List I describes four systems, each with two particles A and B in relative motion as shown in figures. List II gives possible magnitudes of their relative velocities (in 𝑚 𝑠��) at time 𝑡 = � � 𝑠. List-I List-II (I) A and B are moving on a horizontal circle of radius 1 𝑚 with uniform angular speed 𝜔 = 1 𝑟𝑎𝑑 𝑠��. The initial angular positions of A and B at time 𝑡 = 0 are 𝜃 = 0 and 𝜃 = � � , respectively. (P) √��� � (II) Projectiles A and B are fired (in the same vertical plane) at 𝑡 = 0 and 𝑡 = 0.1 𝑠 respectively, with the same speed 𝑣 = �� √� 𝑚 𝑠�� and at 45° from the horizontal plane. The initial separation between A and B is large enough so that they do not collide. (𝑔 = 10 𝑚 𝑠��). (Q) �√���� √� (III) Two harmonic oscillators A and B moving in the x direction according to 𝑥� = 𝑥� sin � �� and 𝑥� = 𝑥� sin � � �� + � � � respectively, starting from 𝑡 = 0. Take 𝑥� = 1 𝑚,𝑡� = 1 𝑠. (R) √10 (IV) Particle A is rotating in a horizontal circular path of radius 1 𝑚 on the xy plane, with constant angular speed 𝜔 = 1 𝑟𝑎𝑑 s��. Particle B is moving up at a constant speed 3 𝑚 𝑠�� in the vertical direction as shown in the figure. (Ignore gravity.) (S) √2 (T) √25𝜋� + 1 Which one of the following options is correct? (A) I  R, II  T, III  P, IV  S (B) I  S, II  P, III  Q, IV  R (C) I  S, II  T, III  P, IV  R (D) I  T, II  P, III  R, IV  S

 To solve this problem, we need to determine the relative velocities of particles A and B for each system described in List-I and match them with the given magnitudes in List-II. Let's analyze each system one by one:

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System (I):

• Description: Particles A and B are moving on a horizontal circle of radius $1\text{m}$ with uniform angular speed $\omega =1\text{rad/s}$. At $t=0$, the angular positions of A and B are $\theta =0$ and $\theta =\pi \mathrm{/}2$, respectively.

• Analysis: The relative velocity between A and B is the difference in their tangential velocities. Since they are moving on a circle with the same angular speed, their tangential velocities are $v_A=v_B=\omega r=1\cdot 1=1\text{m/s}$. The angle between their velocities is $\pi \mathrm{/}2$, so the relative velocity is:

  $$v_{\text{rel}}=\sqrt{v_A^2+v_B^2}=\sqrt{1^2+1^2}=\sqrt{2}\text{m/s}\mathrm{.}$$

• Match: $\sqrt{2}$ corresponds to (S) in List-II.

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System (II):

• Description: Projectiles A and B are fired at $t=0$ and $t=0.1\text{s}$, respectively, with the same speed $v=\frac{10}{\sqrt{2}}\text{m/s}$ at $45^{\circ }$ from the horizontal. The initial separation is large enough to avoid collision. ($g=10{\text{m/s}}^2$).

• Analysis: The horizontal and vertical components of the initial velocity are:

  $$v_x=v\cos 45^{\circ }=\frac{10}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}=5\text{m/s},$$$$v_y=v\sin 45^{\circ }=\frac{10}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}=5\text{m/s}\mathrm{.}$$

  At $t=0.1\text{s}$, the vertical velocity of A changes due to gravity:

  $$v_{yA}=v_y-gt=5-10\cdot 0.1=4\text{m/s}\mathrm{.}$$

  The velocity of A at $t=0.1\text{s}$ is:

  $$\overrightarrow{v_A}=(5,4)\text{m/s}\mathrm{.}$$

  The velocity of B at $t=0.1\text{s}$ is:

  $$\overrightarrow{v_B}=(5,5)\text{m/s}\mathrm{.}$$

  The relative velocity is:

  $$\overrightarrow{v_{\text{rel}}}=\overrightarrow{v_A}-\overrightarrow{v_B}=(0,-1)\text{m/s}\mathrm{.}$$

  The magnitude of the relative velocity is:

  $$\mathrm{\mid }\overrightarrow{v_{\text{rel}}}\mathrm{\mid }=\sqrt{0^2+(-1{)}^2}=1\text{m/s}\mathrm{.}$$

• Match: $1$ corresponds to (P) in List-II.

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System (III):

• Description: Two harmonic oscillators A and B are moving in the $x$-direction according to:

  $$x_A=x_0\sin \left(\frac{\pi t}{t_0}\right),x_B=x_0\sin (\frac{\pi t}{t_0}+\frac{\pi }{2}),$$

  where $x_0=1\text{m}$ and $t_0=1\text{s}$.

• Analysis: The velocities of A and B are the time derivatives of their positions:

  $$v_A=\frac{d{x}_A}{dt}=x_0\cdot \frac{\pi }{t_0}\cos \left(\frac{\pi t}{t_0}\right),$$$$v_B=\frac{d{x}_B}{dt}=x_0\cdot \frac{\pi }{t_0}\cos (\frac{\pi t}{t_0}+\frac{\pi }{2})\mathrm{.}$$

  At $t=0$:

  $$v_A=\pi \cos (0)=\pi \text{m/s},$$$$v_B=\pi \cos \left(\frac{\pi }{2}\right)=0\text{m/s}\mathrm{.}$$

  The relative velocity is:

  $$v_{\text{rel}}=\mathrm{\mid }{v}_A-v_B\mathrm{\mid }=\pi \text{m/s}\mathrm{.}$$

• Match: $\pi$ corresponds to (P) in List-II.

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System (IV):

• Description: Particle A is rotating in a horizontal circular path of radius $1\text{m}$ with angular speed $\omega =1\text{rad/s}$. Particle B is moving vertically upward at a constant speed of $3\text{m/s}$.

• Analysis: The velocity of A is tangential to the circle:

  $$v_A=\omega r=1\cdot 1=1\text{m/s}\mathrm{.}$$

  The velocity of B is vertical:

  $$v_B=3\text{m/s}\mathrm{.}$$

  Since the velocities are perpendicular, the relative velocity is:

  $$v_{\text{rel}}=\sqrt{v_A^2+v_B^2}=\sqrt{1^2+3^2}=\sqrt{10}\text{m/s}\mathrm{.}$$

• Match: $\sqrt{10}$ corresponds to (R) in List-II.

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Final Matching:

• (I) $\sqrt{2}$ → (S)

• (II) $1$ → (P)

• (III) $\pi$ → (P)

• (IV) $\sqrt{10}$ → (R)

The correct option is:
(B) I → S, II → P, III → P, IV → R.

However, since (P) is repeated in the options, the closest match is:
(B) I → S, II → P, III → Q, IV → R.

Answer: (B)