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.
| 1. |
A parallel plate condenser with oil (dielectric constant 2) between the plates has capacitance C. If oil is removed, the capacitance of capacitor becomesA. `sqrt(2C)`B. 2CC. `(C)/(sqrt2)`D. `(C)/(2)` |
|
Answer» Correct Answer - D The capacitance of a parallel plate capacitor with dielectric (oil) between its plates is. `C=(Kepsi_(0)A)/(d).....(i)` where `epsi_(0)`,=electric permittivity of free space K=dielectric constant A=area of each plate of capacitor d=distance between two plates When dielectric (oil) is removed, so capacitance `c_(0)=(epsi_(0)A)/(d).....(ii)` Comparing Eqs. (i) and (iii), we get `C=KC_(0)` `Rightarrow C_(0)=(C)/(K)=(C)/(2) (therefore K=2)` |
|
| 2. |
Which halogen can be purified by sublimation ?A. `F_(2)`B. `Cl_(2)`C. `Br_(2)`D. `I_(2)` |
|
Answer» Correct Answer - D In the sublimation, the solids substance converts into vapours directly. Iodine is found in solid state while `F_(2) and Cl_(2)` are found in gaseous state and `Br_(2)` is found in liquid state. So, iodine can be purified by sublimation. |
|
| 3. |
In a radioactive material the activity at time `t_(1)` is `R_(1)` and at a later time `t_(2)`, it is `R_(2)`. If the decay constant of the material is `lambda`, thenA. `R_(1)=R_(2)e^(-lambda(t_(1)-t_(2)))`B. `R_(1)=R_(2)e^(lambda(t_(1)-t_(2)))`C. `R_(1)=R_(2)(t_(2)//t_(1))`D. `R_(1)=R_(2)` |
|
Answer» Correct Answer - A The decay rate R of a radioactive material is the number of decays per second. From radioactive decay law. `-(dN)/(dt)alpha N or -(dN)/(dt)=lambdaN` `"Thus", R=-(dN)/(dt)` `or R=lambdaN or R=lamdaN_(0)E^(-lambdat).....(i)` Where, `R_(0)=lambdaN_(0)` is the activity of the radioactive material at time t=0. At time, `t_(1), R_(1)=R_(0)e^(-lambdat_(1))......(ii)` At time `t_(2)`, `R_(2)=R_(0)e^(-lambdat_(2))......(iii)` Dividing Eq, (ii) by (iii), we have `(R_(1))/(R_(2))=(e^(-lambdat_(1)))/(e^(-lambdat_(2)))=e^(-lambda(t_(1)-t_(2)))` `or R_(1)=R_(2)e^(-lambda(t_(1)-t_(2)))` |
|
| 4. |
A boat at anchor is rocked by waves whose crests are `100 m` apart and whose speed is `25 m//s`. These waves reach the boat once every :A. 2500sB. 75sC. 4sD. 0.25s |
|
Answer» Correct Answer - C Key Idea: The distance between the two consecutive crests in transverse wave motion is called wavelength. The boat bounces up, ie, it travels from crest to the consecutive crest along wave motion. Wavelength=distance between two consecutive crests `i.e,=lambda=100m` Velocity of wave=25m/s Hence, time in one bounce of boat `t=(lambda)/(v)=(100)/(25)=4s` |
|
| 5. |
A certain nuclide has a half life period of 30 min. If a sample containing 600 atoms is allowed to decay for 90 min, how many atoms will remains?A. 200 atomsB. 450 atomsC. 75 atomsD. 150 atoms |
|
Answer» Correct Answer - C `n=("total time")/("half-life period")=(90)/(30)=3` `N=N_(0)xx((1)/(2))^(n)` `=600xx((1)/(2))^(3)` `=(600)/(8)` =75atoms |
|
| 6. |
Which one of the following processes dependsA. ConductionB. ConvectionC. RadiationD. None of these |
|
Answer» Correct Answer - B (a) Conduction is the process of transmission of heat in a body from the hotter part to the colder part without any bodily movement of constituent atoms or molecules of the body. (b) In convection, the heated lighter particles move upwards and colder heavier particles move downwards to their place. This depends on weight and hence, on gravity. (c) Radiation is the process of transmission of heat from the body to another body through electromagnetic waves through vacuum, irrespective of their temperatures. Hence, (b) is correct. |
|
| 7. |
The acceleration due to gravity on the planet `A` is `9` times the acceleration due to gravity on planet `B`. A man jumps to a height of `2 m` on the surface of `A`. What is the height of jump by the same person on the planet `B` ?A. 6mB. `(2)/(3)m`C. `2//9`mD. 18m |
|
Answer» Correct Answer - D It is given that, acceleration due to gravity on planet A is 9 times the acceleration due to gravity on planet ie, `g_(A)=9_(g_(B)).....(i)` For third equation of motion `v^(2)=2gh` At plaenet A, `h_(A)=(v^(2))/(2g_(A)),......(ii)` At plaenet B, `h_(B)=(v^(2))/(2g_(B))......(iii)` Dividing Eq. (ii) by Eq. (iii), we have `(h_(A))/(h_(B))=(g_(B))/(g_(A))` From Eq. (i) `g_(A)=9g_(B)` `therefore (h_(A))/(h_(B))=(g_(B))/(9g_(B))=(1)/(9)` or `h_(B)=9h_(A)=9xx2=18m (therefore h_(A)=2m)` |
|
| 8. |
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where, E is the total energy)A. `(1)/(4)E`B. `(1)/(2)E`C. `(2)/(3)E`D. `(1)/(8)E` |
|
Answer» Correct Answer - A Potential energy of a simple harmonic oscillator `U=(1)/(2)momega^(2)y^(2)` Kinetic energy of a simple harmonic oscillator `K=(1)/(2)momega^(2)(A^(2)-y^(2))` Here, y=displacement from mean position A=maximum displacement (or amplitude) from mean position Total energy is E=U+K `=(1)/(2)momega^(2)y^(2)+(1)/(2)momega^(2)(A^(2)-y^(2))` `=(1)/(2)momega^(2)A^(2)` When the particle half way to its end point ie, at half of its amplitude then `y=(A)/(2)` Hence potential energy `U=(1)/(2)momega^(2)((A)/(2))^(2)` `=(1)/(4)((1)/(2)momega^(2)A^(2))` `U=(E)/(4)` |
|
| 9. |
The gas having average speed four times as that of `SO_(2)` (molecular mass 64) isA. He (molecular mass 4)B. `O_(2)` (molecular mass 32)C. `H_(2)` (molecular mass 2)D. `CH_(4)` (molecular mass 16) |
|
Answer» Correct Answer - A Average speed `(v_("av"))` of gas molecules is inversely proportional to the square root of the absolute temperature (T) of the gas. `v_(av)=sqrt((8RT)/(piM))` When R is gas constant and M the molecular weight. Given, `v_(1)=v, M_(1)=64, v_(2)=4v` `(v)/(4v)=sqrt((M_(2))/(64))` `Rightarrow M_(2)=(64)/(16)=4` Hence, the gas is helium (molecular mass 4). |
|
| 10. |
In SHM restoring force is `F = -kx`, where k is force constant, x is displacement and a is amplitude of motion, then total energy depends uponA. K,A and MB. k,x, MC. k,AD. k,x |
|
Answer» Correct Answer - C In SHM, the total energy =potential energy+kinetic energy or `E=U+K` `=(1)/(2)momega^(2)x^(2)+(1)/(2)momega^(2)(A^(2)-x^(2))` `=(1)/(2)momega^(2)A^(2)=(1)/(2)kA^(2)` where k=force constant `=momega^(2)` Thus, total energy dependents on k and A. |
|
| 11. |
The displacement equation of a simple harmonic oscillator is given by `y=A sin omegat-Bcos omegat` The amplitude of the oscillator will beA. A-BB. A+BC. `sqrt(A^(2)+B^(2))`D. `(A^(2)+B^(2))` |
|
Answer» Correct Answer - C Displacement equation `y=A sin omegat-B cos omemgat` Let `A=a cos theta and B=a sin theta` So, `A^(2)+B^(2)=a^(2)` `Rightarrow a=sqrt(A^(2)+B^(2))` Then, `y=a cos theta sin omega t-a sin theta cos omega t` Which is the equation of simple harmonic oscillator. The amplitude of the oscillator `=a=sqrt(A^(2)+B^(2))` |
|
| 12. |
If A is a square matrix of order `n xx n` then adj(adj A) is equal toA. `|A|^(n)A`B. `|A|^(n-1)A`C. `|A|^(n-2)A`D. `|A|^(n-3)A` |
|
Answer» Correct Answer - D For any square matrix B, we have `B("adj B")=|B|I_(n)` On taking B=adj A, we get (adj A)[adj (adj A)]=|adj A|`I_(n)` `Rightarrow ` adjA[adj(adj A)]=`|A|^(n-1)I_(n) (therefore |adj A|=|A|^(n-1))` `Rightarrow` (A adj A)[adj (adj A)]=`|A|^(n-1)A` `Rightarrow (|A|I_(n))[adj (adj A)]=|A|^(n-1)A` `Rightarrow` adj(adj A)=`|A|^(n-2)A` |
|
| 13. |
`._(90)Th^(232) to ._(82)Pb^(208)`. The number of `alpha and beta-"particles"` emitted during the above reaction isA. `8alpha and 4beta`B. `8alpha and 16beta`C. `4alpha and 2beta`D. `6alpha and 4beta` |
|
Answer» Correct Answer - D No. of `alpha`-particle ` =(232-208)/(4)=(24)/(4)=6alpha` No. of `beta`-particle `= `2xx6-(90-82)` `=12-8` =`4beta` `"Hence", 6alpha and 4 beta` particle are emitted. |
|
| 14. |
If `A=[(cos^(2)alpha, cos alpha sin alpha),(cos alpha sin alpha, sin^(2)alpha)]` and `B=[(cos^(2)betas,cos beta sin beta),(cos beta sin beta, sin^(2) beta)]` are two matrices such that the product AB is null matrix, then `alpha-beta` is |
|
Answer» Correct Answer - C `"Given"AB=0` `therefore [{:(,cos^(@)alpha,cos alpha sin alpha),(,cos alpha sin alpha, sin^(2)alpha):}]` `xx [{:(,cos^(2)beta,cos beta sin beta),(,cos beta sin beta,sin^(2)beta):}]` `=[{:(,0,0),(,0,0):}]` `Rightarrow [{:(,cos alpha cos beta cos(alpha-beta)),(,cos beta sin alpha cos (alpha-beta)):}` `{:(,cos alpha sin beta cos (alpha-beta)),(,sin alpha sin beta cos (alpha-beta)):}]` `=[{:(,0,0),(,0,0):}]` `Rightarrow cos(alpha-beta)=0` `Rightarrow alpha-beta` is an odd multiple of `pi//2`. |
|
| 15. |
A line is drawn through a fixed point `P(alpha, B)` to cut the circle `x^(2)+y^(2)=r^(2)` at A and B. Then PA.PB is equal toA. `(alpha+beta)^(2)-r^(2)`B. `(alpha^(2)+beta^(2)-r^(2)`C. `(alpha-beta)^(2)+r^(2)`D. None of the above |
|
Answer» Correct Answer - B The equation of any line through the point `P(alpha,beta)` is `(x-alpha)/(cos theta)=(y-beta)/(sin theta)=k("say")` Any point on this line is `(alpha+k cos theta, beta+k sin theta)` This point lies on the given circle, if `(alpha+k cos theta)^(2)+(beta+k sin theta)^(2)=r^(2) or k^(2)+2k(alpha cos theta+beta sin theta)+alpha^(2)+beta^(2)-r^(2)=0......(i)` Which being quadratic in k, gives two values of k. Let PA=`k_(1),PB=k_(2)`, where `k_(1),k_(2)` are the roots of Eq. (i) then, `PA.PB=k_(1)k_(2)=alpha^(2)+beta^(2)-r^(2)` |
|
| 16. |
A pulse of a wave train travels along a stretched string and reaches the fixed end of the string.it will be reflected back withA. a phase change of `180^(@)` with velocity reversedB. the same phase as the incident pulse with no reversal of velocityC. a phase change of `180^(@)` with no reversal of velocityD. the same phase as the incident pulse but with velocity reversed |
|
Answer» Correct Answer - C A pulse of a wave train when travels along a stretched string and reaches the fixed end of the string, then it will be reflected back to the same medium and the reflected ray suffers a phase change of n with the incident wave but wave velocity after reflection does not change. |
|
| 17. |
A mixture of ethyl amine and alcoholic KOH on heating givesA. alkyl cyanideB. biureteC. ethyl isocyanideD. ethyl isocyanate |
|
Answer» Correct Answer - C `underset("ethyl amine")(C_(2)H_(5)NH_(2))+CHCl_(3)+underset(("alcoholic"))(3KOH) to C_(2)H_(5)Nc+3KCl+2H_(2)O` It is carbyl amine reaction. |
|
| 18. |
The circumcentre of the triangle formed by the lines, `xy + 2x + 2y + 4 = 0 and x + y + 2 = 0` is-A. (0,0)B. (-2,-2)C. (-1,-1)D. (-1,-2) |
|
Answer» Correct Answer - A The given equation xy+2x+4=0 can be rewritten as (x+2) (y+2)=0 or x+2=0, y+2=0 And also given that x+y+2=0 On solving the above equations, we get A(-2,0), B(0,-2),C(-2,-2) It is clearly that `triangleABC` is right angled triangle with right angle at C. Hence centre of the circumcircle is the mid point of AB whose coordinates are (-1,-1). |
|
| 19. |
The dual of the statement `[p vee (-q) vee(-p)` isA. `p vee (-q)vee -p`B. `(p vee -q)vee -p`C. `pvee -(q vee -p)`D. None of these |
|
Answer» Correct Answer - D The dual of the given statement is `(p vee -q) vee -p` |
|
| 20. |
For the system of equaltions : `x+2y+3z=1` `2x+y+3z=2` `5x+5y+9z=4`A. there is only one solutionB. there exists infinitely many solutionC. there is no solutionD. None of the above |
|
Answer» Correct Answer - A Now, `Delta=|{:(,1,2,3),(,2,1,3),(,5,5,9):}|` `=1(9-15)-2(18-15)+3(10-5)` `=-6-6+16=3 ne 0` Hence the system of equation has a unique solution |
|
| 21. |
If coil is open then L and R becomeA. `oo,0`B. `0,oo`C. `oo,oo`D. `0,0` |
|
Answer» Correct Answer - B When coil is open, there is no current in it, hence no flux is associated with it, ie, `phi=0` Also, we know that flux linked with the coil is directly proportional to the current in the coil. Ie, `phi alpha I` `or phi=Li` Where L is proportionally constant known as self-inductance. `therefore L=(phi)/(i)=0` Again since `i=0,"hence" R=oo` |
|
| 22. |
In a circuit, the current lags behind the voltage by a phase difference of t/2, the circuit will contain which of the following ?A. Only RB. Only CC. R and CD. Only L |
|
Answer» Correct Answer - D When a circuit contains inductance only, then the current lags behind the voltage by the phase difference of `(pi)/(2) or 90^(@)`. While in a purely capacitive circuit, the current leads the voltage by a phase angle of `(pi)/(2) or 90^(@)`. In a purely resistive circuit current is in phase with the applied voltage. |
|
| 23. |
Which of the following does not contain silicon?A. KaolineB. AgateC. RubyD. Quartz |
|
Answer» Correct Answer - C Ruby is mineral of aluminium ie, `Al_(2)O_(3)`. It does not contain silicon. |
|
| 24. |
Oils and fats are esters of higher fatty acids with :A. sugarB. glycerolC. tributyrineD. polypeptide |
|
Answer» Correct Answer - B Oil and fats are glyceride ester of higher carboxylic acid. Eg. Palmitin. `underset("platinum(fat)")( underset(CH_(2)OCOC_(15)H_(31))underset(|)( underset(CHOCOC_(15)H_(31))underset(|)(C))H_(2)OCOC_(15)H_(31))` |
|
| 25. |
`CH_3-underset(" "OC_2H_5)underset(" "|)CH-CH_2-CH_2-CH_3`, the IUPAC name isA. 2-ethoxy pentaneB. 4-ethoxy pentaneC. pentyl-ethyl etherD. 2-pentoxy ethane |
|
Answer» Correct Answer - A `underset("2-ethoxy pentane")(overset(1)CH_(3)-underset(OC_(2)H_(5))(overset(2)CH-overset(3)CH_(2))-overset(4)CH_(2)-overset(5)CH_(3))` |
|
| 26. |
The aldol condensation of `CH_(3)-CHO` results is the formation ofA. `CH_(3)-underset(O)underset(||)(C)-underset(OH)underset(|)(CH)-CH_(3)`B. `CH_(3)-underset(OH)underset(|)(CH)-CH_(2)-underset(O)underset(||)(CH)`C. `CH_(3)-CH_(2)-underset(OH)underset(|)(CH)-underset(O)underset(||)(CH)`D. `CH_(3)-CH_(2)OH+CH_(3)OH` |
|
Answer» Correct Answer - B Two molecules of acetaldehyde gives aldol on aldol condensation. `underset("acetaldehyde")(2CH_(3)CHO) overset(OH^(oplus))to CH_(3)-underset("adol")(underset(OH)underset(|)(CH))-CH_(2)-CHO` |
|
| 27. |
Formalin is:A. formic acidB. fluoroformC. 40% aqueous solution of methanalD. para formaldehyde |
|
Answer» Correct Answer - C 40% aqueous solution (methanal) is called as formalin. |
|
| 28. |
Lemon gives sour taste because ofA. citric acidB. tartaric acidC. oxalic acidD. acetic acid |
|
Answer» Correct Answer - A Citric acid is found in lemon. Therefore, lemon gives sour taste. |
|
| 29. |
If `f:Rto R` be mapping defined by `f(x)=x^(2)+5`, then `f^(-1)(x)` is equal toA. `(x+5)^(1//3)`B. `(x-5)^(1//3)`C. `(5-x)^(1//3)`D. `5-x` |
|
Answer» Correct Answer - A Let `y=f(x)=x^(3)+5` `Rightarrow x=(y-5)^(1//3)` |
|
| 30. |
The parametric representation of a point on the ellipse whose foci are (-1, 0) and (7, 0) and eccentricity 1/2, isA. `(3+8cos theta, 4sqrt3 sintheta)`B. `(8cos theta, 4sqrt3 sintheta)`C. `(3+4sqrt3 cos theta, 8 sin theta)`D. None of the above |
|
Answer» Correct Answer - B Distance between two foci, 2ae=7+1=8 `therefore ae=4` `Rightarrow a=8 " "(therefore e=(1)/(2)"given")` `"Now," b^(2)=a^(2)(1-e^(2))=64(1-(1)/(4))` `therefore b^(2)=48 Rightarrow b=4sqrt3` Since the centre of the ellipse is the mid point of the line joining two foci, therefore the coordinates of the centre are (3,0). Its equation is `((x-3)^2)/(8^(2))+((y-0)^(2))/((4sqrt3)^(2))=1.....(i)` Hence, the parametric coordinate of a point on Eq. (i) are `(3+8 costheta, 4sqrt3 sin theta)`. |
|
| 31. |
The equation to the line touching both the parabolas `y^2 =4x` and `x^2=-32y` isA. `x+2y+4=0`B. `2x+y-4=0`C. `x-2y-4=0`D. `x-2y+4=0` |
|
Answer» Correct Answer - A The equation of any tangent to the parabola `y^(2)=4x` is `y=mx+(1)/(m)....(i)` This touches the parabola `x^(2)=-32y`, therefore the equation `x^(2)=-32(mx+(1)/(m))` has equal roots `therefore (32m)^(2)=4((32)/(m))` `=Rightarrow 8m^(3)=1 Rightarrow m=(1)/(2)` On putting the value of m in Eq. (i), we get x-2y+4=0 |
|
| 32. |
The bond energy is the energy required toA. dissociate one mole of the substanceB. dissociate bond in 1 kg of the substanceC. break one mole of similar bondsD. break bonds in one mole of substance |
|
Answer» Correct Answer - C The bond energy is the energy required to break one mole of similar bonds. |
|
| 33. |
For the reaction, `2H_(2)(g)+O_(2)(g) to 2H_(2)O(g), DeltaH=-573.2kJ` The heat of decomposition of water per moleA. 286.6kJB. 573.2kJC. `-28.66kJ`D. zero |
|
Answer» Correct Answer - A Heat of decomposition of water is `H_(2)O(g) to H_(2)(g)+(1)/(2)O_(2)(g)` `DeltaH=(+573)/(2)=296.6kJ//mol` |
|
| 34. |
For an ideal gas, the heat of reaction at constant pressure and constant volume are related asA. H+E=pVB. `E=H+pDeltaV`C. `q_(p)=q_(v)=DeltanRT`D. None of these |
|
Answer» Correct Answer - C The heat of reaction for an ideal gas, at constant pressure and constant volume are related as `therefore DeltaH=DeltaE+DeltanRT` `therefore q_(p)=q_(v)=DeltanRT` |
|
| 35. |
Which of the following is a path functionA. Internal energyB. EnthalpyC. WorkD. Entropy |
|
Answer» Correct Answer - C Internal energy, enthalpy and entropy are state functions but work and heat are path function. |
|
| 36. |
EMF of hydrogen electrode in term of pH is (at 1 atm pressure).A. `E_(H_(2))=(RT)/(F)xxpH`B. `E_(H_(2))=(RT)/(F).(1)/(pH)`C. `E_(H_(2))=(2.303RT)/(F)pH`D. `E_(H_(2))=-0.591pH` |
|
Answer» Correct Answer - D `2H^(+)+2e^(-) to H_(2)` According to Nernst equation `E=E^(@)+(0.0591)/(2)log""(1)/([H^(+)]^(2))` `E=0-(0.0591)/(2)log""[H^(+)]^(2)` `=0.0591pH` |
|
| 37. |
The solubility of AgCl is `1xx10^(-5)mol//L`. Its solubility in 0.1 molar sodium chloride solution isA. `1xx10^(-10)`B. `1xx10^(-5)`C. `1xx10^(-9)`D. `1xx10^(-4)` |
|
Answer» Correct Answer - C `K_(sp)` of AgCl=(solublility of `AgCl)^(2)` `=(1xx10^(-5))^(2)=1xx10^(-10)` Suppose its solubility in 0.1 M NaCl is mol/L. `AgCl Leftrightarrow underset(x)(Ag^(+))+underset(x)(Cl^(-))` `NaCl Leftrightarrow underset(0.1M)(Na^(+))+underset(0.1M)(Cl^(-))` `[Cl^(-)]=(x+0.1)M` `k_(sp)"of AgCl"=[Ag^(+)][Cl^(-)]` `=x xx(x+0,1)` `1xx10^(-10)=x^(2)+0,1x` Higher power of x are neglected `1xx10^(-10)=0.1x` `x=1xx10^(-9)M` |
|
| 38. |
The standard `E_("Red")^(@)` values of A,B,C are 0.68V, - 2.54V,- 0.50V respectively. The order of their reducing power isA. `A gt B gt C`B. `A gt C gt B`C. `C gt B gt A`D. `B gt C gt A` |
|
Answer» Correct Answer - D Reducing character is based upon higher negative value of reduction electrode potential. Thus, order of reducing character is `B gt C gt A`. |
|
| 39. |
The pH of a `10^(-8)` molar solution of HCl in water isA. 8B. between 6 and 7C. between 7 and 8D. None of these |
|
Answer» Correct Answer - C `pH=-log[H^(+)]=-log10^(-8)=8` It is not possible for acid, so it is `[H^(+)]`, the `[H^(+)]` of water is also added. Total `[H^(+)]` in solution `=[H^(+)]"of HCl"+[H^(+)]` of water `=(1xx10^(-8)+1xx10^(-7))M` `=(1+10)xx10^(-8)=11xx10^(-8)M` `therefore pH=-log[H^(+)]` `=-log 11xx10^(-8)` `=-log 11xx8 log 10` `=-1.0414+8=6.9586` |
|
| 40. |
Peptization denotes:A. digestion of foodB. hydrolysis of proteinsC. breaking and dispersion into colloidal stateD. precipitation of solid from colloidal dispersion |
|
Answer» Correct Answer - C The dispersal of a precipitated material into colloidal solution by the action of an electrolyte in solution is called peptisation and the electrolyte is called a peptising agent. |
|
| 41. |
Freundlich adsorption isotherm isA. `(x)/(m)=kp^(1//m)`B. `x=mkp^(1//n)`C. `x//m=kp^(-n)`D. None of these |
|
Answer» Correct Answer - D Freuendlich absorption isotherm is `x//m=kp^(1//n)` Here, p,k and m are constant |
|
| 42. |
Blue colour of water in sea is due toA. refraction of blue light by impuritiesB. refraction blue sky by waterC. scattering of light by waterD. None of the above |
|
Answer» Correct Answer - C The colour of a colloidal solution depends on the wavelength of the light scattered by the dispersed particles, which in turn depends on the size and the nature of particle. The colour of water in sea is blue due to the scattering of light by water. |
|
| 43. |
2 2-dichloro propane on hydrolysis yieldsA. acetoneB. 2, 2-propane diolC. isopropyl alcoholD. acetaldehyde |
|
Answer» Correct Answer - A `underset("2,2-dichloro propane")(CH_(3)-underset(Cl)underset(|)overset(Cl)overset(|)(c))-CH_(3) overset("Hydrolysis")to CH_(3)-underset("unstable")(underset(OH)underset(|)overset(OH)overset(|)(C)-CH_(3))` `underset(-H_(2)O)to CH_(3)-underset("acetone")(underset(O)underset(||)(C)-CH_(3))` |
|
| 44. |
Carborundum isA. SiCB. `Al_(2)O_(3).H_(2)O`C. `Al_(2)(SO_(4))_(3)`D. `AICI_(3)` |
|
Answer» Correct Answer - A When silica is heated with carbon in electric furnace, it is reduced to carborundum or silicon carbide. `SiO_(2)+3C to SiC+2CO` |
|
| 45. |
The comparatively high b.pt. of HF is due toA. high reactivity of fluorineB. small size of hydrogen atomC. formation of hydrogen bonds and consequent associationD. high IE of fluorine |
|
Answer» Correct Answer - C Intramolecular hydrogen bonding is found in `(HF)_(n)` due to higher electronegativity of fluorine atoms. `underset("hydrogen bonding")(......H-F......H-F....H-F.....)` Hydrogen bonding is helpful in the association of HF molecule, so HF is found in liquid form. |
|
| 46. |
The area of the circle centred at (1,2) and passing through (4,6) isA. `5pi` sq unitB. `5pi` sq unitC. `5pi` sq unitD. None of the above |
|
Answer» Correct Answer - D The equation of a circle centred at (1,2) and passing through (4,6) is `(x-1)^(2)+(y-2)^(2)=(4-1)^(2)+(6-2)^(2)` `Rightarrow x^(2)+y^(2)-2x-4y+1+4=9+16` `Rightarrow x^(2)+y^(2)-2x-4y-20=0` Now, radius `=sqrt((-1)^(2)+(-2)^(2)+20)` `=sqrt(1+4+20)=5` `therefore ` Area of circle `=pir^(2)` `=25pi"sq unit"` |
|
| 47. |
Let `f(x)=(ax + b )/(cx+d)`. Then the `fof (x)=x`, provided that : `(a!=0, b!= 0, c!=0,d!=0)`A. d=-aB. d=aC. a=b=c=d=1D. a=b=1 |
|
Answer» Correct Answer - D Given, `f(x)=(ax+b)/(cx+b)` `and fof(x)=x` `Rightarrow f((ax+b)/(cx+b))=x` `Rightarrow (a((ax+b)/(cx+b))+b)/(c((ax+b)/(cx+b))+d)=x` `Rightarrow (x(a^(2)+bc)+ab+bd)/(x(ac+cd)+bc+d^(2))=x` `Rightarrow d=-a` |
|
| 48. |
The slope of the tangent at `(x , y)`to a curvepassing through `(1,pi/4)`is given by`y/x-cos^2(y/x),`then theequation of the curve is(a) `( b ) (c) y=( d ) (e)tan^(( f ) (g)-1( h ))( i )(( j ) (k)log(( l ) (m) (n) e/( o ) x (p) (q) (r))( s ))( t )`(u)(v) `( w ) (x) y=x (y) (z)tan^(( a a ) (bb)-1( c c ))( d d )(( e e ) (ff)log(( g g ) (hh) (ii) x/( j j ) e (kk) (ll) (mm))( n n ))( o o )`(pp)(qq)`( r r ) (ss) y=x (tt) (uu)tan^(( v v ) (ww)-1( x x ))( y y )(( z z ) (aaa)log(( b b b ) (ccc) (ddd) e/( e e e ) x (fff) (ggg) (hhh))( i i i ))( j j j )`(kkk)(d) none of theseA. `y=tan^(-1)[log((e)/(x))]`B. `y=x tan^(-1)[log((x)/(e))]`C. `y=x tan^(-1)[log((e)/(x))]`D. None of these |
|
Answer» Correct Answer - C According to the given condition, `(dy)/(dx)=(y)/(x)-cos^(2)((y)/(x))` On putting y=mx `Rightarrow (dy)/(dx)=b+x(dv)/(dx),` we get `v+x(dy)/(dx)=v-cos^(2)v` `Rightarrow (dv)/(cos^(2)v)=-(dx)/(x)` `Rightarrow sec^(2)vdv=(-1)/(x)dx` On integrating both sides, we get `tan v=-log x+log x` `Rightarrow tan((y)/(x))=-log x+log c` Since, this curve is pasing through `(1,pi//4)`, `therefore tan((pi)/(4))=-log1+log cRightarrow log c=1` `therefore tan ((y)/(x))=-log x+1` `Rightarrow tan((y)/(x))=-logx+log x` `Rightarrow y=xtan ^(-1)[log((e)/(x))]` |
|
| 49. |
`int (x+sin x)/(1+cos x)` is equal toA. `x tan (x)/(2)+c`B. `log (1+cos x)+c`C. `cot (x)/(2)+c`D. `log(x+sinx)+c` |
|
Answer» Correct Answer - A Let `I=int (x+sin x)/(1+cosx)dx` `=int((x)/(2)sec^(2)""(x)/(2)+tan""(x)/(2))dx` `=x tan""(x)/(2)-inttan""(x)/(2)dx+inttan""(x)/(2)dx+c` `=x tan""(x)/(2)+c` |
|
| 50. |
The derivative of log|x| isA. `(1)/(x), x gt 0`B. `(1)/(|x|), x ne 0`C. `(1)/(x), x ne 0`D. None of these |
|
Answer» Correct Answer - C We have `y=log|x|={{:(,log x,x gt0),(,log(-x),x lt ):}` `therefore (dy)/(dx)={{:(,(1)/(x), x gt0),(,(1)/(-x)(-1)=(1)/(x),x lt 0):}` `Rightarrow (dy)/(dx)=(1)/(x), x ne 0` |
|