Chemistry-
General
Easy

Question

Two volatile liquids A and B differ in their boiling points by 0 15 C The process which can be used to separate them is

  1. Fractional distillation    
  2. Steam distillation    
  3. Distillation under reduced pressure    
  4. Simple distillation    

The correct answer is: Fractional distillation

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General
Maths-

Consider the ellipse whose equation is fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 For certain pair of diameters of the above ellipse, the product of their slopes is equal to negative fraction numerator b to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction
The equations of two diameters are y equals x and 3 y equals negative 2 x satisfying the condition mentioned above. Then the eccentricity of the ellipse is

Consider the ellipse whose equation is fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 For certain pair of diameters of the above ellipse, the product of their slopes is equal to negative fraction numerator b to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction
The equations of two diameters are y equals x and 3 y equals negative 2 x satisfying the condition mentioned above. Then the eccentricity of the ellipse is

Maths-General
General
Maths-

The smallest possible value of Sequals a subscript 1 end subscript times a subscript 2 end subscript times a subscript 3 end subscript plus b subscript 1 end subscript times b subscript 2 end subscript times b subscript 3 end subscript plus c subscript 1 end subscript times c subscript 2 end subscript times c subscript 3 end subscript text end textwhere text end text a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript comma b subscript 1 end subscript comma b subscript 2 end subscript comma b subscript 3 end subscript comma c subscript 1 end subscript comma c subscript 2 end subscript comma c subscript 3 end subscriptis a permutation of the number 1, 2, 3, 4, 5, 6, 7, 8, and 9 is

The smallest possible value of Sequals a subscript 1 end subscript times a subscript 2 end subscript times a subscript 3 end subscript plus b subscript 1 end subscript times b subscript 2 end subscript times b subscript 3 end subscript plus c subscript 1 end subscript times c subscript 2 end subscript times c subscript 3 end subscript text end textwhere text end text a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript comma b subscript 1 end subscript comma b subscript 2 end subscript comma b subscript 3 end subscript comma c subscript 1 end subscript comma c subscript 2 end subscript comma c subscript 3 end subscriptis a permutation of the number 1, 2, 3, 4, 5, 6, 7, 8, and 9 is

Maths-General
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Maths-

Statement 1 : The second degree equation 4 x to the power of 2 end exponent plus 9 y to the power of 2 end exponent minus 24 x plus 36 y minus 72 equals 0 represents an ellipse Because
Statement 2 : a x to the power of 2 end exponent plus 2 h x y plus b y to the power of 2 end exponent plus 2 g x plus 2 f y plus c equals 0 represents an ellipse if capital delta equals a b c plus 2 f g h minus a f to the power of 2 end exponent minus b g to the power of 2 end exponent minus c h to the power of 2 end exponent not equal to 0 and h to the power of 2 end exponent greater than a b

Statement 1 : The second degree equation 4 x to the power of 2 end exponent plus 9 y to the power of 2 end exponent minus 24 x plus 36 y minus 72 equals 0 represents an ellipse Because
Statement 2 : a x to the power of 2 end exponent plus 2 h x y plus b y to the power of 2 end exponent plus 2 g x plus 2 f y plus c equals 0 represents an ellipse if capital delta equals a b c plus 2 f g h minus a f to the power of 2 end exponent minus b g to the power of 2 end exponent minus c h to the power of 2 end exponent not equal to 0 and h to the power of 2 end exponent greater than a b

Maths-General
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General
Maths-

Statement-1: P is any point such that the chord of contact of tangents from P to the ellipse x to the power of 2 end exponent plus 2 y to the power of 2 end exponent equals 6 touches x to the power of 2 end exponent plus 4 y to the power of 2 end exponent equals 4 The the tangents from P of x to the power of 2 end exponent plus 2 y to the power of 2 end exponent equals 6 are at right angles and
Statement-2: The tangent from any point on the director circle of an ellipse are at right angles

Statement-1: P is any point such that the chord of contact of tangents from P to the ellipse x to the power of 2 end exponent plus 2 y to the power of 2 end exponent equals 6 touches x to the power of 2 end exponent plus 4 y to the power of 2 end exponent equals 4 The the tangents from P of x to the power of 2 end exponent plus 2 y to the power of 2 end exponent equals 6 are at right angles and
Statement-2: The tangent from any point on the director circle of an ellipse are at right angles

Maths-General
General
Maths-

Statement-1: The distance of the tangent through (0, (d) from a focus of the ellipse fraction numerator x to the power of 2 end exponent over denominator 16 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 7 end fraction equals 1 is 5
and
Statement-2: The locus of the foot of the perpendicular from a focus on any tangent to fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 is x to the power of 2 end exponent plus y to the power of 2 end exponent equals a to the power of 2 end exponent

Statement-1: The distance of the tangent through (0, (d) from a focus of the ellipse fraction numerator x to the power of 2 end exponent over denominator 16 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 7 end fraction equals 1 is 5
and
Statement-2: The locus of the foot of the perpendicular from a focus on any tangent to fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 is x to the power of 2 end exponent plus y to the power of 2 end exponent equals a to the power of 2 end exponent

Maths-General
General
Maths-

There are ‘n’ numbered seats around a round table. Total number of ways in which n n1 1 () < n persons can sit around the round table is equal to

There are ‘n’ numbered seats around a round table. Total number of ways in which n n1 1 () < n persons can sit around the round table is equal to

Maths-General
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General
Chemistry-

The IUPAC name of  is

The IUPAC name of  is

Chemistry-General
General
Maths-

If the normal at any given point P on the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 meets its
auxiliary circle at Q and R such that QOR = 90, where O is the centre
of ellipse, then

If the normal at any given point P on the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 meets its
auxiliary circle at Q and R such that QOR = 90, where O is the centre
of ellipse, then

Maths-General
General
Maths-

8-digit numbers are formed using the digits 1, 1, 2, 2, 2, 3, 4, 4. The number of such numbers in which the odd digits do not occupy odd places is

8-digit numbers are formed using the digits 1, 1, 2, 2, 2, 3, 4, 4. The number of such numbers in which the odd digits do not occupy odd places is

Maths-General
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General
Maths-

Given an in equation open vertical bar 1 plus fraction numerator 2 over denominator x end fraction close vertical bar greater than 3 whose solution set is given by left parenthesis a comma 0 right parenthesis union left parenthesis 0 comma b right parenthesis then answer the following questions If solution set for left parenthesis x plus 1 right parenthesis to the power of 2 end exponent less than left parenthesis 7 x minus 3 right parenthesis blankis (c,d), then a+b+c+d=

Given an in equation open vertical bar 1 plus fraction numerator 2 over denominator x end fraction close vertical bar greater than 3 whose solution set is given by left parenthesis a comma 0 right parenthesis union left parenthesis 0 comma b right parenthesis then answer the following questions If solution set for left parenthesis x plus 1 right parenthesis to the power of 2 end exponent less than left parenthesis 7 x minus 3 right parenthesis blankis (c,d), then a+b+c+d=

Maths-General
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Chemistry-

IUPAC name of ethers is

IUPAC name of ethers is

Chemistry-General
General
Chemistry-

The functional group present in acylchlorides is

The functional group present in acylchlorides is

Chemistry-General
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General
Maths-

If Z subscript 1 end subscript comma Z subscript 2 end subscript comma Z subscript 3 end subscript are complex numbers such that A open parentheses Z subscript 1 end subscript close parentheses comma B open parentheses Z subscript 2 end subscript close parentheses comma C open parentheses Z subscript 3 end subscript close parentheses are vertices of a triangle ABC,
angle A equals theta comma fraction numerator A C over denominator A B end fraction equals lambda and open parentheses 1 plus lambda to the power of 2 end exponent minus 2 lambda c o s invisible function application theta close parentheses Z subscript 1 end subscript superscript 2 end superscript plus lambda to the power of 2 end exponent Z subscript 3 end subscript superscript 2 end superscript plus Z subscript 3 end subscript superscript 2 end superscript equals 2 Z subscript 1 end subscript open parentheses open parentheses lambda to the power of 2 end exponent minus lambda c o s invisible function application theta close parentheses Z subscript 2 end subscript equals left parenthesis 1 minus lambda c o s invisible function application theta right parenthesis Z subscript 3 end subscript close parentheses plus 2 lambda c o s invisible function application theta Z subscript 2 end subscript Z subscript 3 end subscript then
If open parentheses 1 plus lambda to the power of 2 end exponent close parentheses Z subscript 1 end subscript superscript 2 end superscript plus lambda to the power of 2 end exponent Z subscript 2 end subscript superscript 2 end superscript plus Z subscript 3 end subscript superscript 2 end superscript equals 2 Z subscript 1 end subscript open parentheses lambda to the power of 2 end exponent Z subscript 2 end subscript plus Z subscript 3 end subscript close parentheses then the triangle is

If Z subscript 1 end subscript comma Z subscript 2 end subscript comma Z subscript 3 end subscript are complex numbers such that A open parentheses Z subscript 1 end subscript close parentheses comma B open parentheses Z subscript 2 end subscript close parentheses comma C open parentheses Z subscript 3 end subscript close parentheses are vertices of a triangle ABC,
angle A equals theta comma fraction numerator A C over denominator A B end fraction equals lambda and open parentheses 1 plus lambda to the power of 2 end exponent minus 2 lambda c o s invisible function application theta close parentheses Z subscript 1 end subscript superscript 2 end superscript plus lambda to the power of 2 end exponent Z subscript 3 end subscript superscript 2 end superscript plus Z subscript 3 end subscript superscript 2 end superscript equals 2 Z subscript 1 end subscript open parentheses open parentheses lambda to the power of 2 end exponent minus lambda c o s invisible function application theta close parentheses Z subscript 2 end subscript equals left parenthesis 1 minus lambda c o s invisible function application theta right parenthesis Z subscript 3 end subscript close parentheses plus 2 lambda c o s invisible function application theta Z subscript 2 end subscript Z subscript 3 end subscript then
If open parentheses 1 plus lambda to the power of 2 end exponent close parentheses Z subscript 1 end subscript superscript 2 end superscript plus lambda to the power of 2 end exponent Z subscript 2 end subscript superscript 2 end superscript plus Z subscript 3 end subscript superscript 2 end superscript equals 2 Z subscript 1 end subscript open parentheses lambda to the power of 2 end exponent Z subscript 2 end subscript plus Z subscript 3 end subscript close parentheses then the triangle is

Maths-General
General
Maths-

The general procedure for solving inequation containing modulus functions is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach.
For example:
i) The inequation f left parenthesis x right parenthesis minus vertical line f left parenthesis x right parenthesis vertical line greater than 0 has no solution.
ii) Solving the inequaiton f left parenthesis x right parenthesis plus vertical line f left parenthesis x right parenthesis vertical line greater than 0 is equivalent to solving the equations 0 less than f left parenthesis x right parenthesis less than infinity
iii) Solving the inequation vertical line f left parenthesis x right parenthesis vertical line minus f left parenthesis x right parenthesis greater than 0is equivalent to solving the equations negative infinity less than f left parenthesis x right parenthesis less than 0
iv) Solving the inequation f left parenthesis x right parenthesis minus vertical line f left parenthesis x right parenthesis vertical line greater or equal than 0is equivalent to solving the equations 0 less or equal than f left parenthesis x right parenthesis less than infinity
v) The inequation f left parenthesis x right parenthesis plus vertical line f left parenthesis x right parenthesis vertical line greater or equal than 0is true for all x element of domain of f left parenthesis x right parenthesis
vi) The inequation vertical line f left parenthesis x right parenthesis vertical line minus f left parenthesis x right parenthesis greater or equal than 0is true for all x element of domain of f left parenthesis x right parenthesis
Set of all real values of x for which fraction numerator 1 over denominator square root of left parenthesis 3 vertical line x minus 7 vertical line minus 6 right parenthesis ∣ negative left parenthesis 3 vertical line x minus 7 vertical line minus 6 right parenthesis end root end fraction is defined, is

The general procedure for solving inequation containing modulus functions is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach.
For example:
i) The inequation f left parenthesis x right parenthesis minus vertical line f left parenthesis x right parenthesis vertical line greater than 0 has no solution.
ii) Solving the inequaiton f left parenthesis x right parenthesis plus vertical line f left parenthesis x right parenthesis vertical line greater than 0 is equivalent to solving the equations 0 less than f left parenthesis x right parenthesis less than infinity
iii) Solving the inequation vertical line f left parenthesis x right parenthesis vertical line minus f left parenthesis x right parenthesis greater than 0is equivalent to solving the equations negative infinity less than f left parenthesis x right parenthesis less than 0
iv) Solving the inequation f left parenthesis x right parenthesis minus vertical line f left parenthesis x right parenthesis vertical line greater or equal than 0is equivalent to solving the equations 0 less or equal than f left parenthesis x right parenthesis less than infinity
v) The inequation f left parenthesis x right parenthesis plus vertical line f left parenthesis x right parenthesis vertical line greater or equal than 0is true for all x element of domain of f left parenthesis x right parenthesis
vi) The inequation vertical line f left parenthesis x right parenthesis vertical line minus f left parenthesis x right parenthesis greater or equal than 0is true for all x element of domain of f left parenthesis x right parenthesis
Set of all real values of x for which fraction numerator 1 over denominator square root of left parenthesis 3 vertical line x minus 7 vertical line minus 6 right parenthesis ∣ negative left parenthesis 3 vertical line x minus 7 vertical line minus 6 right parenthesis end root end fraction is defined, is

Maths-General
General
Maths-

Ten different letters of English alphabet are given. Words with five letters are formed from these given letters. Then, the number of words which have at least one letter repeated is

Ten different letters of English alphabet are given. Words with five letters are formed from these given letters. Then, the number of words which have at least one letter repeated is

Maths-General
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