JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion

A thin bar of length \(L\) has a mass per unit length \(\lambda \), that increases linearly with distance from one end. If its total mass is \(M\) and its mass per unit length at the lighter end is \(\lambda_0\), then the distance of the centre of mass from the lighter end is

A \(\frac{L}{2} - \frac{{{\lambda _0}{L^2}}}{{4M}}\)
B \(\frac{L}{3} + \frac{{{\lambda _0}{L^2}}}{{8M}}\)
C \(\frac{L}{3} + \frac{{{\lambda _0}{L^2}}}{{4M}}\)
D \(\frac{{2L}}{3} - \frac{{{\lambda _0}{L^2}}}{{6M}}\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{L}{3} + \frac{{{\lambda _0}{L^2}}}{{4M}}\)

Step-by-step Solution

Detailed explanation

Mass per unit lengh \(=\lambda_{0}+\mathrm{kx}\) \(\mathrm{M}=\int_{0}^{\mathrm{L}}\left(\lambda_{0}+\mathrm{kx}\right) \mathrm{dx}\) \(M=\lambda_{0} L+\frac{K \times L^{2}}{2}\) \(\frac{2 \mathrm{M}-\lambda_{0} \mathrm{L}}{\mathrm{L}^{2}}=\mathrm{K}\)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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