Step 1: Understanding the Concept:

The given coordinates are expressed in terms of \(\theta\).

The objective is to eliminate the parameter \(\theta\) to find a relationship between \(x\) and \(y\) that fits the form of the given equation.

We observe that the arguments of the tangent functions differ by a constant value.


Step 2: Key Formula or Approach:

Let \(A = \theta + \frac{\pi}{3}\) and \(B = \theta + \frac{\pi}{6}\).

The difference is \(A - B = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}\).

We use the compound angle formula: \[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]

Step 3: Detailed Explanation:

From the given coordinates:
\[ x = 3 \tan A \implies \tan A = \frac{x}{3} \] \[ y = 2 \tan B \implies \tan B = \frac{y}{2} \]
Substitute these values into the tangent identity:
\[ \tan\left(\frac{\pi}{6}\right) = \frac{\frac{x}{3} - \frac{y}{2}}{1 + \left(\frac{x}{3}\right)\left(\frac{y}{2}\right)} \] \[ \frac{1}{\sqrt{3}} = \frac{\frac{2x - 3y}{6}}{\frac{6 + xy}{6}} \] \[ \frac{1}{\sqrt{3}} = \frac{2x - 3y}{6 + xy} \]
Rearranging the terms to match \(xy + \alpha x + \beta y + \gamma = 0\):
\[ 6 + xy = \sqrt{3}(2x - 3y) \] \[ xy - 2\sqrt{3}x + 3\sqrt{3}y + 6 = 0 \]
By comparison, we find:
\(\alpha = -2\sqrt{3}\), \(\beta = 3\sqrt{3}\), and \(\gamma = 6\).


Step 4: Final Answer:

We calculate the required sum of squares:
\[ \alpha^2 + \beta^2 + \gamma^2 = (-2\sqrt{3})^2 + (3\sqrt{3})^2 + (6)^2 \] \[ \alpha^2 + \beta^2 + \gamma^2 = (4 \times 3) + (9 \times 3) + 36 \]
\[ \alpha^2 + \beta^2 + \gamma^2 = 12 + 27 + 36 = 75 \).
Quick Tip: When parametric forms involve \(\tan(A)\) and \(\tan(B)\), check if \((A \pm B)\) is a constant angle. Using \(\tan(A \pm B)\) helps eliminate the parameter quickly.

Step 1: Understanding the Concept:

A complex number \(w\) is purely real if \(w = \bar{w}\) and purely imaginary if \(w + \bar{w} = 0\).

For points on the unit circle \(|z|=1\), we have \(z\bar{z} = 1\), which implies \(\bar{z} = \frac{1}{z}\).


Step 2: Detailed Explanation:

Evaluating (S1):

Let \(w = \frac{z-i}{z+i}\). For \(w\) to be purely real, \(w = \bar{w}\):
\[ \frac{z-i}{z+i} = \frac{\bar{z}+i}{\bar{z}-i} \] \[ (z-i)(\bar{z}-i) = (\bar{z}+i)(z+i) \] \[ z\bar{z} - iz - i\bar{z} - 1 = \bar{z}z + i\bar{z} + iz - 1 \] \[ -i(z + \bar{z}) = i(z + \bar{z}) \implies 2i(z + \bar{z}) = 0 \implies Re(z) = 0 \]
Given \(|z|=1\) and \(Re(z)=0\), \(z\) can be \(i\) or \(-i\).

The domain excludes \(-i\), so \(z=i\) is the only element.

Statement (S1) is incorrect as it contains only one element.


Evaluating (S2):

Let \(w = \frac{z-1}{z+1}\). For \(w\) to be purely imaginary, \(w + \bar{w} = 0\):
\[ \frac{z-1}{z+1} + \frac{\bar{z}-1}{\bar{z}+1} = 0 \] \[ (z-1)(\bar{z}+1) + (\bar{z}-1)(z+1) = 0 \] \[ z\bar{z} + z - \bar{z} - 1 + \bar{z}z - z + \bar{z} - 1 = 0 \] \[ 2|z|^2 - 2 = 0 \implies 2(1) - 2 = 0 \]
This identity holds for all \(z\) such that \(|z|=1\).

Since there are infinitely many such points on the unit circle (excluding \(z=-1\)), the set contains infinitely many elements.

Statement (S2) is correct.


Step 3: Final Answer:

Only statement (S2) is correct. Quick Tip: For any point on the unit circle \(|z|=1\), the expression \(\frac{z-1}{z+1}\) is always purely imaginary because the vector from \(-1\) to \(z\) is perpendicular to the vector from \(1\) to \(z\) (angle in a semicircle).

Step 1: Understanding the Concept:

For a quadratic equation \(ax^2 + bx + c = 0\) to have negative real roots, three conditions must be satisfied:

1. Roots are real: \(D \ge 0\).

2. Sum of roots is negative: \(\frac{-b}{a} < 0\).

3. Product of roots is positive: \(\frac{c}{a} > 0\).


Step 2: Detailed Explanation:

Given: \(x^2 - (p+2)x + (2p+9) = 0\).

Condition 1: \(D \ge 0\)
\[ (p+2)^2 - 4(2p+9) \ge 0 \] \[ p^2 + 4p + 4 - 8p - 36 \ge 0 \implies p^2 - 4p - 32 \ge 0 \] \[ (p-8)(p+4) \ge 0 \implies p \in (-\infty, -4] \cup [8, \infty) \quad \dots(i) \]
Condition 2: Sum \(< 0\)
\[ Sum = p + 2 < 0 \implies p < -2 \quad \dots(ii) \]
Condition 3: Product \(> 0\)
\[ Product = 2p + 9 > 0 \implies p > -4.5 \quad \dots(iii) \]
Intersecting (i), (ii), and (iii):
\(p \in (-4.5, -4]\).

Thus, \(\alpha = -4.5\) and \(\beta = -4\).


Step 3: Final Answer:

\[ \beta - 2\alpha = -4 - 2(-4.5) = -4 + 9 = 5 \).
Quick Tip: Location of roots problems require checking the discriminant, the vertex position, and the value of the function at boundary points. For "negative roots", the boundary point is always \(x=0\).

Step 1: Understanding the Concept:

A system of three linear equations in three variables has infinitely many solutions if the equations are linearly dependent.

This happens when one equation can be expressed as a linear combination of the others.


Step 2: Detailed Explanation:

Let \(L_1: 2x + 3y + 5z = 9\), \(L_2: 7x + 3y - 2z = 8\), and \(L_3: 12x + 3y - (4+\lambda)z = 16 - \mu\).

Subtracting \(L_1\) from \(L_2\):
\[ 5x - 7z = -1 \quad \dots(i) \]
Subtracting \(L_2\) from \(L_3\):
\[ 5x - (2+\lambda)z = 8 - \mu \quad \dots(ii) \]
For infinitely many solutions, equations (i) and (ii) must represent the same plane (or line in \(x-z\) projection):

Comparing coefficients:
\(2 + \lambda = 7 \implies \lambda = 5\).
\(8 - \mu = -1 \implies \mu = 9\).

The center of the circle is \((5, 9)\).


Step 3: Finding the Radius:

The radius \(r\) is the perpendicular distance from \((5, 9)\) to the line \(4x - 3y = 0\):
\[ r = \frac{|4(5) - 3(9)|}{\sqrt{4^2 + (-3)^2}} \] \[ r = \frac{|20 - 27|}{5} = \frac{7}{5} \].


Step 4: Final Answer:

The radius of the circle is \(\frac{7}{5}\). Quick Tip: In systems of equations, if the coefficient of one variable is identical across all equations, subtraction is the fastest way to reduce the system to two variables.

Step 1: Key Formula or Approach:

For an \(m \times m\) matrix \(A\), \(|adj(A)| = |A|^{m-1}\).

Applying this twice: \(|adj(adj(A))| = |A|^{(m-1)^2}\).


Step 2: Detailed Explanation:

Given \(m = 3\) and \(|adj(adj(A))| = 81\):
\[ |A|^{(3-1)^2} = 81 \implies |A|^4 = 3^4 \implies |A| = 3 (taking positive root for base similarity) \]
Now solve the equation for \(n \in S\):
\[ (81)^{\frac{(n-1)^2}{2}} = 3^{(3n^2 - 5n - 4)} \] \[ (3^4)^{\frac{(n-1)^2}{2}} = 3^{(3n^2 - 5n - 4)} \implies 3^{2(n-1)^2} = 3^{(3n^2 - 5n - 4)} \]
Equating powers:
\[ 2(n^2 - 2n + 1) = 3n^2 - 5n - 4 \] \[ 2n^2 - 4n + 2 = 3n^2 - 5n - 4 \] \[ n^2 - n - 6 = 0 \implies (n-3)(n+2) = 0 \implies n = 3, -2 \]
So \(S = \{3, -2\}\).


Step 3: Final Answer:

Calculate the sum:

For \(n = 3\), \(\left| A^{\frac{3^2+3}{2}} \right| = |A|^6 = 3^6 = 729\).

For \(n = -2\), \(\left| A^{\frac{(-2)^2+(-2)}{2}} \right| = |A|^1 = 3^1 = 3\).

Sum \(= 729 + 3 = 732 \).
Quick Tip: Remember \)|adj(adj(A))| = |A|^{(n-1)^2}\( is a standard result. Always convert all bases to the simplest prime factor (like 3 here) to solve exponential equations easily.

Step 1: Key Formula or Approach:

Let the G.P. be \(a, ar, ar^2, ar^3\).

The resulting A.P. is \((a-2), (ar-7), (ar^2-9), (ar^3-5)\).

In an A.P., the common difference is constant.


Step 2: Detailed Explanation:
\(d = (ar-7) - (a-2) = ar - a - 5 \quad \dots(i)\)
\(d = (ar^2-9) - (ar-7) = ar^2 - ar - 2 \quad \dots(ii)\)
\(d = (ar^3-5) - (ar^2-9) = ar^3 - ar^2 + 4 \quad \dots(iii)\)

Equating (i) and (ii): \(a(r-1) - 5 = ar(r-1) - 2 \implies ar(r-1) - a(r-1) = -3 \implies a(r-1)^2 = -3\).

Equating (ii) and (iii): \(ar(r-1) - 2 = ar^2(r-1) + 4 \implies ar^2(r-1) - ar(r-1) = -6 \implies ar(r-1)^2 = -6\).

Dividing the two results:
\(\frac{ar(r-1)^2}{a(r-1)^2} = \frac{-6}{-3} \implies r = 2\).

Substituting \(r=2\) back: \(a(1)^2 = -3 \implies a = -3\).

The numbers are \(x_1 = -3, x_2 = -6, x_3 = -12, x_4 = -24\).


Step 3: Final Answer:

Product \(x_1 x_2 x_3 x_4 = (-3)(-6)(-12)(-24) = 5184\).

Value \(= \frac{5184}{24} = 216 \).
Quick Tip: When a G.P. is transformed into an A.P., utilize the property of common differences. Expressing the common difference in terms of \)a\( and \)r\( often leads to equations where terms can be factored as \)a(r-1)^2\(.

Step 1: Understanding the Concept:

We use the concept of congruences in modular arithmetic.

If \(a \equiv b \pmod{m}\), then \(a^k \equiv b^k \pmod{m}\).


Step 2: Detailed Explanation:

First, evaluate the base 64 modulo 7:
\(64 = 63 + 1 = 9 \times 7 + 1\).

So, \(64 \equiv 1 \pmod{7} \).

Now, raising to the power:

\)(64)^{64 \equiv 1^{64 \pmod{7 \equiv 1 \pmod{7 \).

And finally:
\(((64)^{64})^{64} \equiv 1^{64} \pmod{7} \equiv 1 \pmod{7} \).


Step 3: Final Answer:

The remainder is 1. Quick Tip: For remainder problems, always express the base as \)(mq \pm 1)\(. If the base is \)(mq + 1)\(, any positive integer power of that base will leave a remainder of 1 when divided by \)m\(.

Step 1: Understanding the Concept:

Since a specific batsman (Captain) and a specific bowler (Vice-Captain) must be included, they are fixed.

We need to select the remaining \(10 - 2 = 8\) players from the remaining \(7 - 1 = 6\) batsmen and \(6 - 1 = 5\) bowlers.


Step 2: Detailed Explanation:

Total team requirements: \(\ge 4\) Batsmen and \(\ge 4\) Bowlers.

Including the fixed ones, we need:

1. At least 3 more Batsmen (total \(1+3=4\)).

2. At least 3 more Bowlers (total \(1+3=4\)).

Let \(x\) be the number of additional batsmen and \(y\) be the number of additional bowlers.

We need \(x + y = 8\) where \(x \in \{3, 4, 5, 6\}\) and \(y \in \{3, 4, 5\}\).

Possible Cases \((x, y)\):

- Case 1: \(x=3, y=5 \implies \binom{6}{3} \times \binom{5}{5} = 20 \times 1 = 20\).

- Case 2: \(x=4, y=4 \implies \binom{6}{4} \times \binom{5}{4} = 15 \times 5 = 75\).

- Case 3: \(x=5, y=3 \implies \binom{6}{5} \times \binom{5}{3} = 6 \times 10 = 60\).

(Note: \(x=6, y=2\) is not possible as \(y \ge 3\)).


Step 3: Final Answer:

Total ways \(= 20 + 75 + 60 = 155 \).
Quick Tip: When specific individuals are mandatory, always reduce both the "items to be chosen" and the "pool to choose from" by that amount before forming combinations.

Step 1: Key Formula or Approach:

Mean \(\bar{x} = \frac{\sum x_i}{n}\).

Variance \(\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2\).


Step 2: Correcting the Mean:

Original sum \(= 100 \times 40 = 4000\).

Corrected sum \(= 4000 - 50 + 40 = 3990\).

Correct mean \(\mu = \frac{3990}{100} = 39.9\).


Step 3: Correcting the Standard Deviation:

Given original \(\sigma = 5.1\), so \(\sigma^2 = 26.01\).
\(26.01 = \frac{\sum x_{old}^2}{100} - 40^2 \implies \frac{\sum x_{old}^2}{100} = 1626.01 \implies \sum x_{old}^2 = 162601\).

Corrected \(\sum x_{new}^2 = 162601 - 50^2 + 40^2 = 162601 - 2500 + 1600 = 161701\).

Correct variance \(\sigma_{new}^2 = \frac{161701}{100} - (39.9)^2 = 1617.01 - 1592.01 = 25\).

Correct standard deviation \(\sigma = \sqrt{25} = 5\).


Step 4: Final Answer:
\(10(\mu + \sigma) = 10(39.9 + 5) = 10(44.9) = 449 \).
Quick Tip: When correcting data, always compute the corrected sum (\)\sum x\() and corrected sum of squares (\)\sum x^2\() first. This ensures you account for the change in both mean and variance correctly.

Step 1: Understanding the Concept:

By the definition of a parabola, the distance of a point \((x, y)\) from the focus is equal to its perpendicular distance from the directrix.


Step 2: Detailed Explanation:

Focus \(S = (-2, 1)\), Directrix \(d: 2x + y + 2 = 0\).

Let \((x, y)\) be a point on the parabola. Then:
\[ (x + 2)^2 + (y - 1)^2 = \frac{(2x + y + 2)^2}{2^2 + 1^2} \]
We are given the abscissa \(x = -2\). Substituting this into the equation:
\[ (-2 + 2)^2 + (y - 1)^2 = \frac{(2(-2) + y + 2)^2}{5} \] \[ (y - 1)^2 = \frac{(y - 2)^2}{5} \] \[ 5(y^2 - 2y + 1) = y^2 - 4y + 4 \] \[ 4y^2 - 6y + 1 = 0 \]

Step 3: Final Answer:

The sum of the ordinates (roots of the quadratic in \(y\)) is:
\(Sum = \frac{-b}{a} = \frac{-(-6)}{4} = \frac{3}{2} \).
Quick Tip: For any conic section problem involving "sum of coordinates", first plug the fixed coordinate into the general locus equation and use Vieta's formulas. You don't need to solve for individual coordinates.

Step 1: Understanding the Concept:

The vertices of the triangle can be found by intersecting the given lines. Since B and C lie on the x-axis, their y-coordinates are zero. The orthocentre of a triangle is the point where the three altitudes intersect. Once the coordinates of P, B, and C are found, the area can be calculated using the standard triangle area formula.


Step 2: Key Formula or Approach:

1. Vertices B and C lie on the x-axis (\(y=0\)).

2. Vertex A is the intersection of lines AB and AC.

3. Orthocentre P is the intersection of altitudes.

4. Area of triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is \(\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\).


Step 3: Detailed Explanation:

Find Vertex A: Solve \(3y - x = 2\) and \(x + y = 2\).

Adding the equations: \(4y = 4 \implies y = 1\).

Substituting \(y = 1\) in \(x + y = 2 \implies x = 1\). So, \(A = (1, 1)\).

Find Vertices B and C on x-axis (\(y=0\)):

For B (on AB): \(3(0) - x = 2 \implies x = -2\). So, \(B = (-2, 0)\).

For C (on AC): \(x + 0 = 2 \implies x = 2\). So, \(C = (2, 0)\).

Find Orthocentre P:

Altitude from A to BC (x-axis) is a vertical line passing through \(x = 1\). So, the line is \(x = 1\).

Altitude from C to AB: Slope of AB is \(1/3\), so slope of altitude is \(-3\).

Equation of altitude from \(C(2, 0)\): \(y - 0 = -3(x - 2) \implies y = -3x + 6\).

Intersection of \(x = 1\) and \(y = -3x + 6\):
\(y = -3(1) + 6 = 3\).

Thus, the orthocentre is \(P = (1, 3)\).


Step 4: Final Answer:

Area of \(\triangle PBC\) with vertices \(P(1, 3), B(-2, 0), C(2, 0)\):

Base \(BC = |2 - (-2)| = 4\).

Height = y-coordinate of \(P = 3\).

\[ Area = \frac{1{2 \times \text{base \times \text{height = \frac{1{2 \times 4 \times 3 = 6 \).
Quick Tip: If the base of a triangle lies on a coordinate axis (like the x-axis here), the area is simply \(\frac{1{2} \times (difference in x-coordinates) \times (absolute y-coordinate of the third vertex)\).

Step 1: Understanding the Concept:

A circle in the third quadrant touching both axes with radius \(r\) has its centre at \((-r, -r)\). Since the circles touch externally, the point of contact lies on the line segment connecting the two centres and divides it in the ratio of their radii.


Step 2: Key Formula or Approach:

1. Centre of \(C_1\) is \(O_1(-3, -3)\), radius \(r_1 = 3\).

2. Centre of \(C_2\) is \(O_2(1, 3)\).

3. Distance \(d = O_1O_2 = \sqrt{(1 - (-3))^2 + (3 - (-3))^2}\).

4. Radius of \(C_2\) is \(r_2 = d - r_1\).

5. Point of contact \((\alpha, \beta)\) divides \(O_1O_2\) in ratio \(r_1 : r_2\).


Step 3: Detailed Explanation:

Distance between centres:
\[ d = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \]
Since they touch externally, \(r_2 = d - r_1 = 2\sqrt{13} - 3\).

The point of contact \((\alpha, \beta)\) divides the segment joining \((-3, -3)\) and \((1, 3)\) in the ratio \(3 : (2\sqrt{13} - 3)\).

Using the section formula:
\[ \alpha = \frac{3(1) + (2\sqrt{13} - 3)(-3)}{3 + (2\sqrt{13} - 3)} = \frac{3 - 6\sqrt{13} + 9}{2\sqrt{13}} = \frac{12 - 6\sqrt{13}}{2\sqrt{13}} = \frac{6}{\sqrt{13}} - 3 \] \[ \beta = \frac{3(3) + (2\sqrt{13} - 3)(-3)}{3 + (2\sqrt{13} - 3)} = \frac{9 - 6\sqrt{13} + 9}{2\sqrt{13}} = \frac{18 - 6\sqrt{13}}{2\sqrt{13}} = \frac{9}{\sqrt{13}} - 3 \]
Calculate \((\beta - \alpha)\):
\[ \beta - \alpha = \left( \frac{9}{\sqrt{13}} - 3 \right) - \left( \frac{6}{\sqrt{13}} - 3 \right) = \frac{3}{\sqrt{13}} \] \[ (\beta - \alpha)^2 = \left( \frac{3}{\sqrt{13}} \right)^2 = \frac{9}{13} \]
Here, \(m = 9\) and \(n = 13\). Since \(gcd(9, 13) = 1\), we proceed.


Step 4: Final Answer:
\(m + n = 9 + 13 = 22\).
Quick Tip: When two circles touch externally at point \(P\), the vector \(\vec{OP}\) from a centre is proportional to the difference in coordinates of the centres. In \((\beta - \alpha)^2\) calculation, the constants like \(-3\) cancel out, simplifying the expression significantly.

Step 1: Understanding the Concept:

If a line L passes through a point \(P(1, 1, 1)\) and intersects two skew lines \(L_1\) and \(L_2\), the points of intersection \(Q\) (on \(L_1\)) and \(R\) (on \(L_2\)) must be collinear with \(P\). This means the vectors \(\vec{PQ}\) and \(\vec{PR}\) must be parallel.


Step 2: Key Formula or Approach:
1. General point on \(L_1\): \(Q(1+2\lambda, -1+3\lambda, 1+4\lambda)\).

2. General point on \(L_2\): \(R(3+\mu, 4+2\mu, \mu)\).

3. Vectors \(\vec{PQ} = (2\lambda, -2+3\lambda, 4\lambda)\) and \(\vec{PR} = (2+\mu, 3+2\mu, \mu-1)\).

4. Parallel condition: \(\frac{2\lambda}{2+\mu} = \frac{-2+3\lambda}{3+2\mu} = \frac{4\lambda}{\mu-1}\).


Step 3: Detailed Explanation:

From \(\frac{2\lambda}{2+\mu} = \frac{4\lambda}{\mu-1}\) (assuming \(\lambda \neq 0\)):
\[ \mu - 1 = 2(2 + \mu) \implies \mu - 1 = 4 + 2\mu \implies \mu = -5 \].

Now use \(\frac{2\lambda}{2+\mu} = \frac{-2+3\lambda}{3+2\mu}\) with \(\mu = -5\):
\[ \frac{2\lambda}{-3} = \frac{-2+3\lambda}{-7} \implies -14\lambda = 6 - 9\lambda \implies 5\lambda = -6 \implies \lambda = -6/5 \].

The direction vector of line L is given by \(\vec{PR}\) when \(\mu = -5\):
\(\vec{v} = (2 - 5, 3 - 10, -5 - 1) = (-3, -7, -6)\).

Equivalently, the direction ratios are \((3, 7, 6)\).

Equation of line L: \(\frac{x-1}{3} = \frac{y-1}{7} = \frac{z-1}{6} = k\).

Check options:

For point (B) \((7, 15, 13)\):
\(\frac{7-1}{3} = 2\), \(\frac{15-1}{7} = 2\), \(\frac{13-1}{6} = 2\).

Since all ratios are equal to 2, the point lies on line L.


Step 4: Final Answer:

The point \((7, 15, 13)\) lies on the line L. Quick Tip: To find a line intersecting two lines, the condition of coplanarity of the point and the lines can be used. Alternatively, solving for parameters \(\lambda\) and \(\mu\) by making vectors parallel is a foolproof method.

Step 1: Understanding the Concept:

The shortest distance \(d\) between two skew lines \(\vec{r} = \vec{a}_1 + \lambda \vec{d}_1\) and \(\vec{r} = \vec{a}_2 + \mu \vec{d}_2\) is given by the projection of the vector joining points on the lines onto the vector perpendicular to both lines.


Step 2: Key Formula or Approach: \[ d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{d}_1 \times \vec{d}_2)|}{|\vec{d}_1 \times \vec{d}_2|} \]
where \(\vec{a}_1 = (1, 2, 3)\), \(\vec{d}_1 = (2, 3, 4)\), \(\vec{a}_2 = (0, 0, 5)\), and \(\vec{d}_2 = (1, \alpha, 1)\).


Step 3: Detailed Explanation:
\(\vec{a}_2 - \vec{a}_1 = (-1, -2, 2)\).
\(\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 3 & 4
1 & \alpha & 1 \end{vmatrix} = \hat{i}(3 - 4\alpha) - \hat{j}(2 - 4) + \hat{k}(2\alpha - 3) = (3 - 4\alpha, 2, 2\alpha - 3)\).

Magnitude: \(|\vec{d}_1 \times \vec{d}_2|^2 = (3 - 4\alpha)^2 + 4 + (2\alpha - 3)^2 = 16\alpha^2 - 24\alpha + 9 + 4 + 4\alpha^2 - 12\alpha + 9 = 20\alpha^2 - 36\alpha + 22\).

Dot Product: \((\vec{a}_2 - \vec{a}_1) \cdot (\vec{d}_1 \times \vec{d}_2) = -1(3 - 4\alpha) - 2(2) + 2(2\alpha - 3) = -3 + 4\alpha - 4 + 4\alpha - 6 = 8\alpha - 13\).

Given \(d = 5/\sqrt{6}\):
\[ \frac{|8\alpha - 13|}{\sqrt{20\alpha^2 - 36\alpha + 22}} = \frac{5}{\sqrt{6}} \]
Squaring both sides:
\[ \frac{(8\alpha - 13)^2}{20\alpha^2 - 36\alpha + 22} = \frac{25}{6} \] \[ 6(64\alpha^2 - 208\alpha + 169) = 25(20\alpha^2 - 36\alpha + 22) \] \[ 384\alpha^2 - 1248\alpha + 1014 = 500\alpha^2 - 900\alpha + 550 \] \[ 116\alpha^2 + 348\alpha - 464 = 0 \]
Dividing by 116:
\[ \alpha^2 + 3\alpha - 4 = 0 \].


Step 4: Final Answer:

The sum of values of \(\alpha\) is given by Vieta's formula: \(-(coeff. of \alpha) / (coeff. of \alpha^2) = -3/1 = -3\). Quick Tip: For sum of values in coordinate geometry problems, always look for the final quadratic equation. Sum of roots is simply \(-b/a\), saving time spent on finding individual values of \(\alpha\).

Step 1: Understanding the Concept:

The dot product is distributive. Also, for any two vectors, \((\hat{a} \times \hat{b}) \cdot \hat{a} = 0\) and \((\hat{a} \times \hat{b}) \cdot \hat{b} = 0\) because the cross product is perpendicular to the vectors forming it.


Step 2: Key Formula or Approach:
1. \(\hat{a} \cdot \hat{a} = 1\), \(\hat{b} \cdot \hat{b} = 1\), and \(\hat{a} \cdot \hat{b} = \cos\theta\).

2. Given \(\sin\theta = \sqrt{65}/9\), find \(\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - 65/81} = 4/9\).


Step 3: Detailed Explanation:

Calculate \(\vec{c} \cdot \hat{a}\):
\[ \vec{c} \cdot \hat{a} = (3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})) \cdot \hat{a} = 3(\hat{a} \cdot \hat{a}) + 6(\hat{b} \cdot \hat{a}) + 9((\hat{a} \times \hat{b}) \cdot \hat{a}) \] \[ \vec{c} \cdot \hat{a} = 3(1) + 6\cos\theta + 0 = 3 + 6(4/9) = 3 + 8/3 = 17/3 \].

Calculate \(\vec{c} \cdot \hat{b}\):
\[ \vec{c} \cdot \hat{b} = (3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})) \cdot \hat{b} = 3(\hat{a} \cdot \hat{b}) + 6(\hat{b} \cdot \hat{b}) + 0 \] \[ \vec{c} \cdot \hat{b} = 3\cos\theta + 6(1) = 3(4/9) + 6 = 4/3 + 6 = 22/3 \].

Required value:
\[ 9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b}) = 9(17/3) - 3(22/3) = 3(17) - 22 = 51 - 22 = 29 \].


Step 4: Final Answer:

The value is 29. Quick Tip: Always remember that a cross product vector \((\vec{A} \times \vec{B})\) is orthogonal to both \(\vec{A}\) and \(\vec{B}\). This simplifies scalar products involving \(\vec{c}\) and the unit vectors significantly.

Step 1: Understanding the Concept:

For small \(x\), we use standard limit approximations: \(\tan \theta \approx \theta\), \(\log_e(1+\theta) \approx \theta\), \(\tan^{-1} \theta \approx \theta\), and \(e^\theta - 1 \approx \theta\). This helps in simplifying complex rational expressions involving \(x \to 0\).


Step 2: Detailed Explanation:

Approximations as \(x \to 0^+\):

1. \(\tan(5x^{1/3}) \approx 5x^{1/3}\).

2. \(\log_e(1 + 3x^2) \approx 3x^2\).

3. \(\tan^{-1}(3\sqrt{x}) \approx 3\sqrt{x}\), so \((\tan^{-1} 3\sqrt{x})^2 \approx (3\sqrt{x})^2 = 9x\).

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

Substitute these into the limit expression:
\[ L = \lim_{x \to 0^+} \frac{(5x^{1/3}) (3x^2)}{(9x) (5x^{4/3})} \]
Combine powers of \(x\) in the numerator: \(x^{1/3} \cdot x^2 = x^{7/3}\).

Combine powers of \(x\) in the denominator: \(x^1 \cdot x^{4/3} = x^{7/3}\).

The powers cancel out:
\[ L = \frac{5 \cdot 3}{9 \cdot 5} = \frac{15}{45} = \frac{1}{3} \].


Step 3: Final Answer:

The limit is \(\frac{1}{3}\). Quick Tip: When \(x \to 0\), if the total power of \(x\) in the numerator and denominator matches after using standard approximations, the limit is simply the ratio of the product of coefficients.

Step 1: Understanding the Concept:

Critical points occur where \(f'(x) = 0\). Absolute extrema in a closed interval are found by evaluating the function at critical points within the interval and at the endpoints.


Step 2: Detailed Explanation:
\(f'(x) = 3x^2 + 2ax + \frac{b}{x}\).

Given \(f'(-1) = 0 \implies 3 - 2a - b = 0 \implies 2a + b = 3 \quad \dots(i)\).

Given \(f'(2) = 0 \implies 12 + 4a + \frac{b}{2} = 0 \implies 8a + b = -24 \quad \dots(ii)\).

Subtracting (i) from (ii): \(6a = -27 \implies a = -4.5\).

From (i): \(2(-4.5) + b = 3 \implies b = 12\).

So, \(f(x) = x^3 - 4.5x^2 + 12 \log_e |x| + 1\).

In the interval \([-2, -1/2]\), the critical point \(x = -1\) is included.

Evaluate values:
\(f(-2) = -8 - 4.5(4) + 12 \log_e 2 + 1 = -8 - 18 + 12(0.7) + 1 = -25 + 8.4 = -16.6\).
\(f(-1) = -1 - 4.5 + 0 + 1 = -4.5\).
\(f(-1/2) = -1/8 - 4.5(1/4) + 12 \log_e(1/2) + 1 = -0.125 - 1.125 - 12(0.7) + 1 = -0.25 - 8.4 = -8.65\).

Comparing values: Max \(M = -4.5\), Min \(m = -16.6\).
\(|M + m| = |-4.5 - 16.6| = |-21.1| = 21.1\).

Step 3: Final Answer:
\(|M + m| = 21.1\). Quick Tip: Always check if the critical point lies within the specified domain. If \(x=2\) was the only critical point, you would only check the endpoints of the interval \([-2, -1/2]\).

Step 1: Key Formula or Approach:

Use the "King's property" of definite integrals: \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\).


Step 2: Detailed Explanation:

Let \(I = \int_{0}^{\pi} \frac{(x + 3)\sin x}{1 + 3\cos^2 x} dx \quad \dots(i)\).

Applying \(x \to \pi - x\):
\(I = \int_{0}^{\pi} \frac{(\pi - x + 3)\sin(\pi - x)}{1 + 3\cos^2(\pi - x)} dx = \int_{0}^{\pi} \frac{(\pi - x + 3)\sin x}{1 + 3\cos^2 x} dx \quad \dots(ii)\).

Adding (i) and (ii):
\(2I = \int_{0}^{\pi} \frac{(\pi + 6)\sin x}{1 + 3\cos^2 x} dx = (\pi + 6) \int_{0}^{\pi} \frac{\sin x}{1 + 3\cos^2 x} dx\).

Let \(\sqrt{3}\cos x = t\). Then \(-\sqrt{3}\sin x dx = dt\).

Limits: \(x = 0 \implies t = \sqrt{3}\); \(x = \pi \implies t = -\sqrt{3}\).
\(2I = \frac{\pi + 6}{-\sqrt{3}} \int_{\sqrt{3}}^{-\sqrt{3}} \frac{dt}{1 + t^2} = \frac{\pi + 6}{\sqrt{3}} \int_{-\sqrt{3}}^{\sqrt{3}} \frac{dt}{1 + t^2}\).
\(2I = \frac{\pi + 6}{\sqrt{3}} [ \tan^{-1} t ]_{-\sqrt{3}}^{\sqrt{3}} = \frac{\pi + 6}{\sqrt{3}} [\frac{\pi}{3} - (-\frac{\pi}{3})] = \frac{\pi + 6}{\sqrt{3}} \cdot \frac{2\pi}{3}\).
\(I = \frac{\pi(\pi + 6)}{3\sqrt{3}}\).

Step 3: Final Answer:

The integral is \(\frac{\pi}{3\sqrt{3}}(\pi + 6)\). Quick Tip: When you see an \(x\) multiplied by trigonometric functions with symmetric limits like \(0\) to \(\pi\), the King's property is almost always the intended first step to eliminate the linear \(x\) term.

Step 1: Understanding the Concept:

The area between two curves \(y_1(x)\) and \(y_2(x)\) from \(x=a\) to \(x=b\) is \(\int_a^b |y_1 - y_2| dx\). First, we must find the intersection points of the parabola and the line.


Step 2: Detailed Explanation:

Find intersections: \(4 - \frac{x^2}{4} = \frac{x-4}{2} \implies 16 - x^2 = 2x - 8\).
\(x^2 + 2x - 24 = 0 \implies (x+6)(x-4) = 0 \implies x = -6, 4\).

The parabola \(y = 4 - x^2/4\) is the upper curve in this interval.
\(\alpha = \int_{-6}^{4} (4 - \frac{x^2}{4} - \frac{x-4}{2}) dx = \int_{-6}^{4} (6 - \frac{x}{2} - \frac{x^2}{4}) dx\).
\(\alpha = [6x - \frac{x^2}{4} - \frac{x^3}{12}]_{-6}^{4}\).
\(\alpha = (24 - 4 - \frac{64}{12}) - (-36 - 9 - \frac{-216}{12}) = (20 - \frac{16}{3}) - (-45 + 18) = \frac{44}{3} + 27 = \frac{44 + 81}{3} = \frac{125}{3}\).


Step 3: Final Answer:
\(6\alpha = 6 \cdot \frac{125}{3} = 2 \cdot 125 = 250\). Quick Tip: For a parabola \(y = A - x^2\) and a line, if the limits are roots of the quadratic \(ax^2 + bx + c\), the area is often \(\frac{|a|}{6}(\beta - \alpha)^3\). Here, \(|a| = 1/4\) and \(\beta - \alpha = 10\), so Area \(= \frac{1}{24} \cdot 1000 = 125/3\).

Step 1: Understanding the Concept:

The differential equation is a first-order linear D.E. of the form \(\frac{dy}{dx} + P(x)y = Q(x)\). We solve it using the Integrating Factor (I.F.).


Step 2: Key Formula or Approach: \(\frac{dy}{dx} + \frac{e^x(x-2)}{x(x^2+e^x)} y = \frac{x^2}{x^2+e^x}\).

I.F. \(= e^{\int P(x) dx} = e^{\int \frac{e^x(x-2)}{x(x^2+e^x)} dx}\).


Step 3: Detailed Explanation:

Consider the integral \(\int \frac{e^x(x-2)}{x(x^2+e^x)} dx\). Divide numerator and denominator by \(x^2\):
\(\int \frac{e^x(1/x - 2/x^2)}{1 + e^x/x^2} dx\).

Let \(u = 1 + \frac{e^x}{x^2}\). Then \(du = \frac{x^2 e^x - e^x(2x)}{x^4} dx = \frac{e^x(x-2)}{x^3} dx\).

This doesn't quite match. Let's try another path:

Rearrange D.E.: \(x^3 dy + xe^x dy + xe^x y dx - 2e^x y dx = x^3 dx\).

Divide by \(x^2\): \(x dy + \frac{e^x(x dy + y dx)}{x} - \frac{2e^x y dx}{x^2} = x dx\).
\(x dy + d(\frac{e^x y}{x}) = x dx\). Integrating: \(xy - \int y dx + \frac{e^x y}{x} = \frac{x^2}{2} + C\). (Not helpful).

Let's test \(y = \frac{x^3 - x^2}{x^2 + e^x}\).

At \(x=1\), \(y = 0/(1+e) = 0\). (Matches given point).

Checking if it satisfies the D.E.:
\(y(x^2 + e^x) = x^3 - x^2\). Differentiating:
\((x^2 + e^x) dy + y(2x + e^x) dx = (3x^2 - 2x) dx\).

Multiply by \(x\): \(x(x^2 + e^x) dy + (2x^2 y + xe^x y - 3x^3 + 2x^2) dx = 0\).

Substituting \(x^2 y = x^3 - x^2 - e^x y\):
\(x(x^2 + e^x) dy + (2x^3 - 2x^2 - 2e^x y + xe^x y - 3x^3 + 2x^2) dx = 0\).
\(x(x^2 + e^x) dy + (e^x y(x - 2) - x^3) dx = 0\). This matches the D.E. exactly.


Step 4: Final Answer:

Now find \(y(2)\):
\[ y(2) = \frac{2^3 - 2^2}{2^2 + e^2} = \frac{8 - 4}{4 + e^2} = \frac{4}{4 + e^2} \]. Quick Tip: If the Integrating Factor looks extremely complicated, check if a simple rational function like \(\frac{x^n(x-1)}{x^2+e^x}\) fits the initial conditions and solves the equation.

Step 1: Understanding the Concept:

The problem asks for the number of subsets of a set of \(n\) elements such that no two elements in the subset are consecutive. This is a classic combinatorial problem that leads to a recurrence relation.


Step 2: Key Formula or Approach:

Let \(a_n\) be the number of valid subsets of \(\{1, 2, ..., n\}\).

A valid subset of \(\{1, 2, ..., n\}\) either:

1. Does not contain the element \(n\): In this case, it is a valid subset of \(\{1, 2, ..., n-1\}\). There are \(a_{n-1}\) such subsets.

2. Contains the element \(n\): In this case, it cannot contain \(n-1\). Thus, the remaining elements must form a valid subset of \(\{1, 2, ..., n-2\}\). There are \(a_{n-2}\) such subsets.

This gives the recurrence relation: \( a_n = a_{n-1} + a_{n-2} \), which is the Fibonacci sequence.


Step 3: Detailed Explanation:

We calculate the values step-by-step:

For \( n = 1 \): The valid subsets of \(\{1\}\) are \(\emptyset\) and \(\{1\}\). So, \( a_1 = 2 \).

For \( n = 2 \): The valid subsets of \(\{1, 2\}\) are \(\emptyset, \{1\}, \{2\}\). So, \( a_2 = 3 \).

Using the recurrence relation:
\[ a_3 = a_2 + a_1 = 3 + 2 = 5 \] \[ a_4 = a_3 + a_2 = 5 + 3 = 8 \] \[ a_5 = a_4 + a_3 = 8 + 5 = 13 \]

Step 4: Final Answer:

The number of such subsets for \(n=5\) is \(n(S_5) = 13\).
Quick Tip: The number of subsets of \(\{1, ..., n\}\) with no two consecutive elements is given by the Fibonacci number \(F_{n+2}\), where \(F_1=1, F_2=1, F_3=2...\)

Step 1: Understanding the Concept:

A \(2 \times 2\) matrix \( A = \begin{pmatrix} a & b
c & d \end{pmatrix} \) is singular if its determinant is zero, i.e., \( ad - bc = 0 \implies ad = bc \). The elements \(a, b, c, d\) are chosen from the set \(S = \{2, 3, 6, 9\}\).


Step 2: Key Formula or Approach:

We need to find the number of pairs \((a, d)\) and \((b, c)\) such that their products are equal. We list all possible products \(k = x \cdot y\) where \(x, y \in S\) and count how many pairs result in each product.


Step 3: Detailed Explanation:

Possible products and pairs:

1. \( k = 4 \): (2,2) \(\to\) 1 pair.

2. \( k = 6 \): (2,3), (3,2) \(\to\) 2 pairs.

3. \( k = 12 \): (2,6), (6,2) \(\to\) 2 pairs.

4. \( k = 18 \): (2,9), (9,2), (3,6), (6,3) \(\to\) 4 pairs.

5. \( k = 9 \): (3,3) \(\to\) 1 pair.

6. \( k = 27 \): (3,9), (9,3) \(\to\) 2 pairs.

7. \( k = 36 \): (6,6) \(\to\) 1 pair.

8. \( k = 54 \): (6,9), (9,6) \(\to\) 2 pairs.

9. \( k = 81 \): (9,9) \(\to\) 1 pair.

For each product \(k\), if there are \(N_k\) pairs \((x,y)\) such that \(xy = k\), then there are \( N_k \times N_k \) ways to choose \(a, d, b, c\) such that \(ad = bc = k\).

Total ways = \( \sum N_k^2 \):
\[ Total = 1^2 + 2^2 + 2^2 + 4^2 + 1^2 + 2^2 + 1^2 + 2^2 + 1^2 \] \[ Total = 1 + 4 + 4 + 16 + 1 + 4 + 1 + 4 + 1 = 36 \]

Step 4: Final Answer:

The number of singular matrices is 36.
Quick Tip: Systematically group possible products from the given set. For singular matrices, ensure \(ad = bc\) is satisfied by matching pairs with identical products.

Step 1: Understanding the Concept:

For a standard hyperbola, the foci are at \( (\pm ae, 0) \). We are given one focus \( P(-3, 0) \), so \( ae = 3 \). The other focus is \( F(3, 0) \). The latus rectum through \(F\) consists of points \( (3, \pm b^2/a) \). These subtend \(90^\circ\) at \(P\).


Step 2: Key Formula or Approach:

Let the endpoints of the latus rectum be \( Q(3, b^2/a) \) and \( R(3, -b^2/a) \). Since \( \angle QPR = 90^\circ \), and the configuration is symmetric about the x-axis, the slope of \(PQ\) must be \(1\) (and \(PR\) must be \(-1\)).


Step 3: Detailed Explanation:

Slope of \( PQ = \frac{b^2/a - 0}{3 - (-3)} = \frac{b^2}{6a} \).

For a \(90^\circ\) angle subtended at P, by symmetry:
\[ \frac{b^2}{6a} = 1 \implies b^2 = 6a \]
We also know the relation \( b^2 = a^2(e^2 - 1) = (ae)^2 - a^2 \).

Substituting \( ae = 3 \) and \( b^2 = 6a \):
\[ 6a = 9 - a^2 \implies a^2 + 6a - 9 = 0 \]
Solving for \(a\) (choosing \(a > 0\)):
\[ a = \frac{-6 + \sqrt{36 + 36}}{2} = -3 + 3\sqrt{2} = 3(\sqrt{2} - 1) \]
Now find \( a^2 \) and \( b^2 \):
\[ a^2 = 9(2 + 1 - 2\sqrt{2}) = 9(3 - 2\sqrt{2}) = 27 - 18\sqrt{2} \] \[ b^2 = 6a = 18\sqrt{2} - 18 \]
Calculate \( a^2b^2 \):
\[ a^2b^2 = (27 - 18\sqrt{2})(18\sqrt{2} - 18) = 18^2 (1.5 - \sqrt{2})(\sqrt{2} - 1) \times 2 \]
Or directly: \( a^2b^2 = a^2(6a) = 6a^3 \).
\[ a^2b^2 = 6 \times [3(\sqrt{2}-1)]^3 = 6 \times 27 \times (\sqrt{2}-1)^3 \] \[ (\sqrt{2}-1)^3 = 2\sqrt{2} - 3(2) + 3\sqrt{2} - 1 = 5\sqrt{2} - 7 \] \[ a^2b^2 = 162(5\sqrt{2} - 7) = 810\sqrt{2} - 1134 \]
Comparing with \( \alpha\sqrt{2} - \beta \): \( \alpha = 810 \), \( \beta = 1134 \).


Step 4: Final Answer:
\[ \alpha + \beta = 810 + 1134 = 1944 \] Quick Tip: If an isosceles triangle with base \(2h\) and height \(d\) subtends a right angle at the vertex opposite the base, then \(h = d\). Here, half latus rectum \( b^2/a = 6 \).

Step 1: Understanding the Concept:

A relation \(R\) on \(A\) is:

1. Reflexive: \( (1,1), (2,2), (3,3) \in R \). (3 elements)

2. Transitive: \( (a,b) \in R, (b,c) \in R \implies (a,c) \in R \).

3. Not Symmetric: There exists \( (a,b) \in R \) such that \( (b,a) \notin R \).

4. Includes \( (1,2) \).

5. Total elements \( |R| \le 6 \).


Step 2: Detailed Explanation:

Since \(R\) is reflexive and contains \((1,2)\), we already have 4 elements: \( \{(1,1), (2,2), (3,3), (1,2)\} \). We can add at most 2 more elements from the set of possible pairs \( \{(1,3), (2,1), (2,3), (3,1), (3,2)\} \).

Note that to be transitive with \((1,2)\), if we add \((2,3)\), we must add \((1,3)\).

Case 1: \( |R| = 4 \).
\( R = \{(1,1), (2,2), (3,3), (1,2)\} \). Transitive? Yes. Not symmetric? Yes (\( (2,1) \notin R \)). (1 way)

Case 2: \( |R| = 5 \).

- Add \( (1,3) \): \( \{(1,1), (2,2), (3,3), (1,2), (1,3)\} \). Transitive? Yes. Not symmetric? Yes. (1 way)

- Add \( (3,2) \): \( \{(1,1), (2,2), (3,3), (1,2), (3,2)\} \). Transitive? Yes. Not symmetric? Yes. (1 way)

- Other single additions either violate transitivity or make it symmetric.

Case 3: \( |R| = 6 \).

- Add \( (1,3), (2,3) \): Chain \( 1 \to 2 \to 3 \). Transitive? Yes. Not symmetric? Yes. (1 way)

- Add \( (3,1), (3,2) \): Chain \( 3 \to 1 \to 2 \). Transitive? Yes. Not symmetric? Yes. (1 way)

- Add \( (1,3), (3,2) \): Chain \( 1 \to 3 \to 2 \). Transitive? Yes. Not symmetric? Yes. (1 way)


Step 4: Final Answer:

Total number of such relations = \( 1 + 2 + 3 = 6 \).
Quick Tip: A reflexive and transitive relation is a preorder. Partial orders (antisymmetric preorders) are the most common source of "not symmetric" preorders in such small sets.

Step 1: Understanding the Concept:

Discontinuities of the greatest integer function \( [g(x)] \) occur when \( g(x) \) is an integer, provided the function actually changes value across that point. We check all such integer points in the domain \( [0, 4] \).


Step 2: Detailed Explanation:

Let \( g_1(x) = \frac{x^2}{2} \). For \( x \in [0, 4] \), \( g_1(x) \in [0, 8] \). Integer points for \( g_1(x) \):
\( x^2/2 \in \{1, 2, 3, 4, 5, 6, 7, 8\} \implies x \in \{ \sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}, 4 \} \).

Let \( g_2(x) = \sqrt{x} \). For \( x \in [0, 4] \), \( g_2(x) \in [0, 2] \). Integer points for \( g_2(x) \):
\( \sqrt{x} \in \{1, 2\} \implies x \in \{ 1, 4 \} \).

Candidate points of discontinuity: \( \{ 1, \sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}, 4 \} \).

- For any \( c \in \{1, \sqrt{2}, 2, \dots, \sqrt{14} \} \), either \( g_1(x) \) or \( g_2(x) \) has an integer value while the other is non-integer, resulting in a jump. These 8 points are points of discontinuity.

- Check \( x = 4 \):
\( f(4) = [8] - [2] = 6 \).
\( \lim_{x \to 4^-} f(x) = [8^-] - [\sqrt{4^-}] = 7 - 1 = 6 \).

Since \( Value = Limit \), \( f(x) \) is continuous at \( x = 4 \).

- Check \( x = 0 \):
\( f(0) = 0 \), \( \lim_{x \to 0^+} f(x) = 0 \). Continuous.


Step 4: Final Answer:

The points of discontinuity are \( \{1, \sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}\} \). The total count is 8.
Quick Tip: Always check the endpoints of the interval carefully. For the greatest integer function, left continuity can sometimes "cancel out" jumps from two different terms, though it's rare.

Step 1: Understanding the Concept:

The quantity given is related to Maxwell's modification of Ampere's Law. According to Maxwell's equations, a changing electric flux through a surface gives rise to a "displacement current". This term was added to explain the consistency of the magnetic field near capacitors during charging or discharging.


Step 2: Key Formula or Approach:

From Gauss's Law:
\[ \Phi_E = \frac{q}{\epsilon_0} \]
Rearranging for charge \( q \):
\[ q = \epsilon_0 \Phi_E \]
Differentiating with respect to time \( t \):
\[ \frac{dq}{dt} = \epsilon_0 \frac{d\Phi_E}{dt} \]

Step 3: Detailed Explanation:

1. The term \(\frac{dq}{dt}\) represents the rate of flow of charge, which is the definition of electric current.

2. The quantity \( i_d = \epsilon_0 \frac{d\Phi_E}{dt} \) is specifically known as the displacement current.

3. Since it is a form of current, its physical dimensions must be identical to that of conduction current \( [I] \).

4. Therefore, the dimensions of \(\epsilon_0 \frac{d\Phi_E}{dt}\) are those of electric current.


Step 4: Final Answer:

The dimensions of the given expression correspond to electric current.
Quick Tip: Always remember that in any physics equation, both sides of an equality and terms being added must have the same dimensions. In the Generalized Ampere's Law \( \oint \vec{B} \cdot d\vec{l} = \mu_0(i_c + i_d) \), since \( i_c \) is current, \( i_d = \epsilon_0 \frac{d\Phi_E}{dt} \) must also be current.

Step 1: Understanding the Concept:

When two lenses are separated by a finite distance \( d \), the equivalent focal length \( F \) of the system is not just the sum of individual powers. A correction term accounting for the separation must be used.


Step 2: Key Formula or Approach:

The formula for the equivalent focal length \( F \) of two lenses separated by distance \( d \) is:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \]
The power \( P \) is defined as \( P = \frac{1}{F} \) (with \( F \) in meters).


Step 3: Detailed Explanation:

Given:
\( f_1 = 30 cm = 0.3 m \)
\( f_2 = 10 cm = 0.1 m \)
\( d = 10 cm = 0.1 m \)

Calculate the equivalent power \( P \):
\[ P = P_1 + P_2 - d P_1 P_2 \] \[ P = \frac{1}{0.3} + \frac{1}{0.1} - \frac{0.1}{(0.3)(0.1)} \] \[ P = \frac{10}{3} + 10 - \frac{0.1}{0.03} \] \[ P = \frac{10}{3} + 10 - \frac{10}{3} \] \[ P = 10 D \]

Step 4: Final Answer:

The power of the lens combination is \( 10 D \).
Quick Tip: In numericals involving lens combinations, always ensure that units are converted to meters before calculating power in Diopters (D), or calculate in cm and convert the final \( F \) to meters.

Step 1: Understanding the Concept:

Instantaneous power \( P \) is the dot product of the force vector \( \vec{F} \) and the velocity vector \( \vec{v} \), i.e., \( P = \vec{F} \cdot \vec{v} \). We need to find the velocity by integrating the acceleration derived from Newton's Second Law.


Step 2: Key Formula or Approach:

1. Acceleration \(\vec{a} = \frac{\vec{F}}{m}\).

2. Velocity \(\vec{v} = \int \vec{a} dt\).

3. Power \(P = \vec{F} \cdot \vec{v}\).


Step 3: Detailed Explanation:

Given: \( m = 1000 g = 1 kg \).

Force \(\vec{F} = 2t\hat{i} + 3t^2\hat{j}\).

Acceleration:
\[ \vec{a} = \frac{\vec{F}}{m} = \frac{2t\hat{i} + 3t^2\hat{j}}{1} = 2t\hat{i} + 3t^2\hat{j} \]
Integrating to find velocity (assuming starting from rest at \( t = 0 \)):
\[ \vec{v}(t) = \int (2t\hat{i} + 3t^2\hat{j}) dt = t^2\hat{i} + t^3\hat{j} \]
Now calculate the instantaneous power:
\[ P = \vec{F} \cdot \vec{v} = (2t\hat{i} + 3t^2\hat{j}) \cdot (t^2\hat{i} + t^3\hat{j}) \] \[ P = (2t)(t^2) + (3t^2)(t^3) \] \[ P = 2t^3 + 3t^5 \]

Step 4: Final Answer:

The power generated is \((2t^3 + 3t^5) W\).
Quick Tip: When force is time-dependent, acceleration is also time-dependent. Never use \( v = u + at \) directly if \( a \) is not constant; always use integration.

Step 1: Understanding the Concept:

For a system consisting of multiple continuous objects, the centre of mass (CM) can be found by treating each segment as a point mass located at its individual centre of mass. The total CM is the weighted average of these positions.


Step 2: Detailed Explanation:

Total length = \( 5L \).

One side is \( 2L \), so the other side must be \( 5L - 2L = 3L \).

Let's place the corner (bend) at the origin \((0, 0)\).

1. Horizontal segment (length \( 2L \)): Lies along the x-axis from \((0, 0)\) to \((2L, 0)\).

- Mass \( m_1 \propto 2L \).

- Individual CM \( C_1 = (L, 0) \).

2. Vertical segment (length \( 3L \)): Lies along the y-axis from \((0, 0)\) to \((0, 3L)\).

- Mass \( m_2 \propto 3L \).

- Individual CM \( C_2 = (0, 1.5L) \).

Coordinates of the system CM:
\[ X_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{(2L)(L) + (3L)(0)}{2L + 3L} = \frac{2L^2}{5L} = \frac{2L}{5} \] \[ Y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} = \frac{(2L)(0) + (3L)(1.5L)}{2L + 3L} = \frac{4.5L^2}{5L} = \frac{9L}{10} \]
Substitute \( L = 10 cm \):
\[ X_{cm} = \frac{2 \times 10}{5} = 4 cm \] \[ Y_{cm} = \frac{9 \times 10}{10} = 9 cm \]
The position vector is \( 4\hat{i} + 9\hat{j} \).


Step 4: Final Answer:

The centre of mass is located at \( 4\hat{i} + 9\hat{j} \).
Quick Tip: For uniform rods, the centre of mass is always at the geometric midpoint. Simplifying a complex shape into point masses at their midpoints is the standard trick for such problems.

Step 1: Understanding the Concept:

When a block slides down an inclined plane, the components of gravity act along the plane (\( mg\sin\theta \)) and perpendicular to it (\( mg\cos\theta \)). Friction opposes the motion and is proportional to the normal force.


Step 2: Key Formula or Approach:

Equation of motion along the incline:
\[ mg\sin\theta - f_k = ma \]
Where kinetic friction \( f_k = \mu_k N = \mu_k mg\cos\theta \).


Step 3: Detailed Explanation:

Given: \( \theta = 60^{\circ} \), \( a = \frac{g}{2} \).

Substitute forces into the equation:
\[ mg\sin 60^{\circ} - \mu_k mg\cos 60^{\circ} = m\left(\frac{g}{2}\right) \]
Dividing the entire equation by \( mg \):
\[ \sin 60^{\circ} - \mu_k \cos 60^{\circ} = \frac{1}{2} \]
Substitute trigonometric values \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \) and \( \cos 60^{\circ} = \frac{1}{2} \):
\[ \frac{\sqrt{3}}{2} - \mu_k \left(\frac{1}{2}\right) = \frac{1}{2} \]
Multiply by \( 2 \):
\[ \sqrt{3} - \mu_k = 1 \] \[ \mu_k = \sqrt{3} - 1 \]

Step 4: Final Answer:

The coefficient of kinetic friction is \( \sqrt{3} - 1 \).
Quick Tip: If the block moves with constant velocity (zero acceleration), the equation simplifies to \( \mu_k = \tan\theta \). In this case, since \( a = g/2 \), the friction must be less than \( mg\sin\theta \).

Step 1: Understanding the Concept:

The time of flight \( T \) of a projectile depends on the vertical component of its initial velocity. For projectiles with the same initial speed \( u \), it is proportional to the sine of the projection angle.


Step 2: Key Formula or Approach:

Time of flight formula:
\[ T = \frac{2u\sin\theta}{g} \]

Step 3: Detailed Explanation:

Let the two angles be \( \theta_1 = 45^{\circ} + \alpha \) and \( \theta_2 = 45^{\circ} - \alpha \).

Ratio of times of flight:
\[ \frac{T_1}{T_2} = \frac{\frac{2u\sin\theta_1}{g}}{\frac{2u\sin\theta_2}{g}} = \frac{\sin(45^{\circ} + \alpha)}{\sin(45^{\circ} - \alpha)} \]
Using trigonometric identity \( \sin(A \pm B) = \sin A\cos B \pm \cos A\sin B \):
\[ \frac{T_1}{T_2} = \frac{\sin 45^{\circ}\cos\alpha + \cos 45^{\circ}\sin\alpha}{\sin 45^{\circ}\cos\alpha - \cos 45^{\circ}\sin\alpha} \]
Since \( \sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}} \):
\[ \frac{T_1}{T_2} = \frac{\frac{1}{\sqrt{2}}(\cos\alpha + \sin\alpha)}{\frac{1}{\sqrt{2}}(\cos\alpha - \sin\alpha)} = \frac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha} \]
Divide both numerator and denominator by \( \cos\alpha \):
\[ \frac{T_1}{T_2} = \frac{1 + \tan\alpha}{1 - \tan\alpha} \]

Step 4: Final Answer:

The ratio of their times of flight is \( \frac{1 + \tan\alpha}{1 - \tan\alpha} \).
Quick Tip: Projection angles like \( 45^{\circ} + \alpha \) and \( 45^{\circ} - \alpha \) are complementary. Projectiles at complementary angles have the same range, but their times of flight and maximum heights will differ.

Step 1: Understanding the Concept:

Elongation \( \Delta L \) depends on the material (Young's modulus \( Y \)), the geometry (length and cross-sectional area), and the applied force \( F \).


Step 2: Key Formula or Approach:

Young's Modulus:
\[ Y = \frac{F L}{A \Delta L} \implies \Delta L = \frac{F L}{A Y} \]
Area \( A = \frac{\pi d^2}{4} \), so \( \Delta L \propto \frac{L}{d^2} \) for constant force and same material.


Step 3: Detailed Explanation:

Given:

- \( \frac{L_A}{L_B} = \frac{1}{3} \)

- \( \frac{d_A}{d_B} = 2 \)

- Same force \( F_A = F_B = F \)

- Same material \( Y_A = Y_B = Y \)

Ratio of elongations:
\[ \frac{\Delta L_A}{\Delta L_B} = \frac{\frac{F L_A}{A_A Y}}{\frac{F L_B}{A_B Y}} = \frac{L_A}{L_B} \cdot \frac{A_B}{A_A} \]
Since \( A \propto d^2 \):
\[ \frac{\Delta L_A}{\Delta L_B} = \frac{L_A}{L_B} \cdot \left( \frac{d_B}{d_A} \right)^2 \]
Substitute the given ratios:
\[ \frac{\Delta L_A}{\Delta L_B} = \left( \frac{1}{3} \right) \cdot \left( \frac{1}{2} \right)^2 = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12} \]

Step 4: Final Answer:

The ratio of their respective elongations is \( 1:12 \).
Quick Tip: In proportionality questions, identify constant terms and write the relation for the target variable. Here, \( \Delta L \propto \frac{L}{d^2} \) immediately leads to the answer.

Step 1: Understanding the Concept:

The ratio of specific heats \( \gamma = \frac{C_p}{C_v} \) depends on the degrees of freedom \( f \) of the gas molecules. According to the law of equipartition of energy, \( \gamma = 1 + \frac{2}{f} \).


Step 2: Key Formula or Approach:

1. Monoatomic: \( f = 3 \) (3 translational)

2. Diatomic rigid: \( f = 5 \) (3 trans + 2 rot)

3. Diatomic non-rigid: \( f = 7 \) (3 trans + 2 rot + 2 vib)

4. Triatomic rigid (non-linear): \( f = 6 \) (3 trans + 3 rot)


Step 3: Detailed Explanation:

- A. Triatomic rigid gas: \( f = 6 \). \( \gamma = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3} \). Matches III.

- B. Diatomic non-rigid gas: \( f = 7 \). \( \gamma = 1 + \frac{2}{7} = \frac{9}{7} \). Matches IV.

- C. Monoatomic gas: \( f = 3 \). \( \gamma = 1 + \frac{2}{3} = \frac{5}{3} \). Matches I.

- D. Diatomic rigid gas: \( f = 5 \). \( \gamma = 1 + \frac{2}{5} = \frac{7}{5} \). Matches II.


Step 4: Final Answer:

The correct matching is A-III, B-IV, C-I, D-II.
Quick Tip: Diatomic rigid is the most standard case with \( \gamma = 1.4 \). If a gas is "non-rigid", it simply means you must add vibrational degrees of freedom (usually +2).

Step 1: Understanding the Concept:

Beats occur when two waves of slightly different frequencies superimpose. The resulting wave has a slowly varying amplitude. The equation given represents the product of two cosine terms, which can be decomposed into the sum of two waves.


Step 2: Key Formula or Approach:

Using the identity \( \cos A \cos B = \frac{1}{2} [\cos(B + A) + \cos(B - A)] \):

Here \( A = 1.5t \) and \( B = 50.5t \).


Step 3: Detailed Explanation:

Decomposing the given expression:
\[ x = \frac{a}{2} [\cos(50.5t + 1.5t) + \cos(50.5t - 1.5t)] \] \[ x = \frac{a}{2} [\cos(52t) + \cos(49t)] \]
The two angular frequencies are \( \omega_1 = 52 rad/s \) and \( \omega_2 = 49 rad/s \).

The beat angular frequency is the difference:
\[ \omega_{beat} = |\omega_1 - \omega_2| = 52 - 49 = 3 rad/s \]
The beat period \( T_{beat} \) is given by:
\[ T_{beat} = \frac{2\pi}{\omega_{beat}} = \frac{2 \times 3.14159}{3} \approx \frac{6.283}{3} \approx 2.09 s \]
Rounding to the nearest integer, we get \( 2 s \).


Step 4: Final Answer:

The beat period is approximately \( 2 s \).
Quick Tip: In the modulated wave form \( A \cos(\omega_m t) \cos(\omega_{avg} t) \), the beat frequency is twice the modulation frequency \( \omega_m \). Here \( \omega_{beat} = 2 \times 1.5 = 3 rad/s \).

Step 1: Understanding the Concept:

The root mean square (RMS) value of a signal containing a DC component and multiple AC components is found by taking the square root of the sum of the square of the DC component and the squares of the RMS values of each AC component.


Step 2: Key Formula or Approach:

For \( i = I_{dc} + I_0 \cos(\omega t + \phi) \):
\[ I_{rms}^2 = I_{dc}^2 + \left( \frac{I_0}{\sqrt{2}} \right)^2 \]

Step 3: Detailed Explanation:

Given:

- DC component \( I_{dc} = 5\sqrt{2} Amp \)

- Peak value of AC component \( I_0 = 10 Amp \)

Calculate the squares:

1. \( I_{dc}^2 = (5\sqrt{2})^2 = 25 \times 2 = 50 \)

2. \( I_{ac,rms}^2 = \left( \frac{10}{\sqrt{2}} \right)^2 = \frac{100}{2} = 50 \)

The total RMS value:
\[ I_{rms} = \sqrt{I_{dc}^2 + I_{ac,rms}^2} \] \[ I_{rms} = \sqrt{50 + 50} = \sqrt{100} = 10 Amp \]

Step 4: Final Answer:

The r.m.s value of the current is \( 10 Amp \).
Quick Tip: For a complex wave containing a DC term \( a \) and a sinusoidal term with peak \( b \), the RMS value is always \( \sqrt{a^2 + \frac{b^2}{2}} \). The phase constant \( \phi \) and frequency \( \omega \) do not affect the RMS value.

Step 1: Understanding the Concept:

The triangular pyramid (tetrahedron) consists of 6 segments (edges).

If a single wire of total resistance \( R \) is used to form these 6 segments of equal length, the resistance of each segment is \( r = \frac{R}{6} \).

The problem asks for the equivalent resistance between two vertices \( A \) and \( B \).


Step 2: Key Formula or Approach:

For a symmetric tetrahedral resistor network where each edge has resistance \( r \), the equivalent resistance between any two vertices can be found using nodal analysis or symmetry.

The equivalent resistance between two adjacent vertices is:
\[ R_{eq} = \frac{r}{2} \]

Step 3: Detailed Explanation:

1. Resistance of each edge of the pyramid:

\[ r = \frac{R}{6} \]
2. Let current \( I \) enter at vertex \( A \) and leave at vertex \( B \).

By symmetry, if the other two vertices are \( C \) and \( D \), the potentials at \( C \) and \( D \) are equal because of the balanced nature of the bridge relative to points \( A \) and \( B \).

3. Using the standard formula for a tetrahedral frame where each edge has resistance \( r \), the resistance between any two vertices is \( \frac{r}{2} \).

4. Substituting the value of \( r \):

\[ R_{AB} = \frac{r}{2} = \frac{R/6}{2} = \frac{R}{12} \]
5. Comparing this with the given expression \( R/n \), we identify \( n = 12 \).


Step 4: Final Answer:

The value of \( n \) is 12.
Quick Tip: For a wire frame forming a cube of 12 edges (resistance \( r \)), \( R_{body-diag} = \frac{5}{6}r \). For a tetrahedron of 6 edges, \( R_{vertex-vertex} = \frac{r}{2} \). Memorizing these common symmetric results can save a lot of time in competitive exams.

Step 1: Understanding the Concept:

When a charged particle enters a uniform magnetic field perpendicularly, it moves in a circular path. The centripetal force is provided by the magnetic Lorentz force. We need to find the functional relationship between the radius \( r \) and the kinetic energy \( E \).


Step 2: Key Formula or Approach:

1. Magnetic force: \( F_m = qvB \)

2. Centripetal force: \( F_c = \frac{mv^2}{r} \)

3. Equating them: \( r = \frac{mv}{qB} \)

4. Kinetic energy: \( E = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2E}{m}} \)


Step 3: Detailed Explanation:

Substitute the expression for velocity \( v \) into the radius formula:
\[ r = \frac{m}{qB} \sqrt{\frac{2E}{m}} \] \[ r = \frac{\sqrt{2mE}}{qB} \]
From this equation, we can see that:
\[ r \propto \sqrt{E} \]
This means the relationship between \( r \) and \( E \) is a square root function.

The graph of \( y = \sqrt{x} \) is a parabola opening along the x-axis (kinetic energy axis), which corresponds to a curve that starts at the origin and is concave downwards.


Step 4: Final Answer:

The curve in Option (A) correctly represents \( r \propto \sqrt{E} \).
Quick Tip: Remember the proportionality: \( r \propto \sqrt{m} \), \( r \propto \sqrt{E} \), \( r \propto p \) (momentum), and \( r \propto 1/q \). These relationships are frequently tested in objective physics questions.

Step 1: Understanding the Concept:

The change in potential energy of a system when a charge \( q_3 \) is moved from point \( C \) to point \( D \) is given by \( \Delta U = U_D - U_C \). This can be calculated using the formula for electrostatic potential energy between point charges: \( U = \frac{k Q q}{r} \).


Step 2: Key Formula or Approach:

1. Electrostatic Potential \( V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} \).

2. Potential Energy \( U = qV \).

3. \( \Delta U = q_3 (V_D - V_C) \), where \( V \) is the potential created by \( q_1 \) and \( q_2 \).


Step 3: Detailed Explanation:

Let \( A \) be the origin \((0, 0)\). Charge \( q_1 \) is at \( A \).

Distance \( AB = 30 cm = 0.3 m \). Charge \( q_2 \) is at \( B \).

Charge \( q_3 \) is moved along a circle of radius \( R = 40 cm = 0.4 m \) centered at \( A \).

Initial Position C:

Distance to \( q_1 \): \( r_{1C} = 0.4 m \).

Distance to \( q_2 \): \( r_{2C} = \sqrt{0.3^2 + 0.4^2} = 0.5 m \).

Potential at \( C \): \( V_C = \frac{1}{4\pi\epsilon_0} \left[ \frac{q_1}{0.4} + \frac{q_2}{0.5} \right] \).

Final Position D:

Point \( D \) is on the line \( AB \) (x-axis), at distance \( 0.4 m \) from \( A \).

Distance to \( q_1 \): \( r_{1D} = 0.4 m \).

Distance to \( q_2 \): \( r_{2D} = |0.4 - 0.3| = 0.1 m \).

Potential at \( D \): \( V_D = \frac{1}{4\pi\epsilon_0} \left[ \frac{q_1}{0.4} + \frac{q_2}{0.1} \right] \).

Change in Potential Energy:
\[ \Delta U = q_3 (V_D - V_C) = \frac{q_3}{4\pi\epsilon_0} \left[ \left( \frac{q_1}{0.4} + \frac{q_2}{0.1} \right) - \left( \frac{q_1}{0.4} + \frac{q_2}{0.5} \right) \right] \] \[ \Delta U = \frac{q_3 q_2}{4\pi\epsilon_0} \left[ \frac{1}{0.1} - \frac{1}{0.5} \right] = \frac{q_3 q_2}{4\pi\epsilon_0} [10 - 2] \] \[ \Delta U = \frac{q_3 (8q_2)}{4\pi\epsilon_0} \]
Comparing with the given form \( \frac{q_3 K}{4\pi\epsilon_0} \), we find \( K = 8q_2 \).


Step 4: Final Answer:

The value of \( K \) is \( 8q_2 \).
Quick Tip: Notice that since \( q_3 \) moves along a circular path centered at \( q_1 \), its distance from \( q_1 \) remains constant. Therefore, the work done (and change in energy) due to \( q_1 \) is zero. Only the change in distance relative to \( q_2 \) matters.

Step 1: Understanding the Concept:

The magnetic field inside a solenoid filled with a material of susceptibility \( \chi \) is given by \( B = \mu_0 (1 + \chi) n I \).

The field without the material (vacuum/air) is \( B_0 = \mu_0 n I \).


Step 2: Key Formula or Approach:

1. \( B = B_0(1 + \chi) \)

2. Percentage increase \( = \frac{B - B_0}{B_0} \times 100 = \chi \times 100 % \).


Step 3: Detailed Explanation:

Given susceptibility \( \chi = 1.2 \times 10^{-5} \).

The percentage increase in the magnetic field is:
\[ % Increase = (1.2 \times 10^{-5}) \times 100 % \] \[ % Increase = 1.2 \times 10^{-3} % \]
Expressing \( 1.2 \) as a fraction:
\[ 1.2 = \frac{12}{10} = \frac{6}{5} \]
So, the percentage increase is \( \frac{6}{5} \times 10^{-3} % \).


Step 4: Final Answer:

The percentage increase is \( \frac{6}{5} \times 10^{-3} % \).
Quick Tip: Percentage change in magnetic field is simply \( 100\chi \). Magnesium is paramagnetic (positive but small \( \chi \)), so the field increases slightly.

Step 1: Understanding the Concept:

A charged particle in a magnetic field moves in a circular path with radius \( R = \frac{mv}{qB} \). When it enters from an interface and returns to it, it completes a semi-circle. The displacement along the interface for each semi-circle is equal to the diameter \( 2R \).


Step 2: Key Formula or Approach:

1. Radius \( R = \frac{mv}{qB} \).

2. Displacement for one semi-circle \( = 2R \).


Step 3: Detailed Explanation:

1. In Region 2: The particle enters with velocity \( v \). The radius is \( R_2 = \frac{mv}{qB_2} \). It returns to the interface after moving a distance \( d_2 = 2R_2 = \frac{2mv}{qB_2} \) along the interface from the starting point.

2. In Region 1: The particle enters region 1 at the point where it exited region 2. The magnetic fields are uniform and in the same direction (normal to paper). Due to the sense of rotation, if it moved in one direction along the interface in region 2, it will continue or reverse.

Assuming standard geometry, it completes a semi-circle in Region 1 with radius \( R_1 = \frac{mv}{qB_1} \). It returns to the interface after a further displacement of \( 2R_1 \).

3. Since the particle enters region 1 from the same interface, and the Lorentz force direction results in semi-circles on opposite sides of the interface line, the net displacement from the original starting point is \( |2R_1 - 2R_2| \) if the rotation sense carries it "backwards" relative to the first exit point.

Net Displacement \( D = 2 \left( \frac{mv}{qB_1} - \frac{mv}{qB_2} \right) \).
\[ D = \frac{2mv}{q} \left( \frac{B_2 - B_1}{B_1 B_2} \right) = \frac{2mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right) \]
Which is equivalent to \( \frac{mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right) \times 2 \).


Step 4: Final Answer:

The displacement is \( \frac{mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right) \times 2 \).
Quick Tip: For multiple regions with different \( B \) fields, the net displacement along an interface is typically the sum of the diameters of the arcs, taking direction into account. Here, with \( B_2 > B_1 \), \( R_2 < R_1 \).

Step 1: Understanding the Concept:

When two waves of the same frequency and amplitude \( E_0 \) superpose with a phase difference \( \phi \), the resultant amplitude \( A \) is determined using the phasor addition method or the superposition principle formula.


Step 2: Key Formula or Approach:
\[ A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \phi} \]

Step 3: Detailed Explanation:

Given:
\( A_1 = E_0 \), \( A_2 = E_0 \), and phase difference \( \phi = \pi/3 = 60^\circ \).

Substituting into the formula:
\[ A = \sqrt{E_0^2 + E_0^2 + 2 E_0^2 \cos(60^\circ)} \]
Since \( \cos(60^\circ) = 1/2 \):
\[ A = \sqrt{E_0^2 + E_0^2 + 2 E_0^2(1/2)} \] \[ A = \sqrt{3 E_0^2} = \sqrt{3} E_0 \]
Value of \( \sqrt{3} \approx 1.732 \).

Thus, \( A \approx 1.732 E_0 \), which matches Option (A) \( 1.7 E_0 \).


Step 4: Final Answer:

The amplitude of the resultant wave is \( 1.7 E_0 \).
Quick Tip: For two equal amplitudes \( A \), the resultant is \( 2A \cos(\phi/2) \). Here \( 2 E_0 \cos(30^\circ) = 2 E_0 (\sqrt{3}/2) = \sqrt{3} E_0 \). This is a faster way to solve such problems.

Step 1: Understanding the Concept:

The focal length of a lens changes depending on the surrounding medium's refractive index according to the Lens Maker's Formula.


Step 2: Key Formula or Approach:
\[ \frac{1}{f} = \left( \frac{\mu_{lens}}{\mu_{medium}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Step 3: Detailed Explanation:

Let \( f_a \) be focal length in air (\( \mu_a = 1 \)) and \( f_w \) be focal length in water (\( \mu_w = 1.28 \)).

1. In air:

\[ \frac{1}{12} = (1.6 - 1) K = 0.6 K \quad where K = \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
2. In water:

\[ \frac{1}{f_w} = \left( \frac{1.6}{1.28} - 1 \right) K = (1.25 - 1) K = 0.25 K \]
3. Taking the ratio:

\[ \frac{f_w}{12} = \frac{0.6 K}{0.25 K} = \frac{60}{25} = 2.4 \]
\[ f_w = 12 \times 2.4 = 28.8 cm \]
4. Convert to mm:

\[ f_w = 28.8 \times 10 = 288 mm \]

Step 4: Final Answer:

The focal length in water is 288 mm.
Quick Tip: When a lens is immersed in a liquid of refractive index \( \mu_l \), its focal length usually increases. If the liquid has the same refractive index as the lens, the focal length becomes infinite (it acts as a plane sheet).

Step 1: Understanding the Concept:

The wavelength \(\lambda\) for a transition in a hydrogen atom is given by the Rydberg formula: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
The largest wavelength corresponds to the minimum energy difference, which occurs for a transition from the level immediately above the series base level.


Step 2: Key Formula or Approach:

1. For the Lyman series, the base level is \( n_1 = 1 \). The largest wavelength corresponds to \( n_2 = 2 \).

2. For the Balmer series, the base level is \( n_1 = 2 \). The largest wavelength corresponds to \( n_2 = 3 \).


Step 3: Detailed Explanation:

For the Lyman series (\( \lambda_L \)): \[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \implies \lambda_L = \frac{4}{3R} \]
For the Balmer series (\( \lambda_B \)): \[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36} \implies \lambda_B = \frac{36}{5R} \]
Taking the ratio: \[ \frac{\lambda_L}{\lambda_B} = \frac{4/3R}{36/5R} = \frac{4}{3} \times \frac{5}{36} = \frac{5}{27} \]

Step 4: Final Answer:

The ratio of the largest wavelength of the Lyman series to that of the Balmer series is 5 : 27.
Quick Tip: Remember: Largest wavelength = Smallest frequency = Smallest energy. For any series, this is always the first line of that series.

Step 1: Understanding the Concept:

The energy of the \( n^{th} \) state of a hydrogen-like ion with atomic number \( Z \) is given by: \[ E_n = -13.6 \frac{Z^2}{n^2} eV \]
The \( 2^{nd} \) excitation state refers to the level \( n = 3 \), and the ground state is \( n = 1 \).


Step 2: Detailed Explanation:

The energy difference \( \Delta E \) between the \( 2^{nd} \) excitation state and the ground state is: \[ \Delta E = E_3 - E_1 = 13.6 Z^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]
Given \( \Delta E = 108.8 eV \): \[ 108.8 = 13.6 \times Z^2 \times \left( 1 - \frac{1}{9} \right) \] \[ 108.8 = 13.6 \times Z^2 \times \frac{8}{9} \]
Divide both sides by 13.6: \[ \frac{108.8}{13.6} = \frac{8 Z^2}{9} \implies 8 = \frac{8 Z^2}{9} \] \[ 1 = \frac{Z^2}{9} \implies Z^2 = 9 \implies Z = 3 \]

Step 3: Final Answer:

The atomic number of the ion is 3.
Quick Tip: Always be careful with the terminology: \( n^{th} \) excitation state corresponds to the energy level \( (n+1) \). Thus, 2nd excitation state is \( n=3 \).

Step 1: Understanding the Concept:

In a Zener regulator circuit, the Zener diode maintains a constant breakdown voltage \( V_Z \) across the load resistance \( R_L \), provided the input voltage is high enough to keep it in the breakdown region.


Step 2: Key Formula or Approach:

1. Determine if the Zener diode is in breakdown by calculating the voltage across it as if it were open-circuited.
2. If \( V_{open} > V_Z \), the voltage across the load \( R_L \) is \( V_Z \).
3. The load current (ammeter reading) is \( I_L = \frac{V_Z}{R_L} \).


Step 3: Detailed Explanation:

Input voltage \( V_{in} = 12 V \), Series resistance \( R_s = 100 \Omega \), Load resistance \( R_L = 400 \Omega \).

Potential at the Zener node without the Zener diode: \[ V = V_{in} \left( \frac{R_L}{R_s + R_L} \right) = 12 \left( \frac{400}{100 + 400} \right) = 12 \times \frac{4}{5} = 9.6 V \]
Since \( 9.6 V > 4 V \) (Zener breakdown voltage), the Zener is in breakdown.

The voltage across the \( 400 \Omega \) resistor is fixed at 4 V.

The current through the ammeter is: \[ I_L = \frac{4 V}{400 \Omega} = 0.01 A = 10 mA \]

Step 4: Final Answer:

The reading of the ammeter is 10 mA.
Quick Tip: Always check if the Zener is "on" first. If the potential divider voltage was less than 4 V, the Zener would act as an open circuit, and the current would simply be \( I = V_{in} / (R_s + R_L) \).

Step 1: Understanding the Concept:

The total moment of inertia of the system is the sum of the moments of inertia of its individual components about the common axis PQ. We use the Parallel Axis Theorem: \( I_{axis} = I_{cm} + Md^2 \).


Step 2: Key Formula or Approach:

1. Moment of inertia of disc about diameter \( I = \frac{1}{4}MR^2 \).

2. Moment of inertia of solid sphere about center \( I_s = \frac{2}{5}MR^2 \).

3. Moment of inertia of spherical shell about center \( I_{shell} = \frac{2}{3}MR^2 \).


Step 3: Detailed Explanation:

Axis PQ passes through the center of disc A and is tangent to sphere B and shell C.
1. For Disc A: Axis PQ is its diameter.
\[ I_A = \frac{1}{4}MR^2 \]
2. For Solid Sphere B: Distance from center to PQ is \( R \).
\[ I_B = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2 \]
3. For Spherical Shell C: Distance from center to PQ is \( R \).
\[ I_C = \frac{2}{3}MR^2 + MR^2 = \frac{5}{3}MR^2 \]
Total Moment of Inertia \( I_{PQ} \): \[ I_{PQ} = I_A + I_B + I_C = MR^2 \left( \frac{1}{4} + \frac{7}{5} + \frac{5}{3} \right) \]
Find the LCM of 4, 5, 3, which is 60: \[ I_{PQ} = MR^2 \left( \frac{15 + 84 + 100}{60} \right) = \frac{199}{60}MR^2 \]
Given \( I_{PQ} = \frac{x}{15} I \), where \( I = \frac{1}{4}MR^2 \): \[ \frac{199}{60}MR^2 = \frac{x}{15} \left( \frac{1}{4}MR^2 \right) = \frac{x}{60}MR^2 \]
Comparing both sides, we get \( x = 199 \).


Step 4: Final Answer:

The value of x is 199.
Quick Tip: When multiple bodies are combined, ensure you correctly identify the distance from the axis of rotation to the center of mass of each body before applying the parallel axis theorem.

Step 1: Understanding the Concept:

The power radiated by a hot body is given by Stefan-Boltzmann Law: \[ P = e \sigma A T^4 \]
where \( e \) is emissivity, \( A \) is surface area, and \( T \) is absolute temperature (in Kelvin).


Step 2: Detailed Explanation:

1. Temperature: \( T = 1727 + 273 = 2000 K \).
2. Surface Area \( A \) of a wire: \( A = \pi d l \).
\( d = 0.5 mm = 0.5 \times 10^{-3} m \).
\( l = 10 cm = 0.1 m \).
\( A = 3.14 \times 0.5 \times 10^{-3} \times 0.1 = 1.57 \times 10^{-4} m^2 \).
3. Power Radiation Equation:
\[ 94.2 = e \times (6.0 \times 10^{-8}) \times (1.57 \times 10^{-4}) \times (2000)^4 \]
\[ 94.2 = e \times 6 \times 10^{-8} \times 1.57 \times 10^{-4} \times 16 \times 10^{12} \]
\[ 94.2 = e \times 6 \times 1.57 \times 16 \times 10^0 \]
\[ 94.2 = e \times 150.72 \]
\[ e = \frac{94.2}{150.72} = 0.625 \]
4. Emissivity \( e = \frac{x}{8} \):
\[ 0.625 = \frac{x}{8} \implies x = 0.625 \times 8 = 5 \]

Step 3: Final Answer:

The value of x is 5.
Quick Tip: Always convert temperatures to Kelvin by adding 273 to the Celsius value. Also, check the units of area carefully (surface area of a cylinder is \( \pi d l \)).

Step 1: Understanding the Concept:

In a P-V diagram, the work done in a cyclic process is equal to the area enclosed by the cycle. For a clockwise cycle, the work done is positive.


Step 2: Key Formula or Approach:

The process is shown as a circle on the P-V axes. The area of a circle (or ellipse if scales differ) is: \[ W = Area = \pi \times r_P \times r_V \]
where \( r_P \) and \( r_V \) are the radii along the Pressure and Volume axes respectively.


Step 3: Detailed Explanation:

From the graph:
1. Pressure diameter \( = 500 - 300 = 200 kPa \). Radius \( r_P = 100 kPa = 10^5 Pa \).
2. Volume diameter \( = 350 - 150 = 200 cm^3 \). Radius \( r_V = 100 cm^3 = 100 \times 10^{-6} m^3 = 10^{-4} m^3 \).
Work done \( W \): \[ W = \pi \times (10^5) \times (10^{-4}) = 10\pi J \]
Using \( \pi = 3.14 \): \[ W = 10 \times 3.14 = 31.4 J \]
The question asks for work in terms of \( \_\_\_\_ \times 10^{-1} J \): \[ 31.4 J = 314 \times 10^{-1} J \]
Thus, the value is 314.


Step 4: Final Answer:

The value is 314.
Quick Tip: Note the units on the axes: \( kPa = 10^3 Pa \) and \( cm^3 = 10^{-6} m^3 \). The product of these units gives a factor of \( 10^{-3} \), which is crucial for the Joule calculation.

Step 1: Understanding the Concept:

Potential difference \( V_B - V_A \) depends on the complex impedances of the components in the two branches. The phase of this difference relative to \( V_{in} \) changes with frequency.


Step 2: Key Formula or Approach:

1. \( V_A = V_{in} \frac{1/j\omega C}{R + 1/j\omega C} = \frac{V_{in}}{1 + j\omega RC} \).
2. \( V_B = V_{in} \frac{R}{R + 1/j\omega C} = \frac{j\omega RC V_{in}}{1 + j\omega RC} \).
3. \( V_B - V_A = V_{in} \frac{j\omega RC - 1}{1 + j\omega RC} \).


Step 3: Detailed Explanation:

The phasor ratio is \( \frac{V_B - V_A}{V_{in}} = \frac{j\omega RC - 1}{j\omega RC + 1} \).
The phase \( \phi \) of this expression is: \[ \phi = angle(j\omega RC - 1) - angle(j\omega RC + 1) \]
Let \( \omega RC = \tan \theta \). Then: \[ angle(-1 + j \tan \theta) = \pi - \theta \] \[ angle(1 + j \tan \theta) = \theta \]
So, \( \phi = (\pi - \theta) - \theta = \pi - 2\theta \).
Given \( \phi = 90^{\circ} = \frac{\pi}{2} \): \[ \frac{\pi}{2} = \pi - 2\theta \implies 2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4} \]
Thus, \( \tan \theta = \tan(\pi/4) = 1 \).
So, \( \omega RC = 1 \implies \omega = \frac{1}{RC} \).
Values: \( R = 10^5 \Omega, C = 100 \times 10^{-12} F = 10^{-10} F \). \[ \omega = \frac{1}{10^5 \times 10^{-10}} = \frac{1}{10^{-5}} = 10^5 rad/sec \]
Comparing with \( 10^x \), we get \( x = 5 \).


Step 4: Final Answer:

The value of x is 5.
Quick Tip: For a frequency response function \( \frac{j\omega\tau - 1}{j\omega\tau + 1} \), the magnitude is always 1, but the phase shifts from \( 180^{\circ} \) to \( 0^{\circ} \). At \( \omega\tau = 1 \), the phase shift is exactly \( 90^{\circ} \).

Step 1: Understanding the Concept:

When an object is viewed through multiple layers of refractive media, the total apparent shift is the sum of the shifts caused by each individual layer.


Step 2: Key Formula or Approach:

The apparent shift \( \Delta S \) caused by a medium of thickness \( t \) and refractive index \( \mu \) is: \[ \Delta S = t \left( 1 - \frac{1}{\mu} \right) \]

Step 3: Detailed Explanation:

1. Shift due to the first liquid (\( t_1 = 60 cm, \mu_1 = 1.2 \)):
\[ \Delta S_1 = 60 \left( 1 - \frac{1}{1.2} \right) = 60 \left( 1 - \frac{5}{6} \right) = 60 \times \frac{1}{6} = 10 cm \]
2. Shift due to the second liquid (\( t_2 = H, \mu_2 = 1.6 \)):
\[ \Delta S_2 = H \left( 1 - \frac{1}{1.6} \right) = H \left( 1 - \frac{10}{16} \right) = H \left( 1 - \frac{5}{8} \right) = \frac{3H}{8} \]
Total apparent shift \( \Delta S_{total} = 40 cm \): \[ \Delta S_1 + \Delta S_2 = 40 \implies 10 + \frac{3H}{8} = 40 \] \[ \frac{3H}{8} = 30 \implies 3H = 240 \implies H = 80 cm \]

Step 4: Final Answer:

The value of H is 80 cm.
Quick Tip: Apparent shift is purely additive. This means you can calculate the effect of each layer independently as if the other layers were air.

Step 1: Understanding the Concept:

According to Dalton's Law of Partial Pressures, the partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure. The ratio of partial pressures of two gases is equal to the ratio of their moles.


Step 2: Key Formula or Approach:

Number of moles \( n = \frac{Mass}{Molar Mass} \).

Ratio of partial pressures: \( \frac{P_A}{P_B} = \frac{n_A}{n_B} \).


Step 3: Detailed Explanation:

Assume the total mass of the dry air sample is 100 g.

Mass of \( N_2 = 70.0 g \), Molar mass = \( 28 g/mol \).

Mass of \( O_2 = 27.0 g \), Molar mass = \( 32 g/mol \).

Mass of \( Ar = 3.0 g \), Molar mass = \( 40 g/mol \).


(i) Ratio of partial pressure of \( N_2 \) to \( O_2 \):
\[ \frac{P_{N_2}}{P_{O_2}} = \frac{n_{N_2}}{n_{O_2}} = \frac{70/28}{27/32} = \frac{2.5}{0.84375} \approx 2.96 \]

(ii) Ratio of partial pressure of \( O_2 \) to \( Ar \):
\[ \frac{P_{O_2}}{P_{Ar}} = \frac{n_{O_2}}{n_{Ar}} = \frac{27/32}{3/40} = \frac{0.84375}{0.075} = 11.25 \]

Comparing with options, the values are approximately 2.96 and 11.2.


Step 4: Final Answer:

The calculated ratios are 2.96 and 11.2.
Quick Tip: While calculating ratios of partial pressures, the total pressure value is often redundant because it cancels out in the division. Focus directly on the mole ratio to save time.

Step 1: Understanding the Concept:

Photoelectric emission occurs only if the frequency of incident light (\( \nu \)) is greater than or equal to the threshold frequency (\( \nu_0 \)). The photocurrent is proportional to the intensity (brightness) of light, provided emission is taking place.


Step 2: Key Formula or Approach:

Calculate threshold wavelength \( \lambda_0 = \frac{c}{\nu_0} \).

Incident light wavelength \( \lambda \) must be \( \le \lambda_0 \) for emission.


Step 3: Detailed Explanation:

Threshold frequency \( \nu_0 = 5.16 \times 10^{14} Hz \).

Threshold wavelength \( \lambda_0 = \frac{3 \times 10^8 m/s}{5.16 \times 10^{14} Hz} \approx 581 nm \).


Statement A: Yellow light (approx. 570–590 nm). Since its frequency is very close to and mostly above the threshold for Cs, current is observed. (True)

Statement B: Dimming light reduces intensity, which reduces the number of photoelectrons per second, hence current decreases. (True)

Statement C: Red light (approx. 700 nm) has \( \lambda > \lambda_0 \), so its frequency is below the threshold. No current is produced. (False)

Statement D: Blue light (approx. 450 nm) has \( \lambda < \lambda_0 \), so its frequency is above threshold. Current flows. (True)

Statement E: White light contains blue and yellow components, which have frequencies above the threshold. Current flows. (True)


Step 4: Final Answer:

Statements A, B, D, and E are correct.
Quick Tip: Remember the visible spectrum order VIBGYOR. Moving from Red to Violet, frequency increases and wavelength decreases. Caesium has one of the lowest work functions, allowing even yellow light to cause emission.

Step 1: Understanding the Concept:

The process involves three steps: cooling liquid water to its freezing point, phase change (freezing), and cooling the resulting ice to the final temperature. The total enthalpy change is the sum of these steps.


Step 2: Key Formula or Approach:
\( \Delta H = n C_p \Delta T \) for temperature change.
\( \Delta H = -n \Delta_{fus}H \) for freezing at 0°C.


Step 3: Detailed Explanation:

Total Enthalpy Change \( \Delta H_{total} = \Delta H_1 + \Delta H_2 + \Delta H_3 \).

Step 1: Cooling 1 mol water from 10°C to 0°C (\( \Delta T = -10 K \)):
\[ \Delta H_1 = 1 \times y \times (-10) = -10y J \]
Step 2: Freezing 1 mol water at 0°C:

Given \( \Delta_{fus}H = x kJ/mol = 1000x J/mol \).

For freezing (opposite of fusion): \( \Delta H_2 = -1000x J \).

Step 3: Cooling 1 mol ice from 0°C to -10°C (\( \Delta T = -10 K \)):
\[ \Delta H_3 = 1 \times z \times (-10) = -10z J \]
Total \( \Delta H = -10y - 1000x - 10z = -10(100x + y + z) J \).


Step 4: Final Answer:

The total enthalpy change is \( -10(100x + y + z) \).
Quick Tip: Pay close attention to units. Heat capacities are usually given in Joules, while enthalpies of phase transition are often in kiloJoules. Always convert to a common unit before adding.

Step 1: Understanding the Concept:

When a solution is diluted by adding water, the volume increases and the concentration of solutes decreases. For a strong acid like HCl, pH depends on the final concentration of \( [H^+] \).


Step 2: Key Formula or Approach:
\( [H^+] = 10^{-pH} \).

New concentration \( M_2 = \frac{M_1 V_1}{V_2} \).

Final pH \( = -\log[H^+]_{final} \).


Step 3: Detailed Explanation:

Initial pH = 1.0, so initial \( [H^+] = 10^{-1.0} = 0.1 M \).

Adding an equal volume of water means the final volume \( V_2 = 2V_1 \).

Final concentration \( [H^+]_{final} = \frac{0.1 \times V_1}{2V_1} = 0.05 M \).

Final pH \( = -\log(0.05) = -\log(5 \times 10^{-2}) \).

Final pH \( = 2 - \log 5 = 2 - 0.7 = 1.3 \).


Step 4: Final Answer:

The pH of the solution increases to 1.3.
Quick Tip: Diluting an acidic solution by a factor of 2 increases the pH by \( \log 2 \approx 0.3 \). Diluting by a factor of 10 increases pH by 1 unit.

Step 1: Understanding the Concept:

Mohr's salt is a double salt with the formula \( (NH_4)_2Fe(SO_4)_2 \cdot 6H_2O \). Kohlrausch's Law states that the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions, each multiplied by its stoichiometric coefficient.


Step 2: Detailed Explanation:

Statement I: Mohr's salt dissociates into \( Fe^{2+} \), \( NH_4^+ \), and \( SO_4^{2-} \). Thus, it contains only these three types of ions. (True)

Statement II: From the formula \( (NH_4)_2Fe(SO_4)_2 \), one mole of the salt produces:

1 mole of \( Fe^{2+} \) ions.

2 moles of \( NH_4^+ \) ions.

2 moles of \( SO_4^{2-} \) ions.

According to Kohlrausch's Law:
\[ \Lambda_m^\infty = 1 \cdot \lambda_{Fe^{2+}}^\infty + 2 \cdot \lambda_{NH_4^+}^\infty + 2 \cdot \lambda_{SO_4^{2-}}^\infty \] \[ \Lambda_m^\infty = x_1 + 2x_2 + 2x_3 \]
The statement gives \( x_1 + x_2 + 2x_3 \), which is incorrect because it misses the coefficient for ammonium ions. (False)


Step 3: Final Answer:

Statement I is true but Statement II is false.
Quick Tip: Double salts dissociate completely into their simple ions in aqueous solution. Always write the balanced dissociation equation to determine the correct coefficients for Kohlrausch's Law calculations.

Step 1: Understanding the Concept:

For a first-order gas-phase reaction, the partial pressure of the reactant decreases exponentially over time. The total pressure of the system increases as product moles are greater than reactant moles.


Step 2: Key Formula or Approach:

Initial pressure \( = P_0 \).

At time \( t \): \( A \to P_0 - p \), \( B \to 2p \), \( C \to p \). Total \( P_t = P_0 + 2p \).

At \( t = \infty \): \( A \to 0 \), \( B \to 2P_0 \), \( C \to P_0 \). Total \( P_\infty = 3P_0 \).


Step 3: Detailed Explanation:

1. From \( P_\infty = 240 mm Hg \):
\[ 3P_0 = 240 \implies P_0 = 80 mm Hg (Option A is correct) \]
2. At \( t = 10 min \), \( P_t = 160 mm Hg \):
\[ 80 + 2p = 160 \implies 2p = 80 \implies p = 40 mm Hg \]
Partial pressure of A at \( t = 10 min \):
\[ P_A = P_0 - p = 80 - 40 = 40 mm Hg (Option B is correct) \]
3. Rate constant \( k = \frac{1}{t} \ln\left(\frac{P_0}{P_A}\right) \):
\[ k = \frac{1}{10} \ln\left(\frac{80}{40}\right) = \frac{1}{10} \ln 2 = 0.1 \times 0.693 = 0.0693 min^{-1} \]
Statement C says 1.693, which is incorrect.

4. First-order reactions theoretically take infinite time to reach completion. (Option D is correct)


Step 4: Final Answer:

Statement (C) is incorrect.
Quick Tip: For total pressure problems, express all partial pressures in terms of initial pressure \( P_0 \) and the pressure change \( p \). Using the infinite time data allows you to find \( P_0 \) immediately.

Step 1: Understanding the Concept:

1st order kinetics leads to exponential changes over time. Rate laws relating rate to concentration determine the shape of the rate-vs-concentration graph.


Step 2: Detailed Explanation:

"Before" situation (Growth):

Follows \( 1^{st} \) order kinetics: \( \frac{dN}{dt} = kN \).

Integrated form: \( N = N_0 e^{kt} \implies \frac{N}{N_0} = e^{kt} \).

The graph of \( \frac{N}{N_0} \) vs \( t \) is an exponential growth curve starting from 1 at \( t = 0 \).


"After" situation (Decay):

Given: Rate of decay \( r \propto N^2 \), so \( r = k' N^2 \).

The graph of \( r \) vs \( N \) is a parabola opening upwards (since rate is proportional to the square of concentration).


Step 3: Final Answer:

The matching set is Option (D).
Quick Tip: Identify the variables on the axes carefully. In 1st order growth, concentration increases exponentially with time. If rate \( r \propto N^2 \), the rate-concentration graph is second-order (parabolic).

Step 1: Understanding the Concept:

We use the VSEPR theory to find the number of bond pairs (BP) and lone pairs (LP) around the central atom. The number of valence electrons and the geometry of the molecule are considered.


Step 2: Detailed Explanation:

A. \( ICl_2^- \): Central atom I has 7 valence electrons + 1 from charge = 8. It forms 2 bonds with Cl. Remaining electrons = \( 8 - 2 = 6 \) (3 lone pairs).

BP : LP = 2 : 3. (A-III)


B. \( H_2O \): Central atom O has 6 valence electrons. It forms 2 bonds with H. Remaining electrons = \( 6 - 2 = 4 \) (2 lone pairs).

BP : LP = 2 : 2. (B-IV)


C. \( SO_2 \): Central atom S has 6 valence electrons. It forms double bonds with 2 Oxygen atoms. In VSEPR, each double bond counts as 1 BP region. Remaining electrons = \( 6 - 4 = 2 \) (1 lone pair).

BP : LP = 2 : 1. (Matches C-II if II is intended as 2:1)


D. \( XeF_4 \): Central atom Xe has 8 valence electrons. It forms 4 bonds with F. Remaining electrons = \( 8 - 4 = 4 \) (2 lone pairs).

BP : LP = 4 : 2. (D-I)


Step 3: Final Answer:

The matching is A-III, B-IV, C-II, D-I.
Quick Tip: For VSEPR, treat double or triple bonds as single electron domains when counting bond pairs for molecular geometry, but always subtract all bonding electrons to find the correct number of lone pairs.

Step 1: Understanding the Concept:

Enthalpy of atomization in transition metals depends on the strength of metallic bonding, which is related to the number of unpaired d-electrons available for delocalization. Higher number of unpaired electrons generally leads to higher enthalpy of atomization.


Step 2: Detailed Explanation:

In the 3d transition series, the enthalpies of atomization generally increase towards the middle and then decrease. Among the given metals:

Enthalpy of Atomization (approx. values in kJ/mol):

Cr: 397

Fe: 416

Co: 425

Ni: 430


Among these four, Chromium (Cr) has the lowest enthalpy of atomization. (Note: Manganese is lower than Cr, but not in the options).

Atomic number of Cr = 24.

Electronic configuration: \( [Ar] 3d^5 4s^1 \).

Number of valence electrons (3d + 4s) = \( 5 + 1 = 6 \).


Step 3: Final Answer:

The number of valence electrons is 6.
Quick Tip: Transition metal properties often show a dip in the middle of the series (Mn) and at the end (Zn) due to stable half-filled or full-filled configurations reducing metallic bond strength. Cr is the lowest among common structural transition metals like Fe, Co, Ni.

Step 1: Understanding the Concept:

The first ionization enthalpies (IE\(_1\)) of Group 14 elements (C, Si, Ge, Sn, Pb) generally decrease down the group, but there is an anomaly at the bottom. The values for Sn and Pb are the lowest in the group. Specifically, IE\(_1\) for Sn is 708 kJ/mol and for Pb is 715 kJ/mol. Thus, element A is Tin (Sn) and element B is Lead (Pb).


Step 2: Key Formula or Approach:

The stability of oxidation states in Group 14 is governed by the Inert Pair Effect. For heavier elements, the \(ns^2\) electrons are harder to remove, making the \(+2\) state more stable than the \(+4\) state as we go down the group.


Step 3: Detailed Explanation:

1. For Element A (Sn): The \(+4\) oxidation state is more stable than the \(+2\) state. Therefore, \(Sn^{2+}\) (\(A^{2+}\)) tends to get oxidized to \(Sn^{4+}\) by losing two electrons. Since it undergoes oxidation, it acts as a reducing agent.

2. For Element B (Pb): Due to the strong inert pair effect, the \(+2\) oxidation state is much more stable than the \(+4\) state. Therefore, \(Pb^{4+}\) (\(B^{4+}\)) tends to get reduced to \(Pb^{2+}\) by gaining two electrons. Since it undergoes reduction, it acts as an oxidising agent.


Step 4: Final Answer:

The nature of \(A^{2+}\) is reducing and \(B^{4+}\) is oxidising.
Quick Tip: In Group 14, Sn(II) is a reducing agent (prefers Sn(IV)), while Pb(IV) is a strong oxidising agent (prefers Pb(II)) due to the increasing stability of the lower oxidation state down the group (Inert Pair Effect).

Step 1: Understanding the Concept:

The enthalpy of atomization depends on the number of unpaired electrons available for metallic bonding. In the 3d series, metals like Vanadium (V) and Chromium (Cr) have very high enthalpies of atomization. While Vanadium has the highest experimental value (515 kJ/mol), Chromium is often associated with the highest number of unpaired electrons (6 in \(3d^5 4s^1\)) and is a classic example in transition metal chemistry.


Step 2: Detailed Explanation:

1. Identification of Metal M: Chromium (Cr) and Vanadium (V) both have green-colored aquated ions (\(Cr^{3+}\) and \(V^{3+}\)). However, the nature of Chromium oxides is a very common topic in competitive exams.

2. Color of Ion: The ion \(Cr^{3+}\) in aqueous solution exists as \([Cr(H_2O)_6]^{3+}\), which is characteristic green.

3. Oxide Formation: The oxide corresponding to the \(+3\) oxidation state of Chromium is Chromium(III) oxide (\(Cr_2O_3\)).

4. Nature of Oxide: In the 3d series, lower oxidation states are basic (e.g., \(CrO\)), intermediate ones are amphoteric (e.g., \(Cr_2O_3\)), and higher ones are acidic (e.g., \(CrO_3\)).


Step 3: Final Answer:

The oxide \(Cr_2O_3\) formed by the green \(Cr^{3+}\) ion is amphoteric in nature.
Quick Tip: For transition metals, as the oxidation state increases, the metallic character decreases and the covalent/acidic character of the oxide increases. \(+3\) oxides of Cr and Al are common amphoteric oxides.

Step 1: Understanding the Concept:

Ionisation isomerism occurs when the composition of the coordination sphere and the counter-ions (ionization sphere) are interchanged. This results in the production of different ions in aqueous solution.


Step 2: Detailed Explanation:

1. Isomer A: It gives a white precipitate with \(AgNO_3\), which indicates the presence of free chloride ions (\(Cl^-\)) in the solution. This means \(Cl\) is outside the coordination sphere.

Structure: \([Co(NH_3)_5(SO_4)]Cl\).

Reaction: \([Co(NH_3)_5(SO_4)]Cl + AgNO_3 \rightarrow [Co(NH_3)_5(SO_4)]NO_3 + AgCl \downarrow\) (White ppt).

2. Isomer B: It gives a white precipitate with \(BaCl_2\), which indicates the presence of free sulfate ions (\(SO_4^{2-}\)) in the solution. This means \(SO_4\) is outside the coordination sphere.

Structure: \([Co(NH_3)_5Cl]SO_4\).

Reaction: \([Co(NH_3)_5Cl]SO_4 + BaCl_2 \rightarrow [Co(NH_3)_5Cl]Cl_2 + BaSO_4 \downarrow\) (White ppt).


Step 3: Final Answer:

Since the two isomers differ in the ions they produce in solution due to the exchange of ligands between the coordination and ionization spheres, they are ionisation isomers.
Quick Tip: Precipitation tests are the standard way to identify ionisation isomers. \(BaCl_2\) tests for free \(SO_4^{2-}\) and \(AgNO_3\) tests for free halides like \(Cl^-\).

Step 1: Understanding the Concept:

Ammonium salts react with strong bases like \(NaOH\) to evolve Ammonia gas (\(NH_3\)). Ammonia can be detected using Nessler's Reagent, which forms a characteristic brown precipitate.


Step 2: Detailed Explanation:

1. Identification of Gas X: Many ammonium salts (\(NH_4^+\)) react with \(NaOH\) as follows:
\(NH_4^+ + OH^- \rightarrow NH_3 \uparrow + H_2O\).

Thus, Gas X is Ammonia (\(NH_3\)).

2. Identification of Reagent Y: Nessler's reagent is an alkaline solution of potassium tetraiodomercurate(II), formulated as \(K_2[HgI_4] + KOH\).

3. Reaction with Reagent Y: When \(NH_3\) is passed through Nessler's reagent, it forms a brown precipitate of iodide of Millon's base:
\(2K_2[HgI_4] + NH_3 + 3KOH \rightarrow [HgO \cdot Hg(NH_2)I] \downarrow + 7KI + 2H_2O\).


Step 3: Final Answer:

Gas X is \(NH_3\) and Reagent Y is Nessler's reagent (\(K_2HgI_4 + KOH\)).
Quick Tip: Nessler's reagent test is the most sensitive confirmatory test for the Ammonium (\(NH_4^+\)) radical. The brown precipitate formed is also called "basic mercury(II) amido-iodide".

Step 1: Understanding the Concept:

Solubility of ethers in water depends on their ability to form hydrogen bonds with water molecules. Reactivity of sodium depends on the presence of acidic hydrogens.


Step 2: Detailed Explanation:

1. Statement I: Ethers can form H-bonds with water through their lone pair of electrons on Oxygen. Dimethyl ether (\(CH_3OCH_3\)) is very small and is highly soluble/miscible in water. As the alkyl group size increases (as in diethyl ether, \(C_2H_5OC_2H_5\)), the hydrophobic part increases, significantly reducing the solubility in water (approx. 6g/100mL). Thus, Statement I is true.

2. Statement II: Diethyl ether does not have any acidic hydrogen, so it does not react with sodium metal. Therefore, sodium can be used to remove traces of water from it (dry it). However, ethyl alcohol (\(C_2H_5OH\)) has an acidic hydrogen on the -OH group and reacts vigorously with sodium:
\(2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2 \uparrow\).

Thus, sodium cannot be used to dry alcohols. Statement II is true.


Step 3: Final Answer:

Both statements are true.
Quick Tip: Sodium metal is a standard drying agent for ethers and hydrocarbons but is never used for alcohols, phenols, or acids due to the evolution of hydrogen gas and salt formation.

Step 1: Understanding the Concept:

In IUPAC nomenclature of alkenes with substituents:

1. Select the longest carbon chain containing the double bond.

2. Number the chain to give the double bond the lowest possible locant.

3. If the double bond gets the same number from both ends, number the chain to give the substituents the lowest possible locants.


Step 2: Detailed Explanation:

1. The longest chain containing the double bond has 4 carbons (a butene).

2. The double bond is at position 2 regardless of which end we start from (\(CH_3-CH=C(CH_3)-CH_2Br\)).

3. Now we compare substituent locants:

- If we number from the right (near the Bromine): \(C1\) has Br, \(C2\) has Methyl. Locants = 1, 2.

- If we number from the left: \(C3\) has Methyl, \(C4\) has Br. Locants = 3, 4.

4. The set (1, 2) is lower than (3, 4). Therefore, numbering from the right is correct.

5. Alphabetical order for naming: Bromo comes before Methyl.

Name: 1-Bromo-2-methylbut-2-ene.


Step 3: Final Answer:

The correct IUPAC name is 1-Bromo-2-methylbut-2-ene.
Quick Tip: When numbering an alkene chain, the double bond takes priority over halogens and alkyl groups. Use the "lowest locant set" rule for substituents only if the double bond position is symmetrical.

Step 1: Understanding the Concept:

Preparation of alkenes by elimination typically involves Dehydrohalogenation (using strong base) or Dehydration (using acid or catalysts).


Step 2: Detailed Explanation:

- Reaction A: Bromocyclohexane + NaOEt. This is a dehydrohalogenation (E2 elimination). It produces cyclohexene.

- Reaction B: 2-Bromobutane + KOH (aq). Aqueous KOH favors substitution (\(S_N2/S_N1\)) over elimination. It primarily produces 2-butanol, not an alkene.

- Reaction C: Tert-butyl bromide + NaOMe. Tertiary alkyl halides undergo E2 elimination readily even with standard bases. It produces isobutylene.

- Reaction D: Cyclohexanol + \(Na_2Cr_2O_7/H_2SO_4\). This is an oxidation reaction. It produces cyclohexanone, not an alkene.

- Reaction E: Tert-butyl alcohol + Cu at 573 K. While primary and secondary alcohols undergo dehydrogenation, tertiary alcohols undergo dehydration (elimination) under these conditions to give isobutylene.


Step 3: Final Answer:

Reactions B (substitution) and D (oxidation) do not prepare alkenes by elimination.
Quick Tip: Distinguish between KOH (alc) and KOH (aq). KOH (alc) promotes elimination (alkene), whereas KOH (aq) promotes substitution (alcohol).

Step 1: Understanding the Concept:

Reductive ozonolysis (\(O_3\), then \(Zn/H_2O\)) cleaves the carbon-carbon double bond and adds an oxygen atom to each of the resulting carbon fragments, forming carbonyl compounds.


Step 2: Detailed Explanation:

1. Statement I: cis-2-Butene is \(CH_3-CH=CH-CH_3\). Cleavage of the double bond yields two molecules of acetaldehyde (\(CH_3CHO\)), also known as ethanal. Thus, Statement I is true.

2. Statement II: Let's look at the structure of 3,6-dimethyloct-4-ene:
\(CH_3-CH_2-CH(CH_3)-CH=C(CH_3)-CH_2-CH_2-CH_3\) is the likely structure. Let's name oct-4-ene:
\(C1-C2-C3(Me)-C4=C5-C6(Me)-C7-C8\).

Ozonolysis cleaves at \(C4=C5\).

Product 1 from left fragment: \(CH_3-CH_2-CH(CH_3)-CHO\) (2-methylbutanal). The carbon at position 2 is attached to \(H\), \(CH_3\), \(C_2H_5\), and \(CHO\). This is a chiral carbon atom.

Product 2 from right fragment: \(OHC-CH(CH_3)-CH_2-CH_3\). This is the same chiral molecule.

Since the products contain a chiral carbon, Statement II is false.


Step 3: Final Answer:

Statement I is true, but Statement II is false.
Quick Tip: Ozonolysis is a "cut and paste" reaction. Cut the C=C bond and paste an Oxygen atom on both ends. Then check the resulting structures for asymmetry to identify chiral centers.

Step 1: Understanding the Concept:

The reaction of a compound with aqueous sodium bicarbonate (\( NaHCO_3 \)) to produce \( CO_2 \) (effervescence) is a test for acidity.

A compound will react with \( NaHCO_3 \) if it is a stronger acid than carbonic acid (\( H_2CO_3 \)).

Carbonic acid has a \( pK_a \) of approximately 6.35. Compounds with \( pK_a < 6.35 \) will release \( CO_2 \) gas.


Step 2: Key Formula or Approach:

Acid strength comparison:

Acid + \( NaHCO_3 \rightarrow \) Salt + \( H_2O + CO_2 \uparrow \) (Occurs if Acid is stronger than \( H_2CO_3 \)).


Step 3: Detailed Explanation:

(A) m-Nitrobenzoic acid: Carboxylic acids are generally stronger than \( H_2CO_3 \). Benzoic acid has a \( pK_a \approx 4.2 \), and the nitro group further increases acidity. It will give effervescence.

(B) Anilinium chloride (\( PhNH_3^+Cl^- \)): This is a salt of a weak base and a strong acid. The anilinium ion is a relatively strong acid (\( pK_a \approx 4.6 \)) compared to \( H_2CO_3 \). It can react with bicarbonate.

(C) m-Nitrophenol: Phenols are generally weaker acids than carboxylic acids. Phenol has a \( pK_a \approx 10 \). m-Nitrophenol has a \( pK_a \approx 8.3 \). Since \( 8.3 > 6.35 \), it is a weaker acid than carbonic acid and will not give effervescence.

(D) 2,4,6-Trinitrophenol (Picric Acid): The presence of three strong electron-withdrawing nitro groups makes it exceptionally acidic (\( pK_a \approx 0.38 \)). It is much stronger than \( H_2CO_3 \) and gives a vigorous test.


Step 4: Final Answer:

m-Nitrophenol is the weakest acid among the given options and is least likely to give effervescence.
Quick Tip: Simple phenols and nitrophenols (with 1 or 2 nitro groups) do not react with \( NaHCO_3 \). Only extremely acidic phenols like picric acid (\( 2,4,6 \)-trinitrophenol) give this test.

Step 1: Understanding the Concept:

The carbylamine test (also known as the Isocyanide test) is used to detect the presence of primary (\( 1^\circ \)) amines. Secondary (\( 2^\circ \)) and tertiary (\( 3^\circ \)) amines do not give this test.


Step 2: Key Formula or Approach:

Primary amine + Chloroform + Alcoholic KOH \( \rightarrow \) Isocyanide (Foul smelling)
\( R-NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} R-NC + 3KCl + 3H_2O \)


Step 3: Detailed Explanation:

Let's analyze the given amines:

A. Aniline: It is an aromatic primary (\( 1^\circ \)) amine (\( Ph-NH_2 \)). It will show a positive test.

B. Dimethylamine (\( (CH_3)_2NH \)): It is a secondary (\( 2^\circ \)) aliphatic amine. It will show a negative test.

C. Methylamine (\( CH_3NH_2 \)): It is a primary (\( 1^\circ \)) aliphatic amine. It will show a positive test.

D. Trimethylamine (\( (CH_3)_3N \)): It is a tertiary (\( 3^\circ \)) aliphatic amine. It will show a negative test.

E. N-methylaniline: It is a secondary (\( 2^\circ \)) aromatic amine (\( Ph-NH-CH_3 \)). It will show a negative test.

Thus, only A and C are primary amines.


Step 4: Final Answer:

The amines showing a positive carbylamine test are A and C.
Quick Tip: The carbylamine test is specific only to primary amines. It is a very effective way to distinguish primary amines from secondary and tertiary ones in laboratory settings.

Step 1: Understanding the Concept:

Sucrose is a non-reducing disaccharide composed of \( \alpha \)-\( D \)-glucose and \( \beta \)-\( D \)-fructose linked by a glycosidic bond between \( C1 \) of glucose and \( C2 \) of fructose.


Step 2: Detailed Explanation:

Analysis of Statement I:

Sucrose is formed by the condensation of one molecule of \( D \)-(+)-glucose and one molecule of \( D \)-(-)-fructose with the elimination of water.

Upon hydrolysis, sucrose breaks down into its constituent monosaccharides: glucose and fructose.
\( Sucrose + H_2O \rightarrow D-(+)-glucose + D-(-)-fructose \).

Glucose is dextrorotatory (\( [\alpha]_D = +52.7^\circ \)) and fructose is strongly levorotatory (\( [\alpha]_D = -92.4^\circ \)).

Thus, Statement I is true.


Analysis of Statement II:

Sucrose itself is dextrorotatory (\( [\alpha]_D = +66.5^\circ \)).

During hydrolysis, the sign of optical rotation changes from dextro (+) to levo (-) because the levorotation of fructose is greater than the dextrorotation of glucose.

The resulting mixture is called "Invert Sugar".

Thus, Statement II is true.


Step 3: Final Answer:

Since both statements regarding the composition, hydrolysis, and optical properties of sucrose are correct, the answer is (A).
Quick Tip: Sucrose is unique because it is dextrorotatory, but its hydrolysis products result in a levorotatory mixture. This "inversion" of rotation is why the mixture is named invert sugar.

Step 1: Understanding the Concept:

Passing electricity through the half-cells changes the concentrations of the ions. We first calculate the final concentrations using Faraday's laws and then find the cell potential using the Nernst equation.


Step 2: Key Formula or Approach:

1. Change in moles \( = \frac{Q}{nF} \).

2. \( E_{cell} = E^\circ_{cell} - \frac{0.06}{n} \log Q \).


Step 3: Detailed Explanation:

Part 1: Adjusting Concentrations

For \( Cu^{2+}/Cu \): 1 Faraday passed. Assuming reduction: \( Cu^{2+} + 2e^- \rightarrow Cu \).

Moles of \( Cu^{2+} \) removed \( = \frac{1 F}{2 F} = 0.5 mol \).

Final moles of \( Cu^{2+} = 1.5 - 0.5 = 1.0 mol \).

Concentration \( [Cu^{2+}] = \frac{1.0 mol}{1 L} = 1.0 M \).


For \( Ag^+/Ag \): 0.1 Faraday passed. Assuming reduction: \( Ag^+ + e^- \rightarrow Ag \).

Moles of \( Ag^+ \) removed \( = \frac{0.1 F}{1 F} = 0.1 mol \).

Final moles of \( Ag^+ = 0.2 - 0.1 = 0.1 mol \).

Concentration \( [Ag^+] = \frac{0.1 mol}{1 L} = 0.1 M \).


Part 2: Calculating Cell Potential

Standard cell potential \( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \).

Since \( E^\circ_{Ag^+/Ag} > E^\circ_{Cu^{2+}/Cu} \), Silver is the cathode and Copper is the anode.
\( E^\circ_{cell} = 0.80 - 0.34 = 0.46 V \).

Cell reaction: \( Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s) \). (\( n=2 \))

Applying Nernst Equation:
\[ E_{cell} = 0.46 - \frac{0.06}{2} \log \frac{[Cu^{2+}]}{[Ag^+]^2} \] \[ E_{cell} = 0.46 - 0.03 \log \frac{1.0}{(0.1)^2} = 0.46 - 0.03 \log(100) \] \[ E_{cell} = 0.46 - 0.03 \times 2 = 0.46 - 0.06 = 0.40 V \]
In mV: \( 0.40 \times 1000 = 400 mV \).


Step 4: Final Answer:

The emf of the cell formed is 400 mV.
Quick Tip: Remember that for \( Cu^{2+} \), \( n=2 \). Passing 1 Faraday reduces only 0.5 moles of \( Cu^{2+} \). Miscounting valence factors is a common source of error in electrochemistry.

Step 1: Understanding the Concept:

The van't Hoff factor (\( i \)) represents the ratio of the actual concentration of particles produced when a substance is dissolved to the concentration of the substance as calculated from its mass. It is related to the degree of dissociation (\( \alpha \)).


Step 2: Key Formula or Approach:

For a salt \( MX_n \), the formula is:
\( i = 1 + (n - 1)\alpha \)

where \( n \) is the number of ions produced per formula unit of the salt.


Step 3: Detailed Explanation:

The salt \( MX_3 \) dissociates as:
\( MX_3 \rightleftharpoons M^{3+} + 3X^- \)

Here, one formula unit produces 4 ions (\( n = 1 + 3 = 4 \)).

Given: \( i = 2 \).

Substitute the values into the formula:
\[ 2 = 1 + (4 - 1)\alpha \] \[ 2 = 1 + 3\alpha \] \[ 1 = 3\alpha \] \[ \alpha = \frac{1}{3} \approx 0.3333 \]
Percentage dissociation \( = \alpha \times 100 = 0.3333 \times 100 = 33.33% \).

Rounding to the nearest integer, we get 33.


Step 4: Final Answer:

The percentage dissociation is 33%.
Quick Tip: Degree of dissociation \( \alpha = \frac{i - 1}{n - 1} \). Using this direct formula helps solve van't Hoff factor problems quickly.

Step 1: Understanding the Concept:

Paramagnetism arises from the presence of unpaired electrons. \( d^2sp^3 \) hybridization represents inner-orbital octahedral complexes, which usually occur with strong-field ligands or certain metal ions that can clear the inner d-orbitals.


Step 2: Key Formula or Approach:

1. Identify the oxidation state and d-electron configuration.

2. Determine if the ligand is Strong Field (SFL) or Weak Field (WFL).

3. Determine the electronic arrangement and hybridization.


Step 3: Detailed Explanation:

1. \( [FeF_6]^{3-} \): \( Fe^{3+} \) is \( d^5 \). \( F^- \) is WFL. Hybridization is \( sp^3d^2 \). (Outer orbital, Paramagnetic).

2. \( [Fe(CN)_6]^{3-} \): \( Fe^{3+} \) is \( d^5 \). \( CN^- \) is SFL. Pairing occurs, leaving one unpaired electron. Hybridization is \( \mathbf{d^2sp^3} \). (Paramagnetic).

3. \( [Mn(CN)_6]^{3-} \): \( Mn^{3+} \) is \( d^4 \). \( CN^- \) is SFL. Electrons occupy \( t_{2g} \) orbitals as \( t_{2g}^4 e_g^0 \). Two unpaired electrons remain. Hybridization is \( \mathbf{d^2sp^3} \). (Paramagnetic).

4. \( [Co(C_2O_4)_3]^{3-} \): \( Co^{3+} \) is \( d^6 \). Oxalate is SFL for \( Co^{3+} \). All electrons pair up as \( t_{2g}^6 e_g^0 \). Hybridization is \( d^2sp^3 \). (Diamagnetic).

5. \( [MnCl_6]^{3-} \): \( Mn^{3+} \) is \( d^4 \). \( Cl^- \) is WFL. Hybridization is \( sp^3d^2 \). (Outer orbital, Paramagnetic).

6. \( [CoF_6]^{3-} \): \( Co^{3+} \) is \( d^6 \). \( F^- \) is WFL. Hybridization is \( sp^3d^2 \). (Outer orbital, Paramagnetic).

The complexes that are both \( d^2sp^3 \) and paramagnetic are \( [Fe(CN)_6]^{3-} \) and \( [Mn(CN)_6]^{3-} \).


Step 4: Final Answer:

The count of such complexes is 2.
Quick Tip: Ligands like Oxalate and Ammonia act as strong field ligands for \( Co^{3+} \), causing pairing and resulting in diamagnetic inner-orbital complexes.

Step 1: Understanding the Concept:

In combustion analysis, all the carbon in the organic compound is converted into \( CO_2 \). By measuring the mass of \( CO_2 \) formed, we can determine the mass of carbon in the original sample.


Step 2: Key Formula or Approach:
\[ % C = \frac{12}{44} \times \frac{Mass of CO_2}{Mass of organic compound} \times 100 \]

Step 3: Detailed Explanation:

Given:

Mass of organic compound \( = 500 mg \).

Mass of \( CO_2 \) produced \( = 220 mg \).

Molar mass of \( C = 12 g/mol \).

Molar mass of \( CO_2 = 12 + 2(16) = 44 g/mol \).

Substituting into the formula:
\[ % C = \frac{12}{44} \times \frac{220}{500} \times 100 \] \[ % C = \frac{12}{44} \times \frac{220}{5} \] \[ % C = \frac{12}{1} \times \frac{5}{5} \times \frac{1}{1} (simplifying \frac{220}{44} = 5) \] \[ % C = \frac{12 \times 5}{5} = 12 % (Wait, re-calculating: \frac{12}{44} \times \frac{220}{500} \times 100 = \frac{12}{44} \times \frac{44}{1} = 12? No, \frac{220}{500} \times 100 = 44) \] \[ % C = \frac{12}{44} \times 44 = 12 % \]
Let me check the arithmetic again:
\( 220 / 500 = 0.44 \).
\( 0.44 \times 100 = 44 \).
\( (12/44) \times 44 = 12 \).

The percentage is 12%.


Step 4: Final Answer:

The percentage composition of carbon is 12%.
Quick Tip: Percentage of Carbon \( \approx 27.27 % \) of the mass of \( CO_2 \). Multiplying \( 0.2727 \) by the ratio of masses (\( 220/500 = 0.44 \)) gives the result.

Step 1: Understanding the Concept:

To find the percentage of an element in a compound from its structure, we first determine the molecular formula, calculate the total molar mass, and then find the fraction contributed by that specific element.


Step 2: Key Formula or Approach:
\[ % Element = \frac{Total mass of that element in the molecule}{Molar mass of the compound} \times 100 \]

Step 3: Detailed Explanation:

Determining Molecular Formula:

Counting the atoms in the structure:

Carbons (C): 6 in first ring + 6 in second ring + 3 in the side chain \( = 15 \).

Hydrogens (H): 2 in first ring + 2 in second ring + 4 in side chain + 1 in -OH + 2 in -\( NH_2 \) \( = 11 \).

Iodines (I): 4.

Nitrogen (N): 1.

Oxygens (O): 1 (linkage) + 1 (phenol) + 2 (acid) \( = 4 \).

Molecular Formula: \( C_{15}H_{11}I_4NO_4 \).


Calculating Molar Mass:

Mass of C \( = 15 \times 12 = 180 \)

Mass of H \( = 11 \times 1 = 11 \)

Mass of I \( = 4 \times 127 = 508 \)

Mass of N \( = 1 \times 14 = 14 \)

Mass of O \( = 4 \times 16 = 64 \)

Total Molar Mass \( = 180 + 11 + 508 + 14 + 64 = 777 g/mol \).


Calculating Percentage of Iodine:
\[ % I = \frac{508}{777} \times 100 \approx 65.38 % \]
Rounding to the nearest integer, we get 65.


Step 4: Final Answer:

The percentage of iodine in thyroxine is 65%.
Quick Tip: When counting atoms from a skeletal structure, remember to count the hidden hydrogens on the benzene rings (each carbon has one hydrogen unless it is substituted).