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$E$ denotes electric field in a uniform conductor, $I$ corresponding current through it, ${v_d}$ drift velocity of electrons and $P$ denotes thermal power produced in the conductor, then which of the following graph is incorrect
It is preferable to measure the $e.m.f.$ of a cell by potentiometer than by a voltmeter because of the following possible reasons.
$(i)$ In case of potentiometer, no current flows through the cell.
$(ii)$ The length of the potentiometer allows greater precision.
$(iii)$ Measurement by the potentiometer is quicker.
$(iv)$ The sensitivity of the galvanometer, when using a potentiometer is not relevant.
Which of these reasons are correct?
A resistance wire connected in the left gap of a meter bridge balances a $10\, \Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio $3: 2 .$ If the length of the resistance wire is $1.5 m ,$ then the length of $1\, \Omega$ of the resistance wire is $....... \times 10^{-2}\;m$
A battery of $emf$ $10\,V$ is connected to resistances as shown in the figure. The potential difference between $A$ and $B,\,\,(V_A -V_B)$ is ................ $V$
A battery consists of a variable number $n$ of identical cells (having internal resistance reach) which are connected in series. The terminals of the battery are short-circuited and the current $I$ is measured. Which of the graphs shows the correct relationship between $I$ and $n \,?$
As shown in the figure, a network of resistors is connected to a battery of $2\,V$ with an internal resistance of $3\,\Omega$. The currents through the resistors $R_4$ and $R_5$ are $I_4$ and $I_5$ respectively. The values of $I_4$ and $I_5$ are :
Twelve wires of equal length and same cross-section are connected in the form of a cube. If the resistance of each of the wires is $R$, then the effective resistance between the two diagonal ends would be
Two resistances are joined in parallel whose resultant is $\frac{6}{8}\,ohm$. One of the resistance wire is broken and the effective resistance becomes $2\,\Omega $. Then the resistance in ohm of the wire that got broken was