Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
The galvanometer deflection, when key $K_1$ is closed but $K_2$ is open, equals $\theta_0$ (see figure). On closing $K_2$ also and adjusting $R_2$ to $5\,\Omega $ , the deflection in galvanometer becomes $\frac{{\theta _0}}{5}$. The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]: .................. $\Omega$
The length of a wire of a potentiometer is $100\, cm$, and the $emf$ of its standard cell is $E\,volt$. It is employed to measure the $e.m.f$ of a battery whose internal resistance is $0.5 \,\Omega$. If the balance point is obtained at $l = 30\, cm$ from the positive end, the $e.m.f.$ of the battery is
Figure shows three similar lamps $L_1, L_2$ and $L_3$ connected across a power supply. If the lamp $L_3$ fuses, how will the light emitted by $L_1$ and $L_2$ change ?
Fig. shows rough sketch of meter bridge. $(G)$ deflects zero at length $\ell \, cm$. Now $R_1$ and $R_2$ are interchanged then balancing length increases by $25\, cm$. Find $R_1/R_2$
A uniform metallic wire has a resistance of $18\,\Omega $ and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is .................. $\Omega$
In an experiment to find $emf$ of a cell using potentiometer, the length of null point for a cell of emf $1.5\,V$ is found to be $60\,cm$. If this cell is replaced by another cell of $emf\; E$. the length-of null point increases by $40\,cm$. The value of $E$ is $\frac{x}{10} V$. The value of $x$ is $............$