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Catalogs Discussion Forums -> Mechanics -> Mech -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
3 replies   

see the weight is in the downward direction and the tension in the horizontal direction as u know that horizontal force cannot be balanced by vertical force and vice-versa thus the tension is infinitely large

Catalogs Discussion Forums -> Mechanics -> Mech -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
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max force that can be applied on 3kg block is @mg where @ is the co-efficient of friction


therefore subs the values we 15 N


 

Catalogs Discussion Forums -> Physical Chemistry -> Labelling of H202-In the formula to calcuate volume strength V=N*5.6 -> Go to message
This Post 2 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
5 replies   

eq weight of H2O2 is 17g and V =volume strength or the vol of O2  liberated


H2O2  = H2O  + O


as u can see 34 g contain 16 g of free O OR 11.2 L of O2


therefore 17g contain 8g of  free  O  ( that is the amount of oxygen that is given out )or 5.6 L of O2 


therefore 1N (17 g ) of H2O2  =  5.6 L  of  O2


therefore V=N*5.6


u can rate me if useful

Catalogs Discussion Forums -> Mechanics -> A CHORD IS WOUND................... -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
6 replies   

ahhhhh  a simple sign mistake the accln we get is mg/[m+M]


so the ans is(2hmg/[M +m] )1/2 /R


mukulm it seems that u haven't read my soln it is conceptually correct but thanks as i came to know about my mistake

Catalogs Discussion Forums -> Mechanics -> A CHORD IS WOUND................... -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
6 replies   

ahhhhh  a simple sign mistake the accln we get is mg/[m+M]


so the ans is(2hmg/[M +m] )1/2 /R


mukulm it seems that u haven't read my soln it is conceptually correct but thanks as i came to know about my mistake

Catalogs Discussion Forums -> Mechanics -> Rotation: -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
3 replies   

for the first which end is at wall and which is at floor ?


2) vel of centre of mass is given by = m1v1 + m2v2 / [m1 + m2 ]


here m1 =10 ,v1= 14 , m2= 4 , v2 =0


substituting values we get


vel of com = 10 m/s ans


 

Catalogs Discussion Forums -> Mechanics -> Rotation: -> Go to message
This Post 2 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
1 replies   

in the first case we have both rotational energy ( K.E )as well as tranlational energy ( K . E )


let it fall from height then for 1st case


mgh =1/2 mv 2+ 1/2I $2  where I  = mk2   where k= radius of gyration and $ = angular vel


therefore we have  mgh = m/2 [v 2+k2$2  ]  we also know that r$ = v


so we have mgh = m/2[v2 + k2r2v2]   =mv2/2[1 +r2k] ............(1)


from the given second case mgh = m25v2/32 .................(2)


from 1, 2  we equate and get k = 3r/4  ans


 


 

Catalogs Discussion Forums -> Mechanics -> A CHORD IS WOUND................... -> Go to message
This Post 5 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
6 replies   

from the f.bd  we get


mg -T =ma  ..................(1)


now taking torque about centre of mass of cylinder we get


TR = I% ................(2)  ( where I = MR2 moment of inertia of cylinder about  com , % = angular accln of the cylinder )


since there is no sliding of the string we get


a =R% ....................(3)


solving 1 , 2  we get  a = mg/(m-M)


now using v 2= u2 + 2as  ( where u = 0 is given )


we get v =  [ 2hmg/(m-M) ]1/2and we know that


v =R$ ( $ = nagular vel )


therefore  $ =[ 2hmg/(m-M) ]1/2  / R


 

Catalogs Discussion Forums -> Organic Chemistry -> Conversion -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
4 replies   

C2H5Cl ==(1)======> C2H6 + :CH =====(2)====>C3H8   + O2 ( in  9:1 ratio )====(3)=======>C3H7OH


 


(1) = Zn / HCL


(2) = h.v (ULTRAVIOLET RAYS )


(3) = Cu tube, 200 C, 100 atm

Catalogs Discussion Forums -> Physical Chemistry -> Easy ............ -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
6 replies   

rathin yaar i took the approx now using same method i get the same ans yaar


1C = 6.24 *1018 electrons




















therefore the ans is  18 * 36 * 6.24 * 1020 electrons  = 4.04 * 1023



 


which is equal to 0.671 moles



 





 


rathin both methods are correct

Catalogs Discussion Forums -> Physical Chemistry -> Schrodinger wave equation -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
3 replies   

http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation


 


DeBroglie's matter waves weighing heavy upon his mind, Erwin Schrodinger wanted time to ponder, time to consider all the implications.  Schrodinger, an Austrian physicist noted for his work on the physics of strings, took flight to a villa in the Swiss Alps in 1925, leaving his wife behind and gathering a former Viennese girlfriend.  What would come of this (presumably) quiet period of reflection and thought would forever change the landscape of physics.  Indeed, it would change the way we as a species reckons the universe we live in.


     A sort of microscopic solar system, with electrons orbiting about the nucleus like planets to stars -- the Bohr atomic model was proving to be of limited utility.  For hydrogen atoms, the agreement between predicted and observed behavior was sterling.  However, for atoms with more than one electron -- even helium with only two electrons -- predicted and observed behavior radically diverged.  Schrodinger desired to develop a model that agreed with the experimental evidence.  What came of that illicit vacation to the Swiss Alps was a model that was not derived from any other, a model that can be called an intuitive guess, a leap of imagination, a model that is astonishingly accurate.


     In these pages, we will, for the sake of both brevity and simplicity, only consider the time-independent Schrodinger wave equation in one dimension.  We will not consider the full equation in all of its gruesome splendor.  The time-independent Schrodinger Wave Equation, which could validly be called Schrodinger's law, is given by the differential equation

 











where j (x) is the is the wave function, m is mass,  is Planck's constant divided by 2p, E is the total energy of the particle, and U(x) is the potential energy function of the particle.  As when one ingests something disagreeable and the natural reaction is nausea, so too is the natural reaction to this equation.  However, comfort may be taken if we consider that acceleration is the second derivative of the position function and, therefore, could be written

 











As surely as acceleration simplifies to something more palatable, The Schrodinger wave equation must simplify (a little, at least).


     To find general solutions to this equation, boundary conditions must be established.  The principle conditions that it must adhere to are

 


















1. as 
2. if x is in someplace it is physically impossible to be
3.  j (x) is a continuos function
4.  j (x) is a normalized function


 

In these pages, we will, again for brevity and simplicity, consider the case of a particle in a one-dimensional box of ideal rigidity, such that its walls are impenetrable.  Let the box have length L.  As may be seen in the illustration, the potential



energy function has two states:

 












1.  U(x)=0  for 
2.  U(x)=  for x<0 or x>L


Since it is physically impossible for the particle to be outside of the box, it is the first state that is of interest.  Indeed, this simplifies the wave equation considerably, with the term U(x) dropping out.  Therefore, the wave equation corresponding to the particle in the box is given by

 











 

Before assailing this equation with a display of mathematical acumen, let us ask ourselves what function's second derivative is merely some negative constant -- all of the terms on the right-hand side save j (x) -- multiple of itself?  To simplify, let

 











Therefore, the wave equation becomes

 











It becomes clear, a trigonometric function like sine or cosine would be a good candidate for j (x).  Therefore, our guess for the solutions to the wave equations is

 











By the first above boundary condition, it is known that

 











Therefore,

 











where n=1,2,3,...  When the smoke clears, we have that

 











where A is the function's amplitude.  To determine the amplitude, recall the fourth boundary condition, j (x) is a normalized function.  Mathematically, this means

 











In words, this states that the probability of finding the particle somewhere on the x-axis is one or 100%.  Waving hands a bit to omit the gory details, this gives

 











Gasping for breath, we at last have unearthed the solution to the wave equation for the particle of the nth quantum state in the rigid box.

 












   for  
                   for x<0 or x>L


 

The utility of this solution lies primarily in that the probability of finding the particle at some position x is given by the square of j (x).

 











 

The importance of this relationship is best illustrated graphically.  Consider a particle in the third quantum state.


 
 

 It can be seen that there are regions where the probability of finding the particle is zero -- so-called nodes.  This is not unique to simply the particle in a rigid box model.  It is observed in more sophisticated ones such as the model of an electron orbiting a nucleus...

Catalogs Discussion Forums -> Mechanics -> an easy question -> Go to message
This Post 14 points    (Olaaa!! Perrrfect answer.   in 4 votes )   [?]
2 replies   


33)

first consider the free body diagrams

now for the mass M,

from fbd, T = Mg - Ma

For the combined mass (m+M')

T = (M'+m)a

equating both Mg =(M' + M + m)a

so a = Mg / (M' + M + m)



now consider the fbd of the mass m ,

ma cos x = mg sin x

so a = g tan x

Now equating this a with the formerly obtained a,

Mg/(m+M+M') = g tan x

so M cot x = M + M' +m



 M = (M'+m/ cot x - 1)

ans
Catalogs Discussion Forums -> Physical Chemistry -> Easy ............ -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
6 replies   

q =i*t =18 *3600


now 1 C OF q is approx = 6 * 1018 electrons


therefore the ans is  18 * 36 * 6 * 1020 electrons

Catalogs Discussion Forums -> Mechanics -> Kinematics IIT Question: -> Go to message
This Post 2 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
9 replies   

see  it starts from rest and again comes to rest therefore fist it must have accelerated and then it would have retarded  therefore potion a, d are correct

Catalogs Discussion Forums -> Physical Chemistry -> explain , Dalten's law partial pressure !! -> Go to message
This Post 5 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
4 replies   

the pressure of a mixture of gases is equal to the sum of partial pressures of all the components of the mixture





P = Pa + Pb + Pc  + ................





where Pa , P b  are the partial pressures of the  different gasses present in the mixture which is same the pressure the each gas would exert if they are contained in a seperate contianer if n a, n b....... are the no





ofmoles of diff gases then





P=( na +nb +..............)  ....................1





and partial pressure of A =  Pa =  nART/V  ....................2





TAKING RATIO OF 1,2





P= PA * MOLE FRACTION OF A where





mole fraction A = na/(na + nb +........)





 

 
 
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