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

The correct option is C 3,9

Let old p.d. be V
Old velocity so, $\frac{1}{2}mv{{}_1}^2=eV+v_1=\sqrt{\frac{2eV}{m}}$
Old energy $=eV$
Now,
New p.d. $=9V$
So,
New energy $=9eV$
Now,
New velocity $v_2=\sqrt{\frac{2e9V}{m}}=3v_1$
So,
I) Velocity increases to 3 times
II) The energy increases to 9 times

So, the answer is option (C).