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3-2.Motion in Plane
hard
A huge circular arc of length $4.4$ $ly$ subtends an angle $'4 {s}'$ at the centre of the circle. How long it would take for a body to complete $4$ revolution if its speed is $8 \;AU\;per\, second \;?$
Given : $1\, {ly}=9.46 \times 10^{15} \,{m},$ $\, {AU}=1.5 \times 10^{11}\, {m}$
A$4.1 \times 10^{8} \,{s}$
B$4.5 \times 10^{10} \,{s}$
C$3.5 \times 10^{6}\, {s}$
D$7.2 \times 10^{8} \,{s}$
(JEE MAIN-2021)
Solution
${R}=\frac{\ell}{\theta}$
$\text { Time }=\frac{4 \times 2 \pi {R}}{{v}}=\frac{4 \times 2 \pi}{{v}}\left(\frac{\ell}{\theta}\right)$
$\text { put } \ell=4.4 \times 9.46 \times 10^{15}$
${v}=8 \times 1.5 \times 10^{11}$
$\theta=\frac{4}{3600} \times \frac{\pi}{180} {rad}$
$\text { we get time }=4.5 \times 10^{10} {sec}$
$\text { Time }=\frac{4 \times 2 \pi {R}}{{v}}=\frac{4 \times 2 \pi}{{v}}\left(\frac{\ell}{\theta}\right)$
$\text { put } \ell=4.4 \times 9.46 \times 10^{15}$
${v}=8 \times 1.5 \times 10^{11}$
$\theta=\frac{4}{3600} \times \frac{\pi}{180} {rad}$
$\text { we get time }=4.5 \times 10^{10} {sec}$
Standard 11
Physics
