The position of an object having mass $0.1 \text{ kg}$ as a function of time $t$ is given as $\vec{r} = (10t^2\hat{i} + 5t^3\hat{j}) \text{ m}$. At $t = 1 \text{ s}$, which of the following statements are correct?
$A$. The linear momentum $\vec{p} = (2\hat{i} + 1.5\hat{j}) \text{ kg} \cdot \text{m/s}$.
$B$. The force acting on the object $\vec{F} = (2\hat{i} + 3\hat{j}) \text{ N}$.
$C$. The angular momentum of the object about its origin $\vec{L} = 15\hat{k} \text{ J} \cdot \text{s}$.
$D$. The torque acting on the object about its origin $\vec{\tau} = 20\hat{k} \text{ N} \cdot \text{m}$.
Choose the correct answer from the options given below:

  • A
    $A, B$ and $C$ only
  • B
    $B, C$ and $D$ only
  • C
    $A, C$ and $D$ only
  • D
    $A, B$ and $D$ only

Explore More

Similar Questions

$A$ wagon of $200\, kg$ is moving on a smooth track with a velocity of $2\, m/s$. $A$ man of $80\, kg$ runs in the wagon with a velocity such that the speed of the centre of mass of the system is zero. Find the relative velocity of the man with respect to the wagon in $m/s$.

$A$ solid metallic sphere of radius '$R$' having moment of inertia '$I$' about its diameter is melted and recast into a solid disc of radius '$r$' of uniform thickness. The moment of inertia of the disc about an axis passing through its edge and perpendicular to its plane is also equal to '$I$'. The ratio $\frac{r}{R}$ is

$A$ rotating body has angular momentum $L$. If its frequency is doubled and its kinetic energy is halved,what will be its new angular momentum?

This question has Statement $1$ and Statement $2$. Of the four choices given after the Statements,choose the one that best describes the two Statements.
Statement $1$ : When the moment of inertia $I$ of a body rotating about an axis with angular speed $\omega$ increases,its angular momentum $L$ remains unchanged,but the kinetic energy $K$ decreases if no external torque is applied.
Statement $2$ : $L = I\omega$ and the rotational kinetic energy $K = \frac{1}{2}I\omega^2 = \frac{L^2}{2I}$.

$A$ $2 \, kg$ steel rod of length $0.6 \, m$ is clamped on a table vertically at its lower end and is free to rotate in a vertical plane. The upper end is pushed so that the rod falls under gravity. Ignoring the friction due to clamping at its lower end,the speed of the free end of the rod when it passes through its lowest position is $\ldots \ldots \ldots \ldots \, ms^{-1}$. (Take $g = 10 \, ms^{-2}$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo