Question

a pendulum of length 2m lift at P. When it reaches Q. it losses 10% of itstotal energy due to air resistance. The velocity at Q is
(A) 6m/sec
(B) 1m/sec
(C) 2m/sec
(D) 8m/sec
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Answer 1

$\huge\ANSWER..$

Answer 2

The velocity at Q is (A)  6m/sec

1. At point P only the potential energy is present  is
2. P.E = mgh ,here m is the mass of bob , g is acceleration due to gravity and h is the height
3. now when there is no air resistance all the potential energy is converted into kinetic energy at point Q
4. K.E = 1/2 mV^2 = mgh
5. Now when air resistance is there the energy is lost by 10%
6. now K.E =1/2mV^2=mgh-mgh/10
7. V^2=2 x 10 x 2(1-1/10)=40x9/10= 36
8. V= 6 m/s