
24. Find the acceleration due to gravity in a mine of depth 640 m if the value at the surface is 9.800 m/s². The radius of the earth is 6400 km.
ANSWER: The depth of the mine h = 640 m = 0.640 km
At surface g = 9.800 m/s²
The acceleration due to gravity in a mine is given by
g' = g{1-h/R} Where R =6400 km
→g' = 9.800*{1-0.640/6400} m/s²
=9.800*0.999 m/s²
=9.799 m/s²
25. A body is weighed by a spring balance to be 1.000 kg at the north pole. How much will it weigh at the equator? Account for the earth's rotation only.
ANSWER: The angular speed of the earth
⍵ = 2π/(24*3600) rad/s
Assume R =6400 km = 6400x1000 m = 6400000 m
The net acceleration due to gravity at the equator due to earth's rotation g' = g-⍵²R
=9.800 - ⍵²R =9.80 -4π²*6400000/(24*24*3600*3600)
=9.800 - 4*9.869*640/(24*24*36*36)
=9.800 - 9.869*10/(3*3*9*36)
=9.800 - 98.69/2916
=9.800 - 0.033 m/s²
=9.767 m/s²
So the weight at equator = mg' =1.000*9.767 = 9.767 N =9.767/9.80 kg =0.997 kg
26. A body stretches a spring by a particular length at the earth's surface at the equator. At what height above the south pole will it stretch the same spring by the same length? Assume the earth to be spherical.
ANSWER: The required height will have the acceleration due to gravity same as on the equator so that the weight mg' of the body is the same and it stretches the spring by the same length.
Let the acceleration due to gravity at the South pole = g
At equator g' = g-⍵²R
and at the required height h above the South pole
g' = g[1-2h/R]
Equating the two
g-⍵²R = g[1-2h/R] =g-2gh/R
→2gh/R = ⍵²R
→h = ⍵²R²/2g
=[4π²/24*3600*24*3600]*[64x10⁵*64x10⁵]/{2*9.80}
=4*9.869*64*64x10⁶/{2*9.8*24*36*24*36}
=2*1.007*64*64*10⁶/24*36*24*36 m
=2.014*64*10⁶/9*36*36 m
= 0.011*10⁶ m
=0.011*1000 km
=11 km
≈10 km approx
27. At what rate the earth should rotate so that the apparent g at the equator becomes zero? What will be the length of the day in this situation?
ANSWER: The apparent g at the equator due to the earth's rotation = g' =g-⍵²R, for g' to be zero
⍵²R = g
→⍵ =√(g/R)
=√{9.80/64x10⁵} rad/s
=√{1.531x10⁻⁶} rad/s
=1.237x10⁻³ rad/s
Let the length of the day for this rotation = T hours.
Angular speed for this time = 2π/(T*3600) rad/s
Equating the two angular speeds
2π/(T*3600) = 1.237x10⁻³
→ T = 2π/(3600*1.237x10⁻³) hours
= 2*3.14/3.6*1.237
=1.41 hours
28. A pendulum having a bob of mass m is hanging in a ship sailing along the equator from east to west. When the ship is stationary with respect to water the tension in the string is T₀. (a) Find the speed of the ship due to rotation of the earth about its axis. (b) Find the difference between T₀ and the earth's attraction on the bob. (c) If the ship sails at speed v, what is the tension in the string? The angular speed of earth's rotation is ⍵ and radius of the earth is R.
ANSWER: (a) When the ship is stationary with respect to water its angular speed about the center of the earth will be the same as the earth. So its speed due to rotation of the earth =⍵R
(b) The earth's attraction on the bob = mg
The tension in the string T₀ = mg' {where g' is the apparent acceleration due to gravity at the equator and g' = g-⍵²R}
→T₀ = mg-m⍵²R
→mg-T₀ = m⍵²R
So the difference between T₀ and the earth's attraction on the bob =m⍵²R
(c) If the ship sails from east to west with a speed v, its angular speed about the center of the earth =v/R but it is opposite to ⍵. So net angular speed ⍵' = ⍵-v/R
Now the tension in the string = T = m(g-⍵'²R)
=m{g-(⍵-v/R)²R}
=m{g-(⍵²+v²/R²-2⍵v/R)R}
=m{g-⍵²R-v²/R+2⍵v}
=m{(g-⍵²R)+2⍵v-v²/R}
=m{g'+2⍵v-v²/R}
=mg'+2m⍵v-mv²/R {Since mv²/R is a very small quantity}
=T₀+2m⍵v (Approx.)
29. The time taken by Mars to revolve around the sun is 1.88 years. Find the ratio of the average distance between Mars and the sun to that between the earth and the sun.
ANSWER: Let the average distance between the sun and the earth =R and the sun and the Mars = R'.
The time period of earth's revolution = T = 1 year
The time period of Mars T' = 1.88 year
According to the third law of Kepler, the square of the time period of a planet is proportional to the cube of the radius for a circular orbit.
T² ∝ R³
Hence T'²/T² = R'³/R³
→(1.88/1.0)² = (R'/R)³
→R'/R = ∛1.88² =∛3.534 =1.52
30. The moon takes about 27.3 days to revolve around the earth in a nearly circular orbit of radius 3.84x10⁵ km. Calculate the mass of the earth from these data.
ANSWER: The time period of a satellite T is given by,
T² = 4π²a³/GM {where a is the radius of the orbit and M is the mass of the earth}
→M = 4π²a³/GT²
Given that a = 3.84x10⁵ km =3.84x10⁸ m
T =27.3x24x3600 s
Hence M = 4π²(3.84x10⁸)³/{6.67x10⁻¹¹*(27.3x24x3600)²}
=4*9.869*3.84*3.84*3.84x10²⁴⁺¹¹/6.67*5.564x10¹²
=2235.26x10³⁵⁻¹²/37.11
=60.23x10²³
=6.023x10²⁴ kg
===<<<O>>>===
Links to the chapters -
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CHAPTER- 11 - Gravitation
Questions for Short Answers
OBJECTIVE - I
OBJECTIVE - II
EXERCISES - Q_1 to Q_10
EXERCISES -Q-11 TO Q-20
CHAPTER- 10 - Rotational Mechanics
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OBJECTIVE - I
OBJECTIVE - II
EXERCISES - Q-1 TO Q-15
EXERCISES - Q-16 TO Q-30
EXERCISES Q-31 TO Q-45
EXERCISES Q-46 TO Q-60
EXERCISES Q-61 TO Q-75
EXERCISES Q-76 TO Q-86
CHAPTER- 11 - Gravitation
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EXERCISES - Q_1 to Q_10
EXERCISES -Q-11 TO Q-20
CHAPTER- 10 - Rotational Mechanics
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OBJECTIVE - II
EXERCISES - Q-1 TO Q-15
EXERCISES - Q-16 TO Q-30
EXERCISES Q-31 TO Q-45
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CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
HC Verma's Concepts of Physics, Chapter-8, WORK AND ENERGY
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HC Verma's Concepts of Physics, Chapter-7, Circular Motion
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HC Verma's Concepts of Physics, Chapter-6, Friction
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HC Verma's Concepts of Physics, Chapter-4, The Forces
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