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Work Energy and Power question

2016 · Q144
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Work Energy and Power question

2016 · Q144

NEETPhysicsWork Energy and PowerMCQ+4 / −1
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 enegy of the particle becomes equal to 8 ×\times× 10−-−4 J by the end of the second revoluation after the beginning of the motion ?
  1. A
    0.18 m/s2
  2. B
    0.2 m/s2
  3. C
    0.1 m/s2
  4. D
    0.15 m/s2
View written solutionFree

Correct answer: C

Given: Mass of particle, M = 10g =101000kg = {{10} \over {1000}}kg=100010​kg

radius of circle R = 6.4 cm
Kinetic energy E of particle = 8 × 10–4J
acceleration at = ?

12mv2=E⇒12(101000)v2=8×10−4{1 \over 2}m{v^2} = E \Rightarrow {1 \over 2}\left( {{{10} \over {1000}}} \right){v^2} = 8 \times {10^{ - 4}}21​mv2=E⇒21​(100010​)v2=8×10−4

⇒v2=16×10−2 \Rightarrow {v^2} = 16 \times {10^{ - 2}}⇒v2=16×10−2

⇒v=4×10−1=0.4 m/s \Rightarrow v = 4 \times {10^{ - 1}} = 0.4\,m/s⇒v=4×10−1=0.4m/s

Now, using
v2 = u2 + 2ats (s = 4π\pi πR)

(0.4)2=02+2at(4×227×6.4100){\left( {0.4} \right)^2} = {0^2} + 2{a_t}\left( {4 \times {{22} \over 7} \times {{6.4} \over {100}}} \right)(0.4)2=02+2at​(4×722​×1006.4​)

⇒at=(0.4)2×7×1008×22×6.4=0.1 m/s2 \Rightarrow {a_t} = {\left( {0.4} \right)^2} \times {{7 \times 100} \over {8 \times 22 \times 6.4}} = 0.1\,m/{s^2}⇒at​=(0.4)2×8×22×6.47×100​=0.1m/s2

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