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Center of Mass question

2019 · 12 Apr · Shift 2 · Q63
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Center of Mass question

2019 · 12 Apr · Shift 2 · Q63

JEE MainPhysicsCenter of MassMCQ+4 / −1
Three particles of masses 50 g, 100 g and 150 g are placed at the vertices of an equilateral triangle of side 1 m (as shown in the figure). The (x, y) coordinates of the centre of mass will be : JEE Main 2019 (Online) 12th April Evening Slot Physics - Center of Mass and Collision Question 84 English
  1. A
    (34m,512m)\left( {{{\sqrt 3 } \over 4}m,{5 \over {12}}m} \right)(43​​m,125​m)
  2. B
    (712m,38m)\left( {{7 \over {12}}m,{{\sqrt 3 } \over 8}m} \right)(127​m,83​​m)
  3. C
    (712m,34m)\left( {{7 \over {12}}m,{{\sqrt 3 } \over 4}m} \right)(127​m,43​​m)
  4. D
    (38m,712m)\left( {{{\sqrt 3 } \over 8}m,{7 \over {12}}m} \right)(83​​m,127​m)
View written solutionFree

Correct answer: C

  1. Set up the coordinates

For an equilateral triangle of side 1 m1\,\text{m}1m, take the usual orientation shown in such problems:

  • one side along the xxx-axis,
  • vertices at A(0,0),B(1,0),C(12,32).A(0,0),\quad B(1,0),\quad C\left(\frac12,\frac{\sqrt3}{2}\right).A(0,0),B(1,0),C(21​,23​​).

From the figure/options pattern, the masses are placed as:

  • 50 g50\,\text{g}50g at A(0,0)A(0,0)A(0,0),
  • 100 g100\,\text{g}100g at B(1,0)B(1,0)B(1,0),
  • 150 g150\,\text{g}150g at C(12,32)C\left(\frac12,\frac{\sqrt3}{2}\right)C(21​,23​​).

Total mass: M=50+100+150=300 g.M=50+100+150=300\,\text{g}.M=50+100+150=300g.

  1. Use centre of mass formula

The coordinates of the centre of mass are xcm=∑mixi∑mi,ycm=∑miyi∑mi.x_{cm}=\frac{\sum m_i x_i}{\sum m_i},\qquad y_{cm}=\frac{\sum m_i y_i}{\sum m_i}.xcm​=∑mi​∑mi​xi​​,ycm​=∑mi​∑mi​yi​​.

  1. Compute xxx-coordinate

xcm=50(0)+100(1)+150(12)300x_{cm}=\frac{50(0)+100(1)+150\left(\frac12\right)}{300}xcm​=30050(0)+100(1)+150(21​)​ =0+100+75300=175300=712 m.=\frac{0+100+75}{300}=\frac{175}{300}=\frac{7}{12}\,\text{m}.=3000+100+75​=300175​=127​m.

  1. Compute yyy-coordinate

ycm=50(0)+100(0)+150(32)300y_{cm}=\frac{50(0)+100(0)+150\left(\frac{\sqrt3}{2}\right)}{300}ycm​=30050(0)+100(0)+150(23​​)​ =753300=34 m.=\frac{75\sqrt3}{300}=\frac{\sqrt3}{4}\,\text{m}.=300753​​=43​​m.

  1. Final answer

Therefore, the centre of mass is (712 m,34 m).\boxed{\left(\frac{7}{12}\,\text{m},\frac{\sqrt3}{4}\,\text{m}\right)}.(127​m,43​​m)​.

  1. Option check

This matches Option C.

  1. Comparison with stored answer

Stored correct answer: C
Derived answer: C
So they agree.

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