In the given potentiometer circuit length of the wire $AB$ is $3\, m$ and resistance is $R = 4.5 \, \Omega$ . The length $AC$ for no deflection in galvanometer is
Medium
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since cell connected in secondary circuit is in wrong way so we will not get balancing point.
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A charged particle having drift velocity of $7.5 \times 10^{-4}\, ms ^{-1}$ is an electric field of $3 \times 10^{-10}\, Vm ^{-1}$ has a mobility in $m ^{2} V ^{-1} s ^{-1}$ of
In an aluminium $(A1)$ bar of square cross section, a square hole is drilled and is filled with iron ( $Fe$ ) as shown in the figure. The electrical resistivities of $A 1$ and $Fe$ are $2.7 \times 10^{-8} \ \Omega m$ and $1.0 \times 10^{-7} \ \Omega m$, respectively. The electrical resistance between the two faces $P$ and $Q$ of the composite bar is
In the circuit shown, the reading of the Ammeter is doubled after the switch is closed. Each resistor has a resistance $1\,\Omega$ and the ideal cell has an $e.m.f.$ $10\,V$. Then, the Ammeter has a coil resistance equal to ............ $\Omega$