The problem presents a numerical analogy where the relationship between the first and second numbers is the same as between the third and fourth numbers. We need to find the fifth number based on this established pattern. We will analyze the given pairs to deduce the underlying rule.
A) 234 is the correct answer, derived from applying the pattern \(N^2 + 9\) to the number 15.
This question tests the understanding of the definitions of reduction in chemistry, specifically in terms of oxygen and hydrogen transfer. We need to evaluate each statement based on the classical definitions of oxidation and reduction.
C) Both the statements are true.
The question asks to identify a geothermal energy harnessing site in India from the given options. This requires knowledge of the locations where geothermal energy projects are either operational or under development in India.
Correct Option: B) Manikaran
The question asks to identify the first state in India to launch an 'Air Health Service' in rural areas as of March 2022. This is a factual question related to current affairs and government initiatives in the healthcare sector.
Correct Option: D) Odisha became the first state in India to start 'Air Health Service' in rural areas in March 2022. The service was launched to provide healthcare facilities to people in remote and inaccessible regions, particularly in tribal-dominated districts like Malkangiri.
Find the area of a triangle whose sides are 12 m, 14 m and 16 m.
21\(\sqrt{15}\) m²
7\(\sqrt{15}\) m²
2\(\sqrt{15}\) m²
\(\sqrt{15}\) m²
To find the area of a triangle when all three side lengths are given, we use Heron's formula. Heron's formula requires calculating the semi-perimeter first.
Let the sides of the triangle be \(a\), \(b\), and \(c\).
The semi-perimeter is half the perimeter of the triangle.
\[ s = \frac{a + b + c}{2} \]\[ s = \frac{12 + 14 + 16}{2} \]\[ s = \frac{42}{2} \]\[ s = 21 \text{ m} \]Heron's formula for the area of a triangle (\(A\)) is:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]Factorize the numbers to find perfect squares:
\[ A = \sqrt{(3 \times 7) \times (3 \times 3) \times 7 \times 5} \]\[ A = \sqrt{3^2 \times 7^2 \times 3 \times 5} \]\[ A = \sqrt{3^2 \times 7^2 \times 15} \]Correct Option: A) \(21\sqrt{15}\) m\(^2\)
The question asks about the deity worshipped during the Pongal festival. To answer this, one needs to recall the cultural and religious significance of Pongal, particularly its origins and the primary focus of its celebrations.
D) The Sun — Pongal is a harvest festival primarily dedicated to the Sun God (Surya) as a gesture of gratitude for a good harvest and for providing energy and light essential for agriculture.
This problem involves finding the other root of a quadratic equation given one root. We can use two main properties of quadratic equations: either substitute the given root into the equation to find the unknown coefficient and then solve for the roots, or use Vieta's formulas (sum and product of roots).
Identify the given information:
The quadratic equation is \(x^2 - 12x + k = 0\).
One root is \(x = 3\).
Method 1: Substitute the root to find \(k\), then find the other root.
Since \(x=3\) is a root, it must satisfy the equation. Substitute \(x=3\) into the equation:
Now that we have \(k=27\), the equation becomes \(x^2 - 12x + 27 = 0\).
We can solve this quadratic equation by factoring or using the quadratic formula. By factoring, we look for two numbers that multiply to 27 and add up to -12. These numbers are -3 and -9.
Setting each factor to zero gives the roots:
\[ x - 3 = 0 \implies x = 3 \]\[ x - 9 = 0 \implies x = 9 \]Method 2: Use Vieta's formulas (sum of roots).
For a quadratic equation in the form \(ax^2 + bx + c = 0\), the sum of the roots (\(\alpha + \beta\)) is given by \(-b/a\).
In our equation, \(x^2 - 12x + k = 0\), we have \(a=1\), \(b=-12\), and \(c=k\).
Let the given root be \(\alpha = 3\) and the other root be \(\beta\).
The sum of the roots is \(\alpha + \beta = -(-12)/1 = 12\).
Substitute the known root:
\[ 3 + \beta = 12 \]\[ \beta = 12 - 3 \]\[ \beta = 9 \]Both methods yield the same result for the other root.
Correct Option: B) x = 9
If three numbers are in the ratio 5 : 6 : 8 and the sum of their squares is 1250, then the product of those numbers is:
The problem involves finding the product of three numbers given their ratio and the sum of their squares. We will represent the numbers using a common multiplier based on their ratio, form an equation using the given sum of squares, solve for the multiplier, and then calculate the product of the numbers.
Let the three numbers be \(5x\), \(6x\), and \(8x\), where \(x\) is a common multiplier.
According to the problem, the sum of their squares is 1250. So, we can write the equation:
\[ (5x)^2 + (6x)^2 + (8x)^2 = 1250 \]Calculate the squares:
\[ 25x^2 + 36x^2 + 64x^2 = 1250 \]Sum the terms with \(x^2\):
\[ (25 + 36 + 64)x^2 = 1250 \]\[ 125x^2 = 1250 \]Solve for \(x^2\):
\[ x^2 = \frac{1250}{125} \]\[ x^2 = 10 \]Solve for \(x\):
\[ x = \sqrt{10} \] (Since numbers are usually positive in ratio problems unless specified, we take the positive root.)Now, find the three numbers:
Calculate the product of these numbers:
\[ \text{Product} = (5\sqrt{10}) \times (6\sqrt{10}) \times (8\sqrt{10}) \]\[ \text{Product} = (5 \times 6 \times 8) \times (\sqrt{10} \times \sqrt{10} \times \sqrt{10}) \]\[ \text{Product} = 240 \times (10 \times \sqrt{10}) \]\[ \text{Product} = 240 \times 10\sqrt{10} \]\[ \text{Product} = 2400\sqrt{10} \]Correct Option: B) \(2400\sqrt{10}\)
The question asks to identify the founder of Islam from the given options. This is a direct factual question related to the history and origins of a major world religion.
Correct Option: A) Prophet Muhammad is universally recognized as the founder of Islam. He is considered by Muslims to be the last prophet sent by God to guide humanity, and it was through him that the Quran was revealed.
To determine the most reactive element among the given options, we need to understand their positions in the periodic table and their general chemical properties, particularly their metallic character and tendency to lose electrons.
D) Ca is in Group 2, Period 4. It is an alkaline earth metal. Reactivity of metals increases down a group. Since Ca is below Mg in Group 2, it is more reactive than Mg. It is also more reactive than Al and Pb due to its position in Group 2.