Q103 / 180 · KCET 2021 · Mathematics

A particle starts form rest and its angular displacement (in radians) is given by $\theta=\frac{t^{2}}{20}+\frac{t}{5}$. If the angular velocity at the end of $t=4$ is $k$, then the value of $5 k$ is

  • A
    $0.6$
  • B
    5
  • C
    $5 k$
  • Correct answer
    D
    3

Solution

$\theta=\frac{t^{2}}{20}+\frac{t}{5}$ On differentiating $\theta$ w.r.t. $t$, $\begin{aligned} \frac{d \theta}{d t} &=\frac{t}{10}+\frac{1}{5} \\\left(\frac{d \theta}{d t}\right)_{t=4} &=\frac{4}{10}+\frac{1}{5} \end{aligned}$ $\begin{aligned} \Rightarrow & k &=\frac{3}{5} \\ \therefore & & 5 k &=3 \end{aligned}$

Hence, the correct answer is option: 'D'.

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