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3-2.Motion in Plane
hard
A car goes around uniform circular track of radius $R$ at a uniform speed $v$ once in every $T$ seconds. The magnitude of the centripetal acceleration is $a_c$. If the car now goes uniformly around a larger circular track of radius $2 R$ and experiences a centripetal acceleration of magnitude $8 a_c$. Then, its time period is
A$2 T$
B$3 T$
C$T / 2$
D$3 / 2 T$
(KVPY-2016)
Solution
(c)
When car goes around track of radius $R$, then
Time period, $\quad T=\frac{2 \pi R}{v}$
Centripetal acceleration, $a_c=\frac{v^2}{R}$
When car goes around circular track of radius $2 R$, then
Centripetal acceleration,
$a_c^{\prime}=\frac{v^{\prime 2}}{2 R}=\frac{8 v^2}{R} \Rightarrow v^{\prime}=4 v$
So, time period of car is
$T^{\prime}=\frac{2 \pi R^{\prime}}{v^{\prime}}=2 \pi(2 R)$
$=\frac{1}{2} \times \frac{2 \pi R}{v}=\frac{T}{2}$
When car goes around track of radius $R$, then
Time period, $\quad T=\frac{2 \pi R}{v}$
Centripetal acceleration, $a_c=\frac{v^2}{R}$
When car goes around circular track of radius $2 R$, then
Centripetal acceleration,
$a_c^{\prime}=\frac{v^{\prime 2}}{2 R}=\frac{8 v^2}{R} \Rightarrow v^{\prime}=4 v$
So, time period of car is
$T^{\prime}=\frac{2 \pi R^{\prime}}{v^{\prime}}=2 \pi(2 R)$
$=\frac{1}{2} \times \frac{2 \pi R}{v}=\frac{T}{2}$
Standard 11
Physics
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