In an $NPN$ transistor,$10^{10}$ electrons enter the emitter region in $10^{-6} \, s$. If $2\%$ of electrons are lost in the base region,then the collector current and the current amplification factor $(\beta)$ are,respectively:

$A$ crystal of intrinsic silicon at room temperature has a carrier concentration of $1.6 \times 10^{16} / m^3$. If the donor concentration level is $4.8 \times 10^{20} / m^3$,then the concentration of holes in the semiconductor is

$A$ particle falls from a height $h$ upon a fixed horizontal plane and rebounds. If $e$ is the coefficient of restitution,the total distance travelled before rebounding has stopped is

A positive ion $A$ and a negative ion $B$ has charges $6.67 \times 10^{-19}\ {C}$ and $9.6 \times 10^{-10}\ {C}$, and masses $19.2 \times 10^{-27}\ {kg}$ and $9 \times 10^{-27}\ {kg}$ respectively. At an instant, the ions are separated by a certain distance $r$. At that instant the ratio of the magnitudes of electrostatic force to gravitational force is ${P} \times 10^{-13}$, where the value of $P$ is _________ .
(Take $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\ {Nm}^2 {C}^{-1}$ and universal gravitational constant as $6.67 \times 10^{-11}\ {Nm}^2 {kg}^{-2}$)

If the equilibrium constant for the reaction,$H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$ is $K$,what is the equilibrium constant of $HI_{(g)} \rightleftharpoons \frac{1}{2} H_{2(g)} + \frac{1}{2} I_{2(g)}$?

$y+x^2=\frac{dy}{dx}$ has the solution