$\Rightarrow F \propto \frac{ q }{ m }$
thus $F _{1}: F _{2}: F _{3}=\frac{ q _{1}}{ m _{1}}: \frac{ q _{2}}{ m _{2}}: \frac{ q _{3}}{ m _{3}}$
$=\frac{ e }{ m _{ p }}: \frac{ e }{2 m _{ p }}: \frac{2 e }{4 m _{ p }}$
$=\frac{1}{1}: \frac{1}{2}: \frac{2}{4}$
$=2: 1: 1$
Now for speed calculation
$P = constant \Rightarrow v \propto \frac{1}{ m }$
thus $v _{1}: v _{2}: v _{3}=\frac{1}{ m _{ p }}: \frac{1}{2 m _{ p }}: \frac{1}{4 m _{ p }}$
$=\frac{1}{1}: \frac{1}{2}: \frac{1}{4}$
$=4: 2: 1$
Reason : $I_1 = I_2$ implies that the fields due to the current $I_1$ and $I_2$ will be balanced.
$(A)$ decreasing the number of turns
$(B)$ increasing the magnetic field
$(C)$ decreasing the area of the coil
$(D)$ decreasing the torsional constant of the spring
Choose the most appropriate answer from the options given below.
Reason : The average velocity of free electron is zero.


