Skip to main content

Q1. A hydraulic press has a ram of 15 cm diameter and plunger of 1.5 cm. It is required to lift a mass of 1000kg. The force required on plunger is nearly equal to

 Q1. A hydraulic press has a ram of 15 cm diameter and plunger of 1.5 cm. It is required to lift a mass of 1000kg. The force required on plunger is nearly equal to

(a) 100N 

(b) 1000N

(C) 10000N

(d) 10N

Ans (a)


By pascal law:

F1/A1 = F2/A2

F1/3.14(1.5)^2/4 = 1000×10/3.14(15)^2/4                              

F1 = 100N...................................................................................................(Ans)

*3.14 is value of piπ

*F2= 1000×10= m×g= mass×acceleration due to gravity(9.8) 

*A1= 3.14(1.5)^2/4 =πd2/4 = area of circle 

*A2= 3.14(15)^2/4 = πd2/4

*15 cm diameter value of circle at point2

*1.5 cm diameter value of circle at point1 






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       -------------...