Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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
MathematicsApplication of DerivativesKCETKCET 2021
Options:
  • A $0.6$
  • B 5
  • C $5 k$
  • D 3
Solution:
1979 Upvotes Verified Answer
The correct answer is: 3
$\theta=\frac{t^{2}}{20}+\frac{t}{5}$
On differentiating $\theta$ w.r.t. $t$,
$\begin{aligned} \frac{d \theta}{d t} &=\frac{t}{10}+\frac{1}{5} \\\left(\frac{d \theta}{d t}\right)_{t=4} &=\frac{4}{10}+\frac{1}{5} \end{aligned}$
$\begin{aligned} \Rightarrow & k &=\frac{3}{5} \\ \therefore & & 5 k &=3 \end{aligned}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.