Skip to main content

Q1. Anodising is

 Q1. What is Anodising 

(A) a process of coating of zinc by hot dipping 

(B) a zinc diffusion process

(C) a process used for making thin phosphate coating on steel to act as a base or primer for enamels and paints

(D) an oxidizing process used for aluminum and magnesium articles

Ans (D) an oxidizing process used for aluminum and magnesium articles 

*Anodising:- Anodising is an electrolytic process for producing thick oxide layer(coating), usually on aluminium and its alloys. The oxide layer is typically 5 to 30 micro meter in thickness and is used to give improved surface resistance to wear and corrosion,  or as a decorative layer.



Comments

Popular posts from this blog

Question 1. The reading tA and tB of two centigrade scale thermometers A and B graded at ice point 0°C and steam point 100°C are related by the equation tA=l +mtB+n(tB)², where l, m,n are constants. When both are immersed in oil then thermometer A reads 51°C and thermometer B reads 50 °C. determine the reading on thermometer A when thermometer B reads 25°C? Comment on the question "which thermometer is correct".

Thermodynamics Question solution, Chapter1 Zeroth law of thermodynamics. Question 1. The reading tA and tB of two centigrade scale thermometers A and B graded at ice point 0°C and steam point 100°C are related by the equation tA=l +mtB+n(tB)², where l, m,n are constants. When both are immersed in oil then thermometer A reads 51°C and thermometer B reads 50 °C. determine the reading on thermometer A when thermometer B reads 25°C? Comment on the question "which thermometer is correct". Ans- tA=l+mtB+n(tB)² According to question tA=tB=0°C tA=l+mtB+n(tB)² 0=l+0+0 l=0 Second step tA=tB=100°C tA=l+mtB+n(tB)² 100=0+m×100+n(100)² m+100n=1             ----------------------------(1) For tA=51°C tB=50°C tA=l+mtB+n(tB)² 51=0+M50+n(50)² -----------------------------(2) Now using equation (1) and (2) find m&n 50m+2500n=51 50m+5000n=50 -        -             - ________________ ...

Question 1. A spherical balloon of 1m diameter contains a gas at 150kpa. The gas inside balloon is heated until pressure reaches 459kpa. During the process of heating the pressure of gas inside the balloon is proportional to cube of the diameter of the balloon. Find the work done by the gas.

THERMODYNAMICS QUESTION SOLUTION, CHAPTER. ENERGY INTERACTION. Question 1. A spherical balloon of 1m diameter contains a gas at 150 KPa. The gas inside balloon is heated until pressure reaches 450 KPa. During the process of heating the pressure of gas inside the balloon is proportional to cube of the diameter of the balloon. Find the work done by the gas. Solution: D1= 1 m    P1= 150 KPa                                           V = (4/3)  π R^3 =    πD^3/6        P2 = 450 KPa                                             dV = ( π/6)  (3D^2)    δD P directly proportional D^3                         dV = ( π/2) (D^2)   δD P = k(D^3) ...

Q1.Prove that PV^Y=C for reversible adiabatic process.

 Q1.Prove that PV^Y=C for reversible adiabatic process. Solution:        δQ = dU + PdV Reversible Adiabatic: δQ=0              (:: There is no heat interaction between                                                                   system and surroundings = adiabatic  dU + dV = 0 dU = -PdV m(Cv)dT = -PdV  ----------------------------------------------(1) Now we want 'Y' , so, We know      H = U + PV                  dH = dU + PdV + VdP                  dH =   δQ + VdP (:: There is no heat interaction δQ =0)              dH = VdP       m(Cp)dT = VdP       -------------...