Consider a drop of rain water having mass `1 gram` , 50 `m//s` Take `g` constant with a volume `10 m//s^(2)`. The work done by the
(i) gravitational force and the
(ii) resistive force of air is :

Options

A. (i)-10J,(ii)-8.25J
B. (i)1.25J,(ii)-8.25J
C. (i)100J (ii) 8.75J
D. (i)10 J (ii)-8.75 J

Answer

d

Explanation

(d)Thinking process By work -KE theorem, we have change in KE= work done by all of the forces,
`W_(g)=mgh=10^(-3)xx10xx1xx10^(3)=10J`
Now, from work-KE theorem, we have
`DeltaK=W_("gravity")+W_("air resistance")`
`implies1/(2)xxmv^(2)=mgh+W_("air resistance")`
`W_("air resistance")=(1)/(2)mv^(2)-mgh`
`=10^(-3)((1)/(2)xx50xx50-10xx10^(3))`
` =8.5J`