Question

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Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Mass of ball (m) = 10 grams = 0.01 Kg.
• Radius of circle (r) = 6.4 cm.
• Kinetic energy (K) = 8 × 10^{-4} J.
• Initial velocity (u) = 0 m/s.

$\huge{\bold{\underline{Explanation:-}}}$

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As given Kinetic energy:-

$\large{\boxed{\tt K = \dfrac{1}{2}mv^2}}$

Substituting the values,

$\large{\tt 8 \times 10^{-4} = \dfrac{1}{2} \times 0.01 \times v^2}$

$\large{\tt 8 \times 2 \times 10^{-4} = 0.01 \times v^2}$

$\large{\tt 16 \times 10^{-4} = 0.01 \times v^2}$

$\large{\tt v^2 = \dfrac{16 \times 10^{-4}}{0.01}}$

$\large{\tt v^2 = \dfrac{16 \times 10^{-4}}{10^{-2}}}$

$\large{\boxed{v^2 = 16 \times 10^{-2}}}$

$\large{\tt v^2 = 16 \times 10^{-2}\: ------(1)}$

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As the question is asking acceleration after 2nd revolution,

Then,

Firstly finding the displacement,

$\large{\tt 1 \: revolution = 2 \pi r}$

$\large{\tt 2 \: revolution = 2 \times 2 \pi r}$

Here 2 revolution is equal to the distance travelled by the particle.

$\large{\boxed{\tt S = 4 \pi r}}$

Here we have to find Tangetial acceleration, Denoted by, ${\tt a_T}$

Applying third kinematic equation,

$\large{\boxed{\tt v^2 - u^2 = 2a_TS}}$

Substituting the values, (From equation (1))

[Initial velocity (u) = 0 m/s]

$\large{\tt 16 \times 10^{-2} - 0 = 2 \times a_T \times 4 \pi r}$

$\large{\tt 16 \times 10^{-2} = 8 \pi r \times a_T}$

$\large{\tt a_T = \dfrac{16 \times 10^{-2}}{8 \pi r}}$

Now,

• $\large{\tt \pi = 3.14}$
• $\large{\tt r = 6.4 \: cm = 6.4 \times 10^{-2}\: m}$

Substituting it,

$\large{\tt a_T = \dfrac{16 \times 10^{-2}}{8 \times 3.14 \times 6.4 \times 10^{-2}}}$

Simplifying,

$\large{\tt a_T = \dfrac{\cancel{16} \times \cancel{10^{-2}}}{\cancel{8} \times 3.14 \times 6.4 \times \cancel{10^{-2}}}}$

$\large{\tt a_T = \dfrac{2 }{6.4 \times 3.14}}$

Which nearly comes as,

$\huge{\boxed{\boxed{\tt a_T = 0.1 \: m/s^2}}}$

So, the Tangential acceleration is 0.1 m/s².

Therefore, Option-(2) is correct.

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Answer 2

Answer:

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