Question

A ball of mass $1g$ having a charge of $20\mu C$ is tied to one end of a string of length $0.9m$ can rotate in a vertical plane in a uniform electric field $100N{C}^{-1}$ directed upwards. The minimum horizontal velocity that must be given to the ball at the lowest position so that it completes the vertical circle is (Let, $g=10m{s}^{-2}$ )

• A. $9m{s}^{-1}$
• B. $18m{s}^{-1}$
• C. $36m{s}^{-1}$
• D. $6m{s}^{-1}$

Answer 1

Answer: (D)

Given, mass of ball, $m=1g=\frac{1}{10^3}kg$
charge on the ball, $q=20\mu C=20\times 10^{-6}C$,
length of the string centre of one end, $r=0.9m$
electric field, $E=100N/C$
and acceleration due to gravity, $g=10m/s^2$
Now, according to the question, force exerted by the electric field due to charge on ball is given as, $\therefore$ Force $=$ Charge on the ball $\times$ Electric field
$F=qE\Rightarrow F=20\times 10^{-6}\times 100$
$F_1=2\times 10^{-3}N$
Now, force due to the gravitation,
$F_2=mg=\frac{1}{10^3}\times 10=10\times 10^{-3}N$
Net effective force on the ball,
$F_{eff}=F_2-F_1$
From Eqs. (ii) and (i), we get
$F_{\text{eff }}=10\times 10^{-3}-2\times 10^{-3}\Rightarrow f_{\text{eff }}=8\times 10^{-3}N$
Now, effective gravitational acceleration at lowest position is given as,
$F_{eff}=m{g}_{\text{eff }}$
$g_{\text{eff }}=\frac{F_{\text{eff }}}{m}=\frac{8\times 10^{-3}}{10^{-3}}=8m/s^2$
The minimum horizontal velocity that mut be given to the ball at the lowest position is,
$v=\sqrt{5g_{\text{eff }}r}\Rightarrow v=\sqrt{5\times 8\times \frac{9}{10}}$
$v=\sqrt{4\times 9}=\sqrt{36}\Rightarrow v=6m/s$