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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. |
The equation of a sound wave is `y=0.0015sin(62.4x+316t)` the wavelength of this wave isA. 0.2 unitB. 0.1 unitC. 0.3 unitD. None of these |
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Answer» Correct Answer - B General equation of plane prograssive wale is given by `y=a sin (kx+omegat)" "…(i)` Given equation `y=0.0015sin (62.4x+316t)" "…(ii)` Comparing Eqs. (i) and (ii), we get `k=62.4` `therefore (2pi)/(lamda)=62.4` `lamda=(2pi)/(62.4)=0.1` unit |
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| 2. |
If the temperature increases, then what happens to the frequency of the sound produced by the organ pipe ?A. unchangedB. decreasesC. increasesD. not definite |
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Answer» Correct Answer - C Due to rise in temperature, the speed of sound increases. Since frequency `n=v/lamda and lamda` remains unchanged. Hence frequency n increases. |
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| 3. |
In a resonance tube the first resonance with a tuning fork occurs at 16 cm and second at 49 cm . If the velocity of sound is 330 m/s , the frequency of tuning fork isA. 500B. 300C. 330D. 165 |
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Answer» Correct Answer - A For closed pipe `l_(1)=(v)/(4n)and l_(2)=(3v)/(4n)` `v=2n(l_(2)-l_(1))` `n=(n)/(2(l_(2)-l_(1)))=(330)/(2xx(0.49-0.16))=500Hz` |
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| 4. |
The differential equation whose solution is `(x-h)^2+ (y-k)^2=a^2` is (a is a constant)A. `[1+((dy)/(dx))^(2)]^(3)=a^(2)(d^(2)y)/(dx^(2))`B. `[1+((dy)/(dx))^(2)]^(3)=a^(2)((d^(2)y)/(dx^(2)))^(2)`C. `[1+((dy)/(dx))]^(3)=a^(2)((d^(2)y)/(dx^(2)))^(2)`D. None of the above |
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Answer» Correct Answer - B Given, `(x-h)^(2)+(y-k)^(2)=a^(2)" (i)"` `implies2(x-h)+2(y-k)(dy)/(dx)=0` `implies(x-h)+(y-k)(dy)/(dx)=0" "...(ii)` Again differentiating `(y-k)=-(1+((dy)/(dx))^(2))/(d^(2)y//dx^(2))` Putting in Eq. (ii), we get `x-h =-(y-k)(dy)/(dx)=([1+((dy)/(dx))^(2)](dy)/(dx))/((d^(2)y)/(dx^(2)))` Putting in Eq. (i), we get `([1+((dy)/(dx))^(2)]^(2)((dy)/(dx))^(2))/(((d^(2)y)/(dx^(2)))^(2))+([1+((dy)/(dx))^(2)]^(2))/(((d^(2)y)/(dx^(2)))^(2))=a^(2)` `implies[1+((dy)/(dx))^(2)]^(2)[((dy)/(dx))^(2)+1]=a^(2)((d^(2)y)/(dx^(2)))^(2)` `implies[1+((dy)/(dx))^(2)]^(3)=a^(2)((d^(2)y)/(dx^(2)))^(2)` |
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| 5. |
Find the differential equation of all the circleswhich pass through the origin and whose centres lie on y-axis.A. `(x^(2)-y^(2))(dy)/(dx)-2xy=0`B. `(x^(2)-y^(2))(dy)/(dx)+2xy=0`C. `(x^(2)-y^(2))(dy)/(dx)-xy=0`D. `(x^(2)-y^(2))(dy)/(dx)+xy=0` |
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Answer» Correct Answer - A Equation of a circle is `x^(2)+(y-a)^(2)=a^(2)" "...(i)` `2x+2y(dy)/(dx)-2a(dy)/(dx)=0" "...(ii)` From Eqs. (i) and (ii) `(dy)/(dx)=(2xy)/(x^(2)-y^(2))` `implies(x^(2)-y^(2))(dy)/(dx)-2xy=0` |
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| 6. |
From point `P(8,27)` tangents PQ and PR are drawn to the ellipse `(x^(2))/(4)+(y^(2))/(9)=1.` Then, angle subtended by QR at origin isA. `tan ^(-1)""(2sqrt6)/(65)`B. `tan ^(-1)""(4sqrt6)/(65)`C. `tan^(-1)""(8sqrt2)/(65)`D. None of these |
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Answer» Correct Answer - D Equation of chord of contact QR is `8*x/4+27*y/9=1` `implies2x+3y=1" "...(i)` Now, equation of the pari of lines passing through origin and points Q, R given by `((x^(2))/(4)+(y^(2))/(9))=(2x+3y)^(2)` `implies9x^(2)+4y^(2)=36(4x^(2)+12xy+9y^(2))` `implies135x^(2)+432xy+320y^(2)=0` `therefore` Required angle is `=tan ^(-1)""(2sqrt((216)^(2)-135*320))/(455)` `=tan ^(-1)""(8sqrt(2916-2700))/(455)` `=tan ^(-1)""(8sqrt216)/(455)` `=tan ^(-1)""(48sqrt6)/(455)` |
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| 7. |
In the following sequence of reactions `CH-underset(NH_(2))underset(|)(CH)-CH_(3)overset(HNO_(3))toAoverset("Oxidation")toBunderset((ii)H^(+)//H_(2)O)overset((i)CH_(3)MgI)toC` The compound C formed will beA. butanol-1B. butanol-2C. 2-methyl propanol-1D. 1, 1-dimethyletanol |
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Answer» Correct Answer - D `Ch_(3)-underset(NH_(2))underset(|)CH-CH_(3)overset(HNO_(2))toCH_(3)-underset((A))underset(OH)underset(|)CH-CH_(3)` `overset([o])toCH_(3)-underset((B))underset(O)underset(||)C-Ch_(3)underset((ii)H^(+)//H_(2)O)overset((i)CH_(3)Mgl)tounderset("1,1-dimethylethanol")((CH_(3))_(3)COH)` |
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| 8. |
The equation of the common tangents to the circle`(x-3)^(2)+y^(2)=9` and the parabola `y^(2)=4ax` the x-axis, isA. `sqrt2y=3x+1`B. `sqrt3y=-(x+3)`C. `sqrt3y=x+3`D. `sqrt3y=-(3x+1)` |
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Answer» Correct Answer - C Let the common tangent to the parabola `y^(2)=4xbe,` `y=mx+1/m` It should be also touch the circle `(x-3)^(2)+y^(2)=9` Whose centre is (3,0) and radius =3, then `(|em+1//m|)/(sqrt(1+m^(2)))=3` `implies3m^(2)=1` `impliesm=pm(1)/(sqrt3)` But `m gt 0,` then equatio of common tangent is `y=(1)/(sqrt3)*x+sqrt3` `or sqrt3*y=x+3` |
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| 9. |
The most common oxidation states of cerium areA. `+2,+4`B. `+3,+4`C. `+3,+5`D. `+2,+3` |
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Answer» Correct Answer - B The electronic condiguratin of Ce is as `Ce: [Xe]4f^(2)5d^(0)6s^(2)` The most common oxidation states shown by cerium are +3 and +4. |
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| 10. |
In a sim,le slit differection pattern intensisty and width of fringes areA. unequal widthB. equal widthC. equal width and equal intensityD. unequal width and unequal intensity |
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Answer» Correct Answer - D In single slit diffraction pattern intensity and width of fringes are unequal width and unequal intensity. |
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| 11. |
A plane wave front of wavelength `lamda` is incident on a single slite of width b. What is the angular width for secondary maximum ?A. `(lamda)/(db)`B. `lamda/b`C. `(2lamda)/(b)`D. `(b)/(lamda)` |
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Answer» Correct Answer - B `beta=(lamdaD)/(d)and theta=(beta)/(D)` `therefore theta=(lamda)/(d)` Width of single slit = b SO, `theta=(lamda)/(b)` |
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| 12. |
If the body is moving in a circle of radius r with a constant speed v , its angular velocity isA. `v^(2)//r`B. `vr`C. `v//r`D. r/v |
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Answer» Correct Answer - C Linear velocity `v=r omega` `omega=v/r` |
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| 13. |
A ball of mass 0.25 kg attached to the end of a string of length 1.96 m is moving in a horizontal circle. The string will break if the tension is more than 25 N . What is the maximum speed with which the ball can be movedA. 14 m/sB. 3 m/sC. 3.92 m/sD. 5 m/s |
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Answer» Correct Answer - A Tension of string `T=(mv^(2))/(r) ` `25=(0.25xxv^(2))/(1.96)` `v=14m//s` |
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| 14. |
An acid solution of `pH=6` is diluted `1000` times, the `pH` of the final solution isA. `6.01`B. 9C. `3.5`D. `6.99` |
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Answer» Correct Answer - D `because pH=6` `therefore[H^(+)]=10^(-6)M` After dilution `[H^(+)]=(10^(-6))/(1000)=10^(-9)M` `therefore [H^(+)]`from `H_(2)O` cannot be neglected. Total `[H^(+)]=10^(-9)+10^(-7)` `=10^(-7)(10^(-2)+1)` `=10^(-7)(1.01)` `pH=-log (1.01xx10^(-7))` `=7-0.0043=6.9957` |
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| 15. |
The solubility product of `Ag CrO_(4)` is `32xx10^(-12).` What is the concentration of `CrO_(4)^(2-)` ions in that solution ?A. `2xx10^(-4)`B. `16xx10^(-4)`C. `8xx10^(-4)`D. `12xx10^(-4)` |
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Answer» Correct Answer - A `Ag_(2)CrO_(4)tounderset(2s)2Ag^(+)+underset(s)(CrO_(4)^(2-))` `K_(sp)=[Ag^(+)]^(2)[CrO_(4)^(2-)]` `=(2s)^(2)(s)` `K_(sp)=4s^(3)` `s=((K_(sp))/(4))^(1//3)` `=((32xx10^(-12))/(4))^(1//3)=2xx10^(-4)M` |
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| 16. |
Which of the following salts will not undeergo hydrolysis ?A. `NH_(4)Cl`B. `KCN`C. `KNO_(3)`D. `Na_(2)CO_(3)` |
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Answer» Correct Answer - C Potassium nitrate `(KNO_(3))` is a salt of strong base and strong acid and hence, will not undergo hydrolysis. |
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| 17. |
An aqueous solution of urea containing 18 g urea in 2500 `cm^(3)` of the solution has a density equal to 1.052. if the molecular weight of urea is 60, the molality of the solution isA. `0.200`B. `0.192`C. `0.100`D. `1.200` |
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Answer» Correct Answer - B 1500 cc. solution `=1500xx1.052g =1578g` Weight of solvent `=1560g` Weight of solute =18 g Molality `=("weight of solute")/("molecular weight of solute"xx)` weight of solvent in kg `=(18xx1000)/(60xx1560)=0.192` m |
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| 18. |
S34.2 g of cane sugar is dissolved in 180 g of water. The relative lowering of vapour pressure will beA. `0.0099`B. `1.1597`C. `0.840`D. `0.9901` |
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Answer» Correct Answer - A `(P^(@)-Ps)/(P^(@))=((w_(2))/(m_(2)))/((w_(1))/(m_(1))+(w_(2))/(M_(2)))` `=((34.2)/(342))/((34.2)/(342)+(180)/(18))` |
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| 19. |
What volume of `2MH_(2)SO_(4)` is required to from 0.2 N of 100 mL of solution ?A. 5 mLB. 20 mLC. 10 mLD. 50 mL |
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Answer» Correct Answer - A `because M-H_(2)SO_(4)=2N-H_(2)SO_(4)` `because2M-H_(2)SO_(4)=4N-H_(2)SO_(4)` `N_(1)V_(1)=N_(2)V_(2)` `4xxV_(1)=0.2xx100` `V_(1)=(0.2xx100)/(4)=5mL.` |
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| 20. |
Bactericidal antibiotic anoung the following is :A. ofloxachinB. chloroamphenicolC. erythromycinD. tetracycline |
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Answer» Correct Answer - A Bactericidal antibiotic kills the organism in the body. e.g., penicillin, ofloxacin etc. |
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| 21. |
If X follows the binomial distribution with parameters n=6 and p and 9p(X=4)=P(X=2), then p isA. `1//4`B. `1//3`C. `1//2`D. `2//3` |
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Answer» Correct Answer - A Given, `9P(X=4)=P(X=2)` `therefore9*""^(6)C_(4)p^(4)(1-p)^(2)=""^(6)C_(2)p^(2)(1-p)^(4)` `implies((1-p)^(2))/(p^(2))=9` `implies(1-q)/(p)=3` `impliesp=1//4` |
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| 22. |
Unit of decay constant of radiocative disintegration isA. timeB. `min^(-2)`C. `time^(-1)`D. time `mol^(-1)` |
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Answer» Correct Answer - C For a radioactive disintegration, `k=(2.303)/(t)log ""(N_(0))/(N)` where, k is know as decay constant or disintegration constant. The unit for decay constant is `"time"^(-1).` |
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| 23. |
In the production of beats by two waves of same amplitude and nearly same frequency, the maximum intensity to each of the constituent waves isA. sameB. 2 timesC. 4 timesD. 8 times |
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Answer» Correct Answer - C Beats formation in a case of superposition for which the maximum intensity is 4 time the intensity of either source. |
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| 24. |
A glass slab of thickness `4 cm` contains the same number of waves as `5 cm` of water, when both are traversed by the same monochromatic light. If the refractive index of water is `4//3,` then refractive index of glass isA. `5//3`B. `5//4`C. `16//15`D. `3//2` |
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Answer» Correct Answer - A `""_(a)n_(g)=""_(a)n_(w)xx""_(a)n_(g)` `""_(a)n_(g)=4/3xx(lamda_(w))/(lamda_(g))` `""_(a)n_(g)=4/3xx5/4` `""_(a)n_(g)=5/3` |
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| 25. |
3-methylpentan-3-ol will be prepared fromA. ethyl formate and methyl magnesium bromideB. ethyl ethanoate and ethyl magnesium bromideC. ethyl propanoate and methyl magnesium bromideD. ethyl formate and ethyl magnesium bromide |
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Answer» Correct Answer - B `{:(CH_(3)-C=O+ C_(2)H_(5)MgBrto),(" "| " ""ethyl magnesium"),(" "OC_(2)H_(5) " ""bromide"),(" ""ethyl ethanoate"):}` `{:(" "CH_(2)H_(5) " "CH_(2)CH_(3)),(" "|" "|),(CH_(3)-C-OHgBrunderset(-Mg(OC_(2)H_(5))Br)toCH_(3)C=O),(" "|),(" "OC_(2)H_(5)):}` `{:(" "CH_(2)CH_(3)),(" "|),(overset(C_(2)H_(5)MgBr)toCH-C-OHgBr),(" "|),(" "CH_(2)-CH_(3)):}` `{:(" "CH_(2)CH_(3)),(" "|),(overset(H.OH)toCH_(3)-C-OH),(" "|),(" "CH_(2)CH_(3)),(" ""3-methylpentan-3-ol"):}` |
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| 26. |
`int_(pi//2)^(pi//2)(cosx)/(1+e^(x))dx` is equal toA. 1B. 0C. -1D. None of these |
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Answer» Correct Answer - A `I-int _(-pi)^(pi//2)(cos x)/(1+e^(x))dx" (i)"` `I=int _(-pi)^(pi//2)(cos (pi//2-pi //2-x))/(1+e^((x//2-pi//2-x)))dx` `=int _(-pi)^(pi//2)(cos (-x))/(1+e^(-x))dx` `=int _(-pi)^(pi//2)^(e^(x)cos x)/(1+e^(x))dx` `=int _(-pi)^(pi//2)(e^(x)cos x)/(1+e^(x))dx" "(ii)` On adding Eqs. (i) and (ii), we get `2I=int_(-pi//2)^(pi//2)((1+e^(x))cos )/((1+e^(x)))dx` `int _(-pi//2)^(pi//2)cos x dx` `=2int _(0)^(pi//2)cos x dx` [Since, cos x is an even function.] `therefore 2I=2[sin x]_(0)^(pi//2)=2(1-0)=2` `impliesI=1` |
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| 27. |
A particle moves along a straight line according to the law `s=16-2t+3t^(3)`, where `s` metres is the distance of the particle from a fixed point at the end of `t` second. The acceleration of the particle at the end of `2s` isA. `3.6m//s^(2)`B. `36m//s^(2)`C. `36km//s^(2)`D. `360 m//s^(2)` |
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Answer» Correct Answer - B Given, `s=16-2t+3t^(3)` `implies(ds)/(dt)=-2 +9t^(2)` `implies(d^(2)S)/(dt^(2))=18t` Now, the acceleratio of the particle at the end of `t=2` s is `f=(d^(2)s)/(dt^(2))=18xx2=36m//s^(2)` |
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| 28. |
If `x+y=k`is normal to `y^2=12 x ,`then `k`is3 (b) 9(c) `-9`(d) `-3`A. 3B. 9C. -9D. -3 |
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Answer» Correct Answer - B Let `x+y=k` be normal to `y^(2)=12xx at p(alpha,beta)" "...(i)` `thereforebeta^(2)=12 alpha" "...(ii)` Also, slope of normal at `P(alpha, beta)` is -1 From, Eq. (i), `(dy)/(dx)=6/y` `implies((dy)/(dx))_(p)=(6)/(beta)` `therefore -1 =(-1)/(6//beta)` `impliesbeta=6,alpha =3` `thereforeP` is (3,6) which lies on `x+y=k` `therefore 3+6=k` `impliesk=9` |
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| 29. |
`int[sqrt(cotx)+sqrt(tan x)]dx` ज्ञात कीजिए!A. `sqrt2tan^(-1)((tanx)/(sqrt(2 tan x)))+C`B. `sqrt2tan^(-1)((tanx-1)/(sqrt(2 tan x)))+C`C. `(tanx)/(sqrt2)*tan^(-1)((cot x+1)/(sqrt(2 tanx)))+C`D. `(tanx)/(sqrt2)*tan^(-1)((cotx+1)/(sqrt(tanx)))+C` |
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Answer» Correct Answer - B Let `I=int((sin x+cosx))/(sqrtsinx*cos x)dx` `=int(sqrt2(sinx+cos x))/(sqrt(2sin x*cos x))dx` `=sqrt2 int(sin x+cos x)/(sqrtsin 2x)dx` Put `sinx -cos x=t` `implies(cos x+sin x)dx=dt` Also, `sin 2x=(1-t^(2))` `thereforeI=sqrt2int(dt)/(sqrt(1-t^(2)))` `=sqrt2sin ^(-1)t+c` `=sqrt2sin ^(-1)(sinx-cos x)+c` `=sqrt2 tan ^(-1)((tan x-1)/(sqrt(2 tan x)))+c` |
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| 30. |
x,के सभी वास्तविक मानों के लिए `(1-x+x^2)/(1+x+x^2)` का न्यूनतम मान है : |
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Answer» Correct Answer - B Let `y=(1-x+x^(2))/(1+x+x^(2))=1-(2x)/(1+x+x^(2))` `=1-(2)/((1)/(x)+1+x)` `impliesy=1-2/t` where `t=1/x+1+x` Now, y is minimum, where `2/t` is max `implies` is min. `therefore (dt)/(dx)=-(1)/(x^(2))+1=0` `impliesx=pm1` `(d^(2)t)/dx^(2)=(2)/(x^(2))gt0,` for `x=1` `therefore` Minimum value of y is `1-(2)/(1+1+1)=1-2/3=1/3` |
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| 31. |
`int(x^(2))/((x sinx+cosx)^(2))dx` is equal toA. `(sin x+cos x)/(xsin x+cos x)+C`B. `(x sin x-cos x)/(x sin x+cos x)+C`C. `(sin x-x cos x)/(x sin x+cos c)+C`D. None of these |
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Answer» Correct Answer - C Since, `(d)/(dx)(x sin x+cos x)=x cos x` `therefore I =int (x^(2)dx)/((xsin x+cos x)^(2))` `=int (x)/(cos x).(x cos x)/(cos x(x sin x+cos x)^(2))dx` `=(x)/(cos x)((-1)/(x sin x+cos x))` `-int(cos x-x(-sin x))/(cos ^(2)x)*(-1)/((x sin x+cos x))dx` `=(-x)/(cos x(x sin x+cos x))+int sec^(2)dx` `=(-x)/(cos x(xsin x+cos x))+tan x+C` `=(-x+sin x(x sin x+cos x))/(cos x(x sin x+c os x))+C` `=(-xcos ^(2)x+sin x*c os x)/(cos x(x sin x+cos x))+C` `=((sin x-xcos x)/(cos x+xsin x))+C` |
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| 32. |
`y=logtan(x/2)+sin^(-1)(cosx)`, then `dy/dx` isA. `cosec x-1`B. `cosec x`C. `cosec x+1`D. x |
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Answer» Correct Answer - A `(dy)/(dx)=(1)/(tan x//2)sec ^(2)""(x)/(2)*(1)/(2)+ (1)/(sqrt(1 - cos^(2) x))"(-sin x)"` `(1)/(2sin ""(x)/(2)*cos ""(x)/(2))1=cosecx-1` |
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| 33. |
If ` x = 2 cos t - cos 2t , y = 2 sin t - sin 2t`, then the value of ` |(d^(2) y)/(dx^(2))|_(t= pi//2 ) `isA. `3//2`B. `5//2`C. `5//2`D. `-3//2` |
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Answer» Correct Answer - D `(dx)/(dt)=2cost-2 cos 2t` `therefore (dy)/(dx)=(2 cos t-2 cos 2t)/(-2 sin t+2sin 2t)` `=(cos -cos 2t)/(-2sin 2t-sin t)` `=(cos t-cos 2t)/(sin 2t-sin t)` `implies(2 sin""(3t)/(2)*sin""(t)/(2))/(2cos ""(3t)/(2)*sin ""(t)/(2))` `=tan ""(3t)/(2)` `(d^(2)y)/(dx^(2))=sec ^(2)""(3t)/(2)*3/2*(dt)/(dx)` `=3/2sec^(2)""(3t)/(2)(1)/((2 sin 2t-2sin t))` `implies(d^(2)y)/(dx^(2))|_(t=pi//2)=-3//2` |
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| 34. |
If `x^(p) y^(q) = (x + y)^((p + q)) " then " (dy)/(dx)=` ?A. `y//x`B. `py//qx`C. `x//y`D. `qy//px` |
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Answer» Correct Answer - A Given, `x^(p)y^(q)=(x+y)^(p+q)` Thinking log on both sides, we get `p log x+q log y=(p+q)log (x+y)` `implies(p)/(x)+(q)/(y)(dy)/(dx)=((p+q))/((x+y))(1+(dy)/(dx))` `implies((p)/(x)-(p+q)/(x+y))=((p+q)/(x+y)-(q)/(y))(dy)/(dx)` `implies(dy)/(dx)=y/x` |
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| 35. |
If a, b, c and d are positive then `lim_(x to oo)(1+(1)/(a+bx))^(x+dx)` is eqqualA. `e^(d//b)`B. `e^(c//a)`C. `e^((c+d)//(a+b))`D. e |
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Answer» Correct Answer - A `L=underset(xtooo)lim(1+(1)/(a+bx))^(c+dx)" "["from"(1)^(oo)]` `=e^(underset(x tooo)lim(c+dx)/(a+bx))=e^(underset(x tooo)lim(c//x+d)/(a//x+b))` `=e^((0+d)/(0+d))=e^(d//b)` |
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| 36. |
Let ` f(x) = {{:({1+|sin x|}^(a//|sin x|)", " pi/6 lt x lt 0),(" b, " x = 0 ),(e^(tan 2x//tan 3x) ", "0ltx ltpi/6):}` Determine a and b such that f(x) is continous at x = 0.A. `3//2,^(3//2)`B. `-2//3,e^(-3//2)`C. `2//3,e^(2//3)`D. None of these |
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Answer» Correct Answer - C We have, `underset(x to 0^(-))limf(x)` `=underset(c to 0)lim{1+|sinx|}^((a)/(|sin x|))` `=e^(underset(x to 0)lim|sinx|.(a)/(|sin x|))=e^(a)` and `underset(xto0^(-))limf(x)=underset(xto0)lime^((tan 2x)/(tan3x))` `e^(underset(xto0)lim(tan 2x)/(2x)(3x)/(tan3x)xx2/3)` `e^(2//3)` For f (x) to be continous at x=0, we must have `underset(x to 0^(-))limf(x)=underset(xto0^(+))limf(x)` `=f(0)` `impliese^(a)=e^(2//3)=b` `implies a=2//3` and ` b=e^(2//3)` |
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| 37. |
A gas is compressed at constant temperature. Its molecules gainA. speedB. kinetic energyC. internal energyD. None of these |
| Answer» Correct Answer - D | |
| 38. |
Ziegler-Natta cattalyst catalyse preparation of which of the following compounds ?A. preparation of Ti-metalB. Preparation of low density plasticC. Preparation of high resistance plasticD. Preparation of high density plastic |
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Answer» Correct Answer - D High density polythene (or plastic) is prepared by heating ethylene at about 330-250 K, under a pressure of 1-2 atm and in the presence of Ziegler-Natta catalyst. `nCH_(2)=CH_(2)underset("Ziegler-Natta catalyst")overset(330-350 K, 1-2 atm)to(CH_(2)-CH_(2))n` high density polythen |
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| 39. |
The value of `(1+Delta)(1-Delta)` is |
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Answer» Correct Answer - C We have, `(1+Delta)(1-grad)f(x)` `=(1+Delta){(1-grad)f(x)}` `=(1+Delta)[f(x)-{f(x)-f(x-h))]` `=(1+Delta)f(x-h)=Ef(x-h)` `[because (E=1+Delta)]` `=f(x)=1*f(x)` Thus, `(1+Delta ) (1-grad)f(x)=1*f(x),` for any function `f (x). `therefore (1+Delta)(1+grad)=1` |
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| 40. |
When Zn is treated with excess of NaOH, the product obtained isA. `Zn(OH)_(2)`B. ZnOHC. `Na_(2)ZnO_(2)`D. None of the above |
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Answer» Correct Answer - C When Zn is reacted with excess of NaOH, sodium zincate is obtained. `Zn+2NaOH+2H_(2)Oto Na_(2)underset("soluble")([Zn(OH)_(4)])+H_(2)` or `An+2NaOHto Na_(2)ZnO_(2)+H_(2)` |
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| 41. |
The heat of neutralisation of a strong acid and a strong alkali is `57.0kJ mol ^(-1).` The heat released when 0.2 mole of `HNO_(3)` soluton is mixed with 0.2 mole of KHO isA. `57.0` kJB. `11.4` kJC. `28.5` kJD. `34.9` kJ |
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Answer» Correct Answer - C `because0.2` mole will neutralised `0.2` mole of `HNO_(3).` `therefore` Heat evoled `=57xx0.2=11.4kJ` |
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| 42. |
The standard reduction potential for `Fe^(2+) //Fe` and `Sn^(2+) //Sn` electrodes are `-0.44` and `-0.14 ` volt respectively. For the given cell reaction `Fe^(2+) + Sn rarr Fe + Sn^(2+) `, the standard ` EMF` is.A. `+0.30V`B. `-0.58 V`C. `+0.58V`D. `-0.30V` |
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Answer» Correct Answer - D `Fe^(2+)+SntoFe+Sn^(2+)` `E^(@)=E_(red)^(@)(Fe) +E_(oxi)^(@)(Sn)` `=-0.44+(0.14)` `=-0.30V` |
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| 43. |
The heat of combustion of carbon is `-393.5kJ//mol.` The heat released upon the formation of 35.2 g of `CO_(2)` from carbon and oxygen gas isA. `+135` kJB. `-31.5` kJC. `-315` kJD. `+31.5` kJ |
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Answer» Correct Answer - C `C+O_(2)to CO_(2)` `DeltaH=-393.5kJ` `therefore` Heat released upon the formation of 1 g of `CO_(2)=(-393.5)/(44)kJ` `therefore` Heat released upon the formation of 1 g of `CO_(2)=(-393.5)/(44)xx35.2` `~~-315kJ` |
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| 44. |
As par Bohr model, the minimum energy (in `eV`) required to remove an electron from the ground state of doubly ionized `Li` atom `(Z = 3)` isA. `1.51`B. `13.6`C. `40.8`D. `122.4` |
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Answer» Correct Answer - D `E=-2^(2)xx13.6eV` `E=-9xx1.6eV` `E=-122.e eV` So, inization energy is `+122.4` eV |
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| 45. |
The multiplicative inverse of `A = [(cos theta,-sin theta),(sin theta,cos theta)]`isA. `[{:(-cos theta, sin theta),(-sin theta, -cos theta):}]`B. `[{:(cos theta, sin theta),(-sin theta, cos theta):}]`C. `[{:(-cos theta, -sin theta),(sin theta, -cos theta):}]`D. `[{:(cos theta, sin theta),(sin theta, -cos theta):}]` |
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Answer» Correct Answer - B `|A|=cos ^(2)+sin^(2)theta=1` `adj(A)=[{:(cos theta,sintheta),(-sin theta, cos theta):}]` `A^(-1)=(adj(A))/(|A|)=[{:(cos theta, sin theta),(-sin theta, cos theta):}]` |
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| 46. |
The value of a for which system of equation , `a^3x+(a+1)^3y+(a+2)^3z=0, ax+(a+1)y+(a+2)z=0, x+y+z=0,` has a non-zero solution is:A. 1B. 0C. `-1`D. None of these |
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Answer» Correct Answer - C For non-zero solution, `|{:(a^(3), (a+1)^(3), (a+2)^(3)),(a, (a+1),(a+2)),(1,1,1):}|=0` `implies-|{:(1,1,1),(a,(a+1),(a+2)),(a^(3),(a+1)^(3),(a+2)^(3)):}|=0` `-(a-a-1)(a+1-a-2)(a+2-a)xx(a+a+1+a+2)=0` `implies-2(3a+3)=0` `impliesa=-1` `[because|{:(1,1,1),(x,y,z),(x^(3),y^(3), z^(3)):}|=(x-y)(y-z)(z-x)(x+y+z)]` |
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| 47. |
If one of the lines of the pair `ax^(2)+2hxy+by^(2)=0` bisects the angle between positive direction of the axes, then a, b and h satisfy the relation.A. `a+b=2|h|`B. `a+b=-2h`C. `a-b=2|h|`D. `(a-b)^(2)=4h^(2)` |
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Answer» Correct Answer - B y=x, should satisfy `ax^(2)+2hxy+bt^(2)=0` `impliesa+b=-2h` |
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| 48. |
Find the angle between the following pair of lines: A lines with direction ratios 2,2,1 A line joning (3,1,4)to (7,2,12)A. `cos ^(-1)(2//3)`B. `cos ^(-1)(3//2)`C. `tan^(-1)(-2//3)`D. None of the above |
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Answer» Correct Answer - A Direction ratios of the line joining the point `(3,1,4) and (7,2,12)` are `=lt7-3,2-1,12-4gt` `=lt4,1,8gt` `=lea_(1),a_(2),a_(3)gt (let)` Ant the directin ratio of given line is `=lt2,2,1gt` lt brgt `=ltb_(1),b_(2),b_(3)gt` (let) Let Q be the and between the lines, then `cos theta=(a_(1)b_(1)+a_(2)b_(2)+a_(3)b_(3))/(sqrt(a_(1)^(2)+a_(2)^(1)+a_(3)^(2))sqrt(b_(1)^(2)+b_(2)^(2)+b_(3)^(2)))` `impliescos theta=((4)(2)+(1)(2)+(8)(1))/(sqrt(16+1+64)sqrt(4+4+1))` `impliescos theta=(18)/(sqrt81sqrt9)=(18)/(9xx3)=2/3` `impliestheta=cos ^(-1)((2)/(3))` |
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| 49. |
A black body at a temperature of `227^(@)C` radiates heat energy at the rate of 5 cal/`cm^(2)`-sec. At a temperature of `727^(@)C`, the rate of heat radiated per unit area in cal/`cm^(2)`-sec will beA. 80B. 160C. 250D. 500 |
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Answer» Correct Answer - A `(E_(2))/(E_(1))=((T_(2))/(T_(1)))^(4)` `(E_(2))/(E_(1))=((273+727)/(273+227))` `(E_(2))/(5)=((1000)^(4))/((5000)^(4))=16` `E_(2)=80cal//cm^(2)` |
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| 50. |
Which of the following properties does correspond to the order? `HI lt HBr lt HCl lt HF`A. Thermal stabilityB. Reducting powerC. Ionic characterD. Dipole moment |
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Answer» Correct Answer - B Reducing power increase in the order as H-X bond length increase from HF to HI. `HF lt HCl lt HBrlt HI` |
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