A daniel cell is balanced at 125 cm length of a potentiometer wire. When the cell is circuited by a resistance of 2 Ω , the balance length is obtained at 100 cm. The internal resistance of the Daniel cell is

Question

A daniel cell is balanced at 125 cm length of a potentiometer wire. When the cell is circuited by a resistance of 2 Ω , the balance length is obtained at 100 cm. The internal resistance of the Daniel cell is (A) (4/3)Ω (B) 1.5Ω (C) 1.25Ω (D) 0.5Ω

Tardigrade Admin · Accepted Answer

As we know that the internal resistance of Daniel cell is given as :- r=R((ll/l2) - 1) =2((125/100) - 1) =2((5/4) - 1) =2× (1/4) =0.5 Ω

Q. A daniel cell is balanced at 125 cm length of a potentiometer wire. When the cell is circuited by a resistance of $2 Ω$ , the balance length is obtained at 100 cm. The internal resistance of the Daniel cell is

$\frac{4}{3} Ω$

$1.5Ω$

$1.25Ω$

$0.5Ω$

As we know that the internal resistance of Daniel cell is given as :-
$r = R (\frac{l _{l}}{l _{2}} - 1)$
$= 2 (\frac{125}{100} - 1)$
$= 2 (\frac{5}{4} - 1)$
$= 2 \times \frac{1}{4}$
$= 0.5 Ω$

Q. A daniel cell is balanced at 125 cm length of a potentiometer wire. When the cell is circuited by a resistance of 2Ω , the balance length is obtained at 100 cm. The internal resistance of the Daniel cell is

34​Ω

1.5Ω

1.25Ω

0.5Ω