A wire of resistance $R_{1}$ is drawn out so that its length is increased by twice of its original length.The ratio of new resistance to original resistance is.
A$9: 1$
B$1: 9$
C$4: 1$
D$3: 1$
JEE MAIN 2022, Easy
Download our app for free and get started
A$9: 1$
a $R _{1}=\rho \frac{ L _{1}}{ A _{1}}$
$R _{2}=\rho\left(\frac{3 L _{1}}{ A _{1} / 3}\right)=9 \rho \frac{ L _{1}}{ A _{1}}$
$\therefore \frac{ R _{2}}{R _{1} }=9$
Download our app
and 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.*
Two electric bulbs, rated at $(25\, W, 220\, V)$ and $(100\, W, 220\, V)$, are connected in series acroos a $220\, V$ voltage source. If the $25\, W$ and $100\, W$ bulbs draw powers $P_1$ and $P_2$ respectively, then
A wire of resitance $R$ and length $L$ is cut into $5$ equal part. if these parts are joined parts are joined paralley, than result resistance will be:
In the circuit shown here, what is the value of the unknown resistor $R$ so that the total resistance of the circuit between points $P$ and $Q$ is also equal to $R$