JEE Advance - Physics (2009 - Paper 1 Offline - No. 2)

Look at the drawing given in the figure below which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is $$m$$. the mass of the ink used to draw the outer circle is $$6m$$. The coordinates of the centres of the different parts are: outer circle (0, 0), left inner circle ($$-a,a$$), right inner circle ($$a,a$$), vertical line (0, 0) and horizontal line ($$0,-a$$). The y-coordinate of the centre of mass of the ink in this drawing is

IIT-JEE 2009 Paper 1 Offline Physics - Impulse & Momentum Question 5 English

$$\frac{a}{10}$$
$$\frac{a}{8}$$
$$\frac{a}{12}$$
$$\frac{a}{3}$$

Explanation

The coordinates of centre of mass is defined as

$${R_{CM}} = {{\sum\nolimits_i {{m_i}{r_i}} } \over {\sum\nolimits_i {{m_i}} }}$$

Thus,

$${Y_{CM}} = {{(6m \times 0) + (m \times a) + (m \times a) + (m \times 0) + (m \times - a)} \over {6m + m + m + m + m}} = {a \over {10}}$$

Comments (0)

Advertisement