33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020
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33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

All previous year questions , and these are downloaded from the sources in the internet. I don't own these resources and these are copyrighted by MTG. All previous year questions , and these are downloaded from the sources in the internet. I don't own these resources and these are copyrighted by MTG.

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Telegram @unacademyplusdiscounts100 NEET-AIPMT Chapterwise Topicwise Solutions Physics50. (c) : Efficiency of a Carnot engineη= 1−T2T1where T 1is the temperature of source and T 2is thetemperature of sink respectively.TFor engine A, η A = 1−TFor engine B, η B1T2= 1−TAs per question, h A= h BT T2T T2∴ 1− = 1− ⇒ = or T =T1T T1T51. (c) : Efficiency of an engine, η= 1−TT 1 2T2T1where T 1is the temperature of the source and T 2is thetemperature of the sink.1∴ = − =6 1 T2T25or, ...(i)T1T16When the temperature of the sink is decreased by 62°C(or 62 K), efficiency becomes double.Since, the temperature of the source remains unchanged−−∴ 2 × 1 = −= −6 1 ( T262)13 1 ( T262)or,T1T1or,23T2− 62=T1or, 2T1 = 3T2−186or, 2T3 5 1 = ⎡ T1186⎣ ⎢ ⎤ 6⎦ ⎥ − [using (i)]∴⎡5⎤⎢ − ⎥ = =⎣2 2 ⎦186 T1T1or, 1862or, T 1= 372 K = 99°CT52. (d) : Efficiency of a Carnot engine, 2η= 1−TT12or,T= 1 − 40 3η = 1 −1100= 55 5∴ T1 = × T2= × 300 = 500 K.3 3Increase in efficiency = 50% of 40% = 20%New efficiency, h′ = 40% + 20% = 60%T260 2∴ = 1− =T1′ 100 5T′ 5 51 = × T2= × 300 = 750 K.2 2Increase in temperature of source = T 1′ – T 1= 750 – 500 = 250 KvvvT53. (d) :2 Q21−= 1−⇒ 1− 400 = −T1Q 1 500 1 Q26 × 104 Q2⇒ =5 6 ×4⇒ Q102= 4.8 × 10 4 calNet heat converted into work= 6.0 × 10 4 – 4.8 × 10 4 = 1.2 × 10 4 cal54. (d) : Efficiency of Carnot engineW T2= = 1 −Q T11W 4001W6= − 1 6500= 5⇒ = 5 = 12 . kcal.55. (c) : Efficiency (h) of a carnot engine is given byT2η= 1−T , where T 1is the temperature of the source and1T 2is the temperature of the sink.Here, T 2= 500 K∴ 05 . = 1− 500 ⇒ T1= 1000 K .T1T ′2Now, η= 06 . = 1−(T 1000 2′ is the new sink temperature)⇒ T 2′ = 400 K.56. (a) : Efficiency is maximum in Carnot engine whichis an ideal engine.400 − 300η= × 100% = 25%400\ efficiency 26% is impossible for his heat engine.57. (b) : 1 6 1 T25 T2= − or =T16 T1Now, 1 3 1 T2− 62 5 62= − = 1 − +T1 6 T1T 1= 62 × 6 = 372 K = 99°C and T 2= 310 K = 37°C58. (b) : Reservoir temperature (T 1) = 100°C = 373 Kand T 2= – 23°C = 250 K.The efficiency of a Carnot engineT1−T2373 − 250η= = .T137359. (c) : Efficiency of Carnot engine (h 1) = 40% = 0.4New efficiency (h 2) = 50 % = 0.5.TThe efficiency (h) =2 T21−or = 1−ηT1T1For first case, T 2500 = 1− 04 . or T 2= 300 KFor second case, 300 1 05T= − . or T 1 = 600 K14

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