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QJEEM20Ph28dd: JEE Main 2020 Physics - Class 11 JEEM20Ph28dd Solution: The linear mass density of a thin rod AB of length L varies…

The linear mass density of a thin rod AB of length L varies from A to B as
λ(x)=λ0(1+xL)\lambda \left( x \right) = {\lambda _0}\left( {1 + {x \over L}} \right) , where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is :

Multiple Choice Options

  • A 25ML2{2 \over 5}M{L^2}
  • B 512ML2{5 \over {12}}M{L^2}
  • C 718ML2{7 \over {18}}M{L^2}
  • D 37ML2{3 \over 7}M{L^2}

Step-by-Step Verified Solution

JEE Main 2020 (Online) 6th September Evening Slot Physics - Rotational Motion Question 117 English Explanation

dm = λ\lambda dx

= λ0(1+xL){\lambda _0}\left( {1 + {x \over L}} \right) dx

Integrating both side, we get

∫0Mdm=∫0Lλ0(1+xL)dx\int\limits_0^M {dm} = \int\limits_0^L {{\lambda _0}\left( {1 + {x \over L}} \right)} dx

⇒\Rightarrow M = λ0L+λ0L22L{\lambda _0}L + {{{\lambda _0}{L^2}} \over {2L}} = 3λ0L2{{3{\lambda _0}L} \over 2}

⇒\Rightarrow λ0=2M3L{\lambda _0} = {{2M} \over {3L}} .....(1)

Moment of inertia of small part dx is

dI = dmx2

Integrating both side, we get

∫0IdI=∫0Ldmx2\int\limits_0^I {dI} = \int\limits_0^L {dm{x^2}}

⇒\Rightarrow I = ∫0Lλ0(1+xL)dx x2\int\limits_0^L {{\lambda _0}\left( {1 + {x \over L}} \right)} dx\,{x^2}

= λ0∫0L(x2+x3L)dx{\lambda _0}\int\limits_0^L {\left( {{x^2} + {{{x^3}} \over L}} \right)} dx

= λ0[L33+L34]{\lambda _0}\left[ {{{{L^3}} \over 3} + {{{L^3}} \over 4}} \right]

= 7L3λ012{{7{L^3}{\lambda _0}} \over {12}}

Here by putting the value of λ\lambda 0 form (1)

I = 7L312(2M3L){{7{L^3}} \over {12}}\left( {{{2M} \over {3L}}} \right) = 718ML2{7 \over {18}}M{L^2}

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