When a body is projected vertically up from the ground with certain velocity, its potential energy and kinetic energy at a point $$A$$ are in the ratio $$2 : 3.$$ If the same body is projected with double the previous velocity, then at the same point $$A$$ the ratio of its potential energy to kinetic energy is
A.
$$9:1$$
B.
$$2:9$$
C.
$$1:9$$
D.
$$9:2$$
Answer :
$$1:9$$
Solution :
Let $$E$$ be the total energy then
$$\frac{{P.E}}{{K.E}} = \frac{{mgh}}{{E - mgh}} = \frac{2}{3} \Rightarrow E = \frac{5}{2}mgh$$
When velocity is double then initial energy becomes $$4E.$$
$${\text{So,}}\,\,\frac{{mgh}}{{4E - mgh}} = NL = \frac{{mgh}}{{10mgh - mgh}}$$
On solving we get $$\frac{{P.E}}{{K.E}} = \frac{1}{9}.$$
Releted MCQ Question on Basic Physics >> Work Energy and Power
Releted Question 1
If a machine is lubricated with oil-
A.
the mechanical advantage of the machine increases.
B.
the mechanical efficiency of the machine increases.
C.
both its mechanical advantage and efficiency increase.
D.
its efficiency increases, but its mechanical advantage decreases.
A particle of mass $$m$$ is moving in a circular path of constant radius $$r$$ such that its centripetal acceleration $${a_c}$$ is varying with time $$t$$ as $${a_c} = {k^2}r{t^2}$$ where $$k$$ is a constant. The power delivered to the particles by the force acting on it is:
A.
$$2\pi m{k^2}{r^2}t$$
B.
$$m{k^2}{r^2}t$$
C.
$$\frac{{\left( {m{k^4}{r^2}{t^5}} \right)}}{3}$$
A spring of force-constant $$k$$ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force-constant of-