The neutral temperature of a thermocouple is $350\,^{\circ}C$ when the cold junction is at $0\,^{\circ}C$. When the cold junction is immersed in a bath of $30\,^{\circ}C$,the inversion temperature is .............. $^{\circ}C$.

The magnetic field intensity at the point $O$ of a loop with current $i$,whose shape is illustrated below is

Six point charges,each of magnitude $q$,are arranged in different manners as shown in the image. In each case,a point $M$ and a line $PQ$ passing through $M$ are shown. Let $E$ be the electric field and $V$ be the electric potential at $M$ (potential at infinity is zero) due to the given charge distribution when it is at rest. Now,the whole system is set into rotation with a constant angular velocity about the line $PQ$. Let $B$ be the magnetic field at $M$ and $\mu$ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current. Match the conditions in Column $I$ with the configurations in Column $II$.

Column $I$	Column $II$
$(A)$ $E=0$	$(p)$ Charges at corners of a regular hexagon. $M$ is the centre. $PQ$ is perpendicular to the plane.
$(B)$ $V \neq 0$	$(q)$ Charges on a line perpendicular to $PQ$ at equal intervals. $M$ is the mid-point.
$(C)$ $B=0$	$(r)$ Charges on two coplanar concentric rings. $M$ is the common centre. $PQ$ is perpendicular to the plane.
$(D)$ $\mu \neq 0$	$(s)$ Charges at corners and mid-points of a rectangle. $M$ is the centre. $PQ$ is parallel to the longer sides.
	$(t)$ Charges on two coplanar,identical rings. $M$ is the mid-point between centres. $PQ$ is perpendicular to the line joining centres.

$A$ proton moving with a constant velocity passes through a region of space without any change in its velocity. If $\vec{E}$ and $\vec{B}$ represent the electric and magnetic fields respectively,then the region of space may have :
$(A)$ $E=0, B=0$
$(B)$ $E=0, B \neq 0$
$(C)$ $E \neq 0, B=0$
$(D)$ $E \neq 0, B \neq 0$
Choose the most appropriate answer from the options given below :

$A$ charged particle (electron or proton) is introduced at the origin $(x=0, y=0, z=0)$ with a given initial velocity $\overrightarrow{v}$. $A$ uniform electric field $\overrightarrow{E}$ and magnetic field $\vec{B}$ are given in columns $I, II$ and $III$, respectively. The quantities $E_0, B_0$ are positive in magnitude.

Column $I$	Column $II$	Column $III$
$(I)$ Electron with $\overrightarrow{v}=2 \frac{E_0}{B_0} \hat{x}$	$(i)$ $\overrightarrow{E}=E_0 \hat{z}$	$(P)$ $\overrightarrow{B}=-B_0 \hat{x}$
$(II)$ Electron with $\overrightarrow{v}=\frac{E_0}{B_0} \hat{y}$	$(ii)$ $\overrightarrow{E}=-E_0 \hat{y}$	$(Q)$ $\overrightarrow{B}=B_0 \hat{x}$
$(III)$ Proton with $\overrightarrow{v}=0$	$(iii)$ $\overrightarrow{E}=-E_0 \hat{x}$	$(R)$ $\overrightarrow{B}=B_0 \hat{y}$
$(IV)$ Proton with $\overrightarrow{v}=2 \frac{E_0}{B_0} \hat{x}$	$(iv)$ $\overrightarrow{E}=E_0 \hat{x}$	$(S)$ $\overrightarrow{B}=B_0 \hat{z}$

$(1)$ In which case will the particle move in a straight line with constant velocity?
$(2)$ In which case will the particle describe a helical path with axis along the positive $z$ direction?
$(3)$ In which case would the particle move in a straight line along the negative direction of $y$-axis (i.e., move along $-\hat{y}$)?

The electric charge in uniform motion produces