Chemistry-
General
Easy

Question

Arrange the following compounds in the decreasing order of reactivity for hydrolysis reaction:
1) C subscript 6 end subscript H subscript 5 end subscript C O C l
2)
3)
4)

  1. 2>4>1>3    
  2. 2>4>3>1    
  3. 1>2>3>4    
  4. 4>3>2>1    

The correct answer is: 2>4>1>3

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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

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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

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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.

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
parallel
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
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
parallel

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