From the equation of motion, v = u + gt, we get. The force is along the line joining the centres of two objects. Is your mass more or less than 42 kg? Final velocity of the ball at the maximum height, v = 0. Topics and Sub Topics in Class 9 Science Chapter 10 Gravitation: These solutions are part of NCERT Solutions for Class 9 Science. T. Question 5. he relative density of gold is 19.3. (a) Gravitational Force (b) Electrostatic Force (c) Magnetic Force (d) None. CBSE Test Papers class 9 Science Gravitation. Answer: Question 1. This upward force is the buoyant force. In the school fair, there was a game in which one need to find the heaviest ball without holding them in hand. This will clear students doubts about any question and improve application skills while preparing for board exams. Formulae Handbook for Class 9 Maths and ScienceEducational Loans in India. The large buoyant force on the block is due to its density being smaller than that of water. Therefore, the same mass of gold weighs lesser at the equator than at the poles. Hence the mass reading of 42 kg given by a weighing machine is same as the actual mass of the body. According to the universal law of gravitation, gravitational force (F) acting between two objects is inversely proportional to the square of the distance (r) between them, i.e. The gravitational force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. Initial velocity of the stone, u = 40 m/s, Final velocity of the stone, v = 0 (At the highest point). Balbharati solutions for Science and Technology Part 1 10th Standard SSC Maharashtra State Board chapter 1 (Gravitation) include all questions with solution and detail explanation. Class 9 Science NCERT Textbook – Page 142 (a) The high heel shoes would exert lot of pressure on the loose sand of beach and will sink more in the soil as compared to flat shoes. As Force Mass, therefore, acceleration is constant for a body of any mass. In Fig. Practise the expert solutions to understand the application of the law of gravitation to calculate the weight of an object on the Moon, Earth or other planets. According to the universal law of gravitation, the force of attraction between the Earth and the Sun is given by: R = Average distance between the Earth and the Sun = 1.5 × 1011 m, G = Universal gravitational constant = 6.7 × 10-11 Nm2 kg-2. All rights reserved. This is due to the reason, that the bag of cotton which has more volume (as it has less density) than the iron bar (which has more density), experiences more upthrust due to air. Answer: The density of sea water is more due to dissolved salts in it as compared to the density of river water. However, the mass of the Earth is much larger than the mass of the Moon. So, the force of attraction of Earth on Moon is equal and opposite to the force of attraction of Moon on Earth. In this page find links for free mcq questions for class 9 science. Answer: The iron rod sinks due to high density and less buoyant force exerted by the water on it, but in case of ship the surface area is increased, the upthrust experienced by the body is more. (a) What is lactometer? Do you feel fascinated or intimidated while studying about gravitation in Physics? What is lactometer and hydrometer? Science NCERT Grade 9, Chapter 10, Gravitation as the name suggests, talks about gravitation and universal law of gravitation.The motion of the object under the influence of the gravitational force of the earth is explained in the chapter, Gravitation.The main discussion of the chapter starts by explaining the concept of gravitation by taking the example of the motion of the moon around the earth. Question 3. (b) is least on poles. NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12. Verify your number to create your account, Sign up with different email address/mobile number, NEWSLETTER : Get latest updates in your inbox, Need assistance? Maximum height reached by the stone, h = ? 2. Answer: The pressure of water in dams at the bottom is more, to withstand this pressure the dams have wider walls. The combined displacement (s + s') of both the stones at the meeting point is equal to the height of the tower 100 m. In 4 s, the falling stone has covered a distance given by equation (1) as. 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