Q23.A body of mass (2M) splits into four masses {m, M โm, m, M โm}, which are rearranged to form a square as shown in the figure. The ratio of Mm for which, the gravitational potential energy of the system becomes maximum is x : 1. The value of x is ________.
What This Question Tests
This question requires calculating the gravitational potential energy for a system of four masses and then using calculus to find the ratio of masses for maximum potential energy.
Concepts Tested
Formulas Used
U = -G m1 m2 / r
dU/dm = 0
๐ NCERT Sections This Tests
2.1 โ Two Charges 5 ร 10โ8 C And โ3 ร 10โ8 C Are Located 16 Cm Apart. At
Physics Class 11 ยท Chapter 2
2.1 Two charges 5 ร 10โ8 C and โ3 ร 10โ8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
13.2 โ Obtain The Binding Energy Of The Nuclei 5626Fe And 20983 Bi In Units Of
Physics Class 12 ยท Chapter 13
13.2 Obtain the binding energy of the nuclei 5626Fe and 20983 Bi in units of MeV from the following data: m ( 5626Fe ) = 55.934939 u m ( 20983 Bi ) = 208.980388 u
13.3 โ A Given Coin Has A Mass Of 3.0 G. Calculate The Nuclear Energy That
Physics Class 12 ยท Chapter 13
13.3 A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 6329Cu atoms (of mass 62.92960 u).
๐ Question Details
- Chapter
- Gravitation
- Topic
- Gravitational potential energy of a system of particles
- Year
- 2021
- Shift
- 27 Aug Shift 1
- Q Number
- Q23
- Type
- Numerical
- NCERT Ref
- Class 11 Physics Ch 8: Gravitation
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