-3x^2/8

2,3,5---------3!

2,4,4---------3!/2

2,2,6------------3!/2

3,3,4-----------3!/2

3,6,1-----------3!

4,5,1----------3!

so=27/216=1/8

 

is ans 1/8

 

WHEN  WIRE IS STRETCHED  n TIMES ,

THEN RESISTANCE IS=n²R

WHEN WIRE IS FOLDED 1/n TIMES,

THEN   RESISTANCE=R/n²

  FIRST STRETCH THE CUBE HOLDING THE OPPOSITE DIAGONALS  A&B.

THE BASE & TOP EDGES & ONE OF THE LATERAL EDGES BECOME PARALLEL SO NET RESISTANCE IS R/2

U GET IT AS GIVEN BELOW.    

 

  A ___           ____R/2____________R/2____

                    |                     |                       |

                    |                     |                       | 

                   R                    R/2                  R

                   |                     |                       |

                   |_____R/2____|______R/2____| _______B

 

 now this reduces to

                      __R/2__ __3R/2__

                     |            |              |

                A_ |            R/2          |_B

                     |__3R/2_ |__R/2___|

 

USE KIRCHOFFS  LAW TO FIND  EQ RESISTANCE

 

well, even places should be filled with even nos i.e 2,4,6,8 &odd places with odd no.s

----->fifth place is filled with 5

---->third&fourth places should be filled with mutiples of 4 i.e  16,12,24,28,32,36,

---->sixth,seventh,eighth places should be filled in such a way that the three digit no is a  multiple of 8 so that the no. upto 8 places could be divisible by 8

----->the  sum of first 3 no.s  should be a multiple of 3 

----->the sum of next three no.s  should also be a multiple of 3  bcoz  a no. is divisible by 6 if it is divisible by both 2&3

----->the last place can be filled with any no. that is left out as sum of first 9 no.s  is 45 which is a multiple of 9

i got f(x)=(2x³+1)/3

   f inv.x=(3x-1/2)^1/3

--------->for a given no.  N=(P^m)(Q^n)(R^s)......whereP,Q,R are prime nos     then no. of divisors=(m+1)(n+1)(s+1)

5400=3^3.2^3.5^2

so no of divisors=(3+1)(3+1)(2+1)=4*4*3=48

including the no. & 1

-------->sum of divisors=(1+P^1+P^2+P^3+........P^m)(1+Q^1+Q^2+..........Q^n)(1+R^1+R^2+..............R^s)

5400  sum of divisors==(1+3+3^2+3^3)(1+2+2^2+2^3)(1+5+5^2)

                                =40*15*31=18600

 

LIALH---

carbonyl compds  reduced without redn  double/triple bonds.

note:sometimes =bond is redn  if phenol is at beta carbon

acid der----alcohol

amides{unsubstituted}----------primary amines

            subs---------------------2/3   amines

nitro,azides,oximes,nitriles,----------1 amines

halides--------------hydrocarbon

epoxide------------alcohol

 

 

NaBH4---

milder redn agent than lialh4

more selective

it reduces carbonyl groups of ald,keto,acid chlorides in presence of many reducible groups such as nitro, ester, carboxyl, epoxide, nitrile, double bond{either isolated/conjugated} are not affecte

 

 suppose to red p-nitro benzaldehyde to  p-nitro benzyl alcohol  NaBH4  is used but not LIALH  as it reduces both

this high selectivity makes NaBH4  preffered reagent for redn of carbonyl compds in sensitive  polyfunctional molecules

 

H2/Ni

 

powerfl redn agent

it redcs multiple bonds in aliphatic&alicyclic&aromatic compds

aromatic ethers---------hydc

     Ar0R------------------------>ArH+RH

nitro------------amino

nitriles----------------------ald/amines

 

 

plz rate me