JEE Main 2026 — Jan 24, Shift 1 — Physics solution

JEE Main2026Jan 24, Shift 1Physics

Three masses 200 kg, 300 kg and 400 kg are placed at the vertices of an equilateral triangle with sides 20 m. They are rearranged on the vertices of a bigger triangle of side 25 m and with the same centre. The work done in this process __________ J.

(Gravitational constant $G = 6.7 \times 1 0^{- 11}$ N m $^{2}$ /kg $^{2}$ )

A
$9.86 \times 1 0^{- 6}$
B
$2.85 \times 1 0^{- 7}$
C
$1.74 \times 1 0^{- 7}$
D
$4.77 \times 1 0^{- 7}$

Solution

For three masses at the vertices of an equilateral triangle of side $r$ , the total gravitational potential energy is

$U = - \frac{G}{r} (m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3})$ .

Mass-product sum:

$m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3} = (200) (300) + (300) (400) + (200) (400) = 60000 + 120000 + 80000 = 2.6 \times 1 0^{5}$ kg $^{2}$ .

$U_{i} = - \frac{( 6.7 \times 1 0 ^{- 11} ) ( 2.6 \times 1 0 ^{5} )}{20} = - 8.71 \times 1 0^{- 7}$ J.

$U_{f} = - \frac{( 6.7 \times 1 0 ^{- 11} ) ( 2.6 \times 1 0 ^{5} )}{25} = - 6.968 \times 1 0^{- 7}$ J.

Work done by the external agent equals the change in PE:

$W = U_{f} - U_{i} = (- 6.968 + 8.71) \times 1 0^{- 7} \approx 1.74 \times 1 0^{- 7}$ J.

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