Question 161 of 280

A huge circular arc of length $4.4ly$ subtends an angle $4s$ at the centre of the circle. How long it would take for a body to complete $4$ revolution if its speed is $8AU$ per second?
Given : $1ly=9.46\times {10}^{15}m$
$1AU=1.5\times {10}^{11}m$

A$3.5\times {10}^{6}s$B$4.5\times {10}^{10}s$C$4.1\times {10}^{8}s$D$7.2\times {10}^{8}$

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Answer: (B) $4.5\times {10}^{10}s$

$\ell =4.4ly=4.4\times 9.46\times {10}^{15}$
length of Arc. $=\ell =R\theta$
$4.4\times 9.46\times {10}^{15}=R\theta$
$\theta =4S=4\times 4.843\times {10}^{-6}=1.94\times {10}^{-5}$ rad
$4.4\times 9.46\times {10}^{15}=R\times 1.94\times {10}^{-5}$
$R=2.1455\times {10}^{21}$ meter
Speed $=8Au=8\times 1.5\times {10}^{11}{ms}^{-1}=12\times {10}^{11}{ms}^{-1}$
$4$ revolution means distance $=4\times 2\pi R$ metre
time $=\dfrac{\text{distance}}{\text{speed}}=\dfrac{4\times 2\pi R}{12\times {10}^{11}};$ time $=\dfrac{8\times 3.14\times 2.1455\times {10}^{21}}{12\times {10}^{11}}\Rightarrow 4.5\times {10}^{10}\sec$