SATHEE: Kinematics 2 Question 2

Kinematics 2 Question 2

3. A particle moves in a circle of radius $R$ . In half, the period of revolution its displacement is and distance covered is

(1983, 2M)

4 A particle moves from the point $(2.0 \hat{i} + 4.0 \hat{j}) m$ at $t = 0$ with an initial velocity $(5.0 \hat{i} + 4.0 \hat{j}) m s^{- 1}$ . It is acted upon by a constant force which produces a constant acceleration $(4.0 \hat{i} + 4.0 \hat{j}) m s^{- 2}$ . What is the distance of the particle from the origin at time $2 s$ ?

(2019 Main, 11 Jan II)

(a) $5 m$

(b) $20 \sqrt{2} m$

(d) $15 m$

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Answer:

Correct Answer: 3. (d)

Solution:

Forces on the bob are as shown

For equilibrium,

$\begin{aligned} T \cos θ & = m g \\ T \sin θ & = q E \end{aligned}$

and

$\tan θ = \frac{q E}{m g}$

Here,

$\begin{aligned} q = 5 μ C = 5 \times 10^{- 6} C, E = 2000 V / m, \\ m = 2 g = 2 \times 10^{- 3} k g, g = 10 m s^{- 2} \end{aligned}$

$∴ \tan θ = \frac{5 \times 10^{- 6} \times 2000}{2 \times 10^{- 3} \times 10} = \frac{1}{2} = 0.5$

So, the angle made by the string of the pendulum with the vertical is

$θ = \tan^{- 1} (0.5)$

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Kinematics 2 Question 2

3. A particle moves in a circle of radius $R$ . In half, the period of revolution its displacement is and distance covered 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

Kinematics 2 Question 2

3. A particle moves in a circle of radius R. In half, the period of revolution its displacement is and distance covered 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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3. A particle moves in a circle of radius $R$ . In half, the period of revolution its displacement is and distance covered is