The percentage error in the reading of the voltmeter in the figure shown here is nearly ............ $\%$

To determine the resistance $G$ of a galvanometer by the half-deflection method,a battery of $emf$ $V$ and a series resistance $R$ are used to produce a deflection $\theta$ in the galvanometer. If a shunt resistance $S$ is connected in parallel to the galvanometer to reduce the deflection to $\theta/2$,then $G, R$,and $S$ are related by the equation:

The resistance of a galvanometer is $50\, \Omega$ and the maximum current which can be passed through it is $0.002\, A$. What resistance must be connected to it in order to convert it into an ammeter of range $0 - 0.5\, A$?

$A$ galvanometer,having a resistance of $50 \Omega$,gives a full scale deflection for a current of $0.05 \text{ A}$. The length in metre of a resistance wire of area of cross-section $2.97 \times 10^{-2} \text{ cm}^2$ that can be used to convert the galvanometer into an ammeter which can read a maximum of $5 \text{ A}$ current is: (Specific resistance of the wire $= 5 \times 10^{-7} \Omega\text{-m}$)

When a galvanometer is shunted by a resistance $s$,its current capacity increases $n$ times. If the same galvanometer is shunted by another resistance $s_1$,its capacity will increase to $n_1$ times the original current. The value of $n_1$ is

$A$ galvanometer of resistance $36 \ \Omega$ is converted into an ammeter by using a shunt of $4 \ \Omega$. The fraction $f_0$ of the total current passing through the galvanometer is: