Answer:
The length of the tube is 3.92 m.
Explanation:
Given that,
Electric potential = 100 MV
Length = 4 m
Energy = 100 MeV
We need to calculate the value of 
Using formula of relativistic energy

Put the value into the formula


Here, 



We need to calculate the length
Using formula of length

Put the value into the formula


Hence, The length of the tube is 3.92 m.
If you are pushing the coin across the table at a constant rate, the friction of the table and the horizontal force of your hand pushing are equal, and the coin itself moves at a constant rate. If you push a coin and let it go, there is no horizontal force keeping the coin going. Friction slows the coin to a stop. In both cases, the gravitational downward pull of Earth is equally but oppositely resisted by the upward push of table on the coin.
Kinetic energy is never negative, but potential energy can be.
Potential energy depends on height above some reference level,
and you can pick any level you want as the reference. So, if the
object is below the reference level you pick, then its potential
energy relative to your reference level is negative.
What that means is: You have to lift it / do work on it / give it more
energy than it has now ... in order to move it to the reference level.
(That's exactly the situation with electrons bound to an atom. Their
energy is considered negative, because we have to do work and
give them more energy to rip them away from the atom.)
_____________________________________
Regarding the other choices:
-- Kinetic energy is scalar ... Yes. So is potential energy.
-- Kinetic energy increases with height ...
No. It doesn't, but potential energy does.
-- Kinetic energy depends on position ...
No. It doesn't, but potential energy does.
There are some missing information in the question.
However, since you are talking about magnetic force, I think you refer to the Lorentz force. When a particle of charge q and velocity v is immersed in a magnetic field of intensity B, the force acting on the particle is:

where

is the angle between the magnetic field and the direction of the particle.
Therefore, if force F is doubled, then also the velocity v must be double of its initial value: