The question asks to identify the correct statement regarding aerobic respiration. We need to evaluate each option based on the principles of cellular respiration.
B) In mitochondria, one molecule of pyruvate breaks down to give three molecules of carbon dioxide.
Here's the breakdown:
Thus, for each molecule of pyruvate, a total of \(1 + 2 = 3\) molecules of carbon dioxide are released.
The question asks about a specific initiative launched by the Department of Biotechnology, Ministry of Science and Technology, on March 24, 2022, aimed at eradicating a particular disease. To answer this, one needs to recall current affairs related to health initiatives in India.
B) Tuberculosis is the correct answer. The "Data-Driven Research & Development for TB" initiative was indeed launched by the Department of Biotechnology on March 24, 2022, to accelerate efforts towards eradicating Tuberculosis.
The question asks for the product formed when carbon monoxide (CO) reacts with hydrogen gas (Hâ‚‚) under specific conditions, particularly high pressure. This is a classic industrial synthesis reaction, often catalyzed.
Correct Option: B) \(\text{CH}_3\text{OH}\) (Methanol) is the primary product formed when carbon monoxide reacts with hydrogen gas at high pressure (340 atm) in the presence of appropriate catalysts. This is a major industrial method for producing methanol.
This question tests knowledge about the post-fertilisation changes in flowering plants. Understanding the fate of different floral parts after successful fertilisation is key to answering this question.
Correct Option: B) seed — After fertilisation, the ovule develops into a seed, which contains the embryo and stored food, protected by a seed coat.
The question asks to identify which option is NOT a common route of HIV transmission. We need to recall the known modes of HIV transmission and identify the one that does not facilitate the spread of the virus.
Correct Option: A) Casual contacts like handshakes or hugs with the infected people
A can complete \( \frac{2}{3} \) of a certain work in 12 days and B can complete \( \frac{1}{4} \) of the same work in 3 days. Working together, the number of days they will take to finish the work is:
5\( \frac{1}{5} \)
8\( \frac{1}{5} \)
7\( \frac{1}{5} \)
6\( \frac{1}{5} \)
To solve this problem, we need to determine the individual rates at which A and B complete the work. Once we have their individual rates (work done per day), we can find their combined rate and then calculate the total time taken to finish the entire work together.
Calculate A's daily work rate:
A completes \( \frac{2}{3} \) of the work in 12 days.
So, A completes 1 (whole) work in \( 12 \div \frac{2}{3} \) days.
\[ \text{Time taken by A to complete full work} = 12 \times \frac{3}{2} = 18 \text{ days} \]Therefore, A's one day's work is \( \frac{1}{18} \) of the total work.
Calculate B's daily work rate:
B completes \( \frac{1}{4} \) of the work in 3 days.
So, B completes 1 (whole) work in \( 3 \div \frac{1}{4} \) days.
\[ \text{Time taken by B to complete full work} = 3 \times 4 = 12 \text{ days} \]Therefore, B's one day's work is \( \frac{1}{12} \) of the total work.
Calculate their combined daily work rate:
When A and B work together, their combined one day's work is the sum of their individual daily work rates.
\[ \text{Combined one day's work} = \frac{1}{18} + \frac{1}{12} \]To add these fractions, find a common denominator, which is 36.
\[ \text{Combined one day's work} = \frac{2}{36} + \frac{3}{36} = \frac{2+3}{36} = \frac{5}{36} \]So, together they complete \( \frac{5}{36} \) of the work in one day.
Calculate the total time taken to finish the work together:
If they complete \( \frac{5}{36} \) of the work in one day, then they will complete the entire work (1 unit) in the reciprocal of this fraction.
\[ \text{Total time taken} = \frac{1}{\frac{5}{36}} = \frac{36}{5} \text{ days} \]Convert the improper fraction to a mixed number:
\[ \frac{36}{5} = 7 \frac{1}{5} \text{ days} \]Correct Option: C) 7\( \frac{1}{5} \)
This is a logical reasoning problem involving sequencing events based on given conditions. We need to determine the exam schedule for five individuals (A, B, C, D, E) over five days (Monday to Friday) using the provided clues.
List the days: The exams are held from Monday to Friday.
Place A: "A has the exam on Thursday."
Place D and B: "D has the exam on the day immediately before B." This means D and B must be in consecutive slots (D then B).
Consider D and E: "Only one person has the exam between D and E." This implies a sequence like D _ E or E _ D.
Combine clues for D, B, and E:
If we combine (D, B) with (D, _, E), it means D is followed by B, and there's one person between D and E. This can only happen if B is the person between D and E, leading to the sequence (D, B, E). This sequence requires three consecutive days.
Fit (D, B, E) into the schedule: We need three consecutive empty slots for D, B, E. Looking at our current schedule:
The only three consecutive empty slots are Monday, Tuesday, and Wednesday.
Place the remaining person: The only person left is C, and the only day left is Friday.
Verify all conditions:
All conditions are satisfied.
Identify who has the exam on Friday: From the final schedule, C has the exam on Friday.
Correct Option: A) C
The person who has the exam on Friday is C, as determined by logically placing all individuals based on the given clues.
The question asks to identify the state that declared Urdu as its second official language in 2017. This is a factual recall question related to Indian polity and current affairs.
Correct Option: D) Telangana declared Urdu as its second official language in 2017. This decision was announced by the Chief Minister of Telangana, K. Chandrashekar Rao, to promote the language and ensure its use in official communications and competitive examinations.
Six friends M, N, O, P, Q and R are sitting in a row facing north. Only O is seated between P and M. Only R is seated between N and Q. M is not seated at either of the ends and is not an immediate neighbour of Q. P is sitting at the immediate left of O. Where is N seated?
At one of the extreme ends
2nd from the extreme left end
Immediate right of M
2nd from the extreme right end
This is a linear arrangement puzzle. We need to deduce the seating order of six friends (M, N, O, P, Q, R) based on the given clues. We will build the arrangement step-by-step, eliminating possibilities until a unique order is found.
Clue 1: P is sitting at the immediate left of O.
This gives us a block: P O
Clue 2: Only O is seated between P and M.
Combining with the first clue, this means M must be to the immediate right of O. So the arrangement becomes: P O M
Clue 3: M is not seated at either of the ends.
Since M is in the middle of a block of three, and there are six people in total, this clue is consistent with our current arrangement (P O M). It implies there must be at least one person to M's right.
Clue 4: M is not an immediate neighbour of Q.
This means Q cannot be immediately to the left or right of M. Since O is to M's left, Q cannot be to M's right. So, Q is not next to M.
Clue 5: Only R is seated between N and Q.
This gives us another block: N R Q or Q R N. Since M is not an immediate neighbour of Q, the block cannot be M Q R N or N R Q M. It implies that Q cannot be next to M.
Combining the blocks and clues:
We have P O M as one block. We have N R Q or Q R N as another block. There are six positions. Let's represent the positions as 1 2 3 4 5 6.
If we place P O M, it could be at positions 1 2 3, 2 3 4, 3 4 5, or 4 5 6.
Let's try to fit P O M into the arrangement. Since M is not at an end, P O M cannot be at 4 5 6 (as M would be at position 6) and cannot be at 1 2 3 (as P would be at position 1, and M is not at an end, but this doesn't restrict M from being at position 3). Let's re-evaluate M not being at an end. If M is at position 3, it's not an end. If M is at position 4, it's not an end. If M is at position 5, it's not an end.
Consider the full arrangement. We have P O M. The remaining three people are N, R, Q. They must form the block N R Q or Q R N.
Let's try placing P O M first. The only way to place N R Q or Q R N such that M is not an immediate neighbour of Q is if the two blocks are separated or Q is at an end not adjacent to M.
If the arrangement is P O M _ _ _:
The remaining positions are for N, R, Q. If we put N R Q, it would be P O M N R Q. Here, M is next to N, not Q. This is a valid arrangement.
Let's check all conditions for P O M N R Q:
This arrangement P O M N R Q satisfies all conditions.
Let's consider the other possibility for the N, R, Q block: Q R N. If we try P O M Q R N, then M is an immediate neighbour of Q, which violates a condition. So, Q R N cannot follow P O M directly.
What if the blocks are arranged differently? For example, N R Q P O M. In this case, M is at an end, which is not allowed. So this is not possible.
What if the arrangement is _ P O M _ _? Or _ _ P O M _?
If the arrangement is _ P O M _ _. The block N R Q must fit in the remaining spots. If it's N R Q P O M, M is at an end. If it's P O M N R Q, this is what we found. If it's _ P O M N R Q, then P is not at the start, and there's a blank at the start. This means P is not at an end. But we need 6 people. So the arrangement P O M N R Q is the only valid one.
The final arrangement is: P O M N R Q
Locate N:
In the arrangement P O M N R Q, N is seated immediately to the right of M.
C) Immediate right of M. As derived, the final arrangement is P O M N R Q, where N is immediately to the right of M.
This question deals with the refraction of light as it passes from one medium to another, specifically from air into a glass prism and then from the glass prism back into air. We need to apply Snell's Law and understand the relationship between the angle of incidence and the angle of refraction when light travels from a rarer medium to a denser medium, and vice versa.
Refraction at surface AB (Air to Glass):
When a ray of light enters the prism from the air, it moves from a rarer medium (air) to a denser medium (glass). According to Snell's Law, when light travels from a rarer medium to a denser medium, it bends towards the normal. This means the angle of refraction (\(r_1\)) will be less than the angle of incidence (\(i_1\)).
Mathematically, Snell's Law states:
\[ n_1 \sin(i_1) = n_2 \sin(r_1) \]Here, \(n_1\) is the refractive index of air (approximately 1) and \(n_2\) is the refractive index of glass (greater than 1). Since \(n_2 > n_1\), for the equality to hold, \(\sin(r_1)\) must be less than \(\sin(i_1)\). As angles are acute, this implies \(r_1 < i_1\).
Therefore, statement (a) "The angle of refraction at surface AB is less than the angle of incidence" is correct.
Refraction at surface AC (Glass to Air):
When the ray of light emerges from the prism into the air, it moves from a denser medium (glass) to a rarer medium (air). In this case, the light bends away from the normal. This means the angle of emergence (which is the angle of refraction at surface AC, let's call it \(e\)) will be greater than the angle of incidence inside the prism at surface AC (let's call it \(r_2\)).
Applying Snell's Law at surface AC:
\[ n_2 \sin(r_2) = n_1 \sin(e) \]Here, \(n_2\) is the refractive index of glass and \(n_1\) is the refractive index of air. Since \(n_2 > n_1\), for the equality to hold, \(\sin(e)\) must be greater than \(\sin(r_2)\). This implies \(e > r_2\).
Therefore, statement (b) "The angle of refraction at the surface AC is less than the angle of incidence" is incorrect. The angle of refraction (emergence) at AC is *greater* than the angle of incidence (\(r_2\)) at AC.
C) Only (a) is correct because at surface AB, light goes from air (rarer) to glass (denser), causing it to bend towards the normal, making the angle of refraction less than the angle of incidence.