Messages posted by kartikAIR1

its r square , where r is radius of circle

So what? mi of semicircular disc about perpendicular axis passing through is mrsq. by a2. this is ridiculous that people r supporting wrong ans..

pls help me out..

..cause its a non impulsive force.

EXPANSION is expanded 1¹⁰⁰2⁹⁹3⁹⁸4⁹⁷5⁹⁶6⁹⁵..........98³99²100^1.

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

2⁹⁹4⁹⁷6⁹⁵8⁹³......98³100^1.

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. 2⁵⁰is 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

PRAHALD IS RIGHT..

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

centripital acc= w²r
A = (2*pi*n)²r

CORRECT ANSWER

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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.

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.

WATER(L, O.9 atm, 373K)------>WATER vapour(g, 1atm, 373K)IS above conversion spontaneous?

ANSWER a>1.

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..

above comment for ARJUN VIRMANI post

THE STUPID ANSWER. TOTALLY WRONG.

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

haha...

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.

you get the answer

ANSWER Is MR2{(9pi-16)/6pi}

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