Messages posted by edison

All possible 5 digit odd numbers using 0,1, 4, 5, 4 are

40145, 40415, 41045, 41405, 44105, 44015

40451, 40541, 45041, 45401, 44501, 44051

10445, 14405, 14045,

50441, 54041, 54401

Add all these to obtain the result.

No positrons exist in the nucleus of an atom, however the emission of the positron can result from the conversion of a proton to a neutron.

Positron is infact an antiparticle of electron. Its mass is same as that of electron . Both have the identical spin. But the difference is in charge. The charge of positron is equal and opposite to that of electron.

With in about 10^-9 s, the positron combines with an electron and is converted to gamma rays. Positrons are similar to beta particles in velocity and ionizing effect, however they have very low penetrating power.

Positron and electron never exist (or cannot be found) within nucleus. This can be proved easily using HEISENBERG'S UNCERTAINTY PRINCIPLE.

When an electromagnetic wave changes its medium, then the photons (photon is the quanta of electromagnetic wave) having some frequency

remains unaltered, but wavelength

, changes, also the speed of electromagnetic wave changes depending upon the refractive index of the medium.

The speed of light = c, in vacuum (always remember this).

But when it enters in a medium of refractive index

then its velocity changes and it becomes c/

,

Its frequency in the medium remains same (for example, consider a red light in the vacuum or air. But when it travels through other medium say water it appears red only on the account of its frequency being unaltered, but its speed and wavelength changes)

So wavelength is given by

' = speed of light/ frequency = (c/

)/

or

' = (c /

)

or

' = (

/

)

Thus, energy of the phton on changing the medium is given by

E = h (speed of light / Wavelength)

or, E = h [{c /

}/{

/

}]

or, E = hc/

= h

Which means energy of the photon doesn't change when medium changes.

An isothermal process is a thermodynamic process in which the temperature of the system stays constant: ?T = 0.

This typically occurs when a system is in contact with an outside thermal reservoir (heat bath), and processes occur slowly enough to allow the system to continually adjust to the temperature of the reservoir through heat exchange. An alternative special case in which a system exchanges no heat with its surroundings (Q = 0) is called an adiabatic process.

Consider an ideal gas, in which the temperature depends only on the internal energy, which is a function of the mean translational kinetic energy of the molecules, as given by a Boltzmann distribution; if the internal energy is constant, so is the temperature. Take the number of moles n as a constant.

$\Delta U = n R \Delta T = 0 \,$

but this means, according to the ideal gas law, that

$\Delta (P V) = 0 \,$

so that

$P_i V_i = P V = P_f V_f \,$

where

P i

and

V i

are the pressure and volume of the initial state,

P f

and

V f

are the pressure and volume of the final state, and the variables P and V stand for the pressure and volume of any intermediate state during an isothermal process.

Dear Akash,

It's good to see that you are aiming towards IIT at such an early phase. Indeed it is the most appropriate and ideal time to start as you have got full fledged two years. This gives you sufficient time to prepare and gain confidence. Carry on with your efforts. As you have mentioned about your classroom performance that you claimed to be more than satisfactory.

Your problem is associated with your under performance during phase examinations, may be you are taking stress of examination, that sometimes lead to examination phobia. Often this happens to be a problem of many good scholars. This is quiet normal. You just need to gain more confidence by practising problems and take examination pressure with an ease otherwise your performance may be affected and hampered.

To combat with examination pressure just simulate examination like condition at your home itself. I mean try to create examination like situation and solve the model test papers within stipulated time frame a day or two before the commencement of your actual tests. Just try to keep youself physically as well as mentally fit. Rest all will be fine.

All the best and be in touch with GOIIT comunity by means of our website for any other assistance, guidance and counselling.

I =		CONE (ROTATED ABOUT CENTRAL AXIS) (3/10)mr²

A cone is a series of infinitesimally thin disks of varying radius. If we add up the moments of inertia of all these very, very thin slices we'll get the moment of inertia of the whole cone. Adding up a lot of very small pieces to create a whole is called integration.


I = I_slice dx =	1	m_slicer² dx
I = I_slice dx =	2	m_slicer² dx

Replace mass with density times volume and proceed.

The "trick" to solving this part of the problem is determining how the radius of the slices vary from the vertex (x = 0) to the base (x = h). We need a function that begins at 0, ends at r, and increases linearly.

The law of conservation of energy and mass is not at all violated in nuclear fission. The law is absolutely safe and not in jeopardy. Infact the law is always valid and holds good .

According to Einstien's mass-energy relationship E=(

m)c², (which means that mass and energy are interconvertible) the change in mass of nuclei after and before fission is convertd into binding energy of the nucleus which is used to bind the nucleons (protons and neutrons) by nuclear forces within the nucleus.

The same law is applicable when NUCLEAR WEAPON is detonated. For example when nuclear weapon is detonated, it gives rise to an uncontrolled chain of nuclear fission reactions which in turn gives rise to tremendous amount of energy release that cannot be controlled and thus leads to massive destruction and devastation unlike nuclear reactors where this energy liberation is so by well in control (by means of control rods )and it can be utilized in some pieceful work like production of electricity what we call nuclear power.

If mass of the bars be m₁ and m₂. Mass m₁ be at the left side and m₂ be attached to right side of the spring.

force applied on the system = F on m₁ in the direction left to the spring.

So, acceleration in the system = a = F/(m₁+m₂)

This frame is accelerated and thus non inertial frame of reference, and in order to apply Newton's laws of motion we apply a pseudo force (m₁a) towards left on the block m₁ and m₂a towards left to the block m₂.

We consider the problem from Center of mass frame. In this frame bars move in opposite directions and come to instantaneous rest at some instant. The elongation of the spring will be minimum and maximum at this instant.

Let bar m₁ is displaced by a distance x₁ and the bar m₂ by x₂ from the initial positions.

Applying energy equation in center of mass frame,

T + U = W_ext

Here, Wext = work one by the pseudo forces

U = Potential energy of the spring = k(x₁ +x₂)²/2

T = Kinetic energy = 0

W_ext= m₁F(x₁+x₂)/(m₁+m₂),

So, k(x₁ +x₂)²/2 = m₁F(x₁+x₂)/(m₁+m₂)

or, (x₁+x₂) = 0 or (x₁+x₂) = 2m₁F/k(m₁+m₂)

Hence the maximum seperation b/w the bars equals : L + 2m₁F/k(m₁+m₂)

The minimum separation corresponds to zero elongation and is equal to L.

STRETCHING OF SPRING ON FRICTIONLESS PLANE

This can easily be explained as there is accleration in the mass bar m₂as a result of which there is tension in the spring.

Dear aditi,

I request u to expand ucm to get your query answered.

An electron in en electric circuit continues to flow in the direction opposite to the flow of current no matter whether it is coming out of -ve terminal or entering into +ve. The electron entering into the +ve terminal cannot be distinguished from the electron coming out of the terminal because of "PRINCIPLE OF INDISTINGUISHABILITY" according to which quantum particles such as electrons cannot be distinguished from other electrons as all are having identical mass, charge, spin or other physical parameter associated with them.

ABSOLUTE AND GAUGE PRESSEURE

For gases, pressure is sometimes measured not as an absolute pressure, but relative to atmospheric pressure; such measurements are called gauge pressure. An example of this is the air pressure in an automobile tire, which might be said to be "220 kPa", but is actually 220 kPa above atmospheric pressure. Since atmospheric pressure at sea level is about 100 kPa, the absolute pressure in the tire is therefore about 320 kPa. In technical work, this is written "a gauge pressure of 220 kPa". Where space is limited, such as on pressure gauges, name plates, graph labels, and table headings, the use of a modifier in parentheses, such as "kPa (gauge)" or "kPa (absolute)", is permitted.

Gauge pressure is a critical measure of pressure wherever one is interested in the stress on storage vessels and the plumbing components of fluid systems. However, whenever equation-of-state properties such as densities or changes in densities must be calculated, pressures must be expressed in terms of their absolute values.

As far as problem solving is concerned we always should intend to evaluate Absolute pressure until & otherwise specified

If work function of the metal is W, and the frequency of the falling light should be greater than threshold frequency for Photo electric effect to take place. Say if initially the frequency is f then K.E of electron is

K1 = hf - W, where h = planck's constant

K2 = 2hf - W

So the factor by which kinetic energy increases is given by K2/K1

The increase in intensity will not effect the maximum kinetic energy of electrons released from the surface, but will simply increase the no. of electrons released, which will in turn incerease the current. So if intensity is doubled than this increases the photoelectric current by a factor of 2. But we have also increased the frequency twice so I = dq/dt i.e. rate of flow of charge also increases by some factor. Here the resultant factor by which velocity increases is given by

(K2/K1) as a result of which the factor by which current increases on doubling the frequency and intensity is given by 2

(K2/K1)

To find the general trm of any sequence you can use method of finite differences.

The physical significance of definite integral say _a

^bf(x).dx is area between x-axis from a to b and the curve f(x). So for this the function NEED NOT BE CONTINUOUS as it may be discreet or piecewise continuous like greatest integer value function etc. But for indefinite integral the function should be continuous for its analytical evaluation.

.Q) how to approach aproblem as given below

which of the following sets may represent magnitudes of three vectors adding to zero

a. 2,4,6

b. 4,8,16

c. 1,2,1

Solution) Magnitudes of three vectors will add to zero if the addition or subtration or combination of these gives ZERO as explained below:

a) 2, 4, 6 This is the set which gives zero when 2 + 4 - 6 is performed. So YES

b) 4, 8, 16. No combination of these three will give zero hence NO

c) 1, 2, 1 again 1 + 1 - 2 = 0, so this set may represent magnitudes of three vectors adding to zero.

For the given function to be maximum u need to specify another relation between x and y that will put some constraint on values of x and y.

Answer to your second question is:

Gravitational constant = G, is a Universal constant and it remains same

throughout the universe and thus on moon also the constant is unaltered

G = 6.6742 x 10^-11 N m² kg^-2

Magnitude of gravitational force between earth and 1 kg object on earth's surface is given by

F = GM_em/R²

Where, G = gravitational constant

M_e⁼Mass of earth, m = mass of object, R = radius of earth, m = mass of object

Also, GM_e/R²= g = 9.8m/s²

So, F = mg = 1*9.8 = 9.8 N which is the gravitational force on object and same force is exerted by obect on earth (Newton's third law)

Dear Nilesh, thank you very much for the effort you have made, I appriciate your effort and at the same time would like to thank administrator and moderator that they made avialable the very much required figure. i have tried to give possible approach. Please use the following link

http://www.goiit.com/posts/list/469.htm

Dear Nilesh, your approach is perfectly fine upto equation (3), then it should change as explained below.

let vel. at B = v_b

vel. at P= v_p

We have v_b²=3gr-2g(r-x)

Which boils down to, v_b²=g(r+2x)------------------------------(1)

now, cosµ=h/r

cosµ=r-x/r

r.cosµ=r-x

x=r-rcosµ

x=r(1-cosµ)--------------------------------------------------------(2)

v_b²=g[r+2r(1-cosµ)]----------------------------------from(1)&(2)

=10[20+40(1-cosµ)]

=[200+400(1-cosµ)]

now v²=u²+2as

Upto this point its perfect, Now once the body reaches point B, it will not at all traverse circular path but we follow projectile motion from B to C and this path from B to C should therefore be parabolic and hence apply and use the concept of parabolic motion to get the solution. Moreover the body at B goes tangentially to the circle at B and it will go to the highest point where its component of vertical velocity becomes zero.

Use range R = (v²sin2

)/g = BC

Now to get the condition for maxima differentiate R w.r.t.

and equate to zero.

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