Chemistry-
General
Easy

Question

 The product obtained in above reaction will be :

  1.    
  2.    
  3.    
  4.    

The correct answer is:

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The percentage of nitrogen in urea is:

The percentage of nitrogen in urea is:

Chemistry-General
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In a packet of 40 pens, 12 are red. So, what % age are red pens?

In a packet of 40 pens, 12 are red. So, what % age are red pens?

Maths-General
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Consider the following compounds, which of these will release CO2 with 5% NaHCO3 ?
i)
ii)
iii) 

Consider the following compounds, which of these will release CO2 with 5% NaHCO3 ?
i)
ii)
iii) 

Chemistry-General
parallel
General
Chemistry-

Identify the product' Y' in the following reaction sequence:

Identify the product' Y' in the following reaction sequence:

Chemistry-General
General
Chemistry-

In homogeneous catalytic reactions, there are three alternative paths A,B and C (shown in the figure) which one of the following indicates the relative ease with which the reaction can take place?

In homogeneous catalytic reactions, there are three alternative paths A,B and C (shown in the figure) which one of the following indicates the relative ease with which the reaction can take place?

Chemistry-General
General
Chemistry-

In a spontaneous adsorption process

In a spontaneous adsorption process

Chemistry-General
parallel
General
Chemistry-

Graph between log open parentheses fraction numerator x over denominator m end fraction close parentheses and log P is a straight line at angle 0 45 with intercept OA as shown Hence ,blank open parentheses fraction numerator x over denominator m end fraction close parentheses at a pressure of 2 atm is

Graph between log open parentheses fraction numerator x over denominator m end fraction close parentheses and log P is a straight line at angle 0 45 with intercept OA as shown Hence ,blank open parentheses fraction numerator x over denominator m end fraction close parentheses at a pressure of 2 atm is

Chemistry-General
General
Maths-

If 0 less than alpha less than fraction numerator pi over denominator 2 end fraction and  alpha to the power of pi left parenthesis cosec invisible function application 2 alpha right parenthesis times cos invisible function application 2 alpha plus pi left parenthesis cosec invisible function application 2 alpha sin invisible function application 2 alpha right parenthesis end exponent then

If 0 less than alpha less than fraction numerator pi over denominator 2 end fraction and  alpha to the power of pi left parenthesis cosec invisible function application 2 alpha right parenthesis times cos invisible function application 2 alpha plus pi left parenthesis cosec invisible function application 2 alpha sin invisible function application 2 alpha right parenthesis end exponent then

Maths-General
General
Maths-

Assertion : For a equals fraction numerator 1 over denominator square root of 3 end fraction the volume of the parallel piped formed by vectors i with ˆ on top plus a j with ˆ on top comma a i with ˆ on top plus j with ˆ on top plus k with ˆ on top and j with ˆ on top plus a k with ˆ on top is maximum (The vectors form a right-handed system)
Reason: The volume of the parallel piped having three coterminous edges stack a with ‾ on top comma stack b with ‾ on top and stack c with ‾ on top

For such questions, we should know the formula of a parallelepiped. We should also know how to take a scalar product.

Assertion : For a equals fraction numerator 1 over denominator square root of 3 end fraction the volume of the parallel piped formed by vectors i with ˆ on top plus a j with ˆ on top comma a i with ˆ on top plus j with ˆ on top plus k with ˆ on top and j with ˆ on top plus a k with ˆ on top is maximum (The vectors form a right-handed system)
Reason: The volume of the parallel piped having three coterminous edges stack a with ‾ on top comma stack b with ‾ on top and stack c with ‾ on top

Maths-General

For such questions, we should know the formula of a parallelepiped. We should also know how to take a scalar product.

parallel
General
Maths-

Assertion (A): Let a with rightwards arrow on top equals 3 i with ˆ on top minus j with ˆ on top comma b with rightwards arrow on top equals 2 i with ˆ on top plus j with ˆ on top minus 3 k with ˆ on top. If stack b with rightwards arrow on top equals stack b with rightwards arrow on top subscript 1 end subscript plus stack b with rightwards arrow on top subscript 2 end subscript such that stack b with rightwards arrow on top subscript 1 end subscript is collinear with stack a with rightwards arrow on top and stack b with rightwards arrow on top subscript 2 end subscript is perpendicular to stack a with rightwards arrow on top is possible, then b with rightwards arrow on top subscript 2 equals i with ˆ on top plus 3 j with ˆ on top minus 3 k with ˆ on top.
Reason (R): If stack a with rightwards arrow on top and stack b with rightwards arrow on top are non-zero, non-collinear vectors, then stack b with rightwards arrow on top can be expressed as stack b with rightwards arrow on top equals stack b with rightwards arrow on top subscript 1 end subscript plus stack b with rightwards arrow on top subscript 2 end subscript where stack b with rightwards arrow on top subscript 1 end subscript is collinear with stack a with rightwards arrow on top and stack b with rightwards arrow on top subscript 2 end subscript is perpendicular to stack a with rightwards arrow on top

Assertion (A): Let a with rightwards arrow on top equals 3 i with ˆ on top minus j with ˆ on top comma b with rightwards arrow on top equals 2 i with ˆ on top plus j with ˆ on top minus 3 k with ˆ on top. If stack b with rightwards arrow on top equals stack b with rightwards arrow on top subscript 1 end subscript plus stack b with rightwards arrow on top subscript 2 end subscript such that stack b with rightwards arrow on top subscript 1 end subscript is collinear with stack a with rightwards arrow on top and stack b with rightwards arrow on top subscript 2 end subscript is perpendicular to stack a with rightwards arrow on top is possible, then b with rightwards arrow on top subscript 2 equals i with ˆ on top plus 3 j with ˆ on top minus 3 k with ˆ on top.
Reason (R): If stack a with rightwards arrow on top and stack b with rightwards arrow on top are non-zero, non-collinear vectors, then stack b with rightwards arrow on top can be expressed as stack b with rightwards arrow on top equals stack b with rightwards arrow on top subscript 1 end subscript plus stack b with rightwards arrow on top subscript 2 end subscript where stack b with rightwards arrow on top subscript 1 end subscript is collinear with stack a with rightwards arrow on top and stack b with rightwards arrow on top subscript 2 end subscript is perpendicular to stack a with rightwards arrow on top

Maths-General
General
Maths-

Statement negative 1 colon If a comma b comma c are distinct non-negative numbers and the vectors â plus a j with ˆ on top plus c k with ˆ on top comma i with ˆ on top plus k with ˆ on top and c i with ˆ on top plus c j with ˆ on top plus b k with ˆ on top are coplanar then c is arithmetic mean of a and b.
Statement -2: Parallel vectors have proportional direction ratios.

Statement negative 1 colon If a comma b comma c are distinct non-negative numbers and the vectors â plus a j with ˆ on top plus c k with ˆ on top comma i with ˆ on top plus k with ˆ on top and c i with ˆ on top plus c j with ˆ on top plus b k with ˆ on top are coplanar then c is arithmetic mean of a and b.
Statement -2: Parallel vectors have proportional direction ratios.

Maths-General
General
Maths-

If a with ‾ on top equals i plus j minus k comma b with ‾ on top equals 2 i plus j minus 3 k and stack r with ‾ on top is a vector satisfying 2 stack r with ‾ on top plus stack r with ‾ on top cross times stack a with ‾ on top equals stack b with ‾ on top.
Assertion left parenthesis A right parenthesis colon stack r with ‾ on top can be expressed in terms of stack a with ‾ on top comma stack b with ‾ on top and stack a with ‾ on top cross times stack b with ‾ on top.
Reason left parenthesis R right parenthesis colon r with ‾ on top equals 1 over 7 left parenthesis 7 i plus 5 j minus 9 k plus a with ‾ on top cross times b with ‾ on top right parenthesis

If a with ‾ on top equals i plus j minus k comma b with ‾ on top equals 2 i plus j minus 3 k and stack r with ‾ on top is a vector satisfying 2 stack r with ‾ on top plus stack r with ‾ on top cross times stack a with ‾ on top equals stack b with ‾ on top.
Assertion left parenthesis A right parenthesis colon stack r with ‾ on top can be expressed in terms of stack a with ‾ on top comma stack b with ‾ on top and stack a with ‾ on top cross times stack b with ‾ on top.
Reason left parenthesis R right parenthesis colon r with ‾ on top equals 1 over 7 left parenthesis 7 i plus 5 j minus 9 k plus a with ‾ on top cross times b with ‾ on top right parenthesis

Maths-General
parallel
General
Maths-

If stack a with minus on top comma stack b with minus on top are non-zero vectors such that vertical line stack a with minus on top plus stack b with minus on top vertical line equals vertical line stack a with minus on top minus 2 stack b with minus on top vertical line then
Assertion left parenthesis A right parenthesis : Least value of stack a with ‾ on top times stack b with ‾ on top plus fraction numerator 4 over denominator vertical line stack b with ‾ on top vertical line to the power of 2 end exponent plus 2 end fraction is 2 square root of 2 minus 1
Reason (R): The expression stack a with minus on top times stack b with minus on top plus fraction numerator 4 over denominator vertical line stack b with minus on top vertical line to the power of 2 end exponent plus 2 end fraction is least when magnitude of stack b with minus on top is square root of 2 t a n invisible function application open parentheses fraction numerator pi over denominator 8 end fraction close parentheses end root

If stack a with minus on top comma stack b with minus on top are non-zero vectors such that vertical line stack a with minus on top plus stack b with minus on top vertical line equals vertical line stack a with minus on top minus 2 stack b with minus on top vertical line then
Assertion left parenthesis A right parenthesis : Least value of stack a with ‾ on top times stack b with ‾ on top plus fraction numerator 4 over denominator vertical line stack b with ‾ on top vertical line to the power of 2 end exponent plus 2 end fraction is 2 square root of 2 minus 1
Reason (R): The expression stack a with minus on top times stack b with minus on top plus fraction numerator 4 over denominator vertical line stack b with minus on top vertical line to the power of 2 end exponent plus 2 end fraction is least when magnitude of stack b with minus on top is square root of 2 t a n invisible function application open parentheses fraction numerator pi over denominator 8 end fraction close parentheses end root

Maths-General
General
Maths-

Statement- 1: If a with rightwards arrow on top equals 3 i with ˆ on top minus 3 j with ˆ on top plus k with ˆ on top comma b with rightwards arrow on top equals negative i with ˆ on top plus 2 j with ˆ on top plus k with ˆ on top and c with rightwards arrow on top equals i with ˆ on top plus j with ˆ on top plus k with ˆ on top and d with rightwards arrow on top equals 2 i with ˆ on top minus j with ˆ on top, then there exist real numbers alpha comma beta, gamma such that stack a with rightwards arrow on top equals alpha stack b with rightwards arrow on top plus beta stack c with rightwards arrow on top plus gamma d
Statement- 2: stack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on top comma stack d with rightwards arrow on top are four vectors in a 3 - dimensional space. If stack b with rightwards arrow on top comma stack c with rightwards arrow on top comma stack d with rightwards arrow on top are non-coplanar, then there exist real numbers alpha comma beta comma gamma such that stack a with rightwards arrow on top equals alpha stack b with rightwards arrow on top plus beta stack c with rightwards arrow on top plus gamma stack d with rightwards arrow on top

Statement- 1: If a with rightwards arrow on top equals 3 i with ˆ on top minus 3 j with ˆ on top plus k with ˆ on top comma b with rightwards arrow on top equals negative i with ˆ on top plus 2 j with ˆ on top plus k with ˆ on top and c with rightwards arrow on top equals i with ˆ on top plus j with ˆ on top plus k with ˆ on top and d with rightwards arrow on top equals 2 i with ˆ on top minus j with ˆ on top, then there exist real numbers alpha comma beta, gamma such that stack a with rightwards arrow on top equals alpha stack b with rightwards arrow on top plus beta stack c with rightwards arrow on top plus gamma d
Statement- 2: stack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on top comma stack d with rightwards arrow on top are four vectors in a 3 - dimensional space. If stack b with rightwards arrow on top comma stack c with rightwards arrow on top comma stack d with rightwards arrow on top are non-coplanar, then there exist real numbers alpha comma beta comma gamma such that stack a with rightwards arrow on top equals alpha stack b with rightwards arrow on top plus beta stack c with rightwards arrow on top plus gamma stack d with rightwards arrow on top

Maths-General
General
Maths-

Statement- 1 open parentheses S subscript 1 end subscript close parentheses:If A open parentheses x subscript 1 end subscript comma y subscript 1 end subscript close parentheses comma B open parentheses x subscript 2 end subscript comma y subscript 2 end subscript close parentheses comma C open parentheses x subscript 3 end subscript comma y subscript 3 end subscript close parentheses are non-collinear points. Then every point left parenthesis x comma y right parenthesis in the plane of capital delta to the power of text le  end text end exponent A B C, can be expressed in the form open parentheses fraction numerator k x subscript 1 end subscript plus l x subscript 2 end subscript plus m x subscript 3 end subscript over denominator k plus l plus m end fraction comma fraction numerator k y subscript 1 end subscript plus l y subscript 2 end subscript plus m y subscript 3 end subscript over denominator k plus l plus m end fraction close parentheses
Statement- 2 open parentheses S subscript 2 end subscript close parentheses:The condition for coplanarity of four points A left parenthesis stack a with ‾ on top right parenthesis comma B left parenthesis stack b with ‾ on top right parenthesis comma C left parenthesis stack c with ‾ on top right parenthesis comma D left parenthesis stack d with ‾ on top right parenthesis is that there exists scalars 1 comma m comma n comma p not all zeros such that  l a with ‾ on top plus m b with ‾ on top plus n c with ‾ on top plus p d with ‾ on top equals 0 with minus on top where l plus m plus n plus p equals 0.

Statement- 1 open parentheses S subscript 1 end subscript close parentheses:If A open parentheses x subscript 1 end subscript comma y subscript 1 end subscript close parentheses comma B open parentheses x subscript 2 end subscript comma y subscript 2 end subscript close parentheses comma C open parentheses x subscript 3 end subscript comma y subscript 3 end subscript close parentheses are non-collinear points. Then every point left parenthesis x comma y right parenthesis in the plane of capital delta to the power of text le  end text end exponent A B C, can be expressed in the form open parentheses fraction numerator k x subscript 1 end subscript plus l x subscript 2 end subscript plus m x subscript 3 end subscript over denominator k plus l plus m end fraction comma fraction numerator k y subscript 1 end subscript plus l y subscript 2 end subscript plus m y subscript 3 end subscript over denominator k plus l plus m end fraction close parentheses
Statement- 2 open parentheses S subscript 2 end subscript close parentheses:The condition for coplanarity of four points A left parenthesis stack a with ‾ on top right parenthesis comma B left parenthesis stack b with ‾ on top right parenthesis comma C left parenthesis stack c with ‾ on top right parenthesis comma D left parenthesis stack d with ‾ on top right parenthesis is that there exists scalars 1 comma m comma n comma p not all zeros such that  l a with ‾ on top plus m b with ‾ on top plus n c with ‾ on top plus p d with ‾ on top equals 0 with minus on top where l plus m plus n plus p equals 0.

Maths-General
parallel

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