Question

The p.d. at the ends of a cathode ray tube is increased to 9 times its original potential difference then
i. The velocity of cathode rays increases to …. Times
ii. The energy of cathode rays increases to …. Times
A. 9,3
B. 9,9
C. 3,9
D. 3,3

Answer 1

We know that relation between potential difference and potential energy is equal to $\Delta V = \dfrac{{\Delta PE}}{q}$ so, from this, we can say that change in potential energy is equal to the product of charge and potential difference. $\Delta PE = q\Delta V$. from this formula, we will find the potential energy of cathode rays.

Formula used: According to the work power theorem change in kinetic energy is equal to work done. kinetic energy is given by $KE = \dfrac{1}{2}m{v^2}$.

Complete Answer:
It is given in the question that the p.d. at the ends of a cathode ray tube is increased to 9 times its original potential difference then we have to find the increase in velocity and increase in energy of the cathode rays.
We know that according to the work power theorem the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle. Or in simple words, we can say that a change in the kinetic energy of a particle is equal to work done.
$W = \Delta KE = \dfrac{1}{2}m{v^2}$.
We know that relation between potential difference and potential energy is equal to
$\Delta V = \dfrac{{\Delta PE}}{q}$ or we can say that change in potential energy is equal to the product of charge and potential difference.
$\Delta PE = q\Delta V$
Let us assume that the initial potential difference is V then its initial potential energy is given by
$\Delta PE = q\Delta V$ or we can say it as eV …. (1)
Let us assume that the initial kinetic energy $KE = \dfrac{1}{2}m{v^2}$. Then from the work power theorem, we get $\Delta PE = \Delta KE$
$eV = \dfrac{1}{2}m{v^2}$ …. (2)
From equation 2 we get $v = \sqrt {\dfrac{{2eV}}{m}}$ … (3)
Similarly, finale potential energy is equal to $\Delta PE = q\Delta V$ but here $\Delta V = 9V$ so, the final potential energy of cathode ray $\Delta PE = 9eV$ …. (4)
And final kinetic energy of cathode ray is equal to $v = \sqrt {\dfrac{{2e9V}}{m}}$ … (5)
Now increase in potential energy of the cathode ray is given by the ratio of finale potential energy initial potential energy. Increase in potential energy $= \dfrac{{9eV}}{{eV}} = 9$ times.
An increase in kinetic energy of the cathode ray is given by the ratio of final kinetic energy and initial kinetic energy. Increase in kinetic energy $= \dfrac{{\sqrt {\dfrac{{2e9V}}{m}} }}{{\sqrt {\dfrac{{2eV}}{m}} }} = 3$ times.
Therefore, the kinetic energy of the cathode ray is increased by 3 times, and the potential energy of the cathode ray is increased by 9 times. Thus, option (C) is the correct answer.

Note: One can understand the question wrong and as a result, they may make mistakes in the last step. They subtract the final potential energy of cathode-ray with an initial energy of cathode-ray similarly final kinetic energy with initial kinetic energy. But this is wrong because here we have to calculate the ratio and not the difference. This mistake can be easily avoided by reading and understanding the question carefully before solving it.