The position vectors of two 1 kg particles, (A) and (B), are given by $\overrightarrow{\mathrm{r}}_{\mathrm{A}}=\left(\alpha_{1} \mathrm{t}^{2} \hat{\mathrm{i}}+\alpha_{2} \hat{\mathrm{t}} \hat{\mathrm{j}}+\alpha_{3} t \hat{\mathrm{k}}\right) \mathrm{m}$ and $\overrightarrow{\mathrm{r}}_{\mathrm{B}}=\left(\beta_{1} t \hat{\mathrm{i}}+\beta_{2} \mathrm{t}^{2} \hat{\mathrm{j}}+\beta_{3} t \hat{\mathrm{k}}\right) \mathrm{m}$, respectively; ( $\alpha_{1}=1 \mathrm{~m} / \mathrm{s}^{2}, \alpha_{2}=3 \mathrm{n} \mathrm{m} / \mathrm{s}, \alpha_{3}=2 \mathrm{~m} / \mathrm{s}, \beta_{1}=2 \mathrm{~m} / \mathrm{s}$, $\beta_{2}=-1 \mathrm{~m} / \mathrm{s}^{2}, \beta_{3}=4 \mathrm{p} \mathrm{m} / \mathrm{s}$ ), where t is time, n and p are constants, At $\mathrm{t}=1 \mathrm{~s},\left|\overrightarrow{\mathrm{~V}}_{\mathrm{A}}\right|=\left|\overrightarrow{\mathrm{V}}_{\mathrm{B}}\right|$ and velocities $\vec{V}_{A}$ and $\vec{V}_{B}$ of the particles are orthogonal to each other. At $\mathrm{t}=1 \mathrm{~s}$, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is $\sqrt{L} \mathrm{kgm}^{2} \mathrm{~s}^{-1}$. The value of L is ____.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Rotational Mechanics

The position vectors of two 1 kg particles, (A) and (B), are given by

r_{A} = (α_{1} t^{2} \hat{i} + α_{2} \hat{t} \hat{j} + α_{3} t \hat{k}) m

and

r_{B} = (β_{1} t \hat{i} + β_{2} t^{2} \hat{j} + β_{3} t \hat{k}) m

, respectively;
(

α_{1} = 1 m / s^{2}, α_{2} = 3 nm / s, α_{3} = 2 m / s, β_{1} = 2 m / s

β_{2} = - 1 m / s^{2}, β_{3} = 4 pm / s

), where t is time, n and p are constants, At

t = 1 s, V_{A} = V_{B}

and velocities

V_{A}

and

V_{B}

of the particles are orthogonal to each other. At

t = 1 s

, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is

L kgm^{2} s^{- 1}

. The value of L is ____.

Answer: 90

Differentiating the position vectors gives the velocities

V_{A} = 2 t \hat{i} + 3 n \hat{j} + 2 \hat{k}, V_{B} = 2 \hat{i} - 2 t \hat{j} + 4 p \hat{k} .

t = 1

V_{A} = (2, 3 n, 2)

V_{B} = (2, - 2, 4 p)

.

Orthogonality (

V_{A} \cdot V_{B} = 0

4 - 6 n + 8 p = 0 \Rightarrow 3 n = 2 + 4 p .

Equal magnitudes (

∣ V_{A} ∣ = ∣ V_{B} ∣

8 + 9 n^{2} = 8 + 16 p^{2} \Rightarrow 9 n^{2} = 16 p^{2} \Rightarrow 3 n = \pm 4 p .

Combining with

3 n = 2 + 4 p

, only

3 n = - 4 p

is consistent (the

+

case forces

0 = 2

), so

- 4 p = 2 + 4 p \Rightarrow p = - \frac{1}{4}

, and

n = \frac{1}{3}

.

At

t = 1

V_{A} = (2, 1, 2)

, and the position vectors are

r_{A} = (1, 1, 2)

r_{B} = (2, - 1, - 1)

, giving

r_{A / B} = r_{A} - r_{B} = (- 1, 2, 3) .

The angular momentum of A about B's position is

L = m_{A} (r_{A / B} \times V_{A}) = \hat{i} - 1 2 \hat{j} 21 \hat{k} 32 = \hat{i} (4 - 3) - \hat{j} (- 2 - 6) + \hat{k} (- 1 - 4) = (1, 8, - 5) .

∣ L ∣ = 1 + 64 + 25 = 90 kg m^{2} s^{- 1} .

L = 90

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Integer answer