The question asks to identify the compound that contains only one carbon-carbon single bond. This requires understanding the basic structures of simple alkanes and counting the number of C-C bonds in each.
Ethane has the chemical formula \( \text{C}_2\text{H}_6 \). Its structural formula is \( \text{CH}_3 - \text{CH}_3 \). In this molecule, there is one bond between the two carbon atoms, which is a single bond. Therefore, ethane contains only one carbon-carbon single bond.
Which of the following is equal to 0.1\(\overline{82}\)?
\(\frac{181}{990}\)
\(\frac{81}{500}\)
\(\frac{1}{999}\)
\(\frac{1}{9}\)
To convert a repeating decimal (non-terminating recurring decimal) into a fraction, we use a standard algebraic method. We set the given decimal equal to a variable, multiply it by powers of 10 to align the repeating part, and then subtract the equations to eliminate the repeating part, solving for the variable.
Let the given decimal be \(x\):
\[ x = 0.1\overline{82} \]This means the digits '82' repeat indefinitely:
\[ x = 0.1828282... \quad \text{(Equation 1)} \]First, multiply Equation 1 by 10 to move the non-repeating part (the digit '1') to the left of the decimal point:
\[ 10x = 1.828282... \quad \text{(Equation 2)} \]Next, multiply Equation 1 by 1000 (since there is 1 non-repeating digit and 2 repeating digits, \(10^{(1+2)} = 10^3\)) to move one full cycle of the repeating part to the left of the decimal point:
\[ 1000x = 182.828282... \quad \text{(Equation 3)} \]Now, subtract Equation 2 from Equation 3. This will eliminate the repeating part after the decimal point:
\[ 1000x - 10x = 182.828282... - 1.828282... \]\[ 990x = 181 \]Solve for \(x\):
\[ x = \frac{181}{990} \]A) The calculation shows that \(0.1\overline{82}\) is equal to \(\frac{181}{990}\).
To solve this problem, we need to first identify the original number and its digits. Then, we will arrange these digits in ascending order to form a new number. Finally, we will compare the positions of the digits in the original number with their positions in the new number to find which digits remain unchanged. The sum of these unchanged digits will be the answer.
Correct Option: A) 5
Ozone at the higher levels of the atmosphere is a product of UV radiation acting on _______ molecule.
sulphur dioxide (SO2)
oxygen (O2)
carbon monoxide (CO)
carbon dioxide (CO2)
The question asks about the formation of ozone in the higher atmosphere due to UV radiation. This process is a fundamental aspect of atmospheric chemistry, specifically related to the ozone layer's formation and its role in protecting Earth from harmful UV rays. We need to identify the precursor molecule that reacts with UV radiation to initiate ozone formation.
Correct Option: B) oxygen (O 2 ) — Ozone in the higher atmosphere is formed when ultraviolet (UV) radiation from the sun breaks down oxygen molecules (\(O_2\)) into individual oxygen atoms (\(O\)). These free oxygen atoms then combine with other oxygen molecules to form ozone (\(O_3\)).
The question asks to identify the dam built in the 1970s whose oustees (displaced people) are still struggling to receive promised benefits. This requires knowledge of major dam projects in India, their construction timelines, and associated social issues.
The question asks about the institution under which the Centre of Excellence for Studies in Classical Telugu (CESCT) was established after Telugu received classical language status in 2008. This requires knowledge of the administrative structure related to classical language studies in India.
Correct Option: B) Central Institute of Indian Languages, Mysore
The Centre of Excellence for Studies in Classical Telugu (CESCT) was indeed established under the Central Institute of Indian Languages (CIIL), Mysore, after Telugu was accorded classical language status in 2008. CIIL is the nodal agency for various linguistic research and development activities in India, including those related to classical languages.
This question asks about the fundamental purpose of maintenance processes in living organisms. Life processes involve continuous activity, which can lead to wear and tear. The options provided describe various biological functions, and we need to identify which one directly relates to 'maintenance'.
Correct Option: B) prevent damage and breakdown
Living organisms are constantly undergoing metabolic activities, and their cells and tissues are subject to wear and tear. Maintenance processes, such as cell repair, regeneration, and immune responses, are crucial to prevent damage from accumulating and to repair any breakdown that occurs, ensuring the organism's survival and proper functioning.
The problem involves calculating simple interest for three different schemes and then finding the difference between the principal amounts invested in two of them. The key information is that the simple interest received from all three schemes is the same. We will use the simple interest formula \(SI = \frac{P \times R \times T}{100}\) to establish relationships between the principal amounts and then use the total investment to find the individual principal amounts.
Let the principal amounts invested in schemes A, B, and C be \(P_A\), \(P_B\), and \(P_C\) respectively.
Calculate the simple interest for each scheme:
Scheme A: Rate \(R_A = 10\%\) p.a., Time \(T_A = 3\) years.
\[ SI_A = \frac{P_A \times 10 \times 3}{100} = \frac{30 P_A}{100} \]Scheme B: Rate \(R_B = 9\%\) p.a., Time \(T_B = 4\) years.
\[ SI_B = \frac{P_B \times 9 \times 4}{100} = \frac{36 P_B}{100} \]Scheme C: Rate \(R_C = 8\%\) p.a., Time \(T_C = 5\) years.
\[ SI_C = \frac{P_C \times 8 \times 5}{100} = \frac{40 P_C}{100} \]Given that the simple interest received from all three schemes is the same, let this common interest be \(SI\). So, \(SI_A = SI_B = SI_C = SI\).
\[ \frac{30 P_A}{100} = \frac{36 P_B}{100} = \frac{40 P_C}{100} \]\[ 30 P_A = 36 P_B = 40 P_C \]Find the least common multiple (LCM) of 30, 36, and 40. The LCM is 360.
Divide each term by 360 to find the ratio of \(P_A : P_B : P_C\):
\[ \frac{30 P_A}{360} = \frac{36 P_B}{360} = \frac{40 P_C}{360} \]\[ \frac{P_A}{12} = \frac{P_B}{10} = \frac{P_C}{9} \]This implies that \(P_A : P_B : P_C = 12 : 10 : 9\).
Let \(P_A = 12k\), \(P_B = 10k\), and \(P_C = 9k\) for some constant \(k\).
The total sum invested is ₹47,740.
\[ P_A + P_B + P_C = 47740 \]\[ 12k + 10k + 9k = 47740 \]\[ 31k = 47740 \]\[ k = \frac{47740}{31} \]\[ k = 1540 \]Now, calculate the individual principal amounts:
\(P_A = 12k = 12 \times 1540 = 18480\)
\(P_B = 10k = 10 \times 1540 = 15400\)
\(P_C = 9k = 9 \times 1540 = 13860\)
The question asks for the positive difference between the sums invested in schemes C and A, which is \(|P_C - P_A|\).
\[ |P_C - P_A| = |13860 - 18480| = |-4620| = 4620 \]The question asks about the location of anaerobic respiration in the human body. Anaerobic respiration is a metabolic process that occurs in the absence of oxygen. Understanding the basic cellular processes and the organelles involved is key to answering this question.
Therefore, the cytoplasm is the general site for anaerobic respiration across various cell types capable of it.
Correct Option: C) In the cytoplasm of cells
Anaerobic respiration, such as lactic acid fermentation in humans, begins with glycolysis, which takes place in the cytoplasm. The subsequent conversion of pyruvate to lactic acid also occurs in the cytoplasm. This process does not require oxygen or mitochondria.