In the following common emitter circuit,if $\beta = 100$,$V_{CE} = 7\,V$,$V_{BE} = 0\,V$ (negligible),and $R_C = 2\,k\Omega$,then the value of $I_B$ is.......$mA$.

The quadratic equation whose roots are $\sin ^2 18^{\circ}$ and $\cos ^2 36^{\circ}$ is :

Two liquids $A$ and $B$ are at $32\,^{\circ}C$ and $24\,^{\circ}C$. When mixed in equal masses,the temperature of the mixture is found to be $28\,^{\circ}C$. Their specific heats are in the ratio of:

Five elements $A, B, C, D$ and $E$ have work functions $1.2 \ eV, 2.4 \ eV, 3.6 \ eV, 4.8 \ eV$ and $6 \ eV$ respectively. If light of wavelength $4000 \ \mathring{A}$ is allowed to fall on these elements,then photoelectrons are emitted by:

If $u=e^{x^2-y^2}$, then

The voltage gain of a transistor in common emitter configuration is $160$. The resistances in base and collector sides of the circuit are $1 \text{ k}\Omega$ and $4 \text{ k}\Omega$ respectively. If the change in base current is $100 \mu A$, then the change in output current is: