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 kartikAIR1's messages in the community 1 2 3 ... 14 15 16 GO Go to Page...
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its r square , where r is radius of circle
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So what? mi of semicircular disc about perpendicular axis passing through is mrsq. by a2. this is ridiculous that people r supporting wrong ans..
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pls help me out..
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..cause its a non impulsive force.
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EXPANSION is expanded     1100299398497596695..........9839921001.

terms having even powers are perfect square so keep them aside....it becomes

299497695893......9831001.

now take single numbers from each powers to make powers even and consequently perfect squares.........

it becomes....

= (2.4.6.8....98.100)(298496694892.......982)

now second bracket becomes perfect square so keep it aside.......

we have..

=2.4.6.8...98.100

take 2 common from each term, it becomes,

=(250)(1.2.3.4.5.6.7.......49.50)

here it is. 250 is perfect square and second bracket is 50!

now original expression becomes,

=(perfect square)(50!)

so 50! should be removed for perfect square.

pls RATE IT

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PRAHALD IS RIGHT..
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http://www.4shared.com/get/yNsqsmY7/Halliday-Resnick-Walker_Fundam.html

its FREE

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........
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angular acceleration of particle w = 2*pi*n rad/sec.

• centripital acc= w2r
• A = (2*pi*n)2r

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visit jee website
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swimmer will use all his strength to cross the river i.e perpendicular to river flow.

as seen from ground he has velocity component 5 perpendicular to direc. of flow and 3 km/hr along river which is provided by river downstream.

time taken= 0.5/5 = 0.1 hour.

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I think outer rail will wear out faster..cause of centrifugal force acting outwards will pull train radially outwards....................2)2)tension cant be negative assuming inextensible string.
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WATER(L, O.9 atm, 373K)------>WATER vapour(g, 1atm, 373K)IS above conversion spontaneous?
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if v=0 implies ds/dt=0,

therefore ds=0,

in your equation it becomes indeterminant 0/0 form which doesnt exists.

=> v/ds = 0/0 value doesnt exists.

your equation is not valid at v=0.

...cause you hav derived dis equation by multiplying ds/ds with dv/dt which gives v(ds/dt).

a=dv/dt is always valid. but not yours..

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above comment for ARJUN VIRMANI post

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MI OF A DISC ABOUT CENTRE IS (MR^2)/2.

haha...

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CONSIDER a point O as centre of disc. centre of mass is at a distance 4r/3pi from point O.

now using parallel axis theorem,

let I1 b moment of inertia about P.

let I2 b MI about COM.

let I3 b MI about O.

using parallel axis theorem,

I3=I2 + m(4r/3pi)sq.

MRsq/2 = I2 + m(4r/3pi)sq............(1)

similarly,

I1= I2 + m(r-4r/3pi)sq. ..................(2),

now eliminate I2 from these 2 equations.

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