Explanation
Consider a cube as shown in the fig-1; with uniformly distributed charge within its volume. The potential at one of its vertex P is $${V_0}$$.A cubical portion of half the size (half edge length) of the original cube is cut and removed as shown in the fig-Find the modulus of potential at the point P in the new structure.
A potentiometer has a wire of 100 cm length and its resistance is 10 ohms. It is connected in series with a resistance of 40 ohms and a battery of emf 2 V and negligible internal resistance. If a source of unknown emf E connected in the secondary is balanced by 40 cm length of potentiometer wire, the value of E is:
For the following circuits, the potential difference betweenX and Y in volt is $$\left( {{V_x} - {V_y}} \right)$$
$$\textbf{Step1: Slope of $$V-I$$ graph}$$
Ohm's Law states that $$V=IR$$
So, $$ \dfrac{{{V}_{1}}}{{{I}_{1}}}\ =\tan \theta \ = \ {{R}_{1}} $$
$$\&$$ $$\dfrac{{{V}_{2}}}{{{I}_{2}}}\ =\tan (90-\theta )\ =\cot \theta \ =\ {{R}_{2}} $$
$$\textbf{Step2: Calculation of}$$ $$T_2 -T_1$$
Temperature dependence of resistance is given by:
$$R_1 = R_0(1+\alpha T_1)$$
$$R_2 = R_0(1+\alpha T_2)$$
Subtracting above equations, we get
$$R_2-R_1 = \alpha(T_2-T_1)$$
$$\therefore T_2-T_1$$ is proportional to $$R_2-R_1 $$
$$\Rightarrow$$ $${{T}_{2}}-{{T}_{1}}\ \propto \ \left( \cot \theta -\tan \theta \right)=\dfrac{\cos \theta }{\sin \theta }-\dfrac{\sin \theta }{\cos \theta }=\dfrac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{\sin \theta \cos \theta } $$
$$ {{T}_{2}}-{{T}_{1}}\ \propto \ \dfrac{2\ \cos 2\theta }{\sin 2\theta }\ =\ 2\cot 2\theta $$
Hence, $${{T}_{2}}-{{T}_{1}}\ \propto \ \cot 2\theta $$
Therefore, Option C is correct
Length of potentiometer wire $$l=1\,m$$
Emf $$e={{E}_{1}}$$
Emf of cell $${{E}_{2}}=1.02\,v/s$$
Null point $$=51\,cm=0.51\,m$$
(a)Potential gradient of the wire
$$ V=\dfrac{1.02}{0.51} $$
$$ V=2\,V/m $$
(b) Emf of cell
$$ {{V}_{PQ}}={{E}_{1}} $$
$$ {{E}_{1}}=2\,V $$
(c) When switch S is closed, null point does not shifts because current if not drawn from E2 again.
A circular coil of area $$8{m^2}$$ and number of turns 20 is placed in a magnetic field of 2T with its plane perpendicular to it. It is rotated with an angular velocity of 20rev/s about its natural axis. The emf induced is
Given,
Resistance become $$R=3{{R}_{o}}$$
$$ R={{R}_{0}}\left( 1+\alpha \Delta T \right) $$
$$ 3{{R}_{o}}={{R}_{o}}\left( 1+\alpha \Delta T \right) $$
$$ \Delta T=\dfrac{2}{\alpha }=\dfrac{2}{4\times {{10}^{-3}}}=500{{\,}^{o}}C $$
Hence, when rise in temperature is $$500{{\,}^{o}}C$$
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