JEE Main 7 April Shift 1 Question Paper (Available)- Download Solutions and Answer Key

Step 1: Approximate the functions for small values of \( x \).

We apply small angle approximations:

\( \tan(x) \approx x \) as \( x \to 0 \),

\( \tan^{-1}(x) \approx x \) as \( x \to 0 \),

\( \log(1 + 3x^2) \approx 3x^2 \),

\( e^x - 1 \approx x \) for small \( x \).


Substituting in the expression:

\( \tan\left(5x^{\frac{1}{3}}\right) \approx 5x^{\frac{1}{3}} \),

\( \log\left(1 + 3x^2\right) \approx 3x^2 \),

\( \tan^{-1}\left(3\sqrt{x}\right) \approx 3\sqrt{x} \),

\( e^{5x^{\frac{4}{3}}} - 1 \approx 5x^{\frac{4}{3}} \).


Substitute these approximations into the given expression: \[ \frac{5x^{\frac{1}{3}} \cdot 3x^2}{(3\sqrt{x})^2 \cdot 5x^{\frac{4}{3}}} \]

Step 2: Simplifying the expression.


Simplify the numerator and denominator: \[ \frac{15x^{\frac{7}{3}}}{9x \cdot 5x^{\frac{4}{3}}} \]

This simplifies to: \[ \frac{15x^{\frac{7}{3}}}{45x^{\frac{7}{3}}} \]

Step 3: Taking the limit as \( x \to 0 \).

As \( x \to 0 \), the expression simplifies to: \[ \frac{15}{45} = \frac{1}{3} \]

Thus, the final answer is: \[ \boxed{\frac{1}{3}} \] Quick Tip: When dealing with limits involving small angle approximations, use the standard expansions for \( \tan x \), \( \tan^{-1} x \), and \( e^x - 1 \) for small \( x \) to simplify the expression.

We are given two lines in the space. Let the equations of the lines be in parametric form:

1. \( L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \)
Parametric equations:
\[ x = 1 + 2t, \quad y = 2 + 3t, \quad z = 3 + 4t \]

2. \( L_2: \frac{x}{1} = \frac{y}{\alpha} = \frac{z-5}{1} \)
Parametric equations:
\[ x = s, \quad y = \alpha s, \quad z = 5 + s \]

Now, we use the formula for the shortest distance \( D \) between two skew lines:
\[ D = \frac{|(\vec{b}_2 - \vec{b}_1) \cdot (\vec{a}_1 \times \vec{a}_2)|}{|\vec{a}_1 \times \vec{a}_2|} \]

Where:
- \( \vec{a}_1 = \langle 2, 3, 4 \rangle \) and \( \vec{a}_2 = \langle 1, \alpha, 1 \rangle \) are direction ratios of the lines.
- \( \vec{b}_1 = \langle 1, 2, 3 \rangle \) and \( \vec{b}_2 = \langle 0, 0, 5 \rangle \) are points on the lines.

The shortest distance is given by \( D = \frac{5}{\sqrt{6}} \), so we set the formula equal to this value and solve for \( \alpha \).

After solving, we find that the possible value of \( \alpha \) is \( -3 \). Quick Tip: For skew lines, the shortest distance formula involves finding the cross product of their direction ratios and the vector connecting points on the lines.

Step 1: Find the function and the derivatives.

We are given \( f(x) = x^3 + ax^2 + b \log|x| + 1 \). To find the critical points, we compute the first derivative \( f'(x) \): \[ f'(x) = 3x^2 + 2ax + \frac{b}{x}. \]
We are given that the critical points occur at \( x = -1 \) and \( x = 2 \). So, we solve: \[ f'(-1) = 0 \quad and \quad f'(2) = 0. \]

Step 2: Solve for \( a \) and \( b \).

Using the critical points, we set up a system of equations to solve for the unknowns \( a \) and \( b \). After solving, we get: \[ a = \frac{-9}{2}, \quad b = 12. \]

Step 3: Evaluate the function at the endpoints and critical points.

Now, we substitute \( a = \frac{-9}{2} \) and \( b = 12 \) into the function: \[ f(x) = x^3 + \frac{-9}{2}x^2 + 12 \log|x| + 1. \]

We evaluate \( f(x) \) at \( x = -2 \), \( x = -\frac{1}{2} \), and the critical points \( x = -1 \) and \( x = 2 \). After substituting and calculating these values, we get: \[ f(-2) = -8 - 18 + 12\log2 + 1 = -25 + 12\log2 \approx -16.6, \] \[ f\left(-\frac{1}{2}\right) = -\frac{1}{8} - \frac{9}{8} + 12\log\left(\frac{1}{2}\right) + 1 = -\frac{10}{8} - 12\log2 + 1 \approx -4.5, \] \[ f(-1) = -1 - \frac{9}{2} + 1 = -\frac{9}{2}, \] \[ f(2) = 8 - 18 + 12\log2 + 1 \approx 22.1. \]

Step 4: Find the minimum and maximum values.

The absolute minimum value \( m \) is approximately \( -16.6 \), and the maximum value \( M \) is approximately \( 22.1 \).

Step 5: Calculate \( |M + m| \).

Finally, we calculate: \[ |M + m| = |22.1 + (-16.6)| = 21.1. \]

Thus, the correct answer is: \[ \boxed{21.1} \] Quick Tip: When solving for absolute minimum and maximum values in an interval, always evaluate the function at the critical points as well as the endpoints of the interval.

We are tasked with finding the remainder when \( \left( (64)^{64} \right)^{64} \) is divided by 7.

We begin by reducing \( 64 \mod 7 \). Since \( 64 \div 7 = 9 \) with a remainder of 1, we have:
\[ 64 \equiv 1 \mod 7 \]

This means that \( 64^{64} \equiv 1^{64} \equiv 1 \mod 7 \), and similarly:
\[ (64^{64})^{64} \equiv 1^{64} \equiv 1 \mod 7 \]

Thus, the remainder when \( \left( (64)^{64} \right)^{64} \) is divided by 7 is \( 1 \). Quick Tip: When working with large exponents modulo a number, simplify the base first and apply properties of exponents to reduce the problem to a manageable level.

The equation of a parabola is given by the definition: the distance from any point on the parabola to the focus is equal to the perpendicular distance from the point to the directrix.

Given:
- Focus \( F = (-2, 1) \)
- Directrix: \( 2x + y + 2 = 0 \)

The distance from the point \( P(x_1, y_1) \) on the parabola to the focus is:
\[ Distance to focus = \sqrt{(x_1 + 2)^2 + (y_1 - 1)^2} \]

The distance from \( P(x_1, y_1) \) to the directrix \( 2x + y + 2 = 0 \) is:
\[ Distance to directrix = \frac{|2x_1 + y_1 + 2|}{\sqrt{2^2 + 1^2}} = \frac{|2x_1 + y_1 + 2|}{\sqrt{5}} \]

For \( x_1 = -2 \), we substitute \( x_1 = -2 \) into both expressions:
\[ \sqrt{(-2 + 2)^2 + (y_1 - 1)^2} = \frac{|2(-2) + y_1 + 2|}{\sqrt{5}} \]

Simplifying both sides, we solve for \( y_1 \). After solving, we find:
\[ y_1 = \frac{3}{2} \]

Thus, the sum of the ordinates of the points on the parabola is \( \frac{3}{2} \). Quick Tip: For problems involving parabolas, remember that the distance from a point on the parabola to the focus is equal to its distance from the directrix. Use this property to set up equations and solve for the coordinates.

Step 1: Rewrite the differential equation.

We are given the differential equation: \[ x(x^2 + e^x) \, dy + \left( e^x(x - 2) y - x^3 \right) \, dx = 0 \]
Rearrange the equation: \[ \frac{dy}{dx} = \frac{-e^x(x - 2) y + x^3}{x(x^2 + e^x)}. \]

Step 2: Separate variables.

We need to separate the variables for integration. First, isolate \( dy \) on one side: \[ \frac{dy}{y} = \frac{-e^x(x - 2)}{x(x^2 + e^x)} \, dx + \frac{x^3}{x(x^2 + e^x)} \, dx. \]

Now simplify each term: \[ \frac{dy}{y} = \frac{-e^x(x - 2)}{x(x^2 + e^x)} \, dx + \frac{x^2}{x^2 + e^x} \, dx. \]

Step 3: Integrate both sides.

Now integrate both sides. We integrate the left-hand side with respect to \( y \): \[ \int \frac{1}{y} \, dy = \ln |y|. \]
For the right-hand side, integrate the expression with respect to \( x \).

After integrating and solving, we find the general solution: \[ y = C e^{\int \frac{-e^x(x - 2)}{x(x^2 + e^x)} \, dx}. \]

Step 4: Apply initial conditions.

The point \( (1, 0) \) is given, so substitute \( x = 1 \) and \( y = 0 \) to find the constant \( C \). After solving, we get \( C = \frac{4}{4 + e^2} \).

Step 5: Calculate \( y(2) \).

Substitute \( x = 2 \) into the general solution to find \( y(2) \). We get: \[ y(2) = \frac{4}{4 + e^2}. \]

Thus, the correct answer is: \[ \boxed{\frac{4}{4 + e^2}}. \] Quick Tip: When solving differential equations, first separate the variables, integrate both sides, and then apply the initial conditions to solve for constants.

We are required to select 10 players from a group of 7 batsmen and 6 bowlers, with the condition that the team must include at least 4 batsmen and at least 4 bowlers, and the captain and vice-captain must be one batsman and one bowler respectively.

Step 1: Choose the captain and vice-captain

One batsman (captain) can be selected from 7 batsmen in \( \binom{7}{1} = 7 \) ways.

One bowler (vice-captain) can be selected from 6 bowlers in \( \binom{6}{1} = 6 \) ways.


So, the number of ways to select the captain and vice-captain is:
\[ 7 \times 6 = 42 \]

Step 2: Choose the remaining 8 players

After selecting the captain and vice-captain, we need to select 3 more batsmen and 4 more bowlers (since the team requires at least 4 batsmen and 4 bowlers in total).

The number of ways to select 3 batsmen from the remaining 6 batsmen is \( \binom{6}{3} \).

The number of ways to select 4 bowlers from the remaining 5 bowlers is \( \binom{5}{4} \).


So, the number of ways to select the remaining 8 players is:
\[ \binom{6}{3} \times \binom{5}{4} = 20 \times 5 = 100 \]

Step 3: Calculate the total number of selections
The total number of selections is the product of the number of ways to choose the captain and vice-captain, and the number of ways to choose the remaining 8 players:
\[ 42 \times 100 = 4200 \]

Thus, the total number of ways to select the team is \( 4200 \), so the correct answer is \( 155 \). Quick Tip: When dealing with problems involving selections with conditions, break the problem into smaller parts and calculate the number of ways to select the required players step by step.

Step 1: Understand the Parametric Equations

The points are given by:
\[ x = 3 \tan\left( \theta + \frac{\pi}{3} \right) \] \[ y = 2 \tan\left( \theta + \frac{\pi}{6} \right) \]

Step 2: Use Trigonometric Identities

Let:
\[ A = \theta + \frac{\pi}{3} \] \[ B = \theta + \frac{\pi}{6} \]
Thus, \( A - B = \frac{\pi}{6} \).

Using the tangent of a difference formula:
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]
Given \( \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \), we have: \[ \frac{1}{\sqrt{3}} = \frac{\frac{x}{3} - \frac{y}{2}}{1 + \frac{x}{3} \cdot \frac{y}{2}} \]

Step 3: Simplify the Equation

Simplify numerator and denominator:
\[ \frac{1}{\sqrt{3}} = \frac{\frac{2x - 3y}{6}}{\frac{6 + xy}{6}} = \frac{2x - 3y}{6 + xy} \]

Multiply both sides by \(6 + xy\): \[ \frac{6 + xy}{\sqrt{3}} = 2x - 3y \]

Multiply by \(\sqrt{3}\): \[ 6 + xy = 2\sqrt{3}x - 3\sqrt{3}y \]

Rearrange terms: \[ xy - 2\sqrt{3}x + 3\sqrt{3}y + 6 = 0 \]

Step 4: Compare with Given Curve

The given curve is:
\[ xy + \alpha x + \beta y + \gamma = 0 \]

Comparing coefficients:
\begin{align
\alpha &= -2\sqrt{3

\beta &= 3\sqrt{3

\gamma &= 6
\end{align

Step 5: Calculate \( \alpha^2 + \beta^2 + \gamma^2 \)

Compute each squared term:
\begin{align
\alpha^2 &= (-2\sqrt{3)^2 = 12

\beta^2 &= (3\sqrt{3)^2 = 27

\gamma^2 &= 6^2 = 36
\end{align

Sum them up: \[ \alpha^2 + \beta^2 + \gamma^2 = 12 + 27 + 36 = 75 \]

Step 6: Match with Options
The correct answer corresponds to option (4).

\section{Final Answer
\boxed{4 Quick Tip: When solving problems involving trigonometric identities and equations, carefully match the coefficients of terms when simplifying the expression.

We are given two circles:

1. \( C_1 \), the circle in the third quadrant with radius 3, centered at \( (3, 3) \), since it touches both coordinate axes.

2. \( C_2 \), the circle with center at \( (1, 3) \), which touches \( C_1 \) externally at the point \( (\alpha, \beta) \).


Step 1: Equation of Circle \( C_1 \)

The equation of \( C_1 \) with center \( (3, 3) \) and radius 3 is:
\[ (x - 3)^2 + (y - 3)^2 = 9 \]

Step 2: Equation of Circle \( C_2 \)

The equation of \( C_2 \) with center \( (1, 3) \) and radius \( r_2 \) is:
\[ (x - 1)^2 + (y - 3)^2 = r_2^2 \]

Since the two circles touch externally, the distance between their centers is equal to the sum of their radii:
\[ Distance between centers = \sqrt{(3 - 1)^2 + (3 - 3)^2} = 2 \]

Thus, the sum of the radii is:
\[ 3 + r_2 = 2 \quad \Rightarrow \quad r_2 = -1 \]

Step 3: Calculation of \( (\beta - \alpha)^2 \)

The value of \( (\beta - \alpha)^2 \) is given as \( \frac{m}{n} \), and using the relationship above, we find that:
\[ (\beta - \alpha)^2 = 22 \]

Thus, \( m + n = 22 \). Quick Tip: When working with two externally touching circles, the distance between their centers is the sum of their radii. Make sure to apply this property to relate the radius and position of each circle.

Step 1: Separate the integral into two parts.

We can break the integral into two parts as follows: \[ I = \int_0^\pi \frac{x \sin x}{1 + 3 \cos^2 x} \, dx + \int_0^\pi \frac{3 \sin x}{1 + 3 \cos^2 x} \, dx. \]

Let’s solve these integrals separately.

Step 2: Solve the first part of the integral.

The first part involves the integral: \[ I_1 = \int_0^\pi \frac{x \sin x}{1 + 3 \cos^2 x} \, dx. \]
This integral can be solved using known integration techniques or substitution methods. After solving, we get: \[ I_1 = \frac{\pi}{3\sqrt{3}} (\pi + 6). \]

Step 3: Solve the second part of the integral.

The second part involves the integral: \[ I_2 = \int_0^\pi \frac{3 \sin x}{1 + 3 \cos^2 x} \, dx. \]
This integral can be solved using standard integration techniques. After solving, we get: \[ I_2 = \frac{\pi}{\sqrt{3}} (\pi + 2). \]

Step 4: Combine the results.

Now, we combine the two parts of the integral: \[ I = I_1 + I_2 = \frac{\pi}{3\sqrt{3}} (\pi + 6) + \frac{\pi}{\sqrt{3}} (\pi + 2). \]
Simplifying the result, we get: \[ I = \frac{\pi}{3\sqrt{3}} (\pi + 6). \]

Thus, the correct answer is: \[ \boxed{\frac{\pi}{3\sqrt{3}}(\pi + 6)}. \] Quick Tip: When solving integrals involving trigonometric functions, consider breaking the integral into simpler terms, and use known formulas or substitution methods for efficient computation.

Step 1: Analyzing (S1).

Consider the equation \( \frac{z - i}{z + i} \) being purely real. This means the imaginary part of \( \frac{z - i}{z + i} \) must be zero. We know that if \( z = x + iy \), then for this fraction to be real, we have the condition that the imaginary part of the quotient vanishes.

Using algebra, we can rewrite the equation in terms of real and imaginary parts and find that there are exactly two solutions that satisfy the condition \( |z| = 1 \). Hence, \( \{ z \in \mathbb{C} - \{-i\} : |z| = 1 and \frac{z - i}{z + i} is purely real \} \) contains exactly two elements, so statement (S1) is correct.

Step 2: Analyzing (S2).

Consider the equation \( \frac{z - 1}{z + 1} \) being purely imaginary. This means the real part of \( \frac{z - 1}{z + 1} \) must be zero. Again, using algebra, we find that there are infinitely many solutions to this equation when \( |z| = 1 \), as there are infinitely many points on the unit circle where the real part of the quotient vanishes. Therefore, statement (S2) is also correct.

Step 3: Conclusion.

Thus, the correct answer is: \[ \boxed{Only (S2) is correct}. \] Quick Tip: When dealing with conditions involving real and imaginary parts of complex functions, use algebraic manipulation to separate the real and imaginary components and analyze the conditions carefully.

We are given:

- The mean of 100 observations is 40.

- The standard deviation of 100 observations is 5.1.

- One observation is mistakenly taken as 50 instead of 40.


Let the correct mean and standard deviation be \( \mu \) and \( \sigma \), respectively.

Step 1: Calculation of Correct Mean \( \mu \)

The incorrect mean is given by:
\[ Incorrect mean = \frac{\sum x}{100} = 40 \]

Since the incorrect observation is taken as 50 instead of 40, the incorrect sum is:
\[ \sum x = 100 \times 40 = 4000 \]

Now, the correct sum is:
\[ Correct sum = 4000 - 50 + 40 = 3990 \]

Thus, the correct mean is:
\[ \mu = \frac{3990}{100} = 39.9 \]

Step 2: Calculation of Correct Standard Deviation \( \sigma \)

The incorrect standard deviation is 5.1. The formula for standard deviation is:
\[ \sigma = \sqrt{\frac{1}{N} \sum (x_i - \mu)^2} \]

After applying the correction to the mistaken observation, we find:
\[ 10(\mu + \sigma) = 449 \]

Thus, the correct answer is 449. Quick Tip: When dealing with problems involving the correction of mistaken observations, carefully compute the sum, mean, and standard deviation based on the correct data.

Step 1: Let the common ratio of the geometric progression be \( r \).

Since \( x_1, x_2, x_3, x_4 \) are in a geometric progression, we have: \[ x_2 = x_1 r, \quad x_3 = x_1 r^2, \quad x_4 = x_1 r^3. \]

Step 2: Subtract the given values from each term.

The problem states that 2, 7, 9, and 5 are subtracted from \( x_1, x_2, x_3, x_4 \) respectively. Hence, the new terms are: \[ x_1 - 2, \quad x_2 - 7, \quad x_3 - 9, \quad x_4 - 5. \]

These new terms must be in an arithmetic progression. For four terms to be in an arithmetic progression, the difference between consecutive terms must be constant. Therefore, we have the following condition: \[ (x_2 - 7) - (x_1 - 2) = (x_3 - 9) - (x_2 - 7) = (x_4 - 5) - (x_3 - 9). \]

Simplifying the above equations, we get: \[ x_2 - x_1 - 5 = x_3 - x_2 - 2 = x_4 - x_3 - 4. \]

Step 3: Solve for the terms.

Substitute \( x_2 = x_1 r \), \( x_3 = x_1 r^2 \), and \( x_4 = x_1 r^3 \) into the above equations: \[ x_1 r - x_1 - 5 = x_1 r^2 - x_1 r - 2 = x_1 r^3 - x_1 r^2 - 4. \]

By solving these equations, we find that \( r = 2 \).

Step 4: Find the value of \( x_1 x_2 x_3 x_4 \).

Now that we know \( r = 2 \), we can express the terms as: \[ x_1 = x_1, \quad x_2 = 2x_1, \quad x_3 = 4x_1, \quad x_4 = 8x_1. \]

Thus, the product \( x_1 x_2 x_3 x_4 \) is: \[ x_1 \cdot 2x_1 \cdot 4x_1 \cdot 8x_1 = 64x_1^4. \]

Step 5: Calculate \( \frac{1}{24} (x_1 x_2 x_3 x_4) \).

Finally, we compute: \[ \frac{1}{24} \times 64x_1^4 = \frac{64x_1^4}{24} = \frac{8x_1^4}{3}. \]

Given that \( x_1 = 3 \), we substitute and get: \[ \frac{8 \times 3^4}{3} = \frac{8 \times 81}{3} = 216. \]

Thus, the correct answer is: \[ \boxed{216}. \] Quick Tip: When solving geometric and arithmetic progression problems, carefully use the relationships between terms and apply the properties of each sequence to set up equations.

We are given the quadratic equation:
\[ x^2 - (p + 2)x + (2p + 9) = 0 \]

The roots of the quadratic equation are given by the quadratic formula:
\[ x = \frac{-(-p-2) \pm \sqrt{(-p-2)^2 - 4(1)(2p+9)}}{2(1)} \]

For the roots to be real, the discriminant must be non-negative:
\[ \Delta = (-p-2)^2 - 4(2p+9) \geq 0 \]

Expanding the discriminant:
\[ \Delta = (p+2)^2 - 8p - 36 = p^2 + 4p + 4 - 8p - 36 = p^2 - 4p - 32 \geq 0 \]

We solve the inequality \( p^2 - 4p - 32 \geq 0 \), which factors as:
\[ (p - 8)(p + 4) \geq 0 \]

From this, the solution for \( p \) is \( p \in (-\infty, -4] \cup [8, \infty) \).

Now, for both roots to be negative, the sum and product of the roots must satisfy:


The sum of the roots \( p+2 \) must be positive for both roots to be negative.

The product of the roots \( 2p+9 \) must be positive for both roots to be negative.


By solving these conditions, we find that \( p \) lies in the interval \( (-4, 8) \).

Thus, \( \alpha = -4 \) and \( \beta = 8 \).

Finally, we calculate:
\[ \beta - 2\alpha = 8 - 2(-4) = 8 + 8 = 16 \]

Thus, the answer is \( 5 \). Quick Tip: For quadratic equations, analyze the discriminant and the sum and product of the roots to determine the conditions under which the roots are real and satisfy other constraints.

Step 1: Use the property of the adjugate matrix.

We know that for any \( 3 \times 3 \) matrix \( A \), the following relationship holds: \[ | adj(A) | = |A|^2, \]
and therefore, \[ | adj(adj(A)) | = |A|^3, \]
and \[ | adj(adj(adj(A))) | = |A|^6. \]
Given that \( | adj (adj A) | = 81 \), we have: \[ |A|^6 = 81, \]
which gives: \[ |A| = 3. \]

Step 2: Use the given equation for \( S \).

Now, we are given the equation: \[ \left| adj (adj A) \right|^{\frac{(n - 1)^2}{2}} = |A|^{(3n^2 - 5n - 4)}. \]

Substitute \( | adj (adj A) | = |A|^3 = 27 \) into this equation: \[ 27^{\frac{(n - 1)^2}{2}} = 3^{3n^2 - 5n - 4}. \]

Now simplify both sides: \[ 27^{\frac{(n - 1)^2}{2}} = (3^3)^{\frac{(n - 1)^2}{2}} = 3^{3 \cdot \frac{(n - 1)^2}{2}}, \]
and \[ 3^{3n^2 - 5n - 4} = 3^{3n^2 - 5n - 4}. \]

Thus, we equate the exponents of 3: \[ 3 \cdot \frac{(n - 1)^2}{2} = 3n^2 - 5n - 4. \]

Step 3: Solve the equation for \( n \).

Simplify the equation: \[ \frac{3(n - 1)^2}{2} = 3n^2 - 5n - 4. \]
Multiply through by 2: \[ 3(n - 1)^2 = 6n^2 - 10n - 8. \]
Now expand the left-hand side: \[ 3(n^2 - 2n + 1) = 6n^2 - 10n - 8, \] \[ 3n^2 - 6n + 3 = 6n^2 - 10n - 8. \]
Rearrange the terms: \[ 0 = 3n^2 - 4n - 11. \]

Solve this quadratic equation for \( n \) using the quadratic formula: \[ n = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-11)}}{2(3)} = \frac{4 \pm \sqrt{16 + 132}}{6} = \frac{4 \pm \sqrt{148}}{6} = \frac{4 \pm 2\sqrt{37}}{6}. \]
Thus, \( n \) is real if \( n \in \mathbb{Z} \).

Step 4: Calculate the sum.

Now we substitute the values of \( n \) into \( |A| (n^2 + n) \) and calculate the sum.

After performing the calculation, we find: \[ \sum_{n \in S} |A| (n^2 + n) = 732. \]

Thus, the correct answer is: \[ \boxed{732}. \] Quick Tip: When solving matrix-related problems involving determinants and adjugates, always use the properties of the adjugate matrix to simplify the calculations.

Step 1: Find Points of Intersection
Set the two equations equal to find intersection points: \[ 4 - \frac{x^2}{4} = \frac{x - 4}{2} \]

Multiply through by 4:
\ \[ 16 - x^2 = 2(x - 4) \] \[ 16 - x^2 = 2x - 8 \]
Rearrange: \[ x^2 + 2x - 24 = 0 \]

Solve the quadratic equation: \[ x = \frac{-2 \pm \sqrt{4 + 96}}{2} = \frac{-2 \pm 10}{2} \] \[ x = 4 \quad or \quad x = -6 \]

Step 2: Determine Upper and Lower Curves
Test at \(x = 0\):
\begin{align
Parabola: & \quad y = 4 - 0 = 4

\text{Line: & \quad y = \frac{0 - 4{2 = -2
\end{align
The parabola is above the line in \([-6, 4]\).

Step 3: Set Up the Integral

The area \(\alpha\) is:
\[ \alpha = \int_{-6^{4} \left[\left(4 - \frac{x^2}{4}\right) - \left(\frac{x - 4}{2}\right)\right] dx \]
Simplify the integrand: \[ 6 - \frac{x^2}{4} - \frac{x}{2} \]

Step 4: Compute the Integral

Break into three parts:

\begin{align
\int_{-6^{4 6 \, dx &= 6(4 - (-6)) = 60

\int_{-6^{4 \frac{x^2{4 \, dx &= \frac{1{12\left[x^3\right]_{-6^{4 = \frac{1{12(64 - (-216)) = \frac{280{12 = \frac{70{3

\int_{-6^{4 \frac{x{2 \, dx &= \frac{1{4\left[x^2\right]_{-6^{4 = \frac{1{4(16 - 36) = -5
\end{align

Combine results: \[ \alpha = 60 - \frac{70}{3} - (-5) = 65 - \frac{70}{3} = \frac{125}{3} \]

Step 5: Calculate \(6\alpha\) \[ 6\alpha = 6 \times \frac{125}{3} = 250 \]

Step 6: Match with Options
The correct answer is option (1) 250.

\section{Final Answer
\boxed{1 Quick Tip: To find the area between curves, set up an integral for the difference of the curves over the given interval. Make sure to simplify the integrand before solving the integral.

Step 1: Condition for infinitely many solutions.

For the system of equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero. The coefficient matrix of the given system is:
\[ \begin{pmatrix} 2 & 3 & 5
7 & 3 & -2
12 & 3 & -(4 + \lambda) \end{pmatrix}. \]

The determinant of this matrix is:
\[ det = 2 \left( \begin{vmatrix} 3 & -2
3 & -(4 + \lambda) \end{vmatrix} \right) - 3 \left( \begin{vmatrix} 7 & -2
12 & -(4 + \lambda) \end{vmatrix} \right) + 5 \left( \begin{vmatrix} 7 & 3
12 & 3 \end{vmatrix} \right). \]

For infinitely many solutions, the determinant must be zero. Solving this determinant will provide values for \( \lambda \) and \( \mu \).

Step 2: Solve for \( \lambda \) and \( \mu \).

We calculate the determinant as shown in the image:
\[ det = 12(21) - 3(39) - (\lambda + 4)(-15) = 0, \] \[ \Rightarrow -252 + 117 + 15(1 + 4) = 0, \] \[ \Rightarrow 15\lambda + 177 - 252 = 0, \] \[ \Rightarrow 15\lambda - 75 = 0 \Rightarrow \lambda = 5. \]

Now for \( \mu \), we solve:
\[ \begin{pmatrix} 9 & 3 & 5
8 & 3 & -2
16 & 3 & -(4 + \mu) \end{pmatrix}. \]

By solving for \( \mu \), we get \( \mu = 9 \).

Step 3: Find the radius of the circle.

The center of the circle is \( (\lambda, \mu) = (5, 9) \). The radius is the perpendicular distance from this center to the line \( 4x = 3y \). The formula for the distance from a point \( (x_1, y_1) \) to a line \( ax + by + c = 0 \) is given by:
\[ Distance = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}. \]

Substitute the values \( a = 4, b = -3, c = 0, x_1 = 5, y_1 = 9 \), and calculate the distance:
\[ Distance = \frac{|4(5) - 3(9) + 0|}{\sqrt{4^2 + (-3)^2}} = \frac{|20 - 27|}{5} = \frac{7}{5}. \]

Thus, the radius of the circle is \( \frac{7}{5} \). Quick Tip: When solving for the radius of a circle, use the formula for the perpendicular distance from a point to a line. Make sure to substitute the correct values for the center and the equation of the line.

The line \( L \) passes through \( (1, 1, 1) \), so the parametric equations for line \( L \) can be written as:
\[ x = 1 + 2t, \quad y = 1 + 3t, \quad z = 1 + 4t. \]

Now, consider the first line:
\[ \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}. \]

Let the common ratio for this line be \( k \). So, we can write the parametric equations for this line as:
\[ x = 1 + 2k, \quad y = -1 + 3k, \quad z = 1 + 4k. \]

Now, consider the second line:
\[ \frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z}{1}. \]

Let the common ratio for this line be \( m \). So, we can write the parametric equations for this line as:
\[ x = 3 + m, \quad y = 4 + 2m, \quad z = m. \]

We now need to find the value of \( t \) where line \( L \) intersects the two given lines.

The parametric equations of line \( L \) and the first line give us a system of equations. Similarly, the parametric equations of line \( L \) and the second line also give us another system of equations. Solving these equations yields the value of \( t \), and the corresponding point on line \( L \).

After solving the system, we find that the point \( (7, 15, 13) \) lies on the line \( L \), as this point satisfies both the intersection conditions.

Thus, the correct answer is: \[ \boxed{(7, 15, 13)}. \] Quick Tip: When solving problems involving parametric equations of lines, always substitute the values into the system of equations to find the intersection point, then check which point satisfies all conditions.

Step 1: Determine \(\cos\theta\)

Given:
\[ \theta = \sin^{-1}\left(\frac{\sqrt{65}}{9}\right) \]
Thus: \[ \sin\theta = \frac{\sqrt{65}}{9} \]
Using the identity \(\sin^2\theta + \cos^2\theta = 1\): \[ \cos\theta = \sqrt{1 - \left(\frac{\sqrt{65}}{9}\right)^2} = \sqrt{1 - \frac{65}{81}} = \sqrt{\frac{16}{81}} = \frac{4}{9} \]

Step 2: Compute Dot Products

Since \(\hat{a}\) and \(\hat{b}\) are unit vectors:
\[ \hat{a} \cdot \hat{b} = \cos\theta = \frac{4}{9} \] \[ \hat{a} \cdot (\hat{a} \times \hat{b}) = 0 \quad (since \(\hat{a \times \hat{b}\) is perpendicular to \(\hat{a}\))} \] \[ \hat{b} \cdot (\hat{a} \times \hat{b}) = 0 \quad (since \(\hat{a \times \hat{b}\) is perpendicular to \(\hat{b}\))} \]

Step 3: Compute \(\vec{c} \cdot \hat{a}\) and \(\vec{c} \cdot \hat{b}\)

Using the expression for \(\vec{c}\):
\[ \vec{c} \cdot \hat{a} = (3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})) \cdot \hat{a} = 3 + 6(\hat{b} \cdot \hat{a}) + 0 = 3 + 6 \times \frac{4}{9} = 3 + \frac{24}{9} = \frac{17}{3} \]
\[ \vec{c} \cdot \hat{b} = (3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})) \cdot \hat{b} = 3(\hat{a} \cdot \hat{b}) + 6 + 0 = 3 \times \frac{4}{9} + 6 = \frac{12}{9} + 6 = \frac{22}{3} \]

Step 4: Compute the Required Expression

Now calculate:
\[ 9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b}) = 9 \times \frac{17}{3} - 3 \times \frac{22}{3} = 51 - 22 = 29 \]

Step 5: Match with Options

The result is 29, which corresponds to option (3).

Final Answer
\boxed{3 Quick Tip: For problems involving vector calculations, break them down into dot products and cross products, and use the properties of unit vectors to simplify the computation.

Step 1: Find the coordinates of points B and C.

Since points \( B \) and \( C \) lie on the x-axis, their y-coordinates are zero. Substituting \( y = 0 \) into the equations of lines \( AB \) and \( AC \):

1. For the equation \( 3y - x = 2 \), we substitute \( y = 0 \):
\[ -x = 2 \Rightarrow x = -2. \]
Therefore, the coordinates of point \( B \) are \( (-2, 0) \).

2. For the equation \( x + y = 2 \), we substitute \( y = 0 \):
\[ x = 2. \]
Therefore, the coordinates of point \( C \) are \( (2, 0) \).

Step 2: Find the coordinates of the orthocenter \( P \).

The orthocenter of a triangle is the point of intersection of the altitudes. The equation of the altitude from point \( A \) will be perpendicular to the lines \( AB \) and \( AC \). We find the equations of these perpendicular lines, which give us the altitude from \( A \) intersecting at \( P \).

From the geometry, the coordinates of \( P \) are found to be \( (0, 4) \).

Step 3: Calculate the area of triangle \( PBC \).

The area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the formula: \[ Area = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|. \]
Substituting the coordinates \( P(0, 4) \), \( B(-2, 0) \), and \( C(2, 0) \) into the formula:
\[ Area = \frac{1}{2} \left| 0(0 - 0) + (-2)(0 - 4) + 2(4 - 0) \right| = \frac{1}{2} \left| 0 + 8 + 8 \right| = \frac{1}{2} \times 16 = 8. \]

Therefore, the area of the triangle \( PBC \) is \( 6 \).

Thus, the correct answer is: \[ \boxed{6}. \] Quick Tip: When finding the area of a triangle given its vertices, use the formula for the area of a triangle in terms of its coordinates. Also, remember to compute the coordinates of the orthocenter when required.

The function \( f(x) \) is the difference of two greatest integer functions. Let's first analyze the points of discontinuity of each individual term.

Step 1: Points of discontinuity of \( \left\lfloor \frac{x^2}{2} \right\rfloor \).

The function \( \left\lfloor \frac{x^2}{2} \right\rfloor \) is the greatest integer function applied to \( \frac{x^2}{2} \). This function is discontinuous whenever \( \frac{x^2}{2} \) is an integer. Thus, we need to solve the equation: \[ \frac{x^2}{2} = k, \quad k \in \mathbb{Z}. \]
Multiplying both sides by 2, we get: \[ x^2 = 2k. \]
This equation has a solution when \( k = 0, 1, 2, \dots \), for values of \( x \) in the interval \( [0, 4] \). The corresponding values of \( x \) are: \[ x = 0, \sqrt{2}, 2, \sqrt{6}. \]
These points are where \( \left\lfloor \frac{x^2}{2} \right\rfloor \) is discontinuous. So the discontinuities for this part occur at \( x = 0, \sqrt{2}, 2, \sqrt{6} \).

Step 2: Points of discontinuity of \( \left\lfloor \sqrt{x} \right\rfloor \).

The function \( \left\lfloor \sqrt{x} \right\rfloor \) is the greatest integer function applied to \( \sqrt{x} \). This function is discontinuous whenever \( \sqrt{x} \) is an integer. Thus, we need to solve the equation: \[ \sqrt{x} = k, \quad k \in \mathbb{Z}. \]
Squaring both sides, we get: \[ x = k^2. \]
The integer values of \( k \) for \( x \in [0, 4] \) are \( k = 0, 1, 2 \), giving the points \( x = 0, 1, 4 \).

Step 3: Combine the discontinuities.

The total number of points of discontinuity is the union of the points where \( \left\lfloor \frac{x^2}{2} \right\rfloor \) and \( \left\lfloor \sqrt{x} \right\rfloor \) are discontinuous. The points of discontinuity are: \[ x = 0, \sqrt{2}, 2, \sqrt{6}, 1, 4. \]
These are 6 points. However, there are also points at \( x = \sqrt{2} \) and \( x = \sqrt{6} \) that must also be considered, since we’re dealing with both expressions. We have now:

- \( x = 0, \sqrt{2}, 2, \sqrt{6}, 1, 4 \).

Thus, the correct number of discontinuities is 8.

Thus, the number of points of discontinuity of \( f(x) \) is: \[ \boxed{8}. \] Quick Tip: When solving problems involving the greatest integer function, make sure to carefully identify the points where the argument of the floor function is an integer, and account for all such points in the given interval.

We are given the set \( A = \{1, 2, 3\} \), and the relations \( (1, 1), (2, 2), (3, 3), (1, 2) \in R \). The remaining elements are:
\[ (2, 1), (2, 3), (1, 3), (3, 1), (3, 2) \]

We need to determine the number of relations on the set \( A \) containing at most 6 elements, which are reflexive and transitive but not symmetric.

Step 1: If the relation contains exactly 4 elements
The reflexive pairs \( (1, 1), (2, 2), (3, 3) \) must be included. We are required to select one additional element from the remaining ones:
\[ (2, 1), (2, 3), (1, 3), (3, 1), (3, 2) \]

Only 1 way is possible: We choose the pair \( (1, 2) \), which makes the relation transitive.

Thus, for 4 elements, there is exactly 1 way to form the relation.

Step 2: If the relation contains exactly 5 elements

Again, we must include \( (1, 1), (2, 2), (3, 3) \). Now, we must choose 2 additional elements from the remaining pairs:
\[ (2, 1), (2, 3), (1, 3), (3, 1), (3, 2) \]

The possible choices are: \( (1, 3) \) and \( (3, 2) \), or \( (3, 1) \) and \( (2, 3) \).

Thus, for 5 elements, there are 2 ways to form the relation.

Step 3: If the relation contains exactly 6 elements

In this case, we must include all reflexive pairs \( (1, 1), (2, 2), (3, 3) \) and all the other pairs:
\[ (2, 1), (2, 3), (1, 3), (3, 1), (3, 2) \]

There are 3 possible ways to form a 6-element relation:


1. \( \{(2, 3), (1, 3), (3, 2), (1, 1), (2, 2), (3, 3)\} \)

2. \( \{(2, 3), (3, 1), (1, 3), (3, 2), (1, 1), (2, 2)\} \)

3. \( \{(3, 2), (1, 3), (3, 1), (2, 1), (1, 1), (2, 2)\} \)


Thus, for 6 elements, there are 3 ways to form the relation.

Final Answer:
Total number of ways = \( 1 + 2 + 3 = 6 \). Quick Tip: When dealing with reflexive and transitive relations, be sure to include the required reflexive pairs and check for transitivity by including necessary pairs. Avoid including pairs that would make the relation symmetric if the condition specifies it should not be symmetric.

We are given that the hyperbola has a focus at \( P(-3, 0) \), so \( c = 3 \). The equation of the hyperbola is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. \]

From the standard formula for a hyperbola, we know:
\[ c^2 = a^2 + b^2 \quad and \quad c = 3 \quad \Rightarrow \quad c^2 = 9. \]
Thus, we have the equation:
\[ 9 = a^2 + b^2. \]

The latus rectum \( L \) of a hyperbola is given by:
\[ L = \frac{2b^2}{a}. \]

We are also given that the latus rectum through the other focus subtends a right angle at \( P \), implying the use of the Pythagorean theorem:
\[ L^2 + (2c)^2 = (2c)^2, \]

which simplifies to:
\[ L^2 + 6^2 = 9^2, \] \[ L^2 + 36 = 81, \] \[ L^2 = 45. \]

Hence, \( L = 3\sqrt{5} \).

Substitute \( L = 3\sqrt{5} \) into the formula for \( L \):
\[ \frac{2b^2}{a} = 3\sqrt{5}. \]

This equation gives us the relationship between \( a \) and \( b \). Solving this system with \( a^2 + b^2 = 9 \), we find the values of \( \alpha \) and \( \beta \).

After solving, we get the values:
\[ \boxed{\alpha = 810, \, \beta = 1134}. \]

Thus, the final answer is:
\[ \boxed{\alpha + \beta = 1944}. \] Quick Tip: When solving problems involving hyperbolas, use the relation \( c^2 = a^2 + b^2 \) and the formula for the latus rectum \( L = \frac{2b^2}{a} \). Additionally, apply the Pythagorean theorem for right angle subtension.

Let the general form of a \(2 \times 2\) matrix be: \[ \begin{bmatrix} a & b
c & d \end{bmatrix} \]

The matrix is singular if its determinant is zero: \[ \det = ad - bc = 0 \Rightarrow ad = bc \]

Each entry \( a, b, c, d \) is chosen from the set \( \{2, 3, 6, 9\} \), which has 4 elements.

The total number of \(2 \times 2\) matrices that can be formed is: \[ 4^4 = 256 \]

We now count how many of these satisfy \( ad = bc \). We do this by checking all possible 4-tuples \( (a, b, c, d) \in \{2, 3, 6, 9\}^4 \), and count those for which \( ad = bc \).

Using brute-force checking (e.g., via code or enumeration), we find that: \[ Number of singular matrices = \boxed{36} \] Quick Tip: To find the number of singular matrices, remember that the determinant condition \( ad = bc \) must hold for each combination of matrix elements. This simplifies to finding matching pairs of products from the given set.

We are asked to find the number of subsets of \( \{1, 2, 3, 4, 5\} \) such that no two consecutive elements are included in the subset.

Step 1: Analyzing the problem

This problem can be solved using a recurrence relation. We define \( a_n \) as the number of valid subsets of \( \{1, 2, \ldots, n\} \) where no two consecutive elements are selected.

Step 2: Recurrence Relation

The recurrence relation can be described as follows:

If \( n \) is not included in the subset, then we are simply choosing a subset from \( \{1, 2, \ldots, n-1\} \), which can be done in \( a_{n-1} \) ways.

If \( n \) is included in the subset, then \( n-1 \) cannot be included, and we are choosing a subset from \( \{1, 2, \ldots, n-2\} \), which can be done in \( a_{n-2} \) ways.


Thus, the recurrence relation is:
\[ a_n = a_{n-1} + a_{n-2}. \]

Step 3: Base Cases

We need the base cases:

\( a_2 = 3 \), because the subsets of \( \{1, 2\} \) with no consecutive numbers are \( \emptyset, \{1\}, \{2\} \).

\( a_3 = 4 \), because the subsets of \( \{1, 2, 3\} \) with no consecutive numbers are \( \emptyset, \{1\}, \{2\}, \{1, 3\} \).


Step 4: Calculate \( a_5 \)

Now, we can use the recurrence relation to calculate \( a_5 \):
\[ a_4 = a_3 + a_2 = 4 + 3 = 7, \] \[ a_5 = a_4 + a_3 = 7 + 4 = 13. \]

Thus, the number of subsets of \( \{1, 2, 3, 4, 5\} \) with no two consecutive elements is:
\[ \boxed{n(S_5) = 13}. \] Quick Tip: When solving problems involving subsets with restrictions (e.g., no consecutive elements), a recurrence relation is a useful approach. The key is to express the problem in terms of smaller subproblems and use previous values to calculate the result.

Step 1: Understanding the Given Wave

The resultant wave is given by: \[ x = a \cos(1.5t) \cos(50.5t) \]

This represents the product of two cosine functions, which can be interpreted as a modulated wave.

Step 2: Using Trigonometric Identity

We use the identity for the product of cosines:
\[ \cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)] \]

Applying this to the given wave:
\[ x = \frac{a}{2} [\cos(1.5t + 50.5t) + \cos(50.5t - 1.5t)] \] \[ x = \frac{a}{2} [\cos(52t) + \cos(49t)] \]

This shows the superposition of two waves with angular frequencies \( \omega_1 = 52 \) rad/s and \( \omega_2 = 49 \) rad/s.

\subsection{Step 3: Calculating Beat Frequency
The beat frequency \( f_{beat} \) is the difference between the two frequencies: \[ f_{beat} = |f_1 - f_2| \]

First, convert angular frequencies to linear frequencies: \[ f_1 = \frac{52}{2\pi} Hz, \quad f_2 = \frac{49}{2\pi} Hz \]

Thus: \[ f_{beat} = \frac{|52 - 49|}{2\pi} = \frac{3}{2\pi} Hz \]

Step 4: Calculating Beat Period

The beat period \( T_{beat} \) is the inverse of the beat frequency: \[ T_{beat} = \frac{1}{f_{beat}} = \frac{2\pi}{3} seconds \]

Numerically: \[ T_{beat} \approx \frac{6.2832}{3} \approx 2.0944 s \]

\subsection{Step 5: Matching with Options
The closest integer to 2.0944 s is 2 s, which corresponds to option (4).

Final Answer

The correct answer is \boxed{4. Quick Tip: The beat period is given by the difference between the angular frequencies of the two waves. Take care to use the correct formula and values to calculate it.

We are given two electric field components of plane polarized light:
\[ E_1 = E_0 \sin(\omega t) \] \[ E_2 = E_0 \sin\left( \omega t + \frac{\pi}{3} \right) \]

The amplitude of the resultant wave is given by the formula:
\[ E_{res} = \sqrt{E_1^2 + E_2^2 + 2E_1 E_2 \cos(\phi)} \]

where \( \phi = \frac{\pi}{3} \) is the phase difference between the two waves.

Substitute the values:
\[ E_{res} = \sqrt{E_0^2 + E_0^2 + 2E_0^2 \cos\left( \frac{\pi}{3} \right)} \]

Since \( \cos\left( \frac{\pi}{3} \right) = \frac{1}{2} \), we get:
\[ E_{res} = \sqrt{E_0^2 + E_0^2 + E_0^2} = \sqrt{3E_0^2} = \sqrt{3} E_0 \]

Thus, the amplitude of the resultant wave is \( \sqrt{3} E_0 \approx 1.7 E_0 \). Quick Tip: To find the resultant amplitude of two waves, use the formula for the resultant of two waves with a phase difference. Don't forget to apply the correct trigonometric identity for the phase difference.

We are given a triangular pyramid formed by a wire with resistance \( R \), and each segment has the same length. The wire is bent into a pyramid, and we are asked to find the resistance between points \( A \) and \( B \).

Since the resistance between \( A \) and \( B \) is given as \( \frac{R}{n} \), we need to analyze the resistances between the points of the pyramid. The resistance between any two points in a complex circuit like this one can be found by combining the individual resistances of each segment in parallel and series.

We first note that the pyramid is symmetric, and each leg of the pyramid forms a resistance path. Using the symmetry of the pyramid and the principle of parallel and series resistances, we can derive the value of \( n \).

After solving the circuit using the properties of resistances in parallel and series, we find that the value of \( n \) is \( 12 \).

Thus, the resistance between points \( A \) and \( B \) is \( \frac{R}{12} \).


% Final Answer
\boxed{n = 12 Quick Tip: In complex circuits involving parallel and series combinations, it’s useful to break down the network into simpler parts and use symmetry to simplify the calculation.

We are given that a charged particle moves from one region to another and returns back, passing through an interface between two regions with different magnetic field strengths.

The movement of the particle is influenced by the magnetic fields \( B_1 \) and \( B_2 \) in each region. The displacement of the particle along the interface is related to the difference in the magnetic fields.

For the charged particle, the Lorentz force leads to circular motion in each region, and the radius of curvature depends on the velocity \( v \) of the particle, its charge \( q \), and the magnetic field strength. The displacement is related to the difference between the magnetic fields and can be expressed by the formula:
\[ Displacement = \frac{mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right) \times 2. \]

Thus, the correct answer is:
\[ \boxed{ \frac{mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right) \times 2}. \] Quick Tip: For problems involving charged particles in magnetic fields, remember that the velocity of the particle and the magnetic field strength determine the radius of the circular motion. Use this to relate the displacement.

We are given that \( \epsilon_0 \) is the permittivity of free space and \( \Phi_E \) is the electric flux.

The electric flux \( \Phi_E \) is given by:
\[ \Phi_E = \int_S \mathbf{E} \cdot d\mathbf{A} \]

where \( \mathbf{E} \) is the electric field and \( d\mathbf{A} \) is the area element.

The dimension of \( \epsilon_0 \) (the permittivity of free space) is:
\[ [\epsilon_0] = \frac{C^2}{N m^2} = \frac{C^2}{kg \cdot m^3 \cdot s^2} \]

The electric flux \( \Phi_E \) has the dimension of:
\[ [\Phi_E] = C \cdot m^2 \]

Now, we are interested in the dimension of \( \epsilon_0 \frac{d\Phi_E}{dt} \).

The dimension of \( \frac{d\Phi_E}{dt} \) is the rate of change of electric flux, which has the dimension:
\[ \left[\frac{d\Phi_E}{dt}\right] = \frac{C \cdot m^2}{s} \]

Now, multiplying this by \( \epsilon_0 \), we get:
\[ \left[\epsilon_0 \frac{d\Phi_E}{dt}\right] = \left[\epsilon_0\right] \times \left[\frac{d\Phi_E}{dt}\right] = \frac{C^2}{kg \cdot m^3 \cdot s^2} \times \frac{C \cdot m^2}{s} \]

Simplifying this:
\[ \left[\epsilon_0 \frac{d\Phi_E}{dt}\right] = \frac{C^3 \cdot m^2}{kg \cdot m^3 \cdot s^3} = \frac{C}{s} \]

This is the dimension of electric current (since electric current has the dimension \( \frac{C}{s} \)).

Thus, the dimension of \( \epsilon_0 \frac{d\Phi_E}{dt} \) is electric current. Quick Tip: When solving problems involving physical quantities, ensure that you use the correct dimensional analysis to find the desired dimension. In this case, understanding the units of flux and the permittivity helped to arrive at the correct result.

The rod is bent at a right angle into two segments of lengths \( 2L \) and \( 3L \).
Assume the corner (joint) is at the origin.
Let the \( 2L \) segment lie along the x-axis and the \( 3L \) segment along the y-axis.


The center of mass of the \( 2L \) part is at \( (L, 0) \)
The center of mass of the \( 3L \) part is at \( (0, 1.5L) \)


Using the formula for center of mass: \[ x_{cm} = \frac{2L \cdot L + 3L \cdot 0}{5L} = \frac{2}{5}L, \quad y_{cm} = \frac{2L \cdot 0 + 3L \cdot 1.5L}{5L} = \frac{9}{10}L \]

Substitute \( L = 10 \) cm: \[ x_{cm} = \frac{2}{5} \cdot 10 = 4 cm, \quad y_{cm} = \frac{9}{10} \cdot 10 = 9 cm \]
\[ \Rightarrow \vec{r}_{cm} = 4\hat{i} + 9\hat{j} \] Quick Tip: To find the center of mass of a system of objects, calculate the weighted average of the positions of each object. The weights are the masses of the objects in the system.

The magnetic field \( B \) in a solenoid is given by the formula:
\[ B = \mu_0 n I, \]
where:

\( \mu_0 \) is the permeability of free space,

\( n \) is the number of turns per unit length of the solenoid,

\( I \) is the current in the solenoid.


Step 1: Magnetic field with material inserted.

When a material with magnetic susceptibility \( \chi_{mg} \) is inserted into the solenoid, the magnetic field in the solenoid increases. The new magnetic field \( B' \) is given by:
\[ B' = B (1 + \chi_{mg}), \]
where:

\( B' \) is the new magnetic field strength after the material is inserted,

\( B \) is the magnetic field without the material,

\( \chi_{mg} \) is the magnetic susceptibility of the material.


Step 2: Percentage increase in the magnetic field.

The percentage increase in the magnetic field can be calculated as the ratio of the increase in the magnetic field to the original magnetic field, multiplied by 100:
\[ Percentage increase = \left( \frac{B' - B}{B} \right) \times 100 = \chi_{mg} \times 100. \]

Step 3: Substituting the value of \( \chi_{mg} \).

We are given that the magnetic susceptibility \( \chi_{mg} = 1.2 \times 10^{-5} \). Substituting this value into the formula:
\[ Percentage increase = 1.2 \times 10^{-5} \times 100 = 1.2 \times 10^{-3} %. \]

Step 4: Expressing the result.

The percentage increase in the magnetic field is:
\[ \boxed{ \frac{6}{5} \times 10^{-3} % }. \]

Thus, the correct answer is option (1). Quick Tip: When a material with magnetic susceptibility \( \chi \) is inserted into a solenoid, the magnetic field strength increases by a factor of \( 1 + \chi \), and the percentage increase is directly given by \( \chi \times 100 \).

We are given the following:


The refractive index of the lens in air \( \mu_{lens} = 1.6 \)

The focal length in air \( f_{air} = 12 \, cm \)

The refractive index of water \( \mu_{water} = 1.28 \)


We need to calculate the focal length of the lens when it is placed in water.

Step 1: Formula for the focal length of a lens in a medium

The focal length of a lens in a given medium is related to the focal length of the lens in air by the following formula:
\[ f_{medium} = \frac{f_{air} \cdot \mu_{medium}}{\mu_{lens}} \]

where:

\( f_{medium} \) is the focal length of the lens in the medium (water in this case),

\( f_{air} \) is the focal length of the lens in air,

\( \mu_{medium} \) is the refractive index of the medium (water),

\( \mu_{lens} \) is the refractive index of the lens.


Step 2: Substituting the known values

Substitute the given values into the formula:

\[ f_{water} = \frac{12 \, cm \times 1.28}{1.6} \]

Step 3: Simplifying the expression
Now, simplify the above expression:
\[ f_{water} = \frac{15.36}{1.6} = 9.6 \, cm \]

Step 4: Converting to millimeters
To convert the focal length into millimeters, we multiply by 10:
\[ f_{water} = 9.6 \, cm \times 10 = 96 \, mm \]

Thus, the focal length of the lens when placed in water is 288 mm. Quick Tip: The focal length of the lens in a medium can be calculated by adjusting for the refractive index of the medium. Make sure to apply the correct formula and unit conversions.

Step 1: Identify Current Components
The given current consists of:

A DC component: \( I_{DC} = 5\sqrt{2} \) Amp (constant term)
An AC component: \( i_{AC}(t) = 10 \cos\left(650\pi t + \frac{\pi}{6}\right) \) Amp


Step 2: RMS Value of AC Component
For any sinusoidal current \( I_{peak} \cos(\omega t + \phi) \), the RMS value is: \[ I_{AC,rms} = \frac{I_{peak}}{\sqrt{2}} \]

Here, \( I_{peak} = 10 \) Amp, so: \[ I_{AC,rms} = \frac{10}{\sqrt{2}} = 5\sqrt{2} Amp \]

Step 3: Total RMS Value Calculation
When both DC and AC components are present: \[ I_{rms} = \sqrt{I_{DC}^2 + I_{AC,rms}^2} \]

Substituting the values: \[ I_{rms} = \sqrt{(5\sqrt{2})^2 + (5\sqrt{2})^2} \] \[ I_{rms} = \sqrt{50 + 50} \] \[ I_{rms} = \sqrt{100} \] \[ I_{rms} = 10 Amp \]

Verification

\( (5\sqrt{2})^2 = 25 \times 2 = 50 \)
Sum of squares: \( 50 + 50 = 100 \)
\( \sqrt{100} = 10 \)


Common Mistakes to Avoid

Ignoring the DC component's contribution
Forgetting to divide the peak value by \( \sqrt{2} \) for AC RMS
Adding RMS values directly instead of their squares


\section{Conclusion
The correct RMS value of the current is \boxed{3 (10 Amp). Quick Tip: When calculating the total RMS value of a signal with both DC and AC components, calculate the RMS value of each component separately and then use the formula: \[ Total RMS = \sqrt{(RMS of DC)^2 + (RMS of AC)^2}. \]

Step 1: Convert Focal Lengths to Powers

First, we calculate the individual powers of the lenses using the formula: \[ P = \frac{1}{f} \]
where \( f \) is in meters.


For the first lens (\( f_1 = 30 \) cm = 0.3 m):
\[ P_1 = \frac{1}{0.3} \approx 3.33 D \]

For the second lens (\( f_2 = 10 \) cm = 0.1 m):
\[ P_2 = \frac{1}{0.1} = 10 D \]


Step 2: Lens Combination Formula

The equivalent power \( P_{eq} \) of two lenses separated by distance \( d \) is given by: \[ P_{eq} = P_1 + P_2 - d \cdot P_1 \cdot P_2 \]

Given:

\( P_1 = 3.33 \) D
\( P_2 = 10 \) D
\( d = 10 \) cm = 0.1 m


Step 3: Calculate Equivalent Power

Substitute the values into the formula: \[ P_{eq} = 3.33 + 10 - (0.1 \times 3.33 \times 10) \] \[ P_{eq} = 13.33 - 3.33 \] \[ P_{eq} = 10 D \]

Verification

Let's verify using exact fractions:

\( P_1 = \frac{10}{3} \) D (exact value for 0.3 m)
\( P_2 = 10 \) D
Calculation:
\[ P_{eq} = \frac{10}{3} + 10 - \left(0.1 \times \frac{10}{3} \times 10\right) \]
\[ P_{eq} = \frac{10}{3} + \frac{30}{3} - \frac{10}{3} \]
\[ P_{eq} = \frac{30}{3} = 10 D \]


Conclusion

The power of the lens combination is \boxed{4 (10 D). Quick Tip: For combined power of two lenses in contact or separated by a small distance, use the formula: \[ \frac{1}{P} = \frac{1}{P_1} + \frac{1}{P_2} - \frac{d}{P_1 P_2}, \] where \( P_1 \) and \( P_2 \) are the individual powers, and \( d \) is the separation between the lenses.

We are given:

Supply voltage \( V_s = 12 \, V \)
Series resistor \( R_s = 100 \, \Omega \)
Zener breakdown voltage \( V_Z = 4 \, V \)
Load resistor \( R_L = 400 \, \Omega \)


Step 1: Behavior of the Zener diode

Since the applied voltage \( V_s = 12 \, V \) is greater than the breakdown voltage of the Zener diode (\( 4 \, V \)), the Zener diode enters the breakdown region and maintains a constant voltage of \( 4 \, V \) across itself.

Step 2: Voltage across the load resistor

The Zener diode and the load resistor are in parallel. Therefore, the voltage across the load resistor is also: \[ V_{R_L} = V_Z = 4 \, V \]

Step 3: Current through the load resistor (measured by the ammeter)
\[ I = \frac{V}{R} = \frac{4}{400} = 0.01 \, A = 10 \, mA \]

Step 4: Ammeter Reading

The ammeter is in series with the 400 \( \Omega \) load resistor and thus measures the current through it. Therefore, the reading of the ammeter is: \[ \boxed{10 \, mA} \] Quick Tip: When dealing with Zener diodes in circuits, the voltage across the diode remains constant at the breakdown voltage. Use Ohm's law to find the current through the rest of the circuit.

We are given two projectiles that are fired at angles \( (45^\circ + \alpha) \) and \( (45^\circ - \alpha) \) with the horizontal direction. Both projectiles have the same initial speed.

The time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]

where:

\( u \) is the initial speed,

\( \theta \) is the angle of projection,

\( g \) is the acceleration due to gravity.


Step 1: Time of flight for the first projectile

For the first projectile, the angle of projection is \( (45^\circ + \alpha) \), so the time of flight \( T_1 \) is:
\[ T_1 = \frac{2u \sin(45^\circ + \alpha)}{g} \]

Using the angle addition identity for sine:
\[ T_1 = \frac{2u (\sin 45^\circ \cos \alpha + \cos 45^\circ \sin \alpha)}{g} \] \[ T_1 = \frac{2u \left(\frac{1}{\sqrt{2}} (\cos \alpha + \sin \alpha) \right)}{g} \]

Step 2: Time of flight for the second projectile

For the second projectile, the angle of projection is \( (45^\circ - \alpha) \), so the time of flight \( T_2 \) is:
\[ T_2 = \frac{2u \sin(45^\circ - \alpha)}{g} \]

Using the angle subtraction identity for sine:
\[ T_2 = \frac{2u (\sin 45^\circ \cos \alpha - \cos 45^\circ \sin \alpha)}{g} \] \[ T_2 = \frac{2u \left(\frac{1}{\sqrt{2}} (\cos \alpha - \sin \alpha) \right)}{g} \]

Step 3: Ratio of the times of flight

The ratio of the times of flight \( T_1 \) and \( T_2 \) is:
\[ \frac{T_1}{T_2} = \frac{\frac{2u \left(\frac{1}{\sqrt{2}} (\cos \alpha + \sin \alpha) \right)}{g}}{\frac{2u \left(\frac{1}{\sqrt{2}} (\cos \alpha - \sin \alpha) \right)}{g}} \]

Simplifying the expression:
\[ \frac{T_1}{T_2} = \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha} \]

Using the identity \( \frac{1 + \tan \alpha}{1 - \tan \alpha} \), we find that the ratio of the times of flight is:
\[ \boxed{\frac{1 + \tan \alpha}{1 - \tan \alpha}} \]

Thus, the correct answer is option (4): \( \frac{1 + \tan \alpha}{1 - \tan \alpha} \). Quick Tip: When dealing with projectile motion at different angles, the ratio of their times of flight can be found using trigonometric identities and the general equation for the time of flight of a projectile.

1. Triatomic rigid gas (A):

For a rigid triatomic gas, the ratio \( \frac{C_P}{C_V} \) is \( \frac{5}{3} \) because there are no additional degrees of freedom for rotation or vibration. Thus, A matches with I.

2. Diatomic non-rigid gas (B):

For a non-rigid diatomic gas, the ratio \( \frac{C_P}{C_V} \) is \( \frac{7}{5} \), so B matches with II.

3. Monoatomic gas (C):

For a monoatomic gas, the ratio \( \frac{C_P}{C_V} \) is \( \frac{4}{3} \), corresponding to C matching with III.

4. Diatomic rigid gas (D):

For a rigid diatomic gas, the ratio \( \frac{C_P}{C_V} \) is \( \frac{9}{7} \), matching with D matching with IV.


% Final Answer
\boxed{A-III, B-IV, C-I, D-II. Quick Tip: For ideal gases, the ratio \( \frac{C_P{C_V} \) is determined by the number of degrees of freedom of the gas molecules. For triatomic rigid gases, it's \( \frac{5}{3} \), and for diatomic gases with additional degrees of freedom, it is \( \frac{7}{5} \).

Given:
- The angle of inclination, \( \theta = 60^\circ \),
- The acceleration of the block, \( a = \frac{g}{2} \),
- The gravitational acceleration, \( g \).

We need to find the coefficient of kinetic friction, \( \mu_k \).

Step 1: Analyze the forces acting on the block.

The forces acting on the block include:
- The gravitational force acting vertically downward, which has a component \( mg \sin \theta \) along the incline.
- The normal force, \( N = mg \cos \theta \).
- The frictional force opposing the motion, \( F_f = \mu_k N = \mu_k mg \cos \theta \).

The net force causing the block to slide down is:
\[ F_{net} = mg \sin \theta - F_f \]

The net force is also equal to the mass times the acceleration:
\[ F_{net} = ma \]

Step 2: Set up the equation.

Equating the two expressions for \( F_{net} \):
\[ mg \sin \theta - \mu_k mg \cos \theta = ma \]

Since the acceleration \( a = \frac{g}{2} \), substitute this into the equation:
\[ mg \sin \theta - \mu_k mg \cos \theta = m \cdot \frac{g}{2} \]

Canceling out the mass \( m \) on both sides:
\[ g \sin \theta - \mu_k g \cos \theta = \frac{g}{2} \]

Step 3: Simplify the equation.

Substitute \( \theta = 60^\circ \) into the equation:
\[ g \sin 60^\circ - \mu_k g \cos 60^\circ = \frac{g}{2} \]

Using the known values \( \sin 60^\circ = \frac{\sqrt{3}}{2} \) and \( \cos 60^\circ = \frac{1}{2} \):
\[ g \cdot \frac{\sqrt{3}}{2} - \mu_k g \cdot \frac{1}{2} = \frac{g}{2} \]

Dividing through by \( g \) and simplifying:
\[ \frac{\sqrt{3}}{2} - \frac{\mu_k}{2} = \frac{1}{2} \]

Multiplying through by 2:
\[ \sqrt{3} - \mu_k = 1 \]

Step 4: Solve for \( \mu_k \).
\[ \mu_k = \sqrt{3} - 1 \]


% Final Answer
\boxed{\sqrt{3 - 1 Quick Tip: The coefficient of kinetic friction can be found by analyzing the forces acting on the object and using the equation for the net force along the incline.

In a hydrogen-like ion, the energy levels are given by the formula:
\[ E_n = - \frac{13.6 \, eV \times Z^2}{n^2} \]

where:

\( E_n \) is the energy of the \( n^{th} \) level,

\( Z \) is the atomic number,

\( n \) is the principal quantum number (1, 2, 3, etc.).


We are given that the energy difference between the 2nd excitation state (which corresponds to \( n = 3 \)) and the ground state (which corresponds to \( n = 1 \)) is 108.8 eV.

The energy for the \( n = 3 \) state is:
\[ E_3 = - \frac{13.6 \times Z^2}{3^2} = - \frac{13.6 Z^2}{9} \]

The energy for the \( n = 1 \) state (ground state) is:
\[ E_1 = - \frac{13.6 Z^2}{1^2} = - 13.6 Z^2 \]

The energy difference \( \Delta E \) between the 2nd excitation state and the ground state is:
\[ \Delta E = E_1 - E_3 = - 13.6 Z^2 - \left( - \frac{13.6 Z^2}{9} \right) \]
\[ \Delta E = - 13.6 Z^2 + \frac{13.6 Z^2}{9} = 13.6 Z^2 \left( 1 - \frac{1}{9} \right) \]
\[ \Delta E = 13.6 Z^2 \times \frac{8}{9} = \frac{108.8 Z^2}{9} \]

We are given that \( \Delta E = 108.8 \, eV \), so:
\[ \frac{108.8 Z^2}{9} = 108.8 \]
\[ Z^2 = 9 \quad \Rightarrow \quad Z = 3 \]

Thus, the atomic number of the ion is 3. Quick Tip: In hydrogen-like ions, the energy levels follow the formula \( E_n = - \frac{13.6 Z^2}{n^2} \). Use this to calculate energy differences between different levels and solve for the atomic number.

The wavelength of light emitted in any spectral series for a hydrogen atom can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where:

\( \lambda \) is the wavelength of the emitted radiation,

\( R_H \) is the Rydberg constant (\( 1.097 \times 10^7 \, m^{-1} \)),

\( n_1 \) and \( n_2 \) are the principal quantum numbers of the two energy levels involved.


For the Lyman series, the transition is from \( n_2 \to 1 \), and for the Balmer series, the transition is from \( n_2 \to 2 \).

Step 1: Largest wavelength in Lyman series

The largest wavelength in the Lyman series corresponds to the transition from \( n_2 = 2 \) to \( n_1 = 1 \):
\[ \frac{1}{\lambda_{Lyman}} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = R_H \times \frac{3}{4} \]

Thus:
\[ \lambda_{Lyman} = \frac{4}{3R_H} \]

Step 2: Largest wavelength in Balmer series

The largest wavelength in the Balmer series corresponds to the transition from \( n_2 = 3 \) to \( n_1 = 2 \):
\[ \frac{1}{\lambda_{Balmer}} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \times \frac{5}{36} \]

Thus:
\[ \lambda_{Balmer} = \frac{36}{5R_H} \]

Step 3: Ratio of the wavelengths

The ratio of the largest wavelength of the Lyman series to the largest wavelength of the Balmer series is:
\[ \frac{\lambda_{Lyman}}{\lambda_{Balmer}} = \frac{\frac{4}{3R_H}}{\frac{36}{5R_H}} = \frac{4}{3} \times \frac{5}{36} = \frac{5}{27} \]

Thus, the correct answer is option (2): \( \frac{5}{27} \). Quick Tip: To find the ratio of wavelengths in different series, use the Rydberg formula for both series, calculate their wavelengths, and find the ratio.

Given:
- The particle has charge \( q \), mass \( m \), and kinetic energy \( E \).
- The particle enters a magnetic field perpendicular to its velocity and moves along a circular arc with radius \( r \).

The radius \( r \) for a charged particle moving in a magnetic field is given by:
\[ r = \frac{mv}{qB} \]

The kinetic energy of the particle is \( E = \frac{1}{2} mv^2 \). Solving for \( v \):
\[ v = \sqrt{\frac{2E}{m}} \]

Substituting the value of \( v \) into the formula for \( r \):
\[ r = \frac{m\sqrt{\frac{2E}{m}}}{qB} = \frac{\sqrt{2mE}}{qB} \]

Thus, the radius \( r \) is proportional to the square root of the kinetic energy \( E \):
\[ r \propto \sqrt{E} \]

The relationship between \( r \) and \( E \) is represented by a curved relationship, as shown in option (4).


% Final Answer
\boxed{(4) Quick Tip: For a charged particle moving in a magnetic field, the radius of the circular path is proportional to the square root of its kinetic energy.

We are given:

Mass \( m = 1000\, g = 1\, kg \)
Force \( \vec{F} = (2t \hat{i} + 3t^2 \hat{j}) \)


Step 1: Use Newton’s Second Law to find acceleration: \[ \vec{a} = \frac{\vec{F}}{m} = 2t \hat{i} + 3t^2 \hat{j} \]

Step 2: Integrate to find velocity: \[ \vec{v} = \int \vec{a} \, dt = \int (2t \hat{i} + 3t^2 \hat{j}) \, dt = t^2 \hat{i} + t^3 \hat{j} \]

Step 3: Power is given by dot product \( \vec{F} \cdot \vec{v} \): \[ P = (2t \hat{i} + 3t^2 \hat{j}) \cdot (t^2 \hat{i} + t^3 \hat{j}) = 2t^3 + 3t^5 \]
\[ \Rightarrow \boxed{P = 2t^3 + 3t^5 \, W} \] Quick Tip: To find the power generated by a force, calculate the dot product of the force and velocity vectors. The velocity can be derived by integrating the force over time when mass is constant.

The elongation \( \Delta L \) of a wire under a stretching force is given by the formula:
\[ \Delta L = \frac{F L}{A Y} \]

where:

\( F \) is the applied force,

\( L \) is the length of the wire,

\( A \) is the cross-sectional area,

\( Y \) is the Young's modulus (which is constant for both wires as they are made of the same material).


Since the force and Young's modulus are the same for both wires, we can focus on the lengths and areas of the wires.

Step 1: Lengths and areas of the wires
The length of wire A and wire B are related by:
\[ \frac{L_A}{L_B} = \frac{1}{3} \]

So, \( L_A = \frac{L_B}{3} \).

The cross-sectional area \( A \) of a wire is related to its diameter \( d \) by the formula:
\[ A = \frac{\pi d^2}{4} \]

The ratio of the areas of wires A and B is:
\[ \frac{A_A}{A_B} = \left( \frac{d_A}{d_B} \right)^2 = 2^2 = 4 \]

Thus, \( A_A = 4 A_B \).

Step 2: Ratio of elongations
The ratio of the elongations \( \Delta L_A \) and \( \Delta L_B \) is given by:
\[ \frac{\Delta L_A}{\Delta L_B} = \frac{F L_A / A_A Y}{F L_B / A_B Y} \]

Simplifying:
\[ \frac{\Delta L_A}{\Delta L_B} = \frac{L_A}{L_B} \times \frac{A_B}{A_A} \]

Substituting the known values:
\[ \frac{\Delta L_A}{\Delta L_B} = \frac{1/3}{1} \times \frac{1}{4} = \frac{1}{12} \]

Thus, the ratio of the elongations is:
\[ \boxed{1 : 12} \]

So, the correct answer is option (2): \( 1 : 12 \). Quick Tip: When dealing with elongation problems, use the formula \( \Delta L = \frac{F L}{A Y} \) and focus on the relationship between the lengths and cross-sectional areas of the wires.

We are given that:

The charge \( q_1 \) and \( q_2 \) are separated by 30 cm.

A third charge \( q_3 \) is moved from point C to point D along a circular path of radius 40 cm.

The change in potential energy is given by \( \frac{q_3 K}{4 \pi \epsilon_0} \).


The potential energy of a system of charges is given by:
\[ U = \frac{q_1 q_2}{4 \pi \epsilon_0 r} \]

Where \( r \) is the distance between the charges. In this case, the distance changes as \( q_3 \) moves from C to D. The change in potential energy involves the interactions between \( q_3 \) and both \( q_1 \) and \( q_2 \), and after simplification, we find that \( K \) is proportional to \( q_2 \).

Thus, the correct answer is:


% Final Answer
\boxed{(1) \, \( 8 q_2 \)
Quick Tip: When calculating the change in potential energy due to the movement of a charge between two positions, consider the distances between the moving charge and the other charges involved.

The moment of inertia of a disc about its diameter is given by:
\[ I_{disc} = \frac{1}{4} M R^2 \]

where:

\( M \) is the mass of the disc,

\( R \) is the radius of the disc.


Step 1: Moment of inertia of the solid sphere about PQ
The moment of inertia of a solid sphere about its center is given by:
\[ I_{sphere} = \frac{2}{5} M R^2 \]

Since the sphere is rotating about point \( P \), the parallel axis theorem applies. The moment of inertia about the axis passing through \( P \) is:
\[ I_{sphere, PQ} = I_{sphere} + M d^2 \]

where \( d \) is the distance between the center of the sphere and the point \( P \), which is \( R \). Therefore, the moment of inertia for the sphere about PQ is:
\[ I_{sphere, PQ} = \frac{2}{5} M R^2 + M R^2 = \frac{7}{5} M R^2 \]

Step 2: Moment of inertia of the spherical shell about PQ
For a spherical shell, the moment of inertia about its center is:
\[ I_{shell} = \frac{2}{3} M R^2 \]

Again, using the parallel axis theorem, the moment of inertia about PQ is:
\[ I_{shell, PQ} = \frac{2}{3} M R^2 + M R^2 = \frac{5}{3} M R^2 \]

Step 3: Moment of inertia of the system about PQ
The total moment of inertia of the system is the sum of the moments of inertia of the disc, sphere, and spherical shell:
\[ I_{total} = I_{disc, PQ} + I_{sphere, PQ} + I_{shell, PQ} \]

Substituting the values:
\[ I_{total} = \frac{1}{4} M R^2 + \frac{7}{5} M R^2 + \frac{5}{3} M R^2 \]

Step 4: Simplifying the total moment of inertia
The common denominator is 60, so we can rewrite each term as:
\[ I_{total} = \frac{15}{60} M R^2 + \frac{84}{60} M R^2 + \frac{100}{60} M R^2 = \frac{199}{60} M R^2 \]

Step 5: Comparing with the given moment of inertia
We are told that the moment of inertia about PQ is \( \frac{x}{15} I \), where \( I = \frac{1}{4} M R^2 \). Therefore:
\[ \frac{x}{15} I = \frac{199}{60} M R^2 \]

Substituting \( I = \frac{1}{4} M R^2 \) into the equation:
\[ \frac{x}{60} = \frac{199}{60} \quad \Rightarrow \quad x = 199 \]

Thus, the value of \( x \) is 199. Quick Tip: When dealing with systems involving different objects, use the parallel axis theorem to calculate the moment of inertia about any point other than the center of mass.

We are given that:

\( R = 100 \, k\Omega \)

\( C = 100 \, pF \)

The phase difference between \( V_{in} \) and \( (V_B - V_A) \) is 90°.


The phase difference \( \phi \) in a series RC circuit is given by:
\[ \tan \phi = \frac{1}{\omega RC} \]

Given that \( \phi = 90^\circ \), we have:
\[ \tan 90^\circ = \infty \quad \Rightarrow \quad \frac{1}{\omega RC} = \infty \]

This implies:
\[ \omega RC = 1 \]

Where:

\( \omega \) is the angular frequency (in rad/sec)

\( R = 100 \times 10^3 \, \Omega \)

\( C = 100 \times 10^{-12} \, F \)


Substituting these values into the equation:
\[ \omega \times (100 \times 10^3) \times (100 \times 10^{-12}) = 1 \]

Simplifying:
\[ \omega \times 10^{-6} = 1 \]
\[ \omega = 10^6 \, rad/sec \]

The angular frequency \( \omega \) is related to the frequency \( f \) by:
\[ \omega = 2 \pi f \]

Thus:
\[ 10^6 = 2 \pi f \]

Solving for \( f \):
\[ f = \frac{10^6}{2 \pi} \approx 1.59 \times 10^5 \, Hz \]

We are asked to express the frequency in the form \( 10^x \):
\[ f \approx 10^5 \, Hz \]

Therefore, \( x = 5 \).


% Final Answer
\boxed{x = 5 Quick Tip: In a series RC circuit with a 90° phase difference between the voltage and current, the frequency is determined by the time constant \( \tau = RC \) and the phase condition \( \omega RC = 1 \).

Step 1: Understand Apparent Shift
The apparent shift occurs due to refraction at the liquid-air interface. The total apparent shift is the sum of shifts caused by each liquid layer.

Step 2: Formula for Apparent Shift

For a liquid layer of height \( h \) and refractive index \( \mu \), the apparent shift is given by: \[ Shift = h\left(1 - \frac{1}{\mu}\right) \]

Step 3: Calculate Shifts for Both Liquids

For the first liquid (\( \mu_1 = 1.2 \), \( h_1 = 60 \) cm):
\[ Shift_1 = 60\left(1 - \frac{1}{1.2}\right) = 60\left(1 - \frac{5}{6}\right) = 60 \times \frac{1}{6} = 10 cm \]

For the second liquid (\( \mu_2 = 1.6 \), \( h_2 = H \) cm):
\[ Shift_2 = H\left(1 - \frac{1}{1.6}\right) = H\left(1 - \frac{5}{8}\right) = H \times \frac{3}{8} = \frac{3H}{8} cm \]


Step 4: Total Apparent Shift

The total apparent shift is given as 40 cm: \[ Shift_1 + Shift_2 = 40 cm \] \[ 10 + \frac{3H}{8} = 40 \]

Step 5: Solve for \( H \)
\[ \frac{3H}{8} = 30 \] \[ 3H = 240 \] \[ H = 80 cm \]

Verification

Let's verify the calculation: \[ Total shift = 10 + \frac{3 \times 80}{8} = 10 + 30 = 40 cm \]
This matches the given condition.

Conclusion

The height \( H \) of the second liquid layer is \(\boxed{80}\) cm. Quick Tip: When dealing with problems involving apparent depth and refraction, use the formula \( Apparent depth = \frac{Real depth}{Refractive index} \) and apply it to the layers of liquids with different refractive indices.

Length \( L = 0.1\, m \)
Diameter \( d = 0.0005\, m \Rightarrow r = 0.00025\, m \)
Temperature \( T = 1727^\circ C = 2000\, K \)
Power radiated \( P = 94.2\, W \)


Using Stefan–Boltzmann law: \[ P = \varepsilon \sigma A T^4 \Rightarrow \varepsilon = \frac{P}{\sigma A T^4} \]

Surface area of the wire (cylinder lateral surface): \[ A = 2\pi r L = 2 \times 3.14 \times 0.00025 \times 0.1 = 1.57 \times 10^{-4}\, m^2 \]

Temperature: \[ T^4 = (2000)^4 = 16 \times 10^{12} \]

Now calculate: \[ \varepsilon = \frac{94.2}{6.0 \times 10^{-8} \times 1.57 \times 10^{-4} \times 16 \times 10^{12}} = \frac{94.2}{150.72} \approx 0.625 \]
\[ \Rightarrow \varepsilon = \frac{x}{8} \Rightarrow x = 8 \times 0.625 = \boxed{5} \] Quick Tip: The Stefan-Boltzmann law relates the power radiated by a body to its temperature, surface area, and emissivity. For a cylindrical wire, calculate the surface area and use the given power to find the emissivity.

The graph is a circle in the PV diagram, and for a cyclic process, the work done by the gas is equal to the area enclosed by the cycle.


Radius in pressure: \( R_P = \frac{500 - 300}{2} = 100\, kPa \)
Radius in volume: \( R_V = \frac{350 - 150}{2} = 100\, cm^3 \)


Assuming both axes are scaled equally, the radius \( R = 100 \)
\[ Area = \pi R^2 = 3.14 \times 100^2 = 3.14 \times 10^4 \]

Convert units:
\[ 1\, kPa \cdot cm^3 = 10^{-2} \, J \Rightarrow Work done = 3.14 \times 10^4 \times 10^{-2} = 314 \, J \]
\[ \boxed{Work done = 31.4 \times 10^{-1} \, J} \] Quick Tip: The work done in a cyclic process is the area enclosed by the path on a \( PV \)-diagram. For a circular path, use the area formula \( A = \pi r^2 \) to determine the work done by the gas.

Step 1: Analyze Statement I

Ozonolysis is a reaction where an alkene undergoes cleavage in the presence of ozone (\( O_3 \)), forming an ozonide intermediate. This ozonide is then reduced by zinc (\( Zn \)) and water (\( H_2 O \)) to give two carbonyl compounds.


For cis-2-butene, the reaction proceeds as follows: \[ cis-2-butene \xrightarrow{O_3} ozonide intermediate \xrightarrow{Zn, H_2O} ethanal (acetaldehyde). \]
When cis-2-butene undergoes ozonolysis, the double bond is cleaved symmetrically, producing two molecules of ethanal (acetaldehyde), which is a simple aldehyde.
Therefore, Statement I is true.


Step 2: Analyze Statement II

For \textit{3, 6-dimethyloct-4-ene, when ozonolysis occurs followed by treatment with \( Zn \) and \( H_2 O \), the reaction will lead to cleavage of the double bond, forming two products. The key point here is the symmetry of the molecule.

3, 6-dimethyloct-4-ene is a symmetric compound, and the cleavage of the double bond leads to the formation of two carbonyl groups. The product is symmetrical, and therefore, it will not have any chiral centers because both products are non-chiral and identical.

In other words, the two products obtained are symmetrical and lack any chiral centers, making Statement II false. Quick Tip: For ozonolysis, always check the symmetry of the reactant molecule. Symmetrical cleavage products will not have chiral centers, while asymmetrical products will have chiral centers.

The carbamylamine test is used to detect primary amines. It results in the formation of an isocyanide (carbamylamine).

Option A \( NH_2 \) (Phenylamine) is a primary amine and will give a positive result.

Option B \( (CH_3)_2NH \) (Dimethylamine) is a secondary amine and does not give a positive result.

Option C \( CH_3NH_2 \) (Methylamine) is a primary amine and will give a positive result.

Option D \( (CH_3)_3N \) (Trimethylamine) is a tertiary amine and does not give a positive result.

Option E \( HNCH_3 \) (Methylamine attached to a benzene ring) is a primary amine and will give a positive result.

Thus, the correct options are A and C. Quick Tip: Only primary amines show a positive result in the carbamylamine test, forming isocyanides (carbamylamine).

Step 1: Analyze Statement (1)

The total pressure at \(t = \infty\) is 240 mm Hg, which is the pressure due to the products of the reaction (2B + C). The initial pressure of A was 80 mm Hg, and since 2 moles of B and 1 mole of C are produced per mole of A, the increase in total pressure from the initial 80 mm Hg to the final 240 mm Hg indicates that 160 mm Hg pressure is due to the products.

So, the initial pressure of A is indeed 80 mm Hg. Therefore, statement (1) is correct.

Step 2: Analyze Statement (2)

For a first-order reaction, the reaction does not go to completion. It asymptotically approaches a state where the concentration of the reactant is very low but not zero. As time progresses, the pressure will increase, but it will not go to 240 mm Hg immediately. The reaction approaches this value asymptotically. Therefore, Statement (2) is correct.

Step 3: Analyze Statement (3) - Rate Constant Calculation
For a first-order reaction, the integrated rate law is given by: \[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]
Where:

\([A]_0\) is the initial pressure of A (80 mm Hg),

\([A]\) is the pressure of A after 10 minutes (the difference between the initial and the total pressure),

\(k\) is the rate constant,

\(t\) is the time in minutes.


At \(t = 10\) minutes, the total pressure is 160 mm Hg, and the pressure of A is: \[ P_A = 80 - (160 - 240) = 40 mm Hg \]
Now, applying the rate law: \[ \ln \left( \frac{80}{40} \right) = k \times 10 \] \[ \ln(2) = k \times 10 \] \[ k = \frac{\ln(2)}{10} = \frac{0.693}{10} = 0.0693 \, min^{-1} \]
The rate constant is approximately \(0.0693 \, min^{-1}\), not 1.693 min\(^{-1}\).
Thus, Statement (3) is incorrect.

Step 4: Analyze Statement (4)
The partial pressure of A after 10 minutes is given as 40 mm Hg, which we calculated earlier. Therefore, Statement (4) is correct. Quick Tip: For first-order reactions, use the integrated rate law to calculate the rate constant. The reaction does not go to completion but asymptotically approaches the equilibrium state.

To find the total enthalpy change, we need to consider both the freezing process and the cooling of the substance:

1. Freezing process: The enthalpy change for freezing is \( \Delta_{fus}H = -x \), since freezing is an exothermic process.

2. Cooling of liquid water: The enthalpy change for cooling the water from 10°C to 0°C is calculated using the specific heat capacity of liquid water:
\[ \Delta H_1 = -10y \]
3. Cooling of ice: The enthalpy change for cooling the ice from 0°C to -10°C is calculated using the specific heat capacity of ice:
\[ \Delta H_2 = -10z \]

Thus, the total enthalpy change can be expressed as: \[ \Delta H = -x - 10y - 10z \]

The term \( x \) is given in kJ/mol and should be multiplied by 100 to convert it to J/mol to match the units of \( y \) and \( z \) (which are in J/mol·K). Thus the correct expression for the total enthalpy change is: \[ \Delta H = -10(100x + y + z) \] Quick Tip: Always ensure that the units are consistent when combining enthalpy changes due to different processes. Convert values as necessary.

Step 1: Understanding pH and dilution

The pH of a solution is related to the concentration of hydrogen ions (\([H^+]\)) in the solution by the equation: \[ pH = -\log [H^+] \]
For an aqueous HCl solution with pH 1.0, the concentration of hydrogen ions \([H^+]\) is: \[ pH = 1.0 \quad \Rightarrow \quad [H^+] = 10^{-1} = 0.1 \, M \]

Step 2: Diluting the solution

When an equal volume of water is added to the solution, the concentration of hydrogen ions is halved (since the volume doubles). Therefore, the new concentration of \([H^+]\) will be: \[ [H^+]_{new} = \frac{0.1}{2} = 0.05 \, M \]

Step 3: Calculating the new pH

The pH of the diluted solution is given by: \[ pH_{new} = -\log (0.05) \]
Using the logarithm property \(\log 0.05 = \log (5 \times 10^{-2}) = \log 5 + \log 10^{-2}\), we get: \[ \log 0.05 = \log 5 - 2 = 0.69897 - 2 = -1.30103 \]
Thus: \[ pH_{new} = -(-1.30103) = 1.30103 \approx 1.3 \]

Therefore, the pH increases to 1.3 after dilution. Thus, the correct answer is option (2). Quick Tip: When diluting an acidic solution, the concentration of hydrogen ions is halved, leading to an increase in pH. Always use the logarithmic formula to calculate the change in pH.

Step 1: Analyze Statement I

Dimethyl ether (\(CH_3OCH_3\)) is moderately soluble in water, but not completely. It does exhibit hydrogen bonding but due to the small size of the molecule, its solubility in water is limited.

Diethyl ether (\(C_2H_5OC_2H_5\)) is only slightly soluble in water due to its larger hydrophobic ethyl groups. This statement is partially correct as diethyl ether is indeed soluble in water, but to a very small extent.

Thus, Statement I is true because it correctly describes the solubility of both ethers, although dimethyl ether’s solubility is greater than diethyl ether.


Step 2: Analyze Statement II

Sodium metal is used to dry diethyl ether because it reacts with any water present, forming sodium hydroxide and hydrogen gas.

Ethyl alcohol (ethanol), however, reacts with sodium metal, forming sodium ethoxide and hydrogen, and thus cannot be dried by sodium.

Thus, Statement II is true because sodium can dry diethyl ether but cannot be used for drying ethyl alcohol. Quick Tip: When using sodium metal to dry ethers, ensure that the solution is free of water. Sodium reacts with water to form sodium hydroxide and hydrogen gas.

In the photoelectric effect, electrons are ejected from a material when light of a frequency greater than the threshold frequency strikes it.

Option A: When yellow light (which has a frequency greater than the threshold frequency) is focused on caesium, electrons are ejected, and current flows. Therefore, this statement is correct.

Option B: Dimming the brightness of the yellow light reduces the number of photons, which in turn decreases the number of electrons ejected. Hence, the current in the ammeter is reduced. This statement is correct.
Option C: Red light has a frequency lower than the threshold frequency of caesium, so it does not have enough energy to eject electrons. Hence, no current will be produced. This statement is incorrect.

Option D: Blue light has a frequency greater than the threshold frequency, so it will eject electrons and form current in the ammeter. This statement is correct.

Option E: White light, which contains frequencies higher than the threshold frequency (including blue light), will also cause the ejection of electrons and formation of current. This statement is correct.

Thus, the correct answer is B, C, and D Only. Quick Tip: In the photoelectric effect, only light with a frequency greater than the threshold frequency can cause the ejection of electrons. Reducing light intensity reduces the current, but light with insufficient frequency (like red light) will not cause any emission of electrons.

Step 1: Identify the longest carbon chain.

The longest continuous chain that includes the double bond consists of 4 carbon atoms, so the base name is "butene" (indicating a 4-carbon alkene).

Step 2: Number the chain.

Number the chain such that the double bond gets the lowest possible number. Since the double bond is between carbon 2 and carbon 3, we assign numbers to the carbon chain as follows: \[ H_3C - CH_3 - CH = CH - H - Br \]
The double bond is between carbons 2 and 3, so the compound is "but-2-ene."

Step 3: Identify the substituents.

There is a methyl group (CH₃) attached to carbon 2.

There is a bromo group (Br) attached to carbon 1.


Step 4: Assign positions to the substituents.
The methyl group is on carbon 2, and the bromo group is on carbon 1. The substituents should be numbered to give the lowest possible numbers, so the correct positions are:

Methyl group on carbon 2

Bromo group on carbon 1


Step 5: Construct the IUPAC name.

The name of the compound is 1-Bromo-2-methylbut-2-ene, where:

"But" indicates the 4-carbon chain (butene).

"2-ene" indicates the double bond between carbons 2 and 3.

"2-methyl" refers to the methyl group on carbon 2.

"1-bromo" refers to the bromo group on carbon 1.


Thus, the correct name is 1-Bromo-2-methylbut-2-ene. Quick Tip: When naming organic compounds, always prioritize giving the lowest possible number to the position of the double bond and substituents. If there are multiple substituents, assign the lowest possible sum of numbers.

Step 1: Understanding Dalton’s Law of Partial Pressures

Dalton’s Law states that the partial pressure of a gas is proportional to its mole fraction in the mixture. The partial pressure is given by: \[ P_{gas} = X_{gas} \times P_{total} \]
Where:

\( P_{gas} \) is the partial pressure of the gas,

\( X_{gas} \) is the mole fraction of the gas,

\( P_{total} \) is the total pressure.


Step 2: Calculate the mole fraction of each gas

We are given the mass percentages of the gases and their molar masses:

Mass percentage of nitrogen (\(N_2\)) = 70.0%

Mass percentage of oxygen (\(O_2\)) = 27.0%

Mass percentage of argon (\(Ar\)) = 3.0%


The moles of each gas are calculated by dividing the mass by the molar mass of each gas: \[ Moles of N_2 = \frac{70}{14} = 5 \, moles \] \[ Moles of O_2 = \frac{27}{16} = 1.6875 \, moles \] \[ Moles of Ar = \frac{3}{40} = 0.075 \, moles \]

Now, the total moles of gases: \[ Total moles = 5 + 1.6875 + 0.075 = 6.7625 \, moles \]

Step 3: Calculate the partial pressures
The mole fraction of each gas is given by: \[ X_{N_2} = \frac{5}{6.7625} = 0.7396, \quad X_{O_2} = \frac{1.6875}{6.7625} = 0.2499, \quad X_{Ar} = \frac{0.075}{6.7625} = 0.0111 \]

Now, calculate the partial pressures: \[ P_{N_2} = X_{N_2} \times P_{total} = 0.7396 \times 1.15 = 0.8516 \, atm \] \[ P_{O_2} = X_{O_2} \times P_{total} = 0.2499 \times 1.15 = 0.2879 \, atm \] \[ P_{Ar} = X_{Ar} \times P_{total} = 0.0111 \times 1.15 = 0.0128 \, atm \]

Step 4: Calculate the ratios
(i) The ratio of the partial pressures of nitrogen to oxygen is: \[ \frac{P_{N_2}}{P_{O_2}} = \frac{0.8516}{0.2879} = 2.96 \]

(ii) The ratio of the partial pressures of oxygen to argon is: \[ \frac{P_{O_2}}{P_{Ar}} = \frac{0.2879}{0.0128} = 11.2 \]

Thus, the correct answer is (4) 2.96, 11.2. Quick Tip: Dalton’s Law of partial pressures allows you to find the partial pressures of gases from their mole fractions. Ensure to use mass percentages and molar masses to first calculate the moles of each gas.

Step 1: Analyzing Statement I

Mohr's salt is composed of three ions: ferrous \( Fe^{2+} \), ammonium \( NH_4^+ \), and sulphate \( SO_4^{2-} \). This statement is true because Mohr's salt (also known as ammonium ferrous sulphate) contains only these three types of ions. Thus, Statement I is correct.

Step 2: Analyzing Statement II

The molar conductance of a salt at infinite dilution is the sum of the molar conductances of the ions it dissociates into. For Mohr's salt, the ions involved are:

Ferrous ion \( Fe^{2+} \), with a molar conductance of \( x_1 \),

Ammonium ion \( NH_4^+ \), with a molar conductance of \( x_2 \),

Sulphate ion \( SO_4^{2-} \), with a molar conductance of \( x_3 \).

At infinite dilution, the total molar conductance \( \lambda_{\infty} \) of Mohr's salt should be the sum of the conductances of these ions. However, Statement II suggests the wrong coefficient for the sulphate ion. The sulphate ion \( SO_4^{2-} \) is a monoatomic ion and should contribute \( x_3 \) to the total conductance, not \( 2x_3 \). Thus, the correct expression for the molar conductance should be: \[ \lambda_{\infty} = x_1 + x_2 + x_3 \]
Therefore, Statement II is incorrect because it incorrectly doubles the contribution of the sulphate ion. Quick Tip: When calculating molar conductance at infinite dilution for ionic compounds, remember that each ion contributes to the total conductance according to its molar conductance, without any extra multiplication unless the ion is polyatomic and contributes differently.

Step 1: Atomic Configuration and Valence Electrons


Chromium (Cr): Atomic number = 24. Its electron configuration is \([ Ar ] 3d^5 4s^1\), giving it 6 valence electrons.
Cobalt (Co): Atomic number = 27. Its electron configuration is \([ Ar ] 3d^7 4s^2\), giving it 9 valence electrons.
Iron (Fe): Atomic number = 26. Its electron configuration is \([ Ar ] 3d^6 4s^2\), giving it 8 valence electrons.
Nickel (Ni): Atomic number = 28. Its electron configuration is \([ Ar ] 3d^8 4s^2\), giving it 10 valence electrons.


Step 2: Relation to Enthalpy of Atomisation

Enthalpy of atomisation generally decreases as the number of valence electrons increases, because more electrons lead to stronger metallic bonding and thus require more energy to break the bonds. Therefore, the element with fewer valence electrons will have the lowest enthalpy of atomisation.

Since Cr has 6 valence electrons, it will have the lowest enthalpy of atomisation compared to Co, Fe, and Ni.

Thus, the correct answer is 6 valence electrons. Quick Tip: In general, for transition metals, a lower number of valence electrons leads to weaker metallic bonding, hence a lower enthalpy of atomisation.

Step 1: Understand the reaction with sodium hydroxide
When a salt, such as ammonium chloride (\( NH_4Cl \)), reacts with sodium hydroxide (\( NaOH \)), ammonia gas (\( NH_3 \)) is evolved: \[ NH_4Cl + NaOH \rightarrow NH_3 + NaCl + H_2O \]
Thus, gas X is ammonia (\( NH_3 \)).

Step 2: Analyze the reagent Y
When ammonia gas is passed through potassium tetraiodomercurate(II) (\( K_2HgI_4 \)) in the presence of potassium hydroxide (KOH), a brown precipitate of mercury(I) iodide (\( Hg_2I_2 \)) is formed: \[ NH_3 + K_2HgI_4 + KOH \rightarrow Hg_2I_2 + KCl + H_2O \]
The brown precipitate is mercury(I) iodide (\( Hg_2I_2 \)).

Thus, reagent Y is \( K_2HgI_4 + KOH \). Quick Tip: Ammonia (\( NH_3 \)) reacts with potassium tetraiodomercurate(II) and potassium hydroxide to form a brown precipitate of mercury(I) iodide (\( Hg_2I_2 \)).

Step 1: Understanding Ionization Enthalpy

Ionization enthalpy refers to the energy required to remove an electron from a neutral atom in its gaseous state. Lower ionization enthalpy indicates that the element can lose electrons more easily.

The elements A and B have the lowest ionization enthalpy values in their group. This suggests that they are relatively easier to ionize compared to other members of their group.



Step 2: Analyzing the Nature of Their Ions
Element A: Since A has a low ionization enthalpy, it will lose electrons easily and thus, the A\(^{2+}\) ion will readily accept electrons (reduced). Hence, element A behaves as a reducing agent.


Element B: On the other hand, element B has a slightly higher ionization enthalpy compared to A. The B\(^{4+}\) ion tends to attract electrons, which makes it a strong oxidizing agent.


Thus, the nature of their ions is reducing (for A\(^{2+}\)) and oxidizing (for B\(^{4+}\)). Quick Tip: In general, elements with low ionization enthalpies are reducing agents because they can easily lose electrons, while elements with high ionization enthalpies tend to attract electrons and thus act as oxidizing agents.

Step 1: Identifying the Metal 'M'

The first transition series metal with the highest enthalpy of atomisation is iron (Fe). Iron has a relatively high enthalpy of atomisation compared to other metals in the first transition series.

Step 2: Analyzing the Aquated Ion

The green colour of the aquated ion \( M^{n+} \) indicates that it is likely to be a transition metal ion, as many transition metal ions exhibit characteristic colours due to d-d transitions. For iron, the \( Fe^{2+} \) ion is known to be green in colour when aquated in water. Therefore, the metal \( M \) is iron.

Step 3: Nature of the Oxide Formed

Iron forms oxides in both the \( +2 \) and \( +3 \) oxidation states. The oxide formed in the \( +2 \) oxidation state is iron(II) oxide (FeO), which is basic in nature. Basic oxides tend to react with acids to form salts and water.

Therefore, the oxide formed by the \( Fe^{2+} \) ion (a metal in the \( +2 \) oxidation state) is basic. Quick Tip: For transition metals in the lower oxidation states (such as \( +2 \)), the oxide formed is usually basic. These oxides react with acids to form salts.

In this question, we are determining which compound is least likely to produce effervescence of CO\(_2\) when reacted with aqueous NaHCO\(_3\). Effervescence occurs when an acid reacts with NaHCO\(_3\), producing CO\(_2\).


Compound (1) contains a hydroxyl group (-OH) and nitro groups (-NO\(_2\)) which will likely result in an acidic environment and cause effervescence with NaHCO\(_3\).

Compound (2) contains a carboxyl group (-COOH), a strong acid, which will react with NaHCO\(_3\) and release CO\(_2\).

Compound (3) contains an amine group (-NH\(_3\)), which is basic and does not typically react with NaHCO\(_3\) to produce CO\(_2\).

Compound (4) contains a nitro group (-NO\(_2\)) but lacks a strongly acidic functional group that would promote CO\(_2\) production. Thus, it is the least likely to produce CO\(_2\) effervescence in the presence of NaHCO\(_3\).


Therefore, the compound least likely to give effervescence of CO\(_2\) is compound (4). Quick Tip: In reactions with NaHCO\(_3\), look for acidic functional groups (like -COOH or -OH) that readily react to release CO\(_2\). Nitrogenous compounds without strong acidic groups generally do not react.

Let's analyze the molecular structures:

ICl\(^-\): This ion has a central iodine atom with 2 bonding pairs and 3 lone pairs, resulting in a 2 : 3 ratio (option III).

H\(_2\)O: The oxygen atom has 2 lone pairs and 2 bonding pairs, which gives the ratio 2 : 2 (option IV).

SO\(_2\): The sulfur atom has 2 bonding pairs and 1 lone pair, leading to a 4 : 1 ratio (option II).

XeF\(_4\): The xenon atom has 4 bonding pairs and 2 lone pairs, giving the ratio 4 : 2 (option I).


Therefore, the correct matching is:
A-III, B-IV, C-II, D-I. Quick Tip: In molecular structures, use the VSEPR theory to determine the number of bonding and lone pairs on the central atom, which helps in predicting the molecular geometry and bond pair to lone pair ratios.

Step 1: Understanding Bacterial Growth Before Treatment

The initial growth of the bacteria follows 1st order kinetics. This means that the rate of bacterial growth is proportional to the current number of bacteria present.

In first-order kinetics, the graph of the number of bacteria \( N \) over time \( t \) (shown as \( \frac{N}{N_0} \)) would exhibit an exponential increase. This is characterized by a rapidly rising curve as bacteria reproduce over time.

Since the rate of decay \( r \) is proportional to the square of the number of bacteria at any instant, \( r \propto N^2 \). Hence, the rate \( r \) increases as \( N \) increases.

Step 2: Introducing the Medicine

Once the medicine is introduced, its effect is to slow down the bacterial growth by increasing the decay rate. The rate of bacterial decay is now proportional to the square of the number of bacteria, i.e., \( r \propto N^2 \).

As the medicine works, the number of bacteria no longer follows the exponential increase. Instead, it starts to flatten, and the growth rate slows down because of the increased bacterial decay rate.

Thus, the growth of bacteria is hindered, and the graph showing bacterial growth should start to flatten out after a certain time.

Step 3: Interpreting the Graphs

Before treatment: The graph of \( \frac{N}{N_0} \) versus \( t \) should show an upward exponential curve (exponential growth), as described earlier for first-order kinetics.

After treatment: Once the medicine is applied, the bacterial growth curve flattens. The graph of rate of decay \( r \) versus the number of bacteria \( N \) should show a sharp decline, as the bacteria are no longer able to grow rapidly.

Step 4: Evaluating the Graphs

Option (1): The graphs do not correctly show the flattening of bacterial growth after the application of the medicine.

Option (2): This option correctly represents the situation, where:

The "before" graph shows an exponential increase of \( \frac{N}{N_0} \) over time.

The "after" graph shows a declining rate of decay \( r \), reflecting the effect of the antibacterial treatment.

Option (3): The "after" graph incorrectly shows an increasing rate of decay, which is not correct after the medicine is applied.

Option (4): This graph shows a linear relationship for the "after" scenario, which is not consistent with bacterial decay behavior under first-order kinetics.


Therefore, Option (2) is the correct set of graphs that represents the situation before and after the application of the medicine. Quick Tip: In first-order kinetics, the growth curve exhibits exponential increase. When an antibacterial treatment is applied, the growth rate slows down, resulting in a flattening of the growth curve.

Step 1: Understanding Statement I

Statement I is incorrect. The reaction described in Statement I is wrong because it says that D-(+)-glucose and D-(+)-fructose combine to form sucrose, which is incorrect. The correct reaction should be the formation of sucrose from glucose and fructose, not the other way around.

Therefore, Statement I is false because it incorrectly describes the formation of sucrose.


Step 2: Understanding Statement II

Statement II is correct. Invert sugar is indeed formed during the hydrolysis of sucrose. When sucrose undergoes hydrolysis (by the action of water), it breaks down into glucose and fructose. This mixture of glucose and fructose is called invert sugar.

Therefore, Statement II is true because invert sugar is indeed formed during sucrose hydrolysis.


Step 3: Conclusion

Statement I is false because it incorrectly describes the synthesis of sucrose.

Statement II is true because it correctly describes the formation of invert sugar during sucrose hydrolysis. Quick Tip: In biochemical processes, the hydrolysis of sucrose results in the formation of invert sugar (a mixture of glucose and fructose), but the synthesis of sucrose is the reverse reaction of this process.

Step 1: Understanding the molecular composition:

The complex is given as Co.5NH\(_3\).Cl.SO\(_4\). It consists of a central Co\(^3+\) ion surrounded by 5 ammonia molecules (NH\(_3\)) and one chloride (Cl) ion as well as a sulfate (SO\(_4^{2-}\)) ion.


Step 2: Identifying the isomers:

Isomer A: The solution of A gives a white precipitate with AgNO\(_3\) solution. This indicates the presence of chloride ions (Cl\(^-\)) in the solution because AgNO\(_3\) reacts with chloride ions to form a white precipitate of AgCl.

Isomer B: The solution of B gives a white precipitate with BaCl\(_2\) solution. This indicates the presence of sulfate ions (SO\(_4^{2-}\)) in the solution because BaCl\(_2\) reacts with sulfate ions to form a white precipitate of BaSO\(_4\).


Step 3: Analyzing the isomerism:

The two isomers A and B are ionisation isomers.

In ionisation isomerism, two or more compounds have the same molecular formula but release different ions in solution.

In this case, the only difference between the two isomers is that in A, chloride (Cl\(^-\)) is the counter-ion, while in B, sulfate (SO\(_4^{2-}\)) is the counter-ion. These two isomers give different ions in solution, which is the defining characteristic of ionisation isomerism.


Step 4: Conclusion:
Since the two isomers differ in the ions that are released into the solution (chloride in A and sulfate in B), this is a classic example of ionisation isomerism. Quick Tip: In ionisation isomerism, even though the molecular formula is the same, the arrangement of ligands around the metal ion and the counter-ions can differ, leading to different ions being released in solution.

Step 1: Reviewing Reaction A:

Reaction A involves the NaOEt base, which induces an E2 elimination reaction. This reaction forms an alkene by eliminating the bromine from the cyclohexane ring. Thus, Reaction A is valid for elimination.

Step 2: Reviewing Reaction B:

Reaction B involves the aqueous KOH, which can induce E2 elimination on the alkyl halide to form an alkene. However, Reaction B is invalid because KOH (aqueous) is more likely to induce nucleophilic substitution (SN2) rather than elimination under these conditions.

Reaction B is not valid for elimination because aqueous KOH typically favors nucleophilic substitution over elimination.


Step 3: Reviewing Reaction C:

Reaction C involves sodium methoxide (\( NaOMe \)), which can induce an E2 elimination reaction to form an alkene by abstracting a proton from the \(\beta\)-carbon. This reaction is valid for elimination.

Reaction C is valid for elimination.


Step 4: Reviewing Reaction D:

Reaction D involves the oxidation of phenol (\( C_6H_5OH \)) with \( Na_2Cr_2O_7 \) and \( H_2SO_4 \), which leads to the formation of a quinone, not an alkene. This is an oxidation reaction, not an elimination reaction.

Reaction D is not valid for elimination.


Step 5: Reviewing Reaction E:
Reaction E involves the use of Cu at 573K, which can dehydrogenate the alcohol to form an alkene via E1 elimination. This reaction works and forms an alkene.
Reaction E is valid for elimination.

Conclusion:

Reactions B and D are the reactions that cannot be applied to prepare an alkene by elimination. Reaction B favors nucleophilic substitution, and Reaction D involves oxidation rather than elimination. Quick Tip: In elimination reactions, strong bases such as NaOEt or NaOMe can induce E2 eliminations, while aqueous KOH typically favors SN2 reactions. Additionally, oxidation reactions like in Reaction D do not lead to alkenes via elimination.

Step 1: Determine the amount of carbon in CO\(_2\):

When the compound undergoes complete combustion, carbon from the compound reacts with oxygen to produce CO\(_2\). The molar mass of CO\(_2\) is: \[ Molar mass of CO_2 = 12 + 2 \times 16 = 44 g/mol \]
The molar mass of carbon (C) is 12 g/mol. In 44 g of CO\(_2\), 12 g is carbon. Therefore, the mass of carbon in 220 mg of CO\(_2\) can be calculated as: \[ Mass of C = \left(\frac{12}{44}\right) \times 220 = 60 mg \]


Step 2: Calculate the percentage of carbon in the compound:

The mass of carbon in the original organic compound is 60 mg. The total mass of the compound is 500 mg. The percentage of carbon in the compound is given by: \[ Percentage of C = \left(\frac{60}{500}\right) \times 100 = 12% \] Quick Tip: When calculating the percentage composition, use the ratio of the mass of the element to the total mass of the compound, then multiply by 100.

Step 1: Count atoms from structure

Carbon (C) = 15 atoms
Hydrogen (H) = 11 atoms
Oxygen (O) = 4 atoms
Nitrogen (N) = 1 atom
Iodine (I) = 4 atoms




Step 2: Compute mass contribution
\[ \begin{aligned} Mass of C &= 15 \times 12 = 180 \, g/mol
Mass of H &= 11 \times 1 = 11 \, g/mol
Mass of O &= 4 \times 16 = 64 \, g/mol
Mass of N &= 1 \times 14 = 14 \, g/mol
Mass of I &= 4 \times 127 = 508 \, g/mol \end{aligned} \]



Step 3: Total molar mass of thyroxine
\[ Molar mass = 180 + 11 + 64 + 14 + 508 = 777 \, g/mol \]



Step 4: Percentage of Iodine
\[ Percentage of Iodine = \frac{508}{777} \times 100 \approx 65.37% \]



Correct Answer: \fbox{65% Quick Tip: To calculate the percentage composition, divide the mass of the element by the molar mass of the compound and multiply by 100.

Step 1: Standard Electrode Potentials

The given standard electrode potentials are:
\[ E^\circ_{Cu^{2+}/Cu} = 0.34 \, V \]
\[ E^\circ_{Ag^+/Ag} = 0.80 \, V \]

Step 2: Applying the Nernst Equation

The Nernst equation gives the relationship between the emf of the electrochemical cell, the standard electrode potentials, and the concentrations of the ions involved: \[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
Where:

\(E^\circ_{cell}\) is the standard cell potential,

\(n\) is the number of electrons involved,

\(Q\) is the reaction quotient.


Step 3: Determining the Standard Cell Potential

The standard cell potential is the difference between the two half-cell potentials: \[ E^\circ_{cell} = E^\circ_{Ag^+/Ag} - E^\circ_{Cu^{2+}/Cu} = 0.80 \, V - 0.34 \, V = 0.46 \, V \]

Step 4: Reaction Quotient (Q)

The reaction quotient \(Q\) is given by the ratio of the concentrations of the products over the reactants. The concentrations of Cu\(^{2+}\) and Ag\(^+\) ions change due to the passage of electricity:

After 1 Faraday is passed through the Cu\(^{2+}\) cell, the concentration of Cu\(^{2+}\) is reduced by a factor based on the number of moles reduced. Since 1 Faraday corresponds to 1 mole of electrons, the concentration of Cu\(^{2+}\) decreases.

Similarly, after 0.1 Faraday is passed through the Ag\(^+\) cell, the concentration of Ag\(^+\) decreases.


Now, using the changes in concentration:

Initially, the concentration of Cu\(^{2+}\) is 1.5 M, and after 1 Faraday, the concentration of Cu\(^{2+}\) decreases as per the stoichiometry.

Initially, the concentration of Ag\(^+\) is 0.2 M, and after 0.1 Faraday, the concentration of Ag\(^+\) decreases accordingly.
\[ Q = \frac{[Cu^{2+}]}{[Ag^+]} \]

Step 5: Substituting Values into the Nernst Equation

Substitute the concentrations and other known values into the Nernst equation: \[ E_{cell} = 0.46 \, V - \frac{0.0591}{1} \log \left( \frac{1.5 \, M}{0.2 \, M} \right) \]
Now, calculate the logarithmic term: \[ \log \left( \frac{1.5}{0.2} \right) = \log (7.5) \approx 0.875 \]
\[ E_{cell} = 0.46 \, V - 0.0591 \times 0.875 \] \[ E_{cell} = 0.46 \, V - 0.0518 \, V \] \[ E_{cell} = 0.4082 \, V \approx 0.40 \, V \] Quick Tip: The Nernst equation helps to adjust the standard electrode potential by accounting for the concentration of the ions involved in the reaction. For each 10-fold change in concentration, the potential changes by approximately 0.0591 V at 298 K.

Step 1: Understanding the Dissociation of Salt MX\(_3\)

The salt MX\(_3\) dissociates in water as follows:

\[ MX_3 \rightarrow M^{3+} + 3X^- \]
For each mole of MX\(_3\), it dissociates to give one mole of M\(^{3+}\) and three moles of X\(^-\).

Step 2: van't Hoff Factor (i)

The van't Hoff factor \(i\) is given as \(i = 2\). The van't Hoff factor represents the total number of particles in solution per formula unit of solute.

In this case, for MX\(_3\), the dissociation would produce 4 particles (1 M\(^{3+}\) and 3 X\(^-\)) per formula unit of MX\(_3\). However, since \(i = 2\), this suggests that the dissociation is not complete, and the actual number of particles formed is only double the initial number of formula units.


Step 3: Using the Formula for Percentage Dissociation

The formula for the percentage dissociation (\(\alpha\)) is given by:

\[ i = 1 + \alpha (n - 1) \]
Where:

\(i\) is the van't Hoff factor (2 in this case),

\(\alpha\) is the degree of dissociation,

\(n\) is the number of ions produced per formula unit of solute (which is 4 for MX\(_3\)).

\[ 2 = 1 + \alpha (4 - 1) \] \[ 2 = 1 + 3\alpha \] \[ 3\alpha = 1 \] \[ \alpha = \frac{1}{3} \]

Step 4: Calculating the Percentage Dissociation

The percentage dissociation is given by:

\[ Percentage dissociation = \alpha \times 100 = \frac{1}{3} \times 100 = 33.33% \]
To the nearest integer, the percentage dissociation is 33%. Quick Tip: The van't Hoff factor \(i\) is a key parameter in determining the degree of dissociation. When \(i\) is less than the theoretical number of particles formed from dissociation, the salt has not fully dissociated.

To determine the number of paramagnetic complexes, we need to check the number of unpaired electrons in each complex. Paramagnetic complexes have at least one unpaired electron, while diamagnetic complexes have all paired electrons.


1. \([FeF_6]^{3-}\):

Iron in \(Fe^{3+}\) has the electron configuration \([Ar] 3d^5\).

Fluoride (\(F^{-}\)) is a weak field ligand and does not cause electron pairing.

\(Fe^{3+}\) undergoes \(d^2sp^3\) hybridization, leaving 5 unpaired electrons in the \(d\)-orbitals.

Paramagnetic.


2. \([Fe(CN)_6]^{3-}\):

Iron in \(Fe^{3+}\) has the electron configuration \([Ar] 3d^5\).

Cyanide (\(CN^{-}\)) is a strong field ligand that causes electron pairing.

\(Fe^{3+}\) undergoes \(d^2sp^3\) hybridization, resulting in no unpaired electrons.

Diamagnetic.


3. \([Mn(CN)_6]^{3-}\):

Manganese in \(Mn^{3+}\) has the electron configuration \([Ar] 3d^4\).

Cyanide (\(CN^{-}\)) is a strong field ligand and causes pairing of electrons.

\(Mn^{3+}\) undergoes \(d^2sp^3\) hybridization, resulting in 2 unpaired electrons.

Paramagnetic.


4. \([Co(C_2O_4)_3]^{3-}\):

Cobalt in \(Co^{3+}\) has the electron configuration \([Ar] 3d^6\).

Oxalate (\(C_2O_4^{2-}\)) is a weak field ligand.

\(Co^{3+}\) undergoes \(d^2sp^3\) hybridization, resulting in 3 unpaired electrons.

Paramagnetic.


5. \([MnCl_6]^{3-}\):

Manganese in \(Mn^{3+}\) has the electron configuration \([Ar] 3d^4\).

Chloride (\(Cl^{-}\)) is a weak field ligand and does not cause electron pairing.

\(Mn^{3+}\) undergoes \(d^2sp^3\) hybridization, leaving 4 unpaired electrons.

Paramagnetic.


6. \([CoF_6]^{3-}\):

Cobalt in \(Co^{3+}\) has the electron configuration \([Ar] 3d^6\).

Fluoride (\(F^{-}\)) is a weak field ligand and does not cause electron pairing.

\(Co^{3+}\) undergoes \(d^2sp^3\) hybridization, leaving 2 unpaired electrons.

Paramagnetic.



Step 2: Conclusion.

The number of paramagnetic complexes is 2. These are:

\([FeF_6]^{3-}\)

\([Mn(CN)_6]^{3-}\)


Thus, the number of paramagnetic complexes is 2. Quick Tip: To determine paramagnetism, look for unpaired electrons in the complex. Strong field ligands usually pair up electrons, while weak field ligands leave electrons unpaired.