Half part of the rectangular plate having density \[{{d}_{1}}\] while second half having density \[{{d}_{2}}\]. The length of the plate l then its centre of mass will be
A) \[\left( \frac{{{d}_{1}}+{{d}_{2}}}{{{d}_{1}}{{d}_{2}}} \right)l\]
B) \[\left( \frac{{{d}_{1}}+2{{d}_{2}}}{3{{d}_{1}}+4{{d}_{2}}} \right)l\]
C) \[\left( \frac{3{{d}_{1}}{{d}_{2}}}{{{d}_{1}}+3{{d}_{2}}} \right)l\]
D) \[\left( \frac{{{d}_{1}}+3{{d}_{2}}}{4({{d}_{1}}+{{d}_{2}})} \right)l\]
Correct Answer: D
Solution :
As mass of each part of the plate is assumed to be at the centre of that part. So, the masses \[k\]\[{{d}_{1}}\] and \[k\]\[{{d}_{2}}\] are assumed to be at \[{{C}_{1}}\] and \[{{C}_{2}}\]respectively. Here, \[k\]is a constant. Take the .middle point of the left edge to be the origin. So, the x-coordinate of \[{{C}_{1}}\] is \[l/4\]and that of\[{{C}_{2}}\]is\[3l/4\] Thus, \[{{X}_{cm}}=\frac{(k\,{{d}_{1}})l/4+(k\,{{d}_{2}})3l/4}{k\,{{d}_{1}}+k\,{{d}_{2}}}\] \[=\frac{{{d}_{1}}+3{{d}_{2}}}{4({{d}_{1}}+{{d}_{2}})}l\]
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