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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. |
Transfer of genetic material into a bacterial cell through a viral vector is know as .....A. transformationB. transductionC. transfectionD. translation |
| Answer» Correct Answer - b | |
| 2. |
If the number of chromosomes in an endosperm cell is 27 , what will be the chromosome number in the definitive nucleus ?A. 9B. 18C. 27D. 36 |
| Answer» Correct Answer - b | |
| 3. |
The surface tension of water at ` 0^(@) C` is 75.5 dyn/cm. Calculate the surface tension of water at ` 25^(@)C` . |
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Answer» Given : `T_(0)=75.5` dyne/cm, `alpha=0.0027//^@C, theta = 25^@C` Thus, `T=T_(0)(1=-alphatheta)=75.5[1-(0.0027)(25)]=75.5[1-0.0675]` `75.5(0.9325)` `=70.40` dyne/cm Thus, the surface tensiono of water at `25^@C` is 70.40 dyne/cm. |
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| 4. |
The graph between applied force and change in the length of wire within elastic limit is aA. straight line with positive slopeB. straight line with negative slopeC. curve with positive slope.D. curve the negative slope |
| Answer» Straight line with positive slope | |
| 5. |
A particale in S.H.M. has a perod of 2 seconds and amplitude of 10 cm. Calculate the acceleration when it is at 4 cm from its positive extreme position. |
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Answer» Given : A = 10 cm, T = 2 sec. ltbngt Acceleration, `a=omega^2((2pi)/T)^2x,` `=(2xx22/7xx1/2)"^2,x` Practicle is 4 cm from its positive exterme position. `x=A-4` `=10-4=6cm ` `a=(2xx22/7xx1/2)^2*6` `a=(22xx22)/(49)xx6=2904)/49` `a=59.29`m//s^2` |
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| 6. |
When a longistudinal wave is incident at the bondary of a denser medium, thenA. compression reflects as a compresiono.B. compression reflects as a rarefaction.C. rarefaction reflects as a compressionD. longitudinal wave reflects as trnsverse wave. |
| Answer» Compression reflects as a compression | |
| 7. |
The dimensions of universal gravitational constant are :-A. `[L^1M^0T^0]`B. `[L^2M^1T^0]`C. `[L^(-1)M^(1)T^(-2)]`D. `[L^3M^(-1)T^(-2)]` |
| Answer» Correct Answer - `[L^3M^(-1)T^(-2)]` | |
| 8. |
In circular motion, assuming `barv=barwxxbar r`, obtain an expressio for the resultantj acceleration of a particle in terms of tangential and radial component. |
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Answer» Let the particle be moving around a circular path of constant radiuss r. If it speeds up or slows down, speed `(omega)` and the linear speed (v) change with time. Change with time. Thus, at any instant, `omega` v and r are related as : `vecv=vecomegaxxvecr` On differentiating, we get `(dvecv)/("dt")=d/("dt")(vecr+vecomega)` `(dvecv)/("dt")=d/("dt")xx(vecr+vecomega)xx(dvecr)/("dt")` `(dvecomega)/("dt")=vecalpha"and"(dvecr)/("dt")=vecr` `:." "(dvecv)/("dt") =vecalphaxxvecr+vecomegaxxvecv` `vecalphaxxvecr` is along the tangent of circumference of circular path, it si tangential acceleration `veca_(T)`. `vecomegaxxvecv` is along the radius of circle, pointing towards centre, it is radial accearation `veca_(r)`. ` veca=veca_(T)+veca_(r)` |
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| 9. |
A circular coil of 300 turns and average area `5xx10^(-3)m^(2)` carries a current of 15 A. Calculater the magnitude of magnetic moment associated with coil. |
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Answer» Given : no. of turns, N = 300, area, `A=5xx10^(-3)m^(-2)`, Current, `I=15A` The magnetic moment of the coil, M=NIA `=300xx15xx5xx10^(-3)` `M=22*5"Am"^2`. |
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| 10. |
Two copperr spheres of radii 6 cm and 12 cm respectively are suspended in an evacuated enclosure. Each of them are at a temperature `15^(@)C` above the surroundings. The ratio of their rate of loss of heat isA. `2:1`B. `1:4`C. `1:8`D. `8:1` |
| Answer» Correct Answer - `1:4` | |
| 11. |
What is a Polaroid ? State its two uses. |
| Answer» A polarid is a synthetic dichroic sheet that can polarise a transmitted beam of normal light becaused it is composed of long parallel molecules. It only transmits planepolarised light if these moleculaes are parabllel to the plane of polarisation. It is often used in sunglasses to eliminatte glare and in 3-D motion pictures i.e. holography. | |
| 12. |
Define the following terms : (a) Enthalpy of fusion Enthalpy of atomization |
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Answer» (a) Enthaply of fusion : The enthalpy change or amount of heat absorbed that accompanies the fusion of one mole of a solid at constant temperature and pressure is called enthalpy of fusion. If is denoted by `(Delta_("fus")H)`. (b) Refer to HSC Paper, March 2018, Answer 4. (i) (b). |
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| 13. |
Derive the relation between half life and rate constant for a first order reaction. |
| Answer» Correct Answer - M/A | |
| 14. |
Find the acute angle between the planes `barr*(2hati + hatj - hatk) = 3` and `barr*(hati + 2hatj + hatk) =1`. |
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Answer» Angle between two planes is given by ` 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)))` ` A_(1) = 2, B_(1) = 1, C_(1) = - 1` ` A_(2) = 1, B_(2) = 2, C_(2) = 1` `cos theta = (|(2 xx 1)+(1xx2)+(-1 xx1)|)/(sqrt((2)^(2)+(1)^(2)+(-1)^(2))sqrt((1)^(2)+(2)^(2)+(1)^(2)))` ` cos theta = 3/6 = 1/2 ` ` :. theta = pi/3` ` :. ` Angle between the planes is ` pi/3`. |
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| 15. |
Which are the bacteria responsible for converting organic acids into methane? |
| Answer» Anaerobic bacteria are resposible for converting organic acids to methane . | |
| 16. |
Name the enzyme responsible for delay for ripening of tomato fruit . |
| Answer» Polygalacturonae (PG) is the major enzyme responsible for causing the delay in ripening of tomato fruit . | |
| 17. |
What are antibiotics? Give two examples. |
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Answer» Antibiotics : Antibiotic is drug derived from living matter or micro-organism, used to kill or prevent the growth of other micro-organisms. (i) Broad spectrum antibiotics : Chloramphenicol, Ampicillin. (ii) Narrow spectrum antibiotics : Penicillin |
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| 18. |
Define : (a) Moganetization and (b) Magnetic intensity. |
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Answer» (a) The magnetization of a material may be difined as the net megnetic dipole moment per unit volume of the material. `"Magnetization"=("Net megnetic moment")/("Volume")" "vecM_(Z)" "=vecM_("net")/("Volume")` (b) The magnetic intensity (H) is defined as the product of the number of turns per unit length in a coil (n) and the current that it carries (I). Ist S.I. unit is A/m. Hence, H = n |
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| 19. |
Enlist the various types of cancer . |
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Answer» There are five types of cancer , viz. Carcinoma , sacroma , lymphoma , leukemia and adenoma . 1. Carcinoma is cancer of epithelial cells that cover internal and external parts of the body such as lung , breast and colon cancer. 2 Sarcoma is the cancer of connective tissue such as bone , cartilage , fat , muscle and other supportive tissue . 3 . Lymphoma is the cancer of lymph nodes and immune system tissues . 4 . Leukemia is cancer that begins in the bone marrow and then spreads to the blood cells . 5 . Adenoma is the cancer that arises in the throid , the pituitary gland , the adernal gland and othe r glandular tissues . |
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| 20. |
Which of the following vitamins is water soluble ?A. AB. DC. ED. B |
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Answer» Correct Answer - D Vitamin B is water soluble vitamin. Vitamin A, D and E are fat soluble vitamin. |
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| 21. |
Identify the strongest acid amongst the followings :A. Chloroacetic acidB. Acetic acidC. Trichloroacetic acidD. Dichloroacetic acid |
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Answer» Correct Answer - C The increase in the acid strength of an acid depends on the number of electron with drawing groups. Greater the number of electron withdrawing groups, greater is the strength of acid. Thus, `underset("Trichloroacetic acid")(CH_(3)C-COOH)gtunderset("Dichloroacetic acid Chloroacetic acid Acetic acid")(Cl_(2)CH-COOHgtClCH_(2)-COOH)gtCH_(3)COOH` |
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| 22. |
The solution of the differential equ `(x^2 + y^2) dx = 2 xy dy` is- |
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Answer» Given, `(x^(2)+y^(2)) dx - 2xy dy = 0 ` or ` (x^(2)+y^(2)) dx = 2xy dy` or ` (dy)/(dx) = (x^(2)+y^(2))/(2xy) ` ....(i) Let y = vx Thus `(dy)/(dx) = v + x (dv)/(dx)` Thus ` v+ x (dv)/(dx) = (x^(2)+(vx)^(2))/(2x(vx))` or ` v + x (dv)/(dx) = (1+ v^(2))/(2v)` or ` x (dv)/(dx) = (1+v^(2))/(2v) - v` or ` x (dv)/(dx) = (1+v^(2)-2v^(2))/(2v)` or ` x (dv)/(dx) = (1-v^(2))/(2v)` or ` (dx)/x = (2v)/(1-v^(2)) dv` or ` (dx)/x - (2v)/(1-v^(2)) dv = 0` ...(ii) Integrating both sides, ` log x + log (1-v^(2)) = log C` or ` log x (1 - v^(2)) = log C` or ` x (1 - v^(2)) = C` or ` x (1 - y^(2)/x^(2)) = C` or ` x ((x^(2)-y^(2))/x^(2)) = C` or ` x^(2) - y^(2) = Cx` |
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| 23. |
If ` log _(10)((x^(3)-y^(3))/(x^(3)+y^(3))) = 2`, then show that ` (dy)/(dx) = (-99x^(2))/(101y^(2))` |
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Answer» Given, `log_(10)(x^(3)-y^(3))/(x^(3)+y^(3)) = 2` Or `log_(10)(x^(3)-y^(3)) - log_(10)(x^(3)+y^(3))=2` Now, differentiating with respect to x, we have , ` 1/(x^(3)-y^(3)) (3x^(2)-3y^(2)(dy)/(dx)) - 1/(x^(3)+y^(3))(3x^(2)+3y^(2)(dy)/(dx))` or `(3x^(2))/(x^(3)-y^(3))-(3y^(2))/(x^(3)-y^(3))(dy)/(dx)-(3x^(2))/(x^(3)+y^(3)) - (3y^(2))/(x^(3)+y^(3))(dy)/(dx)=0` or ` - (3y^(2))/(x^(3)-y^(3))(dy)/(dx)-(3y^(2))/(x^(3)+y^(3))(dy)/(dx) = - (3x^(2))/(x^(3)-y^(3))+(3x^(2))/(x^(3)+y^(3))` or ` -3y^(2)(dy)/(dx)[1/(x^(3)-y^(3))+1/(x^(3)+y^(3))]` `=-3x^(2)[1/(x^(3)-y^(3))-1/(x^(3)+y^(3))]` or `y^(2)(dy)/(dx) [1/(x^(3)-y^(3))-1/(x^(3)+y^(3))]` ` = x^(2)[1/(x^(3)-y^(3))-1/(x^(3)+y^(3))]` or `y^(2)(dy)/(dx)[(x^(3)+y^(3)+x^(3)-y^(3))/((x^(3)-y^(3))(x^(3)+y^(3)))]` ` = x^(2)[(x^(3)+y^(3)-x^(3)+y^(3))/((x^(3)-y^(3))(x^(3)+y^(3)))]` or ` y^(2)(dy)/(dx) [(2x^(3))/((x^(3)-y^(3))(x^(3)+y^(3)))]` ` = x^(2)[(2y^(3))/((x^(3)-y^(3))(x^(3)+y^(3)))]` or ` x(dy)/(dx) = y` or ` (dy)/(dx) = y/ x ` ... (i) Also, ` log_(10) . (x^(3)-y^(3))/(x^(3)+y^(3)) = 2 ` or ` (x^(3)-y^(3))/(x^(3)+y^(3)) = 10^(2)` or `(x^(3)-y^(3))/(x^(3)+y^(3)) = 100` or ` x^(3) - y^(3) = 100 (x^(3)+y^(3))` or ` x^(3)-y^(3) = 100 x^(3) + 100 y^(3)` or ` -99 x^(3) = 101 y^(3)` or ` y^(3)/x^(3) = - 99/101` or ` y/x= - (99x^(2))/(101 y^(2))` ...(ii) From (i) and (ii) , we get, ` (dy)/(dx) = - (99x^(2))/(101 y^(2))` |
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| 24. |
If `y=sqrt(sinx+sqrt(sinx+sqrt(sinx+ dotto oo)))`, prove that `(dy)/(dx)=(cosx)/(2y-1)` |
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Answer» Let ` y = sqrt(sin x + sqrt(sin x + sqrt(sin x + ............infty)))` ` y = sqrt(sin x + y ) ` Squaring both sides, we get, ` y^(2) = sin x + y ` Differentiating both sides, we get ` 2 y (dy)/(dx) = (dy)/(dx) + cos x ` ` (dy)/(dx) = (cos x)/((2y-1)) ` |
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| 25. |
`int sqrt(x^2 - a^2) = 1/2 x sqrt (x^2 - a^2) - 1/2 a^2 log ( x + sqrt( x^2 - a^2) + c` |
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Answer» Given, `intsqrt(x^(2) - a^(2)) dx ` ` = sqrt(x^(2)-a^(2))(x)- int (2x)/(2sqrt(x^(2)-a^(2)))x dx ` ` = sqrt(x^(2)-a^(2))(x) - int (x^(2))/(sqrt(x^(2)-a^(2)))dx` ` = sqrt(x^(2)-a^(2))(x) - int (x^(2)-a^(2)+a^(2))/sqrt(x^(2)-a^(2))dx` ` = sqrt(x^(2)-a^(2)) (x) - int (x^(2)-a^(2))/sqrt(x^(2)-a^(2))dx + a^(2) int 1/(sqrt(x^(2)-a^(2)))dx` ` = sqrt(x^(2)-a^(2))(x) - int sqrt(x^(2)-a^(2)) dx + a^(2) int 1/(sqrt(x^(2)-a^(2)) ) dx ` or `2intsqrt(x^(2)-a^(2))dx = sqrt(x^(2)-a^(2))(x) + a^(2) int 1/(sqrt(x^(2)-a^(2)))dx` or ` int sqrt(x^(2)-a^(2))dx = 1/2 sqrt(x^(2)-a^(2))(x)+ 1/2 a^(2) log |x + sqrt(x^(2)-a^(2))|+c` where c is constant term. |
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| 26. |
Find the coordinates of the point on the curve `y=x-(4)/(x)`, where the tangent is parallel to the line `y=2x`. |
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Answer» Equation of line is given by ` y = 2x ` ….(1) We know that ` y = mx + c ` …(2) On comparing equation (2) with (1), We get , m = 2 Given, ` y = x - 4/x` Differentiating w.r.t.x ` (dy)/(dx) = 1 - 4 (-1) * 1/x^(2) ` ` = 1 + 4/x^(2) = m ` ....(3) Slope of the tangent, ` m = 1 + 4/x^(2) ` ` 1 + 4/x^(2) = 2` ` x^(2) + 4 = 2x^(2)` ` x = pm 2` ` :. x_(1) = 2, x_(2) = -2` ` y_(1) = 4, y_(2) = - 4` Co- ordinates of the point of contact are (2, 4) and (-2, -4). |
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| 27. |
Evaluate : `int_(0)^(pi) (x sin x )/(1 + sin x ) dx ` |
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Answer» Let `underset(0)overset(pi) int (x sin x dx)/(1 + sin x ) = I` ....(i) then, ` I =underset(0)overset(pi) int((pi -x)sin (pi -x))/(1+sin (pi -x))dx ` ` = underset(0)overset(pi) int ((pi -x)sin x)/(1 + sin x) dx ` ....(ii) Adding equations (i) and (ii), `2 I = underset(0)overset(pi) int (pi sin x dx)/(1 + sin x ) ` or ` I = pi/2 underset(0)overset(pi) int (sin x)/(1 + sin x) dx ` or ` I = pi/2 underset(0)overset(pi) int((1-sin x) sin x)/((1+sin x)(1-sin x)) dx` or ` I = pi/2 underset(0)overset(pi) int (sin x - sin^(2) x)/(1- sin^(2) x) dx ` or ` I = pi/2 underset(0)overset(pi) int (sin x - sin^(2)x)/(cos^(2)x) dx ` or `I = pi/2 underset(0)overset(pi)int(sin x )/(cos^(2) x) - pi/2 underset(0)overset(pi) int (sin^(2)x)/(cos ^(2) x ) dx ` or ` I = pi/2 underset(0)overset(pi) int tan x sec x dx - pi/2 underset(0)overset(pi)int (1-cos^(2)x)/(cos^(2) x) dx ` or `I = pi/2 [sec x]_(0)^(pi) - pi/2 underset(0)overset(pi)intsec^(2) x dx + pi/2 underset(0)overset(pi)int dx ` or ` I = pi/2 [sec pi - sec 0] - pi/2 [tan x]_(0)^(pi) + pi/2 [x]_(0)^(pi) ` or ` I = pi/2 [-1-1] - pi/2 [tan pi - tan0] + pi/2 [pi - 0]` or ` I = - pi - pi/2 [0-0] + pi/2[pi]` or ` I = pi [pi/2 -1]` Thus, `underset(0)overset(pi)int (x sin x dx)/(1 + sin x ) = pi [(pi -2)/2]` |
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| 28. |
If `y = (tan^(-1))^(2)`, show that `(1+x^(2))(d^(2)y)/(dx^(2)) + 2x(1+x^(2))(dy)/(dx) - 2 = 0` |
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Answer» Given, ` y = (tan^(-1) x)^(2)` Let `tan^(-1) x = z` `x = tan z ` On differentiating both sides or ` dx = sec^(2) z dz` ` = (1 + tan^(2) z) dz` ` = (1 + tan^(2)z) dz` ` = (1 + x^(2)) dz` Now, ` y = z^(2)` or ` dy = 2z dz` ` = 2 tan^(-1) x dz` ` dy = 2 tan^(-1) x * (dx)/((1+x^(2))` Hence, ` (dy)/(dx) = (2 tan^(-1))/((1+x^(2)))` or `(1+x^(2))(dy)/(dx) = 2 tan^(-1) x` or `(1 +x^(2))^(2)((dy)/(dx))^(2) = 4 (tan^(-1)x)^(2)` or ` (1+x^(2))^(2)((dy)/(dx))^(2) - 4y = 0` Differentiating both sides with respect to x, `(1+x^(2))^(2)[2(dy)/(dx)] (d^(2)y)/(dx^(2)) + 2 ((dy)/(dx))^(2) (1+x^(2)) 2x - 4 (dy)/(dx)` = 0 or ` (1 + x^(2))^(2) (d^(2)y)/(dx^(2))+2x (1+x^(2))(dy)/(dx) - 2 = 0` |
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| 29. |
Evaluate : ` int _(0)^(pi//2) 1/(1 + cos x) dx ` |
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Answer» Given, `underset(0)overset(pi/2) int1/(1+cos x) dx ` ` = underset(0)overset(pi/2) int 1/(2 cos^(2). x/2) dx` ` = 1/2 underset(0)overset(pi/2) intsec^(2) (x/2) dx ` `[" Let " x/2 = u " " :. 1/2 dx = du " when" x gt 0, u gt 0 and x gt pi/2, u gt pi/4 ]` ` :. " " = 1/2 underset(0)overset(pi/2) int sec^(2)(x/2) dx ` = `underset(0)overset(pi/2) intsec^(2) u" " du= [tan u]_(0)^(pi/4)` ` = tan pi/4 - tan 0 = 1` |
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| 30. |
A fair coin is tossed 5 times . Find the probability that it shows exactly three times head. |
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Answer» Total no. of possible solutions ` = .^(5)C_(3)` Total no. of solutions ` = 2^(5) = 32` `:. " P(HHH) " = (.^(5)C_(3))/32` ` = 10/32 = 5/16` |
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| 31. |
Let the p.m.f. ( probability mass function ) of random variable x be `P(x) = (4/x)(5/9)^(x)(4/x)^(4-x), x = 0, 1, 2, 3, 4` = 0 , otherwise Find E(x) and Var.(x) |
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Answer» `P(x) = (4/x)(5/9)^(x)(4/x)^(4-x)` ` x = 0, 1, 2, 3, 4` = 0 otherwise Comparing this with `P(x) = .^(n)C_(x) p^(x)q^(n-x)` we get ` n = 4, p = 5/9, q = 4/9` ` E(x) = np = 4 (5/9) = 20/9` ` = 2*22` Var (x) = npq = ` 4 xx 5/9 xx 4/9` `= 80/81 ` `= 0.9876 ` Hence ` E(x) = 2*22 and Var(x) = 0.9876`. |
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| 32. |
Given the probability density functio (p.d.f) of a continuos random variable X as. `f(x)=(x^(2))/(3),-1 lt x lt2` Determine the cumulative distribution function (c.d.f) X and hence find `P(X lt1),P(X gt0), P(1 lt X lt 2)`. |
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Answer» c.d.f. of the continuous variable is given by `f(x) = underset(-1)overset(x) int y^(2)/3 dx ` ` = [y^(3)/9]_(-1)^(x)` ` = (((x^(3)+1))/9) , x in R` Consider ` P (x lt 1) = f(1) = ((1)^(3) + 1)/9 = 2/9` ` P(x le -2) = 0` ` P(x gt 0) = 1 - P (x le 0) ` ` = 1 - f(0) = 1 - (1/9) = 8/9` ` P(1 lt x lt 2) = f(2) - f (1) = 1 - 2/9 = 7/9` |
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| 33. |
The kinetic energy of a rotating body depends uponA. distribution of mass onlyB. ongular speed onlyC. destribution of mass and angular speedD. angular acceleration only |
| Answer» Distribution of mass and angular speed | |
| 34. |
The charge of how many coulomb is required to deposit 1 g of sodium metal (molar mass 23.0 g `mol^(-1)` ) from sodium ions isA. 2098B. 96500C. 193000D. 4196 |
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Answer» Correct Answer - D Now, 1gm `Nato22*99` gm Na. for `Nato22*99xx6.022xx10^(23)Na//mol` for per charge `=22*99xx6*022xx10^(23)xx1*602xx10^(-19)C` `=14196C` |
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| 35. |
Colligative property depends only on . . . . . . . . in a solution.A. Number of solute particlesB. Number of solvent particlesC. NAture of solute particlesD. Nature of solvent particles |
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Answer» Correct Answer - A Colligative properties are the properties of solution that depends on the number of solute particles in solution rather than on their chemica identity. `Na^(+)+e^(-)toNa` |
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| 36. |
An n-type and p-type silicon can be obtained by doping pure silicon with.A. GermaniumB. BoronC. ArsenicD. Antimony |
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Answer» Correct Answer - B Since boron has 3 electrons in its valence shell, it offers 3 of the four electrons that a silicon atom needs. As a result, each silicon center is left with a hole. Semiconductors made in his manner are called p -type. |
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| 37. |
Amongst the following identify the criterion for process to be at equilibriumA. `DeltaGlt0`B. `DeltaGgt0`C. `DeltaS_("total")=0`D. `DeltaSlt0` |
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Answer» Correct Answer - C For a process to be equilibrium, the total entropy of the system should be zero. Therefore, `DeltaS_("total")=0` |
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| 38. |
Describe biprism exper4iment to calcularte the wavelength of a monocharomatic light. Draw the necessary ray diagram. If the critical angle of a medium is `sin^(-1)(3/5)`, find the po0larising angle. |
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Answer» `C=sin^(-1)(3/5)` `:." "sin C = (3/5)` As we know `mu=1/(sin C)` `mu=5/3` `i_(p)=tan^(-1)(5/3)` `i_(p)=59*03^@`. |
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| 39. |
A fine powder of recycled modified plastic is known asA. polyblendB. polytheneC. polyesterD. polymer |
| Answer» Correct Answer - d | |
| 40. |
Appearance of new combinations in `F_(2)` generation in a dihybrid cross proves the law of .....A. domianceB. segregationC. independent assortmentD. purity of gametes |
| Answer» Correct Answer - c | |
| 41. |
What does abbreviation HGP stand for ? |
| Answer» HGP stands for Human Genome Project . | |
| 42. |
What is gene pool ? |
| Answer» The gene pool is the set of all genes , or genetic information , in amy population , usually of a particular species . This also proves to be the basic level at which evolution occurs. | |
| 43. |
In gene therapy , DNAase in used to treat........A. cystic fibrosisB. haemophiliaC. pituitary dwarfismD. insulin dependent diabetes |
| Answer» Correct Answer - a | |
| 44. |
The element that does NOT exhibit allotropy isA. AsB. SbC. BiD. N |
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Answer» Correct Answer - C Allotrophy is defined as the existence of chemical element in two or more forms which may differ in the arrangement of atoms in crystalline solids. All the elements of group 15 except nitrogen (N) and Bismuth (Bi) show allotropy. |
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| 45. |
Arrange the following reducing agents in the order of increasing strength under standard state condition. Justify the answer. `{:("Element",Al(s),Cu(s),Cl(aq),Ni(s)),(E^(@),-1.66V,0.34V,1.36V,-0.26V):}` |
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Answer» Given `Delta=-110kJ` `DeltaS=40JK^(-1)` ltbr? `=0.04kjK^(-1)` Temperature, `T=4000K` `DeltaG=?` Since `DeltaH` is -ve, the reaction is exothermic `DeltaG=DeltaH-TDeltaS` `=-110-400xx0*04` `=-110-16` `=-126kJ` Since, `DeltaG` is negative, the reaction is spontaneous and exothermic. (b) Given `DeltaH=40kJ`, `Delta=-120JK^(-1)` `=-0*12kJK^(-1)` Temperature, `T=250K` `DeltaG=?` Since `DeltaH` is + ve reaction is endothermic `DeltaG=DeltaH-T.DeltaS` `=40-250xx(0*12)` `=40+30` `=70kJ`. Sice `DeltaGgt0`, the reaction is non-spontaneous. `DeltaG=70kJ`, The reaction is endothermic and non-spontaneous. |
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| 46. |
Normal activities of the heart are regulated by.......A. brainB. spinal cordC. modified cardiac musclesD. hormones |
| Answer» Correct Answer - c | |
| 47. |
Linear momentum of an electron in Bohr obrit of H-atom (principal quantum number n) is proporational toA. `1/n^(2)`B. `1/n`C. `n`D. `n^(2)` |
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Answer» Correct Answer - B `1/n` |
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| 48. |
During ....... Type of interaction , both organisms are benefited .A. mutualismB. competitionC. commensalismD. parasitism |
| Answer» Correct Answer - a | |
| 49. |
Write the structures and IUPAC names of the following compounds : (a) Adipic acid (b) `alpha` - methyl butyraldehyde. |
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Answer» Structure and IUPAC name of the compounds are : (a) Adipic acid : Structure : `HO-underset(O)underset(||)(C)-CH_(2)-CH_(2)-CH_(2)-CH_(2)-overset(O)overset(||)(C)-OH` IUPAC name : Hexanedioic acid. (b) `alpha` - methyl butyraldehyde : Structure : `CH_(3)-CH_(2)-underset(CH_(3))underset(|)(CH)-CHO` IUPAC name : 2-methyl butyraldehyde. |
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| 50. |
What is meant by lanthanoids contraction ? What effect does it have on the chemistry of element which follow lanthanoids ? |
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Answer» Refer to HSC Paper, March 2018, Answer 8. Decrease in basicity : (i) In lanthanoids due to lanthanoid contraction, as the atomic number increase, the size of the lanthanoid atoms and their tripositive ions decreases, i.e., from `La^(3+)` to `Lu^(3+)` (ii) As size of the polarizability increases and thus the convalent character of the M-OH bond increases, and ionic character decreases. Therefore the basic nature of the hycroxides decreases. (iii) BAsicity and ionic character decrease in the order La `(OH)_(3)ltCe(OH)_(3)lt . . . . Lu(OH)_(3)` Ionic radii of post -lanthanoids : (i) due to lanthanoid contraction the atomic radii (size) of elements with lanthanum in the 6th period (3rd transition series -Hf, Ta,W, Re) are similar to the elements of the 5th period (4d-series Zr, Nb Mo,Tc). (ii) Due to similarity in their size, post - lanthanoid elements (5d-series) have closely similar properties to the elements of the 2nd transition series (4d-series) which lie immediately above them. (iii) Pairs of elements namely Zr-Hf(Gr-4),Nb-Ta(GR-5), Mo-W(Gr-6), Tc - Re (Gr-7) are called chemical twins since they possess almost indentical sizes and similar properties. |
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