The question asks about the common characteristic of all elements within the same group of the periodic table. We need to recall the organization principles of the periodic table, specifically what defines a group (vertical column).
Correct Option: C) Number of Valence Electrons
The question asks for the specific term used for alcohol that has been treated with colors and other substances to make it unsuitable for drinking, primarily for industrial or laboratory use. This process is done to prevent its consumption and to make it easily identifiable.
Correct Option: C) denatured
Denatured alcohol is ethanol (ethyl alcohol) that has been rendered unfit for human consumption by adding one or more denaturants. These denaturants can be poisonous or impart an unpleasant taste or smell. Coloring agents, like blue dye, are often added to make it easily distinguishable from potable alcohol, thereby preventing its accidental or intentional consumption. This is done to avoid taxes on alcoholic beverages and to ensure it's used for its intended industrial or scientific purposes.
Which country will host the 46th Congress of the International Hockey Federation?
The question asks about the host country for a specific upcoming event, the 46th Congress of the International Hockey Federation (FIH). This requires knowledge of recent sports administration news and international event hosting.
B) India is the correct answer. India is set to host the 46th Congress of the International Hockey Federation (FIH) in 2024. This will be the first time India hosts the FIH Congress.
The question asks to identify the specific acid secreted in the human digestive system. This is a direct knowledge-based question related to human biology and physiology, specifically the digestive process.
D) Hydrochloric acid — Hydrochloric acid (HCl) is secreted by the parietal cells in the stomach lining and is a vital component of gastric juice, essential for protein digestion and protection against pathogens.
The question asks to identify the item that does not belong to the group. This is a classification-based question where we need to find a common characteristic among most items and identify the outlier.
Correct Option: A) D
Ladle is a kitchen utensil, whereas Flour, Pancake, and Cream are all food-related items (ingredients or products).
Choose the wrong statement from the statements given below?
The elements present in the second group are called alkaline earth metals.
The gases in the 18th or zero group are inert
Elements of the first and second group are called s-block elements.
Elements present in group 17 are called alkaline metals.
The question asks to identify the incorrect statement among the given options regarding the classification of elements in the periodic table. We need to evaluate each statement based on standard chemical nomenclature and periodic table groups.
Correct Option: D) Elements present in group 17 are called alkaline metals. This statement is incorrect because Group 17 elements are known as halogens, while alkaline metals are found in Group 1.
A man divides his property among his sons in which one gets half and the other gets one-fourth, third gets \( \frac{1}{5} \) and fourth gets Rs. 1,000. What was the man's total wealth?
To find the total wealth, we need to determine what fraction of the property the given amount (Rs. 1,000) represents. We can do this by summing the fractions given to the first three sons and subtracting this sum from the total property (which is 1, or 100%). The remaining fraction will correspond to Rs. 1,000.
Let the total wealth of the man be \(X\).
The first son gets half of the property, which is \( \frac{1}{2}X \).
The second son gets one-fourth of the property, which is \( \frac{1}{4}X \).
The third son gets \( \frac{1}{5} \) of the property, which is \( \frac{1}{5}X \).
The sum of the fractions received by the first three sons is:
\[ \frac{1}{2} + \frac{1}{4} + \frac{1}{5} \]To add these fractions, find a common denominator. The least common multiple (LCM) of 2, 4, and 5 is 20.
\[ \frac{10}{20} + \frac{5}{20} + \frac{4}{20} = \frac{10 + 5 + 4}{20} = \frac{19}{20} \]This means that \( \frac{19}{20} \) of the total property has been distributed among the first three sons.
The remaining fraction of the property is for the fourth son. This can be found by subtracting the distributed fraction from the total (which is 1):
\[ 1 - \frac{19}{20} = \frac{20}{20} - \frac{19}{20} = \frac{1}{20} \]So, the fourth son gets \( \frac{1}{20} \) of the total property. We are given that the fourth son gets Rs. 1,000.
Therefore, we can set up the equation:
\[ \frac{1}{20}X = 1000 \]To find the total wealth \(X\), multiply both sides by 20:
\[ X = 1000 \times 20 \]\[ X = 20000 \]Thus, the man's total wealth was Rs. 20,000.
Correct Option: D) Rs. 20,000
To find the smallest cube number greater than 1000 that is divisible by 2, 4, and 6, we first need to find the least common multiple (LCM) of 2, 4, and 6. Any number divisible by 2, 4, and 6 must be a multiple of their LCM. Then, we look for a perfect cube that is a multiple of this LCM and is greater than 1000.
The number must be a perfect cube and a multiple of 12. Let the number be \(N\). So, \(N = k^3\) for some integer \(k\), and \(N\) must be divisible by 12. This means \(k^3\) must be divisible by 12.
For \(k^3\) to be divisible by \(12 = 2^2 \times 3^1\), the prime factors in \(k^3\) must have powers that are multiples of 3 and are at least \(2^2\) and \(3^1\).
Let \(k = 2^a \times 3^b \times \dots\). Then \(k^3 = 2^{3a} \times 3^{3b} \times \dots\).
For \(2^{3a}\) to be divisible by \(2^2\), we need \(3a \ge 2\). The smallest integer \(a\) satisfying this is \(a=1\). So, \(2^{3a} = 2^3\).
For \(3^{3b}\) to be divisible by \(3^1\), we need \(3b \ge 1\). The smallest integer \(b\) satisfying this is \(b=1\). So, \(3^{3b} = 3^3\).
Therefore, the smallest perfect cube divisible by 12 must be of the form \( (2^1 \times 3^1 \times m)^3 \) where \(m\) is an integer, such that \( (2^1 \times 3^1 \times m)^3 \) contains at least \(2^2 \times 3^1\). More directly, for \(k^3\) to be divisible by \(2^2 \times 3^1\), \(k\) must contain at least \(2^1 \times 3^1\). To make \(k^3\) divisible by \(2^2\), \(k\) must contain \(2^x\) such that \(3x \ge 2\), so \(x=1\). To make \(k^3\) divisible by \(3^1\), \(k\) must contain \(3^y\) such that \(3y \ge 1\), so \(y=1\). Thus, \(k\) must be a multiple of \(2^1 \times 3^1 = 6\). So, \(k\) must be of the form \(6n\).
The cube number will be \((6n)^3 = 216n^3\).
We need to find the smallest \(n\) such that \(216n^3 > 1000\).
\[ n^3 > \frac{1000}{216} \]
\[ n^3 > 4.629\dots \]
We need to find the smallest integer \(n\) whose cube is greater than 4.629. Since \(1^3 = 1\), \(2^3 = 8\), the smallest integer \(n\) for which \(n^3 > 4.629\) is \(n=2\).
For \(n=2\), the cube number is \((6 \times 2)^3 = 12^3\).
\[ 12^3 = 12 \times 12 \times 12 = 144 \times 12 = 1728 \]
This number 1728 is greater than 1000 and is divisible by 12 (since \(1728 / 12 = 144\)). It is also a perfect cube.
C) 1728
The question asks for the method by which a Copper-T device is placed in the uterus. This requires understanding the purpose and classification of Copper-T.
Copper-T is an Intrauterine Device (IUD) that is inserted into the uterus to prevent pregnancy. The act of preventing pregnancy is known as contraception. Therefore, placing a Copper-T in the uterus is a method of contraception.
To determine which number's square root will be irrational, we need to find the square root of each given option. A number has an irrational square root if it is not a perfect square. We can use the prime factorization method or long division method to find the square root of each number. Alternatively, we can look at the last digit of the number to quickly eliminate some options, as perfect squares can only end in 0, 1, 4, 5, 6, or 9.
We can find the square root by prime factorization or by estimation.
Using prime factorization:
\[ 2704 = 2 \times 1352 = 2 \times 2 \times 676 = 2 \times 2 \times 2 \times 338 = 2 \times 2 \times 2 \times 2 \times 169 = 2^4 \times 13^2 \]So, the square root is:
\[ \sqrt{2704} = \sqrt{2^4 \times 13^2} = 2^2 \times 13 = 4 \times 13 = 52 \]Since 52 is an integer, \(\sqrt{2704}\) is rational.
Using prime factorization:
\[ 2304 = 2 \times 1152 = 2 \times 2 \times 576 = 2 \times 2 \times 2 \times 288 = 2 \times 2 \times 2 \times 2 \times 2 \times 144 = 2^6 \times 12^2 \]Alternatively, we know \(144 = 12^2\), or \(144 = 2^4 \times 3^2\). So,
\[ 2304 = 2^6 \times (2^4 \times 3^2) = 2^{10} \times 3^2 \]So, the square root is:
\[ \sqrt{2304} = \sqrt{2^{10} \times 3^2} = 2^5 \times 3 = 32 \times 3 = 96 \]Since 96 is an integer, \(\sqrt{2304}\) is rational.
Let's find the prime factors of 3264:
\[ 3264 = 2 \times 1632 = 2 \times 2 \times 816 = 2 \times 2 \times 2 \times 408 = 2 \times 2 \times 2 \times 2 \times 204 = 2 \times 2 \times 2 \times 2 \times 2 \times 102 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 51 = 2^6 \times 3 \times 17 \]For a number to be a perfect square, all the exponents in its prime factorization must be even. Here, the exponents of 3 and 17 are 1 (odd). Therefore, 3264 is not a perfect square, and its square root will be irrational.
Using prime factorization:
\[ 1764 = 2 \times 882 = 2 \times 2 \times 441 = 2^2 \times 21^2 \]So, the square root is:
\[ \sqrt{1764} = \sqrt{2^2 \times 21^2} = 2 \times 21 = 42 \]Since 42 is an integer, \(\sqrt{1764}\) is rational.
Correct Option: C) The square root of 3264 is \(\sqrt{3264} = \sqrt{2^6 \times 3 \times 17}\), which is irrational because the prime factors 3 and 17 have odd powers (1).