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A ball of mass \(1 \mathrm{~g}\) having a charge of \(20 \mu \mathrm{C}\) is tied to one end of a string of length \(0.9 \mathrm{~m}\) can rotate in a vertical plane in a uniform electric field \(100 \mathrm{NC}^{-1}\) 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, \(\mathrm{g}=10 \mathrm{~ms}^{-2}\))
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that it completes the vertical circle is (Let, \(\mathrm{g}=10 \mathrm{~ms}^{-2}\))
Solution:
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Verified Answer
The correct answer is:
\(6 \mathrm{~ms}^{-1}\)
Given, mass of ball, \(m=1 \mathrm{~g}=\frac{1}{10^3} \mathrm{~kg}\) charge on the ball, \(q=20 \mu C=20 \times 10^{-6} \mathrm{C}\), length of the string centre of one end, \(r=0.9 \mathrm{~m}\) electric field, \(E=100 \mathrm{~N} / \mathrm{C}\) and acceleration due to gravity, \(g=10 \mathrm{~m} / \mathrm{s}^2\) 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
\(\begin{aligned}
& F=q E \Rightarrow F=20 \times 10^{-6} \times 100 \\
& F_1=2 \times 10^{-3} \mathrm{~N} \quad \ldots (i)
\end{aligned}\)
Now, force due to the gravitation,
\(F_2=m g=\frac{1}{10^3} \times 10=10 \times 10^{-3} \mathrm{~N}\)...(ii)
Net effective force on the ball,
\(F_{\text {eff }}=F_2-F_1\)
From Eqs. (ii) and (i), we get
\(F_{e f f}=10 \times 10^{-3}-2 \times 10^{-3} \Rightarrow f_{e f f}=8 \times 10^{-3} \mathrm{~N}\)
Now, effective gravitational acceleration at lowest position is given as,
\(\begin{aligned}
& F_{e f f}=m g_{e f f} \\
& g_{e f f}=\frac{F_{e f f}}{m}=\frac{8 \times 10^{-3}}{10^{-3}}=8 \mathrm{~m} / \mathrm{s}^2
\end{aligned}\)
The minimum horizontal velocity that must be given to the ball at the lowest position is,
\(\begin{aligned}
& v=\sqrt{5 g_{e f f} r} \quad \Rightarrow v=\sqrt{5 \times 8 \times \frac{9}{10}} \\
& v=\sqrt{4 \times 9}=\sqrt{36} \Rightarrow v=6 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
\(\therefore\) Force \(=\) Charge on the ball \(\times\) Electric field
\(\begin{aligned}
& F=q E \Rightarrow F=20 \times 10^{-6} \times 100 \\
& F_1=2 \times 10^{-3} \mathrm{~N} \quad \ldots (i)
\end{aligned}\)
Now, force due to the gravitation,
\(F_2=m g=\frac{1}{10^3} \times 10=10 \times 10^{-3} \mathrm{~N}\)...(ii)
Net effective force on the ball,
\(F_{\text {eff }}=F_2-F_1\)
From Eqs. (ii) and (i), we get
\(F_{e f f}=10 \times 10^{-3}-2 \times 10^{-3} \Rightarrow f_{e f f}=8 \times 10^{-3} \mathrm{~N}\)
Now, effective gravitational acceleration at lowest position is given as,
\(\begin{aligned}
& F_{e f f}=m g_{e f f} \\
& g_{e f f}=\frac{F_{e f f}}{m}=\frac{8 \times 10^{-3}}{10^{-3}}=8 \mathrm{~m} / \mathrm{s}^2
\end{aligned}\)
The minimum horizontal velocity that must be given to the ball at the lowest position is,
\(\begin{aligned}
& v=\sqrt{5 g_{e f f} r} \quad \Rightarrow v=\sqrt{5 \times 8 \times \frac{9}{10}} \\
& v=\sqrt{4 \times 9}=\sqrt{36} \Rightarrow v=6 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
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