SATHEE: Work Energy And Power Question 249

Work Energy And Power Question 249

Question: . A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to $8 \times 1 0^{- 4} J$ by the end of the second revolution after the beginning of the motion?

Options:

A) $0.1 m / s^{2}$

B) $0.15 m / s^{2}$

C) $0.18 m / s^{2}$

D) $0.2 m / s^{2}$

Show Answer

Answer:

Correct Answer: A

Solution:

[a] Given: Mass of particle,

$M = 10 g = \frac{10}{1000} k g$

radius of circle $R = 6.4 c m$

Kinetic energy E of particle $= 8 \times 1 0^{- 4} J$ acceleration $a_{t} = ?$

$\frac{1}{2} m v^{2} = E \Rightarrow \frac{1}{2} (\frac{10}{1000}) v^{2} = 8 \times 10^{- 4}$

$\Rightarrow v^{2} = 16 \times 10^{- 2} \Rightarrow v = 4 \times 10^{- 1} = 0.4 m / s$

Now, using $v^{2} = u^{2} + 2 a_{t} S (s = 4 π R)$

${(0.4)}^{2} = 0^{2} + 2 a_{t} (4 \times \frac{22}{7} \times \frac{6.4}{100})$

$\Rightarrow a_{t} = {(0.4)}^{2} \times \frac{7 \times 100}{8 \times 22 \times 6.4} = 0.1 m / s^{2}$

Work Energy And Power Question 249

Options:

Answer:

Solution:

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Work Energy And Power Question 249

Question: . A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8×10−4J by the end of the second revolution after the beginning of the motion?

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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