In order to make the parabolas formed by singly ionized ions in one spectrograph and doubly ionized ions in another Thomson's mass spectrograph coincide,the electric fields and magnetic fields are kept in the ratios $1 : 2$ and $3 : 2$ respectively. Then the ratio of the masses of the ions is

  • A
    $3 : 4$
  • B
    $1 : 3$
  • C
    $9 : 4$
  • D
    None of these

Explore More

Similar Questions

Which of the following statements is correct?

Two wires each carrying a steady current $I$ are shown in four configurations in Column $I$. Some of the resulting effects are described in Column $II$. Match the statements in Column $I$ with the statements in Column $II$.
Column $I$Column $II$
$(A)$ Two parallel wires with current in the same direction,$P$ is the midpoint.$(p)$ The magnetic fields $(B)$ at $P$ due to the currents in the wires are in the same direction.
$(B)$ Two coaxial circular loops with current in the same direction,$P$ is the midpoint on the axis.$(q)$ The magnetic fields $(B)$ at $P$ due to the currents in the wires are in opposite directions.
$(C)$ Two coplanar circular loops with current in opposite directions,$P$ is the midpoint.$(r)$ There is no magnetic field at $P$.
$(D)$ Two concentric coplanar circular loops with current in the same direction,$P$ is the common center.$(s)$ The wires repel each other.

$A$ proton moving with a constant velocity passes through a region of space without any change in its velocity. If $\overrightarrow{E}$ and $\overrightarrow{B}$ represent the electric and magnetic fields respectively,then this region of space may have

An $\alpha$ particle of energy $10 \ eV$ is moving in a circular path in a uniform magnetic field. The energy of a proton moving in the same path and the same magnetic field will be [mass of $\alpha$ particle $= 4 \times$ mass of proton]. (in $eV$)

An electron is projected with velocity $v_0$ in a uniform electric field $E$ perpendicular to the field. Again,it is projected with velocity $v_0$ perpendicular to a uniform magnetic field $B$. If $r_1$ is the initial radius of curvature just after entering the electric field and $r_2$ is the initial radius of curvature just after entering the magnetic field,then the ratio $r_1:r_2$ is equal to:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo