A potential difference of $5 \,V$ is applied across a conductor of length $10 \,cm$. If drift velocity of electrons is $2.5 \times 10^{-4} \,m / s$, then electron mobility will be ............ $m ^2 V ^{-1} s ^{-1}$
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(b)
$\mu=\frac{V_D}{E}$
$\mu=\frac{2.5 \times 10^{-4} \times 0.1}{5}$
$\mu=5 \times 10^{-6} \,m ^2 V ^{-1} s ^{-1}$
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$(C)$ If $w _1=2 w _2$ and $d _1= d _2$, then $V _2=2 V _1$
$(D)$ If $w _1=2 w _2$ and $d _1= d _2$, then $V _2= V _1$
$2.$ Consider two different metallic strips ($1$ and $2$) of same dimensions (lengths $\ell$, width w and thickness $d$ ) with carrier densities $n_1$ and $n_2$, respectively. Strip $1$ is placed in magnetic field $B_1$ and strip $2$ is placed in magnetic field $B_2$, both along positive $y$-directions. Then $V_1$ and $V_2$ are the potential differences developed between $K$ and $M$ in strips $1$ and $2$, respectively. Assuming that the current $I$ is the same for both the strips, the correct option$(s)$ is(are)
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$(C)$ If $B _1=2 B _2$ and $n _1= n _2$, then $V _2=0.5 V _1$
$(D)$ If $B_1=2 B_2$ and $n_1=n_2$, then $V_2=V_1$
Give the answer question $1$ and $2.$
