The liner mass density of a thin rod AB of length L varies from A to B as $\lambda(x)=\lambda_{0}\left(1+\frac{x}{L}\right)$, where x is the distance from A. If M is the mass of the rod hen its moment of inertia about an axis passing through A and perpendicular to the rod is:

(B) $\frac{7}{18} \mathrm{ML}^{2}$

(A) $\frac{5}{12} \mathrm{ML}^{2}$

(D) $\frac{3}{7} \mathrm{ML}^{2}$

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Rotational Mechanics

The liner mass density of a thin rod AB of length L varies from A to B as

λ (x) = λ_{0} (1 + \frac{x}{L})

, where x is the distance from A. If M is the mass of the rod hen its moment of inertia about an axis passing through A and perpendicular to the rod is:

(A) $\frac{5}{12} ML^{2}$
(B) $\frac{7}{18} ML^{2}$ ✓ Correct
(C) $\frac{2}{5} M L^{2}$
(D) $\frac{3}{7} ML^{2}$

Answer: (B)

Solution

First the total mass:

M = \int_{0}^{L} λ_{0} (1 + \frac{x}{L}) d x = λ_{0} (L + \frac{L}{2}) = \frac{3 λ _{0} L}{2}

, so

λ_{0} = \frac{2 M}{3 L}

. The moment of inertia about the end

A

I = \int_{0}^{L} x^{2} λ_{0} (1 + \frac{x}{L}) d x = λ_{0} (\frac{L ^{3}}{3} + \frac{L ^{3}}{4}) = \frac{7 λ _{0} L ^{3}}{12}

. Substituting

λ_{0} = \frac{2 M}{3 L}

gives

I = \frac{7}{18} M L^{2}

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Rotational Mechanics

Solution

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