Question:
A uniform thin rod AB of length L has linear mass density \(\mu(x)=a+\frac{bx}{L}\), where x is measured from A. If the CM of the rod lies at a distance of \(\left(\frac{7}{12}L\right)\) from A, then a and b are related as :
A uniform thin rod AB of length L has linear mass density \(\mu(x)=a+\frac{bx}{L}\), where x is measured from A. If the CM of the rod lies at a distance of \(\left(\frac{7}{12}L\right)\) from A, then a and b are related as :
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Use \(x_{cm}=\int x\,dm/\int dm\) with \(dm=\mu\,dx\) and equate to 7L/12.
Updated On: Nov 11, 2026
- \(a=b\)
- \(a=2b\)
- \(2a=b\)
- \(3a=2b\)
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The Correct Option is C
Solution and Explanation
Mass: \(M=\int_0^L\left(a+\frac{bx}{L}\right)dx = aL+\frac{bL}{2}\).
First moment: \(\int_0^L x\left(a+\frac{bx}{L}\right)dx=\frac{aL^2}{2}+\frac{bL^2}{3}\).
\(x_{cm}=\frac{L\left(\frac a2+\frac b3\right)}{a+\frac b2}=\frac{7L}{12}\).
\(12\left(\frac a2+\frac b3\right)=7\left(a+\frac b2\right)\Rightarrow 6a+4b=7a+\frac{7b}{2}\Rightarrow \frac b2=a\).
So \(b=2a\), i.e. \(2a=b\).
Answer: (3) \(2a=b\)
First moment: \(\int_0^L x\left(a+\frac{bx}{L}\right)dx=\frac{aL^2}{2}+\frac{bL^2}{3}\).
\(x_{cm}=\frac{L\left(\frac a2+\frac b3\right)}{a+\frac b2}=\frac{7L}{12}\).
\(12\left(\frac a2+\frac b3\right)=7\left(a+\frac b2\right)\Rightarrow 6a+4b=7a+\frac{7b}{2}\Rightarrow \frac b2=a\).
So \(b=2a\), i.e. \(2a=b\).
Answer: (3) \(2a=b\)
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