Question

A particle has charge +q and particle b has has charge +4 q with each of them having the same mass m. When allowed to fall from rest through the same electrical potential difference, the ratio of their speeds will become

Answer 1

Ka/Kb=q×v/4q×v =1/2mVa^2/1/2mVb^2
Va^2/Vb^2=1/4
Va/VB=1/2
Va:Vb=1:2
It is the ratio between the kinectic energies between the two particles....
Hope this is useful for u guys ....

Answer 2

Answer:

The ratio of the speeds of the particles will become 1/2.

Explanation:

Consider that particle A has charge +q and particle B has charge +4q.

$V_{A}$  and  $V_{B}$ are the speed of particle A and particle B respectively.

We know that in terms of charge and voltage, kinetic energy will be:

$K.E.=qV$

K.E. in terms of mass and velocity, $K.E.=\frac{1}{2}mv^{2}$

For particle A,  $K.E_{A}= qV=\frac{1}{2}mV^{2}_{A}$                    ..............(1)

For particle B,  $K.E_{B}=4qV=\frac{1}{2}mV^{2}_{B}$                   .............(2)

On dividing eq. (1) by eq.(2),  we get,

$\frac{K.E_{A}}{K.E_{B}} =\frac{ qV}{4qV}=\frac{\frac{1}{2}mV^{2}_{A}}{\frac{1}{2}mV^{2}_{B}}$

$\frac{V^{2}_{A}}{V^{2}_{B}} =\frac{1}{4}$

$\frac{V_{A}}{V_{B}} =\frac{1}{2}$

Therefore the ratio of  speeds of two given particles  A and B is equal to 1/2.