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Impulse and Momentum question

2009 · Shift 1 · Q42

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JEE Advanced
Physics
Impulse and Momentum
2009 · Shift 1 · Q42

Impulse and Momentum question

2009 · Shift 1 · Q42

JEE AdvancedPhysicsImpulse and MomentumMCQ+3 / −1

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 7 English

IIT-JEE 2009 Paper 1 Offline Physics - Impulse & Momentum Question 7 English

A
$\frac{a}{10}$
B
$\frac{a}{8}$
C
$\frac{a}{12}$
D
$\frac{a}{3}$

View written solutionFree

Correct answer: A

Identify all parts and their masses

The figure is made of 5 parts:

Outer circle: mass $6m$ , centre at $(0,0)$
Left inner circle: mass $m$ , centre at $(-a,a)$
Right inner circle: mass $m$ , centre at $(a,a)$
Vertical line segment: mass $m$ , centre at $(0,0)$
Horizontal line segment: mass $m$ , centre at $(0,-a)$

We need only the $y$ -coordinate of the centre of mass.

Use the centre of mass formula

If the masses are $m_i$ and their $y$ -coordinates are $y_i$ , then

y_{\text{CM}}=\frac{\sum m_i y_i}{\sum m_i}

Compute total mass

M=6m+m+m+m+m=10m

Compute sum of moments about the $x$ -axis

Contribution of each part:

Outer circle: $6m \cdot 0 = 0$
Left inner circle: $m \cdot a = ma$
Right inner circle: $m \cdot a = ma$
Vertical line: $m \cdot 0 = 0$
Horizontal line: $m \cdot (-a) = -ma$

So,

\sum m_i y_i = ma+ma-ma = ma

Find $y_{\text{CM}}$

y_{\text{CM}}=\frac{ma}{10m}=\frac{a}{10}

Match with options

\boxed{\frac{a}{10}}

So the correct option is A.

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