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well my first reply after 2 years i think i shud take an easy one first,here you go:energy required for water 2 boil=Power supplied* timeQ= (v^2/R) * 5* 60 Q=220^2/R * 5 *60let with 200 volt supply time be t minutes Q1= 200^2/R *t*60Q=Q1 solve it and you'll get the answer :)
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write F as a function of theta first
i'll write o for theta
F= kRo + mgsino + (mu)mgcosO
now at any instant u can write O=x/R where x is arc length on the curve
dw = Fdx = kr*(x/R)dx + mg(sinx/R)dx + (mu)mgcos(x/R)dx
integrate to get the answer
limits for x are x=0 to x=R(alpha)
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hi. i hav a query..,
this year bitsat was also open to biology students, was it also open for them last year too???
if no., then cutoffs for engg. courses may fall down even more as bio students are only applicable under FBS category.., nd i suspect tht bio is certainly more scoring than maths., there may be a good number of candidates >=300 range who r frm bio..
so cutoffs can fall down a bit more for engg courses , if this observation is correct...
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it was really by experimentation, with hexagon, u can only have 3 possible arrangements, so its better to try it out by hit n trial, else u'll need to use a bit of calculus
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well to get min resistance, the calculate res. b/w adjacent corners
u have two resistances of 30 and 6,
R= 30*6/30+6 = 5
and max res when both you take opp ccorner,
2 res of 18 ohms each
R2= 9 ohms..,
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306 on 16th may...
do i have chance of getting mech. in pilani/goa????
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for this type of questions, place the cubes identical to the one given
such that q is common vertex of all the cubes, and is at the centre of the bigger cube formed,
the bigger cube formed will have 8 identical small cubes, try imagining, place the cubes with q as common vertex, nd form a bigger cube)you'll see you need 8 cubes
now flux thru the big cube = q/e ( e---> epsilon)
now q is at centre of cube nd by symmetry, flux thru smaller cubes = q/8e
ans....
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@karthik, rohit
thnx for acknowledging my efforts :)
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is the answer a) 2L/3V
velocity will not remain constant throughout as its a case of variable mass system, nd if you notice tht angular momentum will be conserved about lowest most point, as net torque is zero at this position,
, now as question states neglecting, rolling resistance, etc, so no dissipative forces acting on system so energy of system must be conserved
the lowest most point of roll is point of pure rotation , so total energy can be given by( question also says to neglect PE of system)
E= 1/2 * I*w^2 where I is MI about the point of pure rotation
now,
initial energy = 1/2 * 3/2 * M*R^2*w^2 = 3/4 * M*V^2, { rw=V}
now energy at an instant when x length is unrolled.,
mass per unit length= M/L = k
energy at this instant is 3/4 * (M- x*M/L)*v^2
equating these two we get
v^2 = (L/L-x) * V^2
v= sqrt(L/L-x)*V
sqrt(L-x)*dx = Vsqrt(L)*dt
now integrate, as x varies from 0 to L , L-x varies from L to 0,
on integrating you'll get
t = 2L/3V
hope this is correct ;)
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number of real solutions is 0
y= e^sinx
then you can write this expression as
y^2-4y-1=0
y= 2+sqrt(5) or y=2-sqrt(5)
now 1/e <e^sinx<e { -1<sinx<1}
and both the above obtained values lie outside the specified range
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let radius be 'r'
firstly water will form a film over the hole
now water will begin to enter when
force due to surface tension= force due to liquid above the hole
T.2(pi)r = hdg*(pi)*r^2
r= 2T/hdg
put values you'll be getting answer close to 0.035 mm
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15
v=a[ ]  x
acceleration 'A' = vdv/dx = (ax)* ( a/2x ) = a^2/2
dw = F.dx = ma^2/2 .dx
on integrating
w= mda^2/2
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21 apply energy conservation
enrgy at top of cliff = m*50*50/2 + m*g*40
energy at bottom is purely kinetic
mv^2/2 = m*50*50/2 + m*g*40
now solve from here :)
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identity is an expression which returns the same value whatever be the 'x'
sin^2x + cos^2x =1 is an identity, as it gives answer 1 for any x,
on the other hand if an expression is true for only some values of x the its an equation
x^2 -3x +2 =0 is an equation as it's true for x=2,1 only
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dy/dx gives the slope of the tangent to the curve which passes thru the point considered
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L be length of rod
two force acting on it ((dm)g downwards
nd (dm)v^2/r , radially outwards,
by energy conservation
P.E of dm at 60 degree angle = KE at 37 degree + PE at 37 degree
dmgL(1-cos60) = dm v^2/2 + dmgL(1-cos37)
dmv^2/2=dmgL(cos37-cos60)
dmV^2/L = 2dmg( 4/5 - 1/2)
dmV^2/L = 3dmg/5 this is radially outward
now angle between dmg and dmV^2/L is 37 degree
F = sqrt( ((dmg)^2 + (3dmg/5)^2 + 2*(3dmg/5)^dmg*cos37 ))
solve it to get answer
you shud get 1/5* dmg*sqrt(58)
check for calculation mistakes i might have committed a few, :),
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as things under sqrt shud be positive so
x belongs to [-4,4]
-1<= [1+cosx]<=1
always true except when cosx=-2pi ,0,2pi,...
only point in [-4,4] is 0
so domain is [-4,4]-{0}
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(1+x)^20 = 20C0 + 20C1*x +............. 20C20* x^2
x=1
2^20 = 20c0 + 20c1 +.........+20c11............. + 20c20
2^20= 20c0 + 20c1+............ 20c10 + (20c9 + 20c8 +.......20c0)
2^20 = 2*(20c0+20c1+..... 20c10) - 20c10
(2*20 + 20c10)/2 = 20c0+ 20c1+.... +20c10
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the question is correct yaar
even though a1,a2... a20 be different, there sum will be constant, and so will be there product, its a property of means
sum of a1,a2...., a20 = sigle AM between 3 and 77 = 3+77/2 = 40
by AM >=GM
a1+a2+.....a20/20 >= (a1a2a3....a20)1/20
(40)^20 >=a1a2a3a4...a20
so correct answer is b)
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