JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Rotational Mechanics

ABC is a plane lamina of the shape of an equilateral triangle.

D

E

are mid points of

AB

AC

and G is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through G and perpendicular to the plane ABC is

I_{0}

. If part ADE is removed, the moment of inertia of the remaining part about the same axis is

\frac{NI _{0}}{16}

where N is an integer. Value of N is

Answer: 11

Solution

Take the moment of inertia of the full equilateral lamina about the axis through its centroid

G

, perpendicular to its plane, as

I_{0}

. The small triangle

A D E

cut off by joining the midpoints is similar to the whole with half the side, so it has one quarter of the mass,

M /4

, and its moment of inertia about its own centroidal perpendicular axis scales as (mass)

\times

(side)

^{2}

, which is

(\frac{1}{4}) (\frac{1}{4}) = \frac{1}{16}

I_{0}

. Shifting that to the axis through

G

with the parallel-axis theorem and subtracting it from

I_{0}

leaves the remaining lamina with

I = \frac{11}{16} I_{0}

. Comparing with

\frac{N I _{0}}{16}

gives

N = 11

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Integer answer