If an electric heater is rated at $1000\, W$, then the time required to heat one litre of water from $20\,^oC$ to $60\,^oC$ is
Medium
Download our appand get started for free
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.*
Similar Questions
- 1A bulb rated at ($100\,W$ - $200\,V$) is used on a $100\,V$ line. The current in the bulb isView Solution
- 2Two wires of same metal have the same length but their cross-sections are in the ratio $3:1$. They are joined in series. The resistance of the thicker wire is $10\,\Omega $. The total resistance of the combination will be ............. $\Omega$View Solution
- 3A cell of emf $6\, V$ and resistance $0.5\, ohm$ is short circuited. The current in the cell is ............... $amp$View Solution
- 4Which of the following graphs represent the variation of power loss in the external load with external resistance $R$?View Solution
- 5Two resistances ${R_1}$ and ${R_2}$ are joined as shown in the figure to two batteries of $e.m.f.$ ${E_1}$ and ${E_2}$. If ${E_2}$ is short-circuited, the current through ${R_1}$ isView Solution
- 6In the given circuit diagram, find the current passing through wire $CD$ (in ampere)View Solution
- 7The total current supplied to the circuit by the battery is ............. $A$View Solution
- 8The circuit shown here is used to compare the $e.m.f.$ of two cells ${E_1}$ and ${E_2}({E_1} > {E_2})$. The null point is at $C$ when the galvanometer is connected to ${E_1}$. When the galvanometer is connected to ${E_2}$, the null point will beView Solution
- 9When an ammeter of negligible internal resistance is inserted in series with circuit it reads $1A$. When the voltmeter of very large resistance is connected across $X$ it reads $1V$. When the point $A$ and $B$ are shorted by a conducting wire, the voltmeter measures $10\, V$ across the battery. The internal resistance of the battery is equal to .............. $\Omega$View Solution
- 10Consider a block of conducting material ofresistivity '$\rho$' shown in the figure. Current '$I$' enters at '$A$' and leaves from '$D$'. We apply superp osition principle to find voltage '$\Delta V$ ' developed between '$B$' and '$C$'. The calculation is done in the following steps:View Solution
$(i)$ Take current '$I$' entering from '$A$' and assume it to spread over a hemispherical surface in the block.
$(ii)$ Calculatefield $E(r)$ at distance '$r$' from $A$ by using Ohm's law $E = \rho j$, where j is the current per unit area at '$r$'.
(iii) From the '$r$' dependence of $E(r)$, obtain the potential $V(r)$ at $r$.
(iv) Repeat $(i), (ii)$ and $(iii)$ for current '$I$' leaving '$D$' and superpose results for '$A$' and '$D$'.$\Delta V$ measured between $B$ and $C$ is
