The resistance of a wire is ${10^{ - 6}}\,\Omega $ per meter. It is bend in the form of a circle of diameter $2\,m$. A wire of the same material is connected across its diameter. The total resistance across its diameter $AB$ will be
On solving, $R' = 0.88 \times {10^{ - 6}}\,\Omega $
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Equal potentials are applied on an iron and copper wire of same length. In order to have the same current flow in the two wires, the ratio $r$ (iron)/$r$ (copper) of their radii must be (Given that specific resistance of iron = $1.0 \times {10^{ - 7}}$ $ ohm-m$ and specific resistance of copper = $1.7 \times {10^{ - 8}}\,ohm-m$)
In order to measure the internal resistance $r_1$ of a cell of emf $E$, a meter bridge of wire resistance $R_0=50 \Omega$, a resistance $R_0 / 2$, another cell of emf $E / 2$ (internal resistance $r$ ) and a galvanometer $G$ are used in a circuit, as shown in the figure. If the null point is found at $l=72 cm$, then the value of $r_1=$ . . . . $\Omega$
You are given several identical resistances each of value $R = 10\,\Omega $ and each capable of carrying maximum current of $1\, ampere$. It is required to make a suitable combination of these resistances to produce a resistance of $5\,\Omega $ which can carry a current of $4\, amperes$. The minimum number of resistances of the type $R$ that will be required for this job
A cell of $emf$ $E$ and internal resistance $r$ is connected in series with an external resistacne $nr.$ Then, the ratio of the terminal potential difference to $emf$ is