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