IIT JEE , AIEEE Forum - Physics , Mechanics

waterdemon

Rakesh u are wrong again

plz be calm and think over it once again.

waterdemon

Dude if the question sucks so much then why don't u solve it

Or are u scared of solving

Dude if u don't have the guts to solve the question u don't have any right to Show disrespect to my question

Solve it or just shut up !!!!!!!!!!!!!!!!!!!!!!!

let others solve it

And I am dam sure that u won't be able to solve it.

Such people like u are just skeptics who do all this funking to

Try to look cool but you make a fu of urself only.

waterdemon

The answers will be released tomorrow !!

plZ solve it B4 I solve it

Come on dudes & Gals !!!!!!!!!!!1

waterdemon

Good night everyone meet u tomorrow with the solution

Before I put on the solution

On the notice board try putting ur solutions .

The one who gets it right Gets @ salutes from me.

Cheers !!!!!!!!!@@@@@@@@@!!!!!!!!!!!!!!!!!!!!!!!! :-)

rlukked

can u please give the solution of these

MAYANK1989

WELL I CAN'T GET THE HECK OF IT.

jattpunjabi

i dont think wasting your time on these questions will get u to iit

waterdemon

Helo Jatt punjabi

you aaren't wasting ur time

If u think so then u have already lost the race to Get into IIT

COz such questions help u get ur concepts get better

And are nothing but practice for you However I am putting on the solution I some minutes.

waterdemon

So here is the answer Well it was a difficult one and I tried putting it up for

The GOIITians

Here

= Angle between the line (of R inthe fig. ) and the horizontal '' b ''

First

The pressure that will be acting on the potion aprt from the door will be

dP = P ( a - Rsin

)

As the length of the door will be (a - R Sin

)

dF_x = Pcg(a - R Sin

)R.d

* Sin

dF_y = Pcg (a- Rsin

)Rd

* Cos

Knowing that T = F*R

So the below relations are as per the horizontal and vertical components

And

is the density of the liquid that is inside the Cube.

dT_Fx=

cg ( a -Rin

) d

.Sin

. (R - RCos

)

dT_Fy =

cg (a - Rsin

)d

.Cos

.(RSin

)

Now resolving along the line of resultant "F"

dT = dT_Fx+ dT_Fy

Where T is the torque

=

cg(a - Rsin

)Rd

.Sin. (R - Rcos

) +

cg(a -Rsin

)Rd

.Cos

.Rsin

=

cgR² [ (a-Rsin

)Sin

.(1-Cos

)d

+ (a-Rsin

)Cos

.Sin

.d

]

=

cgR² [ (aSin

- RSin²

)(1 - Cos

)d

+ (aCos

- R.Sin

.cos

)(Sin

)d

=

cgR² [ aSin

- aSin

.Cos

- RSin²

+ RSin²

.Cos

+ aCos

.Sin

- RCos

.Sin²

] d

=

cgR²[ aSin

- RSin²

] d

=

cgR² [ -a [cos

] { from 0 to pi/2 } - R(1 - Cos2

/ 2 ) ] d

=

cgR² [ a - R ( d

/ 2 - Cos2

/ 2 .d

]

=

cgR² [a - R

/4 - (Sin2

/4).d

Sin 2

/4 = 0 as it is from 0 to

/2

=

cgR²[a - R

/4 ]

Therefore we will get

Where F is the force to be applied on point T to avoid leakage of liquid

F*R = T

F*R =

cgR²[a - R

/4]

F =

cgR[a - R

/4]

Hence the answer

plZ rate me for my efforts

Cheers !!!!!!!!!!!!!!!!!!!!!!!!!!

waterdemon

Come on atleast rate me for my efforts yaar !!!!!!!!!!!11

waterdemon

Have a look at this Question it is really good for all the GOIITians .

And plZ rate me.

waterdemon

I cannot understand that why are u all neglecting such a good question

OK don't rate me but plZ this is for ur benefit only.

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