In a region $10^{19}$ $\alpha -$ particels and $10^{19}$ protons per second move to the left, while $10^{19}$ electrons moves to the right per second. The current is
A$3.2\,A$ towards left
B$3.2\,A$ towards right
C$6.4\,A$ towards left
D$6.4\,A$ towards right
Medium
Download our app for free and get started
C$6.4\,A$ towards left
c Current due to protons and $\alpha$ -particles is towards left and as ${e^ - }$ have $-$ ive charge, so current due to electrons is also towards left
$\therefore $ total current $\mathrm{I}=(2 \mathrm{e}+\mathrm{e}+\mathrm{e}) \times 10^{19}$
${=4 \times 1.6 \times 10^{-19} \times 10^{19}} $
${=6.4 \mathrm{\,A}}$
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
$50\,\Omega $ and $100\,\Omega $ resistors are connected in series. This connection is connected with a battery of $2.4\, volts$. When a voltmeter of $100\,\Omega $ resistance is connected across $100\,\Omega $ resistor, then the reading of the voltmeter will be ............. $V$
In the electric network shown, when no current flows through the $4 \Omega$ resistor in the arm EB, the potential difference between the points $A$ and $D$ will be $.............\,V$
In an experiment of potentiometer for measuring the internal resistance of primary cell a balancing length $\ell $ is obtained on the potentiometer wire when the cell is open circuit. Now the cell is short circuited by a resistance $R$. If $R$ is to be equal to the internal resistance of the cell the balancing length on the potentiometer wire will be
In a potentiometer experiment the balancing with a cell is at length $240\, cm$. On shunting the cell with a resistance of $2$ $\Omega$, the balancing length becomes $120\, cm$. The internal resistance of the cell is ................. $\Omega $
The charge flowing through a resistance $R$ varies with time as $Q = 2t - 8t^2$. The total heat produced in the resistance is (for $0 \leq t \leq \frac{1}{8}$)
In order to increase the resistance of a given wire of uniform cross section to four times its value, a fraction of its length is stretched uniformly till the full length of the wire becomes $\frac{3}{2}$ times the original length what is the value of this fraction?