In a potentiometer arrangement, a cell of emf 1.20 V gives a balance point at 36 cm length of wire. This cell is now replaced by another cell of emf 1.80 V. The difference in balancing length of potentiometer wire in above conditions will be cm.

Question

In a potentiometer arrangement, a cell of emf 1.20 V gives a balance point at 36 cm length of wire. This cell is now replaced by another cell of emf 1.80 V. The difference in balancing length of potentiometer wire in above conditions will be cm. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

1.2=( Potential Gradient ) × 36 1.8=( Potential Gradient ) × x On dividing, we get (2/3) =(36/x) x =18 × 3=54 cm Hence difference =54-36=18 cm

Q. In a potentiometer arrangement, a cell of emf $1.20 V$ gives a balance point at $36 c m$ length of wire. This cell is now replaced by another cell of emf $1.80 V$ . The difference in balancing length of potentiometer wire in above conditions will be _______ $c m$ .

$1.2 =$ ( Potential Gradient ) $\times 36$
$1.8 =$ ( Potential Gradient ) $\times x$
On dividing, we get
$\frac{2}{3} = \frac{36}{x}$
$x = 18 \times 3 = 54 c m$
Hence difference $= 54 - 36 = 18 c m$

Q. In a potentiometer arrangement, a cell of emf 1.20V gives a balance point at 36cm length of wire. This cell is now replaced by another cell of emf 1.80V. The difference in balancing length of potentiometer wire in above conditions will be _______cm.

1.2=( Potential Gradient )×36 1.8=( Potential Gradient ) ×x On dividing, we get 32​=x36​ x=18×3=54cm Hence difference =54−36=18cm

Q. In a potentiometer arrangement, a cell of emf $1.20 V$ gives a balance point at $36 c m$ length of wire. This cell is now replaced by another cell of emf $1.80 V$ . The difference in balancing length of potentiometer wire in above conditions will be _______ $c m$ .

$1.2 =$ ( Potential Gradient ) $\times 36$
$1.8 =$ ( Potential Gradient ) $\times x$
On dividing, we get
$\frac{2}{3} = \frac{36}{x}$
$x = 18 \times 3 = 54 c m$
Hence difference $= 54 - 36 = 18 c m$