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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.
| 701. |
The bond order in ` O_(2^(2-))` is-A. 2B. 3C. 1.5D. 1 |
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Answer» Correct Answer - D Bond order of `O_(2^(2-))` Total electron `= 18` Configuration ` = KK sigma (2s)^(2) sigma^(**) (2s )^(2) sigma (2p_(z))^(2) pi(2p_(x))^(2) pi(2p_(y))^(2)pi^(**)(2p_(x))^(2)pi^(**)(2p_(y))^(2)` Bond order `= (N_(b) - N_(a))/(2) = (8-6)/(2) = 1.0` |
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| 702. |
Among the following, the `pi`-acid ligand is-A. `F`B. `NH_(3)`C. `CN^(-)`D. `I^(-)` |
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Answer» Correct Answer - C `CN^(-)` accept electrons from metal ion in vacant `pi^(**)ABMO`. |
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| 703. |
A 25,000 Da protein contains a single binding site for a molecule (ligand), whose molecular weight is 2,500 Da. Assuming high affinity and phsiologically irreversible blinding, the amount of the ligand required to occupy all the binding sites in 10 mg protein will beA. 0.1 mgB. 1 mgC. 10 mgD. 100 mg |
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Answer» Correct Answer - B `(25000)/(2500)=(10)/(x)(x=1mg)` |
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| 704. |
Among the following, the π-acid ligand is- (A) F (B) NH3 (C) CN– (D) I– |
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Answer» Correct Option :- (C) CN– Explanation :- CN– accept electrons from metal ion in its vacant π * ABMO. |
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| 705. |
Consider two sealed jars of equal volume. One contains 2 g of hydrogen at 200 K and the other contains 28 g of nitrogen at 400 K. The gases in the two jars will have - (A) the same pressure (B) the same average kinetic energy (C) the same number of molecules (D) the same average molecular speed |
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Answer» Correct Option :- (C) the same number of molecules Explanation :- 2g H2 ⇒ 1 mole gas at 200 K 28g N2 ⇒ 1 mole gas at 400 K PV = nRT ⇒ P ∝ nT |
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| 706. |
Copper in an alloy is estimated by dissolving in conc. nitric acid. In this process copper is converted to cupric nitrate with the evolution of nitric oxide (NO). The misture when treated with potassium iodide forms cupric iodide. Which is unstable and decomposes to cuprous iodide and iodine. The amount of copper in the alloy is estimated by litrating the libereated iodine with sodium thiosulphate. The reactions are a Cu + b HNO3 → c Cu(NO3)2 + d NO + e H2Of Cul2 → g Cu2I2 + h I2 i Na2S2O3 + j ˆ I2 → k Na2S4O6 + lNaI (Fill up the blanks) (a) The coefficients are : a = ________. b = ________ , c = ________, d = ________ and e = ________ . (b) The coefficients are : f = ________, g = ________ and h = ________ . (c) The coefficients are : i = ________, j = ________, k = ________ and l = ________. (d) If 2.54 g of I2 is evolved from a 2.0 g sample of the alloy, what is the percentage of copper in the alloy ? (atomic, weight of iodine and coper are 127 and 63.5, respectively) |
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Answer» Sol. (a) a = 3, b = 8, c = 3, d = 2 and e = 4 (b) f = 2, g = 1, h = 1. (c) i = 2, j = 1, k = 1, 1 = 2 (d) 2.54 g of I2 = 1/100 mole of I2 = 2/100 gm atom of Cu % Cu = (2/100) × 63.5 / 2) = 63.5% |
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| 707. |
An aqueous solution of HCl has a pH of 2.0 When water is added to increase the pH to 5.0 the hydrogen ion concentration -A. remains the sameB. decrease three-foldC. increases three-foldD. decreases thousand-fold |
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Answer» Correct Answer - D `pH=2 rArr [H^(+)]=10^(-2)` `pH=5 rArr [H^(+)]=10^(-5)` `:. ([H^(+)]"new")/([H^(+)]"old")=(10^(-5))/(10^(-2))=10^(-3)` |
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| 708. |
Copper in an alloy is estimated by dissolving in conc. Nitric acid . In this process copper is converted to cupric nitrate with the evolution of nitric oxide (NO) . The misture when treated with potassium iodide forms cupric iodide. Which is unstable and decomposes to cuprous iodide and iodine.The amount of copper in the alloy is estimated by litrating the libereated iodine with sodium thiosulphate. The reactions are a `Cu + b HNO_(3) rarr c Cu(NO_(3))_(2)+d NO + e H_(2)O` f `Cu l_(2) rarr g Cu_(2)I_(2) + h I_(2)` i `Na_(2)S_(2)O_(3)+ hat (j) I_(2) rarr k Na_(2)S_(4)O_(6)+lNaI` (fill up the blanks) (a) The coefficients are : a = __________ . b = _________ , c = ___________ , d = __________ and e =_________ . (b) The coefficients are : f = _________ , g = _________ and h = ________. (c) The coefficients are : i = ________ , j = ___________ , k = and l = ________. (d) If 2.54 g of `I_(2)` is evolved from a 2.0 g sample of the alloy, what is the percentage of copper in the alloy ? (atomic, weight of iodine and coper are 127 and 63.5, respectively ) |
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Answer» (a) a=3, b=8, c=3, d=2 and e=4 (b) f=2, g=1, h=1. (c) 1=2, j=1, k=1, 1=2 (d) 2.54 g of `I_(2)=1//100` mole of `I_(2)=2//100` gm atom of Cu % Cu = `(2//100)xx63.5//2)=63.5%` |
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| 709. |
Consider two sealed jars of equal volume. One contains 2 g of hydrogen at 200 K and the other contains 28 g of nitrogen at 400 K. The gases in the two jars will have -A. the same pressureB. the same average kinetic energyC. the same number of moleculesD. the same average molecular speed |
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Answer» Correct Answer - C `2g H_(2) rArr ` 1 mole gas at 200 K `28 g N_(2) rArr` 1 mole gas at 400 K PV=nRT `rArr P prop n T` |
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| 710. |
Indentify the stereoisomer pair from the following choice -A. `CH_(3)CH_(2)CH_(2)OH` and `CH_(3)CH_(2)OCH_(3)`B. `CH_(3)CH_(2)CH_(2)Cl` and `CH_(3)CHClCH_(3)`C. `CH_(3)-underset(H)underset(|)C=underset(H)underset(|)C-CH_(3) " and " CH_(3)-underset(H)underset(|)C=overset(H)overset(|)C-CH_(3)`D. |
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Answer» Correct Answer - C `CH_(3)-underset(H)underset(|)C=underset(H)underset(|)C-CH_(3) " and " CH_(3)-underset(H)underset(|)C=overset(H)overset(|)C-CH_(3)` |
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| 711. |
Among the element Li,N,C and Be, one with the largest atomic radius is (A) Li (B) N(C) C (D) Be |
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Answer» Correct Option :- (A) Li Explanation :- as we move left to right in a period atomic radius decrease due to increase in zeff so. greatest radius is of lithium. |
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| 712. |
In the nuclear reaction `._(90)^(234)Th rarr ._(91)^(234)Pa + X`. X is -A. `._(-1)^(0)e`B. `._(1)^(0)e`C. HD. `._(1)^(2)H` |
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Answer» Correct Answer - A `._(90)^(234)Thrarr._(91)^(234)Pa+._(-1)e^(0)` |
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| 713. |
A concentrated solution of copper sulphate, which is dark blue in colour, a mixed at room temperature with a dilute solution of copper sulphate, which is light blue. For this process-A. Entropy change is positive, but enthalpy change is negativeB. Entropy and enthalpy changes are both positiveC. Entropy change is positive and enthalpy does not changeD. Entropy change is negative and enthalpy change is positive |
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Answer» Correct Answer - C Informative |
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| 714. |
A concentrated solution of copper sulphate, which is dark blue in colour, a mixed at room temperature with a dilute solution of copper sulphate, which is light blue. For this process- (A) Entropy change is positive, but enthalpy change is negative (B) Entropy and enthalpy changes are both positive (C) Entropy change is positive and enthalpy does not change (D) Entropy change is negative and enthalpy change is positive |
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Answer» Correct Option :- (C) Entropy change is positive and enthalpy does not change Explanation :- Informative |
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| 715. |
Increasing the temperature increases the rate of reaction but does not increase the- (A) number of collisions (B) activation energy (C) average energy of collisions (D) average velocity of the reactant molecules |
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Answer» Correct Option :- (B) activation energy |
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| 716. |
Consider the following reaction: `2NO_(2)(g) rarr 2NO(g)+ O_(2)(g)` In the figure below, identify the curves `X,Y,` and `Z` associated with the three species in the reaction A. `X = NO, Y = O_(2), Z = NO_(2)`B. `X = O_(2), Y = NO, Z = NO_(2)`C. `X = NO_(3), Y = NO, Z = O_(2)`D. `X = O_(2), Y = NO_(2), Z = NO` |
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Answer» Correct Answer - A `r = - ( 1)/(2)(d[NO_(2)])/(dt) = + 1/2(d[NO])/(dt) = + (d[O_(2)])/(dt)` `:. NO_(2)` is reactant so ` (Z)` rate of dis appearance of ` NO_(2)` = rate of formation of `NO` So NO is (X) |
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| 717. |
which one of the following is a modified leaf ?A. Sweet potatoB. GingerC. OnionD. Carrot |
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Answer» Correct Answer - C in Onion modified leaves are present for food storage |
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| 718. |
which of the following paris are both polysaccharides ?A. Cellulose and glycogenB. Starch and glucoseC. Cellulodse and fructoseD. Ribose and sucrose |
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Answer» Correct Answer - A Cellulose `to` Homopolysachoride of `beta` glucose Glycogen `to ` Homopolysaccharide fo `alpha` glucose |
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| 719. |
Considering the average molecular mass of a base to be 500 Da . What is the molecular mass of a double stranded DNA of 10 base pairs ?A. 500DaB. 5K DaC. 10kDaD. 1 kDa |
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Answer» Correct Answer - C Molecular Mass of base =500Da total No . Of bases = 10bp `=10xx2=20`bases `therefore ` Molecular mass of 20 bases `= 20xx500`da =10000 dalton `=10` kda |
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| 720. |
Considering ABO blood grouping system in humans, during blood transfusion some combinations of blood groups are compatible `(sqrt")` whereas the others are incompatible (X). Which ONE of the following options is COR- RECT ?A. B. C. D. |
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Answer» Correct Answer - D `Orarr"Universal Donor"` `ABrarr"Universal recipient"` |
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| 721. |
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 inter- pretations of the codons assignments for Ser, Leu, Arg, and Glu.A. Ser `rarr` UCU, Leu `rarr` CUC, Arg `rarr` AGA, Glu `rarr` GAGB. Ser `rarr` CUC, Leu `rarr` GAG, Arg `rarr` UCU, Glu `rarr` AGAC. Ser `rarr` AGA, Leu `rarr` UCU, Arg `rarr` GAG, Glu `rarr` CUCD. Ser `rarr` GAG, Leu `rarr` AGA, Arg `rarr` CUC, Glu `rarr` UCU |
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Answer» Correct Answer - A Sequence of 3 nitrogenous base is one codon. |
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| 722. |
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. 10B. 64C. 24D. 32 |
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Answer» Correct Answer - D Time `=(15)/(3)=5` times division occur. No. of bacteria `=2^(5)=32` |
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| 723. |
Using dimensional analysis the resistivity in terms of fundamental constants `h, m_(e), c, e, epsilon_(0)` can be expressed asA. `(h)/(epsilon_(0)m_(e)ce^(2))`B. `(epsilon_(0)m_(e)ce^(2))/(h)`C. `(h^(2))/(m_(e)ce^(2))`D. `(m_(e)epsilon_(0))/(ce^(2))` |
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Answer» Correct Answer - C |
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| 724. |
A pot of the kinetic energy `(1//2mv^(2))` of ejected electrons as a functionof the frequency (v) of incident radiation for four alkali metals `(M_(1),M_(2),M_(3),M_(4))` is shown below. The alkali metals `M_(1),M_(2),M_(3)andM_(4)` are, respectively- A. Li, Na, K, and RbB. Rb, K, Na, and LiC. Na, K, Li and RbD. Rb, Li, Na, and K |
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Answer» Correct Answer - B `overset(Rb,K,Na,Li)toI.Pdarr` |
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| 725. |
The number of moles of `Br_(2)` produced when two moles of potassium permanganate are treated with excess potassium bromide in aqueous acid medium is-A. 1B. 3C. 2D. 4 |
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Answer» Correct Answer - B `KMnO_(4)+KBr+H_(2)SO_(4)toMnSO_(4)+K_(2)SO_(4)+Br_(2)+H_(2)O` Balanced equations `2KMnO_(4)+10KBr+8H_(2)SO_(4)to2MnSO_(4)+5Br_(2)+6K_(2)SO_(4)+8H_(2)O` `Eq. of KMnO_(4)=Eq.ofBr_(2)` `"Mole"xxn="Mole"xxn` `2xx5="Mole"xx2 5 moles of `Br_(2)` are formed |
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| 726. |
A body is born with the normal number and distribution of rods, but no cones in his eyes. We would expect that the body would be-A. coloure blindB. night blindC. blind in both eyesD. blind in one eye |
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Answer» Correct Answer - A Absence of cone cells in eyes is known as total colour blondness or monochromacy. This person views everything as if it was an a block and white television. Monochromacy occurs when 2 or all 3 of cone pigments are missing and color and lightness vision is reduced to one dimension. |
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| 727. |
A baby is born with the normal number and distribution of rods, but no cones in his eyes. We would expect that the baby would be – (A) colour blind (B) night blind (C) blind in both eyes (D) blind in one eye |
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Answer» Correct option (A) colour blind Explanation: Absence of cone cells in eyes is known as total colour blondness or monochromacy. This person views everything as if it were an a black and white television. Monochromacy occurs when 2 or all 3 of cone pigments are missing and color and lightness vision is reduced to one dimension. |
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| 728. |
Let R be the set of all real numbers and let f be a function from R to R such that `f(X)+(X+1/2)f(l-X)=1,` for all `xinR." Then " 2f(0)+3f(1)` is equal to-A. 2B. 0C. -2D. -4 |
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Answer» Correct Answer - C `f(x)+(x+1/2)f(1-x)=1` `f(l-x)+(1-x+1/2)f(1-(1-x))=1` `f(1-x)+(3/2-x)f(x)=1` `(1-f(x))/(x+1/2)+(3/2-x)f(x)=1` `1-f(x)+(3/2x-x^(2)+3/4-x/2)f(x)=x+1/2` `f(x)(x-x^(2)-1/4)=x-1/2` `f(x)(4x-4x^(2)-1)=4x-2` `f(x)=(-2+4x)/(4x-4x^(2)-1)` `2f(0)+3f(1)=2((-2+0)/(0-0-1))+3((-2+4)/(4-4-1))` `=+4+(3(+2))/-1=+4-6=-2` Alternate Put x = 0 `f(0)+1/2f(1)=1rArr2f(0)+f(1)=2` put x = 1 `f(1)+3/2f(0)=1rArr2f(1)+3f(0)=2` solving above `f(0)=2" and "f(1)=-2` `:. 2f(0)+3f(1)=4-6=-2` |
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| 729. |
At 298 K, assuming ideal behaviour, the average kinetic energy of a deuterium molecule is (A) two times that of a hydrogen molecule (B) four times that of a hydrogen molecule (C) half of that of a hydrogen molecule (D) same as that of a hydrogen molecule |
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Answer» Correct option (D) same as that of a hydrogen molecule (K.E.)average = 3/2 kT i.e., average kinetic energy depends only on temperature. |
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| 730. |
A light bulb of resistance `R =16 Omega` is attached in series with an infinite resistor network with identical resistances r as shown below. A 10 V battery drives current in the circuit. What should be the value of r such that the bulb dissipates about 1 W of power. |
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Answer» Correct Answer - A |
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| 731. |
Three equal charges +q are placed at the three vertices of an equilateral triangle centered at the origin. They are held in equilibrium by a restoring force of magnitude F(r) = kr directed towards the origin, where k is a constant. What is the distance of the three charges from the origin ?A. `[(1)/(6piepsilon_(0))(q^(2))/(k)]^(1//2)`B. `[(sqrt(3))/(12piepsilon_(0))(q^(2))/(k)]^(1//2)`C. `[(1)/(6piepsilon_(0))(q^(2))/(k)]^(2//3)`D. `[(sqrt(3))/(4piepsilon_(0))(q^(2))/(k)]^(2//3)` |
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Answer» Correct Answer - B |
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| 732. |
An ideal gas undergoes a circular cycle centered at 4 atm, 4 lit as shown in the diagram. The maximum temperature attained in this process is closed to - A. 30/RB. 36/RC. 24/RD. 16/R |
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Answer» Correct Answer - A |
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| 733. |
Consider the infinite ladder circuit shown below. For which angular frequency ω will the circuit behave like a pure inductance ?A. `(LC)/(sqrt(2))`B. `(1)/(LC)`C. `(2)/(sqrt(LC))`D. `(2L)/(sqrt(C))` |
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Answer» Correct Answer - C |
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| 734. |
A bead of mass m is attached to the mid-point of a taut, weightless string of length l and placed on a frictionless horizontal table. Under a small transverse displacement x, as shown, if the tension in the string is T, then the frequency of oscillation is-A. `(1)/(2pi)sqrt((2T)/(ml)`B. `(1)/(2pi)sqrt((4T)/(ml)`C. `(1)/(2pi)sqrt((4T)/(m)`D. `(1)/(2pi)sqrt((2T)/(m)` |
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Answer» Correct Answer - B |
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| 735. |
A carrot looks orange in colour because of the `beta` carotene molecule in it. This means that the `beta` carotene molecule absorbs light of wavelengthsA. longer than 550 nm.B. shorter than 550 nm.C. longer than 700 nm.D. shorter than 700 nm. |
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Answer» Correct Answer - B |
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| 736. |
Let P be an arbitrary point on the ellipse `(x^(2))/(a^(2)) + (y^(2))/(b^(2)) = 1, a gt b gt 0`. Suppose `F_(1)` and `F_(2)` are the foci are the ellipse. The locus of the centroid of the triangle `PF_(1)F_(2)` as P moves on the ellipse is-(A) a circle (B) a parabola (C) an ellipse (D) a hyperbolaA. a circleB. a parabolaC. an ellipseD. a hyperbola |
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Answer» Correct Answer - C `P rarr a cos theta, b sin theta` ` G rarr ((sumx_(i))/(3),(sumy_(i))/(3))` `F_(i) rarr(ae,0) , F_(2) rarr (-ae,0)` `h = (acostheta + ae - ae)/(3) , cos theta = (3h)/(a)` `k = (b sin theta)/(3), sin theta = (3k)/(b)` `cos^(2)theta + sin^(2)theta = 1` `(9h^(2))/(a^(2)) + (9k^(2))/(b^(2)) = 1 rArr (x^(2))/((a^(2)//9)) + (y^(2))/((b^(2)//9)) = 1` (Ellipse) |
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| 737. |
Suppose a, b are real numbers such that ab ≠ 0. Which of the following four figures represents the curve (y – ax – b)(bx2 + ay2 – ab) = 0 ? |
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Answer» Correct Option :- (B) Explanation :- y = ax + b and x2/a + y2/b= 1 slope = a for the line, y intercept = b Fig.1 for line → a < 0 , b > 0 hence the other fig. cannot be an ellipse Fig.2 a > 0 , b < 0 hence the fig. is a hyperbola Similarly you can check rest 2 options |
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| 738. |
Prove that `(1+tan1^(@))(1+tan2^(@))....(1+tan4 5^(@))=2^(23)`A. `2^(21)`B. `2^(22)`C. `2^(23)`D. `2^(25)` |
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Answer» Correct Answer - C `(1+tantheta)(1+tan(45-theta)) = (1+tantheta) (1+(1-tantheta)/(1+tantheta)) = 2` `(1+tan1^(@))(1+tan44^(@)) = 2 etc` so product `= 2^(22) (1+tan45^(@)) = 2^(23)` |
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| 739. |
Let `I_(n) = int_(0)^(1) (logx)^(n)dx`, where n is a non-negative integer. Then `I_(2001) - 2011 I_(2010)` is equal to -A. `I_(1000) + 999 I_(998)`B. `I_(890) + 890 I_(889)`C. `I_(100) + 100 I_(99)`D. `I_(53) + 54 I_(52)` |
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Answer» Correct Answer - C `I_(n)=overset(e)underset(1)intunderset(II)(1).underset(I)((logx)^(n))dx` ` I_(n) (logx)^(n)x|_(1)^(e)-overset(e)underset(1)int(n(logx)^(n-1))/(x).xdx` `I_(n) = e - 0 - nI_(n-1)` `I_(n) + nI_(n-1) = e` `I_(2001+2011)I_(2010) = e` `I_(100) + 100I_(99) = e` |
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| 740. |
In triangle ABC, we are given that `3 sin A +4 cos B=6` and `4 sin B+3 cos A=1`. Then the measure of the angle C is -A. `30^(@)`B. `150^(@)`C. `60^(@)`D. `75^(@)` |
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Answer» Correct Answer - A Square & add both equations `9+16+24 sin (A+B)=37` `sin (A+B)=1/2 rArr A+B=pi/6 rArr C=(5pi)/6` (wrong) `rArr A+B=(5pi)/6 rArr C=pi/6` because `C=(5pi)/6` does not follow equation `3 sin A + 4 cos B =6` |
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| 741. |
Let `I_(n)=int_(0)^(pi//2)x^(n)cosxdx,` where in is a non-negative integer Then `sum_(n=2)^(oo)((I_(n))/(n!)+(I_(n-2))/((n-2)!))` equals-A. `e^(pi//2)-1-(pi)/(2)`B. `e^(pi//2)-1`C. `e^(pi//2)-(pi)/(2)`D. `e^(pi//2)` |
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Answer» Correct Answer - A `I_(n)=int_(0)^(pi//2)x_(1)^(n)underset(II)cosxdx` `=x^(n)sinx|_(0)^(pi//2)-int_(0)^(pi//2)nx^(n-1)sinxdx` `=((pi)/(2))^(2)-0-(nx^(n-1)(-cosx)|_(0)^(pi//2)-int_(0)^(pi//2)n(n-1)x^(n-2)(-cosx)dx` `=((pi)/(2))^(n)-0-(n-1)int_(0)^(pi//2)x^(n-2)cosxdx` `I_(n)=((pi)/(2))^(n)-n(n-1)I_(n-2)` `underset(n=2)overset(oo)sum((I_(n))/(n!)+(I_(n-2))/((n-2)!))=underset(n=2)overset(oo)sum((((pi)/(2))^(2)-n(n-1)I_(n-2))/(n!)+(I_(n-2))/((n-2)!))` `underset(n=2)overset(oo)sum(((pi)/(2))^(n)(1)/(n!))` `=((pi)/(2))^(2)(1)/(2!)+((pi)/(2))^(3)(1)/(3!)+((pi)/(2))^(4)(1)/(4!)+....` `e^(pi//2)-1-((pi)/(2))` |
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| 742. |
Which of the following intervalsis possible domain of the function `f(x)=(log)_((x))[x]+(log)_([x]){x}`, where [x] is the greatest integer not exceeding `xa n d{x}=x-[x]?``(0,1)`(b) (1,2) (c) (2, 3) (d) (3, 5)A. (0, 1)B. (1, 2)C. (2, 3)D. (3, 5) |
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Answer» Correct Answer - C `x notin I` & `[x] gt 1` `rArr x in (2, 3)` only option satisfy. |
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| 743. |
Let f : `(2, oo) rarr N` be defined by `f (x)=` the largest prime factor of [x]. Then `int_(2)^(8) f(x) dx` is equal to -A. 17B. 22C. 23D. 25 |
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Answer» Correct Answer - B `underset(2)overset(8)(int) f(x) dx=2+3+2+5+3+7=22` |
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| 744. |
Let [x] denote the largest interger not exceeding x and `{x}=x-[x]`. Then `int_(0)^(2012) e^(cos (pi{x}))/(e^(cos (pi{x}))+e^(-cos(pi{x})))dx` is equal to - |
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Answer» Correct Answer - B `I=2012 underset(0)overset(1)(int) e^(cos pi x)/(e^(cos pi x)+e^(-cos pi x))dx` using king property `I =2012 underset(0)overset(1)(int) e^(-cos pi x)/(e^(-cos pi x)+e^(cos pi x))dx rArr 2I=2012 rArr I=1006` |
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| 745. |
Two players play the following game : A writes 3, 5, 6 on three different cards, B writes 8, 9, 10 on three different cards. Both draw randomly two cards from their collections. Then A computes the product of two numbers he/she has drawn, and B computes the sum of two numbers he/she has drawn. The player getting the larger number wins. What is the probability that A wins ?A. `1/3`B. `5/9`C. `4/9`D. `1/9` |
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Answer» Correct Answer - C Foa A to win, A can draw either 3, 6, or 5, 6. If A draws 3, 6 then B can draw only 8 & 9 Prob. `=1/3.1/3=1/9` If A draws 5, 6 then B can draw, any two Probability `=1/3. 1=1/3` Probability `=1/9+1/3=4/9` |
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| 746. |
The value of `lim_(n rarr oo)(1/sqrt(4n^(2)-1)+1/sqrt(4n^(2)-4)+...+1/sqrt(4n^(2)-n^(2)))` is -A. `1/4`B. `pi/12`C. `pi/4`D. `pi/6` |
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Answer» Correct Answer - D `lim_(n rarr oo) sum_(r=1)^(n) (1)/sqrt(4n^(2)-r^(2))=lim_(n rarr oo)1/n sum_(r=1)^(n) (1)/sqrt(4-(r//n)^(2))` `=underset(0)overset(1)(int) (dx)/sqrt(4-x^(2))=(sin^(-1) (x/2))_(0)^(1)=pi/6` |
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| 747. |
Suppose the limit `L= lim_(n to oo)sqrt(n)int_(0)^(1)(1)/((1+x^(2))^(n))dx` exist and is larger than `(1)/(2)`. ThenA. `(1)/(2) lt L lt 2`B. `2 lt L lt 4`C. `3 lt L lt 4`D. `L ge 4 ` |
| Answer» Correct Answer - A | |
| 748. |
Consider the function f(x)= `{{:((x+5)/(x-2),"if " xne 2),(" 1", "if " x=2):}` . Then f(f(x)) is discontinuousA. at all real numbersB. at excatly two values of xC. at ecactly one value of xD. at exactly three values of x |
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Answer» Correct Answer - B discontinuous at x=2 `f(f(x))=f((x+5)/(x-2))` `(((x+5)/(x-2)+5))/(((x+5)/(x-2)-2))=(6x-5)/(-x+9)` `(6x-5)/(9-x)` At x=9 it is discontinuous |
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| 749. |
The value of the limit `lim_(x to -oo) (sqrt(4x^(2) - x )+2x)` isA. `-oo`B. `-(1)/(4)`C. 0D. `(1)/(4)` |
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Answer» Correct Answer - D Rationalise ` lim_(x to -oo) ((sqrt(4x^(2) - x) + 2x) xx (sqrt(4x^(2) - x) - 2x)/(sqrt(4x^(2) -x)- 2x))` `lim_(x to -oo) ((-x)/(|x|sqrt(4 - (1)/(x)) - 2x)) "at" x to -oo |x| = - x ` `lim_(x to -oo) ((-x)/(|x|sqrt(4 - (1)/(x)) - 2x)) = (1)/(2+ 2) = (1)/(4)` |
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| 750. |
The number of solutions to the equations `cos^(4)x+(1)/(cos^(2)x)=sin^(4)x+(1)/(sin^(2)x)` in the interval `[0,2pi]` isA. 6B. 4C. 2D. 0 |
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Answer» Correct Answer - B `cos^(4)x-sin^(4)x=(1)/(sin^(4)x)-(1)/(cos^(2)x)` `(cos^(2)x-sin^(2)x)=((cos^(2)x-sin^(2)x))/(sin^(2)xcos^(2)x)` `cos2x=(4cos2x)/(sin^(2)2x)` `cos 2x=(1-4"cosec"^(2)2x)=0` `cos 2x=0` `2x=2npi+-(pi)/(2)` `x=npi+-(4)/(pi)` At `n=0,x=(pi)/(4)` `n=1, n=(5pi)/(4),(3pi)/(4)` `n=2 ,x=(7pi)/(4)` |
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