$$ r → = 10 t 2 i ^ + 5 t 3 j ^ and mass of object $m=0.1\mathrm{kg}$, then at $t=1s$:

(A) $\overrightarrow{p}=2\hat{i}+1.5\hat{j}$

(B) $\overrightarrow{F}=2\hat{i}+3\hat{j}$

(C) $\overrightarrow{L}=5\hat{k}$

(D) $\overrightarrow{\tau }=20\hat{k}$

Correct Answer: A, B, C, D are correct

Let us analyze each statement step-by-step by calculating the velocity, acceleration, momentum, force, angular momentum, and torque of the object.

The position vector of the object of mass $m=0.1\mathrm{kg}$ at any time $t$ is given by:

$\overrightarrow{r}=10t^2\hat{i}+5t^3\hat{j}$

At time $t=1s$, the position vector is:

$\overrightarrow{r}=10\left(1^2\right)\hat{i}+5\left(1^3\right)\hat{j}=10\hat{i}+5\hat{j}$

Step 1: Finding Velocity and Momentum
The velocity vector $\overrightarrow{v}$ is the first derivative of the position vector with respect to time:

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=\frac{d}{dt}(10t^2\hat{i}+5t^3\hat{j})=20t\hat{i}+15t^2\hat{j}$

At $t=1s$, the velocity is:

$\overrightarrow{v}=20\left(1\right)\hat{i}+15\left(1^2\right)\hat{j}=20\hat{i}+15\hat{j}$

Using the definition of linear momentum $\overrightarrow{p}=m\overrightarrow{v}$:

$\overrightarrow{p}=0.1\times (20\hat{i}+15\hat{j})=2\hat{i}+1.5\hat{j}$

Thus, statement (A) is correct.

Step 2: Finding Acceleration and Force
The acceleration vector $\overrightarrow{a}$ is the first derivative of the velocity vector with respect to time:

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=\frac{d}{dt}(20t\hat{i}+15t^2\hat{j})=20\hat{i}+30t\hat{j}$

At $t=1s$, the acceleration is:

$\overrightarrow{a}=20\hat{i}+30\left(1\right)\hat{j}=20\hat{i}+30\hat{j}$

Using Newton's second law, the force $\overrightarrow{F}=m\overrightarrow{a}$ acting on the object is:

$\overrightarrow{F}=0.1\times (20\hat{i}+30\hat{j})=2\hat{i}+3\hat{j}$

Thus, statement (B) is correct.

Step 3: Finding Angular Momentum
The angular momentum $\overrightarrow{L}$ about the origin is given by the cross product of the position vector and linear momentum:

$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$

Substituting the values at $t=1s$:

$\overrightarrow{L}=(10\hat{i}+5\hat{j})\times (2\hat{i}+1.5\hat{j})$

Using the vector cross product properties ($\hat{i}\times \hat{i}=0$, $\hat{j}\times \hat{j}=0$, $\hat{i}\times \hat{j}=\hat{k}$, $\hat{j}\times \hat{i}=-\hat{k}$):

$\overrightarrow{L}=[10\times 1.5(\hat{i}\times \hat{j}\left)\right]+[5\times 2(\hat{j}\times \hat{i}\left)\right]$
$\overrightarrow{L}=15\hat{k}-10\hat{k}=5\hat{k}$

Thus, statement (C) is correct.

Step 4: Finding Torque
The torque $\overrightarrow{\tau }$ about the origin is given by the cross product of the position vector and the force vector:

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$

Substituting the values at $t=1s$:

$\overrightarrow{\tau }=(10\hat{i}+5\hat{j})\times (2\hat{i}+3\hat{j})$

Expanding the cross product:

$\overrightarrow{\tau }=[10\times 3(\hat{i}\times \hat{j}\left)\right]+[5\times 2(\hat{j}\times \hat{i}\left)\right]$
$\overrightarrow{\tau }=30\hat{k}-10\hat{k}=20\hat{k}$

Thus, statement (D) is correct.

Since statements (A), (B), (C), and (D) are all verified to be correct, the overall correct option is that A, B, C, D are correct.