In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf 1.5 V is found to be 60 cm. If this cell is replaced by another cell of emf E, the length-of null point increases by 40 cm. The value of E is (x/10) V. The value of x is

Question

In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf 1.5 V is found to be 60 cm. If this cell is replaced by another cell of emf E, the length-of null point increases by 40 cm. The value of E is (x/10) V. The value of x is  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(E1/E2)=(l1/l2) (1.5/E2)=(60/60+40)=(6/10)=(3/5) E2=(5/2)=(x/10) x=25

Q. In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf 1.5V is found to be 60cm. If this cell is replaced by another cell of emf E, the length-of null point increases by 40cm. The value of E is 10x​V. The value of x is _______

E2​E1​​=l2​l1​​ E2​1.5​=60+4060​=106​=53​ E2​=25​=10x​ x=25

$\frac{E _{1}}{E _{2}} = \frac{l _{1}}{l _{2}}$
$\frac{1.5}{E _{2}} = \frac{60}{60 + 40} = \frac{6}{10} = \frac{3}{5}$
$E_{2} = \frac{5}{2} = \frac{x}{10}$
$x = 25$