Two cells of emf $E_1$ and $E_2\left(E_1 > E_2\right)$ are connected individually to a potentiometer and their corresponding balancing length are $625 \,cm$ and $500 \,cm$, then the ratio $\frac{E_1}{E_2}$ is ...........
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.*
In an electrolyte $3.2 \times {10^{18}}$ bivalent positive ions drift to the right per second while $3.6 \times {10^{18}}$ monovalent negative ions drift to the left per second. Then the current is
Resistance $n$, each of $r\; ohm$, when connected in parallel give an equivalentresistance of $R\; ohm$. If these resistances were connected in series, the combination would have a resistance in $ohms$, equal to
Two wires of the same dimensions but resistivities ${\rho _1}$ and ${\rho _2}$ are connected in series. The equivalent resistivity of the combination is
A steady current $I$ is set up in a wire whose cross-sectional area decreases in the direction of the flow of the current. Then, as we examine the narrowing region,
The current flowing through a conductor connected across a source is $2\,A$ and $1.2\,A$ at $0^{\circ}\,C$ and $100^{\circ}\,C$ respectively. The current flowing through the conductor at $50^{\circ}\,C$ will be $......\times 10^2\,mA$.
The combination of two identical cells, whether connected in series or parallel combination provides the same current through an external resistance of $2 \,\Omega$. The value of internal resistance of each cell is ............ $\Omega$