NEET-XII-Physics
exam-1 year:2016
- #24A 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^{-4}`` J by the end of the
second revolution after the beginning of the
motion?
(1) 0.1 ``\frac {m}{s^2}``(2) 0.15 ``\frac {m}{s^2}``
(3) 0.18 ``\frac {m}{s^2}`` (4) 0.2 ``\frac {m}{s^2}``digAnsr: 1Ans : 1
Sol. 2
1
mv
2
= E &implies;
21 10 v
2 1000
= 8 × 10-4
&implies; v2 = 16 × 10-2 &implies; v = 4 × 10-1 = 0.4 m/s
Now,
v2 = u2 + 2ats (s = 4R)
&implies;
= +
2
t
16 22 6.4
0 2a 4
100 7 100
&implies; at =
16 7 100
100 8 22 6.4
= 0.1 m/s2