goIIT: IIT JEE AIEEE CET WBJEE VITJEE BITSAT IIIT AICET Engineering Entrance Experts, Student Discussion Community, IIT JEE Preparation Online

E
one two three four five six seven eight

1.
3.2 m/s2

2. compare with y = x tan

- (gx^2)/[2 u^2 (cos

)^2]

we get

= 45 and u =

20
T = 2usin

/g
= (

10)/5

and that javascript error is alpha

nice man anchit!
any more puzzles?

nice explanation and diagrams are given here

http://en.wikipedia.org/wiki/Wavefront

i bet computer001's sequence is a wrong one which doesnt exist!

see the diagram..

velocity of the string = u = dx/dt
velocity of block = v = dy/dt.

applying pythyagoras theorem, we get

x^2 = L^2 + y^2

differentiate wrt 't'

2x dx/dt = 0 + 2y dy/dt
solving we get..
v = dy/dt = u (x/y)

but x/y = sec

according to figure..

so v = u sec

EDIT: enlarge the diagram to see it clearer.

it's u sec

.
wait ill post proof now..

generally, ortho effect increases the acidity of acids.
what happens is that, when a bulky group (like -NO2) is at the ortho position of another grp (ex -COOH in benzoic acid) , the 2 groups go out of the same plane due to steric hindrance. hence, it becomes less stable and can easily donate a H+

no. water will not overflow as it is against the ppl of conservation of energy.
the angle of contact will change to account for it.

edit: ans is 1 only..
do AM- GM
$\frac {a^ab^bc^c + a^bb^cc^a + a^cb^ac^b}{3} \ge (a^{a+b+c}b^{a+b+c}c^{a+b+c})^{\frac 1 3}$
RHS is nothing but $(abc)^{\frac 1 3}$
but max value of abc is when a=b=c in the expression a+b+c = 1
so, max value of abc = 1/27
so max value of RHS = $(\frac 1 {27})^{\frac 1 3}$
$= \frac 1 3$

so $\frac {a^ab^bc^c + a^bb^cc^a + a^cb^ac^b}{3} \ge = \frac 1 3$
$=> {a^ab^bc^c + a^bb^cc^a + a^cb^ac^b} \ge 1$
Smile

potential of both spheres become same.
$V_1 = V_2$

$\frac {kq_1}{r_1} = \frac {kq_2}{r_2}$

$=> \frac {q_1}{q_2} = \frac {r_1}{r_2}$
Smile

$gamma = rac {dV}{V.dT}$
also,
$PV = nRT$ ..... (1)
$=> P.dV = nR.dT$
$=> rac {dV}{dT} = rac {nR}{P}$

so,
$gamma = rac {nR}{PV}$
but, from (1), $rac {nR}{PV} = rac 1 T$
so, $gamma = rac 1 T$
Smile

Note that T is the initial temperature.

i think the temperature of the room will decrease.
b4 the fridge is opened, the temp of the room is obviously > temp inside the fridge.
so, after its opened, heat from the room flows inside the fridge till an equilibrium is attained. As the walls are adiabatic, the heat can't escape outside the room.
so the temperature of the room decreasses..

helloo?!!?!?
did u even BOTHER to read the first 10 pgs of this thread!
the SUM CANT BE SOLVED BY NORMAL METHODS. PERIOD!!

thats y i said D is more appropriate.. but ONLY if it attains F=mg after a while.. otherwise.. the best option is C..