The temperature of the cold junction of a thermocouple is $0\,^{\circ}C$ and the temperature of the hot junction is $T\,^{\circ}C$. The e.m.f. is $E = 16T - 0.04T^2 \, \mu\text{V}$. The temperature of inversion is ................ $^{\circ}C$.

Consider the circuit given here. The potential difference $V_{BC}$ between the points $B$ and $C$ is (in $V$)

$A$ wire of resistance $10 \, \Omega$ is bent to form a circle. $P$ and $Q$ are points on the circumference of the circle dividing it into a quadrant and are connected to a battery of $3 \, V$ and internal resistance $1 \, \Omega$ as shown in the figure. The currents in the two parts of the circle are

The $emf$ of a thermocouple, with one junction kept at $0^{\circ}C$, is given by the equation $e = at + bt^2$. The Peltier coefficient is given by:

For the circuit shown in the figure:
$(A)$ The current $I$ through the battery is $7.5 \text{ mA}$.
$(B)$ The potential difference across $R_L$ is $18 \text{ V}$.
$(C)$ The ratio of powers dissipated in $R_1$ and $R_2$ is $3$.
$(D)$ If $R_1$ and $R_2$ are interchanged,the magnitude of the power dissipated in $R_L$ will decrease by a factor of $9$.

The electrochemical equivalent of a substance is numerically equal to the mass of the substance deposited if a current $I$ flows through the electrolyte for $0.25\,s$. The value of $I$ is ................ $A$.