The question asks to identify the mode of interaction through which the HIV-AIDS virus cannot be transmitted. Understanding the common routes of HIV transmission is crucial for answering this question.
Correct Option: D) Shaking hands is a form of casual contact and does not involve the exchange of bodily fluids (blood, semen, vaginal fluids, breast milk) necessary for HIV transmission. Therefore, the HIV-AIDS virus cannot be transmitted by shaking hands.
The electrolytic decomposition of aqueous sodium chloride involves the electrolysis of a solution containing \(\text{Na}^+\), \(\text{Cl}^-\), \(\text{H}^+\), and \(\text{OH}^-\). We need to determine which species will be oxidized at the anode and which will be reduced at the cathode based on their standard electrode potentials and overpotential effects.
Ions present in aqueous NaCl solution: When sodium chloride (NaCl) is dissolved in water (\(\text{H}_2\text{O}\)), it dissociates into \(\text{Na}^+\) and \(\text{Cl}^-\) ions. Water itself undergoes slight autoionization to produce \(\text{H}^+\) and \(\text{OH}^-\) ions. So, the solution contains \(\text{Na}^+\), \(\text{Cl}^-\), \(\text{H}^+\), and \(\text{OH}^-\) ions, along with \(\text{H}_2\text{O}\) molecules.
At the Cathode (Reduction): Cations move towards the cathode (negative electrode). Here, \(\text{Na}^+\) and \(\text{H}^+\) (or \(\text{H}_2\text{O}\)) compete for reduction. The relevant reduction half-reactions and their standard electrode potentials are:
\(\text{Na}^+(\text{aq}) + \text{e}^- \rightarrow \text{Na}(\text{s})\) \(\text{E}^\circ = -2.71 \text{ V}\)
\(2\text{H}_2\text{O}(\text{l}) + 2\text{e}^- \rightarrow \text{H}_2(\text{g}) + 2\text{OH}^-(\text{aq})\) \(\text{E}^\circ = -0.83 \text{ V}\) (at pH 7)
Since water has a less negative (more positive) reduction potential compared to \(\text{Na}^+\), water is preferentially reduced at the cathode. Therefore, hydrogen gas (\(\text{H}_2\)) is evolved at the cathode.
At the Anode (Oxidation): Anions move towards the anode (positive electrode). Here, \(\text{Cl}^-\) and \(\text{OH}^-\) (or \(\text{H}_2\text{O}\)) compete for oxidation. The relevant oxidation half-reactions (reverse of reduction) and their standard electrode potentials are:
\(2\text{Cl}^-(\text{aq}) \rightarrow \text{Cl}_2(\text{g}) + 2\text{e}^-\) \(\text{E}^\circ = -1.36 \text{ V}\)
\(2\text{H}_2\text{O}(\text{l}) \rightarrow \text{O}_2(\text{g}) + 4\text{H}^+(\text{aq}) + 4\text{e}^-\) \(\text{E}^\circ = -1.23 \text{ V}\)
Based purely on standard electrode potentials, water should be oxidized preferentially as it has a less negative oxidation potential (more positive reduction potential). However, there is a significant overpotential for the evolution of oxygen gas at many electrode surfaces. This overpotential makes the actual potential required for oxygen evolution higher than that for chlorine evolution. Due to this overpotential, chlorine gas (\(\text{Cl}_2\)) is preferentially evolved at the anode in concentrated aqueous NaCl solutions.
Thus, \(\text{Cl}_2\) gas is evolved at the anode and \(\text{H}_2\) gas is evolved at the cathode.
Correct Option: D) Cl₂ gas at the anode and H₂ gas at the cathode.
The question asks to identify P. V. Sindhu's opponent in the final of the Swiss Open Super 300 Badminton Tournament in March 2022, whom she defeated to win the title. This requires knowledge of specific sports event results.
Correct Option: A) Busanan Ongbamrungphan
P. V. Sindhu defeated Thailand's Busanan Ongbamrungphan by 21-16, 21-8 in the final of the Swiss Open Super 300 Badminton Tournament on March 27, 2022, to win the women's singles title.
This problem involves calculations related to a convex mirror. We will use the mirror formula to first find the focal length of the mirror using the initial object and image distances. Once the focal length is known, we will use the mirror formula again with the new object distance to find the new image distance.
The question asks to identify the leading manganese ore-producing state in India for the fiscal year 2018–19, given that it accounted for 33% of the total production. This requires knowledge of the geographical distribution of mineral resources in India.
Correct Option: B) Madhya Pradesh was the leading manganese ore-producing state in India in 2018–19, accounting for 33% of the total production.
The question asks to identify the organelle(s) exclusively found in plant cells. To answer this, we need to recall the distinguishing features of plant cells compared to animal cells and the functions of the given organelles.
Correct Option: B) Plastids are a group of organelles, including chloroplasts, chromoplasts, and leucoplasts, that are characteristic features of plant cells and some algae. Chloroplasts, in particular, are essential for photosynthesis, a process unique to plants and some other autotrophs.
6CO₂(aq) + 12H₂O(l) —(?)→ C₆H₁₂O₆(aq) + 6O₂(aq) + 6H₂O(l). Select the correct reaction conditions for the given reaction.
High temperature
High pressure
Sunlight and chlorophyll
Inert atmosphere
The given chemical equation represents the process of photosynthesis. To identify the correct reaction conditions, one must recall the essential requirements for photosynthesis to occur in plants.
\[ 6\text{CO}_2(\text{aq}) + 12\text{H}_2\text{O}(\text{l}) \xrightarrow{(?)} \text{C}_6\text{H}_{12}\text{O}_6(\text{aq}) + 6\text{O}_2(\text{aq}) + 6\text{H}_2\text{O}(\text{l}) \]
This equation shows carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)) reacting to form glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)), oxygen (\(\text{O}_2\)), and water. This is the balanced chemical equation for photosynthesis.
C) Sunlight and chlorophyll — This option correctly identifies the two crucial conditions required for photosynthesis, which is represented by the given chemical equation. Sunlight provides the energy, and chlorophyll absorbs this light energy to drive the reaction.
In this question, a group of numbers/symbols is coded using letters as per the table given below and the conditions which follow. The correct combination of codes following the conditions is your answer.
| Number/Symbol | 6 | 8 | 4 | + | 5 | # | % | 2 | $ | 1 | ^ | 7 | 9 | * | @ |
| Code | S | U | T | W | N | P | V | D | A | Z | J | M | O | M | Y |
Conditions:
(i) If the first and last elements are numbers, then the last element will be replaced by the symbol @.
(ii) If the first element is a symbol, then the first element will be coded as ©.
(iii) If both the first and second elements are numbers, then the third element will be coded as *.
What will be the code for the following?
6 % $ 8 5
The problem requires us to code a given sequence of numbers and symbols based on a provided coding table and a set of conditions. We need to apply the conditions sequentially and then use the table to find the final code.
D) SVAU@ is the correct code derived after applying the conditions and the coding table.
The given statements are followed by two conclusions. Find out which conclusion is true based on the given statements.
Statements:
J > K = L, M ≥ N < K, O < J < P, N > Q > R
Conclusions:
I. P > M
II. N ≤ P
Only conclusion I is true
Neither conclusion I nor II is true
Only conclusion II is true
Both conclusions I and II are true
This problem involves analyzing a set of inequalities to determine the truthfulness of given conclusions. We need to combine the individual statements to form a consolidated relationship chain and then check if the conclusions logically follow from this chain.
The statements are:
Let's try to link these statements. We can see that \(K\) is common in the first two statements, and \(J\) is common in the first and third statements. \(N\) is common in the second and fourth statements.
From \(J > K\) and \(M \ge N < K\), we get \(J > K > N \le M\). So, \(J > N\).
From \(J > K\) and \(K = L\), we have \(J > K\) and \(J > L\).
From \(M \ge N\) and \(N > Q > R\), we have \(M \ge N > Q > R\).
From \(O < J\) and \(J < P\), we have \(O < J < P\).
Let's try to form a longer chain:
We have \(P > J\), and \(J > K\). So, \(P > J > K\).
We also have \(K > N\). So, \(P > J > K > N\).
And we have \(N \le M\). So, \(P > J > K > N \le M\).
Also, \(N > Q > R\).
And \(O < J\).
A consolidated chain can be written as:
\[ P > J > K = L > N \le M \]\[ P > J > K > N > Q > R \]\[ O < J \]From our consolidated chain, we have \(P > J > K > N \le M\).
This means \(P > N\) and \(N \le M\).
If \(N = M\), then \(P > M\). But if \(N < M\), then \(P > N\) and \(N < M\). In this case, the relationship between \(P\) and \(M\) cannot be definitively determined. For example, if \(P=10, N=5, M=7\), then \(P > M\). But if \(P=10, N=5, M=12\), then \(P < M\). Since the relationship is not always true, Conclusion I is false.
From our consolidated chain, we have \(P > J > K > N\).
Since \(P\) is strictly greater than \(J\), \(J\) is strictly greater than \(K\), and \(K\) is strictly greater than \(N\), it logically follows that \(P\) must be strictly greater than \(N\).
If \(P > N\), then it is also true that \(P \ge N\), which can be written as \(N \le P\).
Therefore, Conclusion II is true.
C) Only conclusion II is true
The question asks why atomic size increases down a group in the Modern Periodic Table. This trend is directly related to the electronic configuration of elements as one moves down a group.
A) number of shells increases. This is the primary reason for the increase in atomic size down a group. Each new period (row) represents the filling of a new electron shell, and elements in the same group but lower down are in higher periods, hence possessing more electron shells.