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what lampard says is true.....

well the only thing we have to make sure is that frequency comes with time and k(wave vector ) with x

Variatio in density with temperature is as follows:-

mass m = constant

m= V*density at initial temperature

m= V(1+

T) * new tempreture

now equate them and get answer

new density = initial density/1+

T

using binomial expansion we get answer that

initial density(1-

T) = new density

now for rolling without slipping te condition that v = r

. Also when released ffrom top of a inclined plane the center of mass will have accleration of mgsin

excellent....

coool...

on the highest point the vertical component of the speed is zero.... and the horizontal = ucos

the velocity of the particle is least at that point.....

secA +tanA=3

(1 + sinA)/cosA=3

sqr both sides

1 + sin^2A + 2sinA=9(cos^2A)

1 + sin^2A + 2sinA=9(1 - sin^2A)

10sin^2A + 2sinA=8

sinA=-1 or 4/5

cos A=0 or 3/5

therefore sec A is 5/3

i tried your sum but other some condition or something should be given....
something is missing.....

5. 2sin

+ 4cos

+ 3..

this is in the form of a cos

+ b sin

+ c ....

so the maximum value will be...

sqrt(a² + b²) + c.....

sqrt(16 + 4) + 3

2

5 + 3....

3.e^{log₁₀tan1' + log₁₀tan 2' + ... + log₁₀tan 89'}

consider the power....

log₁₀tan1' + log₁₀tan 2' + ... + log₁₀tan 89'

so this can be written as: -
log₁₀(tan1*tan2*tan3......tan45......tan87*tan88*tan89)

we know that tan89 can be written as cot1....and so on...

so we can write the equation as
>>> log₁₀(tan1cot1 + tan2cot2.....tan45)

now tan

cot

= 1... and also tan45 is 1...

so it becomes.... e^{log₁₀₍₁₎}

so it is e⁰

so value is 1....Ans

rate me if u find my answers correct....
others i will solve and paste it tomorrow...
they are all very simple sums....

i also do the same..... h c v and d c pandey side by side

see figure 3-Q3....
let the distance from point A to point of mass M block(say point is K), be Z.
now construct a perpendicular to the base where the pulleys are affixed....
name the point as R...
let the distance KR be X....
let the distance RA be Y...
In triangle ARK,
X^2 + Y^2 = Z^2;
differentiating with respect to time....
2Z dZ/dt = 0 + 2X dX/dt...
therefore...
dX/dt= Z/X*dZ/dt...this can be written as....
dZ/dt/X/Z (here / stands for "whole divided by" )
or... u/cos

..........

dZ/dt = u and X/Z= cos

.....

plz...rate me if u find my answer useful.....

see figure 3-Q3....
let the distance from point A to point of mass M block(say point is K), be Z.
now construct a perpendicular to the base where the pulleys are affixed....
name the point as R...
let the distance KR be X....
let the distance RA be Y...
In triangle ARK,
X^2 + Y^2 = Z^2;
differentiating with respect to time....
2Z dZ/dt = 0 + 2X dX/dt...
therefore...
dX/dt= Z/X*dZ/dt...this can be written as....
dZ/dt/X/Z (here / stands for "whole divided by" )
or... u/cos

..........

dZ/dt = u and X/Z= cos

.....

plz...rate me if u find my answer useful.....

hi,
what rajdeep says is correct... Pradeep is a good book and will clear both ur concepts and problem solving.... i have pradeep.... its a brilliant book.... u know wat that around 70% of H.C.Verma sums or sums of that type can be found there...solved.... and more it includes special features like sums for only IIT- Jee aspirants...and more....

hi,

consider that you throw a ball on the ground....
it will bounce up again....
but it will not bounce up to the hieght as when you threw it the first time....
on second bounce it will decrease its hieght in bouncing again....

The ball described here bounces a little low than the previous time it did...

Here it can be considered that the ball collides with ground....

And its velocity after colliding with ground is proportional to the velocity before colliding the ground.....

it is

V_f= e V_i

this e is coefficient of restitution.....

whenever there is elastic collision e = 1 as final velocity is equal to initial velocity....

in inelastic collision e = 0....

for all the other collisions 1 > e > 0;

so coefficient of restitution is that which hampers the velocity of the body in motion and decreases it on its collision....

look i also could not do the sums of REST AND MOTION KINEMATICS directly..... but than when i got the concept of how to do the sums... i tried the sums and they seemed not much tough to me.... the only thing is dat u must read it properly...