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STD 11 - 6. system of particles and rotational motion — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 6. system of particles and rotational motion4 Marks
MCQ
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
    $\frac{L}{3} + \frac{{{\lambda _0}{L^2}}}{{8M}}$
  • ✓
    $\frac{L}{3} + \frac{{{\lambda _0}{L^2}}}{{4M}}$
  • D
    $\frac{{2L}}{3} - \frac{{{\lambda _0}{L^2}}}{{6M}}$
✓

Answer

Correct option: C.
$\frac{L}{3} + \frac{{{\lambda _0}{L^2}}}{{4M}}$
c
Mass per unit lengh $=\lambda_{0}+\mathrm{kx}$

$\mathrm{M}=\int_{0}^{\mathrm{L}}\left(\lambda_{0}+\mathrm{kx}\right) \mathrm{dx}$

$M=\lambda_{0} L+\frac{K \times L^{2}}{2}$

$\frac{2 \mathrm{M}-\lambda_{0} \mathrm{L}}{\mathrm{L}^{2}}=\mathrm{K}$

$\frac{2 \mathrm{M}}{\mathrm{L}^{2}}-\frac{\lambda_{0}}{\mathrm{L}}=\mathrm{K}$

$\frac{\int \mathrm{d} \mathrm{m}(\mathrm{r})}{\int \mathrm{d} \mathrm{m}}=\frac{\int(\lambda \mathrm{d} n) \mathrm{x}}{\mathrm{M}}=\frac{\int_{0}^{\mathrm{L}}\left(\lambda_{0} \mathrm{x}+\mathrm{k} \mathrm{x}^{2}\right) \mathrm{d} \mathrm{x}}{\mathrm{M}}$

$r_{c m}=\frac{\lambda_{0} L+\frac{k L^{2}}{2}}{M}$

substitute $'k"$

$r_{c m}=\frac{2 L}{3}-\frac{\lambda_{0} \ell^{2}}{6 M}$

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