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:

Question

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: <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-xfl3eex0wlnynt5u.jpg" /> (A) ((7/12) m , (√3/4) m) (B) ((7/12) m , (√3/8) m) (C) ((√3/4) m , (5/12) m) (D) ((√3/8) m , (7/12) m)

Tardigrade Admin · Accepted Answer

Coordinates of A:(0 , 0) B:((1/2) m , (√3/2) m) C:(.1 m, 0.) Coordinates of center of mass (xc m , yc m)=((50 × 0 + 150 × (1/2) + 100 × 1/50 + 150 + 100) m , (50 × 0 + 150 × (√3/2) + 100 × 0/50 + 150 + 100) m) =((7/12) m , (√3/4) m)

Q. 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:

$(\frac{7}{12} m, \frac{3}{4} m)$

$(\frac{7}{12} m, \frac{3}{8} m)$

$(\frac{3}{4} m, \frac{5}{12} m)$

$(\frac{3}{8} m, \frac{7}{12} m)$

Q. Three particles of masses 50g , 100g and 150g are placed at the vertices of an equilateral triangle of side 1m (as shown in the figure). The (x,y) coordinates of the centre of mass will be:

(127​m,43​​m)

(127​m,83​​m)

(43​​m,125​m)

(83​​m,127​m)

Coordinates of A:(0,0) B:(21​m,23​​m) C:(1m,0) Coordinates of center of mass (xcm​,ycm​)=(50+150+10050×0+150×21​+100×1​m,50+150+10050×0+150×23​​+100×0​m) =(127​m,43​​m)

Q. 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:

$(\frac{7}{12} m, \frac{3}{4} m)$

$(\frac{7}{12} m, \frac{3}{8} m)$

$(\frac{3}{4} m, \frac{5}{12} m)$

$(\frac{3}{8} m, \frac{7}{12} m)$