Amity University 2006 Diploma Information Technology iit jee with answers - Question Paper

Table 1.
Pressure
(atm) quantity
L
1.0 1.00
1.5 0.67
2.0 0.50
2.5 0.40
3.0 0.33


Experiment two
Experiment one was repeated at room temperature with an initial pressure of one atm and an initial quantity of two L. The outcomes are shown in Table 2.

Pressure
(atm) quantity
L
1.0 2.00
1.5 1.33
2.0 1.00
2.5 0.80
3.0 0.67


Experiment three
In this experiment the entire gas container was insulated to prevent heat loss. The procedures from Experiment one were repeated. It was observed that the temperature of the gas rose as the quantity reduced. Table three indicates the quantity measured at every pressure during the compression.

Pressure
(atm) quantity
L
1.0 1.00
1.5 0.78
2.0 0.66
2.5 0.58
3.0 0.52


The insulation was then removed and the pressure maintained at three atm. As the gas cooled to room temperature, the quantity of the gas slowly reduced from 0.52 to 0.33 L.

Q. 27 How is the design of Experiment one various from that of Experiment 2?


The container is insulated in Experiment 2, but not in Experiment 1.
A various gas is used in Experiment two than in Experiment 1.
The initial quantity of the gas in Experiment one is half that of Experiment 2.
The initial quantity of the gas in Experiment one is twice that of Experiment 2.
Q. 28 A seventeenth-century scientist named Robert Boyle discovered that as the pressure on any confined gas increases (with the temperature held constant), the quantity reduces. The best way to verify these outcomes would be to repeat Experiment one with:


an identical container made of a various material than the original.
an identical quantity of water.
several various gases.
an unsealed container.
Q. 29 If Experiment one is continued and the pressure is increased to four atm and remains fixed at this pressure, the gas would occupy a quantity of approximately:


0.25 L.
0.33 L.
0.50 L.
1.00 L.
Q. 30 If Experiment one had started with 0.5 L of gas at one atm, what quantity would be recorded when the pressure was 2.5 atm?


0.20 L
0.30 L
0.40 L
0.44 L
Q. 31 Which of the subsequent statements best explains why the temperature of the container reduced after the insulation was removed in Experiment 3?


The pressure increased, causing the temperature to reduce.
The quantity decreased, causing the temperature to reduce.
Heat flowed from the surrounding atmosphere, through the container, and into the gas.
Heat flowed from the gas, through the container, and into the surrounding atmosphere.
Q. 32 Suppose that Experiment three was replaced so that the insulation was removed after every pressure increase and the container was allowed to sit at that pressure until its temperature cooled to room temperature. How would the quantity readings be influenced, if at all, by this replaced procedure?


They would be the identical as those in Experiment 1.
They would be the identical as those in Experiment 3.
They would be greater than those in Experiment 2.
They would be smaller than those in Experiment 1. Passage seven

Several methods were investigated to decrease pollution emissions from a steel mill smokestack. Steel is mostly iron (Fe), but it also contains carbon (C). Impurities containing sulfur (S), nitrogen (N), or phosphorus (P) form gaseous compounds with oxygen that become pollutants. The smoke contains these pollutants and also tiny dust particles that pollute the air, are blown around by the wind, and eventually fall to the ground.

Method one
Steel mill smoke was passed across a set of electrified plates in the smokestack. The electrified plates attracted the dust particles in the smoke and removed them from the emissions. The efficiency of dust particle removal, which is dependent on the size of the particles, is shown in Table 1.

Table 1.
Voltage on plates
(V) Particles removed
(%)
100 20
200 50
500 90
1,000 80
3,000 70


Method two
The smoke was passed through filters of various pore size in the smokestack that trapped dust particles and also reduced the amount of smoke leaving the stack. The data are shown in

Table 2.
Filter pore size
(microns) Particles trapped
(%) Amount of time for filters to drop to 50% efficiency
(hours)
1,000 five 2,000
500 20 1,000
200 50 200
100 80 25
50 90 one


Method three
To decrease pollution by chemical means, the smokestack emissions were bubbled through solutions of concentrated alkali (solutions of OH- ions). Table three indicates the percent of the three pollutants removed versus the alkali concentration.

Table 3.
Alkali concentration
(%) S
removed
(%) N
removed
(%) P
removed
(%)
1 80 60 10
  three 90 65 40
  five 91 60 35
  10 92 50 30
  15 93 20 25


Q. 33 If the plant's smoke contained equal amounts of S, N, and P, which of the subsequent alkali concentrations would remove the most total pollution?


1%
3%
5%
10%
Q. 34 Which of the subsequent statements about the alkali concentration is consistent with the outcomes of Method 3?


A higher alkali concentration outcomes in more P removed.
A higher alkali concentration outcomes in more S removed.
A lower alkali concentration requires more filters to remove dust particles.
A lower alkali concentration outcomes in an increased time needed to remove all of the chemical pollutants
Q. 35 Based on the characteristics of the filters used in Method 2, which of the subsequent best defines the practical issue involved in choosing the best kind of smokestack filter? The filter that traps the highest percentage of particles:


requires the highest voltage across the electrified plates.
reacts with alkali solutions.
needs to be changed lowest often.
needs to be changed most often.
Q. 36 If the smokestack of the steel mill was doubled in height, what effect, if any, would this be expected to have on the removal of pollutants?


Filter efficiency would reduce because dust particles could more easily fall back down the stack.
Electrostatic plate efficiency would increase because voltage increases with height.
Dust particles would accumulate into larger pieces and would be more easily removed from the smoke.
It cannot be determined from the provided info.
Q. 37 Which of the subsequent assumptions about reducing pollution emissions is common to both Methods one and 2 ?


Emissions can only be removed by filters.
Emissions can only be removed by electrified plates.
Emissions must be captured in the smokestack to be removed.
Emissions must be captured after they leave the smokestack.
Q. 38 To further investigate the effects of voltage on the removal of dust particles from steel mill smoke, the scientists could use which of the subsequent procedures?


Determining where the particulate matter falls to the ground after leaving the smokestack
Determining what sizes of particles are removed from the smoke at various voltages
Determining how the filters react when an alkali solution is passed through them
Determining how the filters react when an acid solution is passed through them

Earning:   Approval pending.