PHYSICS - OBJECTIVE -6

 13. The shown p-V diagram represents the thermodynamic cycle of an engine,operating with an ideal monoatomic gas. The amount of heat, extracted from the source in a single cycle is 

In the above thermodynamic cycle, input energy is at constant pressure (isentropic) .

So Q1=Heat input ; Q1 = nCpΔT

Cp =5/2 R for monoatomic gas (R is the Avagadro constant)

So, Q1 =5/2nRΔT

Similarly ,Q2 = Heat rejected ; Q2=nCvΔT0

Cp - Cv = R ; So, Cv=5/2R - R =3/2R

Hence, Q2 =3/2nRΔT0

Now , from the graph ΔT = 2ΔT0

Q1 + Q2 = 5/2nR x2ΔT + 3 /2nRΔT = 13/2RΔT

Now , from ideal gas equation PV =nRT

So, RΔT =PΔV

Hence we can deduce , 

Q1 + Q2 = 13/2PV0

So, the answer is (b)

 EXPLANATION: Here, the engine cycle can be approximated to Otto cycle .Here the input is at constant pressure (hence it is isentropic) The output is at constant volume (which is adiabatic).Here the automobile engine undergoes huge pressure difference ,since it is controlled combustion. 

So, my point is the same amount of engine volume has been displaced at the input and output strokes.This shows that the automobile is having fairly low efficiency ( efficiency is low). This problem results in huge pollution since output gases are heavily unfinished(unburned) while undergoing combustion in the piston combustion chamber.This process is completed by addition of muffler to exhaust which reduces the sound of expulsion of output gases ( gases which has particles suspended in it )

PHYSICS - OBJECTIVE -5

12. A block of mass m is on an inclined plane of angle θ.The coefficient of friction between the block and the plane is μ and tan θ > μ.The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive.As P is varied from P1 = mg (sinθ - μ cosθ) to P2 = mg (sinθ + μ cosθ), the frictional force f versus P graph will look like 

 

The above question is from Strength of Materials in engineering.Here there are four choices.One has to find out the correct choice as shown above.

Here we are plotting the stress-strain diagram known as Shear stress diagram .This is of five types. They are as follows:

1. Shear stress diagram of indeterminate nature

Here the Shear (bending moment) is centre-shear .This means the beam is uncontrolled by nature which is end supported with fixed hinges.So here the solution is as in (b)

2. Shear stress diagram of single fixed hinge 

 

Here the solution is a downward straight line in the shear stress diagram 

 

 

3. Shear stress diagram of double fixed type 

 

This type shear stress diagram is two triangles ,both up and down inluded

4. Shear stress diagram with one end fixed and other end roller

Here the shear stress diagram curve is parabolic .

5. Shear stress diagram with both end roller

 

Here one end is hinge roller and other end is full roller.Here is shear stress diagram is extremely complicated 

In the above question of inclined plane one end is fixed hinge and other end is free.This means ,among the above choices diagram which shows shear stress diagram as below X-Y line is correct .The graph should be straight line and not inclined ( inclined is for case 3 where both ends are fixed  and parbolic is for case 4 where one end is fixed and other end is roller supported for the beam type)

So answer is (d)