Answer:

Explanation:
As we know that the electrostatic force is a based upon inverse square law
so we have

now since it depends inverse on the square of the distance so we can say

now we know that


also we know that

now from above equation we have



Answer:
i) 21 cm
ii) At infinity behind the lens.
iii) A virtual, upright, enlarged image behind the object
Explanation:
First identify,
object distance (u) = 42 cm (distance between object and lens, 50 cm - 8 cm)
image distance (v) = 42 cm (distance between image and lens, 92 cm - 50 cm)
The lens formula,

Then applying the new Cartesian sign convention to it,

Where f is (-), u is (+) and v is (-) in all 3 cases. (If not values with signs have to considered, this method that need will not arise)
Substituting values you get,
i) 
f = 21 cm
ii) u =21 cm, f = 21 cm v = ?
Substituting in same equation
v ⇒ ∞ and image will form behind the lens
iii) Now the object will be within the focal length of the lens. So like in the attachment, a virtual, upright, enlarged image behind the object.
Let us first write down the known things .
1660 megawatts = 1660 X 10^6 watt
= 166000 kilowatt
From the above deduction we can conclude that the correct option among all the options that are given in the question is the second option or option "B". I hope that this is the answer that you were looking for and the answer has actually come to your desired help.
Controlled variables are influences that could affect the outcome of an experiment, and so are purposely controlled so that they do not impact the experimental results.