SATHEE: Centre of Mass 1 Question 1

Centre of Mass 1 Question 1

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

(2019 Main, 12 April II)

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

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

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

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Answer:

Correct Answer: 1. (c)

Solution:

The height of equilateral $Δ$ is

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

Thus, coordinates of three masses are $(0, 0), (1, 0)$

and $0.5, \frac{\sqrt{3}}{2}$

Using, $X_{C M} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}$

we have

$\begin{aligned} X_{C M} & = \frac{50 \times 0 + 100 \times 1 + 150 \times 0.5}{50 + 100 + 150} \\ = \frac{175}{300} = \frac{7}{12} m \end{aligned}$

Similarly,

$\begin{aligned} Y_{C M} & = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}} \\ = \frac{50 \times 0 + 100 \times 0 + 150 \times \frac{\sqrt{3}}{2}}{50 + 100 + 150} \\ = \frac{\sqrt{3}}{4} m \end{aligned}$

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Centre of Mass 1 Question 1

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

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Centre of Mass 1 Question 1

1. 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

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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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