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well your answer is given in the following link

http://www.goiit.com/posts/list/662.htm#2020

The answer should be 1:1.

Exactly, I have rectified the same. Thank You very much

Are teh five digits y1, y2, y3, y4, and y5 all different or same or whatever?

I suppose rahul's approach should help u out.

Is your integral this

I = _{[ 0]}

^{[ 2]} e^{cos x} cosx(sinx) dx ?

Do let me know

Dear Shakirshafi12 your approach and solution is perfect and flawless

I have rectified and edited my calculations as above.

thanx for bringing the descrepancy at the earliest.

use the Bhor's theory of Hydrogen atom acoording to which

Radii of stationaty orbits is given by

r_n = (

₀h²/

m e²) n²/Z meter

= (0.53 x 10^-10) n²/Z

Where Z is no. of protons in the nuclei.

Total no. of neucleons = 2 for deuterium, but Z = 1

and nucleon in hydrogen atom = 1, also Z = 1 for hydrogen atom.

To find the remainder when 27¹⁰+7⁵¹ is divided by 10.

Remember whatever is the digit at units palce of 27¹⁰+7⁵¹ is the result when

divided by 10.

So we intend to find the digit at units place of 27¹⁰+7⁵¹

27¹⁰+7⁵¹ = 9¹⁵+ 7 x 49²⁵

= 9 x 81⁷+ 7 x 49²⁵

= 9 x 81⁷+ 343 x (49)²⁴

= 9 x 81⁷+ 343 x (2401)¹²

As 9 x 81⁷ = number with 9 at units place

and 343 x (2401)¹²= number with 3 at units place

so,

9 x 81⁷+ 343 x (2401)¹² = number with digit at units place is carry of (9+3) that is 2

or, digit at units place is 2

So remainder obtained when divided by 10 = 2

here n is such that

n(n+1)/2 = 15 (sum of first n natural numbers)

so n² + n - 30 =0

or (n+6)(n-5) = 0

n = -6, as n can't be negative, so n = 5

thus n = 5 is the solution

No we cannot as the conditions to create are just not feasible at all.

Since the electron and proton are oppositely charged

proton can be assumed to be at rest as mass of proton is 2000 time to that of electron

Now due to coulombic force of attraction electronis accelerated towards the proton and the coulombic force is varying continuously due to change in distance between electron and proton

coulombic force is given by

$F = k_C \frac{q_1 q_2}{r^2}$

where

$k_C = \frac{1}{4 \pi \epsilon_0} \approx$

= 9×10⁹ N m² C^-2 (also m F^-1) is the electrostatic constant

so work done by this force when electron attains velocity twice to that of its initial velocity then

work done by force F is given by

(1/2)mv²₂ - (1/2)mv²₁ =

^r F.dr

where m = mass of electron

here v₂ = 2v₁ and v₁ = 3.24 x 10⁵ m/sec

or (3/2)mv²₁ = ^r F.dr = kq² [-1/r]^r_infinite

or (3/2)mv²₁ = kq²/r -----(1)

here, q = 1.6 x 10^-19 C

Now substitute the values in equation (1) and find the value of r

Agreed with sir cvrmana. Ofcourse speed of light in all the frames whether moving or at rest is c = 300, 000, 000 m/s

I would like to elaborate on the point that photons does not have infinit mass its mass is

m = hf/c^2, where f is frequency of the light of which photon is a quanta.

Momentum is also possessed by photons and momentum of photon is given by P = hf/c

Anything that has momentum ofcourse possess mass and due to momentum only photons exert radiation pressure on the surface on which it falls.

Now coming to ROHITJU's query

anything moving with velocity of light has infinite mass PROVIDED the rest mass of the body is not zero. Thus the concept is not applicable for photons as its rest mass (though photons can never be at rest, remember this and please don't be confused) is zero.

m = m0 / [(1 - v^2/c^2)^2]

where mo is rest mass
and m = mass when body is moving with velocity 'v'

and if mo is NON ZERO then

for v = c or v is comparable to c (limiting case)

m = INFINITE

But the concept fails to explain when applied for photons as

for photons v=c, mo = 0

so m = 0/0 is not defined

Thus we use the concept E = mc^2 = hf

or m = hf/c^2, thus is the mass of photon.

I hope the things are now crystal clear.

Well it happens that one's mind is more inclined towards a particular subject. So is with u.

But at the same time the good thing about u is that u are good at chemistry and mathematics.
Mathematics is ofcourse mother of all the sciences and its application in the physical world is sought in physics. So try to understand the concepts of physics rather than mugging up and also try to correlate and apply these elegant laws to physical world around you.

I am sure the fear will go and you wul love physics.

Try to solve maximum numericals and conceptual physics to gain confidence this will eliminate you fear altogether.

In the above explanation read '?' as

which is solid angle.

Furter i have mentioned above that solid angle is calulated as

= k S/R², where k is dimensionless proportionality constant and thus

is dimensionless as it is ratio of two quantities S(cm²) and R² (cm²) having same dimensions.