Very Short Answer Type Questions (VSAQ) (1) You know that the earth attracts you in the vertically downward direction. Do you attract the ea...
Very Short Answer Type Questions (VSAQ)
(1) You know that the earth attracts you in the vertically downward direction. Do you attract the earth as well? If yes, in which direction?
(2) Write the units of 'G' and 'g'.
(3) Consider two bodies A and B. The body B is heavier than A. Which of them is attract with a greater force by the earth? Which will fall with a greater acceleration?
(4) A coin and a feather are dropped from the roof of a building. Which will fall to the ground first.
(5) What do you mean by the weight of a body on the moon?
(6) The weight of a body on the moon is about one-sixth of its weight on the earth. Is the mass of the body on the moon also one-sixth of its mass on the moon?
(7) How does gravity differ from gravitation?
Ans: The force of attraction between any two objects by virtue of their masses is called gravitational force, whereas the force of gravitation exerted by a huge heavenly body such as earth on a smaller object near its surface is called its gravity or the force of gravity.
(8) Why is the weight of an object on the moon \(\frac{1}{6}th\) of that of the earth.
Ans: This is because the gravity of the moon is \(\frac{1}{6}th\) of that of the earth.
(9) The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?
Ans: The earth attracts the moon with a force equal to the force with which the moon attracts the earth. This is because the action and reaction forces are equal.
(10) Newton's law of gravitation is also called inverse square law. Why is it so called?
Ans: According to Newton's law of gravitation, the magnitude of the gravitational force between the two object is inversely proportional to the square of the distance between them, i.e. \(F \propto \frac{1}{{{r^2}}}\). That is why this law is also called inverse square law.
(11) A boy drops an iron ball and a feather from the top of a tower. He finds that the iron ball reaches the ground much earlier than the feather. Now, he puts the iron ball and the feather in a long jar from which air is completely evacuated. What will be observed when he inverts the jar upside down?
Ans: In the evacuated jar, both iron ball and the feather will fall together at the same time.
Ans: In the evacuated jar, both iron ball and the feather will fall together at the same time.
(12) Suppose gravity of earth suddenly becomes zero, then in which direction will the moon begin to move if no other celestial body affects it?
Ans: The moon will move in the direction of the tangent to the moon's orbit
(13) What do you mean by free fall?
Ans: Whenever an object falls towards the earth under the gravitational pull of the earth alone, the fall is called free-fall.
(14) Give the S.I. unit of G and its value
Ans: S.I. Unit of G is: \(N.{m^2}K{g^{ - 2}}\)
Its value is \(6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}\)
Ans: S.I. Unit of G is: \(N.{m^2}K{g^{ - 2}}\)
Its value is \(6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}\)
(15) What is the centripetal force?
Ans: The force required for a body to move in a circular path is called centripetal force.
Ans: The force required for a body to move in a circular path is called centripetal force.
(16) Define universal gravitational constant.
Ans: The universal gravitational constant is the force of attraction between two point masses of 1 Kg each kept at a distance of 1 metre.
Ans: The universal gravitational constant is the force of attraction between two point masses of 1 Kg each kept at a distance of 1 metre.
(17) State Kepler's third law of planetary motion.
Ans: The cube of the mean distance of a planet from the sun is directly proportional to the square of time it takes to move around the sun.
Ans: The cube of the mean distance of a planet from the sun is directly proportional to the square of time it takes to move around the sun.
(18) State the S.I. unit of acceleration due to gravity.
Ans: The S.I. unit of acceleration due to gravity is \(m.{\sec ^{ - 2}}\)
Ans: The S.I. unit of acceleration due to gravity is \(m.{\sec ^{ - 2}}\)
(19) The earth's gravitational force causes an acceleration of \(5m.{\sec ^{ - 2}}\) in a 1 kg mass somewhere in space. How much will the acceleration of a 3 Kg mass be at the same place?
Ans: The acceleration produced by the gravitational force of earth does not depend on the mass of the object. So, the acceleration produce in the 3 Kg mass will be the same as that produced in the 1 Kg mass. That is, the acceleration produced will be \(5m.{\sec ^{ - 2}}\)
Ans: The acceleration produced by the gravitational force of earth does not depend on the mass of the object. So, the acceleration produce in the 3 Kg mass will be the same as that produced in the 1 Kg mass. That is, the acceleration produced will be \(5m.{\sec ^{ - 2}}\)
(20) What is the relation between mass and weight of a body?
Ans: The relation between mass and weight of a body is
\(W = mg\), Where \(W\) is the weight of that body and m is the mass of the body and V is the acceleration due to gravity.
Ans: The relation between mass and weight of a body is
\(W = mg\), Where \(W\) is the weight of that body and m is the mass of the body and V is the acceleration due to gravity.
(21) The weight of a body is 50 N. What is its mass? (\(g = 9.8m.{\sec ^{ - 2}}\))
Ans: Here \(W = 50N\)
\(g = 9.8m.{\sec ^{ - 2}}\)
we have, \(mass(m) = \frac{{weight(W)}}{{gravity}}\)
or, \(W = \frac{{50N}}{{9.8m.{{\sec }^{ - 2}}}}\)
or, mass = 5.10 kg
(22) A stone and a feather are thrown from a tower. Both the objects should reach the ground at same time, but it does not, give reason.
Ans: As per the motion of object due to 'g' gravitational pull of earth, both the bodies are acted upon by the same force of earth but the stone will fall first and then the feather. Because feather experience the air resistance, being lighter, so it will reach later.
(23) How much would a 70 Kg man weight on the moon? What would be his mass on the earth and on the moon? (Acceleration due to gravity on moon \(1.63m.{\sec ^{ - 2}}\)
Ans: We will first calculate the weight of the man on the moon. Here, mass of the moon on man, \(m = 70Kg\)
Acceleration due to gravity on the moon \(1.63m.{\sec ^{ - 2}}\)
We know \(W = mg\)
or, W=70 Kg \( \times \) \(1.63m.{\sec ^{ - 2}}\)
or, W= 114.1 N
Thus, the man would weigh 114.1 N on the moon. Please note that the mass of a body is constant everywhere in the universe, So, the mass of this man would be the same on earth as well as on the moon, that is, the mass will be 70 Kg on the earth as well as moon also
(24) Calculate the force of gravitation between the earth and the sun, given that the mass of the earth \(6 \times {10^{24}}Kg\) and of the sun \(2 \times {10^{30}}Kg\). The average distance between the two is \(1.5 \times {10^{11}}m\).
Ans: Mass of the earth = \({m_e} = 6 \times {10^{24}}Kg\)
Mass of the Sun (\({m_s}\))= \(2 \times {10^{30}}Kg\)
Average distance between earth & mass is (R) = \(3.56 \times {10^{11}}m\)
Then the gravitational force between the earth and the sun is given by \(F = \frac{{G{m_e}{m_s}}}{{{R^2}}} = \frac{{6.67 \times {{10}^{ - 11}} \times 6 \times {{10}^{24}} \times 2 \times {{10}^{30}}}}{{{{\left( {1.5 \times {{10}^{11}}} \right)}^2}}} = 3.56 \times {10^{22}}N\)
(25) What is the centripetal force? Explain with an example.
Ans: When a body moves around the other body, the force that causes this acceleration and keeps the body moving along a circular path is acting towards the center. This force is called centripetal force. It means centre seeking force.
Example: Tie a stone to a thread and whirl it round as shown in the figure. The stone continuous to move round as long as the string is held. On releasing the thread the the center binding force called as centripetal force is no more acting on it, the stone flies off along the straight line. This straight line will be a tangent to the circular path.
(26) A boy drops a ball from the top of a tower of height 19.6 m. Calculate (i) the velocity of the ball just before it touch the ground.
Ans: Here, \(u = 0\), \(S = - 19.6m\), g= \( - 10m.{\sec ^{ - 2}}\)
Using \({v^2} - {u^2} = 2gS\)
We get, \(v = \pm 19.6m/s\)
Therefore, \(v = 19.6m/s\) [negative sign is retained because of sign convention]
(27) Give the difference between mass and weight.
Ans: Here \(W = 50N\)
\(g = 9.8m.{\sec ^{ - 2}}\)
we have, \(mass(m) = \frac{{weight(W)}}{{gravity}}\)
or, \(W = \frac{{50N}}{{9.8m.{{\sec }^{ - 2}}}}\)
or, mass = 5.10 kg
Short Answer Type Question (SAQ):
(22) A stone and a feather are thrown from a tower. Both the objects should reach the ground at same time, but it does not, give reason.
Ans: As per the motion of object due to 'g' gravitational pull of earth, both the bodies are acted upon by the same force of earth but the stone will fall first and then the feather. Because feather experience the air resistance, being lighter, so it will reach later.
(23) How much would a 70 Kg man weight on the moon? What would be his mass on the earth and on the moon? (Acceleration due to gravity on moon \(1.63m.{\sec ^{ - 2}}\)
Ans: We will first calculate the weight of the man on the moon. Here, mass of the moon on man, \(m = 70Kg\)
Acceleration due to gravity on the moon \(1.63m.{\sec ^{ - 2}}\)
We know \(W = mg\)
or, W=70 Kg \( \times \) \(1.63m.{\sec ^{ - 2}}\)
or, W= 114.1 N
Thus, the man would weigh 114.1 N on the moon. Please note that the mass of a body is constant everywhere in the universe, So, the mass of this man would be the same on earth as well as on the moon, that is, the mass will be 70 Kg on the earth as well as moon also
(24) Calculate the force of gravitation between the earth and the sun, given that the mass of the earth \(6 \times {10^{24}}Kg\) and of the sun \(2 \times {10^{30}}Kg\). The average distance between the two is \(1.5 \times {10^{11}}m\).
Ans: Mass of the earth = \({m_e} = 6 \times {10^{24}}Kg\)
Mass of the Sun (\({m_s}\))= \(2 \times {10^{30}}Kg\)
Average distance between earth & mass is (R) = \(3.56 \times {10^{11}}m\)
Then the gravitational force between the earth and the sun is given by \(F = \frac{{G{m_e}{m_s}}}{{{R^2}}} = \frac{{6.67 \times {{10}^{ - 11}} \times 6 \times {{10}^{24}} \times 2 \times {{10}^{30}}}}{{{{\left( {1.5 \times {{10}^{11}}} \right)}^2}}} = 3.56 \times {10^{22}}N\)
(25) What is the centripetal force? Explain with an example.
Ans: When a body moves around the other body, the force that causes this acceleration and keeps the body moving along a circular path is acting towards the center. This force is called centripetal force. It means centre seeking force.
Example: Tie a stone to a thread and whirl it round as shown in the figure. The stone continuous to move round as long as the string is held. On releasing the thread the the center binding force called as centripetal force is no more acting on it, the stone flies off along the straight line. This straight line will be a tangent to the circular path.
(26) A boy drops a ball from the top of a tower of height 19.6 m. Calculate (i) the velocity of the ball just before it touch the ground.
Ans: Here, \(u = 0\), \(S = - 19.6m\), g= \( - 10m.{\sec ^{ - 2}}\)
Using \({v^2} - {u^2} = 2gS\)
We get, \(v = \pm 19.6m/s\)
Therefore, \(v = 19.6m/s\) [negative sign is retained because of sign convention]
(27) Give the difference between mass and weight.
| Mass | Weight |
|---|---|
| (1) The quantity of matter contained in a body is called the mass of the body | (1) The force with the earth attracts a body towards its centre is called the weight of the body |
| (2) Mass of a body remains constant everywhere | (2) Weight of a change from place to place. |
| (3) Mass of a body is never zero | (3) Weight of a body at the centre of the earth |
| (4) Mass is a scalar quantity | (4) Weight is a vector quantity |
| (5) Mass is measured in Kg | (5) Weight is measure in Kg-wt. or Newton. |
(28) What is the importance of universal laws of gravitation?
(29) A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (i) the maximum height to which it rises. (ii) the total time it takes to return to the surface of the earth?
(30) Give the difference between 'g' and 'G'?
Ans:
| G (Acceleration due to gravity) | G (Universal Gravitational Constant) |
|---|---|
| (i) It is defined as acceleration due to gravity | (i) It is defined as universal gravitational constant |
| (ii) Its value of g is 9.8m.sec−2 | (ii) Its value is 6.67×10−11Nm2kg−2 |
| (iii) It change from place to place at pole of ‘g’ is more than at equator | (ii) It remains constant everywhere |
(31) What is the force of gravity acting on a point mass of 1 kg released from a height of 1 m from the surface of earth? The radius of the earth \(6.4 \times {10^6}m\) and mass of the earth \(6 \times {10^{24}}Kg\)
(32) A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Ans: Using \({v^2} - {u^2} = 2as\)
or, \({\left( v \right)^2} - 0 = 2 \times 9.8 \times 19.6\)
or, \({v^2} = {\left( {19.6} \right)^2}\)
or, \(v = 19.6m/s\)
(33) There are two kind of balance i.e. a beam balance and a spring balance. If both the balance give the same measure of a given body on the surface of the earth, will they give the same measure on the suface of the moon? Explain
Ans: Beam balance measure the mass of a body. Since mass of a body remains constant so the beam balance will give the same measure on the surface of the earth and on the surface of the moon,
On the other hand, spring balance measure the weight of a body. Weight of the body \(W = mg\). It means, weight of a body depends upon the value of 'g'. Since value of 'g' on the moon \( = \frac{1}{6}th\) times the value of 'g' on the earth, so the spring balance shows \(\frac{1}{6}th\) times the weight of the body on the earth at the surface of the moon.
(34) The gravitational force between two objects is F. How will this force change when distance between them is reduced to half?
(35) How does the gravitational force between two bodies changes if the distance between them is tripled?
(36) Two object of masses \({m_1}\) and \({m_2}\) having the same size are dropped simultaneously from height \({h_1}\) and \({h_2}\) respectively. Find out the ratio of time they would take in reaching the ground.
Ans: Using the formula \(S = ut + \frac{1}{2}a{t^2}\), here we can write
\({h_1} = 0 + \frac{1}{2}gt_1^2\)
or, \({t_1} = \sqrt {\frac{{2{h_1}}}{g}} \)
\({h_2} = 0 + \frac{1}{2}gt_2^2\)
or, \({t_2} = \sqrt {\frac{{2{h_2}}}{g}} \)
or, \(\frac{{{t_1}}}{{{t_2}}} = \sqrt {\frac{{{h_1}}}{{{h_2}}}} \)
Long Answer Type Question (LAQ):
(37) An object weighs 294 N on the earth. () What would be its mass on the moon? () What is the acceleration due to gravity on the moon?
(38) Calculate the acceleration due to gravity on the surface of the moon (mass of the moon \(7.4 \times {10^{22}}Kg\) and radius of the moon is \(1.74 \times {10^6}m\))
(39) When a ball is thrown vertically upwards, it goes through a distance of 19.6 m. Find the initial velocity of the ball and the time taken by it to rise to the highest point. (Acceleration due to gravity \(9.8m.{\sec ^{ - 2}}\))
(40) The acceleration of a freely falling body does not depend on the mass of the body. Show this by deriving an expression for the same.
(41) The weight of a body on the surface of earth is 392 N. What will be the weight of this body on a planet whose mass is double than that of the earth and radius is four times the radius of the earth?
(42) A ball thrown up is caught back by the thrower after 6 sec. Calculate (i) the velocity with which the ball was thrown up (ii) the maximum height attained by the ball and (iii) the distance of the ball below the highest point after 2 sec. Take \(g = 10m.{\sec ^{ - 2}}\)
(43) Derive the unit of force using the second law of motion. A force of 5 N produced an acceleration of \(8m.{\sec ^{ - 2}}\) on a mass \({m_1}\) and an acceleration of \(24m.{\sec ^{ - 2}}\) on a mass \({m_2}\). What acceleration would the same force provide if both the masses are tied together?
NCERT SECTION:
(44) State the universal law of gravitation.
(45) Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.
(46) What do you mean by free-fall?
(47) What do you mean by acceleration due to gravity?
(48) What are the difference between the mass of an object and its weight?
(49) Why is the weight of an object on the moon \(\frac{1}{6}th\) its weight on the earth?
(50) How does the force of gravitation between two objects change when the distance between them is reduced to half?
(51) Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?
(52) What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (mass of the earth \(6 \times {10^{24}}Kg\) and radius of the earth \(6.4 \times {10^6}m\) )
(53) The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?
(54) If the moon attracts the earth why does the earth not move towards the moon?
(55) What happens to the force between two objects, if (i) the mass of one object is doubled? (ii) the distance between the objects is doubled and tripled? (iii) the masses of both objects are doubled?
(56) What is the importance of universal law of gravitation?
(57) What is the acceleration of free fall?
(58) What do we call the gravitational force between the earth and an object?
(59) Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, Why?
(60) Gravitational force on the surface of the moon is only \(\frac{1}{6}th\) as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the earth?
(61) A ball is thrown vertically upward with a velocity of 49 m/s. Calculate (i) the maximum height to which it rises (ii) the total time it takes to return to the surface of the earth?
(62) A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
(63) A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking \(g = 10m.{\sec ^{ - 2}}\), finds the maximum height reached by the stone. What is the total distance covered by the stone?
(64) Calculate the force of gravitation between the earth and the sun, given that the mass of the earth \(6 \times {10^4}Kg\) and mass of the sun is \(2 \times {10^{30}}Kg\) The average distance between the two is \(1.5 \times {10^{11}}m\)
(65) A stone is allowed to fall from the top of a tower of 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stone will meet?
(66) A ball thrown up vertically returns to the thrower after 6 sec. Find (i) the velocity with which it was thrown up. (ii) the maximum height it reaches and (iii) its position after 4 sec.
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