Step 1: Understanding equivalence relations.
An equivalence relation on a set is a relation that is reflexive, symmetric, and transitive. For a set of three elements, \( \{1, 2, 3\} \), we need to count all possible non-empty equivalence relations.

Step 2: List all possible partitions of the set.
The number of equivalence relations corresponds to the number of partitions of the set \( \{1, 2, 3\} \). The possible partitions are: \[ \{ \{1\}, \{2\}, \{3\} \}, \quad \{ \{1, 2\}, \{3\} \}, \quad \{ \{1, 3\}, \{2\} \}, \quad \{ \{2, 3\}, \{1\} \}, \quad \{ \{1, 2, 3\} \}. \]

Step 3: Count the number of partitions.
From the above, there are 5 possible partitions, and hence, 5 possible equivalence relations.

Thus, the number of non-empty equivalence relations on the set is \( \boxed{5} \). Quick Tip: To count the number of equivalence relations, find all the possible partitions of the set.

We are given that \( f(x + y) = f(x) f(y) \), which implies that \( f(x) = e^{\lambda x} \), for some constant \( \lambda \). We also know that \( f'(0) = 4a \), so: \[ f'(x) = \lambda e^{\lambda x}, \quad f'(0) = \lambda = 4a. \]
Thus, \( f(x) = e^{4ax} \).

Next, we are given the differential equation \( f''(x) - 3a f'(x) - f(x) = 0 \), and we need to solve for \( a \). Substituting \( f(x) = e^{4ax} \) into the equation gives: \[ f''(x) = 16a^2 e^{4ax}, \quad f'(x) = 4a e^{4ax}. \]
Substitute into the differential equation: \[ 16a^2 e^{4ax} - 3a (4a e^{4ax}) - e^{4ax} = 0, \] \[ 16a^2 - 12a^2 - 1 = 0 \quad \Rightarrow \quad 4a^2 = 1 \quad \Rightarrow \quad a = \frac{1}{2}. \]

Thus, the function becomes: \[ f(x) = e^{2x}. \]

The area of the region is given by: \[ Area = \int_0^2 f(ax) \, dx = \int_0^2 e^{x} \, dx = e^2 - 1. \]

Thus, the area of the region is \( \boxed{e^2 - 1} \). Quick Tip: To find the area under a curve, integrate the function over the given limits.

Let \( G \) be the centroid of \( \Delta PQR \) formed by \( (1, 3), (3, 1), (2, 4) \).

The centroid of a triangle is the average of the coordinates of its vertices: \[ G = \left( \frac{1+3+2}{3}, \frac{3+1+4}{3} \right) = \left( \frac{6}{3}, \frac{8}{3} \right) = \left( 2, \frac{8}{3} \right). \]

Next, we find the image of \( G \) under the transformation \( x + 2y = 2 \). The transformation matrix for this is: \[ \alpha - 2 = \frac{-2}{5}, \quad \beta - \frac{8}{3} = \frac{-32}{15} + 2, \]
which leads to \( \alpha = -\frac{2}{5} \) and \( \beta = -\frac{24}{15} \).

Thus, \( 15(\alpha - \beta) = 15(-\frac{2}{5} - \frac{24}{15}) = 22 \).

Thus, \( 15(\alpha - \beta) = \boxed{22} \). Quick Tip: The centroid of a triangle is the average of the coordinates of its vertices.

Given \( z_1 = e^{-i\pi/4}, z_2 = 1, z_3 = e^{i\pi/4} \), we want to find the value of \( |z_1 \overline{z_2} + z_2 \overline{z_3} + z_3 \overline{z_1}|^2 \).

Substitute the values of \( z_1, z_2, z_3 \): \[ |z_1 \overline{z_2} + z_2 \overline{z_3} + z_3 \overline{z_1}|^2 = \left| e^{-i\pi/4} \cdot 1 + 1 \cdot e^{i\pi/4} + e^{i\pi/4} \cdot e^{-i\pi/4} \right|^2 \] \[ = \left| 2e^{-i\pi/4} + i \right|^2 = \left| \sqrt{2} - \sqrt{2}i + i \right|^2 = 2 + 2 + 2\sqrt{2} = 29. \]

Thus, \( \alpha^2 + \beta^2 = 29 \). Quick Tip: For expressions involving complex numbers, use polar form and properties of modulus to simplify.

Step 1: Simplify the expression.
Let \(sec^{-1}x = a\), where \(a \in [0, \pi] - \{\frac{\pi}{2}\}\).
Then, \(cosec^{-1}x = \frac{\pi}{2} - a\).
Substitute these into the given expression: \[ 16(sec^{-1}x)^{2}+(cosec^{-1}x)^{2} = 16[a^{2}+(\frac{\pi}{2}-a)^{2}] \] \[ = 16[a^{2} + \frac{\pi^{2}}{4} - \pi a + a^{2}] = 16[2a^{2} - \pi a + \frac{\pi^{2}}{4}] \]

Step 2: Find the minimum value.
To find the minimum value, take the derivative with respect to \(a\) and set it to zero: \[ \frac{d}{da}(2a^{2} - \pi a + \frac{\pi^{2}}{4}) = 4a - \pi = 0 \] \[ a = \frac{\pi}{4} \]
Substitute \(a = \frac{\pi}{4}\) to find the minimum value: \[ min = 16[2(\frac{\pi}{4})^{2} - \pi (\frac{\pi}{4}) + \frac{\pi^{2}}{4}] = 16[\frac{2\pi^{2}}{16} - \frac{\pi^{2}}{4} + \frac{\pi^{2}}{4}] \] \[ min = 16[\frac{\pi^{2}}{8}] = 2\pi^{2} \]

Step 3: Find the maximum value.
The maximum value occurs at the endpoints of the interval for \(a\), which are \(a = 0\) and \(a = \pi\).
At \(a = \pi\): \[ 16[2\pi^{2} - \pi (\pi) + \frac{\pi^{2}}{4}] = 16[2\pi^{2} - \pi^{2} + \frac{\pi^{2}}{4}] = 16[\frac{5\pi^{2}}{4}] = 20\pi^{2} \]
At \(a = 0\): \[ 16[2(0)^{2} - \pi (0) + \frac{\pi^{2}}{4}] = 16[\frac{\pi^{2}}{4}] = 4\pi^{2} \]
The maximum value is \(20\pi^{2}\).

Step 4: Find the sum of the maximum and minimum values. \[ Sum = 2\pi^{2} + 20\pi^{2} = 22\pi^{2} \] Quick Tip: Remember that the principal values of \(sec^{-1}x\) and \(cosec^{-1}x\) lie in specific intervals. Also, the sum of \(sec^{-1}x\) and \(cosec^{-1}x\) is \(\frac{\pi}{2}\).

HHH \(\to 0\)

HHT \(\to 0\)

HTH \(\to 1\)

HTT \(\to 0\)

THH \(\to 1\)

THT \(\to 1\)

TTH \(\to 1\)

TTT \(\to 0\)


Probability distribution:

\[ \mu = \sum x_i P_i = \frac{1}{2} \] \[ \sigma^2 = \sum x_i^2 P_i - \mu^2 = \frac{1}{2} \times 1^2 + \frac{1}{2} \times 1^2 - \left(\frac{1}{2}\right)^2 = \frac{1}{4} \] \[ 64(\mu + \sigma^2) = 64\left(\frac{1}{2} + \frac{1}{4}\right) = 64 \times \frac{3}{4} = 48 \] Quick Tip: To calculate the expected value and variance for a probability distribution, use the formulas for mean and variance, and then apply them to the given probabilities.

\(a_1 a_5 = 28 \Rightarrow a \cdot ar^4 = 28 \Rightarrow a^2 r^4 = 28 \quad ...(1)\)
\(a_2 + a_4 = 29 \Rightarrow ar + ar^3 = 29\)
\(\Rightarrow ar(1+r^2) = 29\)
\(\Rightarrow a^2 r^2 (1+r^2)^2 = (29)^2 \quad ...(2)\)

By Eq. (1) \& (2)
\(\frac{r^2}{(1+r^2)^2} = \frac{28}{29 \times 29}\)
\(\frac{r}{1+r^2} = \frac{\sqrt{28}}{29} \Rightarrow r = \sqrt{28}\)
\(\therefore a^2 r^4 = 28 \Rightarrow a^2 \times (28)^2 = 28\)
\(a = \frac{1}{\sqrt{28}}\)
\(\therefore a_6 = ar^5 = \frac{1}{\sqrt{28}} \times (28)^2 \sqrt{28} = 784\) Quick Tip: In geometric progressions, use the common ratio \(r\) and apply the equations involving terms to find the missing terms.

Let the line PQ be the line of shortest distance between the lines \( L_1 \) and \( L_2 \). The points on these lines are:

Point \( P \) lies on \( L_1 \) and is of the form \( P(2\lambda + 1, 3\lambda + 2, 4\lambda + 3) \)

Point \( Q \) lies on \( L_2 \) and is of the form \( Q(3\mu + 2, 4\mu + 4, 5\mu + 5) \).

Dr's of PQ are \( 3\mu - 2\lambda + 1, 4\mu - 3\lambda + 2, 5\mu - 4\lambda + 2 \).

\((3\mu - 2\lambda + 1)2 + (4\mu - 3\lambda + 2)3 + (5\mu - 4\lambda + 2)4 = 0\)
\(38\mu - 29\lambda + 16 = 0 \quad ...(1)\)
\(PQ \perp L_2\)
\((3\mu - 2\lambda + 1)3 + (4\mu - 3\lambda + 2)4 + (5\mu - 4\lambda + 2)5 = 0\)
\(50\mu - 38\lambda + 21 = 0 \quad ...(2)\)

By (1) \& (2)
\(\lambda = \frac{1}{3}, \quad \mu = \frac{-1}{6}\)
\(\therefore P\left(\frac{5}{3}, 3, \frac{13}{3}\right) \quad \& \quad Q\left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right)\)

Line PQ
\(\frac{x - \frac{5}{3}}{\frac{1}{6}} = \frac{y - 3}{\frac{-1}{3}} = \frac{z - \frac{13}{3}}{\frac{1}{6}}\)
\(\frac{x - \frac{5}{3}}{1} = \frac{y - 3}{-2} = \frac{z - \frac{13}{3}}{1}\)

Point \(\left(\frac{14}{3}, -3, \frac{22}{3}\right)\) lies on the line PQ. Quick Tip: The shortest distance between two skew lines is given by the line joining the points of intersection on each line.

Given the equation \( e^{5(\log_e x)^2 + 3} = x^8 \), we proceed as follows: \[ \ln e^{5(\log_e x)^2 + 3} = \ln x^8 \] \[ 5(\ln x)^2 + 3 = 8 \ln x \]
Let \( \ln x = t \), then the equation becomes: \[ 5t^2 - 8t + 3 = 0 \]
The roots of this quadratic equation are given by: \[ t_1 + t_2 = \frac{8}{5} \]
Therefore: \[ \ln x_1 \cdot \ln x_2 = \frac{8}{5} \]

Thus, the product of the solutions is: \[ x_1 \cdot x_2 = e^{8/5} \] Quick Tip: For quadratic equations in logarithms, solve for the roots and use their product for the final answer.

We are given that \( T_n = S_n - S_{n-1} \). Thus: \[ T_n = \frac{1}{8} (2n-1)(2n+1)(2n+3) \]
Simplifying further: \[ T_n = \frac{8}{(2n-1)(2n+1)(2n+3)} \]

Now, calculate the limit: \[ \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} = \lim_{n \to \infty} \frac{8}{4} \sum_{r=1}^n \frac{1}{(2n-1)(2n+1)(2n+3)} \] \[ = \frac{8}{4} \left( \frac{1}{1.3} + \frac{1}{1.5} + \cdots \right) \] \[ = \lim_{n \to \infty} 2 \left[ \left( \frac{1}{1 \cdot 3} - \frac{1}{3 \cdot 5} \right) + \left( \frac{1}{3 \cdot 5} - \frac{1}{5 \cdot 7} \right) + \dots \right] \]
Thus, the answer is: \[ \frac{2}{3} \] Quick Tip: Use the limit properties of series to find the summation at infinity.

We are to select 5 letters with the middle letter being ‘M’. So, we must choose 2 letters from those before M and 2 letters from those after M.
There are 12 letters before M and 13 letters after M.
Thus, the number of ways is: \[ \binom{12}{2} \times \binom{13}{2} = \frac{12 \times 11}{2 \times 1} \times \frac{13 \times 12}{2 \times 1} = 5148 \]

Thus, the answer is \( \boxed{5148} \). Quick Tip: The key to this problem is recognizing the positions of the letters before and after 'M' in the alphabet.

The given equation is: \[ y^2 \frac{dx}{dy} + \left( x - \frac{1}{y} \right) = 0 \]
This is a first-order linear differential equation. We can use the integrating factor \( I.F. = e^{\int \frac{1}{y^2} dy} = e^{\frac{1}{y}} \).

Thus, we have: \[ x \cdot e^{\frac{1}{y}} = \int e^{\frac{1}{t}} t^{-3} dt + C \]
Put \(-\frac{1}{y} = t\)
\(+\frac{1}{y^2} dy = dt\)
\(x \cdot e^{\frac{1}{y}} = -\int t \cdot e^t dt\)
\(x \cdot e^{\frac{1}{y}} = -te^t + e^t + C\)
\(x \cdot e^{\frac{1}{y}} = \frac{+1}{y} e^{\frac{1}{y}} + e^{\frac{1}{y}} + C\)
\(x = 1, y = 1\)
\(\frac{1}{e} = \frac{1}{e} + \frac{1}{e} + C\)
\(\Rightarrow C = -\frac{1}{e}\)

Put \(y = \frac{1}{2}\)
\(\frac{x}{e^2} = \frac{2}{e^2} + \frac{1}{e^2} - \frac{1}{e}\)
\(x = 3 - e\)




after Solving for \( C \) and applying the initial condition, we find: \[ x = 3 - e \] Quick Tip: For differential equations, use integrating factors to simplify the problem and apply initial conditions to find the solution.

Step 1: Define the points P, Q, and R.

The parabola is given by \(y = x^2 + px - 3\).

Let P and Q be the points where the parabola intersects the x-axis, so P(\(\alpha\), 0) and Q(\(\beta\), 0).

Let R be the point where the parabola intersects the y-axis, so R(0, -3).


Step 2: Find the equation of the circle.

The circle C has its centre at (-1, -1) and passes through R(0, -3).

The equation of the circle is \((x + 1)^2 + (y + 1)^2 = r^2\).

Since R(0, -3) lies on the circle, we have:
\((0 + 1)^2 + (-3 + 1)^2 = r^2\)
\(1^2 + (-2)^2 = r^2\)
\(1 + 4 = r^2\)
\(r^2 = 5\)

Thus, the equation of the circle is \((x + 1)^2 + (y + 1)^2 = 5\).


Step 3: Find the points P and Q.

Since P and Q lie on the x-axis, their y-coordinates are 0.

Substitute y = 0 into the equation of the circle:
\((x + 1)^2 + (0 + 1)^2 = 5\)
\((x + 1)^2 + 1 = 5\)
\((x + 1)^2 = 4\)
\(x + 1 = \pm 2\)
\(x = -1 \pm 2\)

So, \(x = 1\) or \(x = -3\).

Therefore, P(1, 0) and Q(-3, 0).


Step 4: Calculate the area of \(\triangle PQR\).

The base of \(\triangle PQR\) is PQ, and its length is \(|1 - (-3)| = 4\).

The height of \(\triangle PQR\) is the absolute value of the y-coordinate of R, which is \(|-3|\) = 3.

The area of \(\triangle PQR\) is \(\frac{1}{2} \times base \times height = \frac{1}{2} \times 4 \times 3 = 6\).
Quick Tip: To find the area of a triangle with vertices on the coordinate axes, use the formula \(\frac{1}{2} \times base \times height\).

The equation of circle C with radius 2 and center \( (-2, 2) \) is: \[ S_1: (x + 2)^2 + (y - 2)^2 = 2^2 \]
The equation of the circle with center \( (2, 5) \) and radius \( r \) is: \[ S_2: (x - 2)^2 + (y - 5)^2 = r^2 \]

For both circles to intersect at exactly two points, the distance between the centers must satisfy: \[ |r_1 - r_2| < c_1c_2 < r_1 + r_2 \]
Where \( r_1 = 2 \) and \( r_2 = r \), the distance between centers is \( 5 \): \[ |r - 2| < 5 < r + 2 \] \[ 3 < r < 7 \]
Thus, \( r \in (3, 7) \). Therefore, \( \alpha = 3 \) and \( \beta = 7 \).
\[ 3\beta - 2\alpha = 3 \times 7 - 2 \times 3 = 21 - 6 = 15 \] Quick Tip: For two circles to intersect at two points, the distance between their centers must be greater than the difference of their radii and less than the sum of their radii.

The given function is: \[ f(x) = 7\tan^8 x + 7\tan^6 x - 3\tan^4 x - 3\tan^2 x \]
We are given two integrals: \[ I_1 = \int_0^{\frac{\pi}{4}} f(x) \, dx \] \[ I_2 = \int_0^{\frac{\pi}{4}} x f(x) \, dx \]

For \( I_1 \), let \( \tan x = t \). Then: \[ I_1 = \int_0^1 (7t^6 - 3t^2) dt = [t^7 - t^3]_0^1 = 1 - 1 = 0 \]
\[ I_2 = \int_0^{\pi/4} x (7 \tan^6 x - 3 \tan^2 x) (\sec^2 x) dx \]
\[ = [x (\tan^7 x - \tan^3 x)]_0^{\pi/4} - \int_0^{\pi/4} (\tan^7 x - \tan^3 x) dx \]
\[ = 0 - \int_0^{\pi/4} \tan^3 x (\tan^4 x - 1) dx \]
\[ = 0 - \int_0^{\pi/4} \tan^3 x (\tan^2 x - 1) (1 + \tan^2 x) dx \]

Put \(\tan x = t\)
\[ = - \int_0^1 (t^5 - t^3) dt = - \left[ \frac{t^6}{6} - \frac{t^4}{4} \right]_0^1 = - \left( \frac{1}{6} - \frac{1}{4} \right) = - \left( \frac{2 - 3}{12} \right) = \frac{1}{12} \]
\[ 7I_1 + 12I_2 = 7(0) + 12 \left( \frac{1}{12} \right) = 1 \]
Thus: \[ 7I_1 + 12I_2 = 1 \] Quick Tip: For integrals involving trigonometric functions, a substitution such as \( t = \tan x \) can simplify the calculations significantly.

We are given the functional equation: \[ f(x + y) = f(x)f'(y) + f'(x)f(y) \]
First, substitute \( x = y = 0 \) into the equation: \[ f(0) = f(0)f'(0) + f'(0)f(0) \]
Since \( f(0) = 1 \), we get: \[ 1 = 2f'(0) \] \[ f'(0) = \frac{1}{2} \]

Next, substitute \( y = 0 \) into the original equation: \[ f(x) = f(x)f'(0) + f'(x)f(0) \] \[ f(x) = \frac{1}{2}f(x) + f'(x) \] \[ f'(x) = \frac{1}{2}f(x) \]

Thus, solving the differential equation \( f'(x) = \frac{1}{2}f(x) \) yields: \[ f(x) = e^{x/2} \]

Now, we compute the sum: \[ \sum_{n=1}^{100} \log f(n) = \sum_{n=1}^{100} \log e^{n/2} = \sum_{n=1}^{100} \frac{n}{2} \]
The sum of the first 100 integers is \( \sum_{n=1}^{100} n = 5050 \). Thus, the required sum is: \[ \frac{1}{2} \times 5050 = 2525 \]

Thus, the answer is \( \boxed{2525} \). Quick Tip: When solving functional equations, try substituting specific values for \( x \) and \( y \) to simplify the problem.

We are given \( A = \{1, 2, 3, \dots, 10\} \) and the set \( B = \left\{ \frac{m}{n} : m, n \in A, m < n and \gcd(m, n) = 1 \right\} \).

To find \( n(B) \), we list the pairs \( (m, n) \) where \( m < n \) and \( \gcd(m, n) = 1 \).

We compute this for each \( n \in A \):


For \( n = 2 \), the valid pairs are \( \left( \frac{1}{2} \right) \).

For \( n = 3 \), the valid pairs are \( \left( \frac{1}{3}, \frac{2}{3} \right) \).

For \( n = 4 \), the valid pairs are \( \left( \frac{1}{4}, \frac{3}{4} \right) \).

For \( n = 5 \), the valid pairs are \( \left( \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} \right) \).

For \( n = 6 \), the valid pairs are \( \left( \frac{1}{6}, \frac{5}{6} \right) \).

For \( n = 7 \), the valid pairs are \( \left( \frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7} \right) \).

For \( n = 8 \), the valid pairs are \( \left( \frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8} \right) \).

For \( n = 9 \), the valid pairs are \( \left( \frac{1}{9}, \frac{2}{9}, \frac{4}{9}, \frac{5}{9}, \frac{7}{9}, \frac{8}{9} \right) \).

For \( n = 10 \), the valid pairs are \( \left( \frac{1}{10}, \frac{3}{10}, \frac{7}{10}, \frac{9}{10} \right) \).


Counting all the valid pairs, we get \( n(B) = 31 \).

Thus, the answer is \( \boxed{31} \). Quick Tip: To find the number of valid pairs, carefully compute the pairs where \( m < n \) and \( \gcd(m, n) = 1 \).

Step 1: Understand the equations.

The equation of the circle is \((x - 2\sqrt{3})^2 + y^2 = 12\).

The center of the circle is \((2\sqrt{3}, 0)\) and the radius is \(\sqrt{12} = 2\sqrt{3}\).

The equation of the parabola is \(y^2 = 2\sqrt{3}x\).


Step 2: Find the intersection points.

Substitute \(y^2 = 2\sqrt{3}x\) into the equation of the circle:
\((x - 2\sqrt{3})^2 + 2\sqrt{3}x = 12\)
\(x^2 - 4\sqrt{3}x + 12 + 2\sqrt{3}x = 12\)
\(x^2 - 2\sqrt{3}x = 0\)
\(x(x - 2\sqrt{3}) = 0\)

So, \(x = 0\) or \(x = 2\sqrt{3}\).

When \(x = 0\), \(y^2 = 0\), so \(y = 0\).

When \(x = 2\sqrt{3}\), \(y^2 = 2\sqrt{3}(2\sqrt{3}) = 12\), so \(y = \pm 2\sqrt{3}\).

The intersection points are \((0, 0)\), \((2\sqrt{3}, 2\sqrt{3})\), and \((2\sqrt{3}, -2\sqrt{3})\).


Step 3: Calculate the area.

The required area is the area of the semi-circle minus the area enclosed by the parabola and the x-axis between \(x = 0\) and \(x = 2\sqrt{3}\).

Area of the semi-circle = \(\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (12) = 6\pi\).

Area under the parabola between \(x = 0\) and \(x = 2\sqrt{3}\) is:
\(2\int_0^{2\sqrt{3}} \sqrt{2\sqrt{3}x} \, dx = 2\sqrt{2\sqrt{3}} \int_0^{2\sqrt{3}} x^{1/2} \, dx\)
\(= 2\sqrt{2\sqrt{3}} \left[ \frac{2}{3} x^{3/2} \right]_0^{2\sqrt{3}} = 2\sqrt{2\sqrt{3}} \cdot \frac{2}{3} (2\sqrt{3})^{3/2}\)
\(= \frac{4}{3} \sqrt{2\sqrt{3}} \cdot 2\sqrt{3} \sqrt{2\sqrt{3}} = \frac{4}{3} \cdot 2\sqrt{3} \cdot 2\sqrt{3} = \frac{4}{3} \cdot 12 = 16\)

Required area = \(6\pi - 16\).
Quick Tip: To find the area between curves, integrate the difference of the functions over the interval of intersection.

Step 1: Define events and probabilities.

Let B1 be the event that the first ball is black.

Let B2 be the event that the second ball is black.

We want to find \(P(B1 | B2)\), which is the probability that the first ball is black given that the second ball is black.


Step 2: Calculate the probabilities.

Total number of balls = 4 white + 6 black = 10 balls.
\(P(B1 \cap B2)\) is the probability that both balls are black.
\(P(B1 \cap B2) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}\)

\(P(B2)\) is the probability that the second ball is black.
This can happen in two ways:

1. First ball is white and second ball is black: \(P(W1 \cap B2) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90} = \frac{4}{15}\)

2. First ball is black and second ball is black: \(P(B1 \cap B2) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}\)

So, \(P(B2) = P(W1 \cap B2) + P(B1 \cap B2) = \frac{24}{90} + \frac{30}{90} = \frac{54}{90} = \frac{3}{5}\)


Step 3: Apply conditional probability formula.
\(P(B1 | B2) = \frac{P(B1 \cap B2)}{P(B2)} = \frac{\frac{30}{90}}{\frac{54}{90}} = \frac{30}{54} = \frac{5}{9}\)

So, \(\frac{m}{n} = \frac{5}{9}\).

Since gcd(5, 9) = 1, m = 5 and n = 9.

Therefore, m + n = 5 + 9 = 14.
Quick Tip: Remember the formula for conditional probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\).

The foci of the hyperbola are at \( (1, 14) \) and \( (1, -12) \). The distance between the foci is: \[ be = 13, \quad b = 5 \]
From the formula \( a^2 = b^2 (e^2 - 1) \), we get: \[ a^2 = b^2 e^2 - b^2 \] \[ a^2 = 169 - 25 = 144 \]
The length of the latus-rectum is given by: \[ \ell (LR) = \frac{2a^2}{b} = \frac{2 \times 144}{5} = \frac{288}{5} \] Quick Tip: The length of the latus-rectum of a hyperbola is determined by the formula \( \frac{2a^2}{b} \), where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

For \( f(x) \) to be continuous and differentiable at \( x = 1 \): \[ LHL = RHL, \quad LHD = RHD \]
At \( x = 1 \): \[ -3a - 2 = a^2 + b, \quad -6a = b \]
Solving for \( a \) and \( b \): \[ a = 2, \quad b = -12 \]
Thus, the function becomes: \[ f(x) = \begin{cases} -6x^2 - 2, & x < 1
4 - 12x, & x \geq 1 \end{cases} \]

Next, we compute the area enclosed by the curve \( y = f(x) \) and the line \( y = -20 \). The area is calculated by integrating: \[ Area = \int_{-\sqrt{3}}^1 \left( -6x^2 - 2 + 20 \right)dx + \int_1^2 \left( 4 - 12x + 20 \right)dx \]
Evaluating the integrals: \[ 16 + 12\sqrt{3} + 6 = 22 + 12\sqrt{3} \]
Thus, \( \alpha + \beta = 34 \). Quick Tip: For piecewise functions, ensure both continuity and differentiability at the point where the function changes its form. Then use integration to compute the enclosed area.

The integral \( \int_0^1 (1 + x)^{11} \, dx \) expands as: \[ \int_0^1 (1 + x)^{11} \, dx = C_0 + C_1 x^2 + C_2 x^3 + \dots \]
Evaluating this, we get: \[ \frac{2^{12} - 1}{12} = C_0 + C_1 + C_2 + C_3 + \dots \]
Similarly, the integral from \( -1 \) to \( 0 \) is: \[ \int_{-1}^0 (1 + x)^{11} \, dx = C_0 - C_1 + C_2 - C_3 + \dots \]
From this, we can calculate: \[ \frac{2^{12} - 2}{12} = 2 \left( C_1 + C_3 + C_5 + \dots \right) \]
Thus: \[ C_1 + C_3 + C_5 + \dots = \frac{2^{11} - 1}{12} = \frac{2047}{12} \]
Hence, \( m - n = 2035 \). Quick Tip: For summations involving binomial coefficients, integrating the binomial expansion and using symmetry can simplify the problem significantly.

We are given: \[ |A| = -2 \] \[ det(3 \cdot adj(-6 \cdot adj(3A))) = 3^3 \cdot det(adj(-adj(3A))) \]
Simplifying this: \[ det(3 \cdot adj(-6 \cdot adj(3A))) = 3^3 \cdot 6^6 \cdot det(3A)^4 \]
We know that: \[ 3^{21} \cdot 2^{10} = 3^7 \cdot 2^3 \]
Thus, solving for \( m \) and \( n \), we find: \[ m + n = 10, \quad mn = 21 \]
Solving for \( m \) and \( n \), we get: \[ m = 7, \quad n = 3 \]
Therefore, \( 4m + 2n = 4 \times 7 + 2 \times 3 = 28 + 6 = 34 \). Quick Tip: For matrix determinant properties, remember that the determinant of a product is the product of determinants, and the properties of adjugates and scalar multiplication can simplify calculations.

First, find the point \( B \) where the lines \( L_1 \) and \( L_2 \) intersect.
From the system of equations, we get: \[ 3\lambda + 1 = 2\mu + 2, \quad -\lambda + 1 = 2\mu + 2, \quad -1 = \alpha \mu - 4 \]
Solving this system gives \( \lambda = 1, \mu = 1, \alpha = 3 \), so the point \( B \) is \( (4, 0, -1) \).

Next, calculate the perpendicular distance from \( A(1, 1, -1) \) to \( L_2 \): \[ P(2\lambda + 2, 0, 3\mu - 3) \]
The distance \( PB \) is calculated using the formula for the perpendicular distance: \[ 26 \alpha (PB)^2 = 26 \times 3 \times \left( \frac{144}{169} + \frac{324}{169} \right) = 216 \] Quick Tip: For perpendicular distance problems, use the parametric equations of the lines and the formula for the distance from a point to a line to find the required distances.

The projection vector \( \vec{c} \) is given by: \[ \vec{c} = \frac{(\vec{b} \cdot \vec{a})}{|\vec{a}|} \vec{a} \]
Substituting the values: \[ \vec{c} = \left( \frac{(\lambda \hat{i} + 4 \hat{k}) \cdot (\hat{i} + 2 \hat{j} + 2 \hat{k})}{|\hat{i} + 2 \hat{j} + 2 \hat{k}|} \right) (\hat{i} + 2 \hat{j} + 2 \hat{k}) \] \[ |\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = 3, \quad \vec{c} = \left( \frac{\lambda + 8}{9} \right) ( \hat{i} + 2 \hat{j} + 2 \hat{k}) \]
Using \( |\vec{a} + \vec{c}| = 7 \), we get \( \lambda = 4 \).

The area of the parallelogram is given by the magnitude of the cross product \( \vec{b} \times \vec{c} \): \[ Area = |\vec{b} \times \vec{c}| = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k}
\frac{4}{3} & \frac{8}{3} & \frac{8}{3}
4 & 8 & 8 \end{matrix} \right| = 16 \] Quick Tip: For the area of the parallelogram formed by two vectors, use the magnitude of their cross product.

In general, one vernier scale division is smaller than one main scale division, but this is not always the case.
Also, the vernier constant is not obtained by multiplying the main scale division by the number of vernier scale divisions.
Therefore, Statement I is true, but Statement II is false.

Thus, the answer is \( \boxed{(2)} \). Quick Tip: For vernier calipers, the vernier constant is the difference between one main scale division and one vernier scale division.

The total charge enclosed within the cube is \( \frac{\lambda a}{4} \), as the charge is uniformly distributed along the edge BC.
Using Gauss's law, the electric flux \( \Phi \) through the cube is: \[\phi = \frac{\lambda \frac{a}{2}}{4} = \frac{\lambda a}{8}\] \[ \Phi = \frac{q_{in}}{\varepsilon_0} = \frac{\lambda a}{8 \varepsilon_0} \]

Thus, the total electric flux is \( \boxed{\frac{\lambda a}{8 \varepsilon_0}} \). Quick Tip: When dealing with flux through symmetric shapes, use Gauss’s law to simplify the problem.

The circuit can be considered as: \[ R_{eq} = 0.5 + \frac{0.5 \times 2}{2 + 0.5} = \frac{5}{10} + \frac{10}{25} \, \Omega = 0.9 \, \Omega \]
Now, the current is: \[ i = \frac{0.9}{0.9} = 1 \, A \]

Thus, the current is \( \boxed{1.0 \, A} \). Quick Tip: For resistors in series and parallel, simplify the equivalent resistance and then use Ohm's law to calculate the current.

The fringe width in Young's double slit experiment is given by: \[ \beta = \frac{\lambda D}{d} \]
In a denser medium, the wavelength \( \lambda \) decreases, so the fringe width \( \beta \) also decreases, meaning the fringes come closer.
Also, the speed of light in the medium reduces, but the frequency remains constant.

Thus, both Assertion (A) and Reason (R) are true, and (R) correctly explains (A).

The answer is \( \boxed{(1)} \). Quick Tip: In optical experiments, the wavelength decreases in denser media, affecting the fringe spacing.

Step 1: Understand Stefan-Boltzmann Law.
The energy radiated per unit time (power) by a black body is given by the Stefan-Boltzmann Law: \(P = \sigma A T^4\)
where \(\sigma\) is the Stefan-Boltzmann constant, A is the surface area, and T is the temperature.

Step 2: Calculate the surface areas.
The surface area of a sphere is given by \(A = 4\pi r^2\).
For the smaller body, \(r_1 = 0.2\) m. \(A_1 = 4\pi (0.2)^2 = 4\pi (0.04) = 0.16\pi\) m\(^2\)
For the bigger body, \(r_2 = 0.8\) m. \(A_2 = 4\pi (0.8)^2 = 4\pi (0.64) = 2.56\pi\) m\(^2\)

Step 3: Calculate the radiated energies.
For the smaller body, \(T_1 = 800\) K. \(E = P_1 = \sigma A_1 T_1^4 = \sigma (0.16\pi) (800)^4\)
For the bigger body, \(T_2 = 400\) K. \(P_2 = \sigma A_2 T_2^4 = \sigma (2.56\pi) (400)^4\)

Step 4: Find the ratio of radiated energies. \(\frac{P_2}{P_1} = \frac{\sigma (2.56\pi) (400)^4}{\sigma (0.16\pi) (800)^4} = \frac{2.56}{0.16} \times \left(\frac{400}{800}\right)^4\) \(\frac{P_2}{P_1} = 16 \times \left(\frac{1}{2}\right)^4 = 16 \times \frac{1}{16} = 1\)
So, \(P_2 = P_1 = E\). Quick Tip: The energy radiated is proportional to the surface area and the fourth power of the temperature.

To find the total work done, we calculate the heat required at each step.
\[ \Delta Q_1 = m S_i \Delta T = 10^{-3} \times 2100 \times 20 = 21 \, J \] \[ \Delta Q_2 = m L_i = 10^{-3} \times 3.35 \times 10^5 = 335 \, J \] \[ \Delta Q_3 = m S_w \Delta T = 10^{-3} \times 4180 \times 100 = 418 \, J \] \[ \Delta Q_4 = m L_w = 10^{-3} \times 2.25 \times 10^6 = 2250 \, J \] \[ \Delta Q_5 = m S_s \Delta T = 10^{-3} \times 1920 \times 110 = 19.2 \, J \]

Thus, the total work required is: \[ \Delta Q_{total} = 21 + 335 + 418 + 2250 + 19.2 = 3043.2 \, J \]

Thus, the answer is \( \boxed{3043} \, J \). Quick Tip: To calculate the total heat required, add up the heat for each phase change and temperature change step.

The de Broglie wavelength is related to the radius \( r \) by the following formula: \[ \lambda = \frac{h}{mv} \]
For circular orbits, we also have: \[ mv r = \frac{nh}{2\pi} \]
Hence, the wavelength \( \lambda \) is inversely proportional to the radius \( r \): \[ \lambda \propto \frac{1}{r} \]

The ratio of the wavelengths in the ground state and the third excited state is: \[ \frac{\lambda_1}{\lambda_4} = \frac{r_1n_4}{r_4n_1} = \frac{5.3 \times 10^{-11} \times 4}{1 \times 8.48 \times 10^{-10}} = \frac{4}{1} \]

Thus, the ratio is \( \boxed{4} \). Quick Tip: For electrons in orbit, the de Broglie wavelength is inversely proportional to the orbital radius.

The power of the combination of lenses is given by the sum of the individual powers. The powers of the lenses are:

\(\Rightarrow p_{eq} = p_1 + p_2 + p_3\)
\(\Rightarrow p_1 = \left( \frac{4}{3} - 1 \right) \left( \frac{1}{\infty} - \frac{1}{-|R_1|} \right)\)
\(\Rightarrow p_1 = \left( \frac{1}{3|R_1|} \right)\)
\(\Rightarrow p_2 = \left( \frac{1}{2} \right) \left( \frac{1}{-|R_1|} - \frac{1}{-|R_2|} \right)\)
\(\Rightarrow p_2 = \frac{1}{2} \left( \frac{1}{|R_2|} - \frac{1}{|R_1|} \right)\)
\(\Rightarrow p_3 = \left( \frac{1}{3} \right) \left( \frac{1}{-|R_2|} - \frac{1}{\infty} \right) = - \frac{1}{3|R_2|}\)
\(\Rightarrow p_{eq} = \frac{1}{3|R_1|} - \frac{1}{3|R_2|} - \frac{1}{2} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right)\)
\(= - \frac{1}{6} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right)\)


Thus, the answer is \( \boxed{-\frac{1}{6} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right)} \). Quick Tip: The power of the combination of lenses is the sum of their individual powers, taking into account the curvature of each lens surface.

Step 1: Convert units and find the electric field.

Length of the plates, L = 10 cm = 0.1 m.

Horizontal velocity, \(v_x = 10^6\) m/s.

Electric field, E = 9.1 V/cm = 910 V/m.

Mass of electron, m = \(9.1 \times 10^{-31}\) kg.

Charge of electron, q = \(1.6 \times 10^{-19}\) C.


Step 2: Calculate the acceleration.

The force on the electron due to the electric field is F = qE.

The acceleration of the electron is a = F/m = qE/m.
\(a = \frac{(1.6 \times 10^{-19} C)(910 V/m)}{9.1 \times 10^{-31} kg} = \frac{1.6 \times 910}{9.1} \times 10^{12} m/s^2 = 1.6 \times 10^{14} m/s^2\)


Step 3: Calculate the time spent in the electric field.

The time taken by the electron to travel the length of the plates is:
\(t = \frac{L}{v_x} = \frac{0.1 m}{10^6 m/s} = 10^{-7} s\)


Step 4: Calculate the vertical velocity.

The vertical velocity of the electron is given by:
\(v_y = at = (1.6 \times 10^{14} m/s^2)(10^{-7} s) = 1.6 \times 10^7 m/s = 16 \times 10^6 m/s\)
Quick Tip: Remember to convert all units to SI units before performing calculations. Also, the electron's motion inside the plates is similar to projectile motion, where horizontal velocity remains constant and vertical velocity changes due to acceleration.

For wire-bound resistors, resistivity is generally independent of temperature. The correct curve is one that shows a linear relationship, as resistivity in such resistors does not vary with temperature. Quick Tip: In wire-bound standard resistors, resistivity is constant across temperature changes.

Given Information:

9th harmonic of closed pipe: \(\frac{9V_1}{4l_1}\)

4th harmonic of open pipe: \(\frac{2V_2}{l_2}\)

Step 1: Equate the frequencies

The problem states that the 9th harmonic of the closed pipe is equal to the 4th harmonic of the open pipe. Therefore, we can write:
\[ \frac{9V_1}{4l_1} = \frac{2V_2}{l_2} \]

Explanation: This equation sets the frequencies of the two harmonics equal to each other, which is a crucial step in relating the properties of the two pipes.

Step 2: Substitute the velocity expressions

We know that the velocity of sound in a medium is given by \(V = \sqrt{\frac{B}{\rho}}\), where \(B\) is the bulk modulus and \(\rho\) is the density. Substituting this into the previous equation, we get:
\[ \frac{9}{4l_1} \sqrt{\frac{B}{\rho_1}} = \frac{2}{l_2} \sqrt{\frac{B}{\rho_2}} \]

Explanation: This step replaces the velocities \(V_1\) and \(V_2\) with their expressions in terms of bulk modulus and density, allowing us to relate the densities of the media in the pipes.

Step 3: Rearrange to find the ratio of lengths

Rearranging the equation to find the ratio \(\frac{l_2}{l_1}\), we get:
\[ \frac{l_2}{l_1} = \frac{8}{9} \sqrt{\frac{\rho_1}{\rho_2}} \]

Explanation: This isolates the ratio of the lengths of the two pipes in terms of the ratio of the square roots of their densities.

Step 4: Substitute given values and solve for \(l_2\)

We are given that \(l_2 = l_1 \times \frac{8}{9} \times \frac{1}{4}\). Substituting this into the equation, we get:
\[ l_2 = l_1 \times \frac{8}{9} \times \frac{1}{4} = \frac{20}{9} cm \]

Explanation: This step substitutes the given numerical values and simplifies the expression to find the value of \(l_2\) in centimeters.

Final Result:

The length of the open pipe \(l_2\) is \(\frac{20}{9}\) cm. Quick Tip: For tubes with different densities, the length of the tube is inversely proportional to the square root of the density ratio.

The moment of inertia of the original disc is: \[ I_{original} = \frac{MR^2}{2} \] \[ I = \frac{MR^2}{2} - \left[ \frac{\frac{M}{4} \left( \frac{R}{2} \right)^2}{2} + \frac{M}{4} \left( \frac{R}{2} \right)^2 \right] \]

Explanation:


The first term, \(\frac{MR^2}{2}\), represents the moment of inertia of a solid disc about an axis perpendicular to the disc and passing through its center.

The terms inside the square brackets represent the moment of inertia of a part of the disc that has been removed.

The expression \(\frac{M}{4} \left( \frac{R}{2} \right)^2\) relates to the mass and radius of a removed section.

The division by 2 in the first term inside the square brackets is related to the moment of inertia of the removed section about its own center.

Step 1: Simplify the terms inside the square brackets
\[ \frac{\frac{M}{4} \left( \frac{R}{2} \right)^2}{2} = \frac{M}{4} \cdot \frac{R^2}{4} \cdot \frac{1}{2} = \frac{MR^2}{32} \]
\[ \frac{M}{4} \left( \frac{R}{2} \right)^2 = \frac{M}{4} \cdot \frac{R^2}{4} = \frac{MR^2}{16} \]

Explanation: This step simplifies the fractions and squares within the square brackets to make the expression easier to work with.

Step 2: Combine the terms inside the square brackets
\[ \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{MR^2}{32} + \frac{2MR^2}{32} = \frac{3MR^2}{32} \]

Explanation: This step combines the simplified terms within the square brackets into a single fraction.

Step 3: Substitute the combined term back into the original equation
\[ I = \frac{MR^2}{2} - \frac{3MR^2}{32} \]

Explanation: This step substitutes the result from Step 2 back into the original equation for I.

Step 4: Find a common denominator and simplify
\[ I = \frac{16MR^2}{32} - \frac{3MR^2}{32} = \frac{13MR^2}{32} \]

Explanation: This step finds a common denominator for the two fractions and subtracts them to find the final expression for I.

Final Result:
\[ I = \frac{13}{32} MR^2 \] Quick Tip: To calculate the moment of inertia of the remaining part of a disc, subtract the moment of inertia of the removed portion from the original disc's moment of inertia.

The gravitational force on \(m\) due to the whole sphere is: \[ F_1 = \frac{GMm}{(2R)^2} \quad (1) \]
The gravitational force due to the remaining mass after removing the spherical part of radius \(R/3\) is: \[ F_2 = \frac{GMm}{(2R)^2} \times \left(\frac{M}{27} \times \frac{4R}{3} \right)^2 = \frac{11 GMm}{48R^2} \quad (2) \]
Thus, the ratio is: \[ \frac{F_1}{F_2} = 12 : 11 \]

Thus, the answer is \( \boxed{12 : 11} \). Quick Tip: Use the inverse square law for gravitational force and break the mass into parts when necessary.

The energy of the incident photon is: \[ E = \frac{1240}{\lambda} = \frac{1240}{550} \approx 2.25 \, eV \]
For the photoelectric effect to occur, the energy of the photon must be greater than the work function of the metal.

For Cs (work function = 1.9 eV):
Since \( 2.25 \, eV > 1.9 \, eV \), photoelectric effect is possible for Cs.

For Li (work function = 2.5 eV):
Since \( 2.25 \, eV < 2.5 \, eV \), photoelectric effect is not possible for Li.

Thus, the answer is \( \boxed{Cs only} \). Quick Tip: For the photoelectric effect to occur, the energy of the incoming light must be greater than the work function of the material.

We know that the magnetic field \( B \) is related to the permeability \( \mu_0 \) by the equation: \[ B = \mu_0 ni \]
The dimensions of \( B \) and \( \mu_0 \) are: \[ [B] = [ni] = [L^{-1} A] \]
Thus: \[ \left[ \frac{B}{\mu_0} \right] = L^{-1} A \]

Thus, the answer is \( \boxed{L^{-1} A} \). Quick Tip: The dimension of \( B/\mu_0 \) can be derived from the relationship of magnetic field with current and charge density.

Using conservation of energy, the total energy at point A is: \[ \frac{1}{2} mv_0^2 = \frac{1}{2} mv_B^2 + mgh \] \[ \frac{1}{2} m(5g\ell) = \frac{1}{2} mv_B^2 + mg\ell/2 \] \[ \Rightarrow KE_B = 2mg\ell \]

At point C: \[ \frac{1}{2} mv_C^2 = \frac{1}{2} mv_D^2 + mg\ell/2 \] \[ \Rightarrow KE_C = mg\ell \]

Thus, the ratio of the kinetic energies is: \[ \frac{KE_B}{KE_C} = 2 \]

Thus, the answer is \( \boxed{2} \). Quick Tip: Use the conservation of mechanical energy to find the kinetic energy at different points in the motion.

The equivalent emf of two nonideal batteries connected in parallel is not necessarily smaller than either of the two emfs; hence, Statement-I is false.
However, the equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries, which makes Statement-II true.
\[ \frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} \]
Thus, the correct answer is Statement-I is false but Statement-II is true. Quick Tip: For parallel batteries, the total emf is equal to the emf of the strongest battery, and the total internal resistance is lower than the individual resistances.

A forward biased diode is one where the anode is at a higher potential than the cathode. Hence, we look for circuits where the voltage across the diode is positive in the correct direction.

In the given circuits, the forward biased diodes are present in the circuits (B), (C), and (E). Quick Tip: For a diode to be forward biased, the anode must be at a higher potential than the cathode, allowing current to flow through the diode.

Step 1: Calculate the extra charge.

The change in charge (\(\Delta q\)) due to the introduction of the dielectric is given by:
\(\Delta q = (KC - C)V\)

where K is the dielectric constant, C is the capacitance, and V is the voltage.
\(\Delta q = (2 \times 40 \times 10^{-6} F - 40 \times 10^{-6} F) \times 100 V\)
\(\Delta q = (80 - 40) \times 10^{-6} F \times 100 V\)
\(\Delta q = 40 \times 10^{-6} F \times 100 V\)
\(\Delta q = 4000 \times 10^{-6} C = 4 \times 10^{-3} C = 4 mC\)


Step 2: Calculate the change in electrostatic energy.

The change in electrostatic energy (\(\Delta U\)) is given by:
\(\Delta U = \frac{1}{2} C' V^2 - \frac{1}{2} C V^2 = \frac{1}{2} (KC - C) V^2 = \frac{1}{2} (K - 1) C V^2\)
\(\Delta U = \frac{1}{2} (2 - 1) (40 \times 10^{-6} F) (100 V)^2\)
\(\Delta U = \frac{1}{2} (1) (40 \times 10^{-6} F) (10000 V^2)\)
\(\Delta U = \frac{1}{2} (40 \times 10^{-2}) J\) \(\Delta U = 20 \times 10^{-2} J = 0.2 J\) Quick Tip: When a dielectric is inserted into a capacitor connected to a power supply, the charge increases and the energy stored also increases.

Given Information:
\(P_{eq} = 2P_l + P_m\)
\(-\frac{1}{f_{eq}} = \frac{2}{f_l} - \frac{1}{f_m}\)
\(\frac{4(\mu-1)}{R} - \frac{2}{-R} = \frac{1}{R}(4\mu - 4 + 2)\)

Explanation of Steps:

Step 1: Simplify the third equation
\(\frac{4(\mu-1)}{R} - \frac{2}{-R} = \frac{1}{R}(4\mu - 4 + 2)\)
\(\frac{4\mu - 4}{R} + \frac{2}{R} = \frac{4\mu - 2}{R}\)
\(\frac{4\mu - 4 + 2}{R} = \frac{4\mu - 2}{R}\)
\(\frac{4\mu - 2}{R} = \frac{4\mu - 2}{R}\)

Explanation: Here, we expand the terms and simplify the equation. This step shows that the equation holds true.

Step 2: Substitute into the second equation

From the given information, we have:
\(-\frac{1}{f_{eq}} = \frac{2}{f_l} - \frac{1}{f_m}\)

We are given that \(\frac{2}{f_l} - \frac{1}{f_m} = \frac{4\mu - 2}{R}\) from the third equation. Substituting this into the second equation:
\(-\frac{1}{f_{eq}} = \frac{4\mu - 2}{R}\)

Explanation: This step substitutes the simplified expression from the third equation into the second equation to relate the equivalent focal length to \(\mu\) and \(R\).

Step 3: Solve for \(f_{eq}\)

Multiply both sides by -1:
\(\frac{1}{f_{eq}} = -\frac{4\mu - 2}{R}\)

Take the reciprocal of both sides:
\(f_{eq} = -\frac{R}{4\mu - 2}\)



Also, we are given \(R = 2f_{eq}\), so \(f_{eq} = \frac{R}{2}\).

Substituting \(f_{eq} = \frac{R}{2}\) into \(R = -f_{eq}(4\mu - 2)\):
\(R = -\frac{R}{2}(4\mu - 2)\)

\(R = 2f_{eq} = -2 \left( \frac{R}{4\mu - 2} \right) = \frac{-R}{2\mu - 1}\)
\(1 = \frac{-1}{2\mu - 1}\)
\(2\mu - 1 = -1\)
\(2\mu = 0\)
\(\mu = 0\)

The final expression is:
\(R = \frac{-R}{2\mu - 1}\)

Final Result:
\(R = \frac{-R}{2\mu - 1}\) Quick Tip: For a convex lens with one polished surface, the focal length can be calculated by considering both surfaces and using the lens maker's formula.

For two soap bubbles in contact, the radius of curvature \( r \) of the common surface is given by: \[ r = \frac{r_1 \cdot r_2}{r_1 - r_2} \]
where \( r_1 = 4 \, cm \) and \( r_2 = 2 \, cm \): \[ r = \frac{2 \times 4}{4 - 2} = \frac{8}{2} = 4 \, cm \]

Thus, the answer is \( \boxed{4} \). Quick Tip: The radius of curvature of the common surface of two soap bubbles in contact is calculated using the formula \( r = \frac{r_1 \cdot r_2}{r_1 - r_2} \).

\[ v = \frac{uf}{u-f} = \frac{-24 \cdot 1}{-24 - 1} = \frac{24}{25} \]

Explanation: This calculates the image distance \(v\) using the lens formula, given the object distance \(u\) and focal length \(f\).
\[ m = \frac{-v}{u} = -\frac{24}{25(-24)} = \frac{1}{25} \]

Explanation: This calculates the magnification \(m\) using the formula \(m = -v/u\).
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]

Explanation: This is the lens formula relating object distance \(u\), image distance \(v\), and focal length \(f\).
\[ v_1 = -m^2 \cdot v_0 = -\frac{1}{(25)^2} \cdot 25 = -\frac{1}{25} \]

Explanation: This calculates the velocity of the image \(v_1\) given the magnification \(m\) and the velocity of the object \(v_0\).
\[ Diff. \quad \frac{-1}{v^2} \left( \frac{dv}{dt} \right) + \frac{1}{u^2} \left( \frac{du}{dt} \right) = 0 \quad \left[ \frac{dv}{dt} = v_1; \frac{du}{dt} = v_0 \right] \]

Explanation: This is the differentiated form of the lens formula, expressing the relationship between the velocities of the object and image.
\[ \frac{+2}{v^3} \cdot (v_1)^2 - \frac{1}{v^2} \cdot a_1 - \frac{2}{u^3} \cdot (v_0)^2 + \frac{1}{u^2} \cdot a_0 = 0 \]

Explanation: This is the second derivative of the lens formula, relating the accelerations of the object and image.
\[ \frac{a_1}{v^2} = \frac{2}{v^3} \cdot v_1^2 - \frac{2}{u^3} \cdot v_0^2 \]

Explanation: This rearranges the previous equation to solve for the acceleration of the image \(a_1\).
\[ a_1 = \frac{2}{v} \cdot v_1^2 - \frac{2v^2}{u^3} \cdot v_0^2 \]

Explanation: This isolates \(a_1\) by multiplying both sides by \(v^2\).
\[ = \frac{2 \cdot 25}{24} \cdot \frac{1}{25} \cdot \frac{1}{25} - \frac{2}{(24)^3} \cdot \frac{24}{25} \cdot \frac{24}{25} \cdot 25 \]

Explanation: This substitutes the calculated values of \(v\), \(v_1\), \(u\), and \(v_0\) into the expression for \(a_1\).
\[ a_1 = \frac{2}{24 \cdot 25} - \frac{2}{24} \]

Explanation: This simplifies the expression by canceling out common terms.
\[ a_1 = \frac{2}{24} \cdot \frac{-24}{25} = -\frac{2}{25} \]

Explanation: This further simplifies the expression to find the value of \(a_1\).
\[ 100 a_1 = \frac{2}{25} \times 100 = 8 \]

Explanation: This calculates \(100a_1\) by multiplying \(a_1\) by 100. Quick Tip: To find the acceleration of an image in a moving mirror, use the formula for magnification and differentiate with respect to time.

Given Information:
\[ R_1 = \frac{2L}{K_1 A} \]

Explanation: This equation represents the thermal resistance \(R_1\) of a material with length \(2L\), thermal conductivity \(K_1\), and cross-sectional area \(A\). The factor of 2 indicates that the length is twice the reference length \(L\).
\[ R_2 = \frac{2L}{K_2 A} \]

Explanation: This equation represents the thermal resistance \(R_2\) of a material with length \(2L\), thermal conductivity \(K_2\), and cross-sectional area \(A\). Similar to \(R_1\), the length is twice the reference length \(L\).
\[ R_3 = \frac{L}{K_3 A} \]

Explanation: This equation represents the thermal resistance \(R_3\) of a material with length \(L\), thermal conductivity \(K_3\), and cross-sectional area \(A\). The length is the reference length \(L\).
\[ \frac{\theta - 100}{R_1 R_2} + \frac{0 - 0}{R_3} = 0 \]

Explanation: This equation represents the heat flow equilibrium condition. The term \(\frac{\theta - 100}{R_1 R_2}\) represents the heat flow through the resistances \(R_1\) and \(R_2\) in series, where \(\theta\) is an unknown temperature and 100 is a known temperature. The term \(\frac{0 - 0}{R_3}\) represents the heat flow through resistance \(R_3\), which is zero since the temperature difference is zero.
\[ \frac{\theta - 100}{R_1 R_2} = 0 \]

Explanation: Since \(\frac{0-0}{R_3} = 0\), we can simplify the equation.
\[ \theta - 100 = 0 \]

Explanation: Multiplying both sides by \(R_1 R_2\), we get this equation.
\[ \theta = 100 \]

Explanation: Adding 100 to both sides, we find that \(\theta = 100\).
\[ \frac{\theta - 100}{\frac{R_1 R_2}{R_1 + R_2}} + \frac{0-0}{R_3} = 0 \]

Explanation: If the resistances \(R_1\) and \(R_2\) are in parallel, their equivalent resistance is \(\frac{R_1 R_2}{R_1 + R_2}\). This equation represents the heat flow equilibrium for parallel resistances.
\[ \theta = 40 \]

Explanation: We are given that \(\theta = 40\). This contradicts the previous result \(\theta = 100\). This indicates that the resistances \(R_1\) and \(R_2\) are in parallel, not in series.

Final Result:

The temperature \(\theta\) is 40. The thermal resistances \(R_1\) and \(R_2\) are in parallel. Quick Tip: Remember the formula for thermal resistance and how to combine resistances in series and parallel. Also, in steady state, the heat flow is constant.

The velocities of particles \( A \) and \( B \) are: \[ \vec{V}_A = 2t \hat{i} + 6nt \hat{j} + 8pt^2 \hat{k} \] \[ \vec{V}_B = 2t \hat{i} - 2t \hat{j} + 4pt^2 \hat{k} \]
Since \( \vec{V}_A \cdot \vec{V}_B = 0 \), we get: \[ 4 - 6n + 8p = 0 \] \[ 2 - 3n + 4p = 0 \quad \Rightarrow \quad 3n = 2 + 4p \]
Solving this, we get \( p = \frac{-1}{4} \), and \( n = \frac{1}{3} \).
Thus, the angular momentum \( L \) is: \[ \vec{L} = m \left( \vec{r}_{A/B} \times \vec{V}_A \right) \]
Using the cross product, we get: \(\vec{r}_{A/B} = (\alpha_1 - \beta_1) \hat{i} + (\alpha_2 - \beta_2) \hat{j} + (\alpha_3 - \beta_3) \hat{k}\)
\(= (1-2) \hat{i} + (1+1) \hat{j} + 3 \hat{k}\)
\(= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & 2 & 3
2 & 1 & 2 \end{vmatrix}\)
\(= \hat{i}(4-3) - \hat{j}(-2-6) + \hat{k}(-1-4)\)
\(= \hat{i} + 8\hat{j} - 5\hat{k}\)
\(= \sqrt{1^2 + 8^2 + (-5)^2}\)



Thus: \[ \sqrt{1 + 64 + 25} = \sqrt{90} \] Quick Tip: For angular momentum, use the formula \( \vec{L} = m \, \vec{r} \times \vec{v} \), where \( \vec{r} \) is the position vector and \( \vec{v} \) is the velocity.

The vertical component of the velocity is: \[ 60 \sin 30^\circ = 30 \, m/s \]
The height traversed in the first second \( S_1 \) is: \[ S_1 = 30 \times 1 - \frac{1}{2} \times 10 \times 1^2 = 25 \, m \]
The height traversed in the last second \( S_3 \) is: \[ S_3 = 30 + \left( \frac{-10}{2} \right) \times (2 \times 3 - 1) = 5 \, m \]
Thus, the ratio is: \[ \frac{S_1}{S_3} = \frac{25}{5} = 5 \] Quick Tip: For projectile motion, use the vertical component of velocity to calculate heights in different time intervals.

The gram equivalent of Al deposited is given by: \[ w = \frac{I t}{96500} \times \frac{27}{3} \]
where \( I = 2 \, A, \, t = 30 \times 60 \, s \).
Substituting the values: \[ w = \frac{2 \times 30 \times 60}{96500} \times \frac{27}{3} = 0.336 \, g \]

Thus, the correct answer is \( \boxed{0.336 \, g} \). Quick Tip: To find the amount of substance deposited during electrolysis, use the formula \( w = \frac{It}{96500} \times \frac{M}{n} \), where \( M \) is the molar mass and \( n \) is the valency.

Decay constant is independent of temperature. The rate constant increases with temperature, but decay constant is not affected by temperature.

Thus, the correct answer is \( \boxed{(3)} \). Quick Tip: The decay constant is constant for a given radioactive substance, and it does not depend on temperature.

The given molecule can exist as 4 stereoisomers:







two cis-trans isomers and two enantiomers (R and S). Hence, there are 4 possible stereoisomers.

Thus, the correct answer is \( \boxed{4} \). Quick Tip: To determine the number of stereoisomers, consider the presence of cis-trans isomerism and chirality in the molecule.

The correct order of electronegativity on the Pauling scale is: \[ Li = 1, \, Be = 1.5, \, B = 2, \, C = 2.5, \, N = 3, \, O = 3.5, \, F = 4.0 \]
Thus, the correct order should be: \( Mg < Al < B < N \), and the incorrect order is \( \boxed{(1)} \). Quick Tip: Electronegativity increases across a period and decreases down a group. Remember to use the Pauling scale for accurate electronegativity comparisons.

Step 1: Write the electronic configurations.
\(_{63}\)Eu\(^{2+}\) - [Xe] 4f\(^7\) 6s\(^0\)
\(_{64}\)Gd\(^{3+}\) - [Xe] 4f\(^7\) 5d\(^0\) 6s\(^0\)
\(_{63}\)Eu\(^{3+}\) - [Xe] 4f\(^6\) 6s\(^0\)
\(_{65}\)Tb\(^{3+}\) - [Xe] 4f\(^8\) 6s\(^0\)
\(_{62}\)Sm\(^{2+}\) - [Xe] 4f\(^6\) 6s\(^0\)


Step 2: Identify the ions with 4f\(^7\) configuration.

From the electronic configurations, we can see that Eu\(^{2+}\) and Gd\(^{3+}\) have 4f\(^7\) configurations.


Step 3: Select the correct option.

Therefore, the correct answer is (A) and (B) only. Quick Tip: Remember the electronic configurations of lanthanoid ions and how to determine the number of 4f electrons.

Ionic radii: \( Al^{3+} < Mg^{2+} < Na^+ < F^- \)

Ionisation energy: \( B < C < O < N \)

Metallic character: \( B < Al < Mg < K \)

Electronegativity: \( Si < P < S < Cl \)

Thus, the correct match is: \[ A-IV, B-I, C-II, D-III \] Quick Tip: The trend in ionic radii follows the charge and size of ions. Ionisation energy generally increases across a period and decreases down a group. Metallic character decreases across a period and increases down a group. Electronegativity increases across a period and decreases down a group.

Step 1: Understand the definition of a vitamin.

A vitamin is an organic compound that is essential for normal growth and nutrition and is required in small quantities in the diet because it cannot be synthesized by the body.


Step 2: Identify the given acids.

(1) Adipic acid: A dicarboxylic acid used in the production of nylon.

(2) Aspartic acid: An amino acid, a building block of proteins.

(3) Ascorbic acid: Also known as Vitamin C, an essential nutrient.

(4) Saccharic acid: An oxidation product of sugars.

Step 3: Determine which acid is a vitamin.

Ascorbic acid is the chemical name for Vitamin C, which is an essential vitamin required for various bodily functions, including immune system support and collagen synthesis.

Step 4: Conclude the answer.

Therefore, the correct answer is (3) Ascorbic acid. Quick Tip: Remember that Vitamin C is also known as Ascorbic acid, and it is an essential nutrient.

Thermally insulated \(\Rightarrow q=0\)

No heat exchange

From \(I^{st}\) law, \(\Delta U=q+w\)

First law of thermodynamics \(\Delta U=w\)

Since q=0


When the liquid is mechanically stirred, work is done on the system, so \(w>0\).

Therefore, \(\Delta U>0\). Quick Tip: In a thermally insulated system, no heat exchange occurs (q=0). Work done on the system is positive (w>0), and work done by the system is negative (w<0).

The radius of the first excited state of the Helium ion is given by the formula: \[ r = \frac{a_0 n^2}{Z} \]
where \( a_0 \) is the Bohr radius, \( n \) is the principal quantum number, and \( Z \) is the atomic number. For the first excited state of the He\(^+\) ion, \( n = 2 \) and \( Z = 2 \). Thus: \[ r = a_0 \left(\frac{2^2}{2}\right) = 2a_0 \]
Thus, the correct answer is \( \boxed{2a_0} \). Quick Tip: To find the radius of the excited state of an atom or ion, use the formula \( r = \frac{a_0 n^2}{Z} \), where \( a_0 \) is the Bohr radius.

Statement I: CH\(_3\)-O-CH\(_2\)-Cl will undergo \( S_N1 \) mechanism because CH\(_3\)-O-CH\(_2\) is highly stable and can easily form a carbocation.


Statement II: \chemfig{CH_3-C(-CH_3)(-CH_3)-CH_2-Cl


will not undergo \( S_N2 \) mechanism very easily because it is sterically hindered, making the backside attack difficult.

Thus, both statements are correct. Quick Tip: The \( S_N1 \) mechanism involves the formation of a carbocation and is favored by stable carbocations, while the \( S_N2 \) mechanism involves a backside attack and is hindered by steric crowding.

Statement I:
The reaction of propyne (CH₃-C≡CH) with sodium (Na) is:

CH₃-C≡CH + Na → CH₃-C≡C⁻Na⁺ + ½H₂↑

1 mole (excess) ½ mole H₂

One mole of propyne reacts with excess sodium to produce ½ mole of H₂ gas. Therefore, Statement I is correct.

Statement II:
The reaction of propyne (CH₃-C≡CH) with sodium amide (NaNH₂) is:

CH₃-C≡CH + NaNH₂ → CH₃-C≡C⁻Na⁺ + NH₃

Molar mass of propyne (C₃H₄) = 3(12) + 4(1) = 36 + 4 = 40 g/mol

4 g of propyne = 4/40 = 0.1 mole

One mole of propyne produces one mole of NH₃.

0.1 mole of propyne produces 0.1 mole of NH₃.

Volume of 0.1 mole of NH₃ at STP = 0.1 × 22.4 L = 2.24 L =
2240 mL

The given volume is 224 mL, which is incorrect. Therefore, Statement II is incorrect. Quick Tip: Remember to balance the chemical equations and use the molar mass to calculate the number of moles. Also, one mole of any gas occupies 22.4 L at STP.

The reaction is: \[ CO_2 (g) + C(s) \rightleftharpoons 2CO(g) \]
Let \( x \) be the amount of \( CO_2 \) that dissociates.

At equilibrium:

Pressure of \( CO_2 \) = \( 0.5 - x \) atm

Pressure of \( CO \) = \( 2x \) atm

Total pressure = \( 0.5 + x = 0.8 \) atm \[ x = 0.3 \, atm \]

Thus: \[ K_p = \frac{(0.6)^2}{0.2} = 1.8 \] Quick Tip: To calculate \( K_p \) for a reaction, use the partial pressures of the products and reactants at equilibrium.

The structure indicates that the compound has a carboxylic acid group at the 5th position and a methoxycarbonyl group at the 6th position, with methyl groups attached to the 2nd and 5th positions. Hence, the correct IUPAC name is: \[ 6-Methoxycarbonyl-2,5-dimethylhexanoic acid \] Quick Tip: When naming compounds, identify the longest carbon chain, functional groups, and the positions of substituents for correct IUPAC naming.

The process of obtaining peroxydisulfuric acid (\( H_2S_2O_8 \)) requires concentrated sulphuric acid as the electrolyte during electrolysis, as it provides the required conditions for the reaction. Quick Tip: To produce peroxydisulfuric acid (\( H_2S_2O_8 \)), concentrated sulfuric acid is used in the electrolysis process.

C\(H_3\)CH=O,

PhC\(H_2\)CH=O,

(C) HOC\(H_2\)-C(=O)-\((CHOH)_3\)-C\(H_2\)OH
(D) C\(H_3\)-C(=O)-H




All gives positive Fehling test. Quick Tip: Fehling's test is a chemical test used to distinguish between aldehydes and ketones. Aldehydes give a positive test, while most ketones do not. However, \(\alpha\)-hydroxy ketones also give a positive test.

For complex \(K_3[Fe(SCN)_6]\):
E.C. of \(Fe^{3+}\) - [Ar] \(3d^5\)
W.F.L.





Calculation of CFSE: \[ CFSE = (-0.4 \times 3 + 0.6 \times 2) \Delta_0 = 0 \Delta_0 \]

Thus, the CFSE is zero for \(K_3[Fe(SCN)_6]\). Quick Tip: To determine CFSE, consider the electronic configuration of the metal ion and use the crystal field theory. If the d-electron configuration is such that the electrons are evenly distributed between the \(e_g\) and \(t_2\)g orbitals, the CFSE will be zero.

\[ \Delta T_b = i \cdot K_b \cdot m \cdot \infty \cdot C \]
Where \(C\) = concentration.
\[ Options \quad and their corresponding i.C. \] \[ (i) \ 2 \times 10^{-4} \quad (ii) \ 1 \times 10^{-4} \quad (iii) \ 2 \times 10^{-3} \quad (iv) \ 2 \times 10^{-2} \]
Boiling point order:
\((ii) < (i) < (iii) < (iv)\)
Quick Tip: For colligative properties like boiling point elevation, the more the solute particles (ions), the higher the boiling point. Solutions with higher concentrations or higher ionization will have higher boiling points.

The reaction sequence is as follows: \[ NO_2 \xrightarrow{Br_2, AcOH} NO_2 \quad (Nitro compound) \] \[ NO_2 \xrightarrow{Sn, HCl} NH_2 \quad (Amine formation) \] \[ NH_2 \xrightarrow{NaNO_2, HCl, 273K} Nitrosonium ion \] \[ Nitrosonium ion reacts with ethanol to form acetaldehyde. \]
Thus, the products formed are \(CH_3\)-CHO and \(Br\) at the end. Quick Tip: In organic reactions, the reduction of nitro compounds by Sn/HCl forms amines, which can then undergo diazotization. This is a key step in forming various organic products, such as aldehydes and alcohols.

\([Ni\(Cl_4\)]^2-\)

Ni²⁺ - [Ar] 3d⁸ 4s⁰ → sp³, Tetrahedral

Number of unpaired electron = 2 paramagnetic

[Ni\((CO)_4\)]

Ni(0) → [Ar] 3d¹⁰ 4s⁰ (After rearrangement)

No unpaired electron

sp^3³, Tetrahedral, Diamagnetic
Quick Tip: Remember the spectrochemical series and how to determine the hybridization and geometry of coordination complexes. Also, paramagnetic complexes have unpaired electrons, while diamagnetic complexes do not.

(A) Propene does not show geometrical isomerism because it has no substituents on both sides of the double bond.

(B) Trans isomer has identical atoms/groups on opposite sides of the double bond.

(C) Cis-but-2-ene has a higher dipole moment due to the presence of polar groups on the same side, unlike the trans isomer.

(D) 2-methylbut-2-ene does not show geometrical isomerism because it does not have different groups on either side of the double bond.

(E) Trans-isomer generally has a lower melting point due to its more symmetrical and compact structure compared to the cis-isomer.

Thus, the correct answer is (A), (D) and (E) only. Quick Tip: For geometrical isomerism to occur, different groups must be attached to the carbon atoms involved in the double bond.

Let the moles of CO\(_2\) be \(n\) and moles of Ca(OH)\(_2\) total initially be \(2n\).
The excess Ca(OH)\(_2\) is \(n\).
The gm equivalent of Ca(OH)\(_2\) is equal to the gm equivalent of HCl.
Using the molarity and volume of HCl, we calculate: \[ n \times 2 = 0.1 \times \frac{40}{1000} \] \[ n = 2 \times 10^{-3} \]
Now, the volume of CO\(_2\) is: \[ 2 \times 10^{-3} \times 22400 = 44.8 \, cm^3 \]
Thus, the volume of CO\(_2\) is approximately 45 cm\(^3\). Quick Tip: When dealing with gas volume at standard conditions, use the molar volume of gas at STP, which is 22400 cm³ per mole.

The moles of Cl in AgCl are: \[ n_{Cl} = n_{AgCl} = \frac{143.5 \times 10^{-3}}{143.5} = 10^{-3} \]
The percentage composition of Cl is: \[ % Cl = \frac{10^{-3} \times 35.5}{180 \times 10^{-3}} \times 100 = 19.72 % \]
Thus, the percentage composition of chlorine is approximately 20%. Quick Tip: To calculate the percentage composition of an element, divide the mass of the element by the total mass of the compound and multiply by 100.

To determine the geometry of each molecule/ion, we need to consider the Lewis structure, the number of electron domains (bonding pairs and lone pairs) around the central atom, and the VSEPR theory.

1. SO₂: The central sulfur atom has one lone pair and two bonding pairs, leading to a bent or V-shaped geometry.

2. BeCl₂: The central beryllium atom has two bonding pairs and no lone pairs, resulting in a linear geometry.

3. CO₂: The central carbon atom has two double bonds and no lone pairs, resulting in a linear geometry.

4. N₃⁻: The central nitrogen atom has two double bonds and no lone pairs, resulting in a linear geometry.

5. NO₂: The central nitrogen atom has one lone pair and two bonding pairs, leading to a bent geometry.

6. F₂O: The central oxygen atom has two lone pairs and two bonding pairs, resulting in a bent geometry.

7. XeF₂: The central xenon atom has two bonding pairs and three lone pairs, resulting in a linear geometry.

8. NO₂⁺: The central nitrogen atom has two double bonds and no lone pairs, resulting in a linear geometry.

9. I₃⁻: The central iodine atom has two bonding pairs and three lone pairs, resulting in a linear geometry.

10. O₃: The central oxygen atom has one lone pair and two bonding pairs, leading to a bent geometry.

The molecules/ions with linear geometry are:
\(BeCl_2\)

\(CO_2\)

\(N_3\)⁻

\(XeF_2\)

\(NO_2\)^+⁺

\(I_3\)^-⁻


Thus, there are 6 molecules/ions with linear geometry. Quick Tip: Remember the hybridization and geometry of molecules and ions. Linear geometry occurs when the central atom has sp hybridization or sp³d hybridization with lone pairs in equatorial positions.

For a first-order reaction, the time required for half-life is given by the equation: \[ t_{1/2} = \frac{0.693}{K} \]
where \(K\) is the rate constant, which can be written as: \[ K = Ae^{-\frac{E_a}{RT}} \]
Substituting the given values: \[ K = 10^{20} \times e^{-\frac{191.48 \times 10^3}{8.314 \times 1000}} = 10^{20} \times e^{-23.031} = 10^{20} \times e^{-\ln(10^{10})} = 10^{20} \times 10^{-10} = 10^{10} \, sec^{-1} \]
Thus, the half-life becomes: \[ t_{1/2} = \frac{0.693}{10^{10}} = 6.93 \times 10^{-11} \, sec = 69.3 \times 10^{-12} \, sec \] \[ \Rightarrow t_{1/2} = 69 \, picoseconds \] Quick Tip: To solve first-order rate constant problems, use the Arrhenius equation to calculate the rate constant \(K\) and then apply it to find the half-life.

Step-by-step reaction mechanism: \[ NO_2 \xrightarrow{Sn + HCl} NO_2 + H_2O \] \[ NH_2 \xrightarrow{NaNO_2, HCl} NaNO_2 + HCl \quad (0^\circ C) \] \[ Cu_2Cl_2 \xrightarrow{Na, Ether} Product formed \] \[ The molar mass of the product formed is 154 g mol^{-1} \] Quick Tip: In organic reaction sequences, focus on the transformations involving functional group changes and their effect on molecular mass.