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@Vinod

\\\int \sqrt{\frac{\sin \left(x-a\right)}{\sin \left(x+a\right)}}dx\\rationalize\;it\\\int \sqrt{\frac{\sin \left(x-a\right)\sin \left(x-a\right)}{\sin \left(x+a\right)\sin \left(x-a\right)}}dx\\\int \frac{\sin \left(x-a\right)}{\sqrt{\sin \:\left(x+a\right)\sin \:\left(x-a\right)}}dx\\\because \sin \left(A+B\right)\sin \left(A-B\right)=\sin ^2A-\sin ^2B\\\int \frac{\sin x\:\cos a-\sin a\:\cos x}{\sqrt{\sin ^2x-\sin ^2a}}dx\\\cos a\int \frac{\sin x\:}{\sqrt{\sin ^2x-\sin ^2a}}dx\:-\sin a\int \:\frac{\:\cos \:x}{\sqrt{\sin \:^2x-\sin \:^2a}}dx\:\:\:

\\\cos a\int \frac{\sin x\:}{\sqrt{\sin ^2x-\sin ^2a}}dx\:\\\cos \:a\int \frac{\sin \:x\:}{\sqrt{1-\cos ^2x-1+\cos ^2a}}dx\:\\\Rightarrow \cos \:a\int \frac{\sin \:x\:}{\sqrt{\cos \:^2a-\cos ^2x}}dx\:\\put\;\cos x=t,and\;solve\\same\;method\;apply\;for\;\sin a\int \:\frac{\:\cos \:x}{\sqrt{\sin \:^2x-\sin \:^2a}}dx\:\:\:

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Let \alpha and \beta   be the roots of equation

x^{2}-6x-2=0.\; if a_{n}=\alpha ^{n}-\beta ^{n},\; for\: n\geq 1,\; then\; the \: value\; of\; \frac{a_{10}-2a_{8}}{2a_{9}}

is equal to:

  • Option 1)

    6

  • Option 2)

    -6

  • Option 3)

    3

  • Option 4)

    -3

Option 1) 6 Option 2) -6 Option 3) 3 Option 4) -3

Compound X will be 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 

2

Find the compound "X" which give the following test:

Neutral FeCl3  -Ve

Fehling Solution  -ve

Iodoform reaction  +ve

Grignard reagent  +ve

 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 

1

Two particles move at right angle to each other de Broglie wavelengths as \lambda_1 and \lambda_2, particle suffers a perfectly inelastic collision. The de Broglie wavelength \lambda of final particle is given by

  • Option 1)

    \lambda=\frac{\lambda _{1}\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}

  • Option 2)

    \lambda=2\frac{\lambda _{1}\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}

  • Option 3)

    \lambda=\frac{\lambda _{1}^2\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}

  • Option 4)

    \lambda=\frac{\lambda _{1}\lambda _{2}}{2\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}

 

option 3

 

An interference experiment ratio of amplitude of coherent waves is . What is the ratio of maximum and minimum intensities of frings?

  • Option 1)

    1

  • Option 2)

    2

  • Option 3)

    3

  • Option 4)

    4

 
Option 1) 1 Option 2) 2 Option 3) 3 Option 4) 4

What is time at which the rate of dissipation is equal to the rate at which magnetic energy stored in the inductor?

  • Option 1)

    1.4 sec

  • Option 2)

    1.9 sec

  • Option 3)

    5 sec

  • Option 4)

    0.03 sec

option 3

 

Plane electromagnetic wave travelled in x-direction has electric field component E = 6V/m which is in y-direction. Corresponding Magnetic field component (B) is-

  • Option 1)

    B_0=2*10^{-8}T (in \ \ z-axis)

  • Option 2)

    B_0=6*10^{-8}T (in \ \ z-axis)

  • Option 3)

    B_0=12*10^{-8}T (in \ \ z-axis)

  • Option 4)

    B_0=2*10^{8}T (in \ \ z-axis)

Option 1)Option 2)Option 3)Option 4)

Bob at pendulum mass 2g, charge , an electric field of intensity 2000v/m. Angle of pendulum with vertical at equilibrium?

  • Option 1)

    \Theta = tan^{-1}{\frac{1}{2}}

  • Option 2)

    \Theta = cot^{-1}{\frac{1}{2}}

  • Option 3)

    \Theta = sin^{-1}{\frac{1}{2}}

  • Option 4)

    \Theta = tan^{-1}{\frac{1}{4}}

 
Option 1) Option 2) Option 3) Option 4)

Consider a Steel wire of radius= 2mm of a load equal to 4 kg to be suspended from a rigid support and its top end, such that it is in hanging in a vertical position. What is its tensile stress due to its own weight? (take g=3.1 \pi ms^{-2} )

  • Option 1)

    6.2*10^{6} Nm^2

  • Option 2)

    3.1*10^{-6} Nm^2

  • Option 3)

    3.1*10^{6} Nm^2

  • Option 4)

    3.1*10^{5} Nm^2

 
Option 1) Option 2) Option 3) Option 4)

If   gas molecules each of mass Kg collide over the area of   in unit second with  Speed , the pressure exerted will be?

  • Option 1)

    3Nm^{2}

  • Option 2)

    2 \ Nm^2

  • Option 3)

    4Nm^{2}

  • Option 4)

    6Nm^{2}

Option 1)Option 2)Option 3)Option 4)

In SI units, the dimension of is

  • Option 1)

  • Option 2)

    M^{-1}L^{2}T^{3}A

  • Option 3)

    M^{-1}L^{2}T^{5}A^{2}

  • Option 4)

    M^{-1}L^{2}T^{-3}A

Option 1)Option 2)Option 3)Option 4)

The major product formed in the reaction given below will be:

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

  Fate of aliphatic diazonium ion - The diazonium ions of aliphatic amines are very unstable and produces carbocation immediately, which can produce different products.  - wherein      Fate of aromatic diazonium ion - Diazonium salts of aromatic amines are comparatively more stable and evolve nitrogen only on heating.These diazonium salts can be isolated at low temperatures. -             As...

If in a parallelogram ABDC, the coordinates of A,B and C are respectively (1,2), (3,4) and (2,5) , then the equation of the diagonal AD is:

  • Option 1)

    5x-3y+1=0

  • Option 2)

    3x-5y+1=0

  • Option 3)

    5x+3y-11=0

  • Option 4)

    3x+5y-13=0

  Mid-point formula -   - wherein If the point P(x,y) is the mid point of line joining A(x1,y1) and B(x2,y2) .     Two – point form of a straight line -   - wherein The lines passes through    and    As BD and AC are parallel ..............................(1) As AB and CD are parallel ..............................(2) Solving (1) and (2) m=4 and n=7              Option 1)Option 2)Option...

50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is :

  • Option 1)

    20 g

  • Option 2)

    4 g

  • Option 3)

    10 g

  • Option 4)

    40 g

 
  Molar Mass - The mass of one mole of a substance in grams is called its molar mass. - wherein  Molar mass of water = 18 g mol-1       Molarity -  Molarity (M) = (Number of moles of solute)/(volume of solution in litres)   - wherein It is defined as the number of moles of the solute in 1 litre of the solution.       Stoichiometry - Stoichiometry deals with measurements of reactants and...

Match the catalysts (Column I) with products (Column II).

             ColumnI                                          Column II

            Catalyst                                               Product

      (A)   V_{2}O_{5}                                          (i) Polyethylene     

      (B)  TiCl_{4}/Al(Me)_{3}                      (ii) ethanal

      (C)   PdCl_{2}                                       (iii)   H_{2}SO_{4}

      (D) Iron\; Oxide                               (iv)   NH_{3}

 

 

  • Option 1)

     (A)-(iii); (B)-(iv);(C)-(i);(D)-(ii)

  • Option 2)

      (A)-(ii);(B)-(iii);(C)-(i);(D)-(iv)

  • Option 3)

     (A)-(iii);(B)-(i);(C)-(ii);(D)-(iv)

  • Option 4)

     (A)-(iv);(B)-(iii);(C)-(ii);(D)-(i)

 
      Contact Process - In this process SO2 obtained by burning of S or iron pyrites which is catalytically oxidised to  in presence of timely divided Pt or  as catalyst -       High density Polythene (HDP) - -  Polymerization of ethene in a hydrocarbon solvent under low pressure in presence of triethylaluminium and titanium tetrachloride (Ziegier. natta catalyst) -  Chain growth,...

The increasing order of nucleophilicity of the following nucleophilies is :

(a)\; CH_{3}CO_{2}^{\ominus }

(b)\; H_{2}O

(c)\; CH_{3}SO_{3}^{\ominus }

(d)\;\overset{\ominus }{O}H

  • Option 1)

    (a)< (d)< (c)< (b)

  • Option 2)

     (b)< (c)< (d)< (a)

  • Option 3)

     (d)< (a)< (c)< (b)

  • Option 4)

    (b)< (c)< (a)< (d)

    Nucleophile - Those e- rich  chemical species having capacity to attack e-deficient portion of substrate. - wherein Nucleophile must have complete octet and atleast one lone pair                      Order of Nucleophilicity is                                                                                                                                                   Both are...

The correct option among the following is :

  • Option 1)

    Colloidal medicines are more effective because they have small surface area.

  • Option 2)

    Addition of alum to water makes it unfit for drinking .

  • Option 3)

    Colloidal particles in lyophobic sols can be precipitated by electrophoresis.

  • Option 4)

     Brownian motion in colloidal solution is faster if the viscosity of the solution is very high.

    Properties of Colloidal Solutions - Coagulation or Precipitation - wherein The stability of the lyophobic sols is due to the presence of charge on colloidal particles. If somehow the charge is removed the particles will come nearer to each other and settle down under gravity.       Properties of Colloidal Solutions - Electrophoresis - wherein The existance of charge on colloidal...

Benzene diazonium chloride on reaction with aniline in the presence of dilute hydrochloric acid gives:

 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

    Coupling reaction of aromatic diazonium - Diazonium ions of aromatic amines also undergo coupling reaction with aromatic rings having a strong activating group to form diazo compounds. - wherein      Option 1)Option 2)Option 3)Option 4)

The metal that gives hydrogen gas upon treatment with both acid as well as base is :

  • Option 1)

    zinc

  • Option 2)

    iron

  • Option 3)

    mercury

  • Option 4)

    magnesium

 
    Oxidation states - Transition elements have a variety of oxidation states but the common oxidation state is +2 for 3d metals.   -     zinc is the metal that gives hydrogen gas on reacting with both acid & base (i)  (ii)  Option (I) is correct Option 1) zinc Option 2) iron Option 3) mercury Option 4) magnesium
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