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Physics · Gravitation

JEE Main 2026 — 24 January, Morning Shift — Question 43

Three masses 200 kg,300 kg200 \mathrm{~kg}, 300 \mathrm{~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=6.7×10−11 N m2/kg2\mathrm{G}=6.7 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{kg}^{2} )

  1. Option A:

    9.86×10−69.86 \times 10^{-6}

  2. Option B:

    2.85×10−72.85 \times 10^{-7}

  3. Option C:

    1.74×10−71.74 \times 10^{-7}

    Correct
  4. Option D:

    4.77×10−74.77 \times 10^{-7}

Answer: C

Step-by-step solution

Work done by external agent :

Wext =ΔUUi=−Gm1 m2ri−Gm2 m3ri−Gm1 m3ri:ri=20 mUf=−Gm1 m2rf−Gm2 m3rf−Gm1 m3rf:rf=25 mUi=−6.67×10−1120[200×300+300×400+200×400]=−6.67×10−1120×26×104=−86.71×10−8 JUf=−6.67×10−110.25[200×300+300×400+200×400]=−6.67×10−110.25×26×104=−693.68×10−9=−69.36×10−8 JΔU=Uf−Ui=1.74×10−7 J\begin{aligned} & \mathrm{W}_{\text {ext }}=\Delta \mathrm{U} & \mathrm{U}_{\mathrm{i}}=-\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}_{\mathrm{i}}}-\frac{\mathrm{Gm}_{2} \mathrm{~m}_{3}}{\mathrm{r}_{\mathrm{i}}}-\frac{\mathrm{Gm}_{1} \mathrm{~m}_{3}}{\mathrm{r}_{\mathrm{i}}}: \mathrm{r}_{\mathrm{i}}=20 \mathrm{~m} & \mathrm{U}_{\mathrm{f}}=-\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}_{\mathrm{f}}}-\frac{\mathrm{Gm}_{2} \mathrm{~m}_{3}}{\mathrm{r}_{\mathrm{f}}}-\frac{\mathrm{Gm}_{1} \mathrm{~m}_{3}}{\mathrm{r}_{\mathrm{f}}}: \mathrm{r}_{\mathrm{f}}=25 \mathrm{~m} & \mathrm{U}_{\mathrm{i}}=\frac{-6.67 \times 10^{-11}}{20}[200 \times 300+300 \times 400+200 \times 400] & =\frac{-6.67 \times 10^{-11}}{20} \times 26 \times 10^{4}=-86.71 \times 10^{-8} \mathrm{~J} & \mathrm{U}_{\mathrm{f}}=\frac{-6.67 \times 10^{-11}}{0.25}[200 \times 300+300 \times 400+200 \times 400] & =\frac{-6.67 \times 10^{-11}}{0.25} \times 26 \times 10^{4}=-693.68 \times 10^{-9} & =-69.36 \times 10^{-8} \mathrm{~J} & \Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=1.74 \times 10^{-7} \mathrm{~J} \end{aligned}
Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
24 January, Morning Shift
Subject
Physics
Chapter
Gravitation
Topic
Gravitational Potential Energy and Potential