Question

A wire of length L and mass M is bent in the form of a rectangle ABCD with AB/BC=2.The Moment of inertia of the wireframe about the side BC is?

Answer 1

solve krke batata hun.

Answer 2

haha abhi hua nhi. thoda confuse ho gya hun. :/

Answer 3

^_^

Answer 4

<3

Answer 5

hmm. 2*m/3(l/6)^2 + m/6(l/3)^2 .. solve kr le. aasan to tha :/

Answer 6

forget him parth, he is mine! :|

Answer 7

\[\LARGE \frac{2m \times 36 }{3L^2}+\frac{9 \times m}{6L^{2}}\]
??

Answer 8

ravi raj kon hai? :O
ye kya kr rha hai? :O
2ml^2/(36*3) + ml^2(6*9)

Answer 9

\[\LARGE \frac{3ML^2}{54}????\]

Answer 10

@yrelhan4 ?:O

Answer 11

ml^2/27 ??

Answer 12

Kaafi easy tha khair.

Answer 13

thik hai?

Answer 14

No :P

Answer 15

lol

Answer 16

oh teri.... kya galti hai?

Answer 17

IDK,answer is weird

Answer 18

\[\LARGE \frac{11}{252}ML^2||\frac{8}{203}ML^2||\frac{5}{36}ML^2||\frac{7}{162}ML^2\]
options^^

Answer 19

\[I_{bc}=\frac{mh^2}{3}+\frac{mw^2}{12} \\ \\ I_{bc}=m(\frac{h^2}{3}+\frac{4h^4}{12}) \\ \\ I_{bc}=m(\frac{h^2}{3}+\frac{h^4}{4})\]
\[L=2w+2h \\ \\ h=\frac{L-2w}{2}\]
hmm

Answer 20

lol

Answer 21

\[I_{bc}=\frac{mh^2}{3}+\frac{mw^2}{12} \\ \\ I_{bc}=m(\frac{h^2}{3}+\frac{4h^2}{12}) \\ \\ I_{bc}=m(\frac{h^2}{3}+\frac{h^2}{3}) \\ \\ I_{bc}=m(\frac{2h^2}{3})\]

\[L=2w+2h \\ \\ h=\frac{L-2w}{2} \\ \\ \text{from}~ w=2h \\ \\ h=\frac{L}{6}\]
\[I=\left( \left(\frac{2}{3} \left(\frac{L}{6}\right)^2\right)\right) m\]
\[I=\frac{mL^2 }{54}\]

Answer 22

Not in options..:(

Answer 23

I(AB) about center of BC = (m/3)(l/3)^2 /12 + (m/3)(l/6)^2 + (m/3)(l/12)^2
I(DC) about center of BC = same as of AB
I(AD) about center of BC = (m/6)(l/6^2) /12 + (m/6)(l/3)^2
I(BC) about center = (m/6)(l/6)^2 /12

I net about BC perp. to the plane = sum of all = 7/144 ml^2

so about BC = 7/288 ml^2

hmm, where did I go wrong ?