F = Mg (2 - 1)

₃[log[x]] dx  =   I

Since, efficiency of the reactor = 10%

 

 and power of the reactor = 1000MW = 10⁹ Watts

 

So, power generated due to the fission of U-235 in order to sustain the 1000MW

 

plant with 10% efficiency = 10¹⁰Watts

 

Now reactor is to function for 10years

 

So during these 10 years Energy generated due to the fission of U-235 is given

 

by E = P*t = 10¹⁰*10*365*24*3600 Joules

 

Energy liberated after fission of 1 nucleus of U-235 = E_f = 200Mev

 

or, E_f = 200*10⁶*1.6*10^-19 Joules

 

hence to produce energy E the no. of U-235 nuclei required N = E/E_f

 

or N = 10¹⁰*10*365*24*3600 /200*10⁶*1.6*10^-19  = 9855*10²⁵

 

Now 1 mole of U-235 = 235 g = 6.023*10²³ nuclei or atoms of U-235

 

So N number of U-235 = N/6.023*10²³  moles

 

or mass of N number of U-235 = [N/6.023*10²³] * 235 grams 

 

                                           = 9855*10²⁵*235/6.023*10²³

                                            = 3843180 g = 38431.80 kg

Here,since at the two slits we have introduced mica and polystyrene strips of

 

thickness 0.50mm = 0.05cm each with and refractive indices 1.58 and 1.55

 

respectively

 

₁ = 1.58

 

₂ = 1.55

 

So this will introduce extra path length of   times thickness

 

so extra path difference introduced is = (₁t -₂t) = (1.58 - 1.55) 0.05 cm

 

or extra path dfference = t = 0.0015cm

 

so in the expressions for young's double slite experiment we will introduce this

 

extra path difference to find fringe width and other parameters

 

so x  = dy/D + t 

 

here d = slit width = 0.12cm

 

 = 590nm

 

D = 1m = 100cm

 

so for constructive interference we use

 

x  = dy/D + t = n

 

and for destructive interference

 

x  = dy/D + t = n/2, where n is any integer

 

 Thus, introduction of 't' in the path lengh does not change the fringe width

 

as width = w = D/d = 100*590*10^-7/0.12 = 49.1 *10^-3cm = 4.91*10^-4cm

 

Now, to find the distance of first maximum from the center, we should

 

remember that the path difference at center is not zero but 't'.

 

so depending on  t = n or n/2 the central point will be maxima or minima or

 

neither of these.

 

Thus using above expressions and conditions we can find the distance of first

 

maximum from the center. 

Dear Kirtana,

could you please verify your problem!!!

I expect one of the function sin or cos to be inverse.

 

You are requested to write the problem briefly so that it will be easy for us,as well as for others. Little effort on your part will help other IIT aspirants to understand what is going on without refering to books. Please don't mind.

PROJECTILE FIRED FROM AN INCLINED PLANE

 

Consider an inclined palne which makes an  with the horizontal. Let a projectile be projected with a velocity 'v' making an angle  with the horizontal.

 

 Let us choose X axis along the inclined plane and Y-axis perpendicular to it.

 

gcos and gsin are the two rectangular components of g.

 

Now,

 

v_x= v cos(-) and v_y = v sin(-).

 

Let T = time of flight of the projectile

 

The displacement of the projectile perpendicular to the inclined plane is clearly

 

Zero.

 

Using S = ut + (1/2) a t², for motion along y-axis, we get

 

0 = v sin(-) T -(1/2) g cos T²

 

or T = 2 v sin(-)/g cos  

 

Range on the inclined plane is

 

R = 2v² sin(-) cos/g cos²  

 

R_max= v²/g (1+sin)

 

H_max= v²sin²(-)/2g cos

 

If  is changed to -, then we get the formula of time of flight and range for

 

'down the inclined plane'. In this case,

 

T = 2v sin(+)/g cos  

 

R = 2v² sin(+) cos/g cos²

 

R_max= v²/g (1-sin)

 

let hole be made at height 'h' from the ground

so, pressure at the level wher hole is made is given by

P = (H-h)g

and F= PA = (H-h)gA

acceleration of water = a = F/m = (H-h)gA/m

Now, time taken by the stream to fall on the ground from height 'h' is

t = 2gh

Therefore, Horizontal distance travelled in time 't' is

x = ut + (1/2)at²

or, x = (1/2) [ (H-h)gA/m] [2gh]

or, x = (H-h)g²Ah/m

so, for x = maximum

dx/dh = 0

dx/dh = -g²Ah/m +(H-h)g²A/m

or (H-h)g²A/m = g²Ah/m

or (H-h) = h

or h = H/2 is the reqiured solution

Problem reads,

A man of mass 60 kg, standing on a light weighing machine kept in a box of mass 30 kg. The box is hanging from a pulley fixed to the ceiling through a light rope, the other end of which is held by the man himself. If the man manages to keep the box at rest, what is the weight shown by the machine? What force Should he exert on the rope to get his correct weight on the machine?

Solution) Let the man exets force F in the state when he manages to keep the box at rest. in this situation his effective weight as shown by the machine will be  = (mg - F), where m = mass of the man = 60 kg

Now, for this we formulate following equation

weight of box+(weight of man - force applied by man)=Force applied by the man

or, 30g + (60g - F) = F

or, 90g = 2F

or F = 45g

Therefore weight of the man as applied on the machine = (mg - F)

                                               = 60g - 45g = 15g

thus, weight applied on the machine = 15g Newton

Which corresponds to mass = 15 kg

 

 

Consider a triangle ABC

Here AB = Vector a

AC=  Vector b

Let angle between a and b be

Drop a perpedicular from point B on AC, say it is BD

Length of the perpendicular = BD = a sin

Therefore, area of the triangle ABC= (1/2) base * altitude = (1/2)*AC * BD

area between vectors a and  b = area of triangle ABC = (1/2)*b*a sin

area between vectors a and  b =(1/2)*b*a sin = (1/2) of magnitude of (a x b)

In equilibrium condition the weight of the cylinder is balanced by force in th spring in upward direction and buoyancy force exerted by the water in upward direction. So we formulate the equation under equilibrium conition as:

 

mg = kx +  Vg

Where, m is mass of the cylinder

since half  of the cylinder is dipped in the water so buoyancy force = Vg

here V is half of the volume of water and  is the density of water which is 1000kg/m^3

 

If an object is floating in some fluid say water, then its portion will be immersed in water and the portion immersed will be such that it displaces liquid of equal volume so as to provide buoyant force to the object and consequently the obect floats.

Under free fall of an object or system there is a phenomenon of WEIGHTLESSNESS. Thus, the body which was floating will continue to float and its none of the fration/portion will be immersed in the water and there will not be any buoyancy.