hey dude congrats

i will suggest u to go to the official web site of FIITJEE

they have uploaded some " sucess check"

enter ur marks there and they will tell ur aprox rank

well wat were ur subject wise marks

feel free to ask mine

all the best

u seem to be new and hence ur knowledge is not enough to prove it 2 u(if u r not testing all of us)

but there r 3 ways by which i can help u

go for these if they seem to b helpful

1st

graph method

if u know graphs of both the curves i.e. y=sinx and y=x

then u can see that y=x is tangent to y= sinx at x=0

hence both of them have very close values at x=0

2nd

series mthod

sinx is just not a function but it is a series

sinx= x  - x^3/3! + x^5/5!....

so sinx/x= 1-x^2/3! + x^4/5!

hence wen we apply lim x tends to zero, sinx / x= 1

3rd

if u know L hopitals rule then u can prove it urself

all the best

see gal one person has already told one way to get to the ans easily

but for beginers i would sugest the folowing way

u hv got 4^87 / 3

= 4.4.4.    ..................4.4.4  87 times divided by 3

=4/3 4^86

=(1+ 1/3) 4^86

=4(1+1/3)4^85

=(4+4/3)4^85

=(4+1+1/3)4^85

=(5+1/3)4^85

=(20+4/3)4^84

=(21+1/3)4^84

...........................

..................

=(k+1/3) where k is a constant

hence remainder is 1

now try this one

its a bit tricky

5555^2222 + 2222^5555 when divided by seven gives remainder = ? ans is 0

hey cant we simply use the graphs

clearly limx-o sinx/x =1- ( say 1-h where h tends to 0) and also limx-0 tanx/x = 1+( say 1+h)

so we get int (1-h)/1+h = 0 [int is greatest integer function]

hey friends i think that none of options given is correct.according to me collision will not take place if driver of car A anyhow reduces his speed to Vb within the distance s so s>= Vb^2 - Va^2/ 2a

"it is a girl:s world"

remain standing

when earth stops attracting objects there is no other massive body which can provide net effective attractive force for the other objectsto cause their motion(acceleration)

7 cars.

since whole of the MNO2 is finished so

according to the equivalent concept

n1 z1 = n2 z2 no of equivalents of MNO2 = no of equivalents of MNO3

 where z1 and z2 are the respective z factors or the n factors

10/molecular mass of MNO2 x 4 =w/molecular mass of MNO3 x 6

on solving we get w = 7.87 gms

i have checked that the answer posted by the forrumn expert and my answer is correct but i think i have an easier method

first substitute the vallue of y=3^1/2 x to get a byquadratic eqn in x as

x^4+ 3^1/2 a x^3 + 3^1/2 b x^2 + (3^1/2 d +c) x+6 = 0

now, the product of the roots of the equation x1 x2 x3 x4 = 6

distance of A from origin ( x1^2 + y1^2)^1/2 now, since x1, y1 lie on y = 3^1/2 x so y1 = 3^1/2 x1

this implies OA = (x1^2 +3 x1^2)^1/2= 2x1

similarly OB = 2 x2 OC = 2 x3 OD = 2 x4

so OA.OB.OC.OD = 16 x1 x2 x3 x4 =16 x 6 = 96

did you get it!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

is the answer 96. if it is then tell me i will post the complete solution

T2>T1 if the masses of the blocks and the spring constants are equal

u might have got the time period of the first block as T1 = 2 pie (m/k)^1/2

in second one if the block is pulled by x distance from the equilibrium position then the pulley will be displaced by x/2 distance as a result an upward force of  -kx/2 is experienced by the pulley. now the tension in the string joining pulley with spring is twice the tension in the string joining the block with the ground. so the force on the block

F = -kx/4

so time period of the second block T2 = 2 pie (4m/k)^1/2 T2 = 2pie x 2 (m/k)^1/2

so T2 = 2T1

answer is 2 kg m^2

take the axis about which moment of inertia is to be found as one of the perpendicular axis and one axus perpndicular to it as another axis and treat them as the perpendicular axis and then apply perpendiculat axis theorem.another way of thinking is that the moment of inertia is dependant only on mass and distribution of mass about the axis which here is same for the axis along the centre in the plane of the ting and about the axis about which moment of inertia is to be found.................

the final time period will be root 2 times the initial time period that is

T- = root 2 T

for solution use t = 2 pie (m/k)^1/2

initially T = 2 pie (M/K)^1/2

now, final time period = 2 pie (2M/k)^1/2

as now the net deforming force is 2Mg = k x-

Tf = (2)^1/2 2 pie (M/k)^1/2

Tf = 2^1/2 Ti = 2^1/2 T

since E=Pt where t is the time for emmisivity of radiant energy.so

E = sigma A absillion T^4 t

where sigma is stefan;s constant

A is the area for radiation

absillion is emissivity

this implies E1/E2 = sigma 4 pie 4^2 (2000)^4 t /sigma 4 pie 1^2 (4000)^4 t

E1/E2= 16x 1/16 = 1

answer is c 2a^2/3a-b

take the equation of shm of the particle as x = A Cos wt-kx

now apply the given condition for the displacemen in first second

for diplacement in second second find total displacement in 2 secods and subtract it from the displacement in first second.

now, solve the obtained equations..........

u will get the answer yourself................

did u get it!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

according to adiabatic expansion,

PV^gamma = constant

using ideal gas equatio the above law can be reduced to

TV^gamma-1 = constant

now, applying the given conditions

T V^gamma-1 = T/2 [4 2^1/2 V]^gamma-1

this implies 2 = [2^5/2]^gamma-1

i.e. 2^1 = 2^(5/2(gamma-1))

this impies 1 = 5/2 (gamma-1)

on solving we get,

gamma = 7/5

equating gamma with 1+2/f, where f is the degree of freedom,

we get f = 5

q/4