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jadavpur would be better .since it is in consideration for iit status,a good stream there would definitely do good

sorry i don't have the diagram.

let r be the radius of sphere

the friction would rotate the sphere

before collision angular momentum due to centre of mass and the friction would oppose each other

before collision = Mvr - Iw = Mvr - ( 2/5)Mvr = (3/5)Mvr

let velocity after collision be u

conserving the angular momentum before and after collision

(3/5)Mvr = ( 2/5)Mur + Mur

u = 3v/7

Thanks Vinu .i have realised my mistake

Mr . rony kindly edit your post

lim {1-(m/n)}^n = e^(-m)
n->i

i means infinity

lim n!/((n-x)!*x!) (m/n)^x*{1-(m/n)}^(n-x)
n->i

lim n!/((n-x)!*x!)* [(m/n)^x*{1-(m/n)}^n )/(1 - ( m/n))^x
n->i

usind

lim {1-(m/n)}^n = e^(-m)
n->i

when n->i { [n!/(n-x)!]/n^x} becomes 1

therefore we get {e^(-m)*m^x}/x!

if a line makes angle a , b ,c with x ,y ,z axis then

(cosa)^2 + (cos b)^2 +(cos c)^2 =1

given cos a =cos b= cos c

therefore cos a = +- [1/ sqrt(3)]

components of vector A

A * ( cos a I + cos b J +cos c K )

I ,J ,K are unit vectors

p = [ (x^3) +2]/x

for maximum value of p dp/dx = 0

3 (x^3) - (x^3) -2 =0

which gives x= 1

therefore maximum value of p= 3

f(x) = x^3 -px +2

for real roots p should be greater than 0

number of values of p 1,2,3

hence number of equations is 3

treat



1-nitro propane is, CH ₃-CH₂ -CH₂ -NO₂ and, 2-nitro propane is, CH₃ -CH-CH₃ I NO₂

with Sn /HCL A WHICH WILL CONVERT IT TO AMINES . THEN WE HAVE TO TREAT WITH DILUTE NITROUS ACID WHICH CONVERT NH2 GROUP TO DIAZONIUM SALT WHICH WHEN TREATED WITH WATER WILL FORM OH GROUP. THEN WE PERFORM VICTOR MEYER'S TEST TO DIFFERENTIATE 1 DEGREE AND 2 DEGREE ALCOHOL

hi AJAYAVSS

considering that the disc is removed from the right end

the centre of mass of the complete disc is at the centre of disc

{ if the disc is removed from the right end the centre of mass of the remaining portion would move towards the left of origin }

Let the centre of mass of the remaining portion move towards the left of origin by a distance 'x '

distance between the centre of removed portion and centre of disc
= 2R -R = R

let the mass per unit area be 'a '

mass of remaining portion = (pie)*a*[ (2R)^2 - R^2]
= (pie)*a*3(R^2)

mass of removed portion= (pie)*a*(R^2)

Taking moments about the origin between the remaining part and removed part

      (pie)*a*3(R^2) * x = (pie)*a*(R^2) * R

               x = R /3

the centre of mass of remaining portion moves towards the left of origin by a distance R/3

[ if the disc is removed from the left end the centre of mass of the remaining portion would move towards the right of origin ,but the distance would remain same ]

numbers in G.P can be taken as a , ar ,a(r^2),where a is the 1st term and r is the common ratio

there are 100 numbers

CASE 1:
let us start by taking the r =2 ,therefore r^2 = 4

possible numbers from which choice of a can be made= 100/4 = 25

CASE 2:
let the next value of r =3 ,therefore r^2 = 9

possible numbers from which choice of a can be made=100/9 =11
{ we take only the maximum number divisible i.e the quotients}

CASE 3:
let the next value of r =4 ,therefore r^2 = 16

possible numbers from which choice of a can be made=100/16 =6

CASE 4:
let the next value of r =5 ,therefore r^2 = 25

possible numbers from which choice of a can be made=100/25 =4

CASE 5:

let the next value of r =6 AND r=7,therefore r^2 = 36 and 49 respectively

possible numbers from which choice of a can be made=100/36 +100/49= 4
{ takin only the quotient of the division}

CASE 6 :

when r=8, 9,10 the maximum numbers divisible is 1 each .since r^2=(64,81,100)

so total number is 1+1+1=3

WE DO NOT TAKE ANY FURTHER VALUES OF r SINCE IT WOULD EXIT THE LIMIT 0F 100

n(E) = event that the numbers in GP is the sum of the abve cases
= 25 + 11 +6 +4 +4+ 3 =53

n(S)= sample space of choosing 3 numbers is 100C3

probability = n(E)/n(S)

= 53/ 100C3

= 53/161700

refractive index (n) = apparent depth/real depth

if displacement with respect to unit time is taken = (9/n)

actual velocity= (9/n) - 3

n is the refractive index of water

_{[ 0]}

^[pi/2]sin^mx.cosⁿx.dx = {G [( m+1)/2] *G [(n+1)/2] } / 2*G[ ( m+n+2)/2]

denoting
beta function by B ALSO KNOWN AS 1ST EULERIAN INTEGRAL
gamma function by G ALSO KNOWN AS 2ND EULERIAN INTEGRAL

B(n,m)= INTEGRATION OF x^(n-1)*[(1-x)^(m-1)] WITH LIMITS FROM 0 TO 1
G(k)= INTEGRATION OF e^(-x)*[(x)^(k-1)] WITH LIMITS FROM 0 TO INFINITY

WHERE n,m,k are all REAL NO.S greater than zero

STANDARD RESULTS

1) G(n+1)=nG(n)
IF n IS A +VE INTEGER THEN G(n+1)=n!
2) G(1)=1
3) B(m,n)=B(n,m)={G(m)G(n)}/G(m+n)
4) G(m)G(1-m)=(PIE)/SIN(mPIE) WITH (0<m<1)

USING THIS STANDARD RESULTS MANY CALCULATIONS CAN BE MADE EASY

considering a number N

N = (p^a)* (q^b)*(r^c)*(s^d)

where p ,q,r,s are odd numbers and a,b,c,d are integers

now for a number N it's divisors are = (1+a)*(1+b)*(1+c)*(1+d)

APPLYING THE METHOD OF CONTINUOS DIVISION

let us start with 1st prime number 2

divide 1000 by 2 , quotient = 500

next prime number is 3 product of 2*3 =6

divide 1000 by 2*3 , quotient = 166

next prime number is 5 product of 2*3*5 =30

divide 1000 by 2*3*5 , quotient = 33

next prime number is 7 product of 2*3*5*7 =210

divide 1000 by 2*3*5*7 , quotient = 4

we cannot further divide as of 2*3*5*7*11 exceeds 1000

we have got the last quotient = 4 and last divisor =210

therefore the largest number divisible is =210*4=840

a = 111111...............81 times

a = (1/9)9999999.............81 tmes

a = (10^81 - 1)/ 9

a = (10^81 -1)/ (10 -1)

a is the sum of G.P whose common ratio is 10 consisting of 81 terms

hence a is divisible by 9

the correct option is a)

i may be wrong

hit_ur_hit whatever i have written is directly picked up from

HALL AND KNIGHT ALGEBRA
CHAPTER - THEORY OF EQUATIONS .CONTEXT NUMBER 557