InterviewSolution
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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. |
In single slit diffraction pattern :A. central fringe has negligible width than othersB. all fringes are of same widthC. central fringes do not existsD. None of the above |
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Answer» Correct Answer - D In single slit diffraction , the central fringe has maximum intensity and has the width double than other fringes . |
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| 2. |
Three capacitor of capacitance `C(mu F)` are connected in parallel to which a capacitor of capacitance C is connected in series. Effective capacitance is 3.75. then capacity off each capacitor isA. `4muF`B. `4 muF`C. ` 6 muF`D. `8 muF` |
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Answer» Correct Answer - B The effective capacitance of three condenser connected in parallel =3C When 3 C is connected in series to C `C_("Result")=(3C xx C)/(3C +C)=3.75` `implies C= 5 muF ` |
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| 3. |
If the line `vec(OR)` makes angles ` theta_(1),theta_(2),theta_(3)` with the planes ` XOY, YOZ, ZOX` respectively , then ` cos^(2)theta_(1)+cos^(2)theta_(2)+cos^(2)theta_(3)` is equal toA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - A If line ` vec(OR)` makes angle ` theta_(1), theta_(2), theta_(3)` with the planes XOY, YOZ, ZOX respectively , then `cos^(2) theta_(1)+ cos^(2) theta_(2)+ cos^(2) theta_(3)=1 ` |
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| 4. |
The IUPAC name of crotonaldehyde isA. butenaldehydeB. butan-1-alC. but-2-en-1-alD. prop-2-en-1-ol |
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Answer» Correct Answer - C The structural formula of crotonaldehyde is `underset(" but-2-en-1-al)(overset(4)(C)H_(3)- overset(3)(C)H=overset(2)(C)Hoverset(1)(C)HO)` |
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| 5. |
Negative inductive effect is shown byA. `-CH_(3)`B. `-CH_(3)CH_(3)`C. `-NH_(2)`D. `(CH_(3))_(2)CH-` |
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Answer» Correct Answer - C Among the given , `-CH_(3),-CH_(3)CH_(3)` and `(CH_(3))_(2)CH-` being electron releasing , increase the electron density , so exhibit positive inductive effect while `-NH_(2)` exhibits negative inductive effect . |
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| 6. |
From a group of 8 boys and 3 girls, a committee of 5 members to be formed. Find the probability that 2 particular girls are included in the committee.A. `(4)/(11)`B. `(2)/(11)`C. `(6)/(11)`D. `(8)/(11)` |
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Answer» Correct Answer - B Total number of ways `=""^(8)C_(2)""^(3)C_(3)+""^(8)C_(3)""^(3)C_(2)+""^(8)C_(4)""^(3)C_(1)+""^(8)C_(5)""^(3)C_(0)` `28xx1+56xx3+70xx3+56xx1`=462 Number of ways in which 2 particular girls are included `""^(9)C_(3)=84` `:.` Required probability `=(84)/(462)=(2)/(11)` |
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| 7. |
In a resonance pipe the first and second resonance are obtained at depths 22.7 cm and 70.2 respectively. What will be the end correction?A. 1.05 cmB. 115.5 cmC. 92.5 cmD. 113.5 cm |
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Answer» Correct Answer - A For end correction x , `(l_(2)+x)/(l_(1)+x)=( 3lambda//4)/(lambda//4)=3 ` ` implies x=(l_(2)-3l_(1))/(2)` `=(70.2-3 xx 22.7)/(2)=1.05 cm` |
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| 8. |
Which of the following is not a colligative property ?A. Elevation in boiling pointB. Lowering of vapour pressureC. Osmotic pressureD. Freezing point |
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Answer» Correct Answer - D Depression in freezing point is a colligative property but freezing point is not a colligative property . |
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| 9. |
At the same conditions of pressure , volume and temperature , work done is maximum for which gas if all gases have equal masses ?A. `NH_(3)`B. `N_(2)`C. `Cl_(2)`D. `H_(2)S` |
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Answer» Correct Answer - A When p, V and T are same and mass is also same , work done depends only uypon molecular mass . `W prop (1)/(M) `( where , M=molecular mass ) Among the given gases , `NH_(3)` has lowest molecular mass , so work done is maximum for it . |
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| 10. |
In ......... process , work is done at the expense of internal energy .A. isothermalB. isochloricC. adiabaticD. isobaric |
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Answer» Correct Answer - C We know that `DeltaE=Q+W` If heat supplied from the surrounding . Q=0 `:. DeltaE=W` i.e., work is done at the expense of only internal energy and Q= 0 for adiabatic process . |
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| 11. |
Miximum depression in freezing point is caused byA. potassium chlorideB. sodium sulphateC. magnesium sulphateD. magnesium carbonate |
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Answer» Correct Answer - B Depression in freezing point is a colligative property , so depends upon the number of particles. Among , potassium chloride (KCl), sodium sulphate `(MgSO_(4))` and magnesium carbonate `(MgCO_(3)),Na_(2)SO_(4)` produces maximum number of particles . `Na_(2)SO_(4) iff underset("3 particles")(ubrace(2Na^(+)+SO_(4)^(2-)))` Thus , it causes maximum depression in freezing point . |
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| 12. |
Hardening of oil is done byA. dehydrogenationB. hydrogenationC. dehydrohalogenationD. dehydration |
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Answer» Correct Answer - B `underset("(unsaturated)")("Oil") overset("Hydrogenation")to underset("(satruated)")("ghee")` This is called the hardening of oil . |
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| 13. |
Which of the following travels with the speed of light ?A. ` alpha`-raysB. `beta`-raysC. `gamma`- raysD. X-rays |
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Answer» Correct Answer - C Since ` gamma`-rays have zero charge and zero mass, thase travel with the speed of light . |
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| 14. |
In a solution , 0.02 M acetic acid is 4% dissociation . The `[OH^(-)]` in the solution isA. `8 xx 10^(-4)`B. `2xx10^(-14)`C. `8xx10^(10)`D. `1.25 xx 10^(-11)` |
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Answer» Correct Answer - D `[H^(+)]=sqrt(alpha^(2)C^(2))=sqrt((0.02)^(2)*(0.04)^(2))=8xx10^(-4)` `[H^(+)][OH^(-)]=10^(-14)` `:. [OH^(-)]=(10^(-14))/(8xx10^(-4))=1.25xx10^(-11)M` |
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| 15. |
The number of electrons required to reduced ` 4.5 xx 10^(-5)g` of Al isA. ` 1.03xx10^(18)`B. `3.01 xx 10^(18)`C. `4.95 xx 10^(26)`D. `7.31 xx 10^(20)` |
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Answer» Correct Answer - B `underset(27)(Al^(3+)) +3e^(-) to underset(27g)(Al)` 27 g of Al is reduced by `=3 xx 6.023xx10^(23)e^(-)s` ` 4.5xx10^(-5)` g of Al will be reduced by `=(3xx6.023xx10^(23) xx 4.5xx10^(-5))/(27)` `=3.01 xx 10^(18)` electrons |
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| 16. |
If `A=[(3,2,4),(1,2,1),(3,2,6)]` and `A_(ij)` are the cofactors of `a_(ij)`, then `a_(11)A_(11)+a_(12)A_(12)+a_(13)A_(13)` is equal toA. 8B. 6C. 4D. 0 |
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Answer» Correct Answer - A `a_(11)A_(11)+a_(12)A_(12)+a_(13)A_(13)` `=3|{:(2,1),(2,6):}|-2|{:(1,1),(3,6):}|+4|{:(1,2),(3,2):}|` `=3(12-2)-2(6-3)+4(2-6)` `=30-6-16` `=8` |
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| 17. |
The radius of hydrogen atom in its ground state is `5.3 xx 10^-11 m`. After collision with an electron it is found to have a radius of `21.2 xx 10^-11 m`. The principal quantum number of the final state of the atom is.A. n=4B. n=4C. n=16D. n=3 |
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Answer» Correct Answer - B ` r prop n^(2)` i.e, ` (r_(f))/( r_(i))=((n_(f))/(n_(i)))^(2)` `implies (21.2 xx 10^(-11))/(5.3xx10^(-11))=[(n)/(1)]^(2)` `implies n^(2)=4` `implies n=2 ` |
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| 18. |
In LED visible light is produced byA. gallium phosphideB. gallium arsenideC. germanium phosphideD. silicon phosphide |
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Answer» Correct Answer - B In material like gallium arsenide the number of photons of light energy is sufficient to produce quite intense visible light . |
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| 19. |
Same current is flowing in two alternating circuits. The first circuit contains only inductances and the other contains only a capacitor, if the frequency of the e.m.f of AC is increased, the effect on the value of the current will beA. increased in the first circuit and decreases in the otherB. increases in both the circuitsC. decreases in both the circuitsD. decreases in the first circuit and increases in the other |
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Answer» Correct Answer - D `I_(L)=(V)/(X_(L)) and I_(C)=(V)/(X_(C))` i.e , `I_(L) prop (1)/(omega) and I_(C) prop omega ` `:.` with increase in ` omega,I_(L)` decrease while `I_C` increase . |
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| 20. |
In photoelectric effect if the intensity of light is doubled then maximum kinetic energy of photoelectrons will becomeA. doubleB. halfC. four timesD. no change |
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Answer» Correct Answer - D `K_("max")` of photoelectrons does not depend upon intesity of incident light . |
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| 21. |
If fringe width is 0.4 mm, the distance between fifth bright and third dark band on same side isA. 1 mmB. 2 mmC. 3 mmD. 4 mm |
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Answer» Correct Answer - A Position of nth bright fringe from central maxima ` x_(n_(1))=(n_(1)lambdaD)/(d)` here `n_(1)=5` `:. x_(n_(1))=(5 lambda D)/(d)` Position of nth dark fringe from central maxima `x_(n_(1))=(n_(1) lambda D)/( d)` here `n_(1)=5 ` ` :. x_(n_(1))=(5)/(2) (lambdaD)/(d)` `x_(n_(1))-x_(n)=(2.5 lambda D)/( d)=2.5 beta ` Given ` beta =0.4 mm` `implies x_(n_(1))-x_(n)=1 mm` |
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| 22. |
If an electron is revolving around the hydrogen nucleus at a distance of 0.1 nm,what should be its speed?A. ` 2.188 xx 10^(6) ms^(-1)`B. `1.094 xx 10^(6) ms^(-1)`C. `4.376 xx 10^(6) ms^(-1)`D. `1.529 xx 10^(6) ms^(-1)` |
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Answer» Correct Answer - D Electrostatic force = centripetal force `(1)/( 4pi epsi_(0))(Ze^(2))/(r^(2))=(mv^(2))/(r)` ` :. V=sqrt(((1)/( 4 pi epsi_(0))(Ze^(2))/(mr)))` `= sqrt((9 xx 10^(9) xx 1 xx(1.6 xx 10^(-19))^(2))/((9.1xx10^(-31))xx(0.1 xx 10^(-9))))=1.59xx10^(6) ms^(-1)` |
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| 23. |
A particle performing SHM has time period ` (2pi)/(sqrt(3))` and path lenght 4 cm. The displacement from mean position at which acceleration is equal to velocity isA. 0 cmB. 0.5 cmC. 1 cmD. 1.5 cm |
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Answer» Correct Answer - C Velocity ` v=omegasqrt(A^(2)-x^(2))` and acceleration ` =omega^(2)x` Given, `omega sqrt(A^(2)-x^(2))=omega^(2)x` or ` sqrt(A^(2)-x^(2))=omega^(2)x` ......(i) Given , ` T=(2pi)/(sqrt(3))` and ` omega=(2pi)/(T)=sqrt(3)` Substituting the value of ` omega ` in Eq(i) , we get ` sqrt(A^(2)-x^(2))=sqrt(3)x` `implies A =2 x ` As amplitude`=("path length")/(2)=2 cm ` `implies x=1 cm ` |
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| 24. |
In hydrogen atom, the electron is making `6.6xx10^(15) rev//sec` around the nucleus in an orbit of radius `0.528 A`. The magnetic moment `(A-m^(2)) will beA. ` 1 xx 10^(-15)`B. `1 xx 10^(-10)`C. `1 xx 10^(-23)`D. `1 xx 10^(-27)` |
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Answer» Correct Answer - C Current , I `=6.6xx10^(15) xx 1.6xx10^(-19)` `=10.5xx10^(-4)A `. Area `A=piR^(2)=3.142 xx (0.528)^(2) xx 10^(-20) m^(2)` So, Magnetic moment `M=IA=10.5xx10^(-4) xx 3.142 xx (0.528)^(2) xx 10^(-20)` `= 10 xx 10^(-24)=10^(-23)` units |
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| 25. |
If a simple pendulum oscillates with an amplitude of 50 mm and time period of 2 sec, then its maximum velocity isA. `0.10 ms^(-1)`B. `0.15 ms^(-1)`C. `0.8 ms^(-1)`D. `0.26 ms^(-1)` |
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Answer» Correct Answer - B `v_("max")=a omega = a xx (2 pi)/(T)=(50xx10^(-3)) xx (2pi)/(2)` =`0.15 ms^(-1) ` |
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| 26. |
Ratio of kinetic energy at mean position to potential energy at A/2 of a particle performing SHMA. ` 2:1`B. `4:1`C. `8:1`D. `1:1` |
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Answer» Correct Answer - B Kinetic energy `K=(1)/(2) momega^(2)(A^(2)-y^(2))` At mean position y=0 `K =(1)/(2) m omega^(2)(A^(2))` .... (i) Potential energy `U=(1)/(2) m omega^(2)y^(2)` `U " at " y=(A)/(2)` `U=(1)/(2)(mA^(2))/(4) omega^(2)` ......(ii) Dividing Eq (i) by Eq (ii) , we get `(KE)/(U)=(4)/(1)` |
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| 27. |
Beats are produced by two waves given by `y_(1)= a sin 2000 pi t` and `y_(2) =a sin 2008 pi t `. The number of beats heard per second isA. zeroB. oneC. fourD. eight |
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Answer» Correct Answer - C Number of beats per second `=n_(1)~n_(2)` `omega_(1)=2000pi=2pin_(1)` `implies n_(1)=1000` and ` omega_(2)=2008 pi=2pin_(2)` `implies n_(2)=1004` Number of beats heard per second `=1004-1000=4` |
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| 28. |
The periodic time of a particle doing simple harmonic motion is 4 second . The time taken by it to go from its mean position to half the maximum displacement (amplitude) isA. `2s`B. 1sC. `(2)/(3)s`D. `(1)/(3)s` |
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Answer» Correct Answer - D `y=A sin""((2pi)/(T))t ` `implies (A)/(2) =A sin""((2pi)/(4))t` `implies (pit)/(2)=(pi)/(6)` `implies t=(1)/(3) s ` |
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| 29. |
The equation of motion of a particle moving along a straight line is `s=2t^(3)-9t^(2)+12t`, where the units of `s` and `t` are centrimetre and second. The acceleration of the particle will be zero afterA. ` (3)/(2)s`B. `(2)/(3)s`C. `(1)/(2) s `D. `1 s ` |
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Answer» Correct Answer - A `(ds)/(dt)=6t^(2)-18t+12` Again , ` (d^(2)s)/(dt^(2))=12t-18`= acceleration If acceleration becomes zero , then `0=12t-18` `implies t=(3)/(2) s ` Hence , acceleration will be zero after `(3)/(2)s `. |
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| 30. |
If the equation given by `hxy+10x+6y+4=0` represents a pair of lines, then h is equal toA. 15B. 30C. 5D. 10 |
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Answer» Correct Answer - A Here , a=0 , b=0 `h=(h)/(2)`, g=5, f=3 and c=4 It represent a pair of lines `:.|{:( 0, h//2, 5),(h//2,0,3),(5,3,4):}|=0` `implies 0-(h)/(2)[4(h)/(2)-15]+5[(3h)/( 2)-0]=0` `implies -h^(2)+(h15)/(2)+(15h)/(2)=0` `implies h^(2)+15h=0` `implies h=0,15` |
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| 31. |
General solution of the differential equation ` (dy)/(dx)=(x+y+1)/(x+y-1)` is given byA. `x+y=log|x+y|+c`B. `x-y=log|x+y|+c`C. `y=x+log|x+y|+c`D. `y=x log|x+y|+c` |
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Answer» Correct Answer - C `(dy)/(dx)=(x+y+1)/(x+y-1)` Put ` x+y=t ` `implies 1+(dy)/(dx)=(dt)/(dx) ` `implies (dy)/(dx)=(dt)/(dx)-1` `:. " "(dt)/(dt)=(t+1+t-1)/(t-1)` `=(dt)/(dx)=(2t)/(t-1)` `implies((t-1)/(2t))dt=dx ` `implies ((1)/(2)-(1)/(2t))dt=dx ` On integrating , we get `(1)/(2)t-(1)/(2) log t=x +c_(1)` `implies t-logt=2x+ 2c_(1)` `implies x+y- log (x+y)=2x+2c_(1)` `implies y=x+ log( x +y)+c ` |
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| 32. |
If `theta` is the angle between the lines `ax^2 +2hxy + by^2 =0` , then angle between `x^2 + 2xy sectheta + y^2 =0` isA. `theta `B. `2 theta `C. `(theta)/(2)`D. `3 theta ` |
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Answer» Correct Answer - A Angle between the lines ` ax^(2)+2hxy+by^(2)=0` is ` tan theta=|(2 sqrt(h^(2)-ab))/(a+b)|` For ` x^(2)+2xy sec theta +y^(2)=0` `h= sec theta , a=b=1` `:. tan phi=|(2sqrt(sec^(2) theta-1))/(1+1)|` `=(2 tan theta )/(2)=tan theta ` `:.` Angle between ` x^(2)+2xy sec theta + y^(2)=0` is theta . |
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| 33. |
Find ` (dy)/(dx) , " if x " = 2 cos theta - cos 2 theta ` and y = 2sin theta - sin 2 theta `.A. `tan""(3 theta )/(2)`B. `- tan""(3 theta )/(2)`C. `cot""(3 theta )/(2)`D. `-cot""(3 theta)/(2)` |
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Answer» Correct Answer - A Given `x=2 cos theta - cos 2 theta ` and `y= 2 sin theta- sin 2 theta ` `(dy)/(d theta )=- 2 sin theta + 2 sin 2 theta ` and `(dy)/(d theta)=(2 cos theta - 2 cos 2 theta )/(-2 sin theta + 2 sin 2 theta )` `=( cos theta - cos 2 theata )/( sin 2 theta - sin theta )` `=( 2 sin""((theta+ 2 theta )/(2)) sin ""((2 theta - theta)/( 2)))/( 2 cos ""((theta+ 2cos theta )/(2)) sin""((2 theta -theta )/(2)))` `=tan""(3 theta )/(2)` |
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| 34. |
`A=[(cos theta, -sin theta),(sin theta, cos theta)]` and `AB=BA=l`, then `B` is equal toA. `[{:(-cos theta , sin theta ),( sin theta, cos theta ):}]`B. `[{:(cos theta , sin theta ),( -sin theta, cos theta ):}]`C. `[{:( - sin theta , cos theta),( cos theta , sin theta):}]`D. `[{:(sin theta ,- cos theta),(- cos theta , sin theta):}]` |
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Answer» Correct Answer - B Given `A=[{:( cos theta , - sin theta ),( sin theta , cos theta ):}]` and `AB=BA=I` `implies B=A^(-1)I=A^(-1)` `=(1)/( cos^(2) theta + sin^(2) theta)[{:( cos theta , sin theta ),(- sin theta , cos theta ):}]` `implies B=[{:( cos theta , sin theta),( - sin theta , cos theta ):}]` |
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| 35. |
The moments of inertia of two rotating bodies A and B are `I_(A)` and `I_(B). (I_(A) gt I_(B))` and their angular momenta are equal. Which one has greater `K.E.` ?A. `K_(A)=K_(B)`B. `K_(A)ne K_(B)`C. `K_(A) lt K_(B)`D. `K_(A)=2K_(B)` |
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Answer» Correct Answer - C Kinetic energy ` E=(L^(2))/(2I)` If angular momenta are equal , then ` E prop (1)/(I)` Kinetic energy E=K ( given in problem ) If ` I_(A) gt I_(B) ` then `K_(A) lt K_(B)` |
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| 36. |
The compound from which formic acid cannot be prepared isA. methyl alcoholB. carbon monoxide + NaOHC. glycerolD. methyl magnesium bromide |
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Answer» Correct Answer - D Formic acid (HCOOH) cannot be prepared from methyl magnesium bromide `(CH_(3)MgBr)` while from all other reagents , it is obtained as (a) `underset(" methyl alcohol ")(CH_(3)OH) overset([O])to HCHO overset([O])to HCOOH` (b) `CO+NaOH to HCOONa underset(Delta)overset(NaHSO_(4))to HCOOH+Na_(2)SO_(4)` (c) Glycerol + Oxalic acid `underset(H_(2)O)overset(383K)toHCOOH`+glycerol |
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| 37. |
If 20 mL of an acidic solution of pH 3 is diluted to 100 mL , the `H^(+)` ion concentration will beA. `1 xx 10^(-3) M`B. `2xx10^(-3)M`C. `2 xx 10^(-4)M`D. `0.02 xx 10^(-4)M` |
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Answer» Correct Answer - C pH=3 (given ) `:.[H^(+)]= 1 xx 10^(-3)M=1 xx 10^(-3)N` `N_(1)V_(1)=N_(2)V_(2)` ` 1 xx 10^(-3) xx 20=N_(2) xx 100` `N_(2)=2 xx 10^(-4)` `:. [H^(+)]= 2 xx 10^(-4)M` |
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| 38. |
Which of the following is true ?A. `int_(0)^(1) e^(x) dx=e `B. `int_(0)^(1) 2^(x) dx=log 2 `C. `int_(0)^(1)sqrt(x)dx=(2)/(3)`D. `int_(0)^(1) x dx =(1)/(3)` |
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Answer» Correct Answer - C (a) `int_(0)^(1) e^(x)dx=[e^(x)]_(0)^(1)=e-1` (b) ` int_(0)^(2) 2^(x) dx=[(2^(x))/(log_(e)2)]_(0)^(1)=(1)/(log 2)*(2-2^(0))=(1)/(log 2)` (c) `int_(0)^(1) sqrt(x)dx=[(x^(3//2))/(3//2)]_(0)^(1)=(2)/(3)` (d) ` int_(0)^(1)x dx =[(x^(2))/(2)]_(0)^(1)=(1)/(2)` |
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| 39. |
The value of ` ((Delta^(2))/(E))x^(3)` at h=1 isA. 8xB. 6xC. `5x^(2)`D. `6x^(2)` |
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Answer» Correct Answer - B `((Delta_(2))/E)x^(3)=(((E-1)^(2))/(E))x^(3)` `=((E^(2)-2E+1)/(E))x^(3)` `=E(x^(3))-2x^(3)+E^(-1)(x^(3))` `=(x+1)^(3)-2x^(3)+(x-1)^(3)` `=x^(3)+1+3x^(2)+3x-2x^(3)+x^(3)-1-3x^(2)+3x` `=6x` |
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| 40. |
Volume of the parallelopiped having vertices at ` O-=(0,0,0) , A-=(2,-2,1),B-=(5,-4,4), and C= (1,-2,4)` isA. 5 cu unitB. 10 cu unitC. 15 cu unitD. 20 cu unit |
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Answer» Correct Answer - B Given , `vec(OA) = 2 hati- 2 hatj+hatk` `vec(OB)=5hati-4hatj+4hatk and vec(OC)=hati-2hatj+4hatk` Volume of parallelopiped`=[vec(OA) vec(OB) vec(OC)]` `=|{:( 2,-2,1),(5,-4,4),(1,-2,4):}|` `=2(-16+8)+2(20-4)+1 (-10+4)` `=10` cu unit |
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| 41. |
`int x log x dx` is equal toA. `(x^(2))/(4) ( 2 log x-1)+c `B. `(x^(2))/(2)( 2 logx-1+c `C. `(x^(2))/(4)(2 log x +1)+c `D. `(x^(2))/(2) (2 log x +1)` |
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Answer» Correct Answer - A `int underset(II)(x) underset(I)(log x)dx=log x*""(x^(2))/(2)-int(1)/(x)*(x^(2))/(2) dx ` `=(x^(2))/(2) log x -(1)/(2)(x^(2))/(2)+c ` `(x^(2))/(4)(2 logx -1)+c ` |
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| 42. |
`int[sin(logx)+cos(logx)]dx`A. ` x cos( log x )+c`B. `cos ( log x )+c`C. `x sin ( logx)+c `D. `sin ( log x)+c ` |
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Answer» Correct Answer - C `int[ sin (log x)+ cos ( log x )] dx ` `= int (d)/(dx){x sin ( log x) } dx ` `= x sin ( log x )+c ` |
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| 43. |
Area bounded between the curve `x^(2)=y` and the line y = 4x isA. `(32)/(3)` sq. unitB. `(1)/(3)` sq unitC. `(8)/(3)` sq unitD. `(16)/(3)` sq unit |
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Answer» Correct Answer - A Given curves are ` x^(2)=y and y=4x` Intersection point are ` (0,0)` and (4,16) `:.` Required are `=int_(0)^(4) (4x-x^(2)) dx` `=[(4x^(2))/(2)-(x^(3))/(3)]_(0)^(4)` `=[32-(64)/(3)]` `=(32)/(3)` sq unit |
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| 44. |
Focus of hyperbola is `(+-3,0)` and equation of tangent is `2x+y-4=0`, find the equation of hyperbola isA. `4x^(2)-5y^(2)=20`B. `5x^(2)-4y^(2)=20`C. `4x^(2)-5y^(2)=1`D. `5x^(2)-4y^(2)=1 ` |
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Answer» Correct Answer - A Given , `(pm ae, 0)=(pm3, 0)` `implies ae=3 ` ` implies a^(2)e^(2)=9` `implies b^(2)+a^(2)=9` .............(i) `:. 2x+y-4=0` `implies y=-2x+4 ` is the tangent to the hyperbola . `:. (4)^(2)=a^(2)(-2)^(2)-b^(2)` `implies 4a^(2)-b^(2)=16` ..............(ii) On solving Eqs . (i) and (ii), we get ` a^(2)=5, b^(2)=4 ` `:.` Equation of hyperbola is ` (x^(2))/(5)-(y^(2))/(4)=1 ` `implies 4x^(2)-5y^(2)=20` |
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| 45. |
If `m_1` and `m_2` are the slopes of tangents to `x^2+y^2= 4` from the point `(3,2)`, then `m_1- m_2` is equal toA. `(5)/(12)`B. `(12)/(5)`C. `(3)/(2)`D. `0` |
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Answer» Correct Answer - B Equation of pair of tangent is ` SS_(1)=T^(2)` `implies (x^(2)+y^(2)-4)(9+4-4)=(3x +2y-4)^(2)` `= 5y^(2)+16y-12xy+ 24 x-52=0` `:. m_(1)+m_(2)=(-2h)/(b)=(12)/(5)` and `m_(1)m_(2)=0` Now, `m_(1)-m_(2)=sqrt((m_(1)+m_(2))^(2)-4m_(1)m_(2))` `sqrt(((5)/(5))^(2)-0)` `=(12)/(5)` |
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| 46. |
`int_(0)^(pi//2) ""(sin x - cos x)/( 1-sin x * cos x) dx ` is equal to |
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Answer» Correct Answer - A Let `I=int_(0)^(pi//2)""(sin x - cos x)/( 1- sinx cos x) dx` .......(i) On putting ` x=((pi)/(2)-x)` in Eq. (i) , we get `I=int_(0)^(pi//2)(sin((pi)/(2)-x)-cos((pi)/(2)-x))/(1-sin((pi)/(2)-x)cos""((pi)/(2)-x))dx ` `implies int_(0)^(pi//2)( cos x- sin x)/( 1- sinx cos s) dx` `implies int_(0)^(pi//2)""((sin x - cos x)/( 1- sinx cosx )) dx` On adding Eqs. (i) and (ii) ,we get `2I=int_(0)^(pi//2) 0 dx=0` `impliesI=0` |
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| 47. |
The equation of the tangent to the curve `y=4xe^(x)` at `(-1,(-4)/e)` isA. y=-1B. `y=-(4)/(e)`C. `x=-1`D. ` x=(-4)/(e)` |
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Answer» Correct Answer - B Given curve is `y=4xe^(x)` `(dy)/(dx)=4e^(x)+4xe^(x)` At `(-1,-(4)/(e))((dy)/(dx))_((-1,-4//e))=4e^(-1)+4(-1)e^(-1)=0` `:.` Equation of tangent is `(y+(4)/(e))=0(x+1)` `implies y=-(4)/(e)` |
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| 48. |
`int_(5)^(10) ""(1)/((x-1)(x-2))dx` is equal toA. `log""(27)/(32)`B. `log""(32)/(27)`C. `log""(8)/(9)`D. `log""(3)/(4)` |
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Answer» Correct Answer - B Let `I=int_(5)^(10)""(1)/((x-1)(x-2)) dx ` `=int_(5)^(10)[(-1)/(x-1)+(1)/(x-2)] dx ` `=[-log (x-1)+ log (x-2)]_(5)^(10)` `=- log 9 +log 8 + log 4- log 3` `= -2 log 3 + 3 log 2 + 2 log 2 - log 3 ` ` =- 3 log 3 + 5 log 2 ` `=- log 27 + log 32` `= log""(32)/(27)` |
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| 49. |
The approximate value of ` root(3)(28)` isA. 3.0037B. 3.037C. 3.0086D. 3.37 |
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Answer» Correct Answer - B Let `y=root(3)(28)` Taking log on both sides , we get `logy=(1)/(3)`log 28 `(1)/(3) xx 1.4472` `=0.4824` `implies y`= antilog (0.4824) = 3.037 ( approximately ) |
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| 50. |
Locus of the point of intersection of perpendicular tangents to the circle ` x^(2)+y^(2)=16` isA. `x^(2)+y^(2)=8`B. `x^(2)+y^(2)=32`C. `x^(2)+y^(2)=64`D. `x^(2)+y^(2)=16` |
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Answer» Correct Answer - B We know that that , if two perpendicular tangents to the circle ` x^(2)+y^(2)=a^(2)` meet at P, then the point P lies on a director circle. `:.` Required locus is ` x^(2)+y^(2)=32` |
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