SATHEE: Laws of Motion 4 Question 1

Laws of Motion 4 Question 1

1. Two particles $A$ and $B$ are moving on two concentric circles of radii $R_{1}$ and $R_{2}$ with equal angular speed $ω$ . At $t = 0$ , their positions and direction of motion are shown in the figure. The relative velocity $v_{A} - v_{B}$ at $t = \frac{π}{2 ω}$ is

given by

(2019 Main, 12 Jan II)

(a) $ω (R_{1} + R_{2}) \hat{i}$

(b) $- ω (R_{1} + R_{2}) \hat{i}$

(d) $ω (R_{2} - R_{1}) \hat{i}$

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Answer:

Correct Answer: 1. (d)

Solution:

Angle covered by each particle in time duration 0 to $\frac{π}{2 ω}$ is

$θ = ω \times t = ω \times \frac{π}{2 ω} = \frac{π}{2} r a d$

So, positions of particles at $t = \frac{π}{2 ω}$ is as shown below;

Velocities of particles at $t = \frac{π}{2 ω}$ are

$v_{A} = - ω R_{1} \hat{i} and v_{B} = - ω R_{2} \hat{i}$

The relative velocity of particles is

$\begin{aligned} v_{A} - v_{B} & = - ω R_{1} \hat{i} - (- ω R_{2} \hat{i}) \\ = - ω (R_{1} - R_{2}) \hat{i} = ω (R_{2} - R_{1}) \hat{i} \end{aligned}$

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अगला

Laws of Motion 4 Question 1

1. Two particles $A$ and $B$ are moving on two concentric circles of radii $R_{1}$ and $R_{2}$ with equal angular speed $ω$ . At $t = 0$ , their positions and direction of motion are shown in the figure. The relative velocity $v_{A} - v_{B}$ at $t = \frac{π}{2 ω}$ is

Answer:

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सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Laws of Motion 4 Question 1

1. Two particles A and B are moving on two concentric circles of radii R1 and R2 with equal angular speed ω. At t=0, their positions and direction of motion are shown in the figure. The relative velocity vA−vB at t=π2ω is

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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1. Two particles $A$ and $B$ are moving on two concentric circles of radii $R_{1}$ and $R_{2}$ with equal angular speed $ω$ . At $t = 0$ , their positions and direction of motion are shown in the figure. The relative velocity $v_{A} - v_{B}$ at $t = \frac{π}{2 ω}$ is