Physics Times – July 2019

5 0

10 80 80

q q q

   

10

5

80

q

 (^)  q 40 C
^ V across^ C 1 40 / 10 4 V
Now just after closing of S 2 charge on each
capacitor remain same.
    10 x 30 40 / 10  y 70 0
30 x  70 y 6 (1)
40 40
5 ( )30 ( ) 100 10
80 80
   x y     x y
(^)   x 30 0
160 130x y 6 0 (2)
y96 / 1510
x0.05amp
Correct option – 2, 3
5.Sol: If (^) h R (^2) r R
0
Q


 clearly from Gauss’ Law
If
8
5
R
h &
3
5
R
r
So for
8
0
5
R
h  
If for h R 2 &
4
5
R
r
(^) enclosed
2
2 2 (1 cos53 )
4 5
Q Q
Q 

     
^
0
2
5
Q



 For
4
2
5
R
h R r  (^) 
0
2
5
Q



If
3
2 &
5
R
h R r 
2 2 (1 cos37 )
4 5
o
enclosed
Q Q
q 

    
^
5.Sol:
6.Sol:
(^50)
Q



6.Sol: When only single material tubes are used
1
(^14)
2 cos 2 0.075 cos 0
7.5cm
1000 2 10 10
o
T
h
rg




 
  
  