JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2022
- Hard
- Numerical
The distance of centre of mass from end \(A\) of a one dimensional rod \(( AB )\) having mass density \(\rho=\rho_{0}\left(1-\frac{ x ^{2}}{ L ^{2}}\right)\,kg / m\) and length \(L\) (in meter) is \(\frac{3 L }{\alpha} m\). The value of \(\alpha\) is \(\ldots \ldots \ldots . .\) (where \(x\) is the distance form end \(A\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(8\)
Step-by-step Solution
Detailed explanation
\(dm =\lambda \cdot dx =\lambda_{0}\left(1-\frac{ x ^{2}}{\ell^{2}}\right)\)
\(X _{ cm }=\frac{\int xdm }{\int dm _{\ell}}\)
\(=\frac{\lambda_{0} \int_{0}^{\ell}\left( x -\frac{ x ^{3}}{\ell^{2}}\right) dx }{\int_{0}^{\ell} \lambda_{0}\left(1-\frac{ x ^{2}}{\ell^{2}}\right) dx }=\frac{\frac{\ell^{2}}{2}-\frac{\ell^{4}}{4 \ell^{2}}}{\ell-\frac{\ell^{3}}{3 \ell^{2}}}=\frac{3 \ell}{8}\)
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