Physics Times 07.2019

2.Sol:

SECTION-3

1.Sol:

x D

m N



(^00)
1
m 1
dx dx
d
C K A m
K A
N


  
   
  
 
 
0 1
dx
x
K A
D


  
 
(^00)
1 1
( )
D
eq
Ddx
d
C C K A D x
  
  

   

0

(^1) ln (2)
eq
D
C K A

0
eq ln 2
K A
C
D

. Therefore  1
1.5sinc1.44sin 90
1.44 24
sin
C 1.50 25
  
24
sin
C 25
x
d
   
25
24
x
d
^ Total length traveled by light
25
9.6 10
24
   m
(^) 
8
8
10
5 10
3 10
1.5
t   

t 50 ns
3.Sol: Let S, L & W are the specific heat of liquid,
latent heat of liquid and water equivalent of
calorimeter respectively
Case-1 If calorimeter is open and after
evaporation liquid escapes
5      S (80 30) 5L W 30 (1)
(^80)    S 20 W 30
(^80)      S 20 5 S 50 5L
(^5) L 1350 S
270
L
S

Case-2 If calorimeter is closed (vapour does
not escape)
Heat gain = Heat loss
5 (80 30) 5S   L W(110 80)
250S + 5L = W × 30 (1)
Now 80gm liquid is poured
Heat gain = Heat loss
Here final temperature  50 oC
80       S 20 5L S 5 30 W 30 (2)
From (1) & (2)
SECTION-3
1.Sol:
2.Sol:
3.Sol:
Case-1
Case-2
4.Sol:
120Ans
L
S

4.Sol: The small work done is dW F dr.
 
dW ydx  2 xdy
A B y dy 1,  0
1
0
WA B ydx  (^1) dx
B C x dx 1,  0
0.5

WB C  2 1  1 dy 2 (0.5)  