in que 1) is the rod hinged about ny point

no.......

just apply 'BODMAS' rule properly , u get the unknown digit as 0.21

perpendicular axis theorem is valid for 2 Dimentional objects onle

it must not necessarly be only x-y plane, but can also be y-z, or x-z planes

yes, in theory coefficient of friction can vary from zero to infinity  !!!!!!!!!!!!

but in practical, it doesnot vary so high, but varies from 0.5 to 2

(1) vertical dir......

initial vel  =  upward

final vel. =  downward

change in momentum =  downward.....

(2)hor. dir

change in momentum = 0 (no change in vel. in horizontal dir.)

rate if useful...............

if initial vel. of projection is same as that in case 1st,then in this case also KE  = k,as KE is  a scalar quantity

vel. with which ball hits the floor =  downward

vel. of ball when it just leaves the ground contact =  upward

impulse = change in momentum = force*time

let mass be 'm'

m(v-(-u)) = mat  (a=acc. at that instant)

v+u = at  (taking upward dir. positive)

 UPWARD......

A HUGE ACCELERATION

SO ANS. IS   'D'

PLEASE RATE IF USEFUL.................

let initial speed be 'u'

max. height of 1st ball =  

max ht. of 2nd ball =

as both r identical, ratio of PE at heighest point will be ratio of their max. height,

by doing D<0 we get,

(n+6)(n-4)>0 & (n-2)>0

we cannot comment on least value of 'n' as it is close to (4+)

but least  integral value of 'n' would be 5

inclined at an angle taninverse 6 with x axis.........

hey shubham,

edit all ur post & write the equation through equation editor,

it is hard to identify which term is divided or multiplied by which one

@ ABHI

u wrote in last line " it attains that maximum value..when applied that much force..!!!! "

so "that maximum value" is given a special name called LIMITING  FRICTION.and it acts only at a perticular time when "that much force " is applied...............so they r not same all the time.......

i dont think static & limiting friction are same.limiting friction is maximum value of static friction & static friction is a self adjustible force.

so ans. must only be limiting friction

friction will act up the incline plane on cyllinder........

we can also think in another way,

when we release the cylinder from rest , the component of wt. of cyllinder acts on it through CM , which tries to

increase its velocity,

now for pure rolling, v = (omega)*r,so a forse causing torque is also needed in anticlock wise direction, which

can increase the angular velocity of cyllinder, which can only be done by frictional force, if it acts up the incline

PLEASE RATE IF USEFUL..........

f(x) is discontinuous & nonderivable for all integral value of x

.

as x,y,z are in HP  yhen 1/x , 1/y , 1/z will be in AP

let the first term & common diff on the AP series be a,d resp,

then

1/x = a+(p-1)d .....(i)

1/y = a+(q-1)d.......(ii)

1/z = a+(r-1)d........(iii)

(1/y) - (1/z) = (q-r)d => (q-r)d/x = (1/xy)-(1/xz) .......(a)

simlarly,

(r-p)d/y = (1/yz)-(1xz)......(b)

(p-q)d/z = (1/xz) - (1/yz)........(c)

adding (a),(b),(c) we get req. result

please rate if useful................................

i think there is a misprint,

(root4 + ax) must be   root (4+ax)

please check it.....