$A$ constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if
A
both the length and the radius of the wire are halved.
B
both the length and the radius of the wire are doubled
C
the radius of the wire is doubled
D
the length of the wire is doubled
Medium
Download our app for free and get started
B
both the length and the radius of the wire are doubled
b Heat developed is
$H=\frac{V^{2}}{R} t$
Heat developed will be doubled when $R$ is halved. Further,
$R=\rho l /\left(\pi r^{2}\right)$ and $H=V^{2} \pi r^{2} t / \rho l .$ So heat produced when both the length and radius of the wire are doubled.
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.*
A cell $E _{1}$ of $emf 6 V$ and internal resistance $2 \Omega$ is connected with another cell $E _{2}$ of $emf 4 V$ and internal resistance $8 \Omega$ (as shown in the figure). The potential difference across points $X$ and $Y$ is............ $V$
In a typical Wheatstone network, the resistances in cyclic order are $A = 10 \,\Omega $, $B = 5 \,\Omega $, $C = 4 \,\Omega $ and $D = 4 \,\Omega $ for the bridge to be balanced