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1.

16 players `P_(1),P_(2),P_(3),….P_(16)` take part in a tennis tournament. Lower suffix player is better than any higher suffix player. These players are to be divided into 4 groups each comprising of 4 players and the best from each group is selected to semifinals. Q. Number of ways in which they can be divided into 4 equal groups if the players `P_(1),P_(2),P_(3)` and `P_(4)` are in different groups, is:A. `((11)!)/(216)`B. `((12)!)/((3!)^(4))`C. `((12)!)/((3!)^(4)(4!))`D. `((11)!)/((3!)^(4))`

Answer» Correct Answer - B
2.

Biodiversity decreases as one moves from equator towards poles. Justify with an example.

Answer»

Species diversity decreases from equator towards poles because

(i) Equator (tropic) regions have less seasonal or more constant environments, which promote niche speciation and thereby more species diversity.

(ii) These regions have more solar energy which contributes to more productivity and greater diversity e.g. The tropics (between 23.5 °N to 23.5 °S) harbours more species than temperate and polar regions. Columbia situated near equator has about 1400 species of birds, while New York (at 41 °N) has about 105 species and Greenland (at 71°) only 56 species.

3.

How are sticky ends formed on DNA strands? Why are they so called?

Answer»

When a restriction enzyme cuts the strands of DNA, a little away from the center of the palindrome site, between the same two bases on the two strands, single-stranded portions are left at the ends.

  • These overhanging single stranded stretches of DNA at the ends, are called ‘sticky ends’.
  • They are so called because they easily form hydrogen bonds with their complementary cut counterparts with the action of enzyme ligase
4.

List any four ways by which GMO's have been useful for enhanced crop output.

Answer»

Some of the potential applications of genetically modified plants in agricultural biotechnology are:

(i) GMO crops have better nutritional quality such as vitamin A-enriched rice

(ii) They are diseases resistant and are more tolerant to abiotic stresses.

(iii) They have pest resistant and hence have reduced the dependency on the chemical pesticides.

(iv) They are better nitrogen fixers

5.

The enthalpies of neutralization to two weak acids `HA` and `HB` with `NaOH` are `-10.1` and `-11.8` `Kcal//g`. eq respectively. Which of the following relation is correct regarding acidic strengths of `HA` and `HB` `:-`A. `HAgtHB`B. `HAltHB`C. `HA=HB`D. Connot be predicted

Answer» Correct Answer - B
6.

Explain the scope of Human Resource Management.

Answer»

1. Human Resource Planning (HRP):

It is the first step of human resource management. HRP is a process through which the company makes an assessment of the present manpower-the number of people employed, number of post vacant, whether there is excess or shortage of staff and prepares plans and policies to deal with them accordingly. It also estimates the future man power requirement. It matches the supply and availability of manpower with the company’s purpose and makes the best person available for the job.

2. Job Analysis:

An important task of HRM is job analysis. It provides the basis for recruitment, training and development of staff. Every job is analysed to identify the inherent requirement of the job – “what the job demands of a worker”. Job analysis provides a detailed picture of the nature of job, the skill and abilities required to perform the job, the duties and responsibilities of the worker required for the job.

This helps in the process of recruitment and selection, to appoint the appropriate person for the job and also facilitates arranging training and development programme for the employees to match their skill and abilities with the job requirement.

3. Job Evaluation:

It forms the basis of wage and salary determination. Different jobs are compared and evaluated to determine the relative importance of different jobs within the organization. The jobs are compared, graded and ranked. Job evaluation determines how much pay to be affixed to the particular job on the basis of their importance in the organisation functioning. This helps to determine a fair wage and salary and to determine a rational pay structure.

4. Recruitment and Selection:

The company on the basis of human resource planning and job analysis identifies the requirement of personnel and explores different internal and external sources from where manpower can be acquired. This process is called recruitment. The right person is selected and put to work.

Performance Appraisal:

It is a systematic method of evaluating the performance of employees. This helps to assess the efficiency, strength and weakness of the employees. It is a controlling process whereby the employees’ performance is matched with the standard level of performance and remedial measures taken in case actual performance does not meet the standard.

It helps the management in their decision-making relating to promotion, pay hike that may be rewarded if the employee meets standard or outperforms the standard. It also helps the management to decide about the training and development needs for employees, who needs up gradation and development of skill.

6. Training and Development:

Human resource management is concerned with arrangement of training and development programme for the newly appointed staff as well as for existing employees. This helps to enhance the qualitative and quantitative performance of the work force.

7. Employee Compensation:

Human resource management performs the function of determining the pay structure for different employees on the basis of qualification, efficiency, experience, nature of the job. It also determines rewards and incentives to be paid for their contribution towards the organisation.

8. Employee Motivation:

Motivation of employees is essential for retention and sustenance of employees within the organisation. Motivation plays an important role in management of personnel. It helps to improve productivity of personnel. There are different ways by which employees may be encouraged. Recognition and reward for performance, promotion, pay-hike, and other monetary and non-monetary rewards motivate the employees.

9. Employee Welfare:

Human resource management focuses on the welfare of employees by providing canteen facilities, creches, rest and lunch rooms, housing, transport, medical assistance, education, health and safety, medical care, sickness benefits, recreation facilities.

It aims at improving the quality of work life by providing a better environment worth working, job security, flexible working hours, eliminating workplace hazards, which will have a positive impact on the productivity and quality of work. Along with monetary benefits provided to employees these non-monetary benefits improves the morale of the employees and motivates them.

10. Industrial Relation:

Labour unrest, conflicts between the work groups and between workers and management over issues like low wage, bad work environment are common problems that must be handled carefully by establishing a balance between the worker’s interest and interest of the organisation and management. Human resource management considers establishment and maintenance of harmonious and cordial relation between employer and employees.

7.

Let the wavelength at which the the spectral emissive power of a black body (at a temperature T) is maximum, be denoted by `lamda_(max)` as the temperature of the body is increased by 1K, `lamda_(max)` decreases by 1 percent. The temperature T of the black body isA. 100 KB. 200 KC. 400 KD. 288 K

Answer» `lamda_(m)T=`const.
`lnlamda_(m)=lnT=C`
`(dlamda_(m))/(lamda_(m))+(dT)/(T)=0" "therefore(dlamda_(m))/(lamda_(m))=(dT)/(T)`
Now `(dlamda_(m))/(lamda_(m))=-1%=-(1)/(100)` (-ve sign indicates decreases)
`dT=1` (given)
`thereforeT=100K`
8.

A block of density `2000kg//m^3` and mass 10kg is suspended by a spring stiffness `100N//m`. The other end of the spring is attached to a fixed support. The block is completely submerged in a liquid of density `1000kg//m^3`. If the block is in equilibrium position.A. the elongation of the spring is 1 cmB. the magnitude of buoyant force acting on the block is 50 NC. the spring potential potential energy is 12.5 JD. magnitude of spring force on the block is greater than the weight of the block

Answer» Correct Answer - B::C

`Kx = 10 xx 10 - (1000)/(2000) xx 10 xx 10`
`Kx = 50 N`
`U_("stored") = (1)/(2) xx(100) ((50)/(100))^(2) = (1)/(2) xx(2500)/(100)=12.5 J`
9.

The number of geometrical isomers for octahedral `[CoCl_(4)(NH_(3))_(2)]^(-)`, square planar `[AgBr_(2)Cl_(2)]^(-)` and `[PtCl_(2)(en)]` are:-A. 2,2,2B. 2,2,0C. 3,2,2D. 2,3,0

Answer» Correct Answer - B
10.

In photoelectric effect explain the effect of(i) intensity of light and photoelectric current,(ii) potential and photoelectric current,(iii) frequency of incidents radiation and stopping potential.Also draw graph in all above case with brief explanation.

Answer»

(i) Intesity of light and photoelectric current.

(ii) potential and photoelectric current.

(iii) frequency of incidents radiation and stopping potential.

Variation of stopping potential with frequency of incident radiation.

11.

Name the type of cell which exert EMF, though both electrodes are same but differ in electrolyte concentration. Two Ag electrodes immersed in \( Ag _{2} SO _{4} \) solution of \( 0.02 M \) and \( 0.042 \) \( M \) at \( 298 K \). Write the cell representation, cell reaction and calculate EMF of the cell.

Answer»

This type of cell is known as electrolyte concentration cell. For this type of cell\(E^0_{cell} =0.\)

At anode

\(2Ag(s) \longrightarrow \underset{c_1 = 0.02 M} {2Ag^+}(aq) + 2e^- \; \text{oxidation}\)

At cathode

\(\underset{c_2 = 0.042 m}{Ag^\oplus_2SO_4}(aq) + 2e^-\longrightarrow 2Ag(s) + SO_4 ^{-2}(aq) \;\text{reduction}\)

Cell representation-

\(Ag(s)/Ag_2SO_4(0.02M) // Ag_2SO_4(0.042M)/Ag(s)\)

Overall cell reaction-

\(Ag^\oplus_2 SO_4 (C_2) \longrightarrow Ag_2SO_4 (C_1)\)

or\(Ag^+(c_2) \longrightarrow Ag^+ (c_1)\)

\(\therefore E_{cell} = E^0_{cell} -\frac{RT}{nf}ln \frac{c_1}{c_2}\)

\(E_{cell} = 0 - \frac{0.0591}1log\frac{0.02}{0.042}\)

\(\therefore E_{cell} = -0.0591 \times log(0.4762)\)

\(= -0.0591 \times (-0.322)\)

\(= 0.019\)

\(E_{cell} = 0.02V\)

\(E_{cell} = 20\, mV\)

12.

\( 80 cm ^{3} \) of methane is mixed with \( 200 cm ^{3} \) of pure oxygen at the same temperature and pressure. The mixture is then ignited. Calculate the composition of the resulting mixture if it is cooled to initial room temperature and pressure. \[ \begin{array}{c} CH _{4}+2 O _{2} \longrightarrow CO _{2}+2 H _{2} O \\ \left(\text { Ans : } CO _{2}=80 cc , O _{2}=40 cc , H _{2} O =\text { negligible }\right) \end{array} \]

Answer»

\(\underset{initially}\,\,\,\;\underset{80cm^3}{CH_4(g) }+ \underset{200cm^3}{2O_2(g)} \longrightarrow CO_2(g) + 2H_2 O(l)\)

∵ Temperature and pressure as constant, from the above chemical equation we can conclude that -

1 volume of CH4 react with 2 volume of pure oxygen (O2) to form 1 volume CO2.

(Volume of H2O(l) is negligible to compare volume of gaseous molecules)

∴ Here, CH4 is limiting reagent.

∴ Volume of pure O2 unreacted = (200 - 160)cm3 = 40 cm3

∴ Volume of CO2 formed = 80 cm3

Hence, mixture contain 40 cm3 O2 and 80 cm3 CO2.

13.

Write chemical properties of amines?

Answer»

Amines are organic compounds which contain and are often actually based on one or more atoms of nitrogen.Structurally amines resemble ammonia in that the nitrogen can bond up to three hydrogens, but amines also have additional properties based on their carbon connectivity.