Maths-
General
Easy

Question

let l ,m,n be the D.Cs of a vector stack r with rightwards arrow on top and a,b,c are three numbers such that a,b,c are proprotionals to l,m,n i.e, fraction numerator l over denominator a end fraction equals fraction numerator m over denominator b end fraction equals fraction numerator n over denominator c end fraction equals k. rightwards double arrow left parenthesis a comma b comma c right parenthesis are direction rations
rightwards double arrow l equals plus-or-minus fraction numerator a over denominator square root of a to the power of 2 end exponent plus b to the power of 2 end exponent plus c to the power of 2 end exponent end root end fraction comma m equals plus-or-minus fraction numerator b over denominator square root of a to the power of 2 end exponent plus b to the power of 2 end exponent plus c to the power of 2 end exponent end root end fraction comma n equals plus-or-minus fraction numerator c over denominator square root of a to the power of 2 end exponent plus b to the power of 2 end exponent plus c to the power of 2 end exponent end root end fraction
text If  end text stack r with rightwards arrow on top equals a i plus b j plus c k be a vector having drs left parenthesis a comma b comma c right parenthesis l equals fraction numerator a over denominator vertical line stack r with rightwards arrow on top vertical line end fraction comma m equals fraction numerator b over denominator vertical line stack r with rightwards arrow on top vertical line end fraction comma l equals n equals fraction numerator c over denominator vertical line stack r with rightwards arrow on top vertical line end fraction the angle between two lines whose DCs are open parentheses l subscript 1 end subscript comma m subscript 1 end subscript comma n subscript 1 end subscript close parentheses text  and  end text open parentheses l subscript 2 end subscript comma m subscript 2 end subscript comma n subscript 2 end subscript close parentheses is c o s invisible function application theta equals sum l subscript 1 end subscript l subscript 2 end subscript
A vector rhas length 21 and drs 2,3,6 the dcs of rwhen stack r with rightwards arrow on topmakes an obtuse angle with x-axis is

  1. 2/7,3/7,-6,/7    
  2. -2/7,-3/7,-6/7    
  3. 2/7,3/7,6/7    
  4. 2/7,-3/7,-6/7    

The correct answer is: -2/7,-3/7,-6/7

Related Questions to study

General
Maths-

integral fraction numerator s i n invisible function application 12 theta minus s i n invisible function application 9 theta over denominator 2 c o s invisible function application 7 theta minus 1 end fraction d theta equals to

integral fraction numerator s i n invisible function application 12 theta minus s i n invisible function application 9 theta over denominator 2 c o s invisible function application 7 theta minus 1 end fraction d theta equals to

Maths-General
General
Maths-

If A = open square brackets table row cell 12 minus 1 end cell row cell negative 112 end cell row cell 2 minus 11 end cell end table close square brackets, then det (adj. (adj. A)) is

If A = open square brackets table row cell 12 minus 1 end cell row cell negative 112 end cell row cell 2 minus 11 end cell end table close square brackets, then det (adj. (adj. A)) is

Maths-General
General
Maths-

Let S be a set containing n elements. Then the total number of binary operations on S is

Let S be a set containing n elements. Then the total number of binary operations on S is

Maths-General
parallel
General
Maths-

fraction numerator 5 over denominator 1.2.3 end fraction plus fraction numerator 7 over denominator 3.4.5 end fraction plus fraction numerator 9 over denominator 5.6.7 end fraction plus... is equal to

fraction numerator 5 over denominator 1.2.3 end fraction plus fraction numerator 7 over denominator 3.4.5 end fraction plus fraction numerator 9 over denominator 5.6.7 end fraction plus... is equal to

Maths-General
General
Maths-

The sum of fraction numerator 1 to the power of 2 end exponent.2 over denominator 1 factorial end fraction plus fraction numerator 2 to the power of 2 end exponent.3 over denominator 2 factorial end fraction plus fraction numerator 3 to the power of 2 end exponent.4 over denominator 3 factorial end fraction plus fraction numerator 4 to the power of 2 end exponent.5 over denominator 4 factorial end fraction plus... infinity equals

The sum of fraction numerator 1 to the power of 2 end exponent.2 over denominator 1 factorial end fraction plus fraction numerator 2 to the power of 2 end exponent.3 over denominator 2 factorial end fraction plus fraction numerator 3 to the power of 2 end exponent.4 over denominator 3 factorial end fraction plus fraction numerator 4 to the power of 2 end exponent.5 over denominator 4 factorial end fraction plus... infinity equals

Maths-General
General
Maths-

If x is nearly equal to 1 such that fraction numerator m x to the power of m end exponent minus n x to the power of n end exponent over denominator m minus n end fraction equals x to the power of lambda end exponent, then the value of l is

If x is nearly equal to 1 such that fraction numerator m x to the power of m end exponent minus n x to the power of n end exponent over denominator m minus n end fraction equals x to the power of lambda end exponent, then the value of l is

Maths-General
parallel
General
Maths-

Greatest term (numerically) in the expansion of open parentheses 5 minus 4 x close parentheses to the power of n end exponent if n equals negative 7 and x equals fraction numerator 1 over denominator 2 end fraction is

Greatest term (numerically) in the expansion of open parentheses 5 minus 4 x close parentheses to the power of n end exponent if n equals negative 7 and x equals fraction numerator 1 over denominator 2 end fraction is

Maths-General
General
Maths-

Co-efficient of x to the power of r end exponent in the expansion of fraction numerator 1 plus 4 x to the power of 2 end exponent plus x to the power of 4 end exponent over denominator open parentheses 1 minus x close parentheses to the power of 4 end exponent end fraction is

Co-efficient of x to the power of r end exponent in the expansion of fraction numerator 1 plus 4 x to the power of 2 end exponent plus x to the power of 4 end exponent over denominator open parentheses 1 minus x close parentheses to the power of 4 end exponent end fraction is

Maths-General
General
Chemistry-

The removal of two atoms or groups, one generally hydrogen open parentheses H to the power of plus end exponent close parentheses and the other a leaving group open parentheses L to the power of minus end exponent close parentheses resulting in the formation of unsaturated compound is known as elimination reaction.

In E subscript 1 end subscript (elimination) reactions, the C minus L bond is broken heterolytic ally (in step 1) to form a carbocation (as in S subscript N to the power of 1 end exponent end subscript reaction) in which open parentheses L to the power of minus end exponent close) is lost (rate determining step). The carbocation (in step 2) loses a proton from the beta-carbon atom by a base (nucleophile) to form an alkene. E subscript 1 end subscript reaction is favoured in compounds in which the leaving group is at secondary open parentheses 2 to the power of ring operator end exponent close parentheses to the power of ´ end exponent or tertiary open parentheses 3 to the power of ring operator end exponent close parentheses position. In E subscript 2 end subscript text end text left parenthesis e l i m i n a t i o n right parenthesis text end text r e a c t i o n s comma text end text t w o text end text s i g m a text end text b o n d s text end text a r e text end text b r o k e n text end text a n d text end text a formed simultaneously. E subscript 2 end subscript reactions occur in one step through a transition state.

E subscript 2 end subscript reactions are most common in haloalkanes (particularly, 1 to the power of ring operator end exponent) and better the leaving group higher is the E subscript 1 end subscript reaction. In E subscript 2 end subscript reactions, both the leaving groups should be antiplanar.
E subscript 1 end subscript cb (Elimination unimolecular conjugate base) reaction involves the removal of proton by. a conjugate base (step 1) to produce carbanion which loses a leaving group to form an alkene (step 2) and is a slow step.
2-Bromobutane on heating with concentrated solution of alcoholic KOH gives major product as:

The removal of two atoms or groups, one generally hydrogen open parentheses H to the power of plus end exponent close parentheses and the other a leaving group open parentheses L to the power of minus end exponent close parentheses resulting in the formation of unsaturated compound is known as elimination reaction.

In E subscript 1 end subscript (elimination) reactions, the C minus L bond is broken heterolytic ally (in step 1) to form a carbocation (as in S subscript N to the power of 1 end exponent end subscript reaction) in which open parentheses L to the power of minus end exponent close) is lost (rate determining step). The carbocation (in step 2) loses a proton from the beta-carbon atom by a base (nucleophile) to form an alkene. E subscript 1 end subscript reaction is favoured in compounds in which the leaving group is at secondary open parentheses 2 to the power of ring operator end exponent close parentheses to the power of ´ end exponent or tertiary open parentheses 3 to the power of ring operator end exponent close parentheses position. In E subscript 2 end subscript text end text left parenthesis e l i m i n a t i o n right parenthesis text end text r e a c t i o n s comma text end text t w o text end text s i g m a text end text b o n d s text end text a r e text end text b r o k e n text end text a n d text end text a formed simultaneously. E subscript 2 end subscript reactions occur in one step through a transition state.

E subscript 2 end subscript reactions are most common in haloalkanes (particularly, 1 to the power of ring operator end exponent) and better the leaving group higher is the E subscript 1 end subscript reaction. In E subscript 2 end subscript reactions, both the leaving groups should be antiplanar.
E subscript 1 end subscript cb (Elimination unimolecular conjugate base) reaction involves the removal of proton by. a conjugate base (step 1) to produce carbanion which loses a leaving group to form an alkene (step 2) and is a slow step.
2-Bromobutane on heating with concentrated solution of alcoholic KOH gives major product as:

Chemistry-General
parallel
General
Chemistry-

The removal of two atoms or groups, one generally hydrogen open parentheses H to the power of plus end exponent close parentheses and the other a leaving group open parentheses L to the power of minus end exponent close parentheses resulting in the formation of unsaturated compound is known as elimination reaction.

In E subscript 1 end subscript (elimination) reactions, the C minus L bond is broken heterolytic ally (in step 1) to form a carbocation (as in S subscript N to the power of 1 end exponent end subscript reaction) in which open parentheses L to the power of minus end exponent close) is lost (rate determining step). The carbocation (in step 2) loses a proton from the beta-carbon atom by a base (nucleophile) to form an alkene. E subscript 1 end subscript reaction is favoured in compounds in which the leaving group is at secondary open parentheses 2 to the power of ring operator end exponent close parentheses to the power of ´ end exponent or tertiary open parentheses 3 to the power of ring operator end exponent close parentheses position. In E subscript 2 end subscript text end text left parenthesis e l i m i n a t i o n right parenthesis text end text r e a c t i o n s comma text end text t w o text end text s i g m a text end text b o n d s text end text a r e text end text b r o k e n text end text a n d text end text a formed simultaneously. E subscript 2 end subscript reactions occur in one step through a transition state.

E subscript 2 end subscript reactions are most common in haloalkanes (particularly, 1 to the power of ring operator end exponent) and better the leaving group higher is the E subscript 1 end subscript reaction. In E subscript 2 end subscript reactions, both the leaving groups should be antiplanar.
E subscript 1 end subscript cb (Elimination unimolecular conjugate base) reaction involves the removal of proton by. a conjugate base (step 1) to produce carbanion which loses a leaving group to form an alkene (step 2) and is a slow step.
Isopropyl chloride on heating with concentrated solution of ethanolic KOH gives mainly:

The removal of two atoms or groups, one generally hydrogen open parentheses H to the power of plus end exponent close parentheses and the other a leaving group open parentheses L to the power of minus end exponent close parentheses resulting in the formation of unsaturated compound is known as elimination reaction.

In E subscript 1 end subscript (elimination) reactions, the C minus L bond is broken heterolytic ally (in step 1) to form a carbocation (as in S subscript N to the power of 1 end exponent end subscript reaction) in which open parentheses L to the power of minus end exponent close) is lost (rate determining step). The carbocation (in step 2) loses a proton from the beta-carbon atom by a base (nucleophile) to form an alkene. E subscript 1 end subscript reaction is favoured in compounds in which the leaving group is at secondary open parentheses 2 to the power of ring operator end exponent close parentheses to the power of ´ end exponent or tertiary open parentheses 3 to the power of ring operator end exponent close parentheses position. In E subscript 2 end subscript text end text left parenthesis e l i m i n a t i o n right parenthesis text end text r e a c t i o n s comma text end text t w o text end text s i g m a text end text b o n d s text end text a r e text end text b r o k e n text end text a n d text end text a formed simultaneously. E subscript 2 end subscript reactions occur in one step through a transition state.

E subscript 2 end subscript reactions are most common in haloalkanes (particularly, 1 to the power of ring operator end exponent) and better the leaving group higher is the E subscript 1 end subscript reaction. In E subscript 2 end subscript reactions, both the leaving groups should be antiplanar.
E subscript 1 end subscript cb (Elimination unimolecular conjugate base) reaction involves the removal of proton by. a conjugate base (step 1) to produce carbanion which loses a leaving group to form an alkene (step 2) and is a slow step.
Isopropyl chloride on heating with concentrated solution of ethanolic KOH gives mainly:

Chemistry-General
General
Chemistry-

Karl Ziegler reported that alkenes react with N-bromosuccinimide (NBS) in presence of light to give products resulting from substitution of hydrogen by bromjne at the allylic position, i.e., the position next to the double bond. Let us consider the halogenation of cyclohexene.

Energy level diagram for allylic, vinylic and alkylic, free radicals is .given below:


Above compound on treatment with NBS gives allylic bromides. How many product(s) will be obtained in this reaction?

Karl Ziegler reported that alkenes react with N-bromosuccinimide (NBS) in presence of light to give products resulting from substitution of hydrogen by bromjne at the allylic position, i.e., the position next to the double bond. Let us consider the halogenation of cyclohexene.

Energy level diagram for allylic, vinylic and alkylic, free radicals is .given below:


Above compound on treatment with NBS gives allylic bromides. How many product(s) will be obtained in this reaction?

Chemistry-General
General
Chemistry-

Karl Ziegler reported that alkenes react with N-bromosuccinimide (NBS) in presence of light to give products resulting from substitution of hydrogen by bromjne at the allylic position, i.e., the position next to the double bond. Let us consider the halogenation of cyclohexene.

Energy level diagram for allylic, vinylic and alkylic, free radicals is .given below:

Which of the following sequences is correct about C-H bond energy?

Karl Ziegler reported that alkenes react with N-bromosuccinimide (NBS) in presence of light to give products resulting from substitution of hydrogen by bromjne at the allylic position, i.e., the position next to the double bond. Let us consider the halogenation of cyclohexene.

Energy level diagram for allylic, vinylic and alkylic, free radicals is .given below:

Which of the following sequences is correct about C-H bond energy?

Chemistry-General
parallel
General
Chemistry-

Karl Ziegler reported that alkenes react with N-bromosuccinimide (NBS) in presence of light to give products resulting from substitution of hydrogen by bromjne at the allylic position, i.e., the position next to the double bond. Let us consider the halogenation of cyclohexene.

Energy level diagram for allylic, vinylic and alkylic, free radicals is .given below:

Consider the three types of C-H bonds in cyclohexene.

Which of the following is/are correctly matched?

Karl Ziegler reported that alkenes react with N-bromosuccinimide (NBS) in presence of light to give products resulting from substitution of hydrogen by bromjne at the allylic position, i.e., the position next to the double bond. Let us consider the halogenation of cyclohexene.

Energy level diagram for allylic, vinylic and alkylic, free radicals is .given below:

Consider the three types of C-H bonds in cyclohexene.

Which of the following is/are correctly matched?

Chemistry-General
General
Chemistry-

Karl Ziegler reported that alkenes react with N-bromosuccinimide (NBS) in presence of light to give products resulting from substitution of hydrogen by bromjne at the allylic position, i.e., the position next to the double bond. Let us consider the halogenation of cyclohexene.

Energy level diagram for allylic, vinylic and alkylic, free radicals is .given below:

In the treatment of cyclohexene with NBS, which of the following products will be least stable?

Karl Ziegler reported that alkenes react with N-bromosuccinimide (NBS) in presence of light to give products resulting from substitution of hydrogen by bromjne at the allylic position, i.e., the position next to the double bond. Let us consider the halogenation of cyclohexene.

Energy level diagram for allylic, vinylic and alkylic, free radicals is .given below:

In the treatment of cyclohexene with NBS, which of the following products will be least stable?

Chemistry-General
General
Chemistry-

The high reactivity of alkyl halides can be explained in terms of nature of C-X bond which is highly polarised covalent bond due to large difference in the electronegativities of carbon and halogen .atom. This polarity is responsible for the nucleophilic substitution reactions of alky- halides which mostly occur by S subscript N to the power of 1 end exponent end subscript and S subscript N to the power of 2 end exponent end subscript mechanisms S subscript N to the power of 1 end exponent end subscript reaction is a two step process . and in the first step, R-X ionises to give carbocation (slow process). In the second step, the nucleophile attacks the carbocation from either side to form the product (fast process). In S subscript N to the power of 1 end exponent end subscript reaction, there can be racemization and inversion. S subscript N to the power of 1 end exponent end subscript reaction is favoured by heavy (bulky) groups on the carbon atom attached to halogens. i.e., R subscript 3 end subscript C minus X greater than R subscript 2 end subscript C H minus X greater than R minus C H subscript 2 end subscript X greater than C H subscript 3 end subscript X
In S subscript N to the power of 2 end exponent end subscript reaction, the strong nucleophile OH- attacks from the opposite side of the chlorine atom to give an intermediate (transition state). which breaks to yield the product (alcohol) and leavj.ng open parentheses X to the power of minus end exponent close parentheses group. The alcohol has a configuration opposite to drat 'of the bromide and is said to proceed with inversion of configuration. S subscript N to the power of 2 end exponent end subscript reaction is favoured by small groups on the carbon atom attached to halogen i.e., C H subscript 3 end subscript minus X greater than R minus C H subscript 2 end subscript X greater than R subscript 2 end subscript C H X greater than R subscript 3 end subscript C minus X
The main product formed in the following reaction is:

The high reactivity of alkyl halides can be explained in terms of nature of C-X bond which is highly polarised covalent bond due to large difference in the electronegativities of carbon and halogen .atom. This polarity is responsible for the nucleophilic substitution reactions of alky- halides which mostly occur by S subscript N to the power of 1 end exponent end subscript and S subscript N to the power of 2 end exponent end subscript mechanisms S subscript N to the power of 1 end exponent end subscript reaction is a two step process . and in the first step, R-X ionises to give carbocation (slow process). In the second step, the nucleophile attacks the carbocation from either side to form the product (fast process). In S subscript N to the power of 1 end exponent end subscript reaction, there can be racemization and inversion. S subscript N to the power of 1 end exponent end subscript reaction is favoured by heavy (bulky) groups on the carbon atom attached to halogens. i.e., R subscript 3 end subscript C minus X greater than R subscript 2 end subscript C H minus X greater than R minus C H subscript 2 end subscript X greater than C H subscript 3 end subscript X
In S subscript N to the power of 2 end exponent end subscript reaction, the strong nucleophile OH- attacks from the opposite side of the chlorine atom to give an intermediate (transition state). which breaks to yield the product (alcohol) and leavj.ng open parentheses X to the power of minus end exponent close parentheses group. The alcohol has a configuration opposite to drat 'of the bromide and is said to proceed with inversion of configuration. S subscript N to the power of 2 end exponent end subscript reaction is favoured by small groups on the carbon atom attached to halogen i.e., C H subscript 3 end subscript minus X greater than R minus C H subscript 2 end subscript X greater than R subscript 2 end subscript C H X greater than R subscript 3 end subscript C minus X
The main product formed in the following reaction is:

Chemistry-General
parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.