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JEE Main 20206 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
6 September
Shift
Evening Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2020 · 6 September · Evening Shift · Physics · Rotational Motion

The linear mass density of a thin rod AB of length L varies from A...

The linear mass density of a thin rod ABAB of length LL varies from AA to BB as λ(x)=λ0(1+xL)\lambda\left(x\right) = {\lambda}_{0}\left(1 + \dfrac{x}{L}\right), where xx is the distance from AA. If MM is the mass of the rod then its moment of inertia about an axis passing through AA and perpendicular to the rod :The linear mass density of a thin rod AB of length L varies from A to B as λ(x) = λ _ 0 (1 + (x)/(L)) , where x is the

Options

  1. A512ML2\dfrac{5}{12}{ML}^{2}
  2. B718ML2\dfrac{7}{18}{ML}^{2}
  3. C25ML2\dfrac{2}{5}{ML}^{2}
  4. D37ML2\dfrac{3}{7}{ML}^{2}

Correct answer

B. 718ML2\dfrac{7}{18}{ML}^{2}

Step-by-step solution

The linear mass density of a thin rod AB of length L varies from A to B as λ(x) = λ _ 0 (1 + (x)/(L)) , where x is the

dm=λ.dvdm = \lambda . dv

dI=dm.x2dI = dm . {x}^{2}

=λ0(1+xL).x2dx= {\lambda}_{0}\left(1 + \dfrac{x}{L}\right) . {x}^{2}dx

I=λ0∫01(x2+x3L)dxI = {\lambda}_{0}{∫}_{0}^{1}\left({x}^{2} + \dfrac{{x}^{3}}{L}\right)dx

=λ0∣x33+x44L∣0L= {\lambda}_{0}{\left|\dfrac{{x}^{3}}{3} + \dfrac{{x}_{4}}{4L}\right|}_{0}^{L}

=λ0[L33+L34]= {\lambda}_{0}\left[\dfrac{{L}^{3}}{3} + \dfrac{{L}_{3}}{4}\right]

I=7λ0L312I = \dfrac{7{\lambda}_{0}{L}^{3}}{12}

m=∫0Lλdx=3λ0L2m = \mathop{∫}_{0}^{L}\lambda dx = \dfrac{3{\lambda}_{0}L}{2}

Im=7L212×32=7L218\dfrac{I}{m} = \dfrac{7{L}^{2}}{12 \times \dfrac{3}{2}} = \dfrac{7{L}^{2}}{18}

⇒⇒ I=718mL2I = \dfrac{7}{18}{mL}^{2}

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