Question

question for genius , complete solution not answer .

Answer 1

hi.

here is your answer

plz refer to the attachment

(B) IS CORRECT ANSWER

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Answer 2

hey mate

here is your answer

Since, the mass of the rod is 4 m and the length is 4 l

so, mass of AB = BO = OC = CD = m

and, length of AB = BO = OC = CD = l

We know, Moment of inertia of a rod about its end = (ml^2)/3

Moment of inertia of AB bout B (I 1) = (ml^2)/3

Moment of inertia of BO about O (I 2) = (ml^2)/3

Moment of inertia of OC about O (I 3)= (ml^2)/3

Moment of inertia of CD about C (I 4)= (ml^2)/3

Now, from parallel axis theorem,

Moment of inertia of AB about O (I 5)

= I 1 + ml^2 ————(parallel axis theorem)

= (ml^2)/3 + ml^2

= (4 ml^2)/3

Similarly, Moment of inertia of CD about O (I 6)

= (4 ml^2)/3

So, moment of inertia of the rod about O

= I 2 + I 3 + I 5 + I 6

= ( ml^2)/3 + ( ml^2)/3 + (4 ml^2)/3 + (4 ml^2)/3

= (10 ml^2)/3

2.1k Views ·

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Mass of the rod will be be equally distributed so

Mass of the section (AB = BO = OC = CD = M)

Length is given (AB = BO = OC = CD = l)

Here i will be using “M.I as moment of inertia”

M.I of BO through O = ml²/3 , similary for OC = ml²/3

For section AB and CD we use perpendicular axis theorem

through B , M.I = ml²/3, shifting it perpendiculary to O,

M.I = ml²/3 + ml² = 4ml²/3

Similarly , M.I of CD through O = 4ml²/3

So M.I OF AB = CD = 4ml²/3

Summing M.I of of AB, BO, OC, CD we get our result

M.I = ml²/3 + ml²/3 + 4ml²/3 + 4ml²/3 = 10ml²/3 ANS.

322 Views · ·

Consider the thin rod be divided into 4 parts 2 on the left side and 2 on the right side. Mass of each part =4M/4= M

Length of each portion =4l/4=l

Considering the extreme left portion

M I about an axis through CG perpendicular to the plane of paper =Ml^2/12

Distance of CG from axis through O=

= √{l² +(l)² /4}

Hence MI about the axis through O

=Ml²/12+M[ √{l²+(l/2)²}] ²=Ml²/12+5Ml²/4=4Ml²/3

MI of second left portion about O=Ml²/3

Hence MI of left part about O =4Ml²/3+Ml²/3

=5Ml²/3

MI of the whole rod About an axis through O

Perpendicular to the plane of paper =2*(5Ml²/3)

=10Ml²/3

Hence (b)

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