2019-03-01_Physics_Times

cos

'

cos

f f v vt

v v

  

  

  

f'f

D

d



 (i)

'

' D

d



  (ii)

Subtract (ii) from (i), we get

 

'

' D D

d



 



 

Substituting the given values, we get

2

5

3

5 10

3 10

1 10

m

m

m

  



 

 



8

6

2

3 10

0.6 10

5 10

 m m









  



 600 nm

As the light is getting red shifted the

wavelength increases.

u

c









u - refractive velocity of the star w.r.to the earth

8

10 3 10 5 10 / 5

6000

c

u m s





  

   

1st and 4th are perpendicular to each other.

Angle between 1 and 2 is 30  and between 2 and 3

is also 30 . The intensity that comes out from the

first prism is I 0 /2. The intensity that comes out

from the last prism is

6

(^0) cos 30 6 0 3 270
2 2 2 128
I I I
I
 
     
 
Let T is the temperature of the interface.
KCopper4 &K KbrassK
As the rate of heat flow is same in two slabs
4. (100 ) ( 0)K A T KA T
l l
 

T  80 C
The equivalent thermal conductivity is
R R1 2R
1 2 1 2
eq 1 2
l l l l
K A K A K A

 
1 2
1 2
1 2
4 5
.
(^43)
2
eq
l l x x
K K
l l x x
K K K K
 
  
 
Hence rate of flow of heat through the given
combination is
 
 
 2 1
2 1
5

. 3
4 5

eq

Q K A T T K A T T

t x x x

 

 



 2  1

1

3

K A T T

x





On comparing it with given equation we get

1

.

3

f

The heat radiated per unit time is

dQ R T2 4

dt



 R T2 4constant

4 2

1 2

2 1

T R

T R

   

   

   

1/

1 2

2 1

T R

T R

 

  

 

1/2 1/

1 1

2 1 1

2 1 /

R R

T T T

R R

   

     

   

25.Sol:

26.Sol:

27.Sol:

28.Sol:

29.Sol:

30.Sol:

31.Sol:

T 2  2 1.414T 1 T 1

We know that

2 0

1 0

ln

T T

t K

T T

  

   

  

From the given information