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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
50 किग्रा. शुद्व दूध में कितना पानी मिलाया जाए कि मिश्रण क्रो शुद्व दूध कै लागत मूल्य पर बेचने पर 10% का लाभ हो। A. `2.5` kgB. `5` kgC. `7.5` kgD. `10` kg |
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Answer» Correct Answer - B (b) Let cost price of 1 kg = Rs 1 cost price of 50kg=Rs 50 `implies` Profit `=(10)/(100)xx50=` Rs. 5 `:.` Qty to added `=(5)/(1)=5 kg` |
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| 2. |
एक व्यक्ति 100 रु में 20 सेब बेचकर 20 % लाभ कमाता है, तो उसने 100 रु में कितने सेब ख़रीदे थे |A. 20B. 22C. 24D. 25 |
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Answer» Correct Answer - C According to question SP of 20 apples = Rs. 100 gained = `20%` `therefore` CP of 20 apples `=(100)/(120)=(250)/(3)` In100 Rs. He buy `=(100)/(25)xx6` = 24 apples |
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| 3. |
10 वस्तुओं का क्रय मूल्य 15 -वस्तुओं र्क विक्रय मूल्य र्क बराबर हैं लाभ या हानि का प्रतिशत कितना हैं?A. 0.255B. 0.35C. 0.1D. 0.333 |
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Answer» Correct Answer - D CP of 10 articles = SP of 15 articles `implies(SP)/(CP)=10/15` `Loss %=5/10xx100=33.3%` |
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| 4. |
एक व्यक्ति 10 रूपये /कप की दर से 100 कप खरीदता है | रास्ते में 20 कप टूट गए | वह बचे हुए कपो को 11 रूपये/कप की दर से बेचता है, तो उसकी प्रतिशत हानि ज्ञात करें |A. 15B. 10C. `17(1)/(2)`D. 12 |
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Answer» Correct Answer - D According to question, CP of 1 cup = Rs. 10 CP of 100 cups `=10xx100` = Rs. 1000 Now 20 cups are broken means (20 कपो के टूट जाने का मतलब है) = 100-20=80cups `=11xx80=Rs.880` `therefore` Loss = CP -SP = 100 - 880=120 `Loss%("Loss")/(CP)xx100` `= (120)/(1000)xx100=12%` |
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| 5. |
एक पंखे पर रू 150 मूल्य अंकित किया गया और 20% की छूट दी गई। तो विक्रय मूल्य क्या होगा ?A. Rs. 180B. Rs. 150C. Rs. 120D. Rs. 110 |
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Answer» Correct Answer - C 20%`=(1)/(5)` M.P S.P 5 4 (After discount) `5rArr150` `1rArr30` S.P=`4xx30=Rs.120` |
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| 6. |
एक टेपरिकॉर्डर का उत्पादन मूल्य ₹ 1500 है | विनिर्माता टेपरिकॉर्डर की मूल्य उत्पादन मूल्य से 20 % बढ़ाकर अंकित करता है और इस प्रकार छूट देता है कि उसे 8 % लाभ होता है, तो छूट का प्रतिशत ज्ञात करें |A. 12B. 8C. 20D. 10 |
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Answer» Correct Answer - D Let CP 100 Profit =8% SP=108 So discount is =120-108=12 `(12)/(120)xx100=10%` |
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| 7. |
एक बईमान विक्रेता अपना लागत मूल्य पर दावा करता है लेकिन किलोग्राम वजन के बदले में 875 ग्राम कै वजन बाँट का करता है। उसका लाभ प्रतिशत कितना तोगा?A. 0.17B. `14(5)/(7)%`C. `14(2)/(7)`D. 0.14 |
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Answer» Correct Answer - C Let dealer purchase 1000 gm in Rs. 1000 but at selling time sold = 875 gm in place of 1000 gm. So `P%(1000-875)/(875)xx100` `=14(2)/(7)%` |
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| 8. |
यदि बिक्री कर को `3(1)/(2)` % से घटाकर `3(1)/(3)` % कर दिया जाता है, तो इससे एक व्यक्ति पर क्या अंतर पड़ेगा जो 8400 रूपये अंकित मूल्य वाली एक वास्तु खरीदता है |A. Rs. 20B. Rs. 15C. Rs. 14D. Rs. 10 |
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Answer» Correct Answer - C According to question, Reduction in S.T. `=7/2-10/3=(21-20)/(6)` `=1/6%=1/600` `therefore` Reduction in martket price at 8400 (8400 अंकित मूल्य पर कमी) `=8400xx(1)/(600)=14` |
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| 9. |
एक यापरी किसी सामान के मूल्य से 20 प्रतिशत बढाकर अंकित करता है वह फिर उसे 20 प्रतिशत छूट देकर बेचता है बिक्री उसे क्या होगाA. No loss or gainB. 4% lossC. 2% gainD. 4% gain |
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Answer» Correct Answer - B Let the C.P. = 100 unit Now, M.R.P. = 120 unit `S.P. =(120xx80)/(100)=96` Loss = 100 - 96 = 4 `Loss %4/100xx100=4%` |
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| 10. |
एक व्यक्ति 50 रु/ कलम की दर से कलम खरीदता है | वह 40 कलमों को 5 % हानि पर बेच देता है | बचे हुए कलमों को कितने % लाभ पर बेचा जाए कि कुल 10 % लाभ हो |A. 15B. 40C. 50D. 70 |
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Answer» Correct Answer - D According to question, CP of 50 pen = Rs. `50xx50` = Rs. 2500 to gain `10%` overall sold at = 2750 Now, 40 pen sold at `5%` loss `therefore"SP of 40 pen" = 40xx47.5` = Rs. 1900 Remaining 10 pens sold to get ocerall profit of `10%` at Rs. 850. (शेष 10 कलमों को 10 प्रतिशत का कुल लाभ कमाते हुए 850 रुपये में बेचा गया) `=(850)/(10)=Rs.85` CP of 1 pen = Rs. 50 profit% of remaining pen `=(35)/(50)xx100=70%` |
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| 11. |
एक वस्तु का क्रय मूल्य 800 रू है। 10% छूट देने पर 12.5% लाभ होता है तो वस्तु का अंकित मूल्य ज्ञात करें।A. `Rs. 1,000`B. `Rs.1,100`C. `Rs.1,200`D. `Rs.1,300` |
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Answer» Correct Answer - A According to question, CP MP (100-Discount)(100+Profit%) `100-10 100+12.5` 90units 112.5 units 90 units `rarr=800` 1 unit `rarr(800)/(90)` 1125 units `rarr(800)/(90)xx(1125)/(10)=1000` `therefore MP =RS. 1000` |
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| 12. |
एक कंपनी रिटेलर को सामान बेचते वक्त अपनी वस्तु के अंकित मूल्य पर 30 % छूट देती है | यदि रिटेलर उन वस्तुओ को अंकित मूल्य पर बेचता है, तो उसका प्रतिशत लाभ ज्ञात करें |A. 0.3B. `(17)/(2)`%C. 0.4D. `42(6)/(7)` % |
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Answer» Correct Answer - D Let the Marked Price =100unit According to question, ` 100(MP)overset(30%"discount")rarr70(SP)` `rarr CP` of retailer CP of retailer =70 Retailer soid at MP=100 Profit=`MP-CP=100-70` =30 units profit Profit `%=(30)/(70)xx 1 00=42( 6)/(7)%` |
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| 13. |
यदि बिक्री मूल्य पर 10 % की हानि है, तो लगत मूल्य पर हानि की दर क्या होनीA. `11(1)/(9)%`B. `9(1)/(11)`C. 0.1D. 0.11 |
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Answer» Correct Answer - B `10%L=1/10{:(larrL),(larrS.P):}` CP = S.P. + L = 10+1=11 Loss at C.P. `=1/11xx100=9(1)/(11)%` |
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| 14. |
एक दुकानदार अंकित मूल्य पर 10% छुट देने के पश्चात 12% लाभ कमाता है तो क्रय मूल्य तथा अंकित मूल्य का अनुपात ज्ञात करें।A. `99:125`B. `25:37`C. `50:61`D. `45:56` |
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Answer» Correct Answer - D According to question CP:MP (100-Discount):(100+profit) `100 -10 :100+12` 90:112 45:56 |
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| 15. |
किसी कमीज का मूल्य 15 की छूट देने के बाद रू119 है । छूट के पहले कमीज का अंकित मूल्य क्या था ?A. रू 129B. रू 140C. रू150D. रू 160 |
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Answer» Correct Answer - B `15%=(3)/(20)` `(MP)/(SP)=(20)/(17)underset(xx7) to 119` `MP=20xx7=140` |
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| 16. |
यदि एक दुकानदार खिलौने पर 20 % की छूट क्रय मूल्य पर देना चाहता है, तो उसे वह ₹ 300 में बेचना पड़ेगा | यदि वह उसे ₹ 405 में बेचता है, तो इसके लाभ का प्रतिशत कितना होगा ?A. `5` %B. `8` %C. `6` %D. `4` % |
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Answer» Correct Answer - B Let CP=100x Discount=20% SP=100x-20%of CP`rArr` 80x According to the question, `80xrarrRs300` `1xrarr(300)/(80)xx100=375` Actual CP=Rs. 375 New SP=Rs 405 Gain percent `=(30)/(375)xx100=8%` |
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| 17. |
Rs. 540 अंकित मुल्य की कोई वस्तु ऑफ सीजन ऑफर में Rs. 496.80 में बेची जाती है तो दी गई छूट की दर कितनी (प्रतिशत में) होगी ?A. `7`B. `7.5`C. `8`D. `10` |
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Answer» Correct Answer - C Discount `=540-496.80=43.2` % discount`=(43.2)/(540)xx100` `=8%` |
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| 18. |
20 % छूट के पश्चात एक कमीज का मूल्य ₹ 64 है, तो उसका वास्तविक मूल्य होगा |A. `76.80`B. `80`C. `88`D. `86.80` |
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Answer» Correct Answer - B According to question, `(MP)/(SP)=(100)/(80) 20%` discount 80units=64 1unit `=(64)/(80)` 100 units `=(64)/(80)xx100= 80` `therefore` Original Price Rs.80 |
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| 19. |
एक वस्तु पर 40 % छूट फिर 30 % और 45 % छूट फिर 20 % छूट के क्रमिक छूटो का अंतर ₹ 12 है | तो वस्तु का अंकित मूल्य ज्ञात करें |A. ₹ 400B. ₹ 800C. ₹ 600D. ₹ 200 |
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Answer» Correct Answer - C प्रथम शर्तानुसार कुल छूट `=a+Bb-(ab)/(100)` `=40+30-(40xx30)/(100)=70-12` प्रथम शर्त में कुल छूट = 58% द्वितीय शर्त में कुल छूट `=45%+20%-(45xx20)/(100)` `rArr` total discount =56% `rArr` According to question, `rArr 50%-56%=Rs.12` `rArr=2%=12=600` `therefore` अंकित मूल्य Rs. 600 |
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| 20. |
एक दुकानदार बराबर मूल्यों पर दो टी.वी. सेट बेचता है। यदि एक पर उसे 20% का लाभ हुआ तथा दूसरे पर उसे 20% की हानि हुई , तो इनमें से कौन सा कथन सत्य है ।A. The shopkeeper makes no net gain or profitB. The shopkeeper loses by 2%C. The shopkeeper gains by 4%D. The shopkeeper loses by 4% |
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Answer» Correct Answer - D Quicker approach Always loss in such type of question (इस तरह के प्रश्नों में हमेशा हानि होती हैं ) Loss%`=(Loss%xxProfit%)/(100)` `=(20xx20)/(100)=4%` loss |
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| 21. |
एक माकन x% की छूट देकर ₹ y में बेचा गया तो उसका सूचि मूल्य क्या था ?A. `(100y)/(1-(x)/(100))`B. `(100x)/(100-y)`C. `(100y)/(100-x)`D. `(100x)/(100-x)` |
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Answer» Correct Answer - C Discount=x% `rArr SP=Rs.y` `rArr` MRP=? `rArr MRPxx(100-x)%=y` `rArr` MRP `=(y)/((100-x)xx(1)/(100))` `rArr` MRP `=(100y)/((100-x))` |
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| 22. |
मोहन ने अंकित मूल्य पर 20 % प्रतिशत की छूट पर एक बैग ख़रीदा | उसने उसे खरीद मूल्य पर 40 प्रतिशत के लाभ पर बेच दिया | अंकित मूल्य पर लाभ का प्रतिशत कितना है ?A. 0.2B. 0.24C. 0.12D. 0.18 |
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Answer» Correct Answer - C According to the question `(MP)/(SP)=(50)/(40)` If S.P.=40 40% लाभ के लिए नया विक्रय मूल्य `40xx(140)/(100)=Rs.55` अंकित मूल्य पर लाभ % `=(6)/(50)xx100=12%` |
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| 23. |
In each of the following number series, a wrong number is given, find out that number.30030, 2310, 210, 21, 6, 21. 23102. 2103. 214. 65. None of these |
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Answer» Correct Answer - Option 3 : 21 Calculation: The series follows following pattern: ⇒ 30030 ÷ 13 = 2310 ⇒ 2310 ÷ 11 = 210 ⇒ 210 ÷ 7 = 30 ⇒ 30 ÷ 5 = 6 ⇒ 6 ÷ 3 = 2 ∴ The wrong number in the series is 30 |
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| 24. |
Assertion : A body can have acceleration even if its velocity is zero at a given instant of time. Reason : A body is momentarily at rest when it reverses its direction of motion.A. Both Assertion and Reason are correct, Reason is the coreect expianation of AssertionB. Bioth Assertion and Reason are Correct but Reason is not the correct expalnation of AssertionC. Assetion is correct and Reason is incorrectD. Assertion is incorrect and Reason is correct |
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Answer» Correct Answer - C |
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| 25. |
Find the torque `(vectau=vecrxxvecF)` of a force `vecF=-3hati+hatj+3hatk` acting at the point `vecr=7hati+3hatj+hatk`A. `14hati-38hatj+16hatk`B. `4hati+4hatj+6hatk`C. `-14hati+38hatj-16hatk`D. `-21hati+3hatj-5hatk` |
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Answer» Correct Answer - A `vectau=vecrxxvecF` |
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| 26. |
A body is fired with a velocity of magnitude `sqrt(gR) lt V lt sqrt(2gR)` at an angle of `30^(@)` with the radius vector of earth. If at the highest point the speed of the body is `V//4`, the maximum height attained by the body is equal to:A. `v^(2)//8g`B. `R`C. `sqrt(2)R`D. None of the above |
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Answer» As the force acting between the body and the earth passes through the centre of the earth. Angular momentum of the system is conserved about the centre of earth. And the velocity at the highest point of the trajectory is perpendicular to the radius vector. So, by conservation of momentum about the centre of earth. `mv sin 30^(@).R=(mv)/(4)(R+h)` `(R )/(2)=(R+h)/(4)rArrh=R` |
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| 27. |
Two identical particles of charge `q` each are connected by a massless spring of force constant `k`. They are placed over a smooth horizontal surface. They are released when the separation between them is `r` and spring is unstretched. If maximum extension of the spring is `r`, the value of `k` is (neglect gravitational effect) A. `(q^(2))/(4piepsilon_(0)r^(2))`B. `(q^(2))/(2piepsilon_(0)r^(3))`C. `(q^(2))/(piepsilon_(0)r^(3))`D. none of these |
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Answer» Conservation of energy gives `(q^(2))/(4piepsilon_(0)r)=(1)/(2)kr^(2)+(q^(2))/(4piepsilon_(0).2r)rArrk=(q^(2))/(4piepsilon_(0)r^(3))` |
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| 28. |
Two statements are given as below I: lb mole is composed of 453.6*6.022*1023 moleculesII: gm mole is composed of 6.022*1023 moleculesNumber of correct statements is/are(a) 0 |
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Answer» The correct answer is (c) 2 The explanation is: A gm mole is composed of 6.022*10^23 molecules and 1 lb mole is 453.6 times a gm mole. |
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| 29. |
What is the potential energy of a box in (ft)(lbf), kept on a wall having a height 20 ft above the surface? Assume that the mass of the box is 100 lb.(a) 1000(b) 2000(c) 3000(d) 4000 |
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Answer» Correct answer is (b) 2000 Explanation: PE = mgh, g = 32.2 ft/s2. |
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| 30. |
How many lb moles of O are there in 22 g of H2SO4?(a) 1.978*10^-4(b) 1.978*10^-3(c) 1.978*10^-2(d) 1.978*10^-1 |
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Answer» Right choice is (b) 1.978*10^-3 The best I can explain: lb mole = 454*gm mole, lb moles of O = 4(lb moles of H2SO4). |
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| 31. |
The theoretical production rate of a chemical is 1µg/(mL)(min.). What is the actual production rate of the chemical in lb mol/ (day)(ft3) with an efficiency of 98%?(a) 0.0880(b) 0.0800(c) 0.0449(d) 0.529 |
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Answer» The correct option is (a) 0.0880 Easy explanation: 1 lb mole = 454 g mole and 1 L = 3.531×10-2 ft3. |
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| 32. |
In a series LCR circuit R = 200 W and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30°. On taking out the inductor from the circuit the current leads the voltage by 30°. The power dissipated in the LCR circuit is(1) 305 W(2) 210 W(3) Zero W(4) 242 W |
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Answer» 4 The given circuit is under resonance as XL = XC Hence power dissipated in the circuit is P = V2/R = 242 W |
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| 33. |
The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 x 10–3 are(1) 5, 1, 2(2) 5, 1, 5(3) 5, 5, 2(4) 4, 4, 2 |
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Answer» The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 x 10–3 are is- (1) 5, 1, 2 |
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| 34. |
A Carnot heat engine has an efficiency of `10%`. If the same engine is worked backward to obtain a refrigerator, then find its coefficient of performance.A. 3B. 5C. 7D. 9 |
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Answer» Correct Answer - D Efficiency of carnot engine is, `eta = 1 - (T_(2))/(T_(1))` `0.1 = 1 - (T_(2))/(T_(1))` `:. (T_(2))/(T_(1)) = 0.9` Coefficient of perfomance (COP) of a refrigerator `= (T_(2))/(T_(1) - T_(2)) = (0.9T_(1))/(T_(1) - 0.9T_(1)) = 9` `:. COP = 9` |
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| 35. |
Two moles of an ideal monoatomic gas undergoes a process `VT =` constant. If temperature of the gas is increased by `DeltaT = 300K`, then the ratio `((DeltaU)/(DeltaQ))` isA. 2B. 3C. 4D. 6 |
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Answer» Correct Answer - B Given, `VT =` constant `PV^(2) =` constant Polytropic constant, `x = 2` `:.` Molar heat capcity, `C = C_(v) + (R )/(1 - x)` `= (3R)/(2) + (R )/(1 - 2)` `= (R )/(2)` `:. DeltaQ = nCdeltaT = 2xx (R )/(2) xx 300 = 300R` `DeltaU = nC_(v)DeltaT = 2 xx (3R)/(2) xx 300 = 900R` `:. (DeltaU)/(DeltaQ) = (900R)/(300R) = 3` |
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| 36. |
What is the number of significant figures in 5.50 × 103 ?(a) 2 (b) 7 (c) 3 (d) 4 |
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Answer» (c) 3 Explanation: Number = 5.50 × 103 power of 10 doesn’t effect number of significant figures ∴ No of significant figures = 3 |
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| 37. |
If ‘M’ is the mass of water that rises in a capillary tube of radius ‘r’, then mass of water which will rise in a capillary tube of radius ‘2r’ is(1) M (2) 4 M (3) 2 M (4) 2 M |
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Answer» Correct option (1) M Explanation: Light will change with change in radius mass remain same |
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| 38. |
In uniform magnetic field, if angle between `vec(v) and vec(B) is 0^(@) lt 0 lt 90^(@)`, the path of particle is helix. Let `v_(1)` be the component of `vec(v) along vec(B) and v_(2)` be the component perpendicular to `vec(B)`. Suppose p is the pitch. T is the time period and r is the radius of helix. Then `T = (2pim)/(qB), r = (mv_(2))/(qB), P = (v_(1))T` Assume a charged particle of charge q and mass m is released from the origin with velocity `vec(v) = v_(0) hat(i) - v_(0) hat(k)` in a uniform magnetic field `vec(B) = -B_(0) hat(k)`. Pitch of helical path described by particle isA. `(2mv_(0))/(qB_(0))`B. `(2pimv_(0))/(qB_(0))`C. `(sqrt(2)pimv_(0))/(qB_(0))`D. `(sqrt(3)pimv_(0))/(qB_(0))` |
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Answer» Correct Answer - B Pitch of helix `P=V_(||) xx T` `P=v_()((2pim)/(qB_(0)))=(2 pi mv_(0))/(qB_(0)` Hence choice (b) is correct. |
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| 39. |
The potential gradient on wire PQ is 10–5 V then the reading of voltmeter is -(1) 3 mV(2) 5 mV(3) 7 mV(4) 9 mV |
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Answer» The correct option is (2) 5 mV. Explanation: Reading = Potential gradient x length = 0.01 x 0.5 = 5 x 10-3 V = 5 mV |
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| 40. |
A spherical balloon contains 1 mole of He at `T_(0)`. The balloon material is such that the pressure inside is alwaus proportional to square of diameter. When volume of balloon becomes 8 times of initial volume, then :-A. Work done by gas `(93RT_(0))/(5)`B. Final temperature of the gas is `32T_(0)`C. Heat taken by gas `65. 1 RT_(0)`D. Heat given by gas` 65.1 RT_(0)` where R is universal gas constant. |
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Answer» Correct Answer - A::B::C Also `=(P prop V^(2//3))/((nRDeltaT)/(1-x)) (P_(0)V_(0))/(RT_(0)) = ((4P_(0))(8V_(0)))/(RT)` `= T=32T_(0)` `= (nR(31T_(0)))/(1+(2)/(3)) = (93RT_(0))/(5)` `DeltaQ = (1) [((3)/(2)R)+((3)/(5))R] (31T_(0))` `= (21)/(10)R (31T_(0)) = 65.1 RT_(0)` |
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| 41. |
From some instruments current measured is `I=10.0amp`, potential different measured is `V=100.0 V`, length of wire is `31.4 cm`, and diameter of wire is `2.00mm` (all in correct significant figure). The resistivity of wire (in correct significant figures)will be (use `pi=3.14`)A. `1.00 xx 10^(-4) Omega-m`B. `1.00 xx 10^(-4) Omega-m`C. `1.00 xx 10^(-4) Omega-m`D. `1.00 xx 10^(-4) Omega-m` |
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Answer» `rho=(piD^(2))/(4L)=(V)/(L)=((3.14)(2.00xx10^(-3))^(2))/(4(0314))((100.0)/(10.0))` and answer should be in three `S.F` so `rho = 1.00 xx 10^(-4)Omega -m` . |
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| 42. |
From ideal instruments, current measured is `I = 10.0` amp, potential difference measured is `V = 100.0` volt, length of wire is `31.4 cm`, and diameter of wire is `2.00 mm`, the resistivity of wire will be (in correct significant figure) `(pi = 3.14)`A. `1.00 xx 10^(-4) Omega - m`B. `1.0 xx 10^(-4) Omega - m`C. `1 xx 10^(-4) Omega - m`D. `1.000 xx 10^(-4) Omega - m` |
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Answer» Correct Answer - C `rho = (pi D^(2))/(4L) (v)/(i) = ((3.14)(2.00 xx 10^(-3))^(-2))/(4(0.314)) ((100.0)/(10.0))` `rho = 1.00 xx 10^(-4) Omega - m` |
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| 43. |
From some instruments current measured is `I=10.0amp`, potential different measured is `V=100.0 V`, length of wire is `31.4 cm`, and diameter of wire is `2.00mm` (all in correct significant figure). The resistivity of wire (in correct significant figures)will be (use `pi=3.14`)A. `1.00xx10^(-4)Omega-m`B. `1.0xx10^(-4)Omega-m`C. `1xx10^(-4)Omega-m`D. `1.000xx10^(-4)Omega-m` |
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Answer» Correct Answer - A `p=(piD^(2))/(4L)=(V)/(I)=((3.14)(2.00xx10^(-3))^(-2))/(4(0.314))((100.0)/(10.0))` and answer should be in three `S.F` so `p=1.00xx10^(-4)Omega-m` |
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| 44. |
A sample of an ideal gas is compressed by a piston from 10 m3 to 5 m3 and simultaneously cooled from 273° C to 0° C. As a result there is: A. an increase in pressure B. a decrease in pressure C. a decrease in density D. no change in density E. an increase in density |
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Answer» E. an increase in density |
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| 45. |
During a slow adiabatic expansion of a gas: A. the pressure remains constant B. energy is added as heatC. work is done on the gas D. no energy enters or leaves as heat E. the temperature is constant |
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Answer» D. no energy enters or leaves as heat |
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| 46. |
A quantity of an ideal gas is compressed to half its initial volume. The process may be adiabatic, isothermal, or isobaric. Rank those three processes in order of the work required of an external agent, least to greatest. A. adiabatic, isothermal, isobaric B. adiabatic, isobaric, isothermal C. isothermal, adiabatic, isobaric D. isobaric, adiabatic, isothermal E. isobaric, isothermal, adiabatic |
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Answer» E. isobaric, isothermal, adiabatic |
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| 47. |
An ideal gas undergoes an isothermal process starting with a pressure of 2 × 105 Pa and a volume of 6 cm3. Which of the following might be the pressure and volume of the final state? A. 1 × 105 Pa and 10 cm3 B. 3 × 105 Pa and 6 cm3 C. 4 × 105 Pa and 4 cm3 D. 6 × 105 Pa and 2 cm3 E. 8 × 105 Pa and 2 cm3 |
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Answer» D. 6 × 105 Pa and 2 cm3 |
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| 48. |
The formation of ice from water is accompanied by: A. absorption of energy as heat B. temperature increase C. decrease in volume D. an evolution of heat E. temperature decrease |
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Answer» A. absorption of energy as heat |
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| 49. |
A system undergoes an adiabatic process in which its internal energy increases by 20 J. Which of the following statements is true? A. 20 J of work was done on the system B. 20 J of work was done by the system C. the system received 20 J of energy as heat D. the system lost 20 J of energy as heat E. none of the above are true |
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Answer» A. 20 J of work was done on the system |
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| 50. |
According to the first law of thermodynamics, applied to a gas, the increase in the internal energy during any process: A. equals the heat input minus the work done on the gas B. equals the heat input plus the work done on the gas C. equals the work done on the gas minus the heat input D. is independent of the heat input E. is independent of the work done on the gas |
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Answer» B. equals the heat input plus the work done on the gas |
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