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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 651. |
A"V" shaped rigid body has two identical uniform arms. What must be the angle between the two arms so that when the body is hung from one end, the other arm is horizontal ?A. `cos^(-1)(1//3)`B. `cos^(-1)(1//2)`C. `cos^(-1)(1//4)`D. `cos^(-1)(1//6)` |
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Answer» Correct Answer - A |
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| 652. |
Physical processes are sometimes described visually by lines. Only the following can cross -A. Streamlines in fluid flowB. Lines of forces in electrostaticsC. Rays in geometrical opticsD. Lines of force in magnetism |
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Answer» Correct Answer - C |
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| 653. |
Uniform ring of radius R is moving on a horizontal surface with speed v and then climbs up a ramp of inclination `30^(@)` to a height h. There is no slipping in the entire motion. Then h isA. `v^(2)//2g`B. `v^(2)//g`C. `3v^(2)//2g`D. `2v^(2)//g` |
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Answer» Correct Answer - B |
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| 654. |
A gas at initial temperature T undergoes sudden expansion from volume V to 2V. Then -A. The process is adiabaticB. The process is isothermalC. The work done in this process is nRT `ln_(e)`(2) where n is the number of moles of the gas.D. The entropy in the process does not change |
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Answer» Correct Answer - A |
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| 655. |
The number of reak rootts of the polynomial equation `x^(4)-x^(2)+2x-1=0` is |
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Answer» Correct Answer - B `x^(4)-x^(2)+2x-1=0` `X^(2)-(xx-1)^(2)=0` `rArr(x^(2)+x-1)(x^(2)-x+1)=0` `x^(2)+x-1=0` has two real roots. |
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| 656. |
Let R be the region of the disc `x^(2)+y^(2) le 1` in the first quadrant. The the area of the largest possible circile contained in R isA. `pi(3-2sqrt(2))`B. `pi(4-3sqrt(2))`C. `(pi)/(6)`D. `pi(2sqrt(2)-2)` |
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Answer» Correct Answer - A Required equation of circle `(x-h)^(2)+(y-h)^(2)=h^(2)` Both circle touch internally `C_(1)C_(2)=|r_(1)-r_(2)|` `sqrt(h^(2)+h^(2))=|h-1|` Solve this `h=sqrt(2)-1` Area `pi(sqrt(2)-1)^(2)=pi(3-2sqrt(2))` |
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| 657. |
The number of continuouss function `f:[0,1] to R ` that satisfy |
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Answer» Correct Answer - B Given equation can be written as `-(1)/(3)=int_(0)^(1)((f(x))/(2)-x)^(2) dx - int_(0)^(1) x^(2) dx` `int_(0)^(1) x^(2)dx(1)/(3)=int_(0)^(1)((f(x))/(2)-x)^(2) dx` `0=int_(0)^(1)((f(x))/(2)-x)^(2) dx` So, no of continuous funtion is 1. |
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| 658. |
Define `g(x)=int_(-3)^(3)f(x-y)f(y)dy,` for all real x, where `f(t)={{:(1","0letle1),(0", elsewhere."):}` ThenA. g(x) is not continuous everywhereB. g(x) is continuous everywhere but differentiable nowhereC. g(x) is continuous everywhere and differentiable everywhere except at x=0,1D. g(x) is continuous everywhere and differentiable everywhere except at x=0,1,2 |
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Answer» Correct Answer - D Definition can be break as `g(x)=int_(0)^(1)f(x-y)dy` `x-y=t,-dy.dt` `g(x)=int_(x-1)^(x)f(t)dt` `g(x)={{:(0,xle0),(x,0lt x lt1),(2-x,1lexle2),(0,xgt2):}` Now, check yourself |
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| 659. |
The integer part of the number `sum_(k=0)^(44)(1)/(cos^(@)cos(k+1)^(@))` isA. 50B. 52C. 57D. 59 |
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Answer» Correct Answer - C `(1)/(cos0^(@) cos 1^(@))+(1)/(cos 1^(@)cos2^(@))+(1)/(cos2^(@) cos3^(@))+.......(1)/(cos 44^(@) cos 45^(@))` multiply & divided by `sin 1 ^(@)` `(1)/(sin1^(@))[(sin1^(@))/(cos0^(@) cos 1^(@))+(sin1^(@))/(cos 1^(@)cos2^(@))+.......(sin1^(@))/(cos 44^(@) cos 45^(@))]` `(1)/(sin1^(@))[(sin(1^(@)-0^(@)))/(cos0^(@) cos 1^(@))+(sin(1^(@)-0^(@)))/(cos 1^(@)cos2^(@))+.......sin(1^(@)-0^(@))/(cos 44^(@) cos 45^(@))]` `(1)/(sin 1^(@))[tan1^(@)-tan0^(@)+tan2^(@)-tan1^(@)+...tan45^(@)-tan44^(@)]` `=(1)/(sin 1^(@))[tan 45^(@)]` `(1)/(0.0174524)=57.2987` Integral part = 57 |
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| 660. |
Let `X_(k)` be real number such that `X_(k)gtk^(4)+k^(2)+1` for `1 le k le 2018`. Denot`N =sum_(k=1)^(2018)k`. Consider the following inequalities: I. `(sum_(k=1)^(2018)kx_(k))^(2)leN(sum_(k=1)^(2018)kx_(k)^(2)) " " II. (sum_(k=1)^(2018)kx_(k))^(2)leN(sum_(k=1)^(2018)k^(2)x_(k)^(2))`A. both I and II are trueB. I is true and II is falseC. I is false and II is trueD. both I and II are false |
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Answer» Correct Answer - A If `x_(1),x_(2),……………,x_(n)` be n numbers then using cauch-schurz theorem - `((x_(1)+x_(2)+.....x_(n))/(n))^(2) le ((x_(1)^(2)+x_(2)^(2)+.....x_(n)^(2))/(n))` Case (i) Consider `x_(1),x_(2),x_(2),x_(3),x_(3),x_(3)..............underset(2018 "times")(ubrace(x_(2018)+x_(2018)+......x_(2018)))` Now using Cauchy-schwarz for above number `((x_(1),x_(2),x_(2)+..............(x_(2018)+x_(2018)+......x_(2018))/(2018 " times"))/(1+2+3......2018))^(2)` `le(x_(1)^(2),x_(2)^(2),x_(2)^(2)+..............(x_(2018)^(2)+x_(2018)^(2)+......)/(2018 " times"))/(1+2+3......2018)` `implies((x_(1)+2x_(2)+3x_(3)+...2018x_(2018))/(sum_(k=1)^(2018)k))^(2) le ((x_(1)^(2)+2x_(2)^(2)+3x_(3)^(2)+...2018x_(2018)^(2))/(sum_(k=1)^(2018)k))^(2)` `(sum_(k=1)^(2018)kx_(k))^(2)lesum_(k=1)^(2018)(sum_(k=1)^(2018)kx_(k^(2)))` `(sum_(k=1)^(2018)kx_(k))^(2)leN(sum_(k=1)^(2018)kx_(k^(2))^(2))` Therefore statement 1 is true. Case (ii) Consider `x_(1),2x_(2),3x_(3)+....2018x_(2018)` Now apply chauchy-Schwarz for above number `((x_(1)+2x_(2)+3x_(3)+.....2018x_(2018))/(2018))^(2) le (x_(1)^(2)+2x_(2)^(2)+.....(2018x_(2018))^(2))/(2018)` `implies(x_(1)+2x_(2)+.....2018_(x2018))^(2) le2018 (x_(1)^(2)+4x_(2)^(2)+...(2018)^(2)x_(2018)^(2))` `implies (sum_(k=1)^(2018)kx_(k))^(2)le 2018(sum_(k=1)^(2018)k^(2)x_(k)^(2))` Since `n=sum_(k=1)^(2018)k=(2018xx2019)/(2)` `:.(sum_(k=1)^(2018)kx_(k))^(2)` is always less than or equal to 2018 `sum_(k=1)^(2018)k^(2)x_(k)^(2)` `:.` It will always be less than `n(sum_(k=1)^(2018)k^(2)x_(k)^(2))` |
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| 661. |
The value of `sum_(n=0)^(1947) 1/(2^(n)+sqrt(2^(1947)))` is equal toA. `487/(sqrt(2^(1945)))`B. `1946/(sqrt(2^(1947)))`C. `1947/(sqrt(2^(1947)))`D. `1948/(sqrt(2^(1947)))` |
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Answer» Correct Answer - A `underset(n=0)overset(1947)sum1/(2^(n)+sqrt(2^(1947)))" Total terms "=1948` `T_(1)=1/(1+sqrt(2^(1947)))` `T_(1948)=1/(2^(1947)+sqrt(2^(1947)))` `T_(1)+T_(1948)=1/(sqrt(2^(1947)))` Similarly, `T_(2)+T_(1947)=1/(sqrt(2^(1947)))=T_(3)+T_(1946)=" and so an".........` `" Total"1948/2=974" pairs"` `:." Sum"=974/sqrt(2^(1947))=974/(sqrt(4xx2^(1945)))=487/(sqrt(2^(1945)))` |
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| 662. |
In a cinema hall, the charge per person is RS. 200. On the first day, only 60% of the seats were filled. The owner decided to reduce the price by 20% abd there was an increasesof 50% in the number of spectators on the next day. The percentage increase in the revenue on the second day was-A. 50B. 40C. 30D. 20 |
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Answer» Correct Answer - D let total seats = 100 Recenue collected first days `=200xx60=12000Rs` Reenue collected seconed day `=90xx160=14400` `:.%" increase "=2400/12000xx100=20%` |
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| 663. |
Let a, b, c be non-zero real numbers such that `a+b+c=0," let "q=a^(2)+b^(2)+c^(2)" and " r=a^(4)+b^(4)+c^(4)`. Then-A. `q^(2)lt2r` alwaysB. `q^(2)=2r` alwaysC. `q^(2)gt2r` alwaysD. `q^(2)-2r` can take both positive and negative values |
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Answer» Correct Answer - B `a+b+c=0` `=a^(2)+b^(2)+c^(2),r=a^(4)+b^(4)+c^(4)` `q^(2)-2r=(a^(2)+b^(2)+c^(2))^(2)-2(a^(4)+b^(4)+c^(4))` `=2a^(2)b^(2)+2b^(2)c^(2)+2a^(2)c^(2)-a^(4)-b^(4)-c^(4)` `=2a^(2)c^(2)+2b^(2)c^(2)-(a^(2)-b^(2))^(2)-c^(4)` `=2c^(2)(a^(2)+b^(2))-c^(2)(a-b)^(2)-c^(4)` `=c^(2)[2a^(2)+2b^(2)-(a-b)^(2)-c^(2)]` `=c^(2)[2ab+a^(2)+b^(2)+c^(2)]` `=c^(2)[(a+b)^(2)-c^(2)]` `=0` |
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| 664. |
Suppose `a_(2),a_(3),a_(4),a_(5),a_(6),a_(7)` are integers such that `5/7=(a_(2))/(2!)+(a_(3))/(3!)+(a_(4))/(4!)+(a_(5))/(5!)+(a_(6))/(6!)+(a_(7))/(7!),` where `0lea_(j)ltj"for "j=2,3,4,5,6,7.` The sum `a_(2)+a_(3)+a_(4)+a_(5)+a_(6)+a_(7)` is-A. 8B. 9C. 10D. 11 |
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Answer» Correct Answer - B `5xx6!=2520a_(2)+840a_(3)+210a_(4)+42a_(5)+7a_(6)+a_(7)` `:. 3600=2520a_(2)+840a_(3)+210a_(4)+42a_(5)+7a_(6)+a_(7)` Now`a_(j)inI{"from "2" to "7}` soabove equation is true if `a_(2)=a_(3)=a_(4)=1` `a_(5)=0` `a_(6)=4,a_(7)=2` |
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| 665. |
The value of `int_(-2012)^(2012)(sin(x^(3))+x^(5)+1)"dx"` is -A. 2012B. 2013C. 0D. 4024 |
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Answer» Correct Answer - D `int_(-2012)^(2012)(sin(x^(3))+x^(5)+1)"dx"=int_(-2012)^(2012)sin(x^(3))"dx"+int_(-2012)^(2012)x^(5)dx+int_(-2012)^(2012)"dx = 4024"` |
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| 666. |
Suppose `a_(1),a_(2),a_(3),……,a_(2012)` are integers arranged on a circle. Each number is equal to the average of its two adjacent numbers. If the sum of all even indexed numbers is 3018, what is the sum of all numbers ? |
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Answer» Correct Answer - D `a_(1),a_(2),a_(3)……,a_(2012)=3018…………….(1)` `(a_(1)+a_(3))/(2)=a_(2)` `2a_(2)+2a_(4)+2a_(6)+.....+2a_(2012)=6036` `(a_(1)+a_(3))+(a_(3)+a_(5))+(a_(5)+a_(7))+....+(a_(2011)+a_(1))=6036` `2(a_(1)+a_(3)+a_(5)+.....+a_(2011))=6036` `a_(1)+a_(3)+a_(5)+.....+a_(2011)=3018...............(2)` Add (1) and (2) Sum of all number `=3018+3018=6036` |
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| 667. |
Let `S_(n)=sum_(k=1)^(n)k` denote the sum of the first n positive integers. The numbers `S_(1),S_(2),S_(3),……S_(99)` are written on 99 cards. The probability of drawing a card with an even number written on it is -A. `1//2`B. `49//100`C. `49//99`D. `48//99` |
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Answer» Correct Answer - C 1,3,6,10,15,21,28,36,45,55,66,78,91,105 till 98 terms 48 terms are even and 48 terms odd `99^(th)` term `=(99xx100)/(2)`=even Total even terms = `48+1=49` Probability `=(49)/(99)` |
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| 668. |
Two rods, one made of copper and the other steel of the same length and cross sectional area are joined together. (The thermal conductivity of copper is `385 J.s^(-1) .m^(-1). K^(-1)` and steel is `50 J.s^(-1).m-^(1).K^(-1)`.) If the copper end is held at `100^(@)C` and the steel end is held at `0^(@)C`, what is the junction temperature (assuming no other heat losses) ?A. 0.12B. 0.5C. 0.73D. 0.88 |
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Answer» Correct Answer - D |
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| 669. |
The capacitor of capacitance C in the circuit shown in fully charged initially. Resistance is R.– After the switch S is closed, the time taken to reduce the stored energy in the capacitor to half its initial value isA. `RC//2`B. `2RC` in 2C. RC ln 2D. `(RC ln 2)/(2)` |
| Answer» Correct Answer - D | |
| 670. |
An ideal monatomic gas expands to twice its volume. If the process is isothermal, the magnitude of work done by the gas is `W_(1)`. If the process is adiabatic, the magnitude of work done by the gas is `W_(a)` . Which of the following is true ?A. `W_(i) = W_(a) gt 0`B. `W_(i) gt W_(a) = 0`C. `W_(i) gt W_(a) gt 0`D. `W_(i) gt W_(a) gt 0` |
| Answer» Correct Answer - B | |
| 671. |
The number of air molecules in a `(5m xx 5m xx 4m)` room at standard temperature and pressure is of the order ofA. `6 xx 10^(23 )`B. `3 xx 10^(24 )`C. `3 xx 10^(27)`D. `6 xx 10^(30 )` |
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Answer» Correct Answer - C |
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| 672. |
The mid-point of the domain of the function `f(x)=sqrt(4-sqrt(2x+5))` for real x is -A. `1//4`B. `3//2`C. `2//3`D. `-2//5` |
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Answer» Correct Answer - B `f(x)=sqrt(4-sqrt(2x+5))` `4-sqrt(2x+5)ge0" "2x+5ge0` `sqrt(2x+5)le4" "xge-5//2` `xle(11)/(2)` `x in[-(5)/(2),(11)/(2)]` mid point `=(-5//2+11//2)/(2)=(3)/(2)` |
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| 673. |
The value `(int_(0)^(pi//2) (sinx)^(sqrt2+1)dx)/(int_(0)^(pi//2)(sinx)^(sqrt2-1)dx)` is -A. `(sqrt2+1)/(sqrt2-1)`B. `(sqrt2-1)/(sqrt2+1)`C. `(sqrt2+1)/(sqrt2)`D. `s-sqrt2` |
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Answer» Correct Answer - D `I_(1)-overset(pi//2)underset(0)int(sin x)^(sqrt2).sin"xdx "I_(2)=overset(pi//2)underset(0)int(sin x)^(sqrt2-1)"dx"` `I_(1)=((sinx)^(sqrt2)intsin"x dx")_(0)^(pi//2)-overset(pi//2)underset(0)int(sqrt2(sinx)^(sqrt2-1)cosx intsin"x dx")` `=-(cos x(sinx)^(sqrt2))_(0)^(pi//2)+sqrt2overset(pi//2)underset(0)int(sinx)^(sqrt2-1)(1-sin^(2)x)dx` `(I_(1))/(I_(2))=(sqrt2)/(1+sqrt2)xx((sqrt2-1))/((sqrt2-1))=2-sqrt2` |
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| 674. |
Let `f:RrarrR` be the function `f(x)=(x-a_(1))(x-a_(2))+(x-a_(2))(x-a_(3))+(x-a_(3))(x-a_(1))` with `a_(1),a_(2),a_(3) in R`. The fix `f(x)ge0` if and only if -A. At least two of `a_(1),a_(2),a_(3)` are equalB. `a_(1)=a_(2)=a_(3)`C. `a_(1),a_(2),a_(3)` are all distinctD. `a_(1), a_(2),a_(3)`, are all positive and distinct |
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Answer» Correct Answer - B Only when `a_(1)=a_(2)=a_(3)` In other cases f(x) will take both positive and negative values |
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| 675. |
Consider the following statements : (I)n All isotopes of an elements have the same number of neutrons. (II) only one isotope of an element can be stable and non-radioactive. (III) All elements have isotopes (IV) All isotopes of Carbon can form chemical compounds with Oxygen - 16A. III and IV onlyB. II ,III and IV onlyC. I, II and III onlyD. I,III and IV only |
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Answer» Correct Answer - A All elements have isotopes. All isotopes of carbon can form chemical compounds with oxygen - 16. |
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| 676. |
The Arrhenius plots of two reactions, I and II are shown graphically- The graph suggest that-A. `E_(I) gt E_(II) and A_(I)gtA_(II)`B. `E_(II) gt E_(I) and A_(II)gtA_(I)`C. `E_(I) gt E_(II) and A_(II)gtA_(I)`D. `E_(II) gt E_(1) and A_(I)gtA_(II)` |
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Answer» Correct Answer - A For plot between In k v/s I/T y-intrercept in In A & slope is `(-E_(a))/(R)` therefore, `E_(II) lt E_(1) and A_(I) gt A_(II)` |
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| 677. |
The element that combines with oxygen to give an emphoteric oxide is-A. NbB. PC. AlD. Na |
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Answer» Correct Answer - C Aluminium form aphoteric oxide with oxygen `(Al_(2)O_(3))` |
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| 678. |
At which phase of the cell cycle, DNA polymerase activity is an its highest?A. Gap 1 (G1)B. Mitotic (M)C. Synthetic (S)D. Gap 2 (G2) |
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Answer» Correct Answer - C S-phase or synthetic phase is significant due to DNA synthesis.is aided by the enzyme DNA polymerases. |
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| 679. |
At which phase of the cell cycle, DNA polymerase activity is at its highest ? (A) Gap 1 (G1) (B) Mitotic (M) (C) Synthetic (S) (D) Gap 2 (G2) |
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Answer» Correct option (C) Synthetic (S) Explanation: S-phase or synthetic phase is significant due to DNA synthesis. DNA synthesis is aided by the enzyme DNA polymerases. |
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| 680. |
When heated in air, brown copper powder turns black. This black powder would turn brown again when heated withA. COB. `O_(2)`C. `H_(2)`D. `NH_(3)` |
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Answer» Correct Answer - C `H_(2)` is reducing agent |
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| 681. |
For a 4p orbital, the number of radial and angular nodes, respectively , areA. 3,2B. 1,2C. 2,4D. 2,1 |
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Answer» Correct Answer - D no. of radial node =n-l-1=4-1-1=2 no. of angular node =l=1 |
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| 682. |
Two different liquids of same mass are kept in two identical vessels, which are placed in a freezer that extracts heat from them at the same rate causing each liquid to transform into a solid. The schematic figure below shows the temperature T vs time t plot for the two materials. We denote the specific heat of metrials in the liquid (solid) states to be CL1 (CS1) and CL2 (CS2) respectively.(A) CL1 < CL2 and CS1 < CS2 (B) CL1> CL2 and CS1 > CS2 (C) CL1 > CL2 and CS1 > CS2 (D) CL1 < CL2 and CS1 > CS2 |
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Answer» Correct Option :- (B) CL1> CL2 and CS1 > CS2 Explanation :- Let Refrigerater extract Q joul/per second Q.t ⇒ ms (Tf – T) Higher the specific heat, Higher the slope |
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| 683. |
Two blocks (1 and 2) of equal mass m are connected by an ideal string (see figure shown) over a frictionless pulley. The blocks are attached to the ground by springs having spring constants `k_(1)` and `k_(2)` such that `k_(1) gt k_(2)` Initially, both springs are unstretched. The block 1 is slowly pulled down a distance x and released. Just after the release the possible values of the magnitude of the acceleration of the blocks `a_(1)` and `a_(2)` can be–A. Either `(a_(1) = a_(2)=((k_(1)+k_(2))x)/(2m))` or `(a_(1)(k_(1)x)/(m)-g and a_(2) = (k_(2)x)/(m)+g)`B. `(a_(1) = a_(2)=((k_(1)+k_(2))x)/(2m))` onlyC. `(a_(1) = a_(2)=((k_(1)-k_(2))x)/(2m))` onlyD. `(a_(1) = a_(2)=((k_(1)-k_(2))x)/(2m))` or `(a_(1) = a_(2)=((k_(1)k_(2))x)/((k_(1)+k_(2))m)-g)` |
| Answer» Correct Answer - B | |
| 684. |
The square root of `(0.75)^(3)/(1-(0.75))+(0.75+(0.75)^(2)+1)` is-A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B say 0.75 is x then `(x^(3))/(1-x)+(x^(2)+x+1)=P` `(x^(3)-(x-1)(x^(2)+x+1))/(1-x)=P` `(""x^(3)-x^(3)+1")")/(1-x)=P` `impliesP=(1)/(1-x)` `impliesP=(1)/(1-(3)/(4))=(4)/(1)` `sqrtP=2` |
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| 685. |
Let `f(x)` be a quadratic polynomial with `f(2)=-2`. Then the coefficient of `x` in `f(x)` is-A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C say `f(x)=ax^(2)+bx+c` `f(x)=ax^(2)+bx+c` `f(2)=4a+2b+c=10` `f(-2)=4a-2b+c=-2` `implies4b=12` b=3 |
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| 686. |
Two friends A and B are 30 km apart and they start simultaneously on motorcycles to meet each other. The speed of A is 3 times that of B. The distance between them decreases at the rate of 2 km per minute. Ten minutes after they start, A's vehicle breaks down and A stops and waits for B to arrive. After how much time (in minutes) A started riding, does B meet A? (A) 15 (B) 20 (C) 25 (D) 30 |
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Answer» Correct option (D) 30 Explanation: Speed of B = V km/hr Speed of A = 3V km/hr Given 4V = 2 × 60 km/hr ⇒ V = 30 km/hr Distance covered by then after 10 min = 2 × 10 = 20 km so remaining distance = (30 – 20) km = 10 km Time taken by B to cover 10 km = 10/30/60 = 20 min. Total time = 20 +10= 30min |
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| 687. |
Consider the following two statement: I. Any pair of consistent linear equations in two variables must have unique solutions. II. There do not exist two consecutive integers, the sum of whose squares is 365. ThenA. Both I and II are trueB. both I and II are falseC. I is true and II falseD. I is false and II is true. |
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Answer» Correct Answer - B Clearly statement 1 is false as they can have infinite solutions statement 2 is also false as `13^(2)+14^(2)=365` |
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| 688. |
find the sum of all three digit natural numbers which are divisible by `7`A. 5497B. 5498.5C. 5499.5D. 5490 |
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Answer» Correct Answer - B Four digit numbers which are divisible by 7 are 1001, 1008, 1015,……,9996 Hence total number of such numbers =1286 `rArr" Medium"=((N/2)^(th)" value"+(N/2+1)^(th)" value")/2=((643)^(th)"value"+(644)^(th)" value")/2` `=1001+(1285xx7)/2` `=1001+4497.5` `=5498.5` |
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| 689. |
The number of polynomials p(x) with integer coefficients such that the curve y = p(x) passes through (2, 2) and (4, 5) is |
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Answer» Correct Answer - A `y=P(X)=a_(0)+a_(1)xxa_(2)X^(2)+……+a_(n)X^(n)" "a_(0),a_(1),a_(2),a_(3),…………a_(n)inl` `2=P(2)" …(1)"` `5=P(4)" …(2)"` By (1) and (2) `rArr3=a_(1)(4-2)+a_(2)(4^(2)-2^(2))+a_(3)(4^(3)-2^(3))+………….+a_(n)(4^(n)-2^(n))` Clearly RHS is even and LHS is odd no polynomial exists. |
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| 690. |
For how many different values of a does the following system have at least two distinct solution ? ax+y=0 x+(a+10)y=0 |
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Answer» Correct Answer - C `(a)/(1)=(1)/((a+10))` `a^(2)+10a-1=0` two value of a |
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| 691. |
Let R be the set of real number and `f:R to R` be defined by `f(x)=({x})/(1+[x]^(2))`, where [x] is the greatest integer less than or equal to x, and {x}=x-[x]. Which of the following statement are true? I. The range of f is a closed interval II. f is continuous on R. III. f is one - one on R.A. I onlyB. II onlyC. III onlyD. None of I, II and III |
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Answer» Correct Answer - D `f{x}=({x})/(1+[x]^(2))` `f(x)={{:((x+1)/(2):-1lex lt0),(x:0lexlt1),((x-1)/(2),1lexlt2),((x-2)/(2),2lexlt3),(" "underset("So-on")(vdots)):}` Now check accordingly |
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| 692. |
Let A denote the matrix `({:(0,i),(i,0):})`, where ` i^(2) = -1` , and let `I` denote the identity matrix `({:(1,0),(0,1):})`. Then `I + A + A^(2) + "….." + A^(2010)` is -A. `({:(0,0),(0,0):})`B. `({:(0,i),(i,0):})`C. `({:(1,i),(i,1):})`D. `({:(-1,0),(0,-1):})` |
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Answer» Correct Answer - C `A = [{:(0,i),(i,0):}] , A^(2) = [{:(-1,0),(0,-1):}], A^(3) = [{:(0,-i),(-i,0):}] , I = [{:(1,0),(0,1):}]` ` I + A + A^(2) + A^(3) = [{:(0,0),(0,0):}] , A^(4) = [{:(1,0),(0,1):}] = 1` `I + A + A^(2) + A^(3) + "......"A^(2010)` ` (I + A + A^(2) + A^(3)) + A^(4)(I + A + A^(2) + A^(3))+"......"+A^(2008)(I+A+A^(2))` `= ({:(1,i),(i,1):})` |
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| 693. |
Three taps A, B, C fill up a tank independently in 10 hr, 20 hr, 30 hr, respectively. Initially the tank is empty and exactly one pair of taps is open during each hour and every pair of taps is open at least for one hour. What is the minimum number of hours required to fill the tank?A. 8B. 9C. 10D. 11 |
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Answer» Correct Answer - A A 10 hr B 20 hr C 30 hr Exactly are pair of taps is open during each hour and every pair of taps is open at least for one hour. First A and B are open for 1 hour then B and C and then C and A `underset("first")((1/10+1/20))+underset("second")((1/20+1/30))+underset("third")((1/30+1/10))=22/60` In three hours the tank will be filled `(22/60)^(th)` part. Now, for minimum time the rest tank must be filled with A and B taps `(1/10+1/20=9/60)` So the rest `(38/60)^(th)` part of tank will take 5 hours more. So the tank will be filled in `8^(th)` hour. |
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| 694. |
Let `x_(n)=(2^(n)+3^(n))^(1//2n)` for all natural number n. ThenA. `lim_(n to oo)x_(n)=oo`B. `lim_(n to oo)x_(n)=sqrt(3)`C. `lim_(n to oo)x_(n)=sqrt(3)+sqrt(2)`D. `lim_(n to oo)x_(n)=sqrt(5)` |
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Answer» Correct Answer - B `lim_( nto oo)(3^(n))^(1//2n)(((2)/(3))^(n)+1)^(1//2n)` Put `lim_(n to oo)sqrt(3)` |
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| 695. |
Let a,b,c,d,e, be real numbers such that `a+b lt c+d , b+c lt d+e, c+d lt e+a, d+e lt a+b`. ThenA. The largest is a and the smallest is bB. The largest is a and the smallest is cC. The largest is c and the smallest is eD. The largest is c and the smallest is b |
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Answer» Correct Answer - A (i) `a+b lt c+d ` (ii) `b+c lt d+e` (iii) ` c+d lt e+a` (iv) ` d+e lt a+b` from (i) & (iii) `a+b lt e+a` `implies b lt e` from (ii) &(iv) `b+c lt a+b` `implies c lt a ` (i)-(ii) `a-c lt c-e` `implies c lt e ` (i)-(iv) `(a-e)+(b-d) lt (c-a)+(d-b)` from thus `d gt b` (i)+(iii)-(ii) `c gt d` overall a is greatest , b is least |
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| 696. |
The diameter of one of the bases of a truncated cone is 100 mm. If the diameter of this base is increased by `21%` such that it still remains a truncated cone with the height and the other base unchanged, the volume also increases by `21%`. The radius the other base (in mm) isA. 65B. 55C. 45D. 35 |
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Answer» Correct Answer - B Let initially 2 bases have radii 5 cm and r cm. Finally base have radii `(1.21 xx 5)` and r Ratios of volumes = `(V_(2))/(V_(1))=1.21` `V_(2)=(pih)/3[(6.05)^(2)+(6.05)r+r^(2)]` `V_(1)=(pih)/3[5^(2)+5r+r^(2)]` `V_(2)/V_(1)=1.21 rArr((6.05)^(2)+(6.05)r+r^(2))/(5^(2)+5r+r^(2))` `rArr r^(2)=(6.3525)/21 ` `rArr r=11/2 cm = 55 mm` |
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| 697. |
One of the solution of the equation ` 8 sin^(3) theta-7 sin theta+sqrt(3) cos theta=0` lies in the intervalA. `(0,10^(@)]`B. `(10^(@),20^(@)]`C. `(20^(@),30^(@)]`D. `(30^(@),40^(@)]` |
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Answer» Correct Answer - B `6 sin theta -2 sin 3theta -7 sin theta +sqrt(3) cos theta=0` `sqrt(3) cos theta- sin theta= 2 sin 3 theta` It can be written as `2( sin (60^(@)- theta))=2 sin 3 theta` ` sin (60^(@)-theta)= sin 3 theta` `60^(@)=4 theta` `implies theta = 15^(@)` is one of the value. |
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| 698. |
The expression `(2^(2)+1)/(2^(2)-1)+(3^(2)+1)/(3^(2)-1)+(4^(2)+1)/(4^(2)-1)+........+((2011)^(2)+1)/((2011)^(2)-1)` lies in the intervalA. `(2010, 2010 1/2)`B. `(2011-1/2011,2011-1/2012)`C. `(2011,2011 1/2)`D. `(2012,2012 1/2)` |
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Answer» Correct Answer - C `(2^(2)+1)/(2^(2)-1)+(3^(2)+1)/(3^(2)-1)+(4^(2)+1)/(4^(2)-1)+......+((2011)^(2)+1)/((2011)^(2)-1)``underset(r=2)overset(2011) sum(r^(2)+1)/(r^(2)-1)=underset(r=2)overset(2011)sum[1+2/((r+1)(r-1))]` = `underset(r=2)overset(2011) sum [1+1/(r+1)-1/(r+1)]` `=2010+[1-1/3+1/2-1/4+1/3-1/5+......+1/(2010)-1/(2012)]` =`2010+1+1/2-1/(2012)-1/(2011)` `= 2011+1/2 -[1/(2011)+1/(2012)]` lies between `(2011, 2011 1/2)` |
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| 699. |
A single bacterium is actively growing in a medium that supports its growth to a number of 100 million. Assuming the division time of the bacterium as 3 hours and the life span of non-dividing bacteria as 5 hours, which ONE of the following represents the maximum number of bacteria that would be present at the end of 15 hour ? (A) 10 (B) 64 (C) 24 (D) 32 |
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Answer» Correct Option :- (D) 32 Explanation :- Time =15/3 = 5 times division occur. No. of bacteria = 25 = 32 |
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| 700. |
In an in vitro tanslation experiment, poly (UC) RNA template produced poly (Ser-Leu), while poly (AG) RNA template produced poly (Arg-Glu) polypeptide. Which ONE of the following options represents correct interpretations of the codons assignments for Ser, Leu, Arg, and Glu.(A) Ser – UCU, Leu – CUC, Arg – AGA, Glu – GAG (B) Ser – CUC, Leu – GAG, Arg – UCU, Glu – AGA (C) Ser – AGA, Leu – UCU, Arg – GAG, Glu – CUC (D) Ser – GAG, Leu – AGA, Arg – CUC, Glu – UCU |
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Answer» Correct Option :- (A) Ser – UCU, Leu – CUC, Arg – AGA, Glu – GAG Explanation :- Sequence of 3 nitrogenous base is one codon. |
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