Q. When a cathode ray tube is operated at 2912 V, the velocity of electrons is $3.2 \times 10^{7}$m/s. Find the velocity of cathode ray if the tube is operated at 5824 V
A) $2.4 \times 10^{7}$ m/s
B) $5.2 \times 10^{7}$ m/s
C) $4.525 \times 10^{7}$ m/s
D) $2.4 \times 10^{6}$ m/s
Explanation
(c) $\ \ $ $v = \sqrt{\frac{2eV}{m}}$
Since e, and m are constant, $\frac{v_{1}}{v_{2}} = \sqrt{\frac{V_{1}}{V_{2}}}$
or $v_{2} = v_{1}\sqrt{\frac{V_{1}}{V_{2}}} = 3.2 \times 10^{7}\sqrt{\frac{5824}{2912}}$
=$4.525 \times 10^{7}$m/s
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.