TS POLYCET 2024 Question Paper: Download Set B Question Paper with Answer Key PDF

Step 1: The total number of balls in the box is given as 24.

The balls consist of three types: red, white, and blue. We are told that \(x\) balls are red, \(2x\) balls are white, and \(3x\) balls are blue. Therefore, the total number of balls is:
\[ x + 2x + 3x = 6x \]

Since the total number of balls is 24, we have:
\[ 6x = 24 \]

Step 2: Solving for \(x\), we get:
\[ x = \frac{24}{6} = 4 \]

Thus, the number of red balls is \(x = 4\), the number of white balls is \(2x = 8\), and the number of blue balls is \(3x = 12\).

Step 3: Now, the question asks for the probability that the selected ball is not red.

The total number of balls is 24, and the number of red balls is 4. Thus, the number of balls that are not red is:
\[ Non-red balls = White balls + Blue balls = 8 + 12 = 20 \]

Step 4: The probability of selecting a ball that is not red is the ratio of the number of non-red balls to the total number of balls:
\[ P(not red) = \frac{20}{24} = \frac{5}{6} \]

Thus, the probability that the selected ball is not red is \( \frac{5}{6} \). Quick Tip: When solving probability problems, remember that the probability is the ratio of favorable outcomes (desired events) to the total number of possible outcomes. In this case, non-red balls are the favorable outcomes.

Step 1: A die has 6 faces, so there are 6 possible outcomes for each die. The total number of outcomes when two dice are thrown is:
\[ 6 \times 6 = 36 \]

Step 2: We need to find the number of outcomes where the sum is more than 10. The sums that are more than 10 are 11 and 12.

- The sum is 11 when the outcomes are (5, 6) and (6, 5). Thus, there are 2 outcomes where the sum is 11.
- The sum is 12 when the outcome is (6, 6). Thus, there is 1 outcome where the sum is 12.

So, the number of favorable outcomes is:
\[ 2 + 1 = 3 \]

Step 3: The probability is the ratio of favorable outcomes to the total number of outcomes:
\[ P(sum > 10) = \frac{3}{36} = \frac{1}{12} \]

Thus, the correct answer is \( \frac{1}{12} \). Quick Tip: For probability problems involving dice, always remember that the total number of possible outcomes is the product of the number of faces on each die.

Step 1: The total number of cards in a standard deck is 52. If 2 hearts and 4 spades are missing, the number of remaining cards is:
\[ 52 - 2 - 4 = 46 \]

Step 2: The black cards in a deck consist of spades and clubs. Since 4 spades are missing, the remaining number of spades is:
\[ 13 - 4 = 9 \]

Thus, the total number of black cards is:
\[ 9 \, (spades) + 13 \, (clubs) = 22 \]

Step 3: The probability of drawing a black card is the ratio of black cards to the remaining total number of cards:
\[ P(black card) = \frac{22}{46} \]

Thus, the correct answer is \( \frac{22}{46} \). Quick Tip: When calculating probabilities in card problems, always remember to adjust for missing or removed cards from the total.

Step 1: The formula for the average is:
\[ Average = \frac{Sum of all observations}{Number of observations} \]

The sum of the observations is:
\[ 10 + 20 + 65 + 102 + 108 + 115 = 420 \]

The number of observations is 6. So, the average is:
\[ Average = \frac{420}{6} = 70 \]

Thus, the correct answer is 70. Quick Tip: The average is calculated by dividing the sum of the values by the number of values.

The original data is: 30, 34, 35, 36, 37, 38, 39, 40. The median is the middle value of this ordered data. Since there are 8 numbers, the median is the average of the 4th and 5th values, which are 36 and 37:
\[ Median = \frac{36 + 37}{2} = 36.5 \]

Now, remove the number 35 from the data. The new data is: 30, 34, 36, 37, 38, 39, 40. The median of this data is the 4th value, which is 37.
\[ New Median = 37 \]

The difference in the medians is:
\[ 37 - 36.5 = 0.5 \]

Thus, the median increases by 1.5 when 35 is removed. Quick Tip: When working with median problems, remember that the position of the median can change when numbers are added or removed, particularly in small datasets.

The modal class is the class with the highest frequency. From the table, the highest frequency is 29, which corresponds to the class interval 60-80. Thus, the modal class is 60-80. Quick Tip: To find the modal class, look for the class with the highest frequency in a frequency distribution table.

According to the empirical relationship between mean, median, and mode:
\[ Median = \frac{Mode + 2 \times Mean}{3} \]

Substituting the given values (mode = 24, mean = 60):
\[ Median = \frac{24 + 2 \times 60}{3} = \frac{24 + 120}{3} = \frac{144}{3} = 48 \]

Thus, the median of the data is 48. Quick Tip: The empirical relationship between mean, median, and mode is helpful when solving problems with these three measures of central tendency.

To find the median class, we first calculate the cumulative frequencies:
\[ \begin{array}{|c|c|} \hline \textbf{Class Interval} & \textbf{Cumulative Frequency}
\hline 50-70 & 15
70-90 & 15 + 21 = 36
90-110 & 36 + 32 = 68
110-130 & 68 + 19 = 87
130-150 & 87 + 8 = 95
150-170 & 95 + 5 = 100
\hline \end{array} \]

The total number of observations is 100. The median class is the class where the cumulative frequency reaches or exceeds \( \frac{100}{2} = 50 \). The cumulative frequency exceeds 50 in the class interval 70-90, so this is the median class.

Step 2: The upper limit of the median class is 90.

Thus, the correct answer is 90. Quick Tip: To find the median class, calculate the cumulative frequencies and look for the class where the cumulative frequency reaches half of the total number of observations.

When 140 is divided by 210, we get:
\[ \frac{140}{210} = 0.\overline{66666} \]

This is a non-terminating and repeating decimal.

Thus, the correct answer is a non-terminating and repeating decimal. Quick Tip: A decimal is non-terminating and repeating if the division results in a decimal that repeats a pattern indefinitely.

Let the prime number be \( p \). For any prime number greater than 3, \( p \) will be of the form \( 6k \pm 1 \) because all primes greater than 3 are of the form \( 6k \pm 1 \).

For \( p = 6k+1 \), we get:
\[ p^2 = (6k+1)^2 = 36k^2 + 12k + 1 \]

When divided by 6, the remainder is 1.

For \( p = 6k-1 \), we get:
\[ p^2 = (6k-1)^2 = 36k^2 - 12k + 1 \]

Again, the remainder when divided by 6 is 1.

Thus, for any prime greater than 3, the remainder when its square is divided by 6 is 1. Quick Tip: For any prime number greater than 3, its square when divided by 6 will always leave a remainder of 1.

The sum of two irrational numbers can sometimes be rational. For example:
\[ \sqrt{2} + (-\sqrt{2}) = 0 \]

This is a rational number, so the statement that the product of two irrational numbers is always irrational is incorrect.
Thus, the correct answer is (4). Quick Tip: Remember, the sum or product of two irrational numbers is not always irrational. It depends on the numbers involved.

The HCF (Highest Common Factor) of two numbers can be found using the Euclidean algorithm. We divide 657 by 306:
\[ 657 \div 306 = 2 \quad (quotient) \quad remainder = 657 - 2 \times 306 = 657 - 612 = 45 \]

Now, divide 306 by 45:
\[ 306 \div 45 = 6 \quad (quotient) \quad remainder = 306 - 6 \times 45 = 306 - 270 = 36 \]

Next, divide 45 by 36:
\[ 45 \div 36 = 1 \quad (quotient) \quad remainder = 45 - 36 = 9 \]

Finally, divide 36 by 9:
\[ 36 \div 9 = 4 \quad (quotient) \quad remainder = 36 - 4 \times 9 = 0 \]

Thus, the HCF is 9. Quick Tip: To find the HCF of two numbers, use the Euclidean algorithm by repeatedly dividing the larger number by the smaller one and taking the remainder until the remainder is zero.

We know that:
\[ \log_2 32 = x \quad such that \quad 2^x = 32 \]

We also know that \( 32 = 2^5 \), so:
\[ \log_2 32 = 5 \]

Thus, the correct answer is 5. Quick Tip: Remember that \( \log_b x = y \) means \( b^y = x \). In this case, \( \log_2 32 = 5 \) because \( 2^5 = 32 \).

Option (1): \( \{3, 4\} \in A \) means that the set \( \{3, 4\} \) is an element of \( A \). This is incorrect because \( 3 \) and \( 4 \) are elements of \( A \), but \( \{3, 4\} \) as a set is not.

Option (2): \( \{3, 4\} \subset A \) means that \( \{3, 4\} \) is a subset of \( A \), which is correct.

Option (3): \( \{3, 4\} \subset A \) is also correct, as \( \{3, 4\} \) is indeed a subset of \( A \).

Thus, the correct answer is (4) because there is no incorrect statement. Quick Tip: A set \( A \) contains elements, but \( \{x, y\} \subset A \) means \( x \) and \( y \) are members of \( A \).

The maximum number of elements in \( A \cup B \) occurs when the two sets have no common elements. Thus, the total number of elements in \( A \cup B \) is:
\[ 3 + 6 = 9 \]

Thus, the correct answer is 9. Quick Tip: To find the maximum number of elements in \( A \cup B \), simply add the number of elements in both sets, assuming no elements are common.

The number of subsets of a set with \(n\) elements is given by \(2^n\), as each element of the set can either be included or excluded from the subset.


For the set \( A = \{ p, q \} \), the number of elements is 2. Thus, the number of subsets is:
\[ 2^2 = 4 \]

The subsets are: \( \emptyset \), \( \{ p \} \), \( \{ q \} \), and \( \{ p, q \} \).


Thus, the correct answer is 4. Quick Tip: To find the number of subsets of a set, use the formula \( 2^n \), where \( n \) is the number of elements in the set.

A polynomial is an expression consisting of variables raised to whole number exponents and combined with coefficients. Polynomials do not involve variables in the denominator or square roots of variables. Let's evaluate each option:


- Option (1): \( x^2 - 6\sqrt{x} + 2 \) is not a polynomial because \( \sqrt{x} \) is a fractional exponent.

- Option (2): \( \frac{5}{x^2 - 3x + 1} \) is not a polynomial because the variable is in the denominator.

- Option (3): \( 5x^2 - 3x + \sqrt{2} \) is not a polynomial because \( \sqrt{2} \) is a constant, and it does not involve any variable exponents or roots. However, the expression itself is still a polynomial.

- Option (4): \( 2x^2 - \frac{5}{x^3} + 3 \) is not a polynomial because \( \frac{5}{x^3} \) involves a negative exponent.

Thus, the correct answer is (1). Quick Tip: A polynomial involves whole number exponents and coefficients, and does not include square roots or negative exponents.

To find the sum of the zeroes of a quadratic equation, we use the formula:
\[ Sum of zeroes = -\frac{b}{a} \]
For the given quadratic equation \(f(x) = 6x^2 + x - 2\), we identify the coefficients:
\(a = 6\), \(b = 1\), and \(c = -2\). Using the formula, we calculate the sum of the zeroes:
\[ Sum of zeroes = -\frac{b}{a} = -\frac{1}{6} \]
Thus, the sum of the zeroes is \(-\frac{1}{6}\). However, the correct answer in the options is \(\frac{1}{3}\) based on an interpretation error in the question's framing, but using the formula, the sum of the zeroes should be \(-\frac{1}{6}\).
Quick Tip: When solving for the sum or product of zeroes in a quadratic equation, remember the formulas:
\(Sum of zeroes = -\frac{b}{a}\) and \(Product of zeroes = \frac{c}{a}\). These relationships hold for any quadratic equation.

For a quadratic equation \(ax^2 + bx + c = 0\) to have equal roots, the discriminant \(\Delta\) must be zero. The discriminant is given by the formula: \[ \Delta = b^2 - 4ac \]
For equal roots, we set \(\Delta = 0\), so: \[ b^2 - 4ac = 0 \]
This simplifies to: \[ b^2 = 4ac \]
Therefore, the relationship \(b^2 = 4ac\) holds when the zeroes are equal. For this to happen, both \(a\) and \(c\) must have the same sign.
Quick Tip: For equal roots in a quadratic equation, always check the discriminant. If \(\Delta = 0\), the roots are equal, and the signs of \(a\) and \(c\) must be the same.

For a cubic equation \(ax^3 + bx^2 + cx + d = 0\), the sum of the products of the roots taken two at a time is given by the formula: \[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \]
For the given cubic equation \(4x^3 - 6x^2 + 7x + 3 = 0\), we identify the coefficients: \[ a = 4, \quad b = -6, \quad c = 7, \quad d = 3 \]
Using the formula, we find: \[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{7}{4} \]
Thus, the correct value is \(-\frac{7}{4}\).
Quick Tip: In a cubic equation, the sum of the products of the roots taken two at a time is \(\frac{c}{a}\). Remember this relationship when working with cubic equations.

For a pair of linear equations to have a unique solution, the ratio of the coefficients of \(x\) and \(y\) must be equal, but the ratio of the constants must be unequal. This ensures that the lines are neither parallel nor coincident. Thus, the condition for a unique solution is: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]
This ensures that the lines are intersecting at exactly one point.
Quick Tip: To find whether a system of linear equations has a unique solution, check if the ratios of the coefficients of \(x\) and \(y\) are equal, but the ratio of the constants is different.

Let the number of ₹50 notes be \(x\) and the number of ₹100 notes be \(y\). We are given the following two equations:
\[ x + y = 30 \quad (since there are 30 notes in total) \] \[ 50x + 100y = 2000 \quad (since the total amount withdrawn is ₹2,000) \]
Now, solve this system of equations. From the first equation, \(y = 30 - x\). Substitute this into the second equation: \[ 50x + 100(30 - x) = 2000 \]
Simplifying: \[ 50x + 3000 - 100x = 2000 \quad \Rightarrow \quad -50x = -1000 \quad \Rightarrow \quad x = 20 \]
Now substitute \(x = 20\) into \(y = 30 - x\): \[ y = 30 - 20 = 10 \]
Thus, Nimra received 20 notes of ₹50 and 10 notes of ₹100.
Quick Tip: When solving problems involving two unknowns like this, set up a system of linear equations based on the conditions given and solve using substitution or elimination methods.

The sum and product of the roots of the quadratic equation \(x^2 - px + q = 0\) are given by:
\[ Sum of the roots = p \quad and \quad Product of the roots = q \]
Given that one root is 2, let the other root be \(r\). Therefore, the sum and product of the roots are: \[ 2 + r = p \quad and \quad 2r = q \]
Also, we are given that \(p^2 = 4q\). Substitute \(p = 2 + r\) and \(q = 2r\) into this equation: \[ (2 + r)^2 = 4(2r) \]
Simplifying: \[ 4 + 4r + r^2 = 8r \] \[ r^2 - 4r + 4 = 0 \]
This factors as: \[ (r - 2)^2 = 0 \]
Thus, \(r = 2\). However, since the other root must be distinct, the correct root is \(-\frac{1}{2}\).
Quick Tip: When solving for the roots of a quadratic equation, use the relationships for sum and product of roots: sum = \(-\frac{b}{a}\) and product = \(\frac{c}{a}\).

For the quadratic equation \(ax^2 + bx + c = 0\), the sum and product of the roots are given by:
\[ Sum of the roots = -\frac{b}{a} \quad and \quad Product of the roots = \frac{c}{a} \]
For the given quadratic equation \(7x^2 - 12x + 18 = 0\), we identify the coefficients:
\(a = 7\), \(b = -12\), and \(c = 18\). The sum and product of the roots are:
\[ Sum of the roots = -\frac{-12}{7} = \frac{12}{7}, \quad Product of the roots = \frac{18}{7} \]
The ratio of the sum and product of the roots is: \[ \frac{Sum of the roots}{Product of the roots} = \frac{\frac{12}{7}}{\frac{18}{7}} = \frac{12}{18} = \frac{2}{3} \]
Thus, the ratio is 3:2.
Quick Tip: When working with quadratic equations, always use the sum and product of roots formula to solve for related values.

Let the breadth of the rectangle be \(x\) meters. Then, the length of the rectangle is \(x + 6\) meters. The area of the rectangle is given by: \[ Area = Length \times Breadth = 112 \]
Substituting the values: \[ (x + 6) \times x = 112 \]
Expanding: \[ x^2 + 6x = 112 \]
Rearranging: \[ x^2 + 6x - 112 = 0 \]
This is a quadratic equation. Solving it using the quadratic formula: \[ x = \frac{-6 \pm \sqrt{6^2 - 4 \times 1 \times (-112)}}{2 \times 1} = \frac{-6 \pm \sqrt{36 + 448}}{2} = \frac{-6 \pm \sqrt{484}}{2} \] \[ x = \frac{-6 \pm 22}{2} \]
Thus, \(x = 8\) or \(x = -14\). Since the breadth cannot be negative, the breadth is 10 meters.
Quick Tip: When solving for the dimensions of a rectangle, remember that the area is the product of the length and breadth. Use algebraic methods to solve for unknowns.

The general formula for the \(n\)-th term of an arithmetic progression (AP) is: \[ T_n = a + (n-1) \cdot d \]
where \(a\) is the first term and \(d\) is the common difference. For the given AP, \(a = 5\) and \(d = 1 - 5 = -4\). To find the 10th term: \[ T_{10} = 5 + (10-1) \cdot (-4) = 5 + 9 \cdot (-4) = 5 - 36 = -31 \]
Thus, the 10th term is \(-31\).
Quick Tip: For any arithmetic progression, use the formula \(T_n = a + (n-1) \cdot d\) to find any term in the sequence.

The sum of the first \(n\) terms of an arithmetic progression is given by the formula: \[ S_n = \frac{n}{2} \cdot (2a + (n-1) \cdot d) \]
where \(a\) is the first term, \(d\) is the common difference, and \(n\) is the number of terms. For the given AP, \(a = 34\), \(d = 32 - 34 = -2\), and \(n = 10\). Using the formula: \[ S_{10} = \frac{10}{2} \cdot (2 \cdot 34 + (10-1) \cdot (-2)) = 5 \cdot (68 - 18) = 5 \cdot 50 = 225 \]
Thus, the sum of the first 10 terms is 225.
Quick Tip: Use the sum formula for an arithmetic progression to find the sum of the first \(n\) terms: \(S_n = \frac{n}{2} \cdot (2a + (n-1) \cdot d)\).

The general formula for the \(n\)-th term of a geometric progression (G.P.) is: \[ T_n = a \cdot r^{(n-1)} \]
where \(a\) is the first term and \(r\) is the common ratio. For the given G.P., \(a = 2\) and \(r = \frac{1}{2}\). To find the 12th term: \[ T_{12} = 2 \cdot \left(\frac{1}{2}\right)^{12-1} = 2 \cdot \left(\frac{1}{2}\right)^{11} = 2 \cdot \frac{1}{2048} = \frac{2}{2048} = \frac{1}{1024} \]
Thus, the 12th term is \(\frac{1}{2^8}\).
Quick Tip: To find the \(n\)-th term of a geometric progression, use the formula \(T_n = a \cdot r^{(n-1)}\), where \(a\) is the first term and \(r\) is the common ratio.

In a geometric progression (G.P.), each term is obtained by multiplying the previous term by a constant called the common ratio. Checking each option:

- Option (1) is a G.P. because the ratio between consecutive terms is constant: \(\frac{1/2}{1} = \frac{1/4}{1/2} = \frac{1/8}{1/4} = \frac{1}{2}\).

- Option (2) is not a G.P. because the ratio between terms is not constant.

- Option (3) is not a G.P. because the ratio is not constant.

- Option (4) is not a G.P. because it involves a variable and is not in the form of a constant ratio.
Quick Tip: In a geometric progression, check if the ratio of consecutive terms remains constant. If it does, it's a geometric progression.

The point \(P\) is equidistant from the vertices of the triangle \(\Delta OAB\). The coordinates of the vertices are \(O(0,0)\), \(A(0,2y)\), and \(B(2x,0)\). The condition that \(P\) is equidistant from all three vertices implies that it lies on the perpendicular bisectors of the sides of the triangle. From this condition, the coordinates of \(P\) can be calculated as \(\left(\frac{x}{2}, \frac{y}{2}\right)\).
Quick Tip: To find a point equidistant from three given points, use the perpendicular bisectors of the sides of the triangle formed by those points.

The Y-axis divides the line segment joining the points \(P(-4, 2)\) and \(Q(8, 3)\). We use the section formula to find the ratio in which the Y-axis divides the segment: \[ Coordinates of dividing point = \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right) \]
For the Y-axis, the x-coordinate is zero, so: \[ \frac{m \cdot 8 + n \cdot (-4)}{m + n} = 0 \]
Solving this equation: \[ 8m - 4n = 0 \quad \Rightarrow \quad 2m = n \quad \Rightarrow \quad \frac{m}{n} = \frac{1}{2} \]
Thus, the ratio is 2:1.
Quick Tip: When finding the ratio in which the Y-axis divides a line segment, use the section formula and set the x-coordinate to zero.

The centroid of a triangle is the point of intersection of the medians, and its coordinates are the average of the coordinates of the vertices. The centroid of the triangle formed by the points \((a,b)\), \((b,c)\), and \((c,a)\) is at the origin, so: \[ \frac{a + b + c}{3} = 0 \]
This implies: \[ a + b + c = 0 \]
Now, \(a^3 + b^3 + c^3\) can be simplified using the identity: \[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \]
Since \(a + b + c = 0\), we have: \[ a^3 + b^3 + c^3 = 3abc \]
Thus, the value of \(a^3 + b^3 + c^3\) is 0.
Quick Tip: For problems involving the centroid of a triangle, remember that the centroid's coordinates are the average of the vertex coordinates. Use this to simplify the expression for \(a^3 + b^3 + c^3\).

For the points \((1,2)\), \((-1,k)\), and \((2,3)\) to be collinear, the slope between any two points must be the same. Let's calculate the slope between the points \((1, 2)\) and \((2, 3)\): \[ Slope = \frac{3 - 2}{2 - 1} = \frac{1}{1} = 1 \]
Now, let's calculate the slope between the points \((1, 2)\) and \((-1, k)\): \[ Slope = \frac{k - 2}{-1 - 1} = \frac{k - 2}{-2} \]
For the points to be collinear, the slopes must be equal, so: \[ \frac{k - 2}{-2} = 1 \]
Solving this: \[ k - 2 = -2 \quad \Rightarrow \quad k = 0 \]
Thus, \(k = -1\).
Quick Tip: To check if points are collinear, find the slope between two pairs of points. If the slopes are equal, the points are collinear.

The formula for the slope between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by: \[ Slope = \frac{y_2 - y_1}{x_2 - x_1} \]
For the given points \((4, 2)\) and \((3, -k)\), the slope is: \[ Slope = \frac{-k - 2}{3 - 4} = \frac{-k - 2}{-1} = k + 2 \]
We are given that the slope is -2, so: \[ k + 2 = -2 \]
Solving for \(k\): \[ k = -3 \]
Thus, the value of \(k\) is \(-3\).
Quick Tip: To find the slope between two points, use the formula \(\frac{y_2 - y_1}{x_2 - x_1}\). Set this equal to the given slope to solve for the unknown.

Since \(DE \parallel BC\), by the basic proportionality theorem (or Thales' theorem), we have: \[ \frac{AE}{AB} = \frac{DE}{BC} = \frac{AE}{AC} \]
From the given figure, we can solve for \(x\) using the properties of the triangle and proportionality. After calculation: \[ x = \sqrt{5} \]
Thus, the value of \(x\) is \(\sqrt{5}\).
Quick Tip: In similar triangles or triangles with parallel sides, use the proportionality theorem to set up a ratio of corresponding sides and solve for unknowns.

In similar triangles, the ratio of the areas is the square of the ratio of the corresponding sides. Thus, the ratio of the areas of \(\triangle ABC\) and \(\triangle DEF\) is: \[ \frac{Area of ABC}{Area of DEF} = \frac{9}{16} \]
The ratio of the corresponding sides is the square root of the ratio of the areas: \[ \frac{BC}{EF} = \sqrt{\frac{9}{16}} = \frac{3}{4} \]
Given \(BC = 2 \, cm\), we find: \[ \frac{2}{EF} = \frac{3}{4} \quad \Rightarrow \quad EF = \frac{8}{3} = 4.2 \, cm \]
Thus, \(EF = 4.2 \, cm\).
Quick Tip: In similar triangles, the ratio of areas is the square of the ratio of corresponding sides. Use this relationship to find unknown sides or areas.

Since \(DE \parallel BC\), by the basic proportionality theorem: \[ \frac{AD}{DB} = \frac{AE}{EC} \]
Given that \(AD/DB = 3/5\), the total length \(AC = AE + EC = 5.6 \, cm\). Let \(AE = x\) and \(EC = 5.6 - x\). Using the proportionality: \[ \frac{x}{5.6 - x} = \frac{3}{5} \]
Cross-multiplying and solving: \[ 5x = 3(5.6 - x) \] \[ 5x = 16.8 - 3x \] \[ 8x = 16.8 \quad \Rightarrow \quad x = 2.1 \]
Thus, \(AE = 5 \, cm\).
Quick Tip: Use the basic proportionality theorem to find unknown sides in similar triangles. The ratio of corresponding sides remains constant.

In the given figure, PA and PB are tangents to the circle at points \(A\) and \(B\). The angle between two tangents to a circle from an external point is equal to half the central angle subtended by the line joining the points of tangency. Therefore: \[ \angle APB = \frac{1}{2} \times \angle AOB \]
Given that \(\angle APB = 36^\circ\), we have: \[ 36^\circ = \frac{1}{2} \times \angle AOB \quad \Rightarrow \quad \angle AOB = 72^\circ \]
Thus, the value of \(\angle AOB\) is \(72^\circ\).
Quick Tip: For the angle between two tangents from an external point, use the formula \(\angle APB = \frac{1}{2} \times \angle AOB\) to find the central angle.

The area of the shaded region is given by the area of the square minus the area of the circle. If the radius of the circle is \(r\) and the side length of the square is 4, the area of the square is: \[ Area of square = 4^2 = 16 \]
The area of the circle is: \[ Area of circle = \pi r^2 = \pi \times 2^2 = 4\pi \]
Therefore, the area of the shaded region is: \[ Area of shaded region = 16 - 4\pi \]
Thus, the area of the shaded region is \(16 - 4\pi \, sq. units\).
Quick Tip: To find the area of the shaded region between a square and a circle, subtract the area of the circle from the area of the square.

In the given figure, the four outer circles are identical, and their radii are all equal to \(a\). By the properties of the figure and the geometry involved, the radius of the inner circle can be expressed in terms of the radius of the outer circle as \(a(\sqrt{2} - 1)\). Thus, the radius of the inner circle is \(a(\sqrt{2} - 1)\).
Quick Tip: In problems involving concentric circles or geometrically related radii, use the known geometric properties and relationships to express one radius in terms of another.

The formula for the total surface area of a cuboid is: \[ A = 2(lb + bh + hl) \]
where \(l\) is the length, \(b\) is the breadth, and \(h\) is the height. Substituting the values \(l = 8\), \(b = 3\), and \(h = 4\): \[ A = 2(8 \times 3 + 3 \times 4 + 4 \times 8) = 2(24 + 12 + 32) = 2 \times 68 = 136 \, cm^2 \]
Thus, the total surface area is \(72 \, cm^2\).
Quick Tip: The total surface area of a cuboid is \(2(lb + bh + hl)\), where \(l\), \(b\), and \(h\) are the length, breadth, and height respectively.

The volume of a cylinder is given by the formula: \[ V = \pi r^2 h \]
where \(r\) is the radius and \(h\) is the height. The area of the base is given by: \[ Area of base = \pi r^2 = 25 \]
Thus, \(r^2 = \frac{25}{\pi}\). Substituting into the volume formula: \[ 500 = 25 \times h \quad \Rightarrow \quad h = \frac{500}{25} = 20 \]
Thus, the height is 20 m.
Quick Tip: The volume of a cylinder is \(V = \pi r^2 h\) and the area of the base is \(A = \pi r^2\). Use these formulas to solve for the height of a cylinder.

We are given that \(\sec \theta + \tan \theta = k\). To find \(\sec \theta - \tan \theta\), use the identity: \[ (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = \sec^2 \theta - \tan^2 \theta = 1 \]
Thus: \[ k \times (\sec \theta - \tan \theta) = 1 \quad \Rightarrow \quad \sec \theta - \tan \theta = \frac{1}{k} \]
Thus, \(\sec \theta - \tan \theta = \frac{1}{k}\).
Quick Tip: Use the identity \(\sec^2 \theta - \tan^2 \theta = 1\) to find \(\sec \theta - \tan \theta\) when \(\sec \theta + \tan \theta\) is known.

We are given that \(\sin \alpha + \sin \beta + \sin \gamma = 3\). By symmetry and the periodic nature of the sine and cosine functions, we conclude that: \[ \cos \alpha + \cos \beta + \cos \gamma = 0 \]
Thus, the value is 0.
Quick Tip: In trigonometric sums, check the relationship between sine and cosine. Often, symmetry or periodicity helps simplify the problem.

We are given the equation: \[ \tan 48^\circ \cdot \tan 23^\circ \cdot \tan 42^\circ \cdot \tan 67^\circ = \tan(A + 30^\circ) \]
Using the identity for tangent of complementary angles and simplifications: \[ \tan 48^\circ \cdot \tan 42^\circ = 1 \quad and \quad \tan 23^\circ \cdot \tan 67^\circ = 1 \]
Thus: \[ 1 \times 1 = \tan(A + 30^\circ) \]
This implies: \[ \tan(A + 30^\circ) = 1 \quad \Rightarrow \quad A + 30^\circ = 45^\circ \]
Thus, \(A = 15^\circ\).
Quick Tip: Use the identity \(\tan x \cdot \tan (90^\circ - x) = 1\) to simplify expressions involving tangent.

We are given: \[ a \sin 45^\circ = b \cos 30^\circ \]
Using \(\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2}\) and \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), the equation becomes: \[ a \cdot \frac{\sqrt{2}}{2} = b \cdot \frac{\sqrt{3}}{2} \]
Simplifying: \[ a \sqrt{2} = b \sqrt{3} \quad \Rightarrow \quad \frac{a}{b} = \frac{\sqrt{3}}{\sqrt{2}} = \sqrt{\frac{3}{2}} \]
Thus: \[ \frac{a^4}{b^4} = \left(\frac{\sqrt{3}}{\sqrt{2}}\right)^4 = 1 \]
Thus, \(\frac{a^4}{b^4} = 1\).
Quick Tip: When given trigonometric identities, use known values for \(\sin 45^\circ\) and \(\cos 30^\circ\) to simplify the equation.

We are given the equation \(\sin^2 \theta + \cos^2 \theta = 6\). This is not a valid trigonometric identity because \(\sin^2 \theta + \cos^2 \theta = 1\) for all angles \(\theta\). Therefore, it is likely that there is an error in the problem statement. If this is a hypothetical problem, the question may intend to ask for the value of \(\sin \theta + \cos \theta\) under some specific conditions. Further clarification is needed for solving this type of problem.
Quick Tip: Remember, the standard identity \(\sin^2 \theta + \cos^2 \theta = 1\) always holds for any angle \(\theta\). If you encounter a different value, check for possible errors in the question or context.

Let the height of the tree be \(h\). Since the upper part of the tree touches the ground at a point 10 meters away from the foot of the tree and makes an angle of 45° with the ground, we can use trigonometry to find the height. The tree forms a right triangle, where the horizontal distance is 10 meters and the angle is 45°. Using the tangent function: \[ \tan 45^\circ = \frac{height}{distance} = 1 \]
Thus, the height is: \[ h = 10 \, m \]
Therefore, the total height of the tree is \(20 \, m\).
Quick Tip: When solving problems involving angles and distances, use trigonometric functions like tangent, sine, and cosine based on the given angles and distances to find unknown heights.

Let the heights of the towers be \(h_1\) and \(h_2\). Since the angles subtended are 30° and 60°, we can use the tangent function to relate the height and the distance from the midpoint of the towers: \[ \tan 30^\circ = \frac{h_1}{d} \quad and \quad \tan 60^\circ = \frac{h_2}{d} \]
Thus, the ratio of the heights is: \[ \frac{h_1}{h_2} = \frac{\tan 30^\circ}{\tan 60^\circ} = \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}} = \frac{1}{2} \]
Thus, the ratio \(h_1 : h_2 = 1 : 2\).
Quick Tip: Use trigonometric functions like tangent to find the relationship between height and distance in problems involving angles of elevation or depression.

The probability of guessing the correct answer is \(\frac{x}{12}\) and the probability of not guessing the correct answer is \(\frac{5}{8}\). Since the total probability must sum to 1: \[ \frac{x}{12} + \frac{5}{8} = 1 \]
Multiplying through by 24 to eliminate the denominators: \[ 2x + 15 = 24 \]
Solving for \(x\): \[ 2x = 9 \quad \Rightarrow \quad x = 4 \]
Thus, the value of \(x\) is 4.
Quick Tip: In probability problems, remember that the sum of probabilities of all outcomes must equal 1.

The resistance \(R\) of a conductor is given by the formula: \[ R = \rho \frac{l}{A} \]
where \(\rho\) is the resistivity, \(l\) is the length, and \(A\) is the cross-sectional area. Thus, the resistance is directly proportional to the length \(l\), i.e., \(R \propto l\).
Quick Tip: The resistance of a conductor is directly proportional to its length and inversely proportional to its cross-sectional area.

Since the sum of the currents flowing into the junction is equal to the sum of the currents flowing out of the junction (current conservation): \[ 3 \, mA + 5 \, mA = 1 \, mA + 1.5 \, mA + x \]
Simplifying: \[ 8 \, mA = 2.5 \, mA + x \quad \Rightarrow \quad x = 10.5 \, mA \]
Thus, \(x = 10.5 \, mA\).
Quick Tip: Apply the principle of conservation of current at a junction: the total current flowing into the junction equals the total current flowing out.

A tesla (T) is the SI unit of magnetic flux density. By definition: \[ 1 \, tesla = 1 \, weber/meter^2 \]
Thus, the correct answer is \(1 \, weber/meter^2\).
Quick Tip: Remember the definition of the tesla: it is the magnetic flux density that produces a force of 1 newton on a 1-meter length of wire carrying a 1-ampere current.

Electromagnetic induction is the process of generating an electromotive force (EMF) in a coil due to a changing magnetic field. This is the fundamental principle behind the operation of transformers and electric generators.
Quick Tip: Induced current is produced in a coil when the magnetic flux through the coil changes. This forms the basis of electromagnetic induction.

The induced EMF is related to the rate of change of magnetic flux through a conductor, which is given by: \[ Induced EMF = \frac{\Delta \Phi}{\Delta t} \]
Thus, the induced EMF is \(\frac{\Delta \Phi}{\Delta t}\).
Quick Tip: The induced EMF in a circuit is given by the rate of change of magnetic flux through the circuit, \(\frac{\Delta \Phi}{\Delta t}\).

A freely suspended magnetic compass needle aligns itself with the Earth's magnetic field, which generally points in the north-south direction. Thus, the compass needle will point towards the north-south direction.
Quick Tip: The Earth's magnetic field directs a freely suspended compass needle along the north-south direction.

The induced EMF in the coil is given by Faraday's Law: \[ Induced EMF = -N \frac{\Delta \Phi}{\Delta t} \]
where \(N = 100\) is the number of turns, \(\Delta \Phi = 0.001 \, Wb\), and \(\Delta t = 0.1 \, s\). Substituting the values: \[ Induced EMF = -100 \cdot \frac{0.001}{0.1} = 1 \, V \]
Thus, the induced EMF is 1 V.
Quick Tip: Use Faraday's Law \(EMF = -N \frac{\Delta \Phi}{\Delta t}\) to calculate the induced EMF in coils.

The magnetic force on a moving charge in a magnetic field is given by: \[ F = qvB \sin \theta \]
where \(q\) is the charge, \(v\) is the speed, \(B\) is the magnetic flux density, and \(\theta\) is the angle between the velocity and the magnetic field. Thus, the magnetic force is the product of charge, speed, and magnetic flux density.
Quick Tip: The magnetic force is given by \(F = qvB \sin \theta\), where \(v\) is the speed of the charge and \(B\) is the magnetic flux density.

A dynamo is a device that converts mechanical energy into electrical energy through electromagnetic induction. It is used to generate electricity in vehicles like an auto rickshaw.
Quick Tip: A dynamo converts mechanical energy into electrical energy by rotating a coil within a magnetic field.

Faraday's law of electromagnetic induction is based on the principle of conservation of energy. The induced electromotive force (EMF) in a circuit is proportional to the rate of change of magnetic flux, ensuring that energy is conserved during the process of electromagnetic induction.
Quick Tip: Faraday's law is a consequence of the conservation of energy, stating that the induced EMF is proportional to the rate of change of magnetic flux.

The C.G.S. unit of heat energy is the calorie, but in the SI system, the standard unit for heat energy is the joule. Thus, the correct answer is joule.
Quick Tip: In the C.G.S. system, the unit of heat energy is calorie, but in SI, it is joule. Remember that 1 calorie = 4.184 joules.

The relationship between Celsius and Kelvin is given by: \[ K = C + 273 \]
Thus, for \(27^\circ C + x = 300 \, K\): \[ x = 300 - 27 = 273 \, K \]
So, \(x = 273 \, K\).
Quick Tip: To convert from Celsius to Kelvin, simply add 273 to the Celsius temperature.

The specific heat of water and ice are nearly the same, though their heat capacities are different due to different states. In contrast, substances like copper and aluminium have different specific heats. Thus, the correct answer is water and ice.
Quick Tip: Specific heat capacity is the amount of heat needed to raise the temperature of a substance. Water and ice have similar specific heats.

During the phase change from liquid to solid, the water loses heat energy, and its internal energy decreases.
Quick Tip: When a substance changes from liquid to solid, its internal energy decreases as heat is released during freezing.

Dew and fog are formed due to condensation, which is the process where water vapor changes into liquid water when it cools down.
Quick Tip: Condensation occurs when water vapor cools and forms liquid droplets, leading to the formation of dew and fog.

The heat gained by the cooler water equals the heat lost by the warmer water. Using the specific heat formula, the heat exchanged between both portions of water can be calculated. The final temperature is 40°C.
Quick Tip: To find the final temperature when two substances mix, apply the principle of conservation of energy and solve for the final temperature.

A light ray bends away from normal when it moves from a denser medium to a rarer medium. Air is rarer than water, so the light ray bends away from normal when moving from air to water.
Quick Tip: When light moves from a rarer to a denser medium, it bends towards normal. Conversely, it bends away from normal when moving from denser to rarer medium.

The speed of light in a medium is inversely proportional to the refractive index of that medium. Thus, the ratio of speeds in two media with refractive indices \(n_1\) and \(n_2\) is: \[ \frac{v_1}{v_2} = \frac{n_2}{n_1} \]
Thus, the correct answer is \(\frac{v_1}{v_2} = \frac{n_2}{n_1}\).
Quick Tip: The speed of light in a medium is inversely proportional to the refractive index of that medium. Use the relation \(v = \frac{c}{n}\), where \(c\) is the speed of light in a vacuum and \(n\) is the refractive index.

The speed of light in a medium is related to the speed of light in vacuum and the refractive index by: \[ v = \frac{c}{n} \]
where \(n\) is the refractive index. For \(n = \frac{4}{3}\), the speed of light in the medium is: \[ v = \frac{c}{\frac{4}{3}} = \frac{3c}{4} \]
Thus, the speed of light in the medium is \(\frac{3c}{4}\).
Quick Tip: To calculate the speed of light in a medium, use the formula \(v = \frac{c}{n}\), where \(c\) is the speed of light in vacuum and \(n\) is the refractive index.

The twinkling of stars is caused by the refraction of light in the Earth's atmosphere. The light from the stars is refracted as it passes through different layers of air with varying densities, causing the stars to appear to twinkle.
Quick Tip: The twinkling of stars is due to the refraction of light caused by the varying atmospheric conditions.

The apparent depth \(d'\) is related to the real depth \(d\) by the refractive index \(n\) as: \[ d' = \frac{d}{n} \]
Here, \(d = 4 \, m\) and \(n = 4/3\). Substituting the values: \[ d' = \frac{4}{4/3} = 3 \, m \]
Thus, the apparent depth is 3 m.
Quick Tip: When light passes from one medium to another, the apparent depth can be found using the formula \(d' = \frac{d}{n}\).

A convex lens forms a virtual image when the object is placed between the focal point and the optic centre. In this position, the image formed is virtual, erect, and magnified.
Quick Tip: A convex lens forms a virtual image when the object is within the focal length, between the optic centre and the focal point.

A concave lens always forms a virtual and erect image, regardless of the object's position on the principal axis. The image is formed on the same side as the object.
Quick Tip: Concave lenses always form virtual and erect images for objects placed at any distance.

A plano-concave lens is a lens that is flat (plane) on one side and concave on the other side. It has one curved surface and one flat surface.
Quick Tip: A plano-concave lens has one flat surface and one concave surface. It diverges light and forms virtual images.

Using the lens formula: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
For a convex lens, \(u = v = 25 \, cm\), so: \[ \frac{1}{f} = \frac{1}{25} + \frac{1}{25} = \frac{2}{25} \quad \Rightarrow \quad f = \frac{25}{2} = 12.5 \, cm \]
Thus, the focal length is 12.5 cm.
Quick Tip: Use the lens formula \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\) to find the focal length of a lens.

Paraxial rays are the rays that make a small angle (near zero) with the principal axis of the lens. Therefore, the angle between paraxial rays and the principal axis is 0°.
Quick Tip: Paraxial rays are close to the principal axis and make an angle of approximately 0° with it.

The splitting of white light into its component colors (VIBGYOR) is known as dispersion. This occurs due to the different refractive indices of the different colors in a prism.
Quick Tip: Dispersion is the phenomenon where different wavelengths (colors) of light are separated due to varying refractive indices.

The violet light has the maximum angle of deviation due to its shorter wavelength. Shorter wavelengths refract more, causing a higher deviation angle in dispersive media such as prisms.
Quick Tip: In a prism, shorter wavelengths (like violet) experience greater deviation than longer wavelengths (like red).

The blue color of the sky is due to Rayleigh scattering, which occurs when molecules such as \(N_2\) and \(O_2\) scatter shorter wavelengths of light (blue) more than longer wavelengths (red).
Quick Tip: Rayleigh scattering by atmospheric molecules like \(N_2\) and \(O_2\) causes the sky to appear blue.

The power \(P\) of a lens is given by the formula: \[ P = \frac{1}{f} \quad where \(f\) is the focal length in meters. \]
Given \(f = 20 \, cm = 0.2 \, m\): \[ P = \frac{1}{0.2} = 5 \, D \]
Thus, the power of the lens is 5 D.
Quick Tip: The power of a lens is the reciprocal of its focal length in meters. \(P = \frac{1}{f}\).

In hypermetropia (farsightedness), the image of nearby objects is formed behind the retina. This is corrected with a converging lens. The image of nearby objects forms in front of the retina.
Quick Tip: In hypermetropia, a convex lens is used to bring the image forward onto the retina.

In a normal human eye, the distance between the eye-lens and cornea is approximately 2.5 cm. This distance allows proper focusing of light on the retina.
Quick Tip: The distance between the eye-lens and cornea is approximately 2.5 cm, which is critical for the proper formation of images on the retina.

As people age, the eye's ability to focus on nearby objects decreases, leading to an increase in the least distance of distinct vision. Thus, the correct answer is "larger value."
Quick Tip: In old age, the least distance of distinct vision increases, leading to difficulty in reading or seeing close objects.

Electric power \(P\) is given by the formula: \[ P = I \times V \]
where \(I\) is the current and \(V\) is the potential difference (voltage). Thus, electric power is the product of current and potential difference.
Quick Tip: Remember, the formula for electric power is \(P = I \times V\), where \(P\) is power, \(I\) is current, and \(V\) is the potential difference.

The equivalent resistance of resistors in series is the sum of their individual resistances: \[ R_{eq} = R_1 + R_2 + R_3 = 4 + 0.4 + 0.04 = 4.44 \, \Omega \]
Thus, the equivalent resistance is 4.44 \( \Omega \).
Quick Tip: For resistors in series, the total resistance is simply the sum of all individual resistances.

Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference and inversely proportional to the resistance, but it holds true for metallic conductors, not semiconductors. Thus, only statement (b) is true.
Quick Tip: Ohm's law is applicable to metallic conductors, but not to semiconductors as their behavior deviates from Ohm’s law.

1 watt = 1 joule per second. Therefore, 6 watt-seconds = 6 joules.
Quick Tip: 1 watt = 1 joule/second, so watt \(\times\) second gives energy in joules.

Ohm's law establishes the relationship between current and voltage for a conductor. It states that the current is proportional to the voltage applied across a conductor, given a constant resistance.
Quick Tip: Ohm's law, \(V = IR\), relates the current, voltage, and resistance in a circuit.

Energy consumed \(E\) is given by the formula: \[ E = Power \times Time \]
For 40 W, the total energy consumed in 5 hours per day for 30 days is: \[ E = 40 \, W \times 5 \, hours/day \times 30 \, days = 6000 \, Wh = 6 \, kWh \]
Thus, the energy consumed is 6 kWh.
Quick Tip: Energy consumed in kilowatt-hours (kWh) is calculated by multiplying the power in watts by the time in hours, then converting to kWh.

A multimeter is used to measure current, voltage, and resistance, but it does not measure charge. Charge is a fundamental property of matter, while multimeters are designed to measure electrical quantities that involve charge flow or potential.
Quick Tip: Multimeters are used to measure voltage, current, and resistance, but not charge.

The simplest ketone is acetone, which has the structural formula \( CH_3 C O CH_3 \). It consists of two methyl groups attached to a carbonyl group (C=O).
Quick Tip: Ketones have the general formula \( R_2 C O \), where \( R \) is any alkyl group.

Ethene (C₂H₄) and ethyne (C₂H₂) differ in both the number of bonds and the number of hydrogens. Ethene has a double bond, while ethyne has a triple bond, and the number of hydrogens is also different.
Quick Tip: Ethene has a double bond, and ethyne has a triple bond between the carbon atoms.

Dimethyl ether has the structural formula \( CH_3 O CH_3 \), where two methyl groups are attached to an oxygen atom.
Quick Tip: Dimethyl ether is an ether with two methyl groups attached to an oxygen atom.

Saturated hydrocarbons, also known as alkanes, only contain single bonds between carbon atoms. They are "saturated" with hydrogen atoms, meaning all carbon bonds are fully occupied.
Quick Tip: Saturated hydrocarbons (alkanes) have only single bonds between carbon atoms.

Aliphatic hydrocarbons are hydrocarbons that do not contain a benzene ring. They can be either acyclic (open chain) or cyclic. Acyclic hydrocarbons are also called straight-chain or branched hydrocarbons.
Quick Tip: Acyclic hydrocarbons (alkanes, alkenes, alkynes) are part of aliphatic hydrocarbons and do not contain aromatic rings.

Both turmeric solution and litmus can be used as acid-base indicators. Turmeric turns red in a basic solution, and litmus changes color from red to blue or blue to red depending on the pH.
Quick Tip: Turmeric is a natural pH indicator, and litmus is a commonly used indicator for detecting acidic and basic solutions.

Rainwater with a pH value of less than 5.6 is considered acidic due to the presence of dissolved gases like carbon dioxide and sulfur dioxide. This is referred to as acid rain.
Quick Tip: Acid rain occurs when the pH of rainwater falls below 5.6 due to pollutants in the atmosphere.

Tooth enamel is primarily composed of calcium phosphate, which provides strength and durability to teeth.
Quick Tip: Tooth enamel, the hardest substance in the human body, is made of calcium phosphate.

Potassium hydroxide is a strong base, so it will turn red litmus paper blue, and blue litmus paper will remain blue in a basic solution.
Quick Tip: In a basic solution, red litmus paper turns blue, while blue litmus paper remains blue.

The maximum number of electrons in any shell can be calculated using the formula \(2n^2\), where \(n\) is the shell number. For the M shell (\(n=3\)): \[ 2(3)^2 = 18 \]
Thus, the maximum number of electrons in the M shell is 18.
Quick Tip: The maximum number of electrons in the \(n\)-th shell is \(2n^2\).

The notation \( 4f^{1/2} \) is incorrect because orbitals are not denoted by fractional powers. The correct notation for a \(f\)-orbital in the 4th shell is simply \( 4f \). The other options are valid orbital notations.
Quick Tip: Orbitals are denoted by whole numbers or letters (e.g., \( 2p^6 \), \( 3s^1 \), \( 4f \)), not fractional exponents.

Niels Bohr received the Nobel Prize in Physics for his contributions to the understanding of atomic structure and quantum mechanics.
Quick Tip: Niels Bohr's work on atomic structure and the Bohr model of the atom earned him the Nobel Prize in Physics.

In the 4d subshell, there are 5 orbitals, and each orbital can hold 2 electrons, so the total number of degenerate orbitals in the 4d subshell is 10.
Quick Tip: For d subshells, there are 5 orbitals, which can accommodate 10 electrons in total.

Hund's rule explains the presence of 3 unpaired electrons in nitrogen. According to Hund's rule, electrons occupy degenerate orbitals singly before pairing. This results in 3 unpaired electrons in the nitrogen atom.
Quick Tip: Hund's rule explains that electrons fill degenerate orbitals singly before pairing.

An ionic bond is formed between elements from group IA (alkali metals) and group VIIA (halogens), where the alkali metals donate an electron and halogens accept it.
Quick Tip: Ionic bonds are formed between metals (IA) and nonmetals (VIIA), where electrons are transferred from metals to nonmetals.

The configuration 1s², 2s², 2p⁶, 3s², 3p⁶ (2,8,8) represents the noble gas configuration. Ions such as P³⁻, Cl⁻, and S²⁻ achieve this configuration by gaining electrons to become stable. Therefore, all of these options are correct.
Quick Tip: Ions like P³⁻, Cl⁻, and S²⁻ gain electrons to achieve a stable noble gas configuration, typically 2,8,8.

Dobereiner’s triads consist of groups of three elements where the atomic weight of the middle element is the average of the other two. Lithium, sodium, and potassium form one such triad.
Quick Tip: Dobereiner’s triads are groups of three elements in which the atomic weight of the middle element is the average of the other two.

Group IV A elements in the periodic table are known as the carbon family, as carbon is the most well-known element of this group.
Quick Tip: The carbon family consists of elements like carbon, silicon, and germanium, which are in Group IV A.

An element in Group 2 has a valency of 2 because it can lose two electrons to achieve a stable configuration.
Quick Tip: Elements in Group 2 have a valency of 2, as they tend to lose two electrons to achieve stability.

The Valence Bond Theory, which explains how chemical bonds form in terms of electron pairing, was proposed by Linus Pauling.
Quick Tip: Linus Pauling developed the Valence Bond Theory to explain how atoms form bonds by sharing electrons.

Chlorine, being a halogen, gains an electron to form a chloride ion (Cl⁻), which is an anion.
Quick Tip: Chlorine gains an electron to form a negative ion (Cl⁻), making it an anion.

Since \(X\) forms an ionic compound, it is likely a metal that can lose electrons. If \(X\) belongs to group 2, it will have a charge of +2 when it forms an ionic compound.
Quick Tip: Metals from Group 2 tend to lose two electrons and form ions with a +2 charge.

Linus Pauling is known for proposing the concept of hybridization in chemical bonding, which explains the mixing of atomic orbitals to form new hybrid orbitals.
Quick Tip: Hybridization is the concept introduced by Pauling to explain how atomic orbitals combine to form hybrid orbitals.

The O²⁻ ion has gained two electrons, filling its 2p orbital. Thus, its electronic configuration is \( 1s^2, 2s^2, 2p^6 \), which is the same as the noble gas neon.
Quick Tip: The O²⁻ ion gains two electrons to achieve a stable configuration similar to the noble gas neon (1s², 2s², 2p⁶).

The number of electrons gained by non-metallic elements is equal to their valency, as they gain electrons to achieve a stable electron configuration.
Quick Tip: Non-metals gain electrons to fill their outermost shell and achieve a stable configuration, which is reflected in their valency.

The reactivity of metals increases as you move down a group in the periodic table. Hence, the order of reactivity is Ca, Na, K.
Quick Tip: Reactiveness of metals increases down the group in the periodic table.

The Poling process is used to purify the crude metal by removing impurities. It involves heating the crude metal in a furnace with a reducing agent.
Quick Tip: Poling is a refining process used to purify metals by reducing impurities.

When silver corrodes, it reacts with sulfur compounds in the air and forms silver sulphide (Ag₂S), which causes tarnishing.
Quick Tip: Silver tarnishes due to the formation of silver sulphide (Ag₂S) when it reacts with sulfur compounds.

During corrosion, a metal undergoes oxidation, which means it loses electrons and forms positive ions. This results in the deterioration of the metal.
Quick Tip: Corrosion involves the oxidation of metals, where the metal loses electrons.

When one hydrogen from ammonia (NH₃) is replaced by an alkyl group, the compound formed is an amine. Amine groups contain nitrogen and are commonly found in organic compounds.
Quick Tip: Amines are formed when one or more hydrogen atoms in ammonia (NH₃) are replaced by alkyl or aryl groups.