A pendulum of length 2 m lift at P . When it reaches Q , it losses 10% of its total energy due to air resistance. The velocity at Q is

Options

A. 2 m/sec
B. 8 m/sec
C. 6 m/sec
D. 1 m/sec

Answer

C

Explanation

When the bob is at P, it possesses only P.E. =mgh and as it is at rest, its total energy =mgh, where h=2 m. When it reaches the lowest point Q, during its oscillation, its
`P.E.=0 and K.E.=(1)/(2)mv^(2)` where v is its velocity at Q. It loses 10% of its total energy in moving from P to Q.
`therefore` By u"sin"g the principle of conservation of energy, we get `(1)/(2)mv^(2)=(9)/(10)(mgh)`
`therefore v^(2)=(9xx2)/(10)xx10xx2=36 therefore v=6 m//s`