Physics Times – July 2019

(Darren Dugan) #1

(^2)
600
20.684
29
H   cm
Case-III:
2 1 2 1 ;
n n n n
v u R

 
3
3
1 3 1
2
H 2 30 300

 
  
;
(^3)
600
19.354
31
H   cm
6.Sol:
1 1 2 2
1 1 2 2
8
5
P P
mix
V V
nC n C
nC n C


 

1 1 2 2
1
PV PV
W




8/5
8/5 0
0 0 (^24)
V
PV P  
 
P P 2 9.2 0
0
0 0 0
0
9.2
(^413)
3/ 5
V
PV P
W RT

  
(^)  W RT (^130)
TV TV1 1γ 1 2 γ 1
T T 2  1 (2) 2.36/5 T 0
Average kinetic energy of gas mixture
nC TVmix 2  23 RT 0
7.Sol:
Let y y y1 2, & 3 are the positions of the three
blocks w.r.t the fixed pulley.
The length of the first string is
l y y 1  1  p (1)
The length of the second string is
l y y y y 2  2  p 3  p (2)
From eq’s 1 & 2
y y y 2  3 2 constant 1 
On differentiating w.r.to t
2 2 2
2 3 1
2 2 2
2
0
d y d y d y
dt dt dt
  
(^) a a 2  3 2( ) 0a 1 
a a a a 1  3  2  1
For other options use m equivalent
1 2
1 2
4( )( ) 4( )(2 ) 8
eq 2 3
m m M M M
m
m m M M
  
 
Case-III:
6.Sol:
7.Sol:
The maximum extension in the spring is
2
0 0
1 8
2 3
Mg
kx  x

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