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

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q29 – JEE-MAIN 2026

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Home Physics Motion in a Straight LineQuestion #8

The position of an object having mass

0.1

kg as a function of time

t

is given as

\vec{r} = (10 t^{2} \hat{i} + 5 t^{3} \hat{j})

m. At

t = 1

s, which of the following statements are correct ? A. The linear momentum

\vec{p} = (2 \hat{i} + 1.5 \hat{j})

kg-m/s. B. The force acting on the object

\vec{F} = (2 \hat{i} + 3 \hat{j})

N. C. The angular momentum of the object about its origin

\vec{L} = 15 \hat{k}

J-s. D. The torque acting on the object about its origin

\vec{τ} = 20 \hat{k}

N-m. Choose the correct answer from the options given below :

Clue 1 of 3: Kinematic Relationships

Recall the definitions of velocity, acceleration, linear momentum, force, angular momentum, and torque in terms of position and mass.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Calculate kinematic quantities at

t = 1

s✦ Active

Given the position vector $\vec{r} (t) = (10 t^{2} \hat{i} + 5 t^{3} \hat{j})$ m and mass $m = 0.1$ kg. First, find the velocity and acceleration vectors by differentiating $\vec{r} (t)$ with respect to time $t$ :

\vec{v} (t) = \frac{d \vec{r}}{d t} = (20 t \hat{i} + 15 t^{2} \hat{j}) m/s

\vec{a} (t) = \frac{d \vec{v}}{d t} = (20 \hat{i} + 30 t \hat{j}) {m/s}^{2}

Now, evaluate these vectors at $t = 1$ s:

\vec{r} (1) = (10 (1)^{2} \hat{i} + 5 (1)^{3} \hat{j}) = (10 \hat{i} + 5 \hat{j}) m

\vec{v} (1) = (20 (1) \hat{i} + 15 (1)^{2} \hat{j}) = (20 \hat{i} + 15 \hat{j}) m/s

\vec{a} (1) = (20 \hat{i} + 30 (1) \hat{j}) = (20 \hat{i} + 30 \hat{j}) {m/s}^{2}

Step 2: Evaluate linear momentum and force○ Expand

Using the calculated velocity and acceleration at $t = 1$ s, and the given mass $m = 0.1$ kg:

A. Linear momentum $\vec{p} = m \vec{v}$ :

\vec{p} (1) = 0.1 (20 \hat{i} + 15 \hat{j}) = (2 \hat{i} + 1.5 \hat{j}) kg-m/s

Statement A is **correct**.

B. Force acting on the object $\vec{F} = m \vec{a}$ :

\vec{F} (1) = 0.1 (20 \hat{i} + 30 \hat{j}) = (2 \hat{i} + 3 \hat{j}) N

Statement B is **correct**.

💡 Teacher's Secret Hint

Remember to use the correct mass and the vectors evaluated at the specific time $t = 1$ s.

Step 3: Evaluate angular momentum and torque about the origin○ Expand

C. Angular momentum about the origin $\vec{L} = \vec{r} \times \vec{p}$ :

\vec{L} (1) = (10 \hat{i} + 5 \hat{j}) \times (2 \hat{i} + 1.5 \hat{j})

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

= 15 \hat{k} + 10 (- \hat{k}) = 5 \hat{k} J-s

Statement C is **incorrect** (it states $15 \hat{k}$ J-s).

D. Torque about the origin $\vec{τ} = \vec{r} \times \vec{F}$ :

\vec{τ} (1) = (10 \hat{i} + 5 \hat{j}) \times (2 \hat{i} + 3 \hat{j})

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

= 30 \hat{k} + 10 (- \hat{k}) = 20 \hat{k} N-m

Statement D is **correct**.

Statements A, B, and D are correct. Therefore, the correct option is 4.

💡 Teacher's Secret Hint

Pay close attention to the cross product rules: $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{i} = - \hat{k}$ .

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Physics Question 29 – JEE-MAIN 2026

Clue 1 of 3: Kinematic Relationships

🧠 Full Solution Path