JEE Main 2022 · 27 June · evening · Single correct · +4 / −1
Four spheres each of mass m from a square of side d (as shown in figure). A fifth sphere of mass M is situated at the centre of square. The total gravitational potential energy of the system is :
- (a) - Gm/d[ (4 + √2 )m + 4√2 M ]
- (b) - Gm/d[ (4 + √2 )M + 4√2 m ]
- (c) - Gm/d[ 3m² + 4√2 M ]
- (d) - Gm/d[ 6m² + 4√2 M ]
Answer and solution
Answer (a) - Gm/d[ (4 + √2 )m + 4√2 M ]
Solution
Physics sir · 3 steps
Five masses make ten pairs; sort them by distance and add up -Gm₁m₂/r for each one.
- Count the pairs. Potential energy of a system is the sum of -Gm₁m₂/r over every pair. Five masses give C(5, 2) = 10 pairs, and they fall into three groups by distance.
Pair type Number of pairs Distance Total energy corner to corner, along a side 4 d -4Gm²/d corner to corner, across a diagonal 2 √2 d -2Gm²/(√2 d) = -(√2 Gm²)/d centre M to a corner m 4 d/√2 -(4√2 GMm)/d - Add and factor. Take -Gm/d out of all three terms. What is left inside the bracket is 4m + √2 m + 4√2 M.
U = -Gm/d[(4+√2) m + 4√2 M]
- So. The total gravitational potential energy matches option (a).
The idea to keep. For system potential energy, list every pair exactly once, group the pairs by distance, and do not forget the diagonals. Then check that every term has units of Gm²/d.
Answer from the NTA final answer key. Solution by PrepWiser.