Question

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Answer 1

Answer:

The Correct option is 2.

$\bigstar\;\boxed{\sf 0.1 m/s^2}}$

Explanation:

Given Data -

Mass of the particle - 10 g = 0.01 kg.

Radius of circle which particle is moving - 6.4 cm.

Kinetic Energy - $\bf 8 \times 10^{-4} J$

So,

$\sf\ \implies \dfrac{1}{2}mv^2 = 8 \times 10^{-4} J$

$\sf \implies v^2 = \dfrac{16 \times 10^{-4}}{0.01}$

$\sf \implies 16 \times 10^{-2}\quad \bf...Equation[1]$

Now, as given -

Kinetic energy of a particle is equal to $\bf 8 \times 10^{-4} J$ by the end of 2nd revolution after the begining of the motion of the particle.

It implies, Initial velocity (u) is 0 m/s at this moment.

Now,

Using Newton's 3rd law of motion,

$\bigstar\;\boxed{\sf v^2+u^2 =2as }}$

Known values in Equation,

$\sf\\ \\\implies v^2 = 2a_ts.\\ \\\implies v^2 = 2a_t (4\pi r)\\ \\\implies a_t = \dfrac{v^2}{8 \pi r}\\ \\\implies \dfrac{16\times 10^{-2}}{8\times 3.14 \times 6.4 \times 10^{-2}}\\ \\ \\ \implies a_t = 0.1 m/s^2$

∴ Option 2 is Correct.

Answer 2

★Answer:-

$\boxed{\tt{Option\;(2)\;0.1 m/s^2}}$

★Explanation:-

m = 10 g = 0.01 kg

$K.E. = 1/2 mv^2 = 8 × 10^-4.........(1)$

$Put\;m = 0.01\;in\;(1)$

$v^2 = 16 × 10^-2...... (2)$

$v^2 = u^2 + 2as...... Newton's\;3rd \;law$

$Initial\;velocity\;u = 0$

$So, v^2 = 2 as$

$v^2 = 2 a (4 \pi r)$

$a = v^2/8 \pi r$

$Substituting\: V^2\; and\;r\; in\;above\;eqns$

$\boxed{\tt{a = 0.1\;m/s^2}}$