In the meter bridge experiment, the length A B of the wire is 1 m. The resistors X and Y have values 5 Ω and 2 Ω respectively. When a shunt resistance S is connected to X, the balancing point is found to be 0.625 m from A. Then, the resistance of the shunt is

Question

In the meter bridge experiment, the length A B of the wire is 1 m. The resistors X and Y have values 5 Ω and 2 Ω respectively. When a shunt resistance S is connected to X, the balancing point is found to be 0.625 m from A. Then, the resistance of the shunt is (A) 5 Ω (B) 10 Ω (C) 7.5 Ω (D) 12.5 Ω

Tardigrade Admin · Accepted Answer

Here in given condition, we have ((b x/b+ x)/2)=(0.625/0.375) (b x/(b +x) 2) =(25/15) (5 b/(b+5) 2) =(5/3) (b/2 b+10) =(1/3) ⇒ 3 b-2 b =10 b=10 Ω

Q. In the meter bridge experiment, the length $A B$ of the wire is $1 m$ . The resistors $X$ and $Y$ have values $5 Ω$ and $2 Ω$ respectively. When a shunt resistance $S$ is connected to $X$ , the balancing point is found to be $0.625 m$ from $A$ . Then, the resistance of the shunt is

$5 Ω$

$10 Ω$

$7.5 Ω$

$12.5 Ω$

Here in given condition, we have
$\frac{\frac{b x}{b + x}}{2} = \frac{0.625}{0.375}$
$\frac{b x}{( b + x ) 2} = \frac{25}{15}$
$\frac{5 b}{( b + 5 ) 2} = \frac{5}{3}$
$\frac{b}{2 b + 10} = \frac{1}{3}$
$\Rightarrow 3 b - 2 b = 10$
$b = 10Ω$

Q. In the meter bridge experiment, the length AB of the wire is 1m. The resistors X and Y have values 5Ω and 2Ω respectively. When a shunt resistance S is connected to X, the balancing point is found to be 0.625m from A. Then, the resistance of the shunt is

5Ω

10Ω

7.5Ω

12.5Ω