wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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 is equal to 8×104 J, by the end of the second revolution (after the beginning of the motion)?

A
0.15 m/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.18 m/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.2 m/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.1 m/s2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 0.1 m/s2
Given m=10 g=102 kg
R=6.4 cm=6.4×102 m,
Final Kinetic energy after two revolutions Kf=8×104 J
12mv2=8×104 J
102v2=16×104
v=0.4 m/s
Final velocity after two revolutions v=0.4 m/s
Given that tangential acceleration is constant, let it be a
Using kinematics equation,
v2u2=2as
here,
Initial velocity u=0
Final velocity v=0.4 m/s
Distance covered for two revolutions is s=4πR

0.42=2a(4πR)
0.16=8πa(0.064)
a=0.160.512π=0.0990.1 m/s2

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon