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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. |
वह न्युनतम संख्या क्या है जिससे 12 तथा 16 से भाग देने पर क्रमशः 5 तथा 9 शेष बचता है?A. 55B. 41C. 39D. 29 |
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Answer» (b) 12-5=7 16-9=7 Remainder always remains same in such question, ( इस प्रकार के प्रश्नो में शेषफल हमेशा बराबर होता है ) `therefore` LCM of (12,16)=48 `therefore` Required result =48-7=41 |
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| 2. |
वह न्यूनतम संख्या क्या है जिससें 5 जोडे देने पर प्राप्त संख्या 24,32,36 तथा 64 प्रत्येक से विभाजित होती है ?A. 869B. 859C. 571D. 427 |
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Answer» (c ) LCM of (24, 32, 36, 64) `rArr` `8xx3xx4xx3xx2=576` required no is = 576 -5 = 571 |
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| 3. |
जब एक संख्या को 15, 20 या 35 से भाग दिया जाता है, तो प्रत्येंक बार 8 शेष बचता है, तो वह न्यूनतम संख्या ज्ञात क्या है ?A. 428B. 427C. 328D. 338 |
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Answer» (a) LCM of (15, 20, 35) = `5xx3xx4xx7=420` required number = 420 +8 = 428 |
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| 4. |
वह न्यूनतम संख्या क्या है, जिससे 5,10,12 तथा 15 से भाग देने पर प्रत्येक स्थिति में 2 शेष बचता है लेकिन 7 से भाग देने पर कोई शेष नहीं बचता है ?A. 189B. 182C. 175D. 91 |
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Answer» (b) LCM of (5, 10, 12, 15) ltBrgt `=5xx2xx6=60` smallest no. divided by (5, 10, 12, 15) leaves remainder 2 and when Divide by 7 leaves no Remainder is `(60K+2)/(7)=(4K+2)/(7)` At k = 3, `(4K+2)/(7)` `rArr` Remainder = 0 No. = 60K+2=60`xx`3+2 = 182 |
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| 5. |
तीन अंकों की दो संख्याओं का म.स. 17 और ल.स. 714 है, तो संख्याओं का योग क्या होगा ?A. 289B. 391C. 221D. 731 |
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Answer» ( c) HCF =17 `therefore` माना कि संख्याएँ = 17x, 17y LCM = 17xy = 714 (given) xy = 42 संभावित संख्याएँ `(17, 714), (34, 357), (51, 238),(102,119)` लेकिन दिया गया है कि संख्याएँ 3 अंको की है| `therefore` numbers are = (102, 119) `therefore` संख्याओं का योग =102+119=221 |
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| 6. |
दो संख्याओं का म0 स0 23 है तथा उनके ल0 स0 के अन्य दो गुणनखण्ड 13 तथा 14 है तो उन संख्याओं में से बड़ी संख्या क्या है?A. 276B. 299C. 345D. 322 |
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Answer» (d) HCF =23 `therefore` Let number are ( माना कि संख्याएँ ) =23x, 23y `therefore` LCM=23xy `implies` Now given that factor of LCM are 13,14 ( दिया गया है कि ल स के गुणज 13 तथा 14 है ) `therefore LCM = 23xx13xx14` number are `=23xx13` =299 and `23xx14` =322 larger =322 |
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| 7. |
दो लगातार धनात्मक सम संख्याओं का म0 स0 तथा ल0 स0 क्रमशः 2 तथा 84 है तो संख्याओं का योग ज्ञात करें?A. `30`B. `26`C. `14`D. `34` |
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Answer» Correct Answer - B माना कि दो लगातार धनात्मक सम संख्याएं क्रमशः `(2x+2)` और `(2x+4)` HCF `=2` (given) common (factor ) LCM `=2(x+1)(x+2)` `darr` HCF (other factors) `implies ` LCM `=84` (given) `:. 2(x+1)(x+2)=84` `(x+1)(x+2)=42` `implies x^(2)+3x+2=42` `x^(2)+3x+2-42=0` `x^(2)+8x-5x-40=0` `implies x(x+8)-5(x+8)=0` `(x-5)(x+8)=0` `(x=+5)` `x=-8` But numbers are even so `(x=5)` `:.` Numbes are `=2xx5+2=12` and `2xx5+4=14` `implies` sum of numbers are `=12+14` `=26` |
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| 8. |
दो संख्याओं का योग 45 है। उनका अंतर योग का `(1)/(9)` है, तो उनका ल. स. ज्ञात करें।A. 200B. 250C. 100D. 150 |
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Answer» (c ) A+B = 45 A-B = `(45)/(9)=5` `therefore` A=25, B= 20 `therefore` LCM of (25, 20)= `5xx5xx4 =100` |
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| 9. |
वह अधिकतम संख्या क्या है जिससे 110 तथा 128 को भाग देने पर प्रत्येक स्थिति में 2 शेष बचता है ?A. 8B. 18C. 28D. 38 |
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Answer» (b) 110-2=108 128-2=126 `therefore` HCF of (108, 126) = 18 |
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| 10. |
दों अंकों की दो संख्याओं का म. स. व.ल.स. क्रमशः 16 तथा 480 है, तो संख्याँए ज्ञात करें ?A. `40,48`B. `60,72`C. `64,80`D. `80,96` |
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Answer» (d) HCF=16 `therefore` numbers are 16x and 16y (माना की संख्याएँ 16x तथा 16y है ।) 16xy=480 xy =30 `therefore` possible pairs (संभावित जोड़े ) =(1,30),(2,15),(6,5) Possible number are संभावित संख्याएँ ) =(16,480) ,(32,240),(80,96) `therefore` (80,96) is the answer in the given option of 2 digit. |
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| 11. |
चार अंकों की वह अधिकतम संख्या क्या होगी जिसे 12, 16 तथा 24 से भाग देने पर क्रमशः 2, 6 तथा 14 शेष बचता है ?A. `9974`B. `9970`C. `9807`D. `9998` |
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Answer» (a) 12-2= 10 16-6 = 10 12-14= 10 ltBrgt LCM of (12, 16, 24) = `6xx2xx4xx1` = 48 4 अंको की अधिकतम संख्या = 9999 ltBrgt `therefore` जब इसे 48 से भाग दिया जाता है तो शेषफल = 15 `rArr` 4 अंको की वह अधिकतम संख्या जो दी गई संख्याओं से पूर्णतः विभाजित है | ltBrgt = 9999-15=9984 `therefore` number is = 9984-10= 9974 |
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| 12. |
वह अधिकतम संख्या क्या है जिसे 5834 में से घटाने पर प्राप्त संख्या 20,28, 32 तथा 35 प्रत्येक से पूर्णकः विभाजित है ?A. `1120`B. `4714`C. `5200`D. `5600` |
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Answer» (b) LCM of (20, 28, 32, 35) `rArr` `4xx5xx7xx8=1120` `therefore` 1120 divided by 20, 28, 32, 35 completely `therefore` Let x be subtracted from 5834, ltBrgt `therefore` 5834 - x = 1120 x = 5834-1120=4714 |
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| 13. |
दो संख्याऐं 12906 और 14818 का म0 स0 478 हैं तो ल0 स0 ज्ञात करें।A. 400086B. 200043C. 600129D. 800172 |
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Answer» (a) HCF =478 Number are =12906 and 14818 `therefore` `LCM xx HCF=12906xx14818` `LCM xx 478=12906xx14818` LCM =400086 |
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| 14. |
12 के दो गुणांको का ल0 स0 1056 है, यदि एक संख्या 132 है, तो दूसरी संख्या ज्ञात करें ?A. 12B. 72C. 96D. 132 |
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Answer» (c ) Let numbers be 12x and 12y respectively LCM = 12xy 12xy = 1056 (given) xy = 88 `therefore` possible pairs are (1, 88) (8, 11) ltBrgt possible numbers are = (12, 1056) (96, 132) ltBrgt given that one number is 132 so other is 96 |
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| 15. |
दो संख्याओं का म0 स0 व ल0 स0 क्रमशः 7 तथा 140 है। यदि संख्याएं 20 तथा 45 है बीच में है तो संख्याओं का योग ज्ञात करें?A. 70B. 77C. 63D. 56 |
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Answer» ( c) HCF of numbers ( संख्याओं का म स )=7 `therefore` Let the numbers are 7x and 7y (माना की संख्याएँ 7x तथा 7y है ।) LCM = 7xy 7xy = 140 (given) xy = 20 `therefore` Possoble co-prime factors of xy ( xy की संभावित असहभाज्य गुणक ) =(1,20),(4,5) `implies` number are between 20 and 45 ( संख्याएं 20 तथा 45 के बीच में है ।) `therefore` Required numbers are `=4xx7=28 " and " 5xx7=35` `implies` sum of number are =28+35=63 |
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| 16. |
दो संख्याओं के म.स.तथा ल.स. का योग 680 है, उनका ल.स.,म. स. का 84 गुणा है। यदि एक संख्या 56 है , तो दूसरी संख्या ज्ञात करें ?A. 84B. 12C. 8D. 96 |
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Answer» (d) Let HCF=x LCM=84x given HCF+LCM =84x+x=680 85x=680 x=8 `therefore` HCF =8 LCM `=84xx8=672` `implies 56 xx y=672xx8` `y=(672xxx)/(56)=96` |
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| 17. |
दो संख्याओं का म.स. व गुणनफल क्रमशः 15 तथा 6300 है, इस तरह के संख्याओं के कितने सभावित जोडे हो सकते है ?A. 4B. 3C. 2D. 1 |
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Answer» (c ) HCF = 15 दो संख्याओं का गुणनफल = 6300 `therefore` Let numbers are 15x, 15y `therefore` 15x `xx` 15y = 6300 (given) xy = `(6300)/(15xx15)` = 28 possible pairs are = (1, 28), (7, 4) `therefore` Total pairs = 2 |
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| 18. |
दो संख्याओं का ल.स. उनके म. स. का 20 गुणा है, उनके म. स. तथा ल. स. का योग 2520 है। यदि एक संख्या 480 है, तो दूसरी संख्या ज्ञात करें ?A. 400B. 480C. 520D. 600 |
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Answer» (d) Let HCF=x `therefore` LCM =20x sum of HCF +LCM=2520 `=x+20x=2520` 21x=2520 x=120 `therefore` HCF =120 LCM `=120xx20=2400` `therefore` one number =480 Let another number (माना की अन्य संख्या ) =y `therefore yxx480=120xx2400 ` `y=(120xx24000)/(480)=600` |
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| 19. |
दो संख्याओं का गुणनफल 216 है, तथा यदि उनका म0 स0 6 है, तो ल0 स0 ज्ञात करें ?A. 72B. 60C. 48D. 36 |
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Answer» (d) संख्याओं का गुणनफल = 216 HCF = 6 LCM = `(216)/(6)` = 36 |
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| 20. |
दो संख्याओं का योग 384 है तथा उनका म0 स0 48 है, तो संख्याओं का अंतर क्या है ?A. 100B. 192C. 288D. 336 |
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Answer» (c ) HCF = 48 `therefore` Let number are 48x & 48y respectively `rArr` 48x+48y = 384 (x+y)= `(384)/(48)` = 8 सह अभाज्य संख्याओं के संभावित जोड़े (1, 7), (3, 5) `therefore` numbers are (48, 336) or (144,240) `therefore` संख्याओं के बीच का अंतर = 336-48 = 288 and 24-144=96 |
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| 21. |
दो संख्याओं का ल0 स0 व म0 स0 क्रमशः 520 तथा 4 है यदि एक संख्या 52 है तो दूसरी संख्या ज्ञात करें?A. 40B. 42C. 50D. 52 |
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Answer» Correct Answer - A LCM `=520` HCF `=4` one number `=52` Let other number is `=y` `:. 52y=4xx520` `y=40` |
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| 22. |
दो संख्याओं का गुणनफल `396 xx 576` व उनका ल0 स0 6336 हैं। तो म0 स0 ज्ञात करें ।A. 36B. 34C. 63D. 46 |
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Answer» (a) We, know that `rArr` (a`xx`b)= (HCF and LCM) 396 `xx` 576 = HCF `xx` 6336 HCF = 36 |
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| 23. |
दो संख्याओं का गुणनफल 20736 व उनका म0 स0 54 है। तो ल0 स0 ज्ञात करें ।A. 685B. 468C. 648D. 384 |
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Answer» (d) Let No. are a and b a`xx`b = 20736 HCF = 54 ltBrgt we know that (a`xx`b)= (HCF `xx` LCM) ltBrgt 20736 = 54 `xx` LCM LCM = `(20736)/(54)=384` |
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| 24. |
दो संख्याओं का अनुपात 4:5 है तथा उनका ल0 स0 120 है तो संख्याऐं ज्ञात करें।A. 30,40B. 40,32C. 24,30D. 36,20 |
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Answer» ( c) Let numvers are A & B respectively ( माना की संख्याये क्रमश: A तथा B है ) `{:(A,:,B),(4x,:,5x):}`(given) `thereforeLCM=4xx5xxx=20x` `therefore20x=120` x=6 `thereforeA =4x=4xx6=24` `B=5x=5xx6=30` |
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| 25. |
दो संख्याओं का ल0 स0 उनके म0 स0 का 44 गुना है , उनके ल0 स0 तथा म0 स0 का योग 1125 है, यदि एक संख्या 25 है तो दूसरी संख्या ज्ञात करें ?A. `1100`B. 975C. 900D. 800 |
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Answer» (a) Let HCF = x LCM = 44x given HCF + LCM = 44x+x=45x 45x= 1125 ltBrgt x= `(1125)/(45)` = 25 `therefore` HCF = 25, ltBrgt LCM = 25`xx`44 ltBrgt `rArr` also given that one number = 25 Let another number = y ltBrgt `therefore` 25y = 25`xx`25`xx`44 y = `(25xx25xx44)/(25)` = 1100 |
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| 26. |
`a,b` का म0 स0 `12` है तथा `(a,b)` का न्यूनतम मान क्या होगा?A. 12,24B. 24,12C. 24,36D. 36,24 |
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Answer» Correct Answer - D HCF (GCD) of a , b number is 12 and `agtbgt12` (given) `:.` smallest value of `a` & `b` are (36, 24) |
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| 27. |
दो संख्याओं का म0 स0 व ल0 क्रमशः 12 तथा 924 है तो इस तरह की संख्याओं के कितने संभावित जोड़े होगें? |
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Answer» Correct Answer - C HCF `=12` `:.` Let numbers are `12x` & `12y` respectively LCM `implies12xy=924` (given) `=xy=77` `implies` possible pairs are `=(1xग77)(7xx11)` `:.` only two pairs are possible |
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| 28. |
दो संख्याओं का योग 36 है तथा उनका म.स.4 है, तो इस तरह की संख्याओं के संभावित जोडों की संख्या क्या होगी ?A. 1B. 2C. 3D. 4 |
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Answer» ( c) HCF= 4 `therefore` Let number are 4x and 4y ( माना की संख्याएँ 4x तथा 4y है ।) given sum =4x=4y =36 and x+y =9 possible pairs (संभावित जोड़े ) `implies` (1+8),(2+7),(4+5), numbers should be co prime. Hence only 3 pairs (असहभाज्य संख्याएँ होनी चाहिए । अतः केवल 3 जोड़े ही संभव है ।) |
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| 29. |
दो संख्याओं का योग 84 है तथा उनका म0स0 12 है तो इस तरह के संख्याओं के जोड़ों की कुल संख्या क्या है ?A. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - B HCF `=12` `:.` माना कि संख्याएं क्रमशः `12x` तथा `12y` हैं। `:.` Given that `(12x+12y)=84` `implies x+y=84/12=7` `:. x+y=7` `implies` संभावित जोड़े हैं। `(1+6), (2+5), (3+4)` `:.` Total numbr of pair are 3 |
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| 30. |
दो संख्याओं का गुणनफल 2028 है तथा म0 स0 13 है तो इस तरह की संख्याओं के कितने जोड़े हो सकते हैं?A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B HCF `=13` (given) माना कि संख्याएं क्रमशः `13x` तथा `13y` है। `:.` also given `12x xx 13y=2028` `13xx13xx xy=2028` `xy=2028/(13xx13)=12` संभावित जोड़े `(1,12),(3,4)` केवल दो जोड़े संभव हैं। |
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| 31. |
दो सहअभाज्य संख्याओं का गुणनफल 117 है तो उनका ल0 स0 ज्ञात करें।A. 117B. 9C. 13D. 39 |
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Answer» Correct Answer - A असहभाज्य संख्याओं का म0 स0 हमेशा 1 होता है। `:.` Let numbers are `=x` and `y` respectively संख्याओं का गुणनफल `=xy` `xy=117` (given) संख्याओं का गुणनफल `=LCM xxHCF` `implies LCM xx1=117` LCM `=117` |
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| 32. |
दो संख्याओं का योग 216 है और उनका म0 स0 27 है। इस तरह की संख्याओं के कितने संभावित जोडे हो सकते है ?A. 1B. 2C. 3D. 0 |
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Answer» (b) HCF = 27 `therefore` Let numbers are 27x and 27y respectively `therefore` 27x + 27y = 216 (given) `rArr` (x+y) = `(216)/(27)=8` ltBrgt संभावित गुणक है | = (1, 7),(3, 5). |
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| 33. |
`2/3, 4/9, 5/6` का ल0स0 क्या होगा?A. `8/27`B. `20/3`C. `10/3`D. `20/27` |
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Answer» Correct Answer - B किसी भिन्न का ल0स0 `implies ("LCM of numerator")/("HCF of denominator")` `implies` LCM `(2/3, 4/9, 5/6)` `implies("LCM" (2,4,5))/("HCF"(3,9,6))=20/3implies20/3` ans. |
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| 34. |
दो संख्याओं का गुणनफल 2160 है और उनका म0स0 12 है तो इस तरह की संख्या के कितने संभावित जोड़े होगें?A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B HCF `=12` `:.` Let numbers are `12x` and `12y` `:.` Product of two number `=12x` `12y=144xy` `=144xy=2160` `impliesxy=15` `:.` Possible pairs are (1,15), (3,5), factors should be co prime. Two pairs are possible. |
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| 35. |
तीन अलग-अलग चौराहों के ट्रेफिक लाइटें क्रमशः 24, 36 तथा 54 सेकेण्ड के अंतराल पर बदलती है यदि वे सभी एक साथ `10:15` AM पद बदलती हों तो वे अगली बार कब बदलेंगी?A. `10:16:54` AMB. 10:18:36AMC. 10:17:02AMD. 10:22:12AM |
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Answer» Correct Answer - B LCM of `(24,36,54)` `implies 12x2x3xx3=216` seconds `implies` वे प्रत्येक 216 सेकेण्ड के पश्चात एक साथ बदलेंगी। `implies 216/60implies 3 36/60` `=` 3 minutes 36 seconds They change 1st at `10:15:00` am So, again they change at `= 10:18:36` am |
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| 36. |
पांच घंटियां एक साथ बजती है और वे क्रमशः 6,7,8,9 तथा 12 सेकेण्डों के अंतराल पर बजती है तो कितने सेकेण्ड के बाद वे पुनः साथ बजेगी?A. 72 secB. 612 secC. 504 secD. 318 sec |
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Answer» Correct Answer - C LCM of `(6,7,8,9,12)` LCM `=(3xx2xx7xx4xx3=504` वे प्रत्येक 504 सेकेण्ड के बाद बजेंगी। |
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| 37. |
चार घंटियां 5,6,7 तथा 9 सेकेण्ड के अंतराल पर बजती है। सभी घंटियां किसी समय एक साथ बजती है तो वे पुनः एक साथ कितने समय बाद बजेगी?A. 6 minutesB. 12 minutesC. 18 minutesD. 24 minutes |
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Answer» (a) LCM of (5,68,9) `=5xx6xx4xx3` =360 seconds `=(360)/(60)`=6 minutes `implies` Bells will ring simultaneosly after every 6 minutes. ( घंटियाँ प्रत्येक 6 मिनट के बाद एक साथ बजेगी ) |
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| 38. |
Solve for x: (i) `10x^(2)-27x+5=0` (ii) `pqx^(2)-(p^(2)+q^(2))x+pq=0` |
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Answer» Correct Answer - (i) `(5)/(2), (1)/(5)` (ii) `(p)/(q), (q)/(p)` (i) `10 x^(2) - 27x +5=0` ltBrgt `x= (-bpm sqrt(b^(2) - 4ac))/(2a) = (27 pm sqrt((-27)^(2) - 4(10)(5)))/(20)` `= (27pm 23)/(20 ) rArr x =(5)/(2), (1)/(5)` (ii) `(pq)x^(2) - (p^(2) + q^(2))x+(pq)=0` `x= ((p^(2) + q^(2)) pm sqrt((p^(2)+ q^(2))^(2) - 4(pq)(pq)))/( 2pq)` `= ((p^(2) + q^(2)) pm (p^(2) - q^(2)))/( 2pq) rArr x = (p)/(q), (q)/(p)` |
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| 39. |
तीन घंटियां 11 am को एक साथ बजती है वे क्रमशः 20 मिनट 30 मिनट तथा 40 मिनट के नियमित अंतरात पर बजती हैं तो वे अगली बार एक साथ कब बजेगीं?A. 2 p.m.B. 1 p.m.C. `1.15 p.m.D. 1.30 p.m. |
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Answer» Correct Answer - B LCM of `(20, 30, 40)` `implies 4xx5xx3xx2=120` minutes `=120/60=2` hours The 1st bell at 11 am. So the again bell after 2 hours at `11+2=1` pm |
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| 40. |
Modern peroidic table helps classify elements according to their similarities.Consider the columns mentioning certain details about elements with comments on their important features. Answer the question that follow : Which of the following combinations is appropriate for Beryllium ?A. (IV)(i)(P)B. (II)(i)(Q)C. (III)(ii)(P)D. (IV)(iii)(R) |
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Answer» Correct Answer - A Be belongs to II period and `2^(nd)` group Be their similarity with AC and is less EN than H. |
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| 41. |
The value of n (i.e. principal qunatum number ) for the valence shell of palladium is (atomic number =46): |
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Answer» Correct Answer - 4 Electron configuration for `._46 Pd` is `[Kr]^36 4d^10` |
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| 42. |
Rydberg atoms are nearly classical atoms in which electron is excited to states corresponding to n = 10 , 2 0 or even 100. There are many difficulties in studying them. 1. The radius of their orbit will be very large. So at ordinary densities, their collisions will allow electrons to jump from one orbit to another without emitting radiation. 2. The energy levels at such value of n will be every close. So experiment on rydberg atoms should be done at very low temperatures and very low pressures. Consider a very dilute gas of hydrogen atoms that are excited to n = 10. Assume for this situation that the average centre to centre distance should be equal to diameter of rydberg atom. [Take wein constant `= 2.88xx10^(-3) mK`] Assuming the atoms to fill up space like footballs filling up a room, how many atoms would fill up in a `1mxx1mxx1m` room. (approx.)A. `10^(20)`B. `10^(28)`C. `10^(24)`D. `10^(32)` |
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Answer» Correct Answer - C `r=r_(0)n^(2)53 Å` N is one line `= (1m)/(2xx53xx10^(-10))~~10^(8)` |
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| 43. |
Write a letter to your friend who is in another town, inviting him to spend the holidays with you and your family. |
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Answer» 15,Pinto Park, Sadar Bazar, Gwalior (M.P.) 25th Nov., 2012 Dear Jai, I hope you have not forgotten your promise to come and spend a few days of your holidays with me. The winter holidays start from 22nd of Dec. to 31st Dec. By the time this letter reaches you, your examination will be over. How have you done in the examination? I hop you will obtain good marks. It will be really such a great pleasure to me and my family if you come over here and stay with us during the holidays. I assure you that you will certainly enjoy not only your visit but the beauty of this place. There are many historical monuments, palaces and gardens. Gwalior Fort, Jai Vilash Palace, Zoo, Museums, Sun Temple and a number of others are very famous places. They all are worth seeing. Besides, there are many places of outing. We will also visit beauty of mountains, temples, springs, etc. I have a number of friends here. They are sure to amuse us. How anxiously I am expecting a reply to this letter! Please let me know when you will come here. Please convey my regards to your parents. Rest is o. k. Sincerely Yours Bharat Kumar |
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| 44. |
Write a letter to your friend Rajan inviting him to spend the summer vacation at your place in Ahmedabad. You are Soham / Sarala. Do not exceed 120-150 words. |
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Answer» 129, Navyug Apartments Pitampura Ahmedabad March 01, 2010 Dear Rajan It’s been a while since I’ve heard from you. Where have you been? I hope this letter finds you in the best of your health. As summers are approaching, I was thinking if we could spend the summer break together at my place in Mumbai. I will introduce you to all my friends and close relatives. I will give you a city tour as well. We will spend some quality time in the afternoons near the sea shore. To add cherry on the cake, the weather here is very pleasant during those days due to sea winds. I am excited even at the thought of you and I spending the summer together after so long. I have to tell you a lot of things and expect the same from you. Give my regards to aunty and uncle! Hope to see you soon. Yours lovingly Soham / Sarala |
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| 45. |
Write a letter inviting your friend to spend summer holidays with you. |
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Answer» 28-09-2019 Bangalore Dear Anuj, I hope you have done well in your final exams. I have also put a lot of effort to get a rank in the exams. If you are not going any where else to spend your holidays, I want you to join me and family here at Bangalore. More over, my cousins are also coming. We will all enjoy playing Cricket and going to swimming pool. I will ask my father to take us to visit some historical places in and around Bangalore. We can also watch some good movies. Let us have a great time. Please come over. I will also ask my father to write to your father to permit you to come to Bangalore. I will be eagerly awaiting your reply. Yours loving friend, XXX To, Anuj / Sb Sri Ranga, No. 17/9, II Main Road, 3rd Stage, MEL Colony Gunj Road. Manga lore |
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| 46. |
The sum of the series `1+ (1)/(4) + (1)/(16) + (1)/(64) + ...........oo` isA. `(8)/(7)`B. `(6)/(5)`C. `(5)/(4)`D. `(4)/(3)` |
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Answer» Correct Answer - 4 `S_(oo) = 1+ (1)/(4) + (1)/(16) + (1)/(64) + ..............= (1)/(1- 1//4)= (4)/(3)` |
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| 47. |
The percentage errors in the measurement of mass and speed are `2% and 3%` , respectively . How much will be the maximum error in the estimation of `KE` obtained by measuring mass and speed?A. `11%`B. `8%`C. `5%`D. `1%` |
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Answer» Correct Answer - B `E=(1)/(2)mv^2` `%` Error in K.E. `=%` error in mass`+2xx%` error in velocity `=2+2xx3=8%` |
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| 48. |
A person in an elevator accelerating upwards with an acceleration of `4 ms^(-2)` tosses a coin vertically. Upwards with a speed of `21 ms^(-1)` with respect to himself. The time at which coin will back into his hand is `(g = 10 ms^(-2))` |
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Answer» Correct Answer - 3 A person …………… `t = (2v)/(g+a) = ((2xx21))/((10+4)) = (42)/(14) = 3 sec` . |
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| 49. |
Two objects of equal volume `V=1 m^(3)` and different densities `d_(1) =500kg//m^(3)` and `d_(2) =1000kg // m^(3)` are gluod to each other so that their contact surface is flat and has an area `A=0.1 m^(2)` When the objects are submerged in a certain liquid they float in stable equilibrium the contact surface being parallel to the surface of the liquid (see the diagram) How deep (H in meters) can the contact surface be in the liquid so that the objects are not torn apart? The maximum force that the glue can with stand is `F=250 N (Neglect atmospheric pressure) |
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Answer» Correct Answer - `3m` At the "critical" depth H, the equilibrium condition for the compound object suggests that the density of the liquid is `d=(d_(1)+d_(2))//2` For the top and the bottom object taken separately, the equilibrium conditions are, respectively `d_(1) Vg+F-(dVg-dgAH)=0` `d_(2)Vg-F-(dVg+dgAH)=0` (For each object, the ter, in parentheses indicates the "effective" buoyancy force) Combining the last two equations yields the answer `H=[(d_(2)-d_(1)) Vg-2F]//[(d_(1)+d_(2)) gA]` |
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| 50. |
A cylindrical massless cord of length `l=10cm` and radius `r=3cm` is slowly extracted from the neck of a bottom opened bottle. If the normal pressure between the botle and un-extracted part of the cork at any instant is constant and equal to `P=(10^(5))/(pi)N//m^(2)`, find the work done (in Joule) in extracting it completely. The coefficient of friction between the cork and bottle is `mu=0.3` |
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Answer» Correct Answer - 9 `W=int_(0)^(l) mu 2pi r x dx P` `=pimurPl^(2)` `=pi(0.3)(3xx10^(-2))(10^(5))/(pi)xx10^(-2)=9J` |
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