\(\text{C-Cl}\) bond in chlorobenzene in comparison to \(\text{C-Cl}\) bond in methyl chloride is :

Aniline reacts as shown below:
\(C_6 H_5 NH_2 \xrightarrow[0-5^{\circ} C]{NaNO_2+HCl} A \xrightarrow[KCN]{CuCN} B\)
The compound \(B\) is:

\(s p^3\) hybridization leads to ______________

The Nernst equation corresponding to the cell reaction given below, is
Ni(s) + 2Ag\(^+\)(aq) \(\rightarrow\) Ni\(^2+\)(aq) + 2Ag(s)

Read the passage carefully and answer the Questions
Alcohols and phenols are obtained when a hydrogen atom in aliphatic and aromatic hydrocarbons respectively, is replaced by \(OH\) group. These compounds find wide applications in industry as well as in our daily life. You may notice that ordinary spirit is used for polishing wooden furniture which is chiefly a compound containing hydroxyl group. An alcohol contains one or more hydroxyl \((OH)\) group(s) directly attached to carbon atom(s), of an aliphatic system \((CH_3OH)\) while a phenol contains \(OH\) group(s) directly attached to carbon atom(s) of an aromatic system \((C_6H_5OH)\). Alcohols are prepared by the hydration of alkenes, fermentation of glucose, and reduction of carbonyl compounds. Alcohols produce alkenes upon dehydration whereas alkenes on acid catalyzed hydration convert into alcohols.
Which of the following compounds lacks phenolic group?

Fehling solution gives red precipitate with.....

Which of the pair(s) are correctly matched?

Colloid Type	Example
(A) Liquid in solids	Gem stones
(B) Solid in liquid	Cell fluids
(C) Solid in Gas	Mist
(D) Gas in liquid	Pumice stone
(E) Liquid in Solid	Butter

Choose the correct answer from the options given below :

Two capacitors of capacitances 2 \( \mu\text{F} \) and 4 \( \mu\text{F} \) are connected in parallel. A charge of 6 \( \mu\text{C} \) is given to the combination which gets shared. The ratio of the charges on the two capacitors, respectively will be

Read the passage carefully and answer the questions:
The degeneracy of the \(d\) orbitals has been removed due to ligand electron-metal electron repulsions in the octahedral complex to yield three orbitals of lower energy, \(t_{2 g}\) set and two orbitals of higher energy, \(e_g\) set. This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting and the energy separation is denoted by \(\Delta_0\). Thus energy of the two \(e_g\) orbitals will increase by \((3 / 5) \Delta_{\circ}\) and that of three \(t_{2 g}\) will decrease by \((2 / 5) \Delta_0\). The crystal field splitting \(\Delta_{ o }\) depends upon the field produced by the ligand and charge on the metal ion. Some ligands are able to produce strong fields, in which case the splitting will be large, whereas others produce weak fields and consequently result in small splitting of \(d\) orbitals. Relative magnitude of crystal field splitting energy \(\Delta_{\circ}\) and pairing energy, \(P\) (energy required for electron pairing in a single orbital) determine the formation of low spin ( \(\Delta_0>P\) ) or high spin ( \(\Delta_0<P\) ) complex. In tetrahedral coordination entity formation, the \(d\)-orbital splitting is inverted and is smaller as compared to the octahedral field splitting. The crystal field theory attributes the colour of the complex to \(d-d\) transition of the electron. The colour of the coordination compounds depends on the crystal field splitting. In the absence of ligand, crystal field splitting does not occur and hence the substance is colourless.
Choose the correct order of the ligands in their increasing crystal field splitting:

Insulin keeps glucose level within the narrow limit. It is released in response to the rapid rise in blood glucose level. The hormone that tends to increase glucose level in the blood is :

In the context of linear programming, Match List I with List II

List - I	List - II
(A) Z = 250x +75y	I. non-trivial constraints
(B) \(5 x+y \leq 100, x-y \leq 4\)	II. feasible solution
(C) \(x, y \geq 0\)	III. objective function
(D) points within and on the boundary of shaded region	IV. Trivial constraints

Read the given paragraph carefully and answer the following questions:
Since the DNA is enclosed within the membranes, we have to break the cell open to release DNA along with other macromolecules such as RNA, proteins, polysaccharides and also lipids. This can be achieved by treating the bacterial cells/plant or animal tissue with enzymes such as lysozyme (bacteria), cellulase (plant cells), chitinase (fungus). There are some basic steps in genetically modifying an organism (i) identification of DNA with desirable genes; (ii) introduction of the identified DNA into the host; (iii) maintenance of introduced DNA in the host and transfer of the DNA to its progeny. Agrobacterium tumifaciens, a pathogen of several dicot plants is able to deliver a piece of DNA known as 'T-DNA' to transform normal plant cells into a tumor and direct these tumor cells to produce the chemicals required by the pathogen. Other molecules can be removed by appropriate treatments and purified DNA ultimately precipitates out after the addition of chilled ethanol. Selection of recombinants due to inactivation of antibiotics is a cumbersome procedure because it requires simultaneous plating on two plates having different antibiotics. Therefore, alternative selectable markers have been developed which differentiate recombinants from non-recombinants on the basis of their ability to produce colour in the presence of a chromogenic substrate. In this, a recombinant DNA is inserted within the coding sequence of an enzyme, ß-galactosidase. This results into inactivation of the gene for synthesis of this enzyme, which is referred to as insertional inactivation.
Which one gives blue colored colonies if the plasmid in the bacteria does not have an insert?

Clarias gariepinus, the African catfish is an example of