Two uniform, thin identical rods each of mass M and length L are joined at middle so as to form a cross as shown. The moment of inertia of the cross about a bisector line E F (in the plane of cross) is

Question

Two uniform, thin identical rods each of mass M and length L are joined at middle so as to form a cross as shown. The moment of inertia of the cross about a bisector line E F (in the plane of cross) is (A) (M L2/6) (B) (M L2/4) (C) (M L2/12) (D) (M L2/3).

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/9fd32f415d4801c650fdac28faa37312-.png" /> We have to find I1 ? I1=I2 ldots (1) I0=(M L2/12)+(M L2/12)=(M L2/6) now I1+I2=I0 ldots (2) From (1) and (2), I1=(I0/2)=(M L2/12)

Q. Two uniform, thin identical rods each of mass $M$ and length $L$ are joined at middle so as to form a cross as shown. The moment of inertia of the cross about a bisector line $EF$ (in the plane of cross) is

$\frac{M L ^{2}}{6}$

$\frac{M L ^{2}}{4}$

$\frac{M L ^{2}}{12}$

$\frac{M L ^{2}}{3}$ .

We have to find $I_{1} ?$
$I_{1} = I_{2} \dots$ (1)
$I_{0} = \frac{M L ^{2}}{12} + \frac{M L ^{2}}{12} = \frac{M L ^{2}}{6}$
now $I_{1} + I_{2} = I_{0} \dots$ (2)
From (1) and (2), $I_{1} = \frac{I _{0}}{2} = \frac{M L ^{2}}{12}$

Q. Two uniform, thin identical rods each of mass M and length L are joined at middle so as to form a cross as shown. The moment of inertia of the cross about a bisector line EF (in the plane of cross) is

6ML2​

4ML2​

12ML2​

3ML2​.