EC
Engineering Classroom
by Himalay Sen

Exams

Prepare for your next test with our collection of exam-based practice sets and question banks designed for effective assessment and revision.

Board Questions

Access past board exam questions organized by year and subject to help you understand patterns, improve preparation, and boost your exam performance.

MCQ
2123. A process in which the temperature of the working substance remains constant during its expansion or compression is called-
Isothermal process
Hyperbolic process
Adiabatic process
Polytrophic process
2124. Workdone in free expansion process is-
Zero
Minimum
Maximum
Positive
ব্যাখ্যা: There is no restraining/opposing force or pressure as expansion occurs against vacuum. So, dw = 0 (in case of free expansion).
2129. In vapour compresion refrigeration system, refrigerant occurs as liquid between-.
condeuser and expäision valve
compressor and evaporator
expansion valve and evaporator
compressor and condenser
2130. Pick up the correct statement about giving up of heat from one medium to other in ammonia absorption system-
strong solution is weak solution
weak solution to strong solution
atozng solution to ammonia vapour
ammonia vapour to weak solution
2133. If the value of n = 0 in the equation PV^n = C, then the process is called-
Constant volume process
Adiabatic process
Constant pressure process
Isothermal process
ব্যাখ্যা: Constant Pressure Process (n= 0) Isothermal Process (n = 1) [BREB-2014] Adiabatic Process (n = y) Polytropic Process (y> n > 1) Constant Volume Process (n = ∞) PV°=C ⇒ P=C [V°=1]
2135. For the same heat input and same compression ration.
Otto cycle and diesel cycle are equally efficient
Otto cycle is less efficient than diesel cycle
Efficiency depends mainly on working substance
None of the above
2136. A path 1-2-3 is given. A system absorbs 100 kj as heat and does 60kj of work while along the path 1-4-3, it does 20kj of work. The heat absorbed during the cycle 1-4-3 is-
-140kj
-80kj
-40kj
+60kj
ব্যাখ্যা: For 1-2-3 cycle. 𝘘1-2-3 = 100 kj W1-2-3 = 60 kj We know, Heat = Internal energy + Work 𝘘1-2-3 = 𝘘1-2-3 + W1-2-3 𝘘1-2-3 = 𝘘1-2-3 - W1-2-3 𝘘1-2-3 = (100-60) kj = 40 kj Again, For 1-4-3 cycle, W1-4-3 = 20kj 𝘘1-2-3 = 𝘘1-4-3 =40kj Again, we know, 𝘘1-4-3 = 𝘘1-4-3 + W1-4-3 = 40 kj+20 kj = 60 kj For 1-2-3 cycle. 𝘘1-2-3 = 100 kj W1-2-3 = 60 kj We know, Heat = Internal energy + Work 𝘘1-2-3 = 𝘘1-2-3 + W1-2-3 𝘘1-2-3 = 𝘘1-2-3 - W1-2-3 𝘘1-2-3 = (100-60) kj = 40 kj Again, For 1-4-3 cycle, W1-4-3 = 20kj 𝘘1-2-3 = 𝘘1-4-3 =40kj Again, we know, 𝘘1-4-3 = 𝘘1-4-3 + W1-4-3 = 40 kj+20 kj = 60 kj
2138. A mixture of gas expands from 0.03m to 0.06^3m at a constant pressure of 1 mpa and absorbs 84kj of heat during the process. The change in internal energy of the mixture is-
30 kj
54 kj
84 kj
114 kj