Answer:
Explanation:
Using the magnification formula.
Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)
M = v/u = H1/H2
v/u = H1/H2...1
3) Given the radius of curvature of the concave lens R = 20cm
Focal length F = R/2
f = 20/2
f = 10cm
Object distance u = 5cm
Object height H2= 5cm
To get the image distance v, we will use the mirror formula
1/f = 1/u+1/v
1/v = 1/10-1/5
1/v = (1-2)/10
1/v =-1/10
v = -10cm
Using the magnification formula
(10)/5 = H1/5
10 = H1
H1 = 10cm
Image height of the peg is 10cm
4) If u = 15cm
1/v = 1/f-1/u
1/v = 1/10-1/15
1/v = 3-2/30
1/v = 1/30
v = 30cm
30/15 = H1/5
15H1 = 150
H1/= 10cm
5) if u = 20cm
1/v = 1/f-1/u
1/v = 1/10-1/20
1/v = 2-1/20
1/v = 1/20
v = 20cm
20/20 = H1/5
20H1 = 100
H1 = 5cm
6) If u = 30cm
1/v = 1/f-1/u
1/v = 1/10-1/30
1/v = 3-1/30
1/v = 2/30
v = 30/2 cm
v =>15cm
15/30 = Hi/5
30H1 = 75
H1 = 75/30
H1 = 2.5cm
Choices A, B, and D are false statements.
I think choice-C is trying to say the right thing, but it
might have gotten copied incorrectly.
Electric fields and electric forces both increase as the distance
decreases, and decrease as the distance increases.
The tank has a volume of 
, where 
is its height and 
is its radius.
At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:
The volume of water in the tank at any given time is
and can be expressed as a function of the water level alone:
Implicity differentiating both sides with respect to time 
gives
We're told the water level rises at a rate of 
at the time when the water level is 
, so the net change in the volume of water 
can be computed:
The net rate of change in volume is the difference between the rate at which water is pumped into the tank and the rate at which it is leaking out:
We're told the water is leaking out at a rate of 
, so we find the rate at which it's being pumped in to be
