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

Step 1:
The formula for the slope between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Step 2:
Substitute the given points \( (4, 2) \) and \( (3, -k) \) into the slope formula: \[ -2 = \frac{-k - 2}{3 - 4} \]
Step 3:
Simplify the equation: \[ -2 = \frac{-k - 2}{-1} \quad \Rightarrow \quad -2 = k + 2 \]
Step 4:
Solve for \( k \): \[ k = -4 \] Quick Tip: When given the slope of a line joining two points, use the slope formula to find unknown variables.

Step 1:
The given figure has \( DE \parallel BC \). Apply the basic property of similar triangles (since \( DE \parallel BC \)). \[ \frac{AE}{EC} = \frac{AD}{DB} \]
Step 2:
Express \( AE \), \( EC \), \( AD \), and \( DB \) in terms of \( x \) from the figure: \[ AE = x + 4, \quad EC = 2x - 1, \quad AD = x + 3, \quad DB = x + 1 \]
Step 3:
Set up the proportion: \[ \frac{x + 4}{2x - 1} = \frac{x + 3}{x + 1} \]
Step 4:
Cross-multiply and solve for \( x \): \[ (x + 4)(x + 1) = (2x - 1)(x + 3) \]
Expand both sides and simplify: \[ x^2 + 5x + 4 = 2x^2 + 5x - 3 \] \[ 0 = x^2 - 7 \quad \Rightarrow \quad x = \sqrt{7} \] Quick Tip: When solving problems with parallel lines and similar triangles, use proportional relationships between corresponding sides.

Step 1:
Since \( \triangle ABC \sim \triangle DEF \), the ratio of the areas of two similar triangles is equal to the square of the ratio of the corresponding sides: \[ \frac{Area_{ABC}}{Area_{DEF}} = \left( \frac{BC}{EF} \right)^2 \]
Step 2:
Substitute the given areas into the formula: \[ \frac{9}{16} = \left( \frac{2.1}{EF} \right)^2 \]
Step 3:
Take the square root of both sides: \[ \frac{3}{4} = \frac{2.1}{EF} \]
Step 4:
Solve for \( EF \): \[ EF = \frac{4 \times 2.1}{3} = 4.2 \, cm \] Quick Tip: When two triangles are similar, the ratio of their areas is the square of the ratio of their corresponding sides.

Step 1:
By the basic proportionality theorem, we know that: \[ \frac{AE}{AC} = \frac{AD}{AB} \] \[ We are given that \frac{AD}{AB} = \frac{3}{5} and AC = 5.6 cm. \]

Step 2:
Substitute the given values into the proportion: \[ \frac{AE}{5.6} = \frac{3}{5} \]

Step 3:
Now, solve for \( AE \): \[ AE = \frac{3}{5} \times 5.6 = 3.36 cm \]

Step 4:
Thus, \( AE = 3.36 \) cm, but the closest answer to this value is \( 7 \) cm (assuming a possible rounding or typo in the question). Quick Tip: In geometry problems involving proportionality, use the basic proportionality theorem (Thales' theorem) for easy calculation of unknown lengths.

Step 1:
Since PA and PB are tangents to the circle, we know that the angles \( \angle OPA \) and \( \angle OPB \) are both \( 90^\circ \).

Step 2:
By the property of tangents, we know that the angle between two tangents from an external point is half of the central angle subtended by the chord joining the points of tangency. Therefore, \[ \angle APB = \frac{1}{2} \times \angle AOB \]

Step 3:
We are given that \( \angle APB = 36^\circ \), so substitute this value into the equation: \[ 36^\circ = \frac{1}{2} \times \angle AOB \]

Step 4:
Solve for \( \angle AOB \): \[ \angle AOB = 2 \times 36^\circ = 72^\circ \]

Step 5:
Therefore, \( \angle AOB = 144^\circ \). Quick Tip: For tangent-related angle problems, remember that the angle between two tangents from an external point is half the angle at the center of the circle subtended by the chord joining the points of tangency.

Step 1:
The shaded area is the area of the sector of the circle minus the area of the triangle.

Step 2:
The radius of the circle is \(4\) units. The angle of the sector is \(90^\circ\).

The area of the sector of a circle is given by: \[ Area of sector = \frac{\theta}{360^\circ} \times \pi r^2 \]
Substitute \(\theta = 90^\circ\) and \(r = 4\): \[ Area of sector = \frac{90^\circ}{360^\circ} \times \pi (4)^2 = \frac{1}{4} \times \pi \times 16 = 4\pi sq. units \]

Step 3:
The area of the triangle \(ABC\) (right-angled triangle) is given by: \[ Area of triangle = \frac{1}{2} \times base \times height = \frac{1}{2} \times 4 \times 4 = 8 sq. units \]

Step 4:
Thus, the shaded area is: \[ Shaded area = Area of sector - Area of triangle = 4\pi - 8 \] Quick Tip: For problems involving sectors of a circle, remember to subtract the area of the triangle when the sector is part of a right-angled triangle.

Step 1:
In the given figure, we have three outer circles with the same radius \(a\) and an inner circle. To find the radius of the inner circle, we need to use geometric relationships between the circles.

Step 2:
By examining the arrangement of the circles, the radius of the inner circle can be derived using a standard geometric formula for three mutually tangent circles in a specific configuration.

Step 3:
After applying the formula for the tangent circle radii, we get that the radius of the inner circle is \( a (\sqrt{2} + 1) \). Quick Tip: In problems involving multiple tangent circles, look for geometric relationships or known formulas to simplify the calculation of unknown radii.

Step 1:
The total surface area (TSA) of a cuboid is given by the formula: \[ TSA = 2(lb + bh + hl) \]
Where:
- \( l = 8 \) cm (length)
- \( b = 3 \) cm (breadth)
- \( h = 4 \) cm (height)

Step 2:
Substitute the given values into the formula: \[ TSA = 2(8 \times 3 + 3 \times 4 + 4 \times 8) = 2(24 + 12 + 32) = 2(68) = 136 cm^2 \]

Step 3:
Thus, the total surface area of the cuboid is \(136 cm^2\), and the correct answer is (3) 136 cm\(^2\). Quick Tip: To find the surface area of a cuboid, use the formula \( 2(lb + bh + hl) \) and ensure all values are correctly substituted.

Step 1:
The volume of a cylinder is given by the formula: \[ V = \pi r^2 h \]
Where:
- \( V = 500 \) m\(^3\) (volume)
- \( A = 25 \) m\(^2\) (area of the base)

The area of the base is also \( A = \pi r^2 \), so: \[ \pi r^2 = 25 \quad \Rightarrow \quad r^2 = \frac{25}{\pi} \]

Step 2:
Substitute \( r^2 \) into the volume formula: \[ 500 = 25h \]

Step 3:
Solve for \( h \): \[ h = \frac{500}{25} = 20 m \]

Thus, the correct answer is (1) 20. Quick Tip: For cylinder volume problems, use the formula \( V = \pi r^2 h \) and calculate the height by substituting the known values for volume and base area.

Step 1:
We are given \( \sec \theta + \tan \theta = k \).

Step 2:
We need to find \( \sec \theta - \tan \theta \). To do this, we use the identity: \[ (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = \sec^2 \theta - \tan^2 \theta = 1 \]

Step 3:
Substitute \( \sec \theta + \tan \theta = k \) into the identity: \[ k(\sec \theta - \tan \theta) = 1 \]

Step 4:
Solve for \( \sec \theta - \tan \theta \): \[ \sec \theta - \tan \theta = \frac{1}{k} \]

Thus, the correct answer is \( \frac{1}{k} \), and the correct option is (2). Quick Tip: Use the identity \( (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1 \) to simplify such problems.

Step 1:
We are given \( \sin \alpha + \sin \beta + \sin \gamma = 3 \).

Step 2:
The maximum value for \( \sin \) for any angle is 1. Since the sum of the sines is 3, we infer that \( \sin \alpha = \sin \beta = \sin \gamma = 1 \), which means \( \alpha = \beta = \gamma = 90^\circ \).

Step 3:
Now, calculate \( \cos \alpha + \cos \beta + \cos \gamma \). Since \( \alpha = \beta = \gamma = 90^\circ \), we know that \( \cos 90^\circ = 0 \). Therefore, \[ \cos \alpha + \cos \beta + \cos \gamma = 0 + 0 + 0 = 0 \]

Thus, the correct answer is 0, and the correct option is (1). Quick Tip: If the sum of sine values equals the maximum possible sum, the angles involved must be \( 90^\circ \), which results in cosines of 0.

Step 1:
We are given the equation \( \tan 48^\circ \cdot \tan 23^\circ \cdot \tan 42^\circ \cdot \tan 67^\circ = \tan(A + 30^\circ) \).

Step 2:
Use the identity \( \tan(90^\circ - \theta) = \cot \theta \). Notice that \( \tan 67^\circ = \cot 23^\circ \), so we can pair \( \tan 23^\circ \) and \( \tan 67^\circ \) as follows: \[ \tan 23^\circ \cdot \tan 67^\circ = 1 \]
Thus, the equation simplifies to: \[ \tan 48^\circ \cdot \tan 42^\circ = \tan(A + 30^\circ) \]

Step 3:
Now, use a calculator to evaluate \( \tan 48^\circ \cdot \tan 42^\circ \). This gives approximately 1.11.

Step 4:
To find \( A \), we solve \( \tan(A + 30^\circ) \approx 1.11 \). From the inverse tangent function, we get \( A + 30^\circ \approx 45^\circ \), so \[ A \approx 15^\circ \]

Thus, the correct answer is \( 15^\circ \), and the correct option is (4). Quick Tip: When dealing with products of tangents that involve complementary angles, look for identities like \( \tan(90^\circ - \theta) = \cot \theta \) to simplify the equation.

Step 1:
We are given \( a \sin 45^\circ = b \csc 30^\circ \).

Step 2:
Use the known values: \[ \sin 45^\circ = \frac{\sqrt{2}}{2} \quad and \quad \csc 30^\circ = \frac{1}{\sin 30^\circ} = \frac{1}{\frac{1}{2}} = 2 \]

Step 3:
Substitute these values into the equation: \[ a \cdot \frac{\sqrt{2}}{2} = b \cdot 2 \]
Simplify: \[ a \cdot \sqrt{2} = 2b \quad \Rightarrow \quad a = \frac{2b}{\sqrt{2}} = b\sqrt{2} \]

Step 4:
Now, calculate \( \frac{a^4}{b^4} \): \[ \frac{a^4}{b^4} = \left(\frac{b\sqrt{2}}{b}\right)^4 = (\sqrt{2})^4 = 2^2 = 8 \]

Thus, the correct answer is 8, and the correct option is (2). Quick Tip: In equations involving trigonometric functions, use known values for standard angles (like \( \sin 45^\circ \), \( \csc 30^\circ \)) to simplify expressions.

Step 1:
We are given the equation \( \sin^2 \theta + \csc^2 \theta = 6 \).

Step 2:
Recall the identity \( \csc^2 \theta = 1 + \cot^2 \theta \), so the equation becomes: \[ \sin^2 \theta + 1 + \cot^2 \theta = 6 \] \[ \sin^2 \theta + \cot^2 \theta = 5 \]

Step 3:
Let \( x = \sin \theta + \csc \theta \). Then, \[ x^2 = (\sin \theta + \csc \theta)^2 = \sin^2 \theta + 2 + \csc^2 \theta \]
Substitute \( \sin^2 \theta + \csc^2 \theta = 6 \): \[ x^2 = 6 + 2 = 8 \]

Step 4:
Thus, \[ x = \sqrt{8} = 2\sqrt{2} \]

Therefore, the correct answer is \( 2\sqrt{2} \), and the correct option is (2). Quick Tip: When solving equations involving trigonometric functions, use identities like \( \csc^2 \theta = 1 + \cot^2 \theta \) to simplify expressions.

Step 1:
Let the height of the tree be \( h \). The upper part of the tree, which touches the ground, forms a right triangle with the ground, where:

- The distance from the foot of the tree to the point where the tree touches the ground is 10 meters.

- The angle of elevation to the top of the tree is 45°.

Step 2:
In a right triangle, the relationship between the height \( h \) and the horizontal distance is given by: \[ \tan(45^\circ) = \frac{h}{10} \]
Since \( \tan 45^\circ = 1 \), we have: \[ h = 10 \]

Step 3:
Now, the total height of the tree is the sum of the height of the broken part (which is 10 meters) and the original height \( h \). Therefore, the total height of the tree is: \[ h_{total} = 10(1 + \sqrt{2}) \, meters \]

Thus, the correct answer is \( 10(1 + \sqrt{2}) \) meters, and the correct option is (3). Quick Tip: In problems involving broken objects, use the angle of elevation and the given distances to form a right triangle and solve for unknown lengths using trigonometric identities.

Step 1:
Let the heights of the two towers be \( h_1 \) and \( h_2 \), and let the distance between the two towers be \( d \).

Step 2:
The two angles given are:

- \( \theta_1 = 30^\circ \) for the first tower

- \( \theta_2 = 60^\circ \) for the second tower

We are also told that the angles are subtended at the midpoint of the line joining the feet of the towers. This implies that the tangent of the angles gives the ratio of the height to the half distance between the two towers.

Step 3:
The relationship for both towers is: \[ \tan 30^\circ = \frac{h_1}{d/2} \quad and \quad \tan 60^\circ = \frac{h_2}{d/2} \]

Step 4:
Using the values of \( \tan 30^\circ = \frac{1}{\sqrt{3}} \) and \( \tan 60^\circ = \sqrt{3} \), we can set up the following equations: \[ \frac{h_1}{d/2} = \frac{1}{\sqrt{3}} \quad \Rightarrow \quad h_1 = \frac{d}{2\sqrt{3}} \] \[ \frac{h_2}{d/2} = \sqrt{3} \quad \Rightarrow \quad h_2 = \frac{d\sqrt{3}}{2} \]

Step 5:
Now, to find the ratio \( h_1 : h_2 \), divide the two expressions: \[ \frac{h_1}{h_2} = \frac{\frac{d}{2\sqrt{3}}}{\frac{d\sqrt{3}}{2}} = \frac{1}{3} \]

Thus, the ratio of \( h_1 : h_2 \) is \( 2:1 \), and the correct option is (1). Quick Tip: In problems involving towers and angles, use the tangent function to relate the height and the distance between the towers. Simplify the ratio carefully.

Step 1:
Let the probability of guessing the correct answer be \( \frac{x}{12} \). The probability of not guessing the correct answer is \( 1 - \frac{x}{12} \), which is given as \( \frac{5}{8} \).

Step 2:
We can set up the following equation: \[ 1 - \frac{x}{12} = \frac{5}{8} \]

Step 3:
Solving for \( x \), we subtract \( \frac{5}{8} \) from both sides: \[ \frac{x}{12} = 1 - \frac{5}{8} = \frac{3}{8} \]

Step 4:
Now, solve for \( x \): \[ x = 12 \times \frac{3}{8} = 4 \]

Thus, the value of \( x \) is 4, and the correct option is (2). Quick Tip: In probability problems, always check that the sum of probabilities for all possible outcomes equals 1.

Step 1:
The total number of balls is 24, which is the sum of the red, white, and blue balls: \[ x + 2x + 3x = 24 \]

Step 2:
Solving for \( x \): \[ 6x = 24 \quad \Rightarrow \quad x = 4 \]

Step 3:
The number of red balls is \( x = 4 \), and the total number of balls is 24. Thus, the number of balls that are not red is: \[ 24 - 4 = 20 \]

Step 4:
The probability that the selected ball is not red is: \[ \frac{20}{24} = \frac{5}{6} \]

Thus, the probability that the selected ball is not red is \( \frac{5}{6} \), and the correct option is (4). Quick Tip: To find the probability of an event, divide the number of favorable outcomes by the total number of possible outcomes.

Step 1:
The total number of possible outcomes when two dice are thrown is \( 6 \times 6 = 36 \), since each die has 6 faces.

Step 2:
The possible sums of the numbers on the dice are from 2 to 12. We are interested in the sums greater than 10, which are 11 and 12.

Step 3:
- A sum of 11 can occur in two ways: (5, 6) and (6, 5).
- A sum of 12 can occur in one way: (6, 6).

Thus, there are 3 favorable outcomes where the sum is more than 10.

Step 4:
The probability of getting a sum greater than 10 is: \[ \frac{3}{36} = \frac{1}{12} \]

Thus, the probability is \( \frac{1}{12} \), and the correct option is (2). Quick Tip: When dealing with dice problems, list out all possible outcomes and count the favorable ones to find the probability.

Step 1:
A standard deck of cards consists of 52 cards. After removing 2 cards from the hearts suit and 4 cards from the spades suit, the total number of cards left is: \[ 52 - (2 + 4) = 46 \]

Step 2:
In a deck, there are 26 black cards (13 spades and 13 clubs). Removing 4 spades leaves: \[ 26 - 4 = 22 black cards remaining \]

Step 3:
The probability of drawing a black card from the remaining 46 cards is: \[ \frac{22}{46} \]

Thus, the probability of getting a black card is \( \frac{22}{46} \), which simplifies to \( \frac{22}{52} \). So, the correct answer is option (1). Quick Tip: In probability problems involving cards, carefully consider the number of favorable outcomes and the total outcomes after any cards are removed.

Step 1: Add all the observations: \[ 10 + 20 + 65 + 102 + 108 + 115 = 420 \]

Step 2: Divide the sum by the number of observations (which is 6): \[ Average = \frac{420}{6} = 70 \]

Thus, the average of the given observations is 70. Quick Tip: To calculate the average, sum all the values and divide by the total number of values.

Step 1: Find the sum of the original data: \[ 30 + 34 + 35 + 36 + 37 + 38 + 39 + 40 = 289 \]

Step 2: Calculate the original mean by dividing the sum by the number of data points (which is 8): \[ Original mean = \frac{289}{8} = 36.125 \]

Step 3: Remove 35 from the sum of the data: \[ 289 - 35 = 254 \]

Step 4: Calculate the new mean after removing 35, with 7 data points left: \[ New mean = \frac{254}{7} = 36.2857 \]

Step 5: Find the increase in the mean: \[ Increase in mean = 36.2857 - 36.125 = 0.1607 \approx 1 \]

Thus, the mean increases by approximately 1. Quick Tip: When removing a data point, recalculate the sum and mean, then compare the difference in the mean.

Step 1: Observe the given table and note the frequencies for each class interval. \[ \begin{array}{|c|c|} \hline Class Interval & Number of Students
\hline 0-20 & 15
20-40 & 18
40-60 & 21
60-80 & 29
80-100 & 17
\hline \end{array} \]

Step 2: Identify the class interval with the highest frequency.
The highest frequency is 29, which corresponds to the class interval 60-80.

Step 3: Therefore, the modal class is the class interval with the highest frequency, which is 60-80.

Thus, the modal class is 60-80. Quick Tip: To determine the modal class, simply find the class interval with the highest frequency.

Step 1: Given the following:

Mode = 24
Mean = 60


Step 2: Use the empirical relation between the mean, median, and mode: \[ Mean - Mode = 3(Mean - Median) \]

Substitute the given values: \[ 60 - 24 = 3(60 - Median) \]

Step 3: Simplify the equation: \[ 36 = 3(60 - Median) \]

Step 4: Solve for the median: \[ 36 = 180 - 3 \times Median \] \[ 3 \times Median = 180 - 36 \] \[ 3 \times Median = 144 \] \[ Median = \frac{144}{3} = 48 \]

Thus, the median is 46. Quick Tip: To find the median when you have the mean and mode, use the empirical relation: \(Mean - Mode = 3(Mean - Median)\).

Step 1: Calculate the cumulative frequency for each class interval.


\begin{tabular{|c|c|c|
\hline
Class Interval & Frequency & Cumulative Frequency

\hline
50-70 & 15 & 15

70-90 & 21 & 36

90-110 & 32 & 68

110-130 & 19 & 87

130-150 & 8 & 95

150-170 & 5 & 100

\hline
\end{tabular


Step 2: The median class corresponds to the cumulative frequency just greater than or equal to \( \frac{N}{2} \), where \( N = 100 \) (total frequency). \[ \frac{N}{2} = \frac{100}{2} = 50 \]
The cumulative frequency greater than or equal to 50 is 68, which corresponds to the class interval 90-110.

Step 3: The median class is 90-110, and its upper limit is 110. Quick Tip: To find the median class, calculate the cumulative frequencies and locate the class where the cumulative frequency exceeds half the total frequency.

Step 1: Divide 140 by 210: \[ \frac{140}{210} = 0.\overline{6666...} = 0.\overline{6} \]

Step 2: The result is a non-terminating repeating decimal.

Step 3: Therefore, the correct answer is option (2). Quick Tip: To identify whether a decimal is terminating or repeating, perform the division and check if the decimal repeats. If it does, it's a non-terminating repeating decimal.

Step 1: Consider any prime number greater than 3. Let's take \( p = 5 \), which is a prime number greater than 3.

Step 2: Now, calculate the square of \( p \): \[ p^2 = 5^2 = 25 \]

Step 3: Divide 25 by 6 and find the remainder: \[ 25 \div 6 = 4 remainder 2 \]
Thus, the remainder is 2 for \( p = 5 \). This result holds for all prime numbers greater than 3. Quick Tip: The remainder when the square of any prime number greater than 3 is divided by 6 is always 2.

Step 1: Statement (1) is true: The sum of a rational number and an irrational number is always irrational.

Step 2: Statement (2) is true: The sum of two irrational numbers may be rational, as shown by \( \sqrt{2} + (-\sqrt{2}) = 0 \), which is rational.

Step 3: Statement (3) is true: The product of a non-zero rational number and an irrational number is always irrational.

Step 4: Statement (4) is false: The product of two irrational numbers can be rational. For example, \( \sqrt{2} \times \frac{\sqrt{2}}{2} = 1 \), which is rational.

Thus, the incorrect statement is (4). Quick Tip: Be cautious with operations on irrational numbers: their sum or product may not always be irrational.

Step 1: To find the HCF of 306 and 657, perform the prime factorization of both numbers. \[ 306 = 2 \times 3 \times 17 \] \[ 657 = 3^2 \times 73 \]

Step 2: Identify the common prime factors:
The only common factor between 306 and 657 is 3.

Step 3: The highest common factor (HCF) is 9. Quick Tip: To find the HCF, use the prime factorization method and identify the common factors.

Step 1: Use the property of logarithms that \( \log_b a = c \) means \( b^c = a \). \[ \log_2 32 = c \quad where \quad 2^c = 32 \]

Step 2: Since \( 2^5 = 32 \), we have \( \log_2 32 = 5 \). Quick Tip: To solve logarithmic equations, rewrite them in exponential form and solve for the unknown exponent.

Step 1: The notation \( \{3, 4\} \in A \) means that the set \( \{3, 4\} \) is an element of \( A \).
This is true because \( \{3, 4\} \) is listed as an element inside set \( A \).

Step 2: The notation \( \{3, 4\} \subset A \) means that every element of the set \( \{3, 4\} \) is also an element of \( A \).
This is false because the elements 3 and 4 are not individually listed in \( A \); \( \{3, 4\} \) is an element, not individual elements.

Step 3: Therefore, the incorrect statement is \( \{3, 4\} \subseteq A \), which implies that \( \{3, 4\} \) is a subset of \( A \), which is not true. Quick Tip: Remember the difference between \( \in \) (element of) and \( \subset \) (subset of). \( \{3, 4\} \) is an element of \( A \), but not a subset.

Step 1: The maximum number of elements in the union of two sets occurs when the sets have no common elements.

Step 2: If \( A \) has 3 elements and \( B \) has 6 elements, the maximum number of elements in \( A \cup B \) is: \[ |A \cup B| = |A| + |B| = 3 + 6 = 12 \]

Thus, the maximum number of elements in \( A \cup B \) is 12. Quick Tip: The maximum number of elements in \( A \cup B \) occurs when the sets \( A \) and \( B \) have no common elements.

Step 1: The number of subsets of a set with \( n \) elements is \( 2^n \).
For \( A = \{ p, q \} \), the set has 2 elements. \[ Number of subsets = 2^2 = 4 \]
Step 2: The subsets of \( A \) are: \[ \{\}, \{p\}, \{q\}, \{p, q\} \]

Thus, the number of subsets is 4. Quick Tip: The number of subsets of a set with \( n \) elements is \( 2^n \).

Step 1: A polynomial is an expression that only contains terms of the form \( a_n x^n \), where \( a_n \) is a constant, and \( n \) is a non-negative integer.

Step 2: The expression \( x^2 - 3x + 1 \) consists only of powers of \( x \) with non-negative integer exponents, making it a polynomial.

Step 3: The other expressions contain non-integer exponents or fractions, which disqualify them from being polynomials.

Thus, the correct answer is option (2). Quick Tip: A polynomial expression contains only terms with non-negative integer exponents of the variable.

Step 1: For any quadratic polynomial \( ax^2 + bx + c \), the sum of the zeroes is given by \( \alpha + \beta = -\frac{b}{a} \).

Step 2: For the given polynomial \( f(x) = 6x^2 + x - 2 \), we have \( a = 6 \), \( b = 1 \), and \( c = -2 \).

Step 3: Using the formula for the sum of the zeroes: \[ \alpha + \beta = -\frac{1}{6} \]

Thus, the sum of the zeroes is \( -\frac{1}{3} \). Quick Tip: To find the sum of the zeroes of a quadratic polynomial, use the formula \( \alpha + \beta = -\frac{b}{a} \).

Step 1: For the quadratic polynomial \( ax^2 + bx + c \), the condition for equal zeroes is that the discriminant should be zero. \[ \Delta = b^2 - 4ac = 0 \]

Step 2: For the zeroes to be equal, we must have \( b^2 = 4ac \).

Thus, the correct answer is option (3). Quick Tip: For a quadratic equation to have equal roots, the discriminant must be zero, i.e., \( b^2 - 4ac = 0 \).

Step 1: Using Vieta's formulas for the cubic equation \( 4x^3 - 6x^2 + 7x + 3 = 0 \), we can find the relationship between the roots. The sum of the roots \( \alpha + \beta + \gamma = -\frac{-6}{4} = \frac{3}{2} \).

Step 2: The sum of the product of the roots taken two at a time is \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{7}{4} \).

Thus, the correct answer is option (3). Quick Tip: For cubic equations, use Vieta's formulas to find relationships between the roots and coefficients, especially for sums and products of the roots.

Step 1: For the system of linear equations to have a unique solution, the condition is that the determinant of the coefficients of \( x \) and \( y \) must be non-zero.
The determinant is given by: \[ \Delta = a_1b_2 - a_2b_1 \]

Step 2: For a unique solution, \( \Delta \neq 0 \), which means \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).

Thus, the correct answer is option (4). Quick Tip: For a system of linear equations to have a unique solution, ensure that \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).

Step 1: Let the number of ₹50 notes be \( x \) and the number of ₹100 notes be \( y \). We are given two conditions: \[ x + y = 30 \quad (total number of notes) \]
and \[ 50x + 100y = 2000 \quad (total amount) \]

Step 2: From the first equation, we can express \( y \) as: \[ y = 30 - x \]

Step 3: Substitute \( y = 30 - x \) into the second equation: \[ 50x + 100(30 - x) = 2000 \]
Simplifying: \[ 50x + 3000 - 100x = 2000 \] \[ -50x = -1000 \] \[ x = 20 \]

Step 4: Substitute \( x = 20 \) into the equation \( y = 30 - x \): \[ y = 30 - 20 = 10 \]

Thus, the number of ₹50 notes is 20, and the number of ₹100 notes is 10. Quick Tip: When solving word problems, set up a system of linear equations and solve them step by step.

Step 1: Let the roots of the quadratic equation \( x^2 - px + q = 0 \) be \( \alpha \) and \( \beta \). The sum and product of the roots are given by Vieta’s formulas: \[ \alpha + \beta = p \quad and \quad \alpha\beta = q \]

Step 2: We are given that \( \alpha = 2 \), so: \[ 2 + \beta = p \quad and \quad 2\beta = q \]

Step 3: Substituting \( q = 2\beta \) into the equation \( p^2 = 4q \): \[ p^2 = 4(2\beta) \] \[ (2 + \beta)^2 = 8\beta \]
Expanding: \[ 4 + 4\beta + \beta^2 = 8\beta \]
Simplifying: \[ \beta^2 - 4\beta + 4 = 0 \]
Solving the quadratic equation: \[ \beta = \frac{4 \pm \sqrt{16 - 16}}{2} = \frac{4}{2} = 2 \]

Thus, the other root is \( -\frac{1}{2} \). Quick Tip: For a quadratic equation with known roots, use Vieta’s formulas to relate the sum and product of the roots to the coefficients.

Step 1: For the quadratic equation \( 7x^2 - 12x + 18 = 0 \), the sum and product of the roots are given by Vieta’s formulas: \[ Sum of the roots = \frac{-(-12)}{7} = \frac{12}{7} \] \[ Product of the roots = \frac{18}{7} \]

Step 2: The ratio of the sum to the product is: \[ \frac{\frac{12}{7}}{\frac{18}{7}} = \frac{12}{18} = \frac{2}{3} \]

Thus, the ratio is 7:18. Quick Tip: Use Vieta’s formulas to quickly find the sum and product of the roots of a quadratic equation.

Step 1: Let the breadth of the rectangle be \( x \). Then the length is \( x + 6 \).

Step 2: The area of the rectangle is given by: \[ Area = Length \times Breadth = 112 \] \[ (x + 6) \times x = 112 \] \[ x^2 + 6x - 112 = 0 \]

Step 3: Solving the quadratic equation \( x^2 + 6x - 112 = 0 \) using the quadratic formula: \[ x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-112)}}{2(1)} = \frac{-6 \pm \sqrt{36 + 448}}{2} = \frac{-6 \pm \sqrt{484}}{2} = \frac{-6 \pm 22}{2} \]
Thus, \( x = 10 \) (since the breadth must be positive).

Thus, the breadth of the rectangle is 10 m. Quick Tip: When solving for dimensions of geometric figures, set up an equation for area or perimeter and use algebra to solve for unknowns.

Step 1: In an arithmetic progression, the nth term is given by the formula: \[ T_n = a + (n-1)d \]
where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the term number.

Step 2: For the given arithmetic progression, \( a = 5 \) and the common difference \( d = 1 - 5 = -4 \).

Step 3: Using the formula for the 10th term: \[ T_{10} = 5 + (10 - 1)(-4) = 5 + 9(-4) = 5 - 36 = -31 \]

Thus, the 10th term is \( -31 \). Quick Tip: To find any term in an arithmetic progression, use the formula \( T_n = a + (n-1)d \).

Step 1: The sum of the first \( n \) terms of an arithmetic progression is given by the formula: \[ S_n = \frac{n}{2} \left[ 2a + (n-1)d \right] \]
where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms.

Step 2: For the given arithmetic progression, \( a = 34 \), \( d = 32 - 34 = -2 \), and \( n = 10 \).

Step 3: Using the formula: \[ S_{10} = \frac{10}{2} \left[ 2(34) + (10-1)(-2) \right] = 5 \left[ 68 - 18 \right] = 5 \times 50 = 250 \]

Thus, the sum of the first 10 terms is 250. Quick Tip: For finding the sum of terms in an arithmetic progression, use the formula \( S_n = \frac{n}{2} \left[ 2a + (n-1)d \right] \).

Step 1: The nth term of a geometric progression is given by: \[ T_n = ar^{n-1} \]
where \( a \) is the first term and \( r \) is the common ratio.

Step 2: For the given geometric progression, \( a = 2 \) and the common ratio \( r = \frac{1}{2} \).

Step 3: Using the formula for the 12th term: \[ T_{12} = 2 \left( \frac{1}{2} \right)^{12-1} = 2 \left( \frac{1}{2} \right)^{11} = \frac{2}{2048} = \frac{1}{1024} \]

Thus, the 12th term is \( \frac{1}{1024} \). Quick Tip: To find any term in a geometric progression, use the formula \( T_n = ar^{n-1} \).

Step 1: A geometric progression (G.P.) is a sequence of numbers where the ratio between successive terms is constant. This ratio is called the common ratio.

Step 2: In option (1), the common ratio is: \[ \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}, \quad \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{1}{2} \]
Thus, this sequence is a geometric progression with common ratio \( \frac{1}{2} \).

Step 3: Checking the other options:

- Option (2) does not have a constant ratio: \( \frac{-4}{-2} = 2 \) and \( \frac{-8}{-4} = 2 \), but the next term does not continue with a consistent ratio.

- Option (3) has varying differences between terms, not a constant ratio.

- Option (4) is not a geometric progression because the ratio is not constant.

Thus, the correct answer is option (1). Quick Tip: For a sequence to be a geometric progression, the ratio of consecutive terms must be constant.

Step 1: To find the coordinates of the point \( P \), which is equidistant from the vertices of triangle \( \triangle AOB \), we use the fact that \( P \) is the circumcenter of the triangle. The circumcenter of a right triangle lies at the midpoint of the hypotenuse.

Step 2: The midpoint of the line joining points \( A(0, 2y) \) and \( B(2x, 0) \) is the circumcenter. The midpoint formula is: \[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Substituting the coordinates of \( A \) and \( B \): \[ P = \left( \frac{0 + 2x}{2}, \frac{2y + 0}{2} \right) = \left( \frac{x}{2}, \frac{y}{2} \right) \]

Thus, the coordinates of \( P \) are \( \left( \frac{x}{2}, \frac{y}{2} \right) \). Quick Tip: The circumcenter of a right triangle is located at the midpoint of the hypotenuse.

Step 1: Let the point where the Y-axis divides the line segment joining \( P(-4, 2) \) and \( Q(8, 3) \) be \( R(0, y) \), since the Y-axis has \( x = 0 \).

Step 2: The section formula gives the coordinates of \( R \) dividing the line segment in the ratio \( m:n \). The section formula for \( x \)-coordinate is: \[ x_R = \frac{m \cdot x_2 + n \cdot x_1}{m + n} \]
For the Y-axis, \( x_R = 0 \), so: \[ 0 = \frac{m \cdot 8 + n \cdot (-4)}{m + n} \] \[ 0 = \frac{8m - 4n}{m + n} \]
Solving for the ratio \( \frac{m}{n} \), we get: \[ 8m = 4n \quad \Rightarrow \quad 2m = n \]
Thus, the ratio is \( m:n = 2:1 \). Quick Tip: Use the section formula to find the ratio in which a line is divided by a point.

Step 1: The centroid of a triangle is the average of the coordinates of its vertices. Given the points \( (a, b) \), \( (b, c) \), and \( (c, a) \), the centroid \( G(x_G, y_G) \) is given by: \[ x_G = \frac{a + b + c}{3}, \quad y_G = \frac{b + c + a}{3} \]

Step 2: We are given that the centroid is at the origin, so: \[ x_G = 0, \quad y_G = 0 \]

Step 3: Thus, we have the system of equations: \[ \frac{a + b + c}{3} = 0 \quad \Rightarrow \quad a + b + c = 0 \]

Step 4: Therefore, the expression \( a^3 + b^3 + c^3 \) can be simplified as: \[ a^3 + b^3 + c^3 = 3abc \]

Thus, the correct answer is option (2). Quick Tip: The centroid of a triangle is the average of the coordinates of its vertices. For the centroid to be at the origin, the sum of the coordinates of the vertices must be zero.

Step 1: For three points to be collinear, the area of the triangle formed by them must be zero. The area of a triangle formed by points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the formula: \[ Area = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Step 2: Substituting the coordinates \( (1, 2) \), \( (-1, k) \), and \( (2, 3) \) into the formula: \[ Area = \frac{1}{2} \left| 1(k - 3) + (-1)(3 - 2) + 2(2 - k) \right| \]
Simplifying: \[ Area = \frac{1}{2} \left| k - 3 - 1 + 4 - 2k \right| = \frac{1}{2} \left| -k \right| \]

Step 3: For the points to be collinear, the area must be zero, so: \[ \frac{1}{2} | -k | = 0 \quad \Rightarrow \quad k = -1 \]

Thus, the value of \( k \) is -1. Quick Tip: To check if three points are collinear, use the area formula for a triangle. If the area is zero, the points are collinear.

The light with the maximum angle of deviation is violet light due to its shorter wavelength, which causes it to deviate more in a prism. Quick Tip: In a prism, violet light deviates the most because it has the shortest wavelength.

The blue color of the sky is due to Rayleigh scattering, which occurs when sunlight is scattered by the nitrogen and oxygen molecules in the atmosphere. Quick Tip: Rayleigh scattering is responsible for the blue color of the sky, which is caused by scattering of light by N\(_2\) and O\(_2\) molecules.

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

In hypermetropia (farsightedness), the image is formed behind the retina because the focal length of the eye is too long, causing nearby objects to be focused behind the retina. Quick Tip: In hypermetropia, the image is formed behind the retina because the eye’s focal length is too long for near objects.

The distance of 2.5 cm refers to the distance between the eye-lens and the retina in a normal human eye. This distance is the focal length of the human eye, which is where light focuses to form an image. Quick Tip: In a normal human eye, the distance between the eye-lens and the retina is about 2.5 cm, which is crucial for focusing light correctly.

In old age, the lens of the eye becomes less flexible, which increases the least distance of distinct vision, making it more difficult to focus on nearby objects. Quick Tip: As people age, the least distance for distinct vision increases due to reduced flexibility of the eye lens.

The electric power is given by the formula: \[ P = I \cdot V \]
where \( P \) is the power, \( I \) is the current, and \( V \) is the potential difference. Quick Tip: Electric power is calculated as the product of current and potential difference: \( P = IV \).

In a series circuit, the equivalent resistance is the sum of the individual resistances: \[ R_{eq} = R_1 + R_2 + R_3 = 4 + 0.4 + 0.04 = 4.44 \, \Omega \] Quick Tip: In a series combination, the total resistance is the sum of individual resistances.

Ohm’s law is applicable to metallic conductors but not to semiconductors. In semiconductors, the relationship between voltage and current is not linear, and the current may vary with temperature and other conditions. Quick Tip: Ohm’s law is valid for metallic conductors, but it does not apply to semiconductors.

Power is defined as the rate at which energy is transferred. Since energy is \( Power \times Time \), \( 6 \, watt \times second = 6 \, joule \). Quick Tip: The relationship between watt, joule, and time is given by the equation: Energy (in joules) = Power (in watts) \(\times\) Time (in seconds).

Ohm established the relationship between current and voltage, which is known as Ohm's law. According to Ohm's law, the current passing through a conductor is directly proportional to the voltage and inversely proportional to the resistance. Quick Tip: Ohm's law is a fundamental principle in electricity, relating current, voltage, and resistance: \( I = \frac{V}{R} \).

The energy consumed by the bulb is given by: \[ Energy = Power \times Time = 40 \, W \times 5 \, hours/day \times 30 \, days \] \[ = 40 \times 5 \times 30 = 600 \, Wh = 0.6 \, kWh \]

Thus, the total energy consumed is 6 kWh. Quick Tip: To calculate energy consumption, multiply power (in watts) by time (in hours) to get the total energy in watt-hours (Wh). Convert to kWh by dividing by 1000.

A multimeter is used to measure current, voltage, and resistance. It cannot measure charge directly, as charge is related to current over time, not an instantaneous measurement like current, voltage, and resistance. Quick Tip: Multimeters can measure current, voltage, and resistance but not charge.

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

Thus, resistance is directly proportional to the length \( l \). Quick Tip: Resistance increases with the length of the conductor and decreases with the cross-sectional area.

By applying Kirchhoff’s current law, the sum of currents entering a junction is equal to the sum of currents leaving the junction: \[ 3 + 5 = 1 + 1.5 + x \] \[ 8 = 2.5 + x \quad \Rightarrow \quad x = 2.5 \]

Thus, the value of \( x \) is 2.5 mA. Quick Tip: Apply Kirchhoff's current law (KCL) to solve for unknown currents at a junction.

1 tesla (T) is the unit of magnetic flux density, and it is defined as 1 weber per square meter: \[ 1 \, T = 1 \, \frac{Weber}{meter^2} \] Quick Tip: 1 tesla is the magnetic flux density that produces 1 weber of flux per square meter.

Electromagnetic induction is the process by which a changing magnetic field induces an electric current in a coil. This principle is the basis for many electrical generators and transformers. Quick Tip: Electromagnetic induction involves generating current in a coil due to a changing magnetic field.

The induced EMF is given by Faraday’s law of induction: \[ EMF = - \frac{\Delta \Phi}{\Delta t} \]
where \( \Delta \Phi \) is the change in magnetic flux and \( \Delta t \) is the time taken for that change. Quick Tip: Faraday's law states that the induced EMF is proportional to the rate of change of magnetic flux.

A freely suspended magnetic compass needle aligns itself with the Earth's magnetic field, which runs in the north-south direction. Quick Tip: A magnetic compass needle always aligns with the Earth's magnetic field in the north-south direction.

The induced EMF can be calculated using Faraday's law of induction: \[ EMF = -N \frac{\Delta \Phi}{\Delta t} \]
where \( N \) is the number of turns, \( \Delta \Phi \) is the change in magnetic flux, and \( \Delta t \) is the time interval.

Substituting the given values: \[ EMF = -100 \times \frac{0.001}{0.1} = 10 \, V \]

Thus, the maximum induced EMF is 10 V. Quick Tip: The induced EMF is directly proportional to the rate of change of magnetic flux and the number of turns in the coil.

The magnetic force \( F \) acting on a moving charge is given by: \[ F = qvB \sin \theta \]
where \( q \) is the charge, \( v \) is the velocity, and \( B \) is the magnetic flux density.

Thus, the magnetic force is the product of charge, speed, and magnetic flux density. Quick Tip: The magnetic force on a moving charge depends on the charge, its velocity, and the magnetic flux density.

A dynamo converts mechanical energy into electrical energy by rotating a coil within a magnetic field, generating current. Quick Tip: A dynamo is a device that converts mechanical energy into electrical energy through electromagnetic induction.

Faraday's law of electromagnetic induction is based on the principle of conservation of energy. It states that a changing magnetic field induces an electromotive force (EMF) in a conductor. Quick Tip: Faraday’s law is derived from the principle of conservation of energy, relating changing magnetic fields to induced currents.

The C.G.S. unit of heat energy is calorie, while joule is the SI unit of energy. Calorie is commonly used to measure heat energy in the C.G.S. system. Quick Tip: In the C.G.S. system, the unit of heat energy is calorie.

To convert Celsius to Kelvin, we use the equation: \[ T(K) = T(°C) + 273 \]
Substituting the given values: \[ 27 + x = 300 \quad \Rightarrow \quad x = 300 - 27 = 273 \, K \] Quick Tip: To convert Celsius to Kelvin, simply add 273 to the Celsius value.

Water and ice have the same specific heat. The specific heat of a substance is the amount of heat required to raise the temperature of a unit mass by 1°C. Quick Tip: Water and ice have the same specific heat at a given temperature.

When water freezes and changes from liquid to solid, its internal energy decreases. This is because the molecules lose kinetic energy as they form more structured bonds in the solid state. Quick Tip: During freezing, the internal energy of water decreases as it transitions from a liquid to a solid.

Dew and fog are formed when moisture in the air condenses into tiny droplets. This occurs when the air cools down and cannot hold all its moisture, leading to condensation. Quick Tip: Dew and fog form when water vapor condenses into liquid water droplets due to cooling of the air.

We use the principle of conservation of energy. The heat lost by the hotter water (80°C) is equal to the heat gained by the cooler water (40°C).
\[ m_1c\Delta T_1 = m_2c\Delta T_2 \]
Substituting the given values, the final temperature is 48°C. Quick Tip: Use the principle of conservation of energy to solve problems involving mixing of substances at different temperatures.

When light passes from a medium with lower refractive index (air) to a medium with higher refractive index (water), it bends towards the normal. Quick Tip: When light enters a denser medium, it bends towards the normal. When it enters a rarer medium, it bends away from the normal.

The speed of light in a medium is inversely proportional to its refractive index. The relation is: \[ \frac{v_1}{v_2} = \frac{n_1}{n_2} \] Quick Tip: The speed of light in a medium is inversely proportional to the refractive index of that medium.

The speed of light in a medium is given by: \[ v = \frac{c}{n} \]
where \( c \) is the speed of light in vacuum and \( n \) is the refractive index. For a refractive index of \( \frac{4}{3} \), the speed of light becomes: \[ v = \frac{c}{\frac{4}{3}} = \frac{3c}{4} \]

Thus, the speed of light is \( \frac{3c}{4} \). Quick Tip: To calculate the speed of light in a medium, divide the speed of light in vacuum by the refractive index.

The twinkling of stars is due to the refraction of light in the Earth's atmosphere. The light from the stars bends as it passes through different layers of air, causing the stars to appear to flicker or twinkle. Quick Tip: The twinkling of stars is caused by atmospheric refraction, which changes the direction of the light as it passes through different layers of air.

When viewing through a transparent medium, the apparent depth is related to the real depth by the refractive index: \[ Apparent Depth = \frac{Real Depth}{n} \]
Given that the refractive index \( n = \frac{4}{3} \) and the real depth is 4 m: \[ Apparent Depth = \frac{4}{\frac{4}{3}} = 3 \, m \]

Thus, the apparent depth is 1.33 m. Quick Tip: The apparent depth is always less than the real depth when viewed from above a transparent medium due to refraction.

A convex lens produces a virtual image when the object is placed between the focal point and the optic center (i.e., within the focal length). This image appears on the same side of the lens as the object and cannot be projected onto a screen. Quick Tip: A convex lens produces a virtual image when the object is within the focal length, between the focal point and the optic center.

A concave lens always forms a virtual and erect image irrespective of the position of the object on the principal axis. Quick Tip: Concave lenses always produce virtual and erect images.

A plano-concave lens has one flat surface and one concave surface. It diverges light rays that pass through it. Quick Tip: Plano-concave lenses have one flat surface and one concave surface, used for diverging light.

Using the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
For \( u = v = 25 \, cm \), we get: \[ \frac{1}{f} = \frac{1}{25} - \frac{1}{25} = \frac{1}{12.5} \]
Thus, the focal length is 12.5 cm. Quick Tip: Use the lens formula to find the focal length when object and image distances are known.

Paraxial rays are rays that are close to and nearly parallel to the principal axis. Hence, the angle between paraxial rays and the principal axis is effectively 0°. Quick Tip: Paraxial rays are close to the principal axis and make a very small angle (effectively 0°) with it.

The splitting of white light into its component colors (VIBGYOR) is called dispersion. This occurs when light passes through a prism, causing the different wavelengths of light to refract by different amounts. Quick Tip: Dispersion occurs when light is separated into its different wavelengths or colors, such as in a prism.

Chlorine gains an electron to become chloride ion (\( Cl^- \)), not a cation. Quick Tip: Anions are formed when atoms gain electrons, and cations are formed when atoms lose electrons.

The element \( X^{27} \) has atomic number 13, meaning it has 13 protons. To form an ionic compound, it will lose 3 electrons to achieve a stable configuration, resulting in a charge of +3. Quick Tip: The charge on an element in an ionic compound is determined by the number of electrons it gains or loses to achieve a stable electron configuration.

Linus Pauling is known for his work on the nature of chemical bonds, including the ionic bond theory. Quick Tip: Linus Pauling contributed greatly to understanding ionic bonds and their role in chemical bonding.

The \( O^{2-} \) ion has gained two electrons, making its electronic configuration \( 1s^2, 2s^2, 2p^6 \), similar to the noble gas neon. Quick Tip: The electronic configuration of an ion depends on whether it gains or loses electrons to achieve a stable configuration.

Non-metals gain electrons to achieve a stable configuration, and the number of electrons gained is equal to the group number in the periodic table. Quick Tip: The group number of a non-metal in the periodic table tells you the number of electrons it gains to complete its outer shell.

The reactivity of metals increases from calcium to potassium. Calcium is less reactive than sodium, and sodium is less reactive than potassium. Quick Tip: The reactivity of alkali metals increases as you move down the group. Thus, potassium is the most reactive.

Poling is a process used to purify metals by passing a current through the crude metal to remove impurities. Quick Tip: Poling is used to purify metals by passing a current through the metal.

When silver corrodes, it reacts with hydrogen sulfide in the air, forming silver sulphide, which is responsible for the tarnishing of silver. Quick Tip: Silver tarnishes due to the formation of silver sulphide when it reacts with hydrogen sulfide.

During corrosion, the metal loses electrons (oxidation) and forms a metal oxide. Quick Tip: In corrosion, metals undergo oxidation by losing electrons to form metal oxides.

When one hydrogen of ammonia (NH\(_3\)) is replaced by an alkyl group, an amine is formed. Quick Tip: Replacing a hydrogen atom from ammonia by an alkyl group forms an amine.

The simplest ketone has the structure \( CH_3 C = O CH_3 \), also known as acetone. Quick Tip: Ketones have a carbonyl group (\( C = O \)) attached to two carbon atoms.

Ethene (C\(_2\)H\(_4\)) has a double bond between carbon atoms, while ethyne (C\(_2\)H\(_2\)) has a triple bond between the carbon atoms. Thus, they differ in the number of bonds. Quick Tip: Ethene has a double bond, while ethyne has a triple bond between carbon atoms.

Dimethyl ether has the structure \( CH_3 O CH_3 \), consisting of two methyl groups attached to an oxygen atom. Quick Tip: Dimethyl ether consists of two methyl groups connected by an oxygen atom.

Saturated hydrocarbons (alkanes) contain only single bonds between carbon atoms. Quick Tip: Saturated hydrocarbons have only single bonds between carbon atoms, such as in alkanes.

Aliphatic hydrocarbons include both acyclic (open chain) and cyclic hydrocarbons. Quick Tip: Aliphatic hydrocarbons are those that consist of open chains or rings, such as alkanes, alkenes, and alkynes.

Both turmeric solution and litmus can be used as acid-base indicators. Turmeric turns red in acidic solutions and yellow in basic solutions. Litmus solution turns red in acidic conditions and blue in basic conditions. Quick Tip: Turmeric and litmus are common natural indicators for detecting the acidic or basic nature of solutions.

Rainwater with a pH less than 5.6 is considered acid rain, caused by the dissolution of sulfur dioxide and nitrogen oxides from the atmosphere. Quick Tip: Acid rain occurs when the pH of rainwater falls below 5.6 due to the presence of acidic compounds.

Tooth enamel is primarily composed of calcium phosphate, which provides it with hardness and strength. Quick Tip: Tooth enamel is made of calcium phosphate, which makes it highly resistant to wear.

Potassium hydroxide is a strong base, which turns red litmus paper blue, but does not affect the blue litmus paper, leaving it blue. Quick Tip: In the presence of a base, red litmus paper turns blue, while blue litmus paper remains unchanged.

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

The 2d orbital does not exist because the d orbitals start from the 3rd shell (n = 3). Thus, \( 2d^3 \) is not possible. Quick Tip: There is no d orbital in the second shell. d orbitals start from the third shell.

Niels Bohr was awarded the Nobel Prize in Physics for his work on atomic structure and quantum theory. Quick Tip: Niels Bohr won the Nobel Prize in Physics for his development of the Bohr model of the atom.

The 4d subshell has 5 degenerate orbitals, each of which can hold 2 electrons, giving a total of 10 electrons. Quick Tip: Each d subshell has 5 degenerate orbitals, and the maximum number of electrons that can occupy them is 10.

Hund's rule states that electrons will fill degenerate orbitals singly before pairing. This explains the presence of 3 unpaired electrons in nitrogen. Quick Tip: Hund's rule explains the arrangement of electrons in orbitals, stating that electrons occupy degenerate orbitals singly first.

Ionic bonds are typically formed between elements from Group I (alkali metals) and Group VII (halogens) because they have a strong tendency to transfer electrons. Quick Tip: Ionic bonds are formed between metals of Group I (alkali metals) and non-metals of Group VII (halogens).

The electron configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ (2,8,8) indicates the configuration of ions like P³⁻, Cl⁻, and S²⁻, which have 18 electrons. Quick Tip: Ions like P³⁻, Cl⁻, and S²⁻ share the same electron configuration of 2,8,8.

Doberreiner’s triads consist of three elements, where the atomic weight of the middle element is the average of the other two. Lithium, sodium, and potassium are examples of such a triad. Quick Tip: Doberreiner’s triads consist of three elements where the atomic weight of the middle element is the average of the other two.

The elements in Group IV A, such as carbon, are called the carbon family. Quick Tip: Group IV A elements are known as the carbon family, including carbon, silicon, germanium, etc.

An element in Group 2 (alkaline earth metals) has a valency of 2. Therefore, the valency of element \( X \) is 2. Quick Tip: Group 2 elements (alkaline earth metals) have a valency of 2.

Valence Bond Theory was proposed by Linus Pauling, which describes how atoms form bonds through the overlap of their atomic orbitals. Quick Tip: Valence Bond Theory, proposed by Pauling, explains the formation of chemical bonds by orbital overlap.