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The circuit shown here is used to compare the emf of two cells $E_1$ and $E_2 (E_1 > E_2)$. The null point is at $C$ when the galvanometer is connected to $E_1$. When the galvanometer is connected to $E_2$, the null point will be
In the adjoining circuit, the battery ${E_1}$ has an $e.m.f.$ of $12\,volt$ and zero internal resistance while the battery $E$ has an $e.m.f.$ of $2\,volt$. If the galvanometer $G$ reads zero, then the value of the resistance $X$ in $ohm$ is
A copper wire of diameter $1.02\, mm$ carries a current of $1.7\, amp$. Find the drift velocity $(v_d)$ of electrons in the wire. Given $n$, number density of electrons in copper $= 8.5 \times 10^{27} /m^3$....................... $mm/sec$
The charge flowing through a resistance $R$ varies with time as $Q = 2t - 8t^2$. The total heat produced in the resistance is (for $0 \leq t \leq \frac{1}{8}$)
The resistances in $WSB$ circuit shown in the diagram all are different and the current through the galvanometer is zero. If all thermal effects are negligible, the current through the galvanometer will not be zero, when
$A$ potentiometer wire has length $10\, m$ and resistance $10\,\Omega$ . It is connected to a battery of $EMF$ $11\, volt$ and internal resistance $1\, \Omega$ , then the potential gradient in the wire is ............... $V/m$