The resistance in the two arms of a meter bridge are $5\,\Omega $ and $R\,\Omega $, respectively. When the resistance $R$ is shunted with an equal resistance, the new balance point is at $1.6\, l_1$. The resistance $‘R’$ is ................. $\Omega$
A$10$
B$15$
C$20$
D$25$
Medium
Download our app for free and get started
B$15$
b $\frac{5}{\mathrm{R}}=\frac{\ell_{1}}{100-\ell_{1}}$ and $\frac{5}{\mathrm{R} / 2}=\frac{1.6 \ell_{1}}{100-1.6 \ell_{1}}$
$\Rightarrow \mathrm{R}=15\, \Omega$
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 wire of length ' $r$ ' and resistance $100 \Omega$ is divided into $10$ equal parts. The first $5$ parts are connected in series while the next $5$ parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:
A potentiometer has uniform potential gradient. The specific resistance of the material of the potentiometer wire is $10^{-7} \, ohm-meter$ and the current passing through it is $0.1\, ampere$; cross-section of the wire is $10^{-6}\, m^2$. The potential gradient along the potentiometer wire is
Resistances $R_1$ and $R_2$ each $60\,\Omega$ are connected in series as shown in figure. The Potential difference between $A$ and $B$ is kept $120$ volt. Then what ............. $V$ will be the reading of voltmeter connected between the point $C$ and $D$ if resistance of voltmeter is $120\,\Omega .$
During lighting, a current pulse, shown in figure, flows from the cloud at a height $1.5\ km$ to the ground. If the breakdown electric field of humid air is about $400\ kVm^{-1}$ , the energy released during lighting would be (in unit of $10^9\ J$ )
The charge flowing through a resistance $R$ varies with time $t$ as $ Q=at-bt^2 $ where $a$ and $b$ are positive constants . The total heat produced in $R$ is