I would like to mark the mistake of my mate "One and only one "

 

As he has taken Paschen Sereis and has made mistake

 

Inactual as it is given that Hydrogen atom is in ground state So we should take

 

n₁ = 1 and thus it has to be Lyman sereis.

 

Since the Hydrogen atom is excited by a ray 

 

With Wavelength of 970.6 A

 

So the nature of the wavelength is " U V Rays "

 

So the value of n₁ = 1 and  n₂ = 2,3,4

 

That is it lies in the Lyman sereis

 

So the maximum value of the wavelength will be when H atom jumps to n₂ = 4

 

1 /  = Rh * Z²  [(1/  n₁²)  - ( 1 / n₂²)]

 

As per the formula

 

1 /  = 109678 [ 1/1 - 1/16 ]

 

1 /  = 109678 [ 15/16 ]

 

1 /  = 102823.13

 

Therefore We will get

 

 = 1 / 102823.13

 

 = 0.0000097

 

 = 97000 A

 

 OR

 

 = 0.0000097 nm.

 

Solved.

 

plZ rate me for my efforts .

 

Cheers !!!!!!!!!!!!!!!!!!!!!

So here is the answer Well it was a difficult one and I tried putting it up for

 

The GOIITians

 

 

Here   = Angle between the line (of R inthe fig. ) and the horizontal '' b ''

 

First

 

The pressure that will be acting on the potion aprt from the door will be

 

dP = P ( a - Rsin )

 

As the length of the door will be (a - R Sin)

 

dF_x = Pcg(a - R Sin)R.d * Sin

 

dF_y = Pcg (a- Rsin)Rd * Cos

 

Knowing that T = F*R

 

So the below relations are as per the horizontal and vertical components

 

And  is the density of the liquid that is inside the Cube.

 

dT_Fx = cg ( a -Rin) d.Sin . (R - RCos)

 

dT_Fy = cg (a - Rsin)d.Cos .(RSin)

 

Now resolving along the line of resultant "F"

 

dT = dT_Fx + dT_Fy

 

Where T is the torque

 

= cg(a - Rsin)Rd.Sin. (R - Rcos) + cg(a -Rsin)Rd.Cos.Rsin

 

=cgR² [ (a-Rsin)Sin.(1-Cos)d  +  (a-Rsin)Cos.Sin.d ]

 

=cgR² [ (aSin - RSin²)(1 - Cos)d + (aCos - R.Sin.cos)(Sin)d

 

=cgR² [ aSin - aSin.Cos - RSin² + RSin².Cos + aCos.Sin 

 - RCos.Sin² ] d

 

=cgR² [ aSin - RSin² ] d

 

=cgR² [ -a [cos] { from 0 to pi/2 } - R(1 - Cos2 / 2 ) ] d

 

=cgR² [ a - R ( d / 2 - Cos2 / 2 .d ]

 

=cgR² [a - R/4 - (Sin2/4).d

 

Sin 2/4 = 0 as it is from 0 to /2

 

=cgR² [a - R/4 ]

 

Therefore we will get

 

Where F is the force to be applied on point T to avoid leakage of liquid

 

F*R = T

 

F*R = cgR²[a - R/4]

 

F = cgR[a - R/4]

 

Hence the answer

 

plZ rate me for my efforts

Cheers !!!!!!!!!!!!!!!!!!!!!!!!!!

I Have attached the diagram along with this for explanation

 

Ther is a ring with a small element .

 

resistivity =

 

Taking (b-a) << L

 

R = L / A

 

R =  (x /  r²⁾

 

R (Eq) =  R

 

R (Eq) = _{[0 ]}^{[L ]}  dx /  r²

 

Tan  = (b-a)/L

 

Above "r" = a + xTan

 

R (Eq) =   dx /  (a+ xTan)²

 

R (Eq) = ( /  Tan) [ 1 / (a + xTan)] From 0 to L

 

R (Eq) =  /  Tan  [ 1 /( a + xTan)]  From 0To L

 

Therefore From above Tan = (b-a)/L

 

R (Eq) =  /  Tan [ (-1 / a) + (1/ a + b - a )

 

R (Eq) =  /  Tan ( b - a / ab )

 

R (Eq) = [ /  (b - a)] * [(b -a) / ab]

 

(b - a) gets cancelled and we get

 

R (Eq) = L /  ab

 

Hence solved

 

plZ rate me for my efforts.

 

Cheers !!!!!!!!!!!!!!@@@@@@@@!!!!!!!!!!!!!!!!!!!!!!!!!

 

As Tan = (b-a) / L and (b-a