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Test Index
IIT JEE Advanced 2009 Paper 1
Show Para
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Section:
Physics
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© examsnet.com
Question : 1 of 60
Marks:
+1
,
-0
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
[JEE Adv 2009 P1]
a
10
\frac{a}{10}
10
a
​
a
8
\frac{a}{8}
8
a
​
a
12
\frac{a}{12}
12
a
​
a
3
\frac{a}{3}
3
a
​
Validate
Solution:
The coordinates of centre of mass is defined as
R
C
R_C
R
C
​
=
∑
i
m
i
r
i
∑
i
m
i
\frac{\sum\limits_{i} m_i r_i}{\sum\limits_i m_i}
i
∑
​
m
i
​
i
∑
​
m
i
​
r
i
​
​
Thus,
Y
C
M
Y_{CM}
Y
CM
​
=
(
6
m
×
0
)
+
(
m
×
a
)
+
(
m
×
a
)
+
(
m
×
0
)
+
(
m
×
(
−
a
)
)
6
m
+
m
+
m
+
m
+
m
\frac{(6m \times 0) + (m \times a) + (m \times a) + (m \times 0) + (m \times (-a))}{6m+m+m+m+m}
6
m
+
m
+
m
+
m
+
m
(
6
m
×
0
)
+
(
m
×
a
)
+
(
m
×
a
)
+
(
m
×
0
)
+
(
m
×
(
−
a
))
​
=
a
10
\frac{a}{10}
10
a
​
© examsnet.com
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