As we know that

here we know that


now from above equation we have


so image will form on left side of lens at a distance of 15 cm
This image will be magnified and virtual image
Ray diagram is attached below here
Solution: The correct order is: C, A, B
The statement of the problem:
How can we prove Earth is round and calculate its circumference?
Hypotheis:
If the sun casts shadows at different angles at the same time of day in different places, we can determine how much Earth curves.
If the Earth was flat, the angle measured at different places at the same time of the day would be same.
Observation:
In Syene, the sun's rays are vertical at noon. At the same time in Alexandria, the rays are 7.2 degrees from the vertical.
Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.
At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .
At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .
At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .
Answer:
Given that
V= 0.06 m³
Cv= 2.5 R= 5/2 R
T₁=500 K
P₁=1 bar
Heat addition = 15000 J
We know that heat addition at constant volume process ( rigid vessel ) given as
Q = n Cv ΔT
We know that
P V = n R T
n=PV/RT
n= (100 x 0.06)(500 x 8.314)
n=1.443 mol
So
Q = n Cv ΔT
15000 = 1.433 x 2.5 x 8.314 ( T₂-500)
T₂=1000.12 K
We know that at constant volume process
P₂/P₁=T₂/T₁
P₂/1 = 1000.21/500
P₂= 2 bar
Entropy change given as

Cp-Cv= R
Cp=7/2 R
Now by putting the values


a)ΔS= 20.79 J/K
b)
If the process is adiabatic it means that heat transfer is zero.
So
ΔS= 20.79 J/K
We know that

Process is adiabatic




Answer:
( a ) The specific volume by ideal gas equation = 0.02632 
% Error = 20.75 %
(b) The value of specific volume From the generalized compressibility chart = 0.0142 
% Error = - 34.85 %
Explanation:
Pressure = 1 M pa
Temperature = 50 °c = 323 K
Gas constant ( R ) for refrigerant = 81.49 
(a). From ideal gas equation P V = m R T ---------- (1)
⇒
= 
⇒ Here
= Specific volume = v
⇒ v = 
Put all the values in the above formula we get
⇒ v =
×81.49
⇒ v = 0.02632 
This is the specific volume by ideal gas equation.
Actual value = 0.021796 
Error = 0.02632 - 0.021796 = 0.004524 
% Error =
× 100
% Error = 20.75 %
(b). From the generalized compressibility chart the value of specific volume
= v = 0.0142 
The actual value = 0.021796 
Error = 0.0142 - 0.021796 = 
% Error =
× 100
% Error = - 34.85 %