InterviewSolution
Saved Bookmarks
InterviewSolution
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.
| 751. |
Let R be a relation on the set of all natural numbers given by a R `hArr` a divides `b^(2)` .Which of the following properties does R satisfy? I . Reflexivity II. Symmetry III. TransitivityA. I onlyB. III onlyC. I and III onlyD. I and II only |
|
Answer» Correct Answer - A (I) This relation is reflexive relation because every natural no. divides square of itself a R a `hArr` a divides `a^(2)` (II) not symmetric eg. 5 R 10 `cancel implies 5` Divide 100 But 10 R 5 `cancel implies` 10 Divide 25 Not transitivity for example if 8 R 4 & 4 R 2 `cancelimplies` 8 R 2 only (I)Option |
|
| 752. |
Let a `gt` 0 be real number. Then the limit `lim_(xto2) (a^(x)+a^(3-x)-(a^(2)+a))/(a^(3-x)-a^(x//2))` is-A. 2 log aB. `-(4)/(3)` aC. `(a^(2)+a)/(2)`D. `(2)/(3)(1-a)` |
|
Answer» Correct Answer - D `underset(xto2)lim(a^(x)+a^(3-x)-(a^(2)+a))/(a^(3-x)-a^(x//2))((0)/(0))` Apply L hospital rule `underset(xto2)lim(a^(x)lna-a^(3-x)lna)/(-a^(3-x)lna-(1)/(2)a^(x//2lna))` `(a^(2)lna-alna)/(-alna-(a)/(2)lna)=(a^(2)-a)/(-(3a)/(2))=(2)/(3)(1-a)` |
|
| 753. |
The number of real solutions of the equation 2 sin 3x + sin 7x - 3 = 0 which lie in the interval `[-2pi , 2pi]` isA. 1B. 2C. 3D. 4 |
|
Answer» Correct Answer - B only possible when sin 3x = 1 & sin 7x = 1 sin 3x = 1 sin3x = sin(4n+ 1) `(pi)/(2) n in I` ` 3 x = (4n+1)(pi)/(2)implies x=(4n+1)(pi)/(6)` `sin 7x= sin(4m+1)(pi)/(2), m in I` `x=(4m+1)(pi)/(14)` for common solution `(4n+1)(pi)/(6)=(4m+1)(pi)/(14)` Solving these `1=3m-7n` First solution ios m=5,n=2 Second solution is m=12,n=5 So two solutions are possible |
|
| 754. |
Suppose p,q,r and real number such that `q=p(4-p),r=q(4-q),p=r(4-r)`. The maximum possible value of p+q+r is |
|
Answer» Correct Answer - C Add all these `p+q+r=(p^(2)+q^(2)+r^(2))/(3)` For maximum value p=3, q=3, r=3 Answer is 9. |
|
| 755. |
Consider the set `A_(n)` of point (x,y) such that `0 le x le n, 0 le y le n` where n,x,y are integers. Let `S_(n)` be the set of all lines passing through at least two distinct points from `A_(n)`. Suppose we choose a line l at random from `S_(n)`. Let `P_(n)` be the probability that l is tangent to the circle `x^(2)+y^(2)=n^(2)(1+(1-(1)/(sqrt(n)))^(2))`. Then the limit `lim_(n to oo)P_(n)` is |
|
Answer» Correct Answer - A Equation of line passing through `(x_(1),y_(1)) " and " (x_(2),y_(2))` is `(y-y_(1))/(x-x_(1))=(y_(2)-y_(1))/(x_(2)-x_(1))` `implies (x_(2)= - x_(1))y +(y_(1)-y_(2))x+y_(1)(x_(1)-x_(2))+x_(1)(y_(2)-y_(1))=0` `implies ax+by+c=0 " where a, b, c " inI` `a=x_(2)-x_(1),b=y_(1)-y_(2),c=y_(1)(x_(1)-x_(2))+x_(1)(y_(2)-y_(1))` square of distance of (0,0) from `((c)/(sqrt(a^(2)+b^(2))))^(2)=(c6(2))/(a^(2)+b^(2))=` rational Case -1: if n is not perfect square And square of radius `=n^(2)(1+(1-(1)/(sqrt(n)))^(2))`= irrational `implies r^(2) ne(c^(2))/(a^(2)+b^(2))s` `implies ax +by+x=0` never be tangent to given circle `implies lim_(n to oo) P_(n)=0` Case - 2: if n is perfect square In this case number of tangents passing through two points from given set are few, but total number of lines are in much quantity when n approaches to infinite . `implies lim_(n to oo) P_(n)=0` |
|
| 756. |
The number of solution x of the equation `sin (x+x^(2))- sin(x^(2))= sin x` in the interval [2,3] is |
|
Answer» Correct Answer - C `2 cos(x+(2x^(2))/(2)).sin ""(x)/(2)= 2 sin ""(x)/(2). cos ""(x)/(2)` `implies sin ""(x)/(2)[cos((x+2x^(2))/(2))= cos "" (x)/(2)]=0` `sin ""(x)/(2)=0 " " or " "2 sin((2x+2x^(2))/(4)).sin((2x^(2))/(4))=0` `(x)/(2)=0,pi,2pi " "or " "sin((x+x^(2))/(2))=0 " "or" " sin ""(x^(2))/(2)=0` `X=0,2pi,4pi " " or " " (x+x^(2))/(2)=0,pi,2pi` `(x+x^(2))/(2)=pi " " x^(2)=2pi` `implies x^(2)+x-2pi=0" " impliesx=sqrt(2pi)` `x=(-1pmsqrt(1+8pi))/(2)" "2 gtsqrt(2pi) lt3` `sqrt(1+8pi)~~sqrt(25.14)` `" "~~5.2` `:. x=(5.2-1)/(2)=(4.2)/(2)=2.1` `:.` total number of solution lies between (2,3)=2 |
|
| 757. |
The number of different possible values for the sum x+y+z, where x,y,z are real numbers such that `x^(4)+4y^(4)+16z^(4)+64=32 xyz` is(A) 1 (B) 2 (C) 4 (D) 8A. 1B. 2C. 4D. 8 |
|
Answer» Correct Answer - C Applying Am`ge`gm. `(x^(4)+4y^(4)+16z^(4)+64)/(4) ge(4^(6)x^(4)y^(4)z^(4))^(1//4)` `x^(4)+4y^(4)+16z^(4)+64ge32|xyz|` So equal when each term is equal `:. X^(4)+4y^(4)=16z^(4)=64` `impliesx=pm2sqrt(2)` `y=pm2` `x= pmsqrt(2)` For x.y.z For `X^(4)+4y^(4)+16z^(4)_64=32xyz` Either each of x,y,z is (+)ve`to` 1 case/ Or two of x,y,z are (-)ve `to ` 3 cases `:.` 4 cases of different (x,y,z) triplets `:.` 4 possible x+y+z values (as `x ne y ne z`) |
|
| 758. |
Let m (respectively, n) be the number of 5-digit integers obtained by using the digits 1,2,3,4 ,5 with repetitions (respectively, without repetitions) such that the sum of any two adjacent digits is odd. Then `m/n` is equal toA. 9B. 12C. 15D. 18 |
|
Answer» Correct Answer - C Digits are 1,2,3,4,5 `{:(,"Even digits = 2,4",,,"number of Even digits = 2"),(,"Totalmarks = 2375",,,"number of Odd digits = 3"):}` Sum of any 2 adhacent digits is odd `rArr` Odd and even digits will alternate Case I For n, Repetition is not allowed `rArrOEOEO` is the only possibility of arrangement of digits, Where O=Odd digit, Where O = Odd digit, E = Even digit. So number of Arrangements `n=3/Oxx2/Exx2/Oxx1/Exx1/O=12` Case II For m, Repetition is allowed ltbrlt `rArr` Tow possioilities (a) OEOEO Number of such arrangements `=3/Oxx2/Exxx3/Oxx2/Exx3/O=108` (b) EOEOE Number of such arrangements `=2/Exx3/Oxx2/Exx3/Oxx2/E=72` So m = 108 + 72 = 180 `m/n=180/12=15` |
|
| 759. |
If parents have free ear lobes and the offspring has attached ear lobes, then the parents must be (A) homozygous (B) heterozygous (C) co-dominant (D) nullizygous |
|
Answer» The Correct option is (B) heterozygous |
|
| 760. |
Suppose n is a natural number such that `|i + 2i^2 + 3i^3 +...... + ni^n|=18sqrt2` where `i` is the square root of `-1`. Then n isA. 9B. 18C. 36D. 72 |
|
Answer» Correct Answer - C `{:(S=i+2i^(2)+3i^(3)+....+ni^(n)),(ul(iS=" "i^(2)+2i^(3)+....+(n-1)i^(n)+ni^(n+1)" ")),(S(1-i)=i+i^(2)+i^(3)+....+i^(n)-ni^(n+1)),(S(1-i)=(i(1-i^(n)))/(1-i)-ni^(n+1)rArr S=(1-i^(n))/(-2i)-(ni^(n+i))/(1-i)),(" "darr" " darr),(" "Z_(1)"(say) "z_(2) "(say)"):}` `|z_(1)|=1/sqrt(2)` or `0 |z_(2)|=n/sqrt(2)=n/2 sqrt(2)` `n/2=18 rArr n=36` |
|
| 761. |
Ribonucleic Acids (RNA) that catalyze enzymatic reations are called ribozymes. Which one of the following acts as a ribozyme ?A. RibosomeB. AmylaseC. tRNAD. Riboflavin |
| Answer» Correct Answer - A | |
| 762. |
A mixture of toluene and benzene boils at `100^(@)C`. Assuming ideal behaviour, the mole fraction of toluene in the mixture is closest to [Vapour presure of pure toluene and pure benzene at `100^(@)C` are 0.742 and 1.800 bar respectively. 1 atm = 1.013 bar]A. 0.824B. 0.744C. 0.544D. 0.624 |
|
Answer» Correct Answer - B `1.013=0.742X_(t)+1.8 X_(b)` `1.013=0.742(1-X_(b))+1.8 X_(b)` `X_(b)=0.256` `:. X_(t)=1-X_(b)=0.744` |
|
| 763. |
Stoke’s law states that the viscous drag force F experienced by a sphere of radius a, moving with a speed V through a fluid with coefficient of viscosity `eta`, is given by `F = 6pi"na"v`. If this fluid is flowing through a cylindrical pipe of radius r, length l and a pressure difference of P across its two ends, then the volume of water V which flows through the pipe in time t can be written as c`(V)/(t)=k((P)/(l))eta^(b)r^(c)` , where k is a dimensional constant. Correct values of a, b and c areA. a = 1, b = –1, c = 4B. a = `–1`, b = 1, c = 4C. a = 2, b =` –1`, c = 3D. a = 1, b =` –2, c = –4` |
|
Answer» Correct Answer - A |
|
| 764. |
The monarch butterfly avoids predators such as birds byA. Changing color frequentlyB. Flying away from the predator swiftlyC. Producing a chemical obnoxious to the predatorD. Producing ultrasonic waves |
|
Answer» Correct Answer - C Pray may have some defence mechanism to protect itself from predator like producing toxic substance. |
|
| 765. |
An electron in an electron microscope with initial velocity `v_(0)hati` enters a region of a stray transverse electric field` E_(0)hatj` . The time taken for the change in its de-Broglie wavelength from the initial value of `lambda` to `lambda//3` is proportional toA. `E_(0)`B. `(1)/(E_(0))`C. `(1)/(sqrt(E_(0)))`D. `sqrt(E_(0))` |
|
Answer» Correct Answer - B |
|
| 766. |
A bird sitting on a single high tension wire does not get electrocuted becauseA. the circuit is not complete.B. the bird feet has an insulating covering.C. capacitance of the bird is too small and the line frequency is too small.D. resistance of the bird is too high |
|
Answer» Correct Answer - C |
|
| 767. |
A transverse wave of frequency 500 Hz and speed 100 m/s is traveling in the positive x direction on a long string. At time t = 0 s the displacements at `x = 0.0` m and at `x = 0.25` m are `0.0` m and `0.02` m, respectively. The displacement at `x = 0.2` m at `t = 5xx10^(-4) s` isA. `-0.04` mB. `-0.02` mC. `0.04` mD. `0.02` m |
|
Answer» Correct Answer - D |
|
| 768. |
A body moves in a circular orbit of radius R under the action of a central force. Potential due to the central force is given by V(r) = kr (k is a positive constant). Period of revolution of the body is proportional to-A. `R^(1//2)`B. `R^(-1//2)`C. `R^(3//2)`D. `R^(-5//2)` |
|
Answer» Correct Answer - A |
|
| 769. |
Suppose two perpendicular tangents can be drawn from the origin to the circle `x^(2)+y^(2)-6x-2py+17=0`, for some real p. then `|p|`= |
|
Answer» Correct Answer - C `(x-3)^(2)+(y-p)^(2)=9-17+p^(2)` Director circle is `(x-3)^(2)+(y-p)^(2)=2(p^(2)-8)` Presses through (0, 0) `9+p^(2)=2p^(2)-16` `p^(2)=25rArrp=pm5rArr|p|=5` |
|
| 770. |
The maxinum value M of `3^(x)+5^(x)-9^(x)+15^(x)-25^(x)`, as x varies over reals, satisfies-A. 3 lt M lt 5B. 0 lt M lt 2C. 9 lt M lt 25D. 5 lt M lt 9 |
|
Answer» Correct Answer - A::B `Mlea+b-ab` `Mlt1-(a-1)(b-1)" min is zero"` hence `0ltMlt2` |
|
| 771. |
The equation `sqrt(x+3-4sqrt(x-1))+sqrt(x+8-6sqrt(x-1))=` hsA. No solutionB. Exactly two distinct solutionsC. Exactly four distinct solutionsD. Infinitely may solutions |
|
Answer» Correct Answer - D `sqrt(x+3-4sqrt(x-1))+sqrt(x+8-6sqrt(x-1))=1,Xge1` `sqrt((x+1)-2xx2sqrt(x-1)+4)+sqrt((x-1)-6sqrt(x-1)+9)=1` `|sqrt(x-1)-2|+|sqrt(x-1)-3|=1` Case - 1 `sqrt(x-1)-2+sqrt(x-1)-3=1" "xge10` `2sqrt(x-1)=6` x = 10 Case - II `sqrt(x-1)-2-sqrt(x-1)+3=1" "5lexle10` Case - III `-sqrt(x-1)+2-sqrt(x-1)+3=1" "1lexle5` `2sqrt(x-1)=4` x = 5 |
|
| 772. |
Three children, each accompanied by a guardian, seek admission in a school. The principal want to interview all the 6 persons one after the other subject to the condition that no child is interviewed before its guradian. In how many ways can this be done-A. 60B. 90C. 120D. 180 |
|
Answer» Correct Answer - B `(6!)/((2!)^(3))=90` |
|
| 773. |
If the speed (v) of the bob in a simple pendulum is plotted against the tangential acceleration (a), the correct graph will be represented by A. IB. IIC. IIID. IV |
|
Answer» Correct Answer - A |
|
| 774. |
Given the fact that histone birds DNA, it should be rich inA. arginine, lysineB. cysteine, methionineC. glutamate, aspartateD. isloeucine, leucine |
|
Answer» Correct Answer - A Histone is basic protein and it is rich in lysine and arginine |
|
| 775. |
A horse has 64 chromosomes and a donkey has 62. Mules reuslt from crossing a horse and a donkey. State which of the following is INCORRECT ?A. Mules can have either 64, 63 or 62 chromosomesB. Mules are infertileC. Mules have well defined gender (male/female)D. Mules have 63 chromosomes |
| Answer» Correct Answer - A | |
| 776. |
The reducing ability of the metals K, Au, Zn and Pb follows the orderA. `KgtPbgtAugtZn`B. `PbgtKgtZngtAu`C. `ZngtAugtKgtPb`D. `KgtZngtPbgtAu` |
|
Answer» Correct Answer - D Theoritical |
|
| 777. |
The correct order of energy of 2s orbitals in H,Li,Na and K, isA. `KltNaltLiltH`B. `NaltLiltKltH`C. `NaltKltHltLi`D. `HltNaltLiltK` |
|
Answer» Correct Answer - A As the atomic number incrreases the of orbitals decteases. |
|
| 778. |
The correct order of energy of 2s orbitals in H, Li, Na and K, is (A) K < Na < Li < H (B) Na < Li < K < H (C) Na < K < H < Li (D) H < Na < Li < K |
|
Answer» Correct option (A) K < Na < Li < H Explanation: As we go down the group IA, there is increase in shell so size of atom increases and energy of 2s orbital decreases. |
|
| 779. |
What fraction of the assimilated energy is used in respiration by the herbivores ?A. `~ 10` percentB. `~60` percentC. `~30` percentD. `~80` percent |
| Answer» Correct Answer - C | |
| 780. |
The oxidation state of cobalt in the following molecule is – (A) 3 (B) 1 (C) 2 (D) 0 |
|
Answer» Correct :- (D) 0 |
|
| 781. |
The oxidation state of cobalt in the following molecule is - A. 3B. 1C. 2D. 0 |
| Answer» Correct Answer - D | |
| 782. |
Usain Bolt, an Olympic runner, at the end of a 100 meter sprint, will have more of which of the following in his muscles ? (A) ATP (B) Pyruvic acid (C) Lactic acid (D) Carbon dioxide |
|
Answer» Correct option (C) Lactic acid Explanation: A two meter sprint takes less than 10 sec. to complete. During this very short period the major driving forces are stored high energy phosphates and anaerobic glycol sis which produces lactic acid. |
|
| 783. |
The most stable conformation of 2, 3-dibromobutane is -A. B. C. D. |
| Answer» Correct Answer - C | |
| 784. |
Typical electronic energy gaps in molecules are about 1.0 eV. In terms of temperature, the gap is closed to – (A) 102 K (B) 104 K (C) 103 K (D) 105 K |
|
Answer» Correct :- 105 K Explanation :- 3/2 KT = 1.6×10-19 3/2 ×1.38×10-23 = 1.6 × 10-19 T=105K |
|
| 785. |
Using Bolt, an Olympic runner, at the end of a 100 meter sprint, will have more of which of the following in his muscles ?A. ATPB. Pyruvic acidC. Lactic acidD. Carbon dioxide |
|
Answer» Correct Answer - C A two meter sprint takes less than 10 sec to complete. During this very short period the major driving forces are stored high energy phosphates and anarerobic glycol sis which produces lactic acid. |
|
| 786. |
For a one-electron atom, the set of allowed quantum numbers is – (A) n = 1, l = 0, m1 = 0, ms = +½ (B) n = 1, l = 1, m1 = 0, ms = +½ (C) n = 1, l = 0, m1 = –1, ms = –½ (D) n = 1, l = 1, m1 = 1, ms = –½ |
|
Answer» Correct :- (A) n = 1, l = 0, m1 = 0, ms = +½ |
|
| 787. |
Desert temperature often varies between 0 to `50^(@)C`. The DNA polymerase isolated from a camel living in the desert will be able to synthesize DNA most efficiently at-A. `0^(@)C`B. `37^(@)C`C. `50^(@)C`D. `25^(@)C` |
|
Answer» Correct Answer - B The optimum temperature for DNA polymerase is 37 degree Celcius. |
|
| 788. |
Among the following, the species with the highest bond order is-A. `O_(2)`B. `F_(2)`C. `O_(2)^(+)`D. `F_(2)` |
|
Answer» Correct Answer - C (A) `O_(2)`,B.O=2 (B) `F_(2)`, B.O=1 (C) `O_(2)^(+)`B.O=2.5 (D) `F_(2)^(-)` B.O=0.5 |
|
| 789. |
The maximum number of genotypes of the pollens produced by a tall pea plant with round, yellow seeds of the genotype TtRrYY, if the three loci are unlinked, would be :A. 1B. 2C. 4D. 8 |
| Answer» Correct Answer - C | |
| 790. |
Methane is greenhouse gas because -A. it absorbs longer wavelengths of the electromagnetic spectrum while transmitting shorter wavelengths.B. it absorbs shorter wavelengths of the electromagnetic spectrum while transmitting longer wavelengthsC. it absorbs all wavelengths of the electromagnetic spectrumD. it transmits all wavelengths of the electromagnetic spectrum |
|
Answer» Correct Answer - A |
|
| 791. |
An electrode is placed in the axioplasm of a mammalian axon and another electrode is placed just outside the axon. The potential difference measured will be- (A) 0 (B) –70 mV (C) –70 µV(D) +70 µV |
|
Answer» Correct Option :- (B) –70 mV |
|
| 792. |
At which stage of Meiosis I does crossing over occur ? (A) lepoptene (B) zygotene (C) pachytene (D) diplotene |
|
Answer» Correct Option :- (C) pachytene |
|
| 793. |
In photosynthesis, oxygen is produced byA. photosystem I from carbon dioxide.B. photosystem II from carbon dioxideC. photosystem I from waterD. photosystem II from water |
|
Answer» Correct Answer - B phtosynthesis at Ps II in lumen of thylakoid |
|
| 794. |
The rigidity of cellulose is due toA. coiled structure of glucose polymerB. `beta(1 to 4)` glycosidic linkageC. hydrogen bonding with adjacent glucose polymerD. cross-linking between glucose and peptides |
|
Answer» Correct Answer - C Bundle of cellulose fibre and rigid to H-bonds. |
|
| 795. |
Proteins P,Q, and R are associated with intact organellar membrane in a cell. If the intact organellel is treated with a high ionic strength buffer, only protein R remained associated with the membrane fraction. Based on this, one could conclude thatA. P and Q are per peripheral membrane proteinB. R is a peripheral membrane proteinC. P and Q are integral membrane bound proteins.D. P is peripheral and Q is integral membrane protein. |
|
Answer» Correct Answer - A Peripheral proteins can be detached earily. |
|
| 796. |
Compared to the atmospheric air, the alveolar air hasA. more `pO_(2)` and less `pCO_(2)`B. less `pO_(2)` and `pCO_(2)`C. more `pO_(2)` and more `pCO_(2)`D. less `pO_(2)` and less `pCO_(2)` |
|
Answer» Correct Answer - B Pressure gradient is essental for movement of gases in and out of the body. |
|
| 797. |
Two genetic loci controlling two different traits are linked. During the inheritance of these traits, the Mendelian laws that would be affected is/areA. Law of dominance, law of segregation and law of independent assortmentB. Law of segregation and Law of independent assortmentC. Only law of independent assortmentD. Only Law of segregation |
|
Answer» Correct Answer - C Linkage in exception of law of independent assortment |
|
| 798. |
Antigen-antibody reactionA. always result in precipitation of the complexB. depend only on covalent interactions.C. are irreversibleD. depend on ionic and hydrophobic interactions. |
|
Answer» Correct Answer - D Antibody-antigen interacton is essentially Non-Covalent, Electrostatic interaction, Hydrogen Bonds,Vander walls forces and Hydrophobic interactions are all known to be involved depending on the interaction sites. |
|
| 799. |
For a particular gene that determines the coat color in a diploid organism, there are three different allesles that are codominant. How may different skin colors are possible in such an organism.A. 9B. 6C. 4D. 3 |
|
Answer» Correct Answer - B `(n(n+1))/(2)=(3(3+1))/(2)=6` |
|
| 800. |
The genetic distance between genes A and B is 10 cm. An organism with Ab combination of the alleles is crossed with the organism with aB combination of alleles. What will be the percentage of the gametes with AB allesl combinatory by an F1 inidividual ?A. 1B. 5C. 10D. 50 |
|
Answer» Correct Answer - B Recombinants formed = 10% (AB 5% & ab 5%) |
|