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3-2.Motion in Plane
medium
A particle is revolving in a circle of radius $2\ m$ with angular velocity $\omega = t^2 -4t + 8\ rad/s$ . The time when speed of the particle becomes $8\ m/s$ is ......... $\sec$
A$1$
B$2$
C$3$
D$4$
Solution
$\mathrm{v}=8 \mathrm{m} / \mathrm{s}=\mathrm{R} \omega$
$\Rightarrow \omega=4=t^{2}-4 t+8$
$t^{2}-4 t+4=0$
$\Rightarrow t=2 \sec$
$\Rightarrow \omega=4=t^{2}-4 t+8$
$t^{2}-4 t+4=0$
$\Rightarrow t=2 \sec$
Standard 11
Physics
