From a uniform disc of radius R, an equilateral triangle of side `sqrt(3)R` is cut as shown in the figure. The new position of centre of mass is :

Options

A. `(0,0)`
B. `(0,R)`
C. `(0,(sqrt(3)R)/(2))`
D. none of these

Answer

B

Explanation

`m_1` = mass of remaining part.
`m_2` = mass of cut part and `y_1 y_2 cm` of `m_1 m_2`
respectively for complete system, we have
`y_(cm) = (m_1y_1 + m_2y_2)/((m_1 + m_2))" "`(with respect to centre of the circle, `y_(cm) = 0` moreover centre of mass of the `Delta` is also at this centre of the circle, so `y_2` is also 0.
`0 = (sigma(pi R^2 - (3Rsqrt(3))/4)y_1 + sigma(3R^2sqrt(3))/4 xx (0))/(sigma(pi R^2)) implies y_1 = 0 "so" (0,0)`.