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System of Particles and Rotational Motion

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Three particles of masses 50 {~g}, 100 {~g} and 150 {g} are pl

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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

(a) $(\frac{3}{4} m, \frac{5}{12} m)$
(b) $(\frac{7}{12} m, \frac{3}{8} m)$
(c) $(\frac{7}{12} m, \frac{3}{4} m)$
(d) $(\frac{3}{8} m, \frac{7}{12} m)$

Name: Video solution 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 (a) (43 m,125 m)(b) (127 m,83 m)(c) (127 m,43 m)(d) (83 m,127 m)
Uploaded: 2022-09-22T13:30:34.000Z
Duration: 14 min
Description: Video solution 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 (a) (43 m,125 m)(b) (127 m,83 m)(c) (127 m,43 m)(d) (83 m,127 m)

Views: 6,133 students

Updated on: Feb 28, 2023

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2019

Found 4 tutors discussing this question

Asher Discussed

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

(a)

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

(b)

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

(c)

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

(d)

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

10 mins ago

Discuss this question LIVE

10 mins ago

Text solutionVerified

E x p \cdot (c)

The height of equilateral

Δ

h = y_{3} = (1)^{2} - (0.5)^{2} = 3 /2 m

Thus, coordinates of three masses are

(0, 0), (1, 0)

and

(0.5, \frac{3}{2})

Using,

X_{CM} = \frac{m _{1} x _{1} + m _{2} x _{2} + m _{3} x _{3}}{m _{1} + m _{2} + m _{3}},

we have

X_{CM} = \frac{50 \times 0 + 100 \times 1 + 150 \times 0.5}{50 + 100 + 150}

= \frac{175}{300} = \frac{7}{12} m

Similarly,

Y_{CM} = \frac{m _{1} y _{1} + m _{2} y _{2} + m _{3} y _{3}}{m _{1} + m _{2} + m _{3}}

= \frac{50 \times 0 + 100 \times 0 + 150 \times \frac{3}{2}}{50 + 100 + 150} = \frac{3}{4} m

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Question Text	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 (a) $(\frac{3}{4} m, \frac{5}{12} m)$ (b) $(\frac{7}{12} m, \frac{3}{8} m)$ (c) $(\frac{7}{12} m, \frac{3}{4} m)$ (d) $(\frac{3}{8} m, \frac{7}{12} m)$
Updated On	Feb 28, 2023
Topic	System of Particles and Rotational Motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 2
Upvotes	148
Avg. Video Duration	10 min

Text solutionVerified

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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 (a) (43​​ m,125​ m)(b) (127​ m,83​​ m)(c) (127​ m,43​​ m)(d) (83​​ m,127​ m)

Text solutionVerified

Filo tutor solutions (2)

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