Saved Bookmarks
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. |
A,B औार C तीन क्रिकेट टीमों के खिलाडियों की संख्या का अनूपात 2:5:3 है। यदि तीनों टीमों A,B और C के प्रत्येक खिलाडी के औसत रनों का अनुपात 30: 17: 25 है तो सामूहिक रूप से सभी तीन टीमों में प्रत्येक खिलाडी दवारा बनाए गए रनों की ओसत संख्या क्या होगी ?A. 20B. 21C. 22D. 23 |
|
Answer» Correct Answer - C A:B:C=2:5:3 Avg`=(30xx2+17xx5+25xx3)/10` `=(60+85+75)/10=22` |
|
| 2. |
यदि 7,5,13,x, तथा 9 का औसत 10 हो, तो x का मान क्या होगा ?A. 10B. 12C. 14D. 16 |
|
Answer» Correct Answer - D Arithmetic mean =`("Total sum")/("Total number")` According to the question , `10=(7+5+13+x+9)/5` 50=x+34 x=50-34 x=16 |
|
| 3. |
8 विद्यर्थियो के अंको का औसत 51 है तथा अन्य 9 विद्यर्थियो के अंको का औसत 68 है । सभी विद्यर्थियो के अंको का औसत ज्ञात कीजिये ?A. 59B. 59.5C. 60D. 60.5 |
|
Answer» Correct Answer - C Total marks of 8 students = 51 x 8= 408 Total marks of 9 students = 68 x 9 =612 Total marks of 17 students =408+612=1020 Average of 17 students =`1020/17` = 60 |
|
| 4. |
किसी विद्यालय में, 50 विद्यर्थियो की कक्षा में 30 छात्राओ का औसत भार 16 kg तथा अन्य छात्रों का औसत भार 15.5 kg है । पूरी कक्षा का औसत भार ज्ञात करे?A. 15.2 kgB. 15.8 kgC. 15.4 kgD. 15.6 kg |
|
Answer» Correct Answer - B According to the question Average `=(30xx16+20xx15.5)/50` `=(480+310)/50=790/50=15.8` kg |
|
| 5. |
5 प्राकृतिक संख्याओं का औसत M हैं। यदि अगले तीन प्राकृतिक संख्या को सम्मिलित किया जाये तो नया औसत M से कितना अधिक हैं ?A. 2B. 1C. 1.4D. 1.5 |
|
Answer» Correct Answer - D माना की लागातर 5 संख्या = 1,2,3,4,5 `Average of no m=(1+2+3+4+5)/5 =3` 8 संख्या का औसत =`(1+2+3+4+5+6+7+8)/8` `=36/8=4.5` 3+x=4.5 x=1.5 the average of no of increased by 1.5 |
|
| 6. |
एक सप्ताह के पहले चार दिनों का औसत उच्च तापमान `25^(@)C` है जबकि अंतिम चार दिनों को औसत `25.5^(@)C` है। यदि पुरे सप्ताह का औसत तापमान `25.2^(@)C` है तो चौथे दिन का तापमान है-A. `25^(@)C`B. `25.2^(@)C`C. `25.6^(@)C`D. `25.5^(@)C` |
|
Answer» Correct Answer - C According to question `implies` Total temperature of first four days Mon + Tues + Wed + Thu =25 x 4= `100^@`…(i) `implies` Total temperature of Last Four days Thu + sat + sun =25 x4 = `100^@`…(i) `implies` Total temperature of Last four days Thr + Fri + sat + sun = 25.5 x 4 =`102^@`...(ii) `implies` Total temperature of week `=25.2^@ xx 7=176.4^@` After adding euation (i) + (ii) Mon + Tue + Wed + 2 x Thu + Fri + Sat + sun =`202^@` ...(iv) after subtracting equation (iv) - (iii) `202-176.4^@=25.6^@ C` `implies` Temperatrue of 4th day `=25.6^@C` |
|
| 7. |
सेट A में 27,28,30 और 33 पूर्णाक है। यदि पूर्णाक K को सैट में सम्मिलित किया जाता है, तो सेट A के औसत में 30 प्रतिशत वृदिध होगी । पूर्णाक K का मान क्या है ?A. 68B. 79C. 73.75D. 75.25 |
|
Answer» Correct Answer - C Average of set A =`(27+28+30+33)/4=188/4=29.5` वृद्धि के बाद औसत `=(29.5xx130)/100=38.35` Now `("sum of first four numbers +k")/5` New Average 118+k=38.35 x 5 188+k = 191.75 K=191.75 - 118 =73.75 |
|
| 8. |
मंगलवार, बुधवार और वीरवार का औसत तापमान 41 डिग्री और बुधवार, गुरूवार और शुक्रवार का औसत तापमान 40 डिग्री था यदि शुक्रवार को वास्तव में 39 डिग्री तापमान था, तो मंगलवार का तापमान क्या था ?A. 42 degrees`//`डिग्रीB. 46 degrees`//`डिग्रीC. 23 degrees`//`डिग्रीD. 26 degrees`//`डिग्री |
|
Answer» Correct Answer - A Tue + Wed + Thu = `41^@ xx 3 =123^@`…(i) Wed + Thu+Fri = `40^@ xx 3 =120^@`…(ii) (i)-(ii) After solving both , we get Tue - Fri =`3^@` Tue = `3^@`+Fri `=3^@ + 39^@ =42^@` |
|
| 9. |
पाँच संख्यओं का औसत 7 है। यदि तीन नई संख्याऐं जोडी जाए तो नया औसत 8.5 हो जाता है। उन तीनों नई संख्याओं का औसत कितना है ?A. 9B. 10.5C. 11D. 11.5 |
|
Answer» Correct Answer - C 5 संख्याओ का योग =7 x 5 =35 8 संख्याओ का योग =8.5 x 8 =68 जोड़ी गयी तीन संख्याओ का योग = 68-35=33 `33/3=11` |
|
| 10. |
छात्रों को बचत की औसत (अंकगणितीय मध्य) राशि रूपय 600 है। 3 छात्र ऐसे हैं जिनकी कोई बचत नहीं हैं और अन्य छात्रों में से प्रत्येक की कम से कम रूपय 250 की बचत है और निहार की रूपय 1300 की बचत है। बताइए बचत की सबसे बडी राशि (में रूपय) क्या हो सकती है ?A. 3250B. 3450C. 3650D. 3850 |
|
Answer» Correct Answer - B Average amount of saving of 10 students is = 600 total saving of 10 students = 600 x 10 =6000 Nihar has =1300 3 has no saving so , remaining student is = 10-4=6 these 6 student at least has 250 so , these student total amount =6 x 250 =1500 amount after including Nihar is = 1500+1300 = 2800 Remaining saving 2800 = 3200 this 3200 amount could have. and 250 has . so, greatest amount is =3200+250 =Rs.3450 |
|
| 11. |
9 क्रमागत विषम संख्याओं का औसत 53 है। सबसे छोटी विषम संख्या ज्ञात करें ?A. 22B. 27C. 85D. 45 |
|
Answer» Correct Answer - D माना की 9 लगातार संख्या x,x+2,x+4,x+6,x+8,x+10,x+12,x+14,x+16 `therefore (x+x+2+x+4+x+6+x+8+x+10+x+12+x+14+x+16)/9=53` 9x+72=477 9x=405 x=45 `therefore` सबसे छोटी विषम संख्या =45 |
|
| 12. |
दो संख्याओं का औसत 8 है और अन्य तीन संख्याओं का औसत 3 है। पाँचो संख्याओं का औसत बताइए।A. 2B. 3C. 5D. 6 |
|
Answer» Correct Answer - C Sum of two no = 8 x 2 =16 Sum of other three no = 3 x 3 =9 Total sum = 25 Avg = `25/5` =5 |
|
| 13. |
5 वर्ष पहले A,B,C, तथा D की औसत आयु 45 वर्ष थी। जब E सम्मिलित हो जाता है, तो पाँचों की औसत आयु 49 वर्ष हो जाती है, E की आयु ज्ञात करें ?A. 25 yearsB. 40 yearsC. 45 yearsD. 64 years |
|
Answer» Correct Answer - C माना की E की आयु = x years According to the question , (5 x 4)+4 x 45 + x =5 x 49 20 + 180 + x =245 x = 245 -200 x = 45 years therefore, age of E = 45 years |
|
| 14. |
किक्रेट क किसी खेले में पहले 10 ओवरों में रन-रेट 3.2 था। शेष 40 ओवरों में क्या रन-रेट होना चाहिए कि 282 रन के लक्ष्य तक पँहुचा जा सके ?A. 6.4B. 6.3C. 6.25D. 6.5 |
|
Answer» Correct Answer - C Total run till 10 overs is =3.2 x 10 =32 Remaining = runs = 282-32 =250 250 runs make in 40 overs so average runs per overs = `250/40=6.25` |
|
| 15. |
एक समूह में उपस्थित विघार्थियों का औसत 52 है। 20% मेधावी छात्रों का औसत 80 है एवं 25% मन्दबुदिध छात्रों का औसत 31 है। बचे हुए 55% छात्रों का औसत ज्ञात करें ।A. 45B. 50C. 51.4D. 54.6 |
|
Answer» Correct Answer - C माना की छात्रो की कुल संख्या `=100` According to question `20xx80+25xx31+55xx x=52 xx 100` 1600 + 775 +55x =5200 55x=5200-1600-775 55x=2825 x=51.4 |
|
| 16. |
प्रथम तीन संख्याओं का औसत चौथी संख्या का तीन गुना है। यदि चारों संख्याओं का औसत 5 हो तो चतुर्थ संख्या ज्ञात करें ।A. 4.5B. 5C. 2D. 4 |
|
Answer» Correct Answer - C Let the fourth no `=x` the average of first three no = 3x According to the question Sum of four no = 5 x 4 =20 also, x +3 x 3x =20 10x = 20 x =2 `therefore` Fourth no =2 |
|
| 17. |
दी गई श्रृंखाला में पहले तीन पूर्ण संख्याओं का औसत 24 है और शेष बचे हुए पूर्ण संख्याओं औसत 18 है। इस श्रृंखाला की सभी संख्याओं का औसत क्या होगा ?A. 18 से कमB. 18 और 24 के बीचC. 24 से अधिकD. निर्धारित नहीं किया जा सकता |
|
Answer» Correct Answer - B Let the no of remaining number be x then, Total avg `implies(24xx3+18x)/(x+3)` `(72+18x)/(x+3)` Now put values of x For x = 0 , Avg. = 24 for x=1, Avg, = 22.5 for x = 2 , Avg, = 21.6 इसलिया के किसी मान के लिया मान की लिया नया औसत हमेसा 18 & 24 के बीच में होगा |
|
| 18. |
कोई कार A से B तक 40 किमी़./घंटा की रफतार से यात्रा करती है और B से A तक 60 किमी./घंटा की रफतार से लौटती है। सम्पूर्ण यात्रा में कार की औसत रफतार बताइए ?A. 48 km/ h / 48 किमी./ घंटाB. 50 km/ h / 50 किमी./ घंटाC. 45 km/ h / 45 किमी./ घंटाD. 60 km/ h / 60 किमी./ घंटा |
|
Answer» Correct Answer - A Let distance = LCM (40,60) = 120 A to B , `T_1=120/40=3` hours `B to A , T_2=120/60=2` hours Average speed = `("Total distance")/("Total time")` `=(120+120)/(3+2)=240/5=48` km/h |
|
| 19. |
तीन संख्याओं का औसत 60 है। यदि प्रथम संख्या सभी संख्याओं के योग का एक- चौथाई है। प्रथम संख्या है।A. 30B. 36C. 42D. 45 |
|
Answer» Correct Answer - D तीन संख्याओ का योग `= 6xx3=180` पहली संख्या`=1/4xx180`=45 |
|
| 20. |
60 विधार्थियों की एक कक्षा में प्राप्त अंको का औसत 65 है। उनमे से आधे विधार्थियों दवारा औसतन 85 अंक प्राप्त किए जाते है शेष विधार्थियों दवारा प्राप्त औसत अंक ज्ञात कीजिये ?A. 35B. 45C. 55D. 65 |
|
Answer» Correct Answer - B Let us take 30 students average marks = x then, 30x + (3 x 85 )= 60 x 65 x = 130-85 x=45 |
|
| 21. |
किसी एक घंटे लम्बी परीक्षा में 60 प्रश्न होते हैं। यदि कोई विघार्थी 25 मिनट में 30 प्रश्न कर लेता है तो शेष प्रश्नों में से प्रत्येक को पूरा करने में औसतन उसे कितने सेकण्ड का समय लगेगा ?A. 70 seconds / 70 सेकण्डB. 50 seconds / 50 सेकण्डC. 40 seconds / 40 सेकण्डD. 30 seconds / 30 सेकण्ड |
|
Answer» Correct Answer - A Remaining question = 60-30 = 30 Remaining minutes = 60-25 =35 Per question time is =`35/30` min `=35/30xx60` sec =70 seconds |
|
| 22. |
35 बच्चों की एक कक्षा के औसत अंक 35 हैं। 35 अंक प्राप्त करने एक विघार्थी के अंक गलती से 65 लिखे गए। कक्षा की सही औसत क्या है ?A. 33.76B. 34.14C. 35.24D. 36.5 |
|
Answer» Correct Answer - B Avg. marks of 35 children is 35 एक छात्र का गलत अंक =65 तब अतिरिक्त संख्या =65-35=30 प्रत्येक छात्र का घटा हुआ औसत अंक =`30/35=6/7=0.867` `therefore` प्रत्येक छात्र का सही औसत =35-0.857 =34.14 |
|
| 23. |
एक विघार्थी से x का मान ज्ञात करने को कहा गया’ यदि दिये गये 12 संख्याओं 3.11.7.9.15.13.8.19.17.21.14. एवं x का समान्तर माध्य 12 है ।A. 3B. 7C. 17D. 31 |
|
Answer» Correct Answer - B `(3+11+7+9+15+13+8+19+17+21+14+x)/12` =12 `implies` 137+x = 12 x 12 x=144-137 =7 |
|
| 24. |
1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6,6,6,6,6,6, तथा 7,7,7,7,7,7,7 का समान्तर माध्य ज्ञात करे ?A. 4B. 5C. 14D. 20 |
|
Answer» Correct Answer - B According to question Avg. `=(1+2+2+3+3+3+4+4+4+4+5+5+5+5+5+6+6+6+6+6+6+7+7+7+7+7+7+7)/28` Avg. `(1+2xx2+3xx3+4xx4+5xx5+6xx6+7xx7)/28` Avg.=`(1+4+9+16+25+36+49)/28` Avg. =5 Alternate : we know that sum of `n^2` terms `=(n(n+1)(2n+1))/6` `=(7xx8xx15)/6=(7xx150)/6=140` here terms are 28. `therefore` Average =`140/28=5` |
|
| 25. |
9 व्यक्तियो में से 8 व्यक्ति प्रत्येक खाने पर रु 30 खर्च करता है। 9 वा व्यक्ति औसत की अपेक्षा रु 20 अधिक खर्च करता है। सभी व्यक्तियो द्वारा खर्च कुल धन ज्ञात करे ?A. Rs.260B. Rs.290C. Rs.292.50D. Rs.400.50 |
|
Answer» Correct Answer - C let the average of 9 people expenditure = Rs. x According to the question `(30xx8+x+20)/9=x` 240+20+x=9x 260=8x `x=260/8` x=32.5 Total expenditure are = 32.5 x 9 = Rs . 292.5 |
|
| 26. |
रु 510 में 30 पेन एव 75 पेंसिल खरीदे गए। यदि एक पेंसिल का औसत मूल्य रु2 है, तो पेन का औसत मूल्य ज्ञात करे?A. Rs.9B. Rs.10C. Rs.11D. Rs.12 |
|
Answer» Correct Answer - D According to question =Rs. 2 =2 x 175 =Rs. 150 `therefore` =510-150=Rs.360 `therefore` =`360/30=Rs 12` |
|
| 27. |
12. क्या समांतर श्रेढ़ी \( 18,15 \frac{1}{2}, 13, \ldots \ldots . . \) का एक पद \( -47 \) है? यदि हाँ तो कौनसा पद है? |
|
Answer» \(18, 15\frac12, 13, ....\) \(\because a= a_1 = 18\) & \(d = a_2 - a_1 = 15\frac 12 - 18 = \frac{31}2 - 18 = \frac{31 - 36}2 = \frac{-5}2\) or \(d = a_3 - a_2 = 13 - 15 \frac 12 = 13 - \frac{31}2 = \frac {26 - 31} 2 = \frac {-5}2\) Let \(a_n = -47\) \(\therefore a + (n - 1)d = -47\) ⇒ \(18 + (n -1)(\frac{-5}2) = -47\) ⇒ \(-\frac 52 (n - 1) = - 47 - 18 = -65\) ⇒ \(n - 1 = -65 \times\frac{-2}5 = 26\) ⇒ \(n = 26 + 1 = 27\) Hence, 27th term of given A.P. will be -47. |
|
| 28. |
Calculate the no. Of oxygen molecules in 12.8 g of SO2. |
|
Answer» Molecular weight of SO2 = 64 g/mol ∴ Number of moles of SO2 in 12.8 g = \(\frac{12.8}{64} \) = 0.2 mol ∵ 1 molecule of SO2 contains = 1 molecule of O2 ∴ 0.2 moles of SO2 contains = 0.2 moles O2 molecule ∴ Number of O2 molecule in 12.8g SO2 = 0.2 x 6.022 x 1023 molecule = 1.2 x 1023 molecules |
|
| 29. |
How can we test acetic acid without indicator |
|
Answer» Take a small amount of acetic acid in a test tube and dip a strip of blue litmus paper in it. Blue litmus turns red, It is acidic in nature. |
|
| 30. |
Write the balanced chemical equations of the following word equation:Aluminium + sodium hydroxide + water → sodium meta aluminate + hydrogen |
|
Answer» 2Al+2NaOH+2H2O→2NaAlO2+3H2. |
|
| 31. |
Which of the following is not correctly matched ?A. `Ni^(2+)-` Red precipitate with `DMG `and ammonia solutionB. `Zn^(2+)-` While precipitate with `H_(2)S` gas in presence of `NH_(4)OH` and `NH_(4)Cl`C. `Al^(3+)-` Infusible green mass is obtained in cobalt nitrate test.D. `Pb^(2+)-` Yellow precipitate with both `KI` and `K_(2)CrO_(4)` reagents. |
|
Answer» Correct Answer - 3 `Al_(2)O_(3)+CoOoverset(Delta)rarrAl_(2)O_(3).CoO(` blue mass `). ZnO.CoO` is green in colour. |
|
| 32. |
Brass is an alloy ofA. copper and tin onlyB. copper and zinc onyC. copper , zinc and nickel onlyD. copper, tin and zinc only |
|
Answer» Correct Answer - 2 Brass `," "C=60"to "80%,Zn=40"to "20%` Bronze `," "Cu=75"to"90%,Sn=25"to "10%` German Silver `," "Cu=56%,Zn=24%,Ni=20%` Gun metal `," "Cu87%,Zn=3%,Sn=10%` |
|
| 33. |
Which of the following statements is incorrect about the `f-` block elements ?A. `f-` block consists of two series, lanthanoids and actinoids.B. Lanthanoids shown in general stable common oxidation state, `+3`C. The actinoids are radioactive elements.D. Actinoids contractions is smaller from element to element then lanthanoid contraction. |
|
Answer» Correct Answer - 4 In addition to 4f electrons, the 5f electrons themselves provide poor shielding from element to element in the series so, actionoid contraction is greater than lanthanoid contraction. |
|
| 34. |
In `[Ni(CO)_(4)]` the `Ni-C` bond possesses `:`A. `pi` characer onlyB. both `sigma` and `pi` characterC. `sigma `character onlyD. no`sigma` and `pi` characters |
|
Answer» Correct Answer - 2 `Niunderset(sigma)overset(pi)hArrCO` |
|
| 35. |
Flux is used to removeA. silicaB. metal oxideC. silica and metal oxideD. impurities from ore |
|
Answer» Correct Answer - 4 The additional substance used to remove impurities `(` gangue `)` from concentrated ore by chemically reacting with it and converting in into fusible mass `(` slag `)` is called flux. |
|
| 36. |
The clouds consist of charged particles of water dispersed in air.Some of them are +vely charged, others are -vely charged.When +vely charged clouds come closer they cause lightening and thundering whereas when +ve and -ve charged colloids come closer they cause heavy rain by aggregration of minute particles.It is possible to cause artifical rain by throwing electrified sand or silver iodide from an aeroplane and thus coagulating the mist hangind in air. Agl helps in artificial rain because :A. it helps in condensation processB. It helps in dispersion processC. It helps in coagulationD. all of them |
|
Answer» Correct Answer - C Agl coagulates mist hanging in air |
|
| 37. |
What products are expected from the disproportionation reaction of chlorine with hot concentrated solution of sodium hydroxide ?A. `NaCl,NaClO_(3)`B. `NaCl,NaClO`C. `NaCl,NaClO_(3)`D. `NaCl,NaClO_(2)` |
|
Answer» Correct Answer - 1 `6NaOH+3Cl_(2)rarr5NaCl+NaClO_(3)+3H_(2)O` |
|
| 38. |
At 2 atm pressure the value of `x/m` will be :(log 2=0.3010) |
|
Answer» Correct Answer - 4 `x/m=K P^(1//n)` `"log"x/m="log"K+1/n"log"P` `:. K=2 & 1/n=1 or n=1` On putting in `eq^n(1)` `x/m=2xx2=4` |
|
| 39. |
The catalyst used in the manufacture of sulphuric acid by contact process is `:A. `Fe_(2)O_(3)`B. `CoCl_(2)`C. `V_(2)O_(6)`D. `CuCl_(2)` |
|
Answer» Correct Answer - 3 `2SO_(2)(g)+O_(2)(g)overset(V_(2)O_(5))rarr2SO_(3)(g)` |
|
| 40. |
Match list I (Colloidal system ) with list II (Example) `{:("Column-I","Column-II"),((A)"Emulsifier",(p)"Colloidal solution"),((B)"Xerogel",(q)"Soaps "),((C )"Colloidal electrolyte",(r)"Foil of cellophane"),((D)"Purple of Cassius",(s)"Dextrin"),(,(t)"Gold sol"):}` |
|
Answer» Correct Answer - A`to`s; B`to`r ; C`to` q ; D`to`p,t (A)Dextrin is an example of emulsifier. (B)Foil of cellophane is an example Xerogel ( C)Soaps are colloidal electrolyte (D)Purple of cassius (gold sol.) is colloidal solution. |
|
| 41. |
The colloidal particles are electrically charged as a indicated by their migration towards cathode or anode under the applied electric field. In a particular colloidal system, all particles carry either positive charge or negative charge. The electric charge on colloidal particles orginate in several ways. According to preferential adsorption theory, the freshly obtained precipitate particles adsorb ions from the dispersion medium, which are common to their lattice and acquire the charge of adsorbed ions. For example, For example, freshly obtained `Fe(OH)_(3)` precipitated is dispersed, by a little `FeCl_(3)`, into colloidal solution owing to the adsorption of `Fe^(3+)` ions in preference. Thus sol particles will be positively charged. In some cases the colloidal particles are aggregates of cations or anions having ampiphilic character. When the ions posses hydrophobic part (hydrocarbon end) as well as hydrophilic part (polar end group), they undergo association in aqueous solution to form particles having colloidal size. The formation of such particles, called micelles plays a very important role in the solubilization of water insoluble substances, (hydrocarbon, oils, fats, grease etc.). In micelles, the polar end groups are directed towards water and the hydrocarbon ends into the centre. The charge on sol particles of proteins depends on the pH. At low pH, the basic group of protein molecule is ionized (protonated) and at higher pH (alkaline medium), the acidic group is ionized. At isoelectric pH, characteristic to the protein, both basix and acidic groups are equally ionized. The stability of colloidal solution is attributed largely to the electric charge of the dispersed particles. This charge causes them to be coagulated or precipitated. On addition of small amount of electrolytes, the ions carrying oppiste charge are adsorbed by sol particles resulting in the neutralization of their charge. When the sol particles either with no charge or reduced charge, come closer due to Brownian movement, they coalesce to form bigger particles resulting in their separation from the dispersion medium. This is what is called coagulating or precipitation of the colloidal solution. The coagulating power of the effective ion, which depend on its charge, is expressed in terms of its coagulating value, defined as its minimum concentration (m mol/L) needed to precipitate a given sol. A gelatin sol at pH less than the isoelectric value is subjected to an electric field. The sol particles migrate toward :A. AnodeB. CathodeC. Both anode and cathodeD. Neither anode nor cathode |
|
Answer» Correct Answer - B At low pH the basic group will be ionized (protonated) so will have positive charge and hence sol particles will move towards cathode. |
|
| 42. |
Match list I with list II and select the correct answer : `{:("Column-I","Column-II"),((A)"Coagulation",(p)"Scattering of light"),((B)"Dialysis",(q)"Washing of precipitates"),((C )"Peptization",(r)"Purification of colloids"),((D)"Tyndall effect",(s)"Electrolyte"),(,(t)"Visibility of tall of comets"):}` |
|
Answer» Correct Answer - A`to`s; B`to`r ; C`to` q ; D`to`p,t (A)Coagulation is known as accumulation of collidal sols. (B)Dialysis is purification of colloids. ( C)Peptization is formation of colloidal solution from precipitates. (D)Tyndall effect is scattering of light by colloidal and also visibility of tail of comets is due to Tyndall effect. |
|
| 43. |
The colloidal particles are electrically charged as a indicated by their migration towards cathode or anode under the applied electric field. In a particular colloidal system, all particles carry either positive charge or negative charge. The electric charge on colloidal particles orginate in several ways. According to preferential adsorption theory, the freshly obtained precipitate particles adsorb ions from the dispersion medium, which are common to their lattice and acquire the charge of adsorbed ions. For example, For example, freshly obtained `Fe(OH)_(3)` precipitated is dispersed, by a little `FeCl_(3)`, into colloidal solution owing to the adsorption of `Fe^(3+)` ions in preference. Thus sol particles will be positively charged. In some cases the colloidal particles are aggregates of cations or anions having ampiphilic character. When the ions posses hydrophobic part (hydrocarbon end) as well as hydrophilic part (polar end group), they undergo association in aqueous solution to form particles having colloidal size. The formation of such particles, called micelles plays a very important role in the solubilization of water insoluble substances, (hydrocarbon, oils, fats, grease etc.). In micelles, the polar end groups are directed towards water and the hydrocarbon ends into the centre. The charge on sol particles of proteins depends on the pH. At low pH, the basic group of protein molecule is ionized (protonated) and at higher pH (alkaline medium), the acidic group is ionized. At isoelectric pH, characteristic to the protein, both basix and acidic groups are equally ionized. The stability of colloidal solution is attributed largely to the electric charge of the dispersed particles. This charge causes them to be coagulated or precipitated. On addition of small amount of electrolytes, the ions carrying oppiste charge are adsorbed by sol particles resulting in the neutralization of their charge. When the sol particles either with no charge or reduced charge, come closer due to Brownian movement, they coalesce to form bigger particles resulting in their separation from the dispersion medium. This is what is called coagulating or precipitation of the colloidal solution. The coagulating power of the effective ion, which depend on its charge, is expressed in terms of its coagulating value, defined as its minimum concentration (m mol/L) needed to precipitate a given sol. Which of the following ions would have the minimum coagulating value for sol obtained on peptizing `Sn(OH)_(4)` by little NaOH solution?A. `Cl^-`B. `SO_4^(2-)`C. `K^+`D. `Ba^(2+)` |
|
Answer» Correct Answer - D Minimum coagulating value will be for the ion with maximum charge and since the sol particles are negatively charged, hence positively charged ions are required for coagulations. |
|
| 44. |
Calculate the perimeter of given in `HCP` unit cell `(` Given that radius of atoms `=RÅ)`. A. `6.437R`B. `15.32R`C. `16R`D. `8R` |
|
Answer» Correct Answer - 1 Perimeater of plane is `2C+8R=(2xx4sqrt(2)R)/(3)+8R=6.437R` |
|
| 45. |
Which is the correct statement in case of milk?A. Milk is an emulsion of fat in waterB. Milk is an emulsion of protein in waterC. Milk is stabilized by proteinD. Milk is stabilized by fat |
|
Answer» Correct Answer - 1 Milk is an emulsion of fat in water. |
|
| 46. |
The clouds consist of charged particles of water dispersed in air.Some of them are +vely charged, others are -vely charged.When +vely charged clouds come closer they cause lightening and thundering whereas when +ve and -ve charged colloids come closer they cause heavy rain by aggregration of minute particles.It is possible to cause artifical rain by throwing electrified sand or silver iodide from an aeroplane and thus coagulating the mist hangind in air. Smoke screens consist ofA. fine particles of `TiO_2` dispersed in air by aeroplanesB. fine particles of Agl dispersed in air by aeroplanesC. fine partilces of `Al_2O_3` dispersed in air by aeroplanesD. None of these |
|
Answer» Correct Answer - A Smoke screen is cloud of smoke. It consists of fine particles of `TIO_2` |
|
| 47. |
The colloidal particles are electrically charged as a indicated by their migration towards cathode or anode under the applied electric field. In a particular colloidal system, all particles carry either positive charge or negative charge. The electric charge on colloidal particles orginate in several ways. According to preferential adsorption theory, the freshly obtained precipitate particles adsorb ions from the dispersion medium, which are common to their lattice and acquire the charge of adsorbed ions. For example, For example, freshly obtained `Fe(OH)_(3)` precipitated is dispersed, by a little `FeCl_(3)`, into colloidal solution owing to the adsorption of `Fe^(3+)` ions in preference. Thus sol particles will be positively charged. In some cases the colloidal particles are aggregates of cations or anions having ampiphilic character. When the ions posses hydrophobic part (hydrocarbon end) as well as hydrophilic part (polar end group), they undergo association in aqueous solution to form particles having colloidal size. The formation of such particles, called micelles plays a very important role in the solubilization of water insoluble substances, (hydrocarbon, oils, fats, grease etc.). In micelles, the polar end groups are directed towards water and the hydrocarbon ends into the centre. The charge on sol particles of proteins depends on the pH. At low pH, the basic group of protein molecule is ionized (protonated) and at higher pH (alkaline medium), the acidic group is ionized. At isoelectric pH, characteristic to the protein, both basix and acidic groups are equally ionized. The stability of colloidal solution is attributed largely to the electric charge of the dispersed particles. This charge causes them to be coagulated or precipitated. On addition of small amount of electrolytes, the ions carrying oppiste charge are adsorbed by sol particles resulting in the neutralization of their charge. When the sol particles either with no charge or reduced charge, come closer due to Brownian movement, they coalesce to form bigger particles resulting in their separation from the dispersion medium. This is what is called coagulating or precipitation of the colloidal solution. The coagulating power of the effective ion, which depend on its charge, is expressed in terms of its coagulating value, defined as its minimum concentration (m mol/L) needed to precipitate a given sol. How would you obtain a sol of AgI, the particles of which migrate towards cathode under the electric field?A. By adding little excess of KI to `AgNO_3` solutionB. By adding little excess of `AgNO_3` to KI solutionC. By mixing equal volumes of 0.010 M `AgNO_3` and 0.010 M KID. None of these |
|
Answer» Correct Answer - B We want to prepare sol of AgI having positively charged particles so a little excess of `Ag^+` should be added to KI |
|
| 48. |
Insulin `(C_(2)H_(10)O_(5))_(n)` is dissolved in a suitable solvent and the osmotic pressure `pi` of the solution of various concentration ( in `kg//m^(3))` is measured at `20^(@)C`. The slpe of a plot of `pi` against c is found to be `8.134xx10^(-3)` ( SI units ). The molecular weight of the insulin ( in `kg//mol)` is `:`A. `4.8xx10^(5)`B. `9xx10^(5)`C. `293xx10^(3)`D. `8.314xx10^(5)` |
|
Answer» Correct Answer - C `pi=CRT` `pi=(c)/(M)RT" "C=` moles`//`liter,`c=kg//m^(3)` `(pi)/(c)=(RT)/(M)` `M=(RT)/(pi //c)" "[pi//c=8.314xx10^(-3)]` `[T=293 k]` `M=(8.314xx293)/(8.314xx10^(-3))=293xx10^(3)` |
|
| 49. |
Adsorption is the presence of excess concentration of any particular component at the surface of liquid a solid phase as compared to bulk.This is due to presence of residual forces at the surface of body.In the adsorption of hydrogen gas over a sample of charcoal, `1.12cm^3` of `H_2(g)` measured over S.T.P was found to absorb per gram of charcoal.Consider only monolayer adsorption.Density of `H_2` is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100 ml of 0.5M `CH_3COOH` to form monolayer and thereby the molarity of `CH_3COOH` reduces to 0.49. `((3xx47.43)/(4pi))^(1//3)=2.24` `(2.24)^2=5` What is the radius of adsorb hydrogen molecule-A. `2.24xx10^(-7) cm`B. `2.24xx10^(-8) cm`C. `2.24xx10^(-9) cm`D. `2.24xx10^(-10) cm` |
|
Answer» Correct Answer - B Density of `H_2=0.07 gm//c c` 0.07 gm `H_2` has volume = 1ml 2gm `H_2` has volume`=1/0.07xx2` ml So`N_A` molecules of `H_2` have volume `=1/0.07xx2`ml 1 molecules of `H_2` has volume`=2/(0.07xx6.023xx10^23)` assuming `H_2` to be spherical molecule `=2/(0.07xx6.023xx10^23)=4/3pir^3 implies r_3=(3xx47.437xx10^(-24))/(4pi)implies r=2.24xx10^(-8) cm` |
|
| 50. |
Adsorption is the presence of excess concentration of any particular component at the surface of liquid a solid phase as compared to bulk.This is due to presence of residual forces at the surface of body.In the adsorption of hydrogen gas over a sample of charcoal, `1.12cm^3` of `H_2(g)` measured over S.T.P was found to absorb per gram of charcoal.Consider only monolayer adsorption.Density of `H_2` is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100 ml of 0.5M `CH_3COOH` to form monolayer and thereby the molarity of `CH_3COOH` reduces to 0.49. `((3xx47.43)/(4pi))^(1//3)=2.24` `(2.24)^2=5` Number of molecules of hydrogen over 1 gram charcoal are-A. `3.01xx10^(18)`B. `3.01xx10^(19)`C. `3.01xx10^(21)`D. `3.01xx10^(22)` |
|
Answer» Correct Answer - B Volume of `H_2` over 1 gm charcoal `=1.12 cm^2` at STP mole of `H_2=1.12/22400=5xx10^(-5)` mole Number of molecules of `H_2=5xx10^(-5)xx6.023xx10^(23)=3.01xx10^19` molecules |
|