In a meter bridge, as shown in the figure, it is given that resistance Y=12.5 Ω and that the balance is obtained at a distance 39.5 cm from end A (by Jockey J). After interchanging the resistances X and Y, a new balance point is found at a distance l2 from end A. What are the values of X and l2 ?

Question

In a meter bridge, as shown in the figure, it is given that resistance Y=12.5 Ω and that the balance is obtained at a distance 39.5 cm from end A (by Jockey J). After interchanging the resistances X and Y, a new balance point is found at a distance l2 from end A. What are the values of X and l2 ? (A) 8.16 Ω and 60.5 cm (B) 19.15 Ω and 39.5 cm (C) 8.16 Ω and 39.5 cm (D) 19.15 Ω and 60.5 cm

Tardigrade Admin · Accepted Answer

The balanced condition of Wheatstone bridge is

X (100 - l_{1}) = Y \times l_{1}

Given

l_{1} = 39.5 c m; Y = 12.5 □

. Therefore,

X (100 - 39.5) = 12.5 (39.5)

\Rightarrow X (60.5) = 12.5 (39.5)

\Rightarrow X = \frac{12.5 \times 39.5}{60.5} = 8.16Ω

Now, if

X

and

Y

are interchanged then balanced condition of Wheatstone bridge becomes

Y (100 - l_{2}) = X l_{2}

In this condition,

Y = 12.5 □; X = 8.16

. Therefore,

12.5 (100 - l_{1}) = 8.16 l_{2}

\Rightarrow 1250 - 12.5 l_{2} = 8.16 l_{2}

\Rightarrow 1250 = 20.66 l_{2}

\Rightarrow l_{2} = \frac{1250}{20.66} = 60.6 c m

$8.16 Ω$ and $60.5 c m$

$19.15 Ω$ and $39.5 c m$

$8.16 Ω$ and $39.5 c m$

$19.15 Ω$ and $60.5 c m$

8.16Ω and 60.5cm

19.15Ω and 39.5cm

8.16Ω and 39.5cm

19.15Ω and 60.5cm

$8.16 Ω$ and $60.5 c m$

$19.15 Ω$ and $39.5 c m$

$8.16 Ω$ and $39.5 c m$

$19.15 Ω$ and $60.5 c m$