Messages posted by akshay_4691

x²+y²=L²/4 where L is the length of the rod.........

consider a function which is strictly decreasing in a given interval...........such as f(x)=cosx

in [0,pi] its range is [-1,1] its domain is [0,pi]

i.e [cos inverse(1),cos inverse(-1)]

well, im absolutely shocked that your reason for preparing for iits is just because you want to work in america!
i've given jee 2009 and irrespective of the rank i get, i can surely say that these two years have been the most enjoyable of my life....

secondly, the purpose of existence of IITs is not to serve the american economy....!

personally, i find it disgusting that several students who study in IITs availing subsidized educational facilities eventually go abroad and assist foreign economies...............

well, i personally feel that desire to live in america is no reason to study for JEE.................u may as well enter some private college paying donation.................

on a related topic, id like to know if i stand a chance of getting a good rank with 48-chem,78-maths,68-phy total 194,.........will i get selected at all?

tangential velocity of the points on the rim of the disc after one second=, where R is the radius of the disc and alpha the angular acceleration (given)...................

...................now, tangential accn.=(constant).

now ,as the disc is rotating, the points on the rim of the disc are executing circular motion in a circle of radius R.............therefore, they must be acted upon by some .centripetal force, which produces some centripetal acceleration, given by;

A=V^2/R, where V=tangential velocity of the points on the rim...................................................

...................so,

also,the tangential acceleration is given by:

the total acceleration is the vector sum of these two accelerations;

A(result.)=sqrt(A^2+R^2alpha^2)

it comes out to be approximately 1.04m/s^2

pl correct if wrong......................

think the question posted by hell is a rolle's theorem one.....

if f(a)=f(b)= for a function continuous and differentiable in (a,b) and [a,b] respectively...then there is atleast one value of x between a and b for which derivative of function is zero.

now, f(x)=ax^+bx+c

now, we integrate the function

so we have

say g(x)=ax^3/3 +bx^2/2+cx+d (d is an arbitrary constant)

now, for f to have a root bet 1 and 2, g(2)=g(1)

ie

a/3 +b/2+c+d=2c+8a/3+2b+d

so

a+3b/2+3c=6c+8a+6b

2a+3b+6c=12c+16a+12b

14a+9b+6c=0

Sec(270+20)+[1/rt(3)].cosec(270-20) (note)

= cosec20-1/sqrt(3).sec20

=cosec20-[1/rt3]cosec70

=1/sin20-1/rt(3)sin70

={root(3)sin70-sin20}/rt(3)sin20.sin70

=multiply and divide by ½. So that

the expression becomes.

Sin(60-20)/sin60.sin70.sin20

=2sin20.cos20/sin60.sin70.sin20=2/sin60(since cos20=sin70)

which is 4/rt(3)

approximately 2.31....as the person above has said.........

please correct if wrong!

x-y=sqrt(xy)

we have

x-y/sqrt(xy)=1

so?

sqrt(x/y)-sqrt(y/x)=1

let sqrt(x/y)=t

therefore.

t-1/t=1

so

t^2-t-1=0

solve for t

the roots will bet={1+sqrt(5)}/2

and 1-sqrt(5)/2

now square

1+5+2sqrt(5)/4

= 3+-sqrt(5)/2 considering the negative root

anyone got a shorter method?

A shortcut to finding the hybridization state is;

Let X=no. of sigma bonds+no. of lone pairs

if X=2, then sp

if X=3, then sp2

and so on....

for instance.....

hybridization of carbon in ethyne is sp

Carbons form a triple bond with each other,and a single bond with a hydrogen each.....

therefore they've formed 2 sigma bonds with zero lone pairs as all their electrons are used in bonding.....

so,

X=2+0

so sp hybridization...

this could prove quite useful......

85. Akshay...

1.Velocity of an electron in a Bohr orbit

=2.18 X Z/n X 10^8 cm/s

where:Z=atomic number

n=no. of orbit

2. Radius of a bohr atom:

0.529 X n^2/Z Angstrom units

Where n and Z have the same meaning as in the first case...

C2H5

|

C2H5-C-C2H5

|

(C2H5)3C-C-CH3

|

C2H5-C-C2H5

|

C2H5

Here,

The parent chain is in bold

The complex alkyl substituent is in maroon

The other substituents are in.......shades of blue......

@balganesh....

im afraid your structure is wrong......tune CH2 aur CH3 ko ulta kar diya hai....lol..

isn't the answer 3,3,5,5 tetraethyl-4-(1',1'diethyl propyl)-4-methyl heptane?

do correct if wrong, please.....

because, when you consider the situation in which no two girls are together

..........the gap method is preferred...........

total no of arrangements:10!

now when no two girls are together...........

the boys can be arranged in 5! ways.........

now.consider the below arrangement

GBGBGBGBGBG

the girls have to be arranged in six gaps

no of permutations=6P5=6!

so..no. of arrangements of this kind possible are=6!5!

therefore,

total nof permutations in which atleast two girls are together=10!-5!6!

do correct if wrong,please

@joyfrancis.

\

well, isn't the answer 10!-(5!.6!)?

Navi Mumbai,

Maharashtra.............

@urvashishami

i beg to differ.............the work done would have been zero if the boy was not displaced..........work is zero if the displacement of the body on which the force was applied were zero......isn't it?

here's what i think...................when the boy climbs the ladder, he exerts a downward force on the ladder........the ladder exerts an equal and opposite force on the boy...........now, when he climbs up, there are two forces performing work......gravity(Wg) and the ladder.(Wl)......

apply the work energy theorem

between the ground and the top of the ladder..........

we have

Wg+Wl=0(as initial KE=final KE=0)

now calculate work done by gravity, which is negative..........

mass=20

height=8

g=10

theta=180 degree

so

Wg=20.8.10.cos(180)

= -1600J

so

Wl=-(Wg)

= +1600 J

i don't think the answer is zero.........but then again, how can hcv sir be wrong>?

this is also observed in case of frictional force........instance; when a body moves on a rough surface, accleration due to friction is always opposite to direction of motion........... .....isn't it so?