Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Intermolecular forces between two benzene molecules are nearly of same strength as those between two toluene molecules. For a mixture of benzene and toluene, which of the following are not true?

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`Delta_(mix)H-zero`
`Delta_(mix)V=zero`
These will FORM minimum BOILING azeotrope.
These will not form ideal solution.

Solution :are not ture statemnents
2.

Intermolecular attractions among the moleules of any gas are minimum at

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`-10^(@)C and 5` ATM
`-10^(@)C and 3` atm
`5^(@)C and 2` atm
`100^(@)C and 1` atm

Answer :D
3.

Intermolecular dehydration of alcohols gives :

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Alkenes
Ketones
Alkynes
Ethers

Answer :D
4.

Intermolecular dehydration of alcohol gives

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Alkenes
Ethers
Alkynes
Aldehydes

Answer :B
5.

Intermediates formed during the reaction of RCONH_2 with Br_2and KOH are :

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RNHCOBr and RNCO
RNHBr and RCONHBR
`RCONBr_2`
RCONHBr and RNCO

Answer :D
6.

Intermediate product of hydrolysis of cyanide is

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`RCOONH_4`
`RCONH_2`
`R-UNDERSET(OH)underset(|)C=NH`
`R-C-=overset(o+)NH`

ANSWER :B
7.

Intermediate.

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Solution :The species formed during the COURSE of a reaction, which has definite life- TIME and can be isolated. It has energy LOWER than that of the preceing transition STATE, but higher than the REACTANTS or products is called an intermediate.
8.

Intermediate product formed in the acid catalysed dehydration of n-propyl alcohol is:

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`CH_3 - CH_2 - CH_3`
`CH_3 - CH - CH_2^+`
`CH_3 - CH^+ - CH_3`
`CH_3 - CH_2 - CH_2`

ANSWER :C
9.

Intermediate compound formation theory is unable to explain the mechanism of ...............

Answer»

SOLUTION :HETEROGENEOUS CATALYSIS
10.

Intermediate involved in Reimer- Tiemann reaction is

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CARBOCATION
carbanion
carbene
free RADICAL.

SOLUTION :Carbene, ACTUALLY dichlorocarbene is the electrophile INVOLVED in Reimer-Tiemann reaction.
11.

Interhalogen compounds are more reactive than the corresponding halogens. Explain.

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TWO halogens are present in place of one
They are more ionic
Their bond POLARITY is more than that of the HALOGEN molecule
They carry more energy

Answer :C
12.

Interhalogen compounds are more reactive than the constituent halogens except fluorine - Explain.

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Two halogens PRESENT in place of ONE
They are more ionic
Their bond POLARITY is more than that of the halogen molecule
They carry more energy

Answer :C
13.

Interhalogen compounds are more reactive than the individual halogen because:

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two halogens are present in place of one
their bond ENERGY is less than the bond energy of the HALOGEN molecule
they have polar bond
they CARRY more energy

Answer :C
14.

Interhalogen compounds are compounds formed by combination of different halogen atoms. Which are more reactive, halogens or interhalogen compounds? Give reason

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Solution :Interhalogen compounds are more reactive than HALOGENS `(except F_2)`. This is because the COVALENT BOND X-X' in interhalogen compound is WEAKER. than that in halogens,
15.

Interhalogen compounds are

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IONIC compounds
Co-ordinated compounds
Moleculor compounds
Covalent compounds

Answer :D
16.

Interferon is a product of biotechnology and is used against :

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VIRAL diseases
Diabetes
Sickle CELL anaemia
Haemorrhage

Answer :A
17.

Interfering radicals interfere the testof

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Group III radicalsonly
Group III radicals or downward
Cation which are present in group II FITRATE
NONEOF these

Answer :B,C
18.

Interferon is :

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TONIC
Virus
Carbohydrate
Ore of iron

Answer :B
19.

Intereferon is connected with

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Carbohydrate
Virus
Tonic
Ore of iron

Answer :B
20.

interaction between dispersed phase and dispersion medium colloids classified on the basis ?

Answer»

SOLUTION :Classification on the basis of nature of interactions between dispersed phase and DISPERSION medium. These may be classified as
(i) LYOPHILIC SOLS and (II) lyophobic sols.
21.

Intermolecular hydrogen bonds are not present in :

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`CH_3COOH`
`C_2H_5NH_2`
`CH_3CH_2OH`
`CH_3OCH_3`

ANSWER :D
22.

Inter halogens are more reactive than halogens why?

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SOLUTION :The BOND in the INTER halogen X-A is weaker than A-A bond in HALOGENS. This is an account of less effective overlapping between orbitals of dissimilar atoms than those of similar atoms.
23.

Intensity of the scattered light depends upon the difference of which of the following property of the dispersed phase and the dispersion medium ?

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densities
viscosities
SURFACE tension
refractive indices

Solution :Intensity of scattered LIGHT DEPENDS upon the DIFFERENCE of refactive indices of the dispersed phase and the dispersion medium.
24.

Integrated velocity equation for first order reaction is

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`[A]_(o) = [A]e^(-kt)`
`K = [A]_(o) e^(-A//t)`
`Kt = 2.303 LOG ``([A]_(o))/([A])`
log `([A]_(o))/([A]) = -2.303 Kt`

Solution :INTEGRATED velocity EQUATION for first order REACTION is
`k = (2.303)/(t)` log `((A)_(0))/((A))`.
25.

Integrated rate expression for rate constant of first order reaction is given by k=2.303/tlog[[R]]_0/[[R]]for general reaction RtoP.Derive an expression for half life period of first order reaction.

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SOLUTION :`k=2.303/tlog[[R]]_0/[[R]]`At half life PERIOD, t=t_(1/2)`and `[R]=[R]_0\2`
therefore`t_(1/2)`=2.303/klog`([[R]]_0/[[R]])/2t_(1/2)2.303/klog2=0.693/k`
26.

Insulin production and its action in human body are responsible for the level of diabetes. This compound belongs to which of the following categories ?

Answer»

an ENZYME
A HORMONE
A co-enzyme
An antibiotic.

Answer :B
27.

Insulin regulate the metabolism of

Answer»

minerals
amino acids
glucose
vitamins

Answer :C
28.

Insulin regulates the metabolism of

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minerals
amino acids
GLUCOSE
vitamins

Solution :Insulin is ESSENTIAL for metabolism of carbohydrates in the body. It facilitates the entry of glucose into the cells by INCREASING the penetration of CELL membranes and phosphorylation of glucose.This DECREASES glucose concentration in blood.
29.

Insulin is dissolved in a suitable solvent and the osmotic pressure (pi) of solutions of various concentration (g//cm​^3​) C is measured at 20^@C. The slope of a plote of pi against C is found to be 4.65xx10^(-3) ​. The molecular weight of the insulin is :-

Answer»

`3xx10^5`
`9xx10^5`
`4.5xx10^5`
`5.16xx10^6`

SOLUTION :`pi=w/mxx(RT)/V`
slope = `(RT)/M`
`M=(RT)/"slope"=(0.0821xx293)/(4.65xx10^(-3)xx10^(-3))=5.16xx10^6` g
30.

Insulin is dissolved in a suitable solvent and the osmotic pressure (TT) of solutions of various concentration (g//cm^(3)) C is measured at 20^(@)C. The slope of a plote of TT against C is found to be 4.65xx10^(-3). The molecular weight of the insulin is

Answer»

`3xx10^(5)`
`9xx10^(5)`
`4.5xx10^(5)`
`5.16xx10^(6)`

Solution :`pi=w/mxx(RT)/V`
slope`=(RT)/M`
`M=(RT)/("slope")=(0.0821xx293)/(4.65xx10^(-3)xx10^(-3))=5.16xx10^(6)G`
31.

Insulin is secreted from

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thyroid
adrenal body
pancreas
liver

Answer :C
32.

Insulin is secreted from :

Answer»

Thyroid
Pancreas
Adrenal body
Cellular membrane.

Solution :INSULIN is SECRETED from pancreas.
33.

Insulin is dissolved in a suitable solvent and the osmotic pressure pi in atm of solutions of various concentrations C in g//cm^(3) is measured at 27^(@) C. The slope of plot of pi against C is found to be 4.1 xx 10^(-3) atm cm^(3)g^(-1). The molecular mass of insulin is:

Answer»

`6 xx 10^(3)`
`3 xx 10^(6)`
`6 xx 10^(6)`
`3 xx 10^(3)`

ANSWER :C
34.

Insulin is an example of ___ hormone.

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PARACRINE
ENDOCRINE
AUTOCRINE
NONE of the above

SOLUTION :endocrine
35.

Insulin is a protein whichplays the role of ………. .

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an ANTIBODY
a hormone
an ENZYME
a TRANSPORTING AGENT

Answer :B
36.

Insulin is a

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Non STEROIDAL, PEPTIDE hormone
Steroidal, peptide hormone
Non steroidal ,amino acid hormone
Steroidal amino acid derivative hormone

ANSWER :A
37.

Insulin is a homopolymer of _____.

Answer»


ANSWER :FRUCTOSE
38.

Insulin is

Answer»

POLYMER of `ALPHA`-AMINO acid
Monomer of `alpha`-amino acid
Polymer of `BETA`-amino acid
Monomer of `beta`-amino acid

Answer :A
39.

Insulin has 51 amino acids in two polypeptide chains which are linked by :

Answer»

ONE SULPHIDE bond
One DISULPHIDE bond
Two disulphide bonds
Three disulphide

Answer :C
40.

Insulin contains 3.4% sulphur. Then, the minimum molecular mass of the insulin in about.

Answer»

940amu
9400amu
3600amu
970amu

Solution :3.4g SULPHUR is present in 100g INSULIN
`THEREFORE 32G` sulphur will be present in `(100)/(3.4)xx32g` insulin=940
`therefore` Molar mass of insulin is about 940 amu
41.

Insulincontains 3.4%sulphur. The minimummolecular weightof insulin is

Answer»

350
470
560
940

Solution :Minimum mas of sulphur `= wt.` of its one atom `= 32`
`:' 3.4 GM` of sulphur PRESENT in `100 gm`.
`:' 32 gm` of sulphur present in `= (100xx 32)/(3.4) = 940 gm`.
42.

Insulin contains 3.4% sulphur. The minimum molecular weight of insulin is

Answer»

944
941.176
945.27
None

Answer :B
43.

Insulin (C_(2)H_(10)O_(5))_(n) is dissolved in a suitable solvent and the osmotic pressure (pi) of solutions of various concentrations (C) is measured at 20^(@)C. The slope of the plot of pi (atm) versus C (in g//cm^(3)) is found to be 4.65xx10^(-3). The molecular weight of insulin is

Answer»

`3.17xx10^(6)`
`4.17xx10^(6)`
`5.17xx10^(6)`
`6.17xx10^(6)`

Solution :`pi=C'RT`. HENCE CONCENTRATION C' is in `"MOL L"^(-1)` To convert `"g cm"^(-3)` into `"mol L"^(-1)`, wehave
Thus, C' in the formula should be replaced by this factor, i.e.,
`pi=(Cxx1000)/(M_(2))RT=(1000RT)/(M_(2))C`
Thus, a plot of `pi` VS C will be linear with SLOPE
`=(1000RT)/(M_(2))=4.65xx10^(-3)"(Given)"`
`therefore (1000xx0.0821xx293)/(M_(2))=4.65xx10^(-3)`
`"or"M_(2)=5.17xx10^(6)`
44.

Insulin contain 3.4% sulphur. Calculate the minimum mass of insullin.

Answer»

Solution :Minimum MOLECULAR MASS of INSULIN will be the mass containing at least ONE ATOM of sulphur.
But one atoms of S = 32 amu
Given that 3.4 amu of sulphur is present in 100 amu of insulin
`therefore"32 amu of sulphur will be present in insulin"=(100)/(3.4)xx32=941.2amu`
Hence, minimum mass of insulin = 941.2 amu.
45.

Insulin conatains3.4% sulphur .The minimum mol. Weight of insulin is :

Answer»

941.176
944
945.27
None

Answer :A
46.

Insulin, (C_(2)H_(10)O_(5))_(n) is dissolved in a suitable solvent and the osmotic pressure (pi) of solutions of various cocentrations ("g cm"^(-3)) is measured at 20^(@)C. The slope of the plot of pi against C is found to be 4.65xx10^(-3). Calculate the molecular mass of insulin.

Answer»

Solution :`pi="CRT where C = MOLAR conc. "("mol L"^(-1))`
Given values of C (say C') are in `"g cm"^(-3)`. To CONVERT then into `"mol L"^(-1),` we have
`C=(C')/(M_(2))xx1000`
`therefore""pi=(1000C')/(M_(2))xxRT.` It is like `y=mx`
Hence, plot of `pi " vs " C'` will be LINEAR with slope `=(1000RT)/(M_(2))`
`therefore(1000RT)/(M_(2))=4.65xx10^(-3)"or"M_(2)=(1000RT)/(4.65xx10^(-3))=(1000xx0.0821xx293)/(4.65xx10^(-3))=5.17xx10^(6)`
47.

Insulin (C_2H_10O_5)_n is dissolved in a suitable solvent and osmotic pressure (pi) of solutions of various concentrations (g cm^-3) C is measured at 20^@C. The slope of a plot of pi against C is found to be 4.65xx10^-3. The molecular weight of the insulin is:

Answer»

`4.8xx10^5`
`9xx10^5`
`3xx10^5`
`5.16xx10^6`

ANSWER :D
48.

Insulin , a protein acts as

Answer»

An antibody
A hormone
An enzyme
A TRANSPORT AGENT

ANSWER :B
49.

Insulin, a hormonechemically is ………… ..

Answer»

TACGAACT
TCCGAACT
TACGTACT
TACGRAGT

Answer :A
50.

Instead of principal quantum number (n), azimuthal quantum number (l) & magnetic quantum number (m), a set of new quantum number s, t & u was introduced with similar logic but different values as defined below s = 1, 2, 3,.......oo all positive integral values. t = (s^(2) - 1^(2)), (s^(2) - 2^(2)), (s^(2) - 3^(2)).........No negative value u = - ((t + 1))/(2) "to" + ((t + 1))/(2) (including zero, if any) in integral steps. Each orbital can have maximum four electrons. (s + t) rule is defined, similar to (n + l) rule. Number of electrons foe which s = 2, t = 3 for an element with atomic number 24

Answer»

8
4
0
None of these

Solution :`{:(s = 1,t = 0,u = - (1)/(2)","(1)/(2),),(s = 2,t = 0,u = - (1)/(2)","(1)/(2),),(,t = 3,u = - 2"," -1"," 0","+1","+2 ,),(s = 3,t = 0,u = - (1)/(2)","(1)/(2),),(,t = 5,u = -3"," -2"," -1"," 0"," +1"," +2"," +3,),(,t = 8,u = - (9)/(2) "," - (7)/(2)"," - (5)/(2)"," - (3)/(2)"," - (1)/(2)"," (1)/(2)"," (3)/(2)"," (5)/(2)"," (7)/(2)"," (9)/(2),):}`
According to `(s + t)` rule INCREASING order of ENERGY
1s 2s 3s 4s 2f 5s
`{:((i),"NUMBER of electron in " s = 2,7 xx 4 = 28,,),(,"Number of electron in " s = 3,19 xx 4 = 76,"Hence (B)",):}`
`{:((ii),"For " Z = 24,1s^(8) 2s^(8) 3s^(8) " i.e. number electrons (s) in " s = 2,t = 3 ""[2f^(@)] = 0,),(,"Hence" (C),,,):}`