The question asks to identify a Buddhist temple among the given options. This requires knowledge of prominent religious sites in India and their associated religions.
Correct Option: A) Mahabodhi Temple
The Mahabodhi Temple Complex, located in Bodh Gaya, Bihar, is one of the most sacred sites in Buddhism. It is believed to be the place where Gautama Buddha attained enlightenment under the Bodhi tree. It is a UNESCO World Heritage Site and a prime example of brick architecture from the Gupta period, with later additions. Its primary purpose is as a Buddhist place of worship and pilgrimage.
This problem involves simple interest. We are given a scenario where an increase in the interest rate leads to a specific increase in the total interest earned over a fixed period. We can use the simple interest formula to set up an equation and solve for the principal amount (sum invested).
Thus, the sum invested was Rs 6,250.
A) 6250 — This is the correct sum calculated based on the given conditions.
The question asks for the SI unit of the power of a lens. We need to recall the definition of the power of a lens and its associated unit in the International System of Units (SI).
Correct Option: C) dioptre — The dioptre (D) is the SI unit for the power of a lens. One dioptre is equivalent to one reciprocal meter \((1 \text{ D} = 1 \text{ m}^{-1})\).
The question asks about the language with the maximum number of speakers in India according to the 2011 Census data among the given options. To answer this, one needs to recall or refer to the linguistic data from the Census of India 2011.
B) Bengali has approximately 97.2 million speakers according to the Census 2011, making it the language with the maximum number of speakers among the given options.
To balance a chemical equation, we need to ensure that the number of atoms of each element is the same on both the reactant side and the product side. We will balance the elements one by one, starting with elements other than hydrogen and oxygen, then hydrogen, and finally oxygen.
Identify the unbalanced equation:
\[ \text{aHNO}_3\text{(aq)} + \text{bCa(OH)}_2\text{(aq)} \rightarrow \text{Ca(NO}_3\text{)}_2 + \text{cH}_2\text{O} \]
Balance Calcium (Ca) atoms:
On the product side, there is \(1\) Ca atom in \(\text{Ca(NO}_3\text{)}_2\).
On the reactant side, there is \(1\) Ca atom in \(\text{Ca(OH)}_2\).
So, Ca is already balanced. We can set \(b=1\).
\[ \text{aHNO}_3\text{(aq)} + 1\text{Ca(OH)}_2\text{(aq)} \rightarrow \text{Ca(NO}_3\text{)}_2 + \text{cH}_2\text{O} \]
Balance Nitrate (NO₃) or Nitrogen (N) atoms:
On the product side, there are \(2\) \(\text{NO}_3\) groups in \(\text{Ca(NO}_3\text{)}_2\).
On the reactant side, there is \(1\) \(\text{NO}_3\) group in \(\text{HNO}_3\).
To balance the \(\text{NO}_3\) groups, we need to place a coefficient of \(2\) in front of \(\text{HNO}_3\). So, \(a=2\).
\[ 2\text{HNO}_3\text{(aq)} + 1\text{Ca(OH)}_2\text{(aq)} \rightarrow \text{Ca(NO}_3\text{)}_2 + \text{cH}_2\text{O} \]
Balance Hydrogen (H) atoms:
On the reactant side:
From \(2\text{HNO}_3\), there are \(2 \times 1 = 2\) H atoms.
From \(1\text{Ca(OH)}_2\), there are \(1 \times 2 = 2\) H atoms.
Total H atoms on the reactant side = \(2 + 2 = 4\) H atoms.
On the product side:
From \(\text{cH}_2\text{O}\), there are \(c \times 2\) H atoms.
To balance H atoms, \(2c = 4\), which means \(c = 2\).
\[ 2\text{HNO}_3\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow \text{Ca(NO}_3\text{)}_2 + 2\text{H}_2\text{O} \]
Balance Oxygen (O) atoms (as a final check):
On the reactant side:
From \(2\text{HNO}_3\), there are \(2 \times 3 = 6\) O atoms.
From \(1\text{Ca(OH)}_2\), there are \(1 \times 2 = 2\) O atoms.
Total O atoms on the reactant side = \(6 + 2 = 8\) O atoms.
On the product side:
From \(\text{Ca(NO}_3\text{)}_2\), there are \(2 \times 3 = 6\) O atoms.
From \(2\text{H}_2\text{O}\), there are \(2 \times 1 = 2\) O atoms.
Total O atoms on the product side = \(6 + 2 = 8\) O atoms.
The oxygen atoms are balanced.
Final balanced equation:
\[ 2\text{HNO}_3\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow \text{Ca(NO}_3\text{)}_2 + 2\text{H}_2\text{O} \]
The coefficients are \(a=2\), \(b=1\), and \(c=2\).
Correct Option: C) 2, 1 and 2, respectively
Refer to the following letter, number, symbol series and answer the question.
(Left) E 7 8 # K 2 L ? = R 3 @ W * X 5 C H D 4 (Right)
If all numbers are dropped from the series, which of the following will be seventh from the left?
The problem asks us to identify the seventh element from the left in a given series after all numbers have been removed. We need to first filter out the numbers and then count the elements from the left to find the seventh one.
C) R — As determined in the step-by-step analysis, after removing all numbers, 'R' is the seventh element from the left.
The question asks to identify the incorrectly matched event and year from modern Indian history. We need to recall the correct years for each of the given historical events.
Based on the analysis, Option A is the incorrectly matched pair.
Correct Option: A) Dandi March by Mahatma Gandhi – 1910
The Dandi March, also known as the Salt Satyagraha, was initiated by Mahatma Gandhi on March 12, 1930, and concluded on April 6, 1930. The year 1910 is incorrect for this event.
The power dissipated by an electrical component is determined by the voltage across it and the current flowing through it. We will use the formula for power in terms of voltage and resistance.
Therefore, if no voltage is applied across the bulb, no power is dissipated.
Correct Option: A) 0 W
The problem is a number analogy question. We need to find the relationship between the first and second terms, and the fifth and sixth terms, and then apply the same relationship to the third term to find the missing fourth term.
Let the first term be \(N_1 = 15\). The second term is \(N_2 = 4913\).
We need to find a mathematical operation that transforms 15 into 4913. Let's try operations involving cubes or squares, as 4913 is a relatively large number.
Consider \((N_1 + 2)^3\):
\[ (15 + 2)^3 = (17)^3 \]\[ 17 \times 17 = 289 \]\[ 289 \times 17 = 4913 \]So, the relationship is \(N_2 = (N_1 + 2)^3\).Let the fifth term be \(N_5 = 11\). The sixth term is \(N_6 = 2197\).
Apply the same relationship: \((N_5 + 2)^3\)
\[ (11 + 2)^3 = (13)^3 \]\[ 13 \times 13 = 169 \]\[ 169 \times 13 = 2197 \]The relationship holds true for the third pair as well.Let the third term be \(N_3 = 8\). We need to find the missing fourth term, \(N_4\).
Using the established relationship \(N_4 = (N_3 + 2)^3\):
\[ N_4 = (8 + 2)^3 \]\[ N_4 = (10)^3 \]\[ N_4 = 10 \times 10 \times 10 \]\[ N_4 = 1000 \]C) 1000 — The pattern established is that the second term is the cube of (the first term plus 2). Applying this to 8, we get \((8+2)^3 = 10^3 = 1000\).
Find the mode for the following data of student ages:
16, 17, 15, 17, 16, 15, 14, 14, 13, 17, 13, 12, 12, 16, 10, 14, 17, 10, 11.
To find the mode of a dataset, we need to identify the value that appears most frequently. This involves counting the occurrences of each unique data point.
C) 17 is the correct answer because it appears 4 times, which is the highest frequency among all the given ages.