2019-03-01_Physics_Times

3 [100 ] [ 0]

KA KA

 T T

 

300 3 T T 300 4T

300

75

4

  T C

As temperature of copper decreases its entropy

also decreases. As heat is taken by the pond so its

entropy increases. But

Spond SCu

Suniverse    Spond SCu 0

The power genetated through the wire is equal

to the energy loss through radiation.

(^22)  (^2)  4
L
I e rL T
r
  


2
4
2 2 3
I
T
r e

 

So T is independent on L.
Heat capacity is more at high temperature so for
a given amount of heat the charge in temperature
is small for the body at high temperature.
Let C (^1) avg and C (^2) avg are the heat capacities at lower
and higher temperatures.
C T (^1) avg   0  C T (^2) avg  100 0
avg
avg
2
1
1
100
C T
C T
 

T     100 T T 50 c
Let T be the temperature at the junction.
(^50)  (^100)  (^400)  (^0) 
0.1 0.2
  A T   A T

(^)   T 20 C
Let T is the temperature of the black body
The rate of radiation emitted by the body is
(^4)   1
dQ
e AT e
dt
  
The total radiation energy supplied by it in five
minutes is e AT^45 
e AT^45  mc T
(^) 
4
1 1
4
2 2
AT i C
A T T



Given that 2 1 , 2 2 1
2
A
A  T T
     T 8 C Tf 18 C
(^) P AT     ^44 R T^24
 
4
2
' 4 2
2
T
P    R   
 
 '
4
P
P
The general equation of a progressive wave is
y A kx t sin( )
v f
100 1
500 5
  m
At t s y0 , 0.02 at x0.25
2 1
0.02 sin.
4
y A


 
   
 
(at t s 0 )
5
0.02 sin
2
y A
 
   
 
 A 0.02m
 y 0.02sin(kx t)
0.02sin(10 0.2 1000 5 10 ]    ^4
0.02sin[2 0.5 ] 
3
0.02sin 0.02
2
m
 
  
 
The intensities on the two scales are shown in
the graph.
3.Sol:
4.Sol:
5.Sol:
6.Sol:
7.Sol:
8.Sol:
Waves on String
1.Sol:
Sound WavesWaves
1.Sol:
From the graph it can be found that at 5 KHZ the
intensity in decibles is
60 20
40
9 5 5 1
x x
x dB
 
  
 