The bob of a pendulum was released from a horizontal position. The length of the pendulum is $$10$$ m. If it dissipates $$10\%$$ of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is: [Use, $$g = 10 \text{ m s}^{-2}$$]

Given,
bob of pendulum is released from horizontal position,
Length of Pendulum ,L= $$10$$ m
$$g = 10 \text{ m s}^{-2}$$
At horizontal position, the bob is at height h=L above the lowest point.

Energy of bob at Initial Position is Ei = P.E = mgh

where: m = mass of the bob - $$g = 10 \text{ m s}^{-2}$$ (acceleration due to gravity) - h= $$10$$ m (length of the pendulum) So,

                                 P.E= m$$\times\ $$g$$\times\ 10$$

Now Lets Calculate Energy Dissipated:-

                                 K.E=$$\ \frac{\ 1}{2}\times\ m\times\ v^2$$
 Since 10% of the energy is dissipated due to air resistance, the energy available for conversion to kinetic energy (KE) is:

                                 K.E(final)=$$90\ \%\ $$ of $$m\times\ g\times\ h$$ = $$\ \frac{\ 1}{2}\times\ m\times\ v^2$$

                                 K.E(final)=$$\ \frac{\ 90}{100}\times\ m\times\ 10\times\ 10$$= $$\ \frac{\ 1}{2}\times\ m\times\ v^2$$
                                 On cancelling m on both sides,

                                 K.E(final)=$$\ \frac{\ 90}{100}\times\ 10\times\ 10$$= $$\ \frac{\ 1}{2}\times\ v^2$$         

                                 $$v^2=2\times\ \ \frac{\ 90}{100}\times\ 10\times\ 10$$

                                  $$v^2=2\times\ \ \frac{\ 90}{100}\times\ 100$$

                                  $$v^2=180$$

                                   $$v=\sqrt{\ 180}$$

                                   $$v=6\sqrt{\ 5}$$

Therefore, Final velocity v=$$6\sqrt{\ 5}$$
                  Correct answer is A