Search any question & find its solution
Question:
Answered & Verified by Expert
Assume proton is rotating along a circular path of radius $1 \mathrm{~m}$ under a centrifugal force of $4 \times 10^{-12} \mathrm{~N}$. If the mass of proton is $1.6 \times 10^{-27} \mathrm{~kg}$, then its angular velocity of rotation is
Options:
Solution:
2055 Upvotes
Verified Answer
The correct answer is:
$5 \times 10^7 \mathrm{rad} / \mathrm{s}$
Given, $r=1 \mathrm{~m}$
$\begin{aligned}
& F_{\text {centrifugal }}=4 \times 10^{-12} \mathrm{~N} \\
& m=1.6 \times 10^{-27} \mathrm{~kg} \\
& \omega=?
\end{aligned}$
$\begin{aligned} & \because \quad F_{\text {centrifugal }}=m \omega^2 r \Rightarrow \omega^2=\frac{F_{\text {centrifugal }}}{m r} \\ & \omega=\sqrt{\frac{F_{\text {centrifugal }}}{m r}}=\sqrt{\frac{4 \times 10^{-12}}{1.6 \times 10^{-27} \times 1}}=\sqrt{\frac{10^{-12}}{0.4 \times 10^{-27}}} \\ & =\sqrt{25 \times 10^{15}}=\sqrt{25 \times 10^{14}}=5 \times 10^7 \mathrm{rad} / \mathrm{s} \\ & \end{aligned}$
$\begin{aligned}
& F_{\text {centrifugal }}=4 \times 10^{-12} \mathrm{~N} \\
& m=1.6 \times 10^{-27} \mathrm{~kg} \\
& \omega=?
\end{aligned}$
$\begin{aligned} & \because \quad F_{\text {centrifugal }}=m \omega^2 r \Rightarrow \omega^2=\frac{F_{\text {centrifugal }}}{m r} \\ & \omega=\sqrt{\frac{F_{\text {centrifugal }}}{m r}}=\sqrt{\frac{4 \times 10^{-12}}{1.6 \times 10^{-27} \times 1}}=\sqrt{\frac{10^{-12}}{0.4 \times 10^{-27}}} \\ & =\sqrt{25 \times 10^{15}}=\sqrt{25 \times 10^{14}}=5 \times 10^7 \mathrm{rad} / \mathrm{s} \\ & \end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.