Prove that if the earth attracts two bodies placed at same distance from the centre of the earth with the same force, then their masses are equal. Answer. Answer: (f) Acceleration = 10 ms-2 The mass of the Earth M is 5.98 x 1024 kg and the radius of the Earth R is 6.38 x 106 m. (3 marks) Ask your doubt of gravitation and get answer from subject experts and students on TopperLearning. Calculate the average density of the earth in terms of g, G and R. Thus a big stone will fall with the same acceleration as a small stone. Q1. Numericals A comprehensive database of gravitation quizzes online, test your knowledge with gravitation quiz questions. Wm = \(\frac{G M_{m} m}{R_{m}^{2}}\) …(1) Both C and D; [(Kg-m/Sec 2) = N] Q.19 Weight of free fall object is. i.e., weight will increase by 10%. A student thought that two bricks tied together would fall faster than a single one under the action of gravity. Why does formation of tides takes place in sea or ocean? It also implies that a body orbiting in space has zero weight with respect to a spaceship. Answer: The entire NCERT textbook questions have been solved by best teachers for you. Answer: Give reasons. Downward motion The tides in the sea formed by the rising and falling of water level in the sea are due to the gravitational force of attraction which the sun and the moon exert on the water surface in the sea. That is, for the second stone, t = (3.16 – 1)s = 2.16s v2 = 2gs = 2 × 9.8 × 100 = 1960 Does the apple also attract the earth? Here, g = 1.67 ms-2, R = 1.74 × 106 m and G = 6.67 × 10-2 Nm2 kg-2, Question 2. Class 9 | Science | Chapter 10 |Gravitation| NCERT Solutions. (d) What is the ball’s acceleration at point C? Free Question Bank for NEET Physics Gravitation. Answer: Question 3. Therefore, the packets fall slowly at equator in comparison to the poles. Original weight, W0 = mg = mG \(\frac{M}{R^{2}}\) the time taken by the second stone to reach the ground is one second less than that taken by the first stone as both the stones reach the ground at the same time. [NCERT Exemplar] The value of ‘g’ at the equator of the earth is lesser than that at poles. (i) Original weight = \(\frac{G M m}{r^{2}}\), where M is the mass of the earth. The time taken for a body is less if the acceleration due to gravity is more when the initial velocities and the distance travelled are the same. Answer: The radius of the earth at the poles is 6357 km, the radius at the equator is 6378 km. Answer: This solution contains questions, answers, images, step by step explanations of the complete Chapter 10 titled Gravitation of Science taught in class 9. Answer: From a cliff of 49 m high, a man drops a stone. For both the stones New mass, M’=M + 10% of M Mass is the quantity of matter contained in the body. Suppose gravity of earth suddenly becomes zero, then which direction will the moon begin to move if no other celestial body affects it? (b) time of fall of the body? Enter OTP. Find out the ratio of time they would take in reaching the ground. and radius of the moon, Rm = 1740 km The radius of the earth is 6400 km and g = 10 m/s². What does a small value of G indicate? The mass of the Sun is 2 x 10 30 kg and that of the Earth is 6 x 10 24 kg. Very Short Answer Type Questions. Find the percentage change in the weight of a body when it is taken from the equator to the poles. Is weight a force? ∴ v = 0 + (-10) × 2 = -20 ms-1 Answer: (e) Acceleration = -10 ms-2 We know from Newton’s second law of motion that the force is the product of mass and acceleration. Page No: 134. Question 4. So, v2 = u2 + 2gs, Question 16. Answer: Let P and Q be the two bodies, Weightlessness is a state when an object does not weigh anything. Free PDF download of Important Questions with solutions for CBSE Class 9 Science Chapter 10 - Gravitation prepared by expert Science teachers from latest edition of CBSE(NCERT) books. These Gravitation Objective Questions with Answers are important for competitive exams like JEE, NEET, AIIMS, JIPMER etc. Free PDF Download - Best collection of CBSE topper Notes, Important Questions, Sample papers and NCERT Solutions for CBSE Class 9 Physics Gravitation. Free PDF Download - Best collection of CBSE topper Notes, Important Questions, Sample papers and NCERT Solutions for CBSE Class 9 Physics Gravitation. When hypothetically M becomes 4 M and R becomes \(\frac{R}{2}\). A stone is dropped from a cliff. FQ = \(\frac{G \times M_{e} \times m_{2}}{R^{2}}\) ……. Question 7: Law of gravitation gives the gravitational force between (a) the earth and a point mass only Register online for Science tuition on Vedantu.com to score more marks in your examination. Why is ‘G’ called the universal gravitational constant? The earth attracts an apple. Gravitation Class 9 Extra Questions Numericals. State the universal law of gravitation. ∴ For the second stone, ⇒ t2 = \(\frac{98}{9.8}\) = 10 […] Answer: = M + \(\frac{10}{100}\)M = M + \(\frac{M}{10}\) = \(\frac{11M}{10}\) = 1.1 M What would happen to your weight? On the earth, a stone is thrown from a height in a direction parallel to the earth’s surface while another stone is simultaneously dropped from the same height. Answer: Gravitation is the force of attraction between any two bodies while gravity refers to attraction between any body and the earth. Question 1. Fp = \(\frac{G \times M_{e} \times m_{1}}{R^{2}}\) …..(1) Weight is not a constant quantity. 6.7 × 10-11 Nm2 kg-2) everywhere in the universe. This test will determine your knowledge of the subject. New weight becomes 1.1 times. When does an object show weightlessness? (ii) Weight = \(\frac{G M m}{r^{2}}\), where r is the radius of the earth. Fill in the blanks : _____ of an object is the force of gravity acting on it.. Class 9 - Physics - Gravitation . Which stone would reach the ground first and why? Why? (b) At which point is the bal 1 stationary? But we know, acceleration ∝ 1/m. Question 5. Question 1. MCQ quiz on Gravitation multiple choice questions and answers on gravitation MCQ questions quiz on gravitation objectives questions with answer test pdf. How does the force of attraction between the two bodies depend upon their masses and distance between them? Answer: or D = \(\frac{g R^{2}}{G \frac{4}{3} \pi R^{3}}=\frac{3 g}{4 \pi G R}\), Question 6. Comment. \(\frac{W_{m}}{W_{e}}=\frac{M_{m}}{M_{e}} \times \frac{R_{e}^{2}}{R_{m}^{2}}\) …. At what place on the earth’s surface is the weight of a body minimum? ∴ 49 = 0 × t + \(\frac{1}{2}\) × 9.8 × t2 or, v2 – 02 = 2(-10) × (-40) When a body is dropped from a height, what is its initial velocity? radius of the earth, Re = 6400 km Show that the weight of an object on the moon is \(\frac{1}{6}\) th of its weight on the earth. How is gravitation different from gravity? Here, if the masses mx and m2 of the two bodies are of 1 kg and the distance (r) between them is 1 m, then putting m1 = 1 kg, m2 = 1 kg and r = 1 m in the above formula, we get The accepted value of G is 6.67x 1CT-11 Nm 2 kg-2. ∴ New weight becomes 4 times. F ∝ \(\frac{m_{1} \times m_{2}}{r^{2}}\) Will this ratio remain the same if (i) one of the objects is hollow and the other one is solid; and (ii) both of them are hollow, size remaining the same in each case? Question 4. Question 6. In order that a body of 5 kg weighs zero at … Question 18. If the small and big stones are dropped from the roof of a house simultaneously, they will reach the ground at the same time. Zero. (e) What is the ball’s acceleration at point A? But the acceleration due to gravity is represented by the symbol g. Therefore, we can write = \(\frac{G M m}{4 r^{2}}=\frac{W}{4}\) [NCERT Exemplar] Differences between g andG. Calculate its speed after 2 s. Also find the speed with which the stone strikes the ground. 3. Q.17 Does the gravitational force same for two objects inside and outside the water? In a hypothetical case, if the diameter of the earth becomes half of its present value and its mass becomes four times of its present value, then how would the weight of any object on the surface of the earth be affected? Answer: State the universal law of gravitation. Question 15. where Me = mass of earth and Re = radius of earth. Answer: Gravitational constant is numerically equal to the force of attraction between two masses of 1 kg that are separated by a distance of 1 m. Given m1 = 3 kg; m2 = 12 kg What is the source of centripetal force that a planet requires to revolve around the sun? Similarly, the force of attraction between the earth and the body Q is given by Answer: The universal law of gravitation was able to explain successfully. They both hit the ground at the same time. How much do you know about gravitation? Weight, W = mg, m = \(\frac{W}{g}\), m = \(\frac{9.8}{9.8}\) = 1 kg [NCERT Exemplar] u = 0 + gt2 Then force acting on m3 due to m1, is equal and opposite to the force acting on m3 due to m2. Answer: Previous: Contents: Next: How does the weight of an object vary with respect to mass and radius of the earth? We = \(\frac{G M_{e} m}{R_{e}^{2}}\) …(2) Do you agree with his hypothesis or not? Here, initial velocity, u = 0 Because the value of acceleration due to gravity (g) on the moon’s surface is nearly l/6th to that of the surface of the earth. (f) What is the bal l’s acceleration at point B? Answer: We know that G = 6.67 × 10-11 Nm2 kg-2 A stone is dropped from the top of a 40 m high tower. Justify your answer. 0 = 0.5 – 9.8t Professionals, Teachers, Students and Kids Trivia Quizzes to test your knowledge on the subject. Answer: Answer: (2) The moon will begin to move in a straight line in the direction in which it was moving at that instant because the circular motion of moon is due to centripetal force provided by the gravitational force of the earth. or h = \(\frac{0.25}{19.6}\) = 0.0127 m When M changes to new mass M’ So, the earth does not fall into the sun. Then weight becomes Wn = mG \(\frac{4 M}{\left(\frac{R}{2}\right)^{2}}\) = (16 m G) \(\frac{M}{R^{2}}\) = 16 × W0 Solution: The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. If the mass of a body is 9.8 kg on the earth, what would be its mass on the moon? (ii) Putting the values in the formula v = u + gt At the equator. On the moon’s surface, the acceleration due to gravity is 1.67 ms-2. A force of 20 N acts upon a body whose weight is 9.8 N. What is the mass of the body and how much is its acceleration? Answer: mg = \(G \frac{m M}{R^{2}}\) or g = \(\frac{G M}{R^{2}}\) …(3) Question Answers, Page 134. The force (F) of gravitational attraction on a body of mass m due to earth of mass M and radius R is given by mass of the moon, Mm = 7.4 × 1022 kg where G is a constant known as universal gravitational constant. ... plz answer this questions. As we know s = ut + \(\frac{1}{2}\) gt2 On what factors does that force depend? Answer: When a body is thrown vertically upwards, what is its final velocity? i.e., weight will be reduced to one-fourth of the original. What will be its speed when it has fallen 100 m? The way the world works is exciting. In case of free fall acceleration does not depend upon mass and size. Question 3. This will clear students doubts about any question and improve application skills while preparing for board exams. As u = 0, h1 = \(\frac{1}{2} g t_{1}^{2}\) [NCERT Exemplar] \(\frac{\iota_{1}}{t_{2}}=\sqrt{\frac{h_{1}}{h_{2}}}\) This force depends on the product of the masses of the planet and sun and the distance between them. Yes. 1. It is the measure of inertia of the body. Therefore, the body appears to be floating weightlessly. One second later, he throws another stone. A ball is thrown up with a speed of 0.5 m/s. 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