Monday, August 23, 2021

Vector Algebra: Mathematics MCQ

 

Q.1. The vector  straight a with rightwards arrow on top equals straight alpha straight i with hat on top plus 2 straight j with hat on top plus straight beta straight k with hat on top lies in the plane of the vectors straight b with rightwards arrow on top equals straight i with hat on top plus straight j with hat on top space and space straight c with rightwards arrow on top equals straight j with hat on top plus straight k with hat on top and bisect the angle between b with rightwards arrow on top space a n d space c with rightwards arrow on top . Then which one of the following gives possible values of straight alpha space and space straight beta    (AIEEE 2008) 

A) alpha equals 1 comma space beta equals 2

B) alpha equals 2 comma space beta equals 1

C) alpha equals 1 comma space beta equals 1

D) alpha equals 2 comma space beta equals 2 


Q2. If 4i+7j+8k, 2i+3j+4k and 2i+5j+7k are the position vectors of the vertices A,B and C respectively of triangle ABC. The position vectors of the point where the bisectors of angle A meets BC is... (Pb. CET 2004)

A) 1 third left parenthesis 6 i plus 13 j plus 18 k right parenthesis

B) 2 over 3 left parenthesis 6 i _ 12 j minus 8 k right parenthesis

C) 1 third left parenthesis negative 6 i minus 8 j minus 9 k right parenthesis

D) 2 over 3 left parenthesis negative 6 i plus 12 j plus 8 k right parenthesis 


Q.3. Direction of a reciprocal vector of a vector a with rightwards arrow on top is    ( MP PET 2013)

a) Same as that of a with rightwards arrow on top

B) Opposite to a with rightwards arrow on top

C) Perpendicular to a with rightwards arrow on top

D) None of these


Q.4. If top enclose a equals i with hat on top plus j with hat on top plus k with hat on top comma space top enclose b equals 2 i with hat on top plus lambda j with hat on top plus k with hat on top comma space top enclose c equals i with hat on top minus j with hat on top plus 4 k with hat on top space space a n d space top enclose a. left parenthesis top enclose b cross times top enclose c right parenthesis equals 10 , then 

A) 6

B) 7

C) 9

D) 10

Explanation-

open square brackets a with bar on top space b with bar on top space c with bar on top close square brackets equals 10
open vertical bar
1 space space space space 1 space space space space space space 1
2 space space space space lambda space space space space space space 1
1 space space space minus 1 space space space space 4 space close vertical bar equals 10
left parenthesis 4 lambda plus 1 right parenthesis minus left parenthesis 8 minus 1 right parenthesis plus left parenthesis negative 2 minus lambda right parenthesis equals 10
4 lambda plus 1 minus 7 minus 2 minus lambda equals 10
3 lambda equals 10 plus 8 equals 18
lambda equals 6.             


Q.5. If the position vector of three points are straight a minus 2 straight b plus 3 straight c comma space 2 straight a plus 3 straight b minus 4 straight c comma negative 7 straight b plus 10 straight c comma then the three points are 

A) Collinear 

B) Non-coplanar

C) Non-collinear

D) Non of these 

Explanation-

Let P,Q, R be the given points. Then, we have 

                   QR=-2PQ

Hence, P,Q and R are collinear.


Q.6. If the vectors negative straight i with hat on top plus 3 straight j with hat on top minus 4 straight k with hat on top space and space 2 straight i with hat on top minus 6 straight j with hat on top plus straight lambda straight k with hat on top are collinear then find the value of straight lambda 

A) 4

B) 6

C) 8

D) 10

Explanation-

Let space top enclose straight alpha equals negative straight i with hat on top plus 3 straight j with hat on top minus 4 straight k with hat on top space and space top enclose straight b equals 2 straight i with hat on top minus 6 straight j with hat on top plus straight lambda straight k with hat on top
As space top enclose straight alpha space and space top enclose straight b space are space collinear space comma space top enclose straight alpha cross times straight b with minus on top equals 0
therefore space open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row cell negative 1 end cell 3 cell negative 4 end cell row 2 cell negative 6 end cell straight lambda end table close vertical bar equals 0
therefore space straight i with hat on top open parentheses 3 straight lambda minus 24 close parentheses minus straight j with hat on top open parentheses negative straight lambda plus 8 close parentheses plus straight k with hat on top open parentheses 4 minus 6 close parentheses equals top enclose 0
rightwards double arrow space 3 open parentheses straight lambda minus 8 close parentheses straight i with hat on top plus open parentheses straight lambda minus 8 close parentheses straight j with hat on top plus 0 straight k with hat on top equals 0 straight i with hat on top plus 0 straight j with hat on top plus 0 straight k with hat on top
rightwards double arrow space straight lambda minus 8 equals 0 space space space space space space space space space space space space rightwards double arrow straight lambda equals 8 


Q.7. If the origin is the centroid of the triangle whose vertices are A (a, 2, -1) , B (5, b, 2) and C (-3, 1, (C) then find the values of a , b and c .

A) a = 2 , b = 3 , c = 1

B) a = 1 , b = 2 , c = 3

C) a = - 2 , b = - 3 , c = 1

D) a = 2 , b = 3 , c = 1

Explanation-

Let space top enclose straight alpha equals straight alpha straight i with hat on top plus 2 straight j with hat on top minus straight k with hat on top comma space space space space space top enclose straight b equals 5 straight i with hat on top plus straight b straight j with hat on top plus 2 straight k with hat on top space space space and space space top enclose straight c equals negative 3 straight i with hat on top plus straight j with hat on top plus straight c straight k with hat on top.
If space straight O open parentheses top enclose straight o close parentheses space is space the space centriod space of space the space triangle space then space by space centriod space formula space.
top enclose straight o equals fraction numerator top enclose straight alpha plus top enclose straight b plus top enclose straight c over denominator 3 end fraction
space space equals fraction numerator open parentheses straight alpha plus 2 close parentheses straight i with hat on top plus left parenthesis 3 plus straight b right parenthesis straight j with hat on top plus open parentheses 1 plus straight c close parentheses straight k with hat on top over denominator 3 end fraction
space space equals fraction numerator open parentheses straight alpha straight i with hat on top plus 2 straight j with hat on top minus straight k with hat on top close parentheses plus open parentheses 5 straight i with hat on top plus straight b straight j with hat on top plus 2 straight k with hat on top close parentheses plus open parentheses negative 3 straight i with hat on top plus straight j with hat on top plus straight c straight k with hat on top close parentheses over denominator 3 end fraction
top enclose straight o space equals open parentheses fraction numerator straight alpha plus 2 over denominator 3 end fraction close parentheses straight i with hat on top plus open parentheses fraction numerator 3 plus straight b over denominator 3 end fraction close parentheses straight j with hat on top plus open parentheses fraction numerator 1 plus straight c over denominator 3 end fraction close parentheses straight k with hat on top
rightwards double arrow space fraction numerator straight alpha plus 2 over denominator 3 end fraction equals 0 space comma space fraction numerator 3 plus straight b over denominator 3 end fraction equals 0 comma space fraction numerator 1 plus straight c over denominator 3 end fraction equals 0
rightwards double arrow space straight alpha equals negative 2 comma space straight b equals negative 3 comma space straight c equals negative 1 space 


Q.8. If O be the origin and the position vector of A be 4 i +5 j , then a unit vector parallel to stack O A with rightwards arrow on top is

A) fraction numerator 3 over denominator square root of 41 end fraction open parentheses straight i plus 4 straight j close parentheses

B) fraction numerator 4 over denominator square root of 41 end fraction open parentheses straight i minus 6 straight j close parentheses

C) fraction numerator 1 over denominator square root of 41 end fraction open parentheses 4 space straight i space plus 5 space straight j close parentheses

D) fraction numerator 1 over denominator square root of 31 end fraction open parentheses 4 space straight i space minus 5 space straight j close parentheses 

Explanation-

OA with hat on top equals fraction numerator 4 straight i plus 5 straight j over denominator square root of 16 plus 25 end root end fraction equals fraction numerator 1 over denominator square root of 41 end fraction open parentheses 4 straight i plus 5 straight j close parentheses 


Q.9. If top enclose straight a space equals space 5 straight i with hat on top space minus space 3 straight j with hat on top space plus space 4 straight k with hat on top space comma space top enclose straight b space equals space space straight i with hat on top space minus space 2 straight j with hat on top space plus space straight k with hat on top comma space top enclose straight c space equals space 3 straight i with hat on top space plus space 5 straight j with hat on top then find top enclose straight a space. space left parenthesis top enclose straight b space cross times space top enclose straight c right parenthesis 

A) 8

B) 10

C) 14

D) 20

Explanation-

We have, top enclose straight a space equals space 5 straight i with hat on top space minus space 3 straight j with hat on top space plus space 4 straight k with hat on top space comma space top enclose straight b space equals space straight i with hat on top space minus space 2 straight j with hat on top space plus space straight k with hat on top space comma space top enclose straight c space equals space 3 straight i with hat on top space plus space 5 straight j with hat on top space

therefore space top enclose straight a space. space left parenthesis top enclose straight b space cross times space top enclose straight c space right parenthesis space equals space left square bracket top enclose straight a space top enclose straight b space top enclose straight c right square bracket
equals space open vertical bar table row 5 cell negative 3 end cell 4 row 1 cell negative 2 end cell 1 row 3 5 0 end table close vertical bar
equals space 5 space left parenthesis 0 minus 5 right parenthesis space plus space 3 space left parenthesis 0 minus 3 right parenthesis space plus 4 space left parenthesis 5 space plus 6 right parenthesis
equals space minus 25 minus space 9 space plus space 44 space equals 10 


Q.10. Find the vector of magnitude 3 units which is parallel to 4 straight i with hat on top plus 3 straight j with hat on top .

A) 2/5 (4 straight i with hat on top plus 3 straight j with hat on top)

B) 3/5(4 straight i with hat on top plus 3 straight j with hat on top)

C) 7/5(4 straight i with hat on top plus 3 straight j with hat on top)

D) 3/5(4 straight i with hat on top plus 3 straight j with hat on top)

Explanation-

top enclose straight a equals 4 straight i with hat on top plus 3 straight j with hat on top equals open vertical bar top enclose straight a close vertical bar equals square root of 4 squared plus 3 squared end root equals square root of 16 plus 9 end root equals square root of 25 equals 5
We comma space have comma space straight a with hat on top equals fraction numerator top enclose straight a over denominator open vertical bar top enclose straight a close vertical bar end fraction equals fraction numerator 4 straight i with hat on top plus 3 straight j with hat on top over denominator 5 end fraction
therefore The space required space vector space is space 3 top enclose straight a equals 3 divided by 5 left parenthesis 4 straight i with hat on top plus 3 straight j with hat on top right parenthesis  


                              


Hydrogen: chemistry


  • Answers are in BOLD Option name.

 Q.1.  When COblank subscript 2  is bubbled through a solution of barium peroxide in water 

A) Oblank subscript 2 is released 

B) Carbonic acid is formed 

C) Hblank subscript 2Oblank subscript 2 is formed 

D) no reaction occurs 

Explanation-

BaO subscript 2 plus CO subscript 2 plus straight H subscript 2 straight O rightwards arrow BaCO subscript 3 downwards arrow plus straight H subscript 2 straight O subscript 2 


Q.2.  The hydrogen molecule is not very reactive under normal conditions because 

A) Hydrogen has a large number of isotopes 

B) The corresponding triangle S degree value is very low 

C) The bond dissociation energy of  the hydrogen molecule is high 

D) The hydrogen molecule has a low activation energy 

Explanation-

High bond dissociation energy of hydrogen molecule explains its less reactive nature.


Q.3.  Decomposition of H subscript 2 O subscript 2  can be prevented by addition of

A) Acetanilide

B) Benzene

C) Alkali metal oxide 

D) MnOblank subscript 2 

Explanation-

Acetanilide acts as stabilizer for straight H subscript 2 straight O subscript 2.


Q.4. Which is the correct order of boiling point of VA, VIA and VIIA hydride ?

A) NH subscript 3 space greater than space SbH subscript 3 space greater than space AsH subscript 3 space greater than space PH subscript 3

B) straight H subscript 2 straight O space greater than space straight H subscript 2 Te space greater than space straight H subscript 2 Se space greater than space straight H subscript 2 straight S

C) HF > HI > HBr > HCI

D) All of the above are correct 


Q.5.   HCI is added to following oxides . Which one would give H subscript 2 O subscript 2 ?

A) MnOblank subscript 2

B) PbOblank subscript 2

C) BaO

D) None of the above 

Explanation-

Only peroxides on action of dil. acid gives straight H subscript 2 straight O subscript 2 


Q.6.  Which of the following ions will cause hardness in water sample ?

A) Nablank to the power of plus

B) Kblank to the power of plus 

C) Brblank to the power of minus 

D) Cablank to the power of 2 plus end exponent 

Explanation-

Soluble Salts of calcium can cause hardness in water.


Q.7.  In context with the industrial preparation of hydrogen from water gas (CO+Hblank subscript 2) , which of the following is the correct statement ?

A) CO is oxidized to COblank subscript 2 with steam in the presence of a catalyst followed by absorption of COblank subscript 2 in alkali .

B) CO and Hblank subscript 2 are fractionally separated using differences in their densities.

C) CO is removed by absorption in aqueous Cublank subscript 2Clblank subscript 2 solution .

D) Hblank subscript 2 is removed through occlusion with Pd


Q.8.  A dilute solution of straight H subscript 2 straight O subscript 2  can be concentrated by 

A) Drying it over anhydrous CaClblank subscript 2

B) Drying it over concentrated Hblank subscript 2SOblank subscript 4

C) Drying it over anhydrous MgSOblank subscript 4 

D) Heating it under reduced pressure 

Explanation-

A dilute solution of straight H subscript 2 straight O subscript 2 can be concentrated by heating (distillation) under reduced pressure.


Q.9. Hydrides of metals are named like 

A) alkane 

B) alkene

C) alkyne

D) None is correct


Q.10. Consider the following statements : 

I : Rate of transfer  of Dblank to the power of plus  from Dblank subscript 2O  is  slower then that of Hblank to the power of minus b from Hblank subscript 2O

II :  Kblank subscript a for CHblank subscript 3COOH rightwards harpoon over leftwards harpoon CHblank subscript 3COOblank to the power of minus  + Hblank to the power of plus

III : Tritium is a  radioactive isotope . Select the correct statements .

A) I , II 

B) II , III 

C) I , III 

D) I, II , III