The complete question in the attached figure
Let
c ------------------- > is the speed of light
v ------------------- > is the speed in medium
n ------------------- > is the refractive index of medium
we know that
c/v = n
n = (3 x 10^8)/(2.04 x 10^8) = 1.47
1.47 is the refractive index of glycerol.
therefore
the answer is (4) glycerol
They have scales and they rub off easily
x = V<em>x</em> * t
given V<em>x</em> = 10m/s n t = 3.53s
x = 10 * 3.53
= 35.3m
Answer:
Final velocity of electron,
Explanation:
It is given that,
Electric field, E = 1.55 N/C
Initial velocity at point A, 
We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :
........(1)
a is the acceleration, 
We know that electric force, F = qE

Use above equation in equation (1) as:


v = 647302.09 m/s
or

So, the final velocity of the electron when it reaches point B is
. Hence, this is the required solution.