Physics Times 07.2019

match is

(a)I S II R III Q IV T ,  ,  , 

(b) I Q II R III P IV U ,  ,  , 

(c) I Q II S III R IV U ,  ,  , 

(d) I Q II R III S IV U ,  ,  , 

SECTION-1

1. (c,d) 2. (b,c&d) 3. (a,b) 4. (b)


2. (a,d) 6. (a,c,d) 7. d 8. (a,c)

SECTION-2

1. 0.69 2. 1 3. 1.5 4. 135


2. 4 6. 0.63

SECTION-3

1. b 2. c 3. d 4. d

SECTION-1

SECTION-2

SECTION-3

SECTION-1
1.Sol: R>> dipole size
So the circle is equipotential
So,

2.Sol:

EnetShould beto surface so

1/3

0 0

3 0

(^404) 0 0
p p
E r
r E
 
   
 
At point B net electric field will be zero.
EB 0
(^3000)
2
( )A Net 3
kp
E E E
r
  
Electric field at point A 0
(^3) ˆ ˆ
[ ]
A 2
E  E i j

( ) 0EB Net
  K U 0
2
0
1
2
I  U
2
(^12)
2 3 4
ML L
    Mg 
 
3
2
g
L


2
radial 0
3 3
2 2 2 4
L g L g
a I
L
     
2
sin60 3 3
2
4
3
L
Mg g
L L
M


 
The acceleration of CM of the rod along
vertical direction is
sin60^2 cos60
v 2 2
L L
 a    
 
^
3 3 3 3
v 8 2 8
g g
a  
^
SECTION-1
1.Sol:
2.Sol:
15
v 16
g
a 