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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.
| 1. |
Describe anomalous behaviour of fluorine with the other elements of group 17 with reference to : (a) Hydrogen bonding (b) Oxidation state (c) Polyhalide ions. |
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Answer» Anomalous behaviour of flurone with the other elements of group 17 : Fluorine shows anomalous behaviour with the other elements of group 17 due to : (1) small size, (2) high electronegativity and (3) non-availability of - orbitals in its valence shell. (a) Hydrogen bonding : O account of high electronegativity of fluorine atom, extensive hydrogen bonding occurs in HF molecule while it is absent in HCl, HBr and HI. The presence of hydrogen bonding in HF is explained as under hydrogen bonding in HF is explained as under : (1) HCl,HBr and HI are gaseous while HF is a liquid with an abnormally high boiling points. (2) HF forms a number of compounds containing `HF_(2)^(-)` ion (e.g., `KHF_(2)`) while such compounds are not given by other HX molecules. `HF* * * * * * * * H-F * * * * * * * *underset("H-bond")underset(uarr)(H)-F" "(HF)_(n)` (b) Oxidation state : Due to maximum electronegativity of fluorine, it shows only a negative oxidation state of - 1. It does not show any positive oxidation state. The other members of group 17 show negative as well positive oxidation states of `+1,+3," "+5and +7`. (c) Polyhalide ions : Because of the absence of d-orbitals in its valence shell, fluorine does not combine with `F^(-)` ions to give polyhalide ions (like `F_(3)^(-)` ) while other members of group 17 give such polyhalide ions like `Cl_(3)^(-),Br_(3)^(-),I_(3)^(-),I_(5)^(-)` etc., because they contain d-orbilats and also there electronegativity is low. |
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| 2. |
If `y=1-costheta,x=1-sintheta,"then "(dy)/(dx)" at "theta=(pi)/(4)" is"`A. `-1`B. 1C. `(1)/(2)`D. `(1)/(sqrt2)` |
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Answer» Correct Answer - A `y=1-cos theta rArr (dy)/(d theta)=sin theta,` `x=1-sin theta rArr (dx)/(d theta)=-cos theta` `therefore" "(dy)/(dx)=- tan theta` `" "theta=(pi)/(4)` `therefore" "[(dy)/(dx)]_(theta=(pi)/(4))=-tan.(pi)/(4)=-1` |
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| 3. |
If the function f(x) is continuous in the interval `[-2, 2].` find the values of a and b where `{:(f(x)=(sinax)/(x)-2," ,for "-2 le x lt0),(=2x+1," ,for "0 le x le1),(=2bsqrt(x^(2)+3)-1," ,for " 1 lt x le 2):}` |
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Answer» `underset(xrarr0^(-))(lim)f(x)=underset(xrarr0^(-))(lim)(sinax)/(x)-2` `=underset(xrarr0^(-))(lim)(a(sinax)/(ax)-2)` `=axx1-2 (because underset(xrarr0)(lim)(sinx)/(x)=1)` `=a-2` `underset(xrarr0^(+))(lim)f(x)=underset(xrarr0^(+))(lim)(2x+1)` `=1` Function is continuous at x = 0 `underset(xrarr0^(-))(lim)f(x)=underset(xrarr0^(+))(lim)f(x)` `a-2=1` `a = 3` Again, `underset(xrarr1^(-))(lim)f(x)=underset(xrarr1^(-))(lim)(2x+1)` = 3 `underset(xrarr1^(+))(lim)f(x)=underset(xrarr1^(+))(lim)2bsqrt(x^(2)+3-1)` `=2b sqrt4-1=4b-1` Function is continuous at x = 1 Then `4b-1=3` `rArr" "4b=4` `rArr" "b=1.` |
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| 4. |
An open box is to be made out of a piece of a square card board of sides 18 cm by cutting off equal squares from the corners and turning up the sides. Find the maximum volume of the box. |
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Answer» Let each side of the square cut off from each corner be x cm. Then volume of box `V=(18-2x)(18-2x)x` `V=(18-2x)^(2)x` `V=4x^(3)+324x-72x^(2)" …(i)"` Differentiating w.r.to x, we get `(dV)/(dx)=12x^(2)+324-144x` `(dV)/(dx)=12(x^(2)-12x+27)" ...(ii)"` For maximum volume, `(dV)/(dx)=0` `rArr" "12(x^(2)-12x+27)=0` `rArr" "x^(2)-9x-3x+27=0` `rArr" "(x-9)(x-3)=0` `rArr" "x=9,3` Again differentiating, we get `(d^(2)V)/(dx^(2))=2x-12" ...(iii)"` at x = 9 `(d^(2)V)/(dx^(2))=+ve` `therefore " V is minimum at x = 9"` at x = 3 `(d^(2)V)/(dx^(2))=-ve` `therefore" V is maximum at x = 3"` `therefore" Maximum volume V "=(18-6)(18-6)xx3` `" "=12xx12xx3=432cm^(3)` |
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| 5. |
`intsec^nxtanxdx,n!=0` |
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Answer» `I=int sec^(n)x tan x dx` `I=intsec^(n-1)x sec x tanx dx` Let `sec x= t rArr sec x tan x dx=dt` `therefore" "I=int l^(n-1)dt` ltbr. `=(t^(n))/(n)+c` `=(sec^(n)x)/(n)+c` |
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| 6. |
Solve the following differential equations : (1) ` (dy)/(dx) = (y + sqrtx^(2)+y^(2))/x ` |
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Answer» `(dy)/(dx)=(y+sqrt(x^(2)y^(2)))/(x)" …(i)"` Putting `y=vx rArr (dy)/(dx)=v+x(dv)/(dx)` `therefore" "v=(y)/(x)` `therefore" "` Equation (i) becomes, `v+x(dv)/(dx)=(vx+sqrt(x^(2)+v^(2)x^(2))/(x))` `therefore" "v+x(dv)/(dx)=v+sqrt(1+v^(2))` `rArr" "x(dv)/(dx)=sqrt(1+v^(2))` `rArr" "(1)/(sqrt(1+v^(2)))dv=(1)/(x)dx` On integration, we get `int(1)/(sqrt(1+v^(2)))dv=int(1)/(x)dx` `rArr" "log lv+sqrt(1+v^(2))|=log|x|+logx` `rArr" "log|(y)/(x)+sqrt(1+(y^(2))/(x^(2)))|=log|cx|` `rArr " "y+sqrt(x^(2)+y^(2))=cx^(2)` This is the general solution. |
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| 7. |
The probability mass function (p. m. f.) of X is given below : `|{:(X=x,1,2,3),(P(X=x),(1)/(5),(2)/(5),(2)/(5)):}|` find `E(X^(2))` |
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Answer» `E(X^(2))=SigmaX^(2)P(X)` `E(X^(2))=1^(2)xx(1)/(5)+2^(2)xx(2)/(5)+3^(2)xx(2)/(5)` `=(1)/(5)+(8)/(8)+(18)/(5)=(27)/(5)` |
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| 8. |
Evaluate `int(x^2+1)/((x^2+2)(2x^2+1))dx` |
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Answer» `I=int(x^(2))/((x^(2)+2)(2x^(2)+1))dx` Let`" "(x^(2))/((x^(2)+2)(2x^(2)+1))=(A)/(x^(2)+2)+(B)/(2x^(2)+1)` Put `x^(2)=y` `(y)/((y+2)(2y+1))+(A)/(y+2)+(B)/(2y+1)` `y=A(2y+1)+B(y+2)` Put `y+2 =0 rArr y=-2` `-2 = a(-3)+0` `therefore" "A=(2)/(3)` Put `2y+1=0 rArr y=-(1)/(2)` `-(1)/(2)=0+B(-(1)/(2)+2)` `-(1)/(2)=B((3)/(2))` `therefore" "B=-(1)/(3)` `therefore" "(y)/((y+2)(2y+1))=(2)/(3(y+2))-(1)/(3(2y+1))` Then `(x^(2))/((x^(2)+2)(2x^(2)+1))=(2)/(3(x^(2)+2))-(1)/(3(2x^(2)+1))` `int(x^(2))/((x^(2)+2)(2x^(2)+1))dx=int(2)/(3(x^(2)+2))dx-int(dx)/(3(2x^(2)+1))` `=(2)/(3)int(dx)/(x^(2)+(sqrt2)^(2))-(1)/(3)int(dx)/(2x^(2)+1)` `=(2)/(3)xx(1)/(sqrt2)tan^(-1).(x)/(sqrt2)-(1)/(3xx2)int(dx)/(x^(2)+((1)/(sqrt2))^(2))` `=(sqrt2)/(3)tan^(-1).(x)/(sqrt2)-(1)/(6)xx(1)/(1//sqrt2) tan^(-1).(x)/(1//sqrt2)+c` `=(sqrt2)/(3)tan^(-1).(x)/(3sqrt2)tan^(-1)sqrt2 x+c` |
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| 9. |
Obtion the differential equation by elininating arbitrary constants A, B from the equation - `y=A cos (logx)+B sin (logx)` |
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Answer» `y=A cos (logx)+Bsin(logx)` Differentiating w.r. to x, we get `(dy)/(dx)=-A sin(logx)xx(1)/(x)+B cos (logx)xx(1)/(x)` `x(dy)/(dx)=-Asin(logx)+B cos (logx)` Again differentiating, we get `x(d^(2)y)/(dx^(2))+(dy)/(dx)=-Acos(logx)xx(1)/(x)-B sin (log x)xx(1)/(x)` `x^(2)(d^(2)y)/(dx^(2))+x(dy)/(dx)=-[A cos (logx)+B sin (logx)]` `x^(2)(d^(2)y)/(dx^(2))+x(dy)/(dx)=-y` `x^(2)(dy)/(dx)+y=0` |
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| 10. |
Find the equation of the tangent to the curve `y=3x^(2)-x+1` at P(1, 3).A. `y=5x+2`B. `y=5x-2`C. `y=(1)/(5)x+2`D. `y=(1)/(5)x-2` |
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Answer» Correct Answer - B `y=3x^(2)-x+1` `therefore" "(dy)/(dx)=6x-1` `" "((dy)/(dx))_("(1, 3)")=6-1=5` `therefore" "m=5` Equation of tangent `y=mx+c` `y=5x+c" …(i)"` Put (1, 3) in equaiton (i), `3=5xx1+c` `therefore" "c=-2` Equation of tangent `y=5x-2` |
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| 11. |
Examine the continuity of the funciton `f(x)=sinx - cos x, " for "x ne 0` `=-1" ,for "x =0` at the point x = 0 |
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Answer» `" L.H.L."=underset(xrarr0^(-))(lim)f(x)` `=underset(xrarr0^(-))(lim)(sinx - cosx)` `=0-1` `=-1` `" R.H.L."=underset(xrarr0^(+))(lim)f(x)` `=underset(xrarr0^(+))(lim)(sinx-cosx)` `=-1` `underset(xrarr0)(lim)f(x)=-1" (Given)"` `"L.H.L. = R.H.L. "=underset(xrarr0)f(x)` So, function is continuous at x = 0. Hence proved. |
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| 12. |
A random variable X has the following probability distribution : (a) Find k (b) Find P (O |
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Answer» (a) `K+3K+5K+7K+9K+11K+13K=1` `therefore" "49K=1` `therefore" "K=(1)/(49)` (b) `P(0 lt x lt 4)=P(1)+P(2)+P(3)` `=(3)/(49)+(10)/(49)+(21)/(49)` `=(34)/(49)` (c) `c.d.f = (2x+1)K,, 0 le x` |
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| 13. |
If `x^(p) y^(q) = (x + y)^((p + q)) " then " (dy)/(dx)=` ? |
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Answer» `x^(p)y^(q)=(x+y)^(p+q)` take log on both sides, we get `plog x+q log y = (p+q)log(x+y)` differentiate w.r.t. x `(p)/(x)+(q)/(y)(dy)/(dx)=(P+q)/(x+y)(1+(dy)/(dx))` `(dy)/(dx)((q)/(y)-(p+q)/(x+y))=(P+q)/(x+y)-(p)/(x)` `(dy)/(dx)((qx-py)/(y(x+y)))= (qx-py)/(x(x+y))` `(dy)/(dx)=(y)/(x)` Hence proved . |
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| 14. |
Name any two edible varieties of mushroms. Give nutritional values of these mushrooms. |
| Answer» The two edible varieties of Mushrooms are Agaricus bisporus and Lentinula edodes. The mushrooms are rich in protein and poor in fat content. The protein values range from `1.75-3.63%` of the total fresh weight. They are also a good source or the total fresh weight. They are also a good source of iron, phosphorus and vitamins. Mushrooms are rich in essential amino acids. | |
| 15. |
The equation of plane passing through the line of intersection of planes `2x-y+z=3,4x-3y-5z+9=0` and parallel to the line `(x+1)/(2)=(y+3)/(4)=(z-3)/(5)` is |
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Answer» Equation of the plane passing through the line of intersection of planes `(2x-y+z)-3=0` and `4x-3y+5z+9=0` is `(2x-y+z-3)+lambda (4x-3y+5z+9)=0" …(i)"` `(2+4lambda)x-(1+3lambda)y+(1+5lambda)z-3+9lambda=0` Since this plane is parallel to the line `(x+1)/(2)=(y+3)/(4)=(z-3)/(5)` whose direction ratio are 2, 4, 5. `therefore `The normal to the plane is perpendicualr to this line. `therefore (2+4lambda)2+[-(1+3lambda)]4+(1+5lamba)5=0` `rArr 4+8lambda-4-12lambda+5lambda25lambda=0` `rArr" "5+21lambda =0` `rArr" "lambda=(-5)/(21)` Substituting `lambda =(-5)/(21)` in equation (i), we get `(2x-y+z-3)+((-5)/(21))(4x-3y+5z+9)=0` `l42x-21y+21z-63-20x+15y-25z-45=0` `22x-6y-4z-108=0` `2(11x-3y-2z-54)=0` `11x-3y-2z-54=0` |
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| 16. |
To get `n`-type doped semiconductor, impurity to be added to silicon should have the following number of valence electronsA. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - A The outer shell of Si have `3p^(2)` electrons, addition of impurity with 2 valence electrons leads to half - filled `3p^(3)` orbital and an extra electron which conducts electricity on passage of current through Si semiconductor. Since the current flowing in the semiconductor after doping is due to electron, it will be n-type semiconductor. |
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| 17. |
For formation of 50 seeds, how many minimum meiotic divisions are necessary?A. 25B. 50C. 75D. 63 |
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Answer» Correct Answer - A 25 |
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| 18. |
From the visible spectrum of light, which component is reflected by the green leaves ?A. BlueB. RedC. GreenD. Orange |
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Answer» Correct Answer - C Green |
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| 19. |
How is methoxyethane prepared from : (a) Methyl iodide (b) Diazomethane |
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Answer» (a) Methyl Iodide `to` Methoxyethane `underset("ethoxide")underset("Methyl")(CH_(3)I)+underset("ethoxide")underset("Potassium")(C_(2)H_(5)O^(-)K^(+))tounderset()underset("Methoxyethane")(C_(2)H_(5)-O)-CH_(3)+KI` (b) Diazomethane `to` Methoxythane `underset("(Diazomethane)")(CH_(2)N_(2))+underset("(Ethanol)")(C_(2)H_(5)OH)underset(Delta)overset(BF_(3))tounderset("Methoxyethane")(C_(2)H_(5)-O-CH_(3))` |
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| 20. |
IUPAC name of `K_(4)|Fe(CN)_(6)|` isA. Tetrapotassium ferrocyanideB. Potassium ferricyanideC. Potassium ferrocyanideD. Potassium hexacyanoferrate |
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Answer» Correct Answer - D The IUPAC name of `K_(4)[Fe(CN)_(6)_(6)]`, Potassium hexacyanoferrate. |
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| 21. |
Carbon atom in methyl carbocation contains how many pairs of electrons ?A. 8B. 4C. 3D. 5 |
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Answer» Correct Answer - C In methl carbocation `(.^(+)CH_(3))`, the central, carbon atom is `sp^(2)` hybridized with empty p-orbital A. methyl group has six electrons in the three carbon - hydrogen bond. Therefore carbon atom in methyl carbocation contains 3 pairs of electrons. |
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| 22. |
Which of the following is not a colligative property?A. Vapour pressureB. Depression in freezing pointC. elevation in boiling pointD. Osmotic pressure |
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Answer» Correct Answer - A The four colligative properties are - (i) Relative lowering of vapour pressure, (ii) Elevation of boiling points, (iii) Depression of freezing point and (iv) Osmotic pressure. |
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| 23. |
A system absorbs 640 J heat and does work of 260 J, the change in internal energy of the system will be :A. `+380J`B. `-380J`C. `+900J`D. `-900J` |
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Answer» Correct Answer - B The system has been given internal energy of 640 J and the work system does on surroundings is 260 J. Therefore, by first law of thermodynamics the internal energy of system increases by `-q=DeltaU-W=(-640+260)=-360J` |
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| 24. |
The rate of reaction of certain reaction is expressed as : `(1)/(3)(d[A])/(dt)=-(1)/(2)(d[B])/(dt)=-(d[C])/(dt)` The reaction is :A. `3Ato2B+C`B. `2Bto3A+C`C. `2B+Cto3A`D. `3A+2BtoC` |
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Answer» Correct Answer - C For a reaction `aA+bBtoc C`, the rate of reaction is expressed as - Rate of reaction `=(-1)/(a)(d[A])/(dT)=(-1)/(b)(d[B])/(dT)=(+1d[C])/(c" "dT)` Thus, for a given rate expression the reaction is `2B+Cto3A` |
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| 25. |
What is molecular formula of oleum ?A. `H_(2)SO_(3)`B. `H_(2)SO_(4)`C. `H_(2)S_(2)O_(7)`D. `H_(2)S_(2)O_(8)` |
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Answer» Correct Answer - C The molecular formula of oleum is `H_(2)S_(2)O_(7)`. |
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| 26. |
Number of faradays of electricity required to liberate 12 g of hydrogen is :A. 1B. 8C. 12D. 16 |
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Answer» Correct Answer - C A Faraday is defined as the amount of electric charge in one mole of electrons. Since 1F of electricity is required to liberate 1g of hydrogen thus, 12 Faraday of electricity is required to liberate 12 g of hydrogen. |
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| 27. |
Identify the weakest base amongst the following :A. p-methoxyanilineB. o-totuidineC. Benzene-1, 4-diamineD. 4-aminobenzoic acid |
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Answer» Correct Answer - A p-methoxyaniline group, donates electrons to the benzene ring thereby increasing group `(-OCH_(3))` being the electron donating group, donates electrons to the benzene ring thereby increasing the concentration of electrons on the ring and thus the attack of nucleophile becomes difficult. |
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| 28. |
Gene flow. |
| Answer» Gene Flow, also called migration, is the movement of genes from a genetically different population to another and is important for genetic variations. | |
| 29. |
Why do sex linked traits appear more in males than in females ? |
| Answer» The sex linked traits are recessive. A male having a X-linked recessive allele expresses the trait as his Y-chromosome cannot hide the expression of X-linked disorder due to the absence of counteracting allele. | |
| 30. |
What are jumping genes ? |
| Answer» Some repetitive DNA sequences change their positions in DNA. These are jumping genes or Transposons. | |
| 31. |
Pregnancy in second trimester is maintained by ………………………..A. LH (lutenizing hormone)B. progesteroneC. estrogenD. HCG (human chorionic gonadotropin) |
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Answer» Correct Answer - B progesterone |
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| 32. |
Find the measure of a acute angle between the line direction ratios are 5, 12, -13 and 3, -4, 5. |
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Answer» `a_(1)=5, b_(1)=12, c_(1)=-13` `a_(2)=3, b_(2)=-4, c_(2)=5` Acute angle, `cos theta=|(a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2))/(sqrt(a_(1)^(2)+b_(1)^(2)+c_(1)^(2))sqrt(a_(2)^(2)+b_(2)^(2)+c_(2)^(2)))|` `rArr" "cos theta=|(15-48-65)/(sqrt(25+144+169)sqrt(9+16+25))|` `rArr" "cos theta=|(-98)/(13sqrt2xx5xxsqrt2)|` `rArr" "cos theta=|(-98)/(13xx5xx2)|` `rArr" "cos theta =|-(49)/(65)|` `rArr" "cos theta =(49)/(65)` `therefore" "theta=cos^(-1)((49)/(65))` |
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| 33. |
Write the dual of the following statements : (i) `(p vvq)^^T` (2) Madhuri has curly hair and brown eyes. |
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Answer» Dual statements are (1) `(p ^^ q) vv T` (2) Madhuri has curly hair or brown eyes. |
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| 34. |
write balanced chemical equations for action of potassium permanganate on : (a) Hydrogen (b) Warm conc. Sulphuric acid Explain why `Mn^(2+)` ion is more stable than `Mn^(3+)` ? (given : `MntoZ=25` ) |
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Answer» (a) When potassium permanganate `KMnO_(4)`, is heated in a current of `H_(2)` , KOH, MnO and steam are formed. `2KMnO_(4)+5H_(2)to2KOH+underset("oxide")underset("Managanese")(2MnO)+4H_(2)O`. (b) When `KMnO_(4)` is heated with warm concentrated `H_(2)SO_(4),K_(2)SO_(4),MnSO_(4)` and five O-atoms rendered available from `MnO_(4)^(-)` ions are formed. `2KMnO_(4)+underset(("Warm conc."))(3H_(2)SO_(4))toK_(2)SO_(4)+2MnSO_(4)+3H_(2)+5O` `Mn^(2+)` more stable than `Mn^(3+)` : The electronic configuration of `Mn_(25)` is `[Ar]_(18)4s^(2)` `3d^(5),Mn_(23)^(2+)` is `[Ar]_(18)4s^(0)3d^(5)andMn_(22)^(3+)to[Ar]_(18)4s^(0)` `3d^(4)`. We know, half filled or full electronic configuration is more stable than partially filled electronic configuration. `Mn^(2+)` has partially filled electronic configuration. `Mn^(2+)` has half filled d-orbital whereas `Mn^(3+)` has partially filled d-orbital. Hence, `Mn^(2+)` is more stable than `Mn^(3+)`. |
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| 35. |
Bakelite is a polymer ofA. Benzaldehyde and phenolB. Acetaldehyde and phenolC. Formaldehyde and phenolD. Formaldehyde and benzyl alcohol |
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Answer» Correct Answer - C Bakelite is cross-linked condensation polymer of phenol `(C_(6)H_(5)-OH)` and formaldehyde (HCHO). It is thermosetting plastic used for making electrical switches and switch boards. |
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| 36. |
Formalin is `40%` aqueous solution of :A. MethanalB. Methanoic acidC. MethanolD. Methanamine |
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Answer» Correct Answer - A Methanal or formaldehyde is soluble in water, alcohol and ether. Its `40%` aqueous solution is marketed as formalin. |
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| 37. |
The general solution of the trigonometric equation `tan^2theta=1` isA. `theta=n pi pm(pi)/(3), n in z`B. `theta = npi pm(pi)/(6), n in z`C. `theta=n pi pm (pi)/(4), n in z`D. `theta = n pi, n in z` |
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Answer» Correct Answer - C `tan^(2)theta = tan alpha` `theta= n pi pm alpha` `tan^(2)theta=1` `tan^(2) theta=tan^(2).(pi)/(4)` `therefore" "theta=n pi pm(pi)/(4)` |
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| 38. |
The cartesian equation of a line is `(x - 6) /(2) = ( y + 4)/(7) = (z -5)/(3)` , find its vector equation . |
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Answer» Equation of line `(x-6)/(2)=(y+4)/(7)=(z-5)/(3)` `therefore `The line is passing through the point `(6,-4,5)` and having direction ratios `2, 7, 3.` `"so "veca=6hati - 4hatj+5hatk` `"and "vec b=2hati+7hatj+3hatk` `therefore "The vector eqaution of line is " vecr=veca+lambda vecb` `rarr vecr = (6hati-4hatj+5hatk)+lambda(2hati+7hatj+3hatk)` |
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| 39. |
If `bar(a),bar(b),bar(c)`are the position vectors of the points A, B, C respectively and `2bar(a)+3bar(b)-5bar(c)=bar(0)`, then find the ratio in which the point C divides line segment AB. |
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Answer» Let the ration be `lambda :1` Position vector of point (C) `vecc=(lambda veca+vecb)/(lambda+1)` Put the value of `vecc" in "2veca +3 vecb-5vecc=0` `2veca+3vecb-5xx((lambda veca+vecb)/(lambda+1))=0` `rArr" "((lambda+1)(2veca+3vecb)-5lambda veca- 5vecb)/((lambda+1))=0` `rArr 2veca lambda+3vecb lambda+2veca +3vecb-5lambda veca-5vecb=0` `rarr" "3vecb lambda-3vec a lambda+2veca-vecb=0` `rArr" "3lambda(vecb-veca)=2vecb-2veca` `rArr" "3lambda =(2(vecb-veca))/((vecb-veca))` `rArr" "(lambda)/(1)=(2)/(3)` Hence the point C divides line segment AB in ratio `2:3` |
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| 40. |
There is a hole in the ozone layer. What do you understand by this? |
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Answer» 1. The Ozone layer is found in the upper layer of the atmosphere, the stratosphere. 2. This layer acts as a shield absorbing UV radiations from the sun. 3. The Ozone gas is formed by the action of UV rays on molecular oxygen and also degraded into molecular oxygen in the stratosphere. 4. For the Ozone layer to be intact there has to be balance between the production and degradation, this has been disturbed by the CFCs (Chlorofloro carbons) used in refrigerators. 5. The CFCs react with UV rays to form Cl atoms. They act as a catalysts that degrade the ozone releasing molecular oxygen. 6. The Cl has a permanent and continuing effect on the ozone levels as it is not consumed thus resulting in the thining out of the ozone layer also known as ozone hole. |
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| 41. |
The permanent removal of forests and woodlands is called………………..A. reforestationB. afforestationC. deforestationD. agroforestry |
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Answer» Correct Answer - C deforestation |
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| 42. |
Abundance of phosphate, causing algal overgrowth, resulting in depletion of oxygen and killing other aquatic life is known as …………….A. ecological successionB. eutrophicationC. guano depositsD. greenhouse effect |
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Answer» Correct Answer - B eutrophication |
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| 43. |
BIOMAGNIFICATION |
| Answer» Biomagnification is the increase in concentration of a pollutant that is non-biodegradable in each higher trophic level of a food chain. Example : DDT and Mercury. | |
| 44. |
Why is zona pellucida retained around the egg till it reaches uterus? |
| Answer» Zona pellucida is retained around the egg till it reaches the uterus to prevent implatation of the blastocyst at an abnormal site and it also protects the ovum. | |
| 45. |
The biological scissors is ………………….A. restriction endonucleaseB. gyraseC. DNA ligaseD. helicase |
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Answer» Correct Answer - A restriction endonuclease |
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| 46. |
Write a note on desert adaptations. |
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Answer» 1. The animals living in desert store water in muscles, connective tissue of the hump and water cells of stomach. 2. Animals show physiological adaptations such as losing minimum water through urine. 3. In camel, which is a desert animal, such types of adaptations are seen. 4. In the kidney, there are longer loops of Henle in the nephrons for more reabsorption of water. |
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| 47. |
Give the ecomoic importance of fisheries |
| Answer» Economic importance of fisheries : | |
| 48. |
Dead and dried cell mass of microbes having nutritive value is also known as ………………..A. BGA (blue - green algae)B. SCP (single cell protein)C. STP (sewage treatment plant)D. VAM (vesicular arbuscular mycorrhizae) |
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Answer» Correct Answer - B SCP (single cell protein) |
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| 49. |
The most common types of fossils are ……………A. mouldsB. castsC. actual remainsD. models |
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Answer» Correct Answer - C actual remains |
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| 50. |
Give applications of a vaccine. |
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Answer» Applications of a vaccine : 1. Vaccines provide immunity against a particular disease by evoing the immune system of human and animals. 2. Vaccines for rabies, malaria, hepatitis B, polio virus and also small pox are produced using biotechnology. |
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