The resistance of the meter bridge AB is given figure is 4
. With a cell of emf ε=0.5 V and rheostat resistance Rh= 2 
the null point is obtained at some point J. When the cell is replaced by another one of emf ε=
the same null point J is found for Rh=6 
. The emf 
is ;
(a)
0.6 V
(b)
0.5 V
(c)
0.3 V
(d)
0.4 V
The resistance of the meter bridge AB is given figure is 4. With a cell of emf ε=0.5 V and rheostat resistance Rh= 2 the null point is obtained at some point J. When the cell is replaced by another one of emf ε= the same null point J is found for Rh=6 . The emf is ;
(a)
0.6 V
(b)
0.5 V
(c)
0.3 V
(d)
0.4 V
Question ID - 50312 | SaraNextGen Top Answer
The resistance of the meter bridge AB is given figure is 4. With a cell of emf ε=0.5 V and rheostat resistance Rh= 2 the null point is obtained at some point J. When the cell is replaced by another one of emf ε= the same null point J is found for Rh=6 . The emf is ;
(a)
0.6 V
(b)
0.5 V
(c)
0.3 V
(d)
0.4 V
|
The resistance of the meter bridge AB is given figure is 4
|
|||
|
(a) |
0.6 V |
(b) |
0.5 V |
|
(c) |
0.3 V |
(d) |
0.4 V |
Potential gradient with Rh =2
is 
×
=
L=100cm
Let null point be at 
cm
Thus 
=0.5V=
……(1)
Now with Rh = 6
and at null point
…...(2)
Dividing equation (1) by (2) we get
thus 