JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Conservation of Momentum

WhatsApp Email

Look at the drawing given in the figure 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

6 m

. 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

(A) $\frac{a}{10}$ ✓ Correct
(B) $\frac{a}{8}$
(C) $\frac{a}{12}$
(D) $\frac{a}{3}$

Answer: (A)

Solution

The

y

-coordinate of the center of mass is given by

Y_{cm} = \frac{\sum m _{i} y _{i}}{\sum m _{i}}

. The outer circle (mass

6 m

) is at

y = 0

, left inner circle (mass

m

) is at

y = a

, right inner circle (mass

m

) is at

y = a

, vertical line (mass

m

) is at

y = 0

, and horizontal line (mass

m

) is at

y = - a

. Total mass is

10 m

. Therefore,

Y_{cm} = \frac{6 m \cdot 0 + m \cdot a + m \cdot a + m \cdot 0 + m \cdot ( - a )}{10 m} = \frac{a}{10}

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Single correct option

Practise more: all JEE Advanced PYQs on Conservation of Momentum →

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Conservation of Momentum

Solution

More JEE Advanced questions on Conservation of Momentum