A 10kg mass is at a distance of 1m from a 100kg mass.Find the gravitational force of attraction when a) 10kg mass exerts force on the 100kg mass b) 100 kg mass exerts force on the 10kg mass Physics class 11th problems Physics class 11th numericals First year physics problems First year physics numericals Gravitation problems
00:00Question says, a 10 kg mass is at a distance of 1 meter from a 100 kg mass, find the gravitational
00:16force of attraction when a 10 kg mass exerts force on the 100 kg mass, b 100 kg mass exerts
00:28force on the 10 kg mass, so first of all, we will make a data, this problem ko solve
00:39karne se pehle, hum data form karenghe, data ko banayenghe taakhe is ko hum aasani se solve
00:47kar sakhe, so for data, is m1 is given, which is 10 kg, m2 is given is 100 kg, and the distance
01:15between m1 and m2 is 1 meter, so r is 1 meter, what we have to find, we have to find force
01:35of 10 kg mass exerts on 100 kg mass, and force of 100 kg mass exerted on 10 kg mass.
01:50So, the force on m2 by m1 is given as f1 on 2 is equal to watts, and force exerted on m1
02:15by m2 is given as f2 on 1, which is also equal to watts. So, we have to calculate these forces.
02:29So, for solution, take a suppose, there are two bodies, this is m1 and this is m2, the mass
02:50mass of m1 is 10 kg, and the mass of m2 is 100 kg, as the distance directed from body 2 means
03:05m2 to m1 is given by r2 on 1, and the distance directed from body 1 to body 2 is given by rs,
03:191 on 2. So, first of all, we will find f1 on 2 is according
03:40to Newton's law of universal gravitation, the force 1 on 2 is given by g m1 m2 upon rs, r squares,
04:01so by placing the given values, here g is universal gravitational constant, so its value is constant,
04:11which is 6.67 into 10 raised power minus 11 Newton into meter square per kilogram square. So, by putting the given values,
04:27force 1 on m2 by m1 is given as, is the value of g is 6.67 into 10 raised power minus 11 multiplied by m1 is 10 kg multiplied with m2 which is 100 kg,
04:54so the square of 1, so the square of 1, so the square of 1. So, force exerted on m2 by m1 is given as, after simplification we will get the answer is, 6.67 into 10 raised power minus
05:018 Newton. So, this is our first answer is 10 raised power minus 8 Newton. So, this is our first answer,
05:08now force on m1 by m2 by m2 by m2 by m2 by m1 is given as, after simplification we will get the answer is 6.67 into 10 raised power minus 8 Newton. So, this is our first answer, now force on m1 by m2 by m2.
05:37m2 is given as, by g m2 by m2 by r2 so, by putting the given values, force on m1 by m2 is equal to the value of
05:432 upon R square. So, by putting the given values force on M1 by M2 is equal to the value
06:00of capital G is 6.67 into 10 raised power minus 11 multiplied with the value of M1 which
06:09is 10 kilogram into the value of M2 which is hundreds kilogram divided by R square which
06:17is 1 meter. So, the square on 1. So, after simplifying this we will also get the answer
06:26is 6.67 into 10 raised power minus 8 Newton. So, this is our second answer. Finally, we
06:37can say that the force is same in both the conditions.
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