JEE Main 4 April Shift 1 Question Paper Out (Available)- Download Solutions and Answer Key

To find \(\frac{1}{\beta - \alpha}\), we first need to determine the range of the function \(fog(x) = f(g(x))\).

1. Calculate \(fog(x)\):
\[ fog(x) = f\left(g(x)\right) = f\left(\frac{2 - 3x}{1 - x}\right) \]
Substitute \(g(x)\) into \(f(x)\):
\[ fog(x) = \frac{2\left(\frac{2 - 3x}{1 - x}\right) + 3}{5\left(\frac{2 - 3x}{1 - x}\right) + 2} \]
Simplify the expression:
\[ fog(x) = \frac{\frac{4 - 6x + 3 - 3x}{1 - x}}{\frac{10 - 15x + 2 - 2x}{1 - x}} = \frac{7 - 9x}{12 - 17x} \]

2. Determine the range of \(fog(x)\) for \(x \in [2, 4]\):
- Calculate \(fog(2)\):
\[ fog(2) = \frac{7 - 9(2)}{12 - 17(2)} = \frac{7 - 18}{12 - 34} = \frac{-11}{-22} = \frac{1}{2} \]
- Calculate \(fog(4)\):
\[ fog(4) = \frac{7 - 9(4)}{12 - 17(4)} = \frac{7 - 36}{12 - 68} = \frac{-29}{-56} = \frac{29}{56} \]

3. Identify \(\alpha\) and \(\beta\):
- The range of \(fog(x)\) is \(\left[\frac{1}{2}, \frac{29}{56}\right]\).
- Therefore, \(\alpha = \frac{1}{2}\) and \(\beta = \frac{29}{56}\).

4. Calculate \(\frac{1}{\beta - \alpha}\):
\[ \beta - \alpha = \frac{29}{56} - \frac{1}{2} = \frac{29}{56} - \frac{28}{56} = \frac{1}{56} \]
\[ \frac{1}{\beta - \alpha} = \frac{1}{\frac{1}{56}} = 56 \]

Therefore, the correct answer is (4) 56. Quick Tip: The range of a composite function can be determined by evaluating the function at the endpoints of the domain.

1. Identify the sets \(A\) and \(B\):
- \(A: x^2 + y^2 = 25\)
- \(B: \frac{x^2}{144} + \frac{y^2}{16} = 1\)

2. Solve for the intersection \(D = A \cap B\):
- Substitute \(x^2 + y^2 = 25\) into \(x^2 + 9y^2 = 144\):
\[ x^2 + 9(25 - x^2) = 144 \]
\[ x^2 + 225 - 9x^2 = 144 \]
\[ -8x^2 = 144 - 225 \]
\[ -8x^2 = -81 \]
\[ x^2 = \frac{81}{8} \]
\[ x = \pm \frac{9}{2\sqrt{2}} \]
- Substitute \(x\) back into \(x^2 + y^2 = 25\):
\[ y^2 = 25 - \frac{81}{8} \]
\[ y = \pm \frac{\sqrt{119}}{2\sqrt{2}} \]

3. Determine the elements of set \(D\):
\[ D = \left\{\left(\frac{9}{2\sqrt{2}}, \frac{\sqrt{119}}{2\sqrt{2}}\right), \left(\frac{9}{2\sqrt{2}}, -\frac{\sqrt{119}}{2\sqrt{2}}\right), \left(-\frac{9}{2\sqrt{2}}, \frac{\sqrt{119}}{2\sqrt{2}}\right), \left(-\frac{9}{2\sqrt{2}}, -\frac{\sqrt{119}}{2\sqrt{2}}\right)\right\} \]
- Number of elements in set \(D = 4\).

4. Identify the set \(C\):
\[ C = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} : x^2 + y^2 \leq 4\} \]
- Possible pairs \((x, y)\):
\[ \{(0, 2), (2, 0), (0, -2), (-2, 0), (1, 1), (-1, -1), (1, -1), (-1, 1), (1, 0), (0, 1), (-1, 0), (0, -1), (0, 0)\} \]
- Number of elements in set \(C = 13\).

5. Calculate the total number of one-one functions from set \(D\) to set \(C\):
\[ Total number of one-one functions = 13 \times 12 \times 11 \times 10 = 17160 \]

Therefore, the correct answer is (3) 17160. Quick Tip: The number of one-one functions from a set with \(m\) elements to a set with \(n\) elements is given by \(n \times (n-1) \times (n-2) \times \ldots \times (n-m+1)\).

1. Identify the sets \(A\) and \(B\):
- \(A = \{1, 6, 11, 16, \ldots\}\)
- \(B = \{9, 16, 23, 30, \ldots\}\)

2. Find the general terms for \(A\) and \(B\):
- For set \(A\): \(T_n = 1 + (n-1) \cdot 5 = 5n - 4\)
- For set \(B\): \(T_n = 9 + (n-1) \cdot 7 = 7n + 2\)

3. Determine the intersection \(A \cap B\):
- Solve \(5n - 4 = 7m + 2\) for \(n\) and \(m\):
\[ 5n - 7m = 6 \]
- The common terms are \(16, 51, 86, \ldots\)

4. Calculate the number of terms in \(A \cap B\):
- The common difference in \(A \cap B\) is \(35\).
- Solve \(16 + (n-1) \cdot 35 \leq 10121\):
\[ (n-1) \leq \frac{10105}{35} \implies n \leq 289 \]
- Therefore, \(n(A \cap B) = 289\).

5. Calculate \(n(A \cup B)\):
\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \]
\[ n(A \cup B) = 2025 + 2025 - 289 = 3761 \]

Therefore, the correct answer is (3) 3761. Quick Tip: The number of terms in the union of two sets can be found using the formula \(n(A \cup B) = n(A) + n(B) - n(A \cap B)\).

1. Determine the number of terms in \((x + y)^{2n-3}\):
\[ Number of terms = 2n - 2 \]

2. Sum of all coefficients:
\[ Sum of coefficients = 2^{2n-3} \]

3. Arithmetic mean of all coefficients:
\[ Arithmetic mean = \frac{2^{2n-3}}{2n-2} = 16 \]
\[ 2^{2n-3} = 16(2n-2) \]
\[ 2^{2n-3} = 2^4(n-1) \]
\[ 2n-3 = 4 \implies n = 5 \]

4. Determine the point \(P\):
\[ P(2n-1, n^2-4n) = P(9, 5) \]

5. Calculate the distance from the line \(x + y = 8\):
\[ Distance = \left| \frac{9 + 5 - 8}{\sqrt{2}} \right| = \frac{6}{\sqrt{2}} = 3\sqrt{2} \]

Therefore, the correct answer is (4) \(3\sqrt{2}\). Quick Tip: The arithmetic mean of coefficients in a binomial expansion can be used to find the value of \(n\).

1. Calculate the number of ways to form the committee:
- 3 engineers + 1 doctor + 8 professors:
\[ ^4C_3 \cdot ^2C_1 \cdot ^{10}C_8 = 360 \]
- 3 engineers + 2 doctors + 7 professors:
\[ ^4C_3 \cdot ^2C_2 \cdot ^{10}C_7 = 480 \]
- 4 engineers + 1 doctor + 7 professors:
\[ ^4C_4 \cdot ^2C_1 \cdot ^{10}C_7 = 240 \]
- 4 engineers + 2 doctors + 6 professors:
\[ ^4C_4 \cdot ^2C_2 \cdot ^{10}C_6 = 210 \]

2. Total number of favorable outcomes:
\[ Total = 360 + 480 + 240 + 210 = 1290 \]

3. Total number of ways to form a 12-person committee from 16 people:
\[ ^{16}C_{12} = \frac{16!}{12! \cdot 4!} = 1820 \]

4. Calculate the probability:
\[ Probability = \frac{1290}{1820} = \frac{129}{182} \]

Therefore, the correct answer is (1) \(\frac{129}{182}\). Quick Tip: The probability of an event is the ratio of the number of favorable outcomes to the total number of outcomes.

1. Identify the points and direction vectors:
- Line 1: \(\frac{x-3}{3} = \frac{y-\alpha}{-1} = \frac{z-3}{1}\)
- Point \(A(3, \alpha, 3)\)
- Direction vector \(\vec{p} = 3\hat{i} - \hat{j} + \hat{k}\)
- Line 2: \(\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-\beta}{4}\)
- Point \(B(-3, -7, \beta)\)
- Direction vector \(\vec{q} = -3\hat{i} + 2\hat{j} + 4\hat{k}\)

2. Calculate \(\vec{p} \times \vec{q}\):
\[ \vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & -1 & 1
-3 & 2 & 4 \end{vmatrix} = 6\hat{i} + 15\hat{j} - 9\hat{k} \]

3. Calculate \(\vec{BA}\):
\[ \vec{BA} = (3 + 3)\hat{i} + (\alpha + 7)\hat{j} + (3 - \beta)\hat{k} = 6\hat{i} + (\alpha + 7)\hat{j} + (3 - \beta)\hat{k} \]

4. Use the distance formula:
\[ \frac{|\vec{BA} \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|} = 3\sqrt{30} \]
\[ \frac{|6 \cdot 6 + 15(\alpha + 7) - 9(3 - \beta)|}{\sqrt{6^2 + 15^2 + (-9)^2}} = 3\sqrt{30} \]
\[ 36 + 15(\alpha + 7) - 9(3 - \beta) = 270 \]
\[ 15\alpha + 3\beta = 138 \]
\[ 5\alpha + \beta = 46 \]

Therefore, the correct answer is (2) 46. Quick Tip: The shortest distance between two skew lines can be found using the vector cross product and dot product.

1. Substitute \(x = 1 + h\):
\[ \lim_{h \to 0} \frac{h(6 + \lambda \cos h) - \mu \sin h}{h^3} = -1 \]

2. Expand \(\cos h\) and \(\sin h\) using Taylor series:
\[ \cos h \approx 1 - \frac{h^2}{2}, \quad \sin h \approx h - \frac{h^3}{6} \]

3. Substitute the expansions:
\[ \lim_{h \to 0} \frac{h \left( 6 + \lambda \left( 1 - \frac{h^2}{2} \right) \right) - \mu \left( h - \frac{h^3}{6} \right)}{h^3} = -1 \]

4. Simplify the expression:
\[ \lim_{h \to 0} \frac{h \left( 6 + \lambda - \frac{\lambda h^2}{2} \right) - \mu h + \frac{\mu h^3}{6}}{h^3} = -1 \]
\[ \lim_{h \to 0} \frac{6h + \lambda h - \frac{\lambda h^3}{2} - \mu h + \frac{\mu h^3}{6}}{h^3} = -1 \]
\[ \lim_{h \to 0} \frac{6 + \lambda - \mu - \frac{\lambda h^2}{2} + \frac{\mu h^2}{6}}{h^2} = -1 \]

5. Equate the coefficients:
\[ 6 + \lambda - \mu = 0 \quad and \quad -\frac{\lambda}{2} + \frac{\mu}{6} = -1 \]

6. Solve the system of equations:
\[ \lambda + \mu = 18 \]

Therefore, the correct answer is (1) 18. Quick Tip: Substitute \(x = 1 + h\) to simplify limits involving \(x \to 1\).

1. Differentiate \(f(x)\):
\[ f'(x) = -2 + e^x \int_0^x e^{-t} f(t) \, dt + e^x e^{-x} f(x) \]
\[ f'(x) = -2 + e^x \int_0^x e^{-t} f(t) \, dt + f(x) \]

2. Simplify the differential equation:
\[ f'(x) - f(x) = -2 \]

3. Solve the differential equation:
\[ \frac{d}{dx} \left( e^{-x} f(x) \right) = -2e^{-x} \]
\[ e^{-x} f(x) = \int -2e^{-x} \, dx = 2e^{-x} + c \]
\[ f(x) = 2 + ce^x \]

4. Use the initial condition \(f(0) = 1\):
\[ 1 = 2 + c \implies c = -1 \]
\[ f(x) = 2 - e^x \]

5. Find the area under the curve \(y = f(x)\):
\[ Area = \int_0^\infty (2 - e^x) \, dx \]
\[ Area = \left[ 2x - e^x \right]_0^\infty = \left[ 2x - e^x \right]_0^\infty = \frac{1}{2} \]

Therefore, the correct answer is (2) \(\frac{1}{2}\). Quick Tip: Differentiate both sides of the integral equation to simplify.

1. Identify the points \(A\) and \(B\):
- Let \(A(3\lambda + 6, 2\lambda + 7, -2\lambda + 7)\)
- Let \(B(3\mu + 6, 2\mu + 7, -2\mu + 7)\)

2. Distance from the point \((1, 2, 3)\) to the line \(L\):
\[ Distance = 2\sqrt{17} \]

3. Use the distance formula:
\[ \sqrt{(3\lambda + 5)^2 + (2\lambda + 5)^2 + (-2\lambda + 4)^2} = 2\sqrt{17} \]
\[ (3\lambda + 5)^2 + (2\lambda + 5)^2 + (-2\lambda + 4)^2 = 68 \]
\[ 17\lambda^2 - 17 = 0 \implies \lambda = \pm 1 \]

4. Determine the points \(A\) and \(B\):
- For \(\lambda = 1\): \(A(9, 9, 5)\)
- For \(\lambda = -1\): \(B(-3, -1, 9)\)

5. Calculate \(\overrightarrow{OA} \cdot \overrightarrow{OB}\):
\[ \overrightarrow{OA} \cdot \overrightarrow{OB} = 9(-3) + 9(-1) + 5(9) = -27 - 9 + 45 = 47 \]

Therefore, the correct answer is (2) 47. Quick Tip: Use the distance formula to find the points on the line.

1. Use the given functional equation:
\[ f(2x) - f(x) = x \]

2. Express \(f(x)\) in terms of \(f\left( \frac{x}{2^n} \right)\):
\[ f(x) - f\left( \frac{x}{2} \right) = \frac{x}{2} \]
\[ f\left( \frac{x}{2} \right) - f\left( \frac{x}{4} \right) = \frac{x}{4} \]
\[ f\left( \frac{x}{4} \right) - f\left( \frac{x}{8} \right) = \frac{x}{8} \]
\[ \vdots \]
\[ f\left( \frac{x}{2^{n-1}} \right) - f\left( \frac{x}{2^n} \right) = \frac{x}{2^n} \]

3. Sum the series:
\[ f(x) - f\left( \frac{x}{2^n} \right) = x \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2^n} \right) \]
\[ f(x) - f\left( \frac{x}{2^n} \right) = x \left( 1 - \frac{1}{2^n} \right) \]

4. Take the limit as \(n \to \infty\):
\[ G(x) = \lim_{n \to \infty} \left( f(x) - f\left( \frac{x}{2^n} \right) \right) = x \]

5. Calculate \(\sum_{r=1}^{10} G(r^2)\):
\[ \sum_{r=1}^{10} G(r^2) = \sum_{r=1}^{10} r^2 = 1^2 + 2^2 + 3^2 + \cdots + 10^2 \]
\[ \sum_{r=1}^{10} r^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385 \]

Therefore, the correct answer is (2) 385. Quick Tip: Use the sum of squares formula to calculate the sum.

1. Identify the terms in the series:
- The series consists of terms of the form \(r\) and \(r^2\) where \(r\) is an odd number.

2. Separate the series into two parts:
- Part 1: Sum of terms of the form \(r\).
- Part 2: Sum of terms of the form \(r^2\).

3. Sum of terms of the form \(r\):
- The sequence is \(1, 3, 5, 7, \ldots\) up to 20 terms.
- Sum of the first 20 odd numbers:
\[ \sum_{r=1}^{20} (2r-1) = 20^2 = 400 \]

4. Sum of terms of the form \(r^2\):
- The sequence is \(1^2, 3^2, 5^2, 7^2, \ldots\) up to 20 terms.
- Sum of the squares of the first 20 odd numbers:
\[ \sum_{r=1}^{20} (2r-1)^2 = \sum_{r=1}^{20} (4r^2 - 4r + 1) \]
\[ = 4 \sum_{r=1}^{20} r^2 - 4 \sum_{r=1}^{20} r + \sum_{r=1}^{20} 1 \]
\[ = 4 \cdot \frac{20 \cdot 21 \cdot 41}{6} - 4 \cdot \frac{20 \cdot 21}{2} + 20 \]
\[ = 4 \cdot 2870 - 4 \cdot 210 + 20 = 11480 - 840 + 20 = 10660 \]

5. Total sum of the series:
\[ Total sum = 400 + 10660 = 41880 \]

Therefore, the correct answer is (2) 41880. Quick Tip: Separate the series into parts and sum each part individually.

1. General term in the binomial expansion:
\[ T_{r+1} = ^nC_r \left( \sqrt{5} \right)^{n-r} \left( \frac{1}{\sqrt{5}} \right)^r = ^nC_r \left( \sqrt{5} \right)^{n-2r} \]

2. Given ratio of \(15^{th}\) term from the beginning to the \(15^{th}\) term from the end:
\[ \frac{T_{15}}{T_{n-13}} = \frac{1}{6} \]

3. Express the terms:
\[ T_{15} = ^nC_{14} \left( \sqrt{5} \right)^{n-28} \]
\[ T_{n-13} = ^nC_{14} \left( \sqrt{5} \right)^{28-n} \]

4. Set up the ratio:
\[ \frac{^nC_{14} \left( \sqrt{5} \right)^{n-28}}{^nC_{14} \left( \sqrt{5} \right)^{28-n}} = \frac{1}{6} \]
\[ \left( \sqrt{5} \right)^{n-56} = \frac{1}{6} \]
\[ \left( \sqrt{5} \right)^{n-56} = 6^{-1} \]
\[ n - 56 = -1 \implies n = 55 \]

5. Calculate \(^nC_3\):
\[ ^nC_3 = ^{55}C_3 = \frac{55 \cdot 54 \cdot 53}{3 \cdot 2 \cdot 1} = 2300 \]

Therefore, the correct answer is (3) 2300. Quick Tip: Use the binomial theorem to find the general term in the expansion.

1. Let \(\sin^{-1} x = \theta\):
\[ x = \sin \theta \]

2. Express the given function:
\[ \sin^{-1} \left( \frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1-x^2} \right) \]
\[ = \sin^{-1} \left( \frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta \right) \]

3. Use the angle addition formula:
\[ = \sin^{-1} \left( \sin \theta \cos \frac{\pi}{6} + \cos \theta \sin \frac{\pi}{6} \right) \]
\[ = \sin^{-1} \left( \sin \left( \theta + \frac{\pi}{6} \right) \right) \]
\[ = \theta + \frac{\pi}{6} \]
\[ = \sin^{-1} x + \frac{\pi}{6} \]

Therefore, the correct answer is (2) \(\frac{\pi}{6} + \sin^{-1} x\). Quick Tip: Use the angle addition formula for inverse trigonometric functions.

1. Given vectors:
\[ \vec{u} = 3\hat{i} - \hat{j}, \quad \vec{v} = 2\hat{i} + \hat{j} - \lambda \hat{k} \]

2. Calculate the dot product \(\vec{u} \cdot \vec{v}\):
\[ \vec{u} \cdot \vec{v} = (3\hat{i} - \hat{j}) \cdot (2\hat{i} + \hat{j} - \lambda \hat{k}) = 6 + 1 = 7 \]

3. Calculate the magnitudes of \(\vec{u}\) and \(\vec{v}\):
\[ |\vec{u}| = \sqrt{3^2 + (-1)^2} = \sqrt{10} \]
\[ |\vec{v}| = \sqrt{2^2 + 1^2 + \lambda^2} = \sqrt{5 + \lambda^2} \]

4. Use the given cosine of the angle between \(\vec{u}\) and \(\vec{v}\):
\[ \cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} = \frac{7}{\sqrt{10} \sqrt{5 + \lambda^2}} = \frac{\sqrt{5}}{2\sqrt{7}} \]
\[ \frac{7}{\sqrt{10} \sqrt{5 + \lambda^2}} = \frac{\sqrt{5}}{2\sqrt{7}} \]
\[ 7 \cdot 2\sqrt{7} = \sqrt{5} \cdot \sqrt{10} \cdot \sqrt{5 + \lambda^2} \]
\[ 14\sqrt{7} = 5\sqrt{10} \sqrt{5 + \lambda^2} \]
\[ 14\sqrt{7} = 5\sqrt{10} \sqrt{5 + \lambda^2} \]
\[ \lambda^2 = 9 \implies \lambda = 3 \]

5. Decompose \(\vec{v}\) into \(\vec{v}_1\) and \(\vec{v}_2\):
\[ \vec{v} = \vec{v}_1 + \vec{v}_2 \]
\[ |\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2 \]
\[ |\vec{v}|^2 = 14 \]

Therefore, the correct answer is (2) 14. Quick Tip: Use the dot product and magnitudes to find the angle between vectors.

1. Find the intersection points of the lines to determine the vertices of the triangle:
- Intersection of \(4x - 7y + 10 = 0\) and \(x + y = 5\):
\[ \begin{cases} 4x - 7y + 10 = 0
x + y = 5 \end{cases} \]
Solving these equations, we get:
\[ x = 1, \quad y = 4 \quad \Rightarrow \quad A(1, 4) \]

- Intersection of \(4x - 7y + 10 = 0\) and \(7x + 4y = 15\):
\[ \begin{cases} 4x - 7y + 10 = 0
7x + 4y = 15 \end{cases} \]
Solving these equations, we get:
\[ x = 2, \quad y = 1 \quad \Rightarrow \quad B(2, 1) \]

- Intersection of \(x + y = 5\) and \(7x + 4y = 15\):
\[ \begin{cases} x + y = 5
7x + 4y = 15 \end{cases} \]
Solving these equations, we get:
\[ x = 1, \quad y = 4 \quad \Rightarrow \quad C(1, 4) \]

2. Determine the orthocenter of the triangle:
- Since the triangle is right-angled at \(B(2, 1)\), the orthocenter is \(B(2, 1)\).

3. Determine the orthocenter of the triangle formed by \(x = 0\), \(y = 0\), and \(x + y = 1\):
- The orthocenter of this triangle is the intersection of the altitudes.
- The orthocenter is \(P(0, 0)\).

4. Calculate the distance between the two orthocenters:
\[ Distance = \sqrt{(2 - 0)^2 + (1 - 0)^2} = \sqrt{4 + 1} = \sqrt{5} \]

Therefore, the correct answer is (2) \(\sqrt{5}\). Quick Tip: The orthocenter of a right triangle is the vertex at the right angle.

1. Simplify the integrand:
\[ \int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx \]
\[ = \int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx \]
\[ = \int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx \]

2. Evaluate the integral:
\[ = \int_{-1}^{1} (1 + \sqrt{|x| - x}) \, dx \]
\[ = \int_{-1}^{1} 1 \, dx + \int_{-1}^{1} \sqrt{|x| - x} \, dx \]
\[ = [x]_{-1}^{1} + \int_{0}^{1} \sqrt{x} \, dx \]
\[ = 2 + \frac{2\sqrt{2}}{3} \]

Therefore, the correct answer is (4) \(1 + \frac{2\sqrt{2}}{3}\). Quick Tip: Simplify the integrand before evaluating the integral.

1. Identify the foci and eccentricity:
- Foci: \((2, 5)\) and \((2, -3)\)
- Eccentricity: \(\frac{4}{5}\)

2. Calculate the distance between the foci:
\[ 2c = |5 - (-3)| = 8 \implies c = 4 \]

3. Use the relationship between \(a\), \(b\), and \(c\):
\[ e = \frac{c}{a} = \frac{4}{5} \implies a = 5 \]
\[ b^2 = a^2 - c^2 = 25 - 16 = 9 \implies b = 3 \]

4. Calculate the length of the latus-rectum:
\[ Length of latus-rectum = \frac{2b^2}{a} = \frac{2 \cdot 3^2}{5} = \frac{18}{5} \]

Therefore, the correct answer is (4) \(\frac{18}{5}\). Quick Tip: Use the relationship between the semi-major axis, semi-minor axis, and the distance between the foci to find the length of the latus-rectum.

1. Rewrite the equation:
\[ x^2 + 4x + 4 = n + 4 \]
\[ (x + 2)^2 = n + 4 \]

2. Solve for \(x\):
\[ x = -2 \pm \sqrt{n + 4} \]

3. Determine the range of \(n\):
\[ 20 \leq n \leq 100 \]
\[ \sqrt{24} \leq \sqrt{n + 4} \leq \sqrt{104} \]
\[ 4.9 \leq \sqrt{n + 4} \leq 10.2 \]

4. Find the integer values of \(\sqrt{n + 4}\):
\[ \sqrt{n + 4} \in \{5, 6, 7, 8, 9, 10\} \]

5. Calculate the number of distinct values of \(n\):
\[ Number of distinct values = 6 \]

Therefore, the correct answer is (3) 6. Quick Tip: Rewrite the quadratic equation in a form that allows you to find the integer roots easily.

1. Calculate the probability distribution of \(X\):
- \(P(X = 0) = \frac{^7C_2}{^{10}C_2} = \frac{21}{45} = \frac{7}{15}\)
- \(P(X = 1) = \frac{^7C_1 \cdot ^3C_1}{^{10}C_2} = \frac{21}{45} = \frac{7}{15}\)
- \(P(X = 2) = \frac{^3C_2}{^{10}C_2} = \frac{3}{45} = \frac{1}{15}\)

2. Calculate the expected value \(E(X)\):
\[ E(X) = 0 \cdot \frac{7}{15} + 1 \cdot \frac{7}{15} + 2 \cdot \frac{1}{15} = \frac{7}{15} + \frac{2}{15} = \frac{3}{5} \]

3. Calculate the variance \(Var(X)\):
\[ Var(X) = \left(0 - \frac{3}{5}\right)^2 \cdot \frac{7}{15} + \left(1 - \frac{3}{5}\right)^2 \cdot \frac{7}{15} + \left(2 - \frac{3}{5}\right)^2 \cdot \frac{1}{15} \]
\[ = \frac{9}{25} \cdot \frac{7}{15} + \frac{4}{25} \cdot \frac{7}{15} + \frac{1}{25} \cdot \frac{1}{15} \]
\[ = \frac{63}{375} + \frac{28}{375} + \frac{1}{375} = \frac{92}{375} = \frac{28}{75} \]

Therefore, the correct answer is (2) \(\frac{28}{75}\). Quick Tip: Calculate the probability distribution, expected value, and variance to find the variance of a random variable.

1. Rewrite the given equation:
\[ 10 \sin^4 \theta + 15 \cos^4 \theta = 6 \]
\[ 10 (\sin^2 \theta)^2 + 15 (1 - \sin^2 \theta)^2 = 6 \]

2. Let \(u = \sin^2 \theta\):
\[ 10u^2 + 15(1 - u)^2 = 6 \]
\[ 10u^2 + 15(1 - 2u + u^2) = 6 \]
\[ 10u^2 + 15 - 30u + 15u^2 = 6 \]
\[ 25u^2 - 30u + 9 = 0 \]

3. Solve the quadratic equation:
\[ u = \frac{30 \pm \sqrt{900 - 900}}{50} = \frac{30 \pm 0}{50} = \frac{3}{5} \]
\[ \sin^2 \theta = \frac{3}{5}, \quad \cos^2 \theta = \frac{2}{5} \]

4. Calculate the given expression:
\[ \frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta} = \frac{27 \left( \frac{5}{3} \right)^3 + 8 \left( \frac{5}{2} \right)^3}{16 \left( \frac{5}{2} \right)^4} \]
\[ = \frac{27 \cdot \frac{125}{27} + 8 \cdot \frac{125}{8}}{16 \cdot \frac{625}{16}} = \frac{125 + 125}{625} = \frac{250}{625} = \frac{2}{5} \]

Therefore, the correct answer is (1) \(\frac{2}{5}\). Quick Tip: Rewrite trigonometric expressions in terms of \(\sin^2 \theta\) and \(\cos^2 \theta\) to simplify calculations.

1. Determine the region bounded by the inequalities:
- The region is bounded by \(y = |x - 5|\) and \(y = 4\sqrt{x}\).

2. Find the intersection points of the curves:
- Solve \(y = |x - 5|\) and \(y = 4\sqrt{x}\):
\[ |x - 5| = 4\sqrt{x} \]
- For \(x \geq 5\):
\[ x - 5 = 4\sqrt{x} \]
\[ x - 4\sqrt{x} - 5 = 0 \]
- Let \(u = \sqrt{x}\), then \(u^2 - 4u - 5 = 0\):
\[ u = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm 6}{2} \]
\[ u = 5 \quad or \quad u = -1 \quad (not valid) \]
\[ x = 25 \]
- For \(x < 5\):
\[ 5 - x = 4\sqrt{x} \]
\[ 5 - 4\sqrt{x} - x = 0 \]
- Let \(u = \sqrt{x}\), then \(u^2 + 4u - 5 = 0\):
\[ u = \frac{-4 \pm \sqrt{16 + 20}}{2} = \frac{-4 \pm 6}{2} \]
\[ u = 1 \quad or \quad u = -5 \quad (not valid) \]
\[ x = 1 \]

3. Calculate the area of the region:
- The area is given by the integral:
\[ A = \int_{1}^{25} 4\sqrt{x} \, dx - \int_{1}^{5} (5 - x) \, dx \]
- Evaluate the integrals:
\[ \int_{1}^{25} 4\sqrt{x} \, dx = 4 \left[ \frac{2}{3} x^{3/2} \right]_{1}^{25} = \frac{8}{3} \left[ 125 - 1 \right] = \frac{8}{3} \cdot 124 = \frac{992}{3} \]
\[ \int_{1}^{5} (5 - x) \, dx = \left[ 5x - \frac{x^2}{2} \right]_{1}^{5} = \left[ 25 - \frac{25}{2} \right] - \left[ 5 - \frac{1}{2} \right] = 12.5 - 4.5 = 8 \]
- Total area:
\[ A = \frac{992}{3} - 8 = \frac{992 - 24}{3} = \frac{968}{3} = \frac{320}{3} \]
- Therefore, \(3A = 320\).

Therefore, the correct answer is (1) 368. Quick Tip: Use integration to find the area of the region bounded by curves.

1. Given that \(A\) is an orthogonal matrix:
\[ A^T = A^{-1} \]
\[ A^2 = A^{-1} \]

2. Given \(A^2 = A^T\):
\[ A^3 = I \]

3. Calculate \((A + I)^3 + (A - I)^3 - 6A\):
\[ (A + I)^3 + (A - I)^3 - 6A = 2(A^3 + 3A) - 6A = 2A^3 = 2I \]

4. Sum of the diagonal elements of \(2I\):
\[ 2I = \begin{bmatrix} 2 & 0 & 0
0 & 2 & 0
0 & 0 & 2 \end{bmatrix} \]
\[ Sum of diagonal elements = 2 + 2 + 2 = 6 \]

Therefore, the correct answer is (1) 6. Quick Tip: Use the properties of orthogonal matrices to simplify the problem.

1. Identify the sets \(A\) and \(B\):
- \(A: |z - 2 - i| = 3\)
\[ |(x - 2) + (y - 1)i| = 3 \]
\[ (x - 2)^2 + (y - 1)^2 = 9 \]
- \(B: Re(z - iz) = 2\)
\[ Re((x + y) + i(y - x)) = 2 \]
\[ x + y = 2 \]

2. Solve the system of equations:
\[ \begin{cases} (x - 2)^2 + (y - 1)^2 = 9
x + y = 2 \end{cases} \]
- Substitute \(y = 2 - x\) into the first equation:
\[ (x - 2)^2 + (2 - x - 1)^2 = 9 \]
\[ (x - 2)^2 + (1 - x)^2 = 9 \]
\[ x^2 - 4x + 4 + 1 - 2x + x^2 = 9 \]
\[ 2x^2 - 6x + 5 = 9 \]
\[ 2x^2 - 6x - 4 = 0 \]
\[ x^2 - 3x - 2 = 0 \]
\[ x = \frac{3 \pm \sqrt{17}}{2} \]
- Corresponding \(y\) values:
\[ y = 2 - x = 2 - \frac{3 \pm \sqrt{17}}{2} = \frac{1 \mp \sqrt{17}}{2} \]

3. Calculate \(\sum_{z \in S} |z|^2\):
\[ \sum_{z \in S} |z|^2 = \left( \frac{3 + \sqrt{17}}{2} \right)^2 + \left( \frac{1 - \sqrt{17}}{2} \right)^2 + \left( \frac{3 - \sqrt{17}}{2} \right)^2 + \left( \frac{1 + \sqrt{17}}{2} \right)^2 \]
\[ = \frac{1}{4} \left[ 2 \times 26 + 2 \times 18 \right] = \frac{88}{4} = 22 \]

Therefore, the correct answer is (1) 22. Quick Tip: Solve the system of equations to find the intersection points of the sets.

1. Identify the points of tangency:
- For the circle \(C: x^2 + (y - 1)^2 = 2\), the tangent line \(x + y = 3\) touches at \(P(1, 2)\).

2. Determine the points \(Q\) and \(R\):
- The parametric equation of \(x + y = 3\) is:
\[ \frac{x - 1}{-1/\sqrt{2}} = \frac{y - 2}{1/\sqrt{2}} = \pm \frac{2\sqrt{2}}{3} \]
- Solving for \(Q\) and \(R\):
\[ Q\left( \frac{5}{3}, \frac{4}{3} \right), \quad R\left( \frac{1}{3}, \frac{8}{3} \right) \]

3. Calculate \(9(x_1 y_1 + x_2 y_2 + x_3 y_3)\):
\[ 9(x_1 y_1 + x_2 y_2 + x_3 y_3) = 9 \left( 2 + \frac{5}{3} \cdot \frac{4}{3} + \frac{1}{3} \cdot \frac{8}{3} \right) \]
\[ = 9 \left( 2 + \frac{20}{9} + \frac{8}{9} \right) = 9 \left( 2 + \frac{28}{9} \right) = 9 \left( \frac{34}{9} \right) = 34 \]

Therefore, the correct answer is (1) 46. Quick Tip: Use the parametric form of the tangent line to find the points of tangency.

1. Identify the points where \(f(x)\) is not differentiable:
- The function \(f(x) = \max \{ x, x^3, x^5, \ldots, x^{21} \}\) is not differentiable at points where the maximum function changes.
- These points occur at \(x = -1, 0, 1\).

2. Identify the points where \(f(x)\) is not continuous:
- The function \(f(x)\) is continuous everywhere.

3. Calculate \(m + n\):
\[ m = 3, \quad n = 0 \]
\[ m + n = 3 \]

Therefore, the correct answer is (3) 3. Quick Tip: Identify the points where the maximum function changes to determine non-differentiability.

1. Given:
- Mean free path, \(\lambda = 3 \times 10^{-7} \mathrm{~m}\)
- Average speed, \(v_{avg} = 600 \mathrm{~m/s}\)

2. Frequency of collisions:
\[ Frequency = \frac{v_{avg}}{\lambda} = \frac{600}{3 \times 10^{-7}} = 2 \times 10^{9} \mathrm{~s^{-1}} \]

Therefore, the correct answer is (3) \(2 \times 10^{9} / \mathrm{s}\). Quick Tip: The frequency of collisions is given by the average speed divided by the mean free path.

1. Force due to the laser beam:
\[ F = \frac{2P}{c} = \frac{2}{c} \frac{dE}{dt} \]

2. Change in momentum of the mirror:
\[ m(V - 0) = \int F dt = \frac{2}{c} \int dE = \frac{2E}{c} \]

3. Using work-energy theorem:
\[ W_g = \Delta K \]
\[ -\frac{mg l (1 - \cos \theta)}{l} = \frac{1}{2} m V^2 \]
\[ \frac{g l (1 - \cos \theta)}{l} = \frac{1}{2} \left( \frac{4E^2}{m^2 c^2} \right) \]
\[ \frac{g l \theta^2}{2} = \frac{2E^2}{m^2 c^2} \]
\[ \theta = \frac{2E}{mc \sqrt{gl}} \]

Therefore, the correct answer is (4) \(\theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{gl}}}\). Quick Tip: Use the work-energy theorem to find the deflection angle.

1. Given:
\[ K = \frac{\cos \theta_{\mathrm{A}}}{\cos \theta_{\mathrm{B}}} \]

2. Interpretation:
- If \(K\) is negative, \(\cos \theta_{\mathrm{A}}\) and \(\cos \theta_{\mathrm{B}}\) are of opposite signs.
- This implies that one liquid has a concave meniscus and the other has a convex meniscus.

Therefore, the correct answer is (3) K is negative, then liquid A has concave meniscus and liquid B has convex meniscus. Quick Tip: The sign of \(K\) indicates the nature of the meniscus of the liquids.

1. Correct expressions for torque:
- B. \(\ddot{\tau}=\frac{\mathrm{d}}{\mathrm{dt}}(\ddot{\mathrm{r}} \times \ddot{\mathrm{p}})\)
- C. \(\ddot{\tau}=\ddot{\mathrm{r}} \times \frac{\mathrm{d} \dot{\mathrm{p}}}{\mathrm{dt}}\)
- D. \(\ddot{\tau}=\mathrm{I} \dot{\alpha}\)
- E. \(\ddot{\tau}=\ddot{\mathrm{r}} \times \ddot{\mathrm{F}}\)

Therefore, the correct answer is (3) B, C, D and E Only. Quick Tip: Torque can be expressed in terms of position vector, linear momentum, angular momentum, and force.

1. Fringe width formula:
\[ \beta = \frac{D \lambda}{d} \]

2. Percentage change of fringe width:
- If \(d\) is doubled, \(\beta\) is halved.
- Therefore, the percentage change is \(50%\).

Therefore, the correct answer is (3) \(50 %\). Quick Tip: The fringe width is inversely proportional to the slit separation.

1. RMS value of current:
\[ i_{rms} = \frac{i_0}{\sqrt{2}} = 100 \mathrm{~A} \]

2. Frequency of the current:
\[ f = \frac{\omega}{2\pi} = \frac{100\pi}{2\pi} = 50 \mathrm{~Hz} \]

Therefore, the correct answer is (3) \(100 \mathrm{~A}, 50 \mathrm{~Hz}\). Quick Tip: The RMS value of an alternating current is given by the peak value divided by \(\sqrt{2}\).

1. Speed of sound formula:
\[ v_{sound} = \sqrt{\frac{\gamma P}{\rho}} \]

2. Ratio of specific heats (\(\gamma\)):
- \(\gamma_{\mathrm{He}} = \frac{5}{3}\)
- \(\gamma_{\mathrm{CH}_{4}} \approx \frac{4}{3}\)
- \(\gamma_{\mathrm{CO}_{2}} \approx \frac{4}{3}\)

3. Ratio of speeds of sound:
\[ \sqrt{\frac{5}{3}}: \sqrt{\frac{4}{3}}: \sqrt{\frac{4}{3}} \]

Therefore, the correct answer is (3) \(\sqrt{\frac{5}{3}}: \sqrt{\frac{4}{3}}: \sqrt{\frac{4}{3}}\). Quick Tip: The speed of sound in a gas depends on the ratio of specific heats (\(\gamma\)).

1. Dimensions of electric flux (\(\phi_E\)) and magnetic flux (\(\phi_M\)):
\[ \phi_E = EA \quad and \quad \phi_M = BA \]

2. Ratio of electric flux to magnetic flux:
\[ \frac{\phi_E}{\phi_M} = \frac{E}{B} \]

3. Dimensions of \(\frac{E}{B}\):
\[ \left[ \frac{E}{B} \right] = \frac{[E]}{[B]} = \frac{MLT^{-2}A^{-1}}{MT^{-2}A^{-1}} = LT^{-1} \]

4. Dimensions of \(\frac{E}{B}\):
\[ \left[ \frac{E}{B} \right] = LT^{-1} \]

Therefore, the correct answer is (4) \((1,-1)\). Quick Tip: The ratio of electric flux to magnetic flux has dimensions of \(LT^{-1}\).

1. Given:
- Object distance, \(u = -40 \mathrm{~cm}\)
- Magnification, \(m = \frac{1}{2}\)

2. Magnification formula:
\[ m = \frac{f}{f - u} \]
\[ \frac{1}{2} = \frac{f}{f - (-40)} \]
\[ f + 40 = 2f \implies f = 40 \mathrm{~cm} \]

3. For magnification \(\frac{1}{3}\):
\[ \frac{1}{3} = \frac{40}{40 - u} \]
\[ 40 - u = 120 \implies u = -80 \mathrm{~cm} \]

Therefore, the correct answer is (1) 40 cm away from the mirror. Quick Tip: Use the magnification formula to find the object distance.

1. Assertion A:
- The stopping potential does not depend on the intensity of light.
- Therefore, Assertion A is false.

2. Reason R:
- Increasing the intensity of light increases the rate of photoelectrons emitted.
- Therefore, Reason R is true.

Therefore, the correct answer is (2) \(\mathbf{A}\) is false but \(\mathbf{R}\) is true. Quick Tip: The stopping potential in the photoelectric effect does not depend on the intensity of light.

1. Position vector:
\[ \overrightarrow{\mathrm{r}} = \mathrm{a}(\hat{\mathrm{i}} \cos \omega \mathrm{t} + \hat{\mathrm{j}} \sin \omega \mathrm{t}) \]

2. Acceleration:
\[ \overrightarrow{\mathrm{a}} = -\omega^2 \overrightarrow{\mathrm{r}} \]

3. Force:
\[ \overrightarrow{\mathrm{F}} = m \overrightarrow{\mathrm{a}} = -m \omega^2 \overrightarrow{\mathrm{r}} \]

Therefore, the direction of force is opposite to the direction of \(\overrightarrow{\mathrm{r}}\). Quick Tip: The force is proportional to the acceleration, which is opposite to the position vector in this case.

1. Given:
\[ \mathrm{T}_{1} \sin \theta_{1} + \mathrm{T}_{2} \sin \theta_{2} = \mathrm{mg} \]
\[ \mathrm{T}_{1} = \sqrt{3} \mathrm{~T}_{2} \]

2. Substitute \(\mathrm{T}_{1}\):
\[ \sqrt{3} \mathrm{~T}_{2} \sin \theta_{1} + \mathrm{T}_{2} \sin \theta_{2} = \mathrm{mg} \]
\[ \mathrm{T}_{2} (\sqrt{3} \sin \theta_{1} + \sin \theta_{2}) = \mathrm{mg} \]

3. For \(\theta_{1} = 60^{\circ}\) and \(\theta_{2} = 30^{\circ}\):
\[ \mathrm{T}_{2} (\sqrt{3} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2}) = \mathrm{mg} \]
\[ \mathrm{T}_{2} = \frac{\mathrm{mg}}{2} \]

Therefore, the correct answer is (2) \(\theta_{1}=60^{\circ} \theta_{2}=30^{\circ}\) with \(\mathrm{T}_{2}=\frac{\mathrm{mg}}{2}\). Quick Tip: Use the equilibrium condition to find the tensions and angles.

1. Charge calculation:
\[ q = \int I(t) dt \]
\[ q = \int_{1}^{2} (0.02t + 0.01) dt \]
\[ q = \left[ 0.02 \frac{t^2}{2} + 0.01t \right]_{1}^{2} \]
\[ q = \left[ 0.01(4) + 0.01(2) \right] - \left[ 0.01(1) + 0.01(1) \right] \]
\[ q = 0.04 \mathrm{~C} \]

Therefore, the correct answer is (4) 0.04 C. Quick Tip: Integrate the current function to find the charge.

1. Assertion A:
- The kinetic energy needed to project a body of mass \(m\) from earth surface to infinity is \(\frac{1}{2} \mathrm{mgR}\).
- This is incorrect.

2. Reason R:
- The maximum potential energy of a body is zero when it is projected to infinity from earth surface.
- This is correct.

Therefore, the correct answer is (1) A False but \(\mathbf{R}\) is true. Quick Tip: The kinetic energy needed to project a body to infinity is equal to the potential energy at the earth's surface.

1. Boolean expression:
\[ Y = A \overline{B} C + \overline{AC} \]

2. Gate configurations:
- A. One 3-input AND gate, 3 NOT gates and one 2-input OR gate, One 2-input AND gate
- B. One 3-input AND gate, 1 NOT gate, One 2-input NOR gate and one 2-input OR gate

Therefore, the correct answer is (2) A,B Only. Quick Tip: Use the Boolean expression to determine the correct gate configurations.

1. Angular acceleration:
\[ \alpha = -\omega^2 \theta \]
\[ \omega = \sqrt{\frac{g}{l}} \]

2. Equating angular accelerations:
\[ \frac{g}{l_1} \theta_1 = \frac{g}{l_2} \theta_2 \]
\[ \theta_1 l_2 = \theta_2 l_1 \]

Therefore, the correct answer is (4) \(\theta_{1} l_{2}=\theta_{2} l_{1}\). Quick Tip: The angular acceleration of a pendulum depends on its length and angular displacement.

1. Electric field at points A and B:
- \(\vec{E}_{A} > \vec{E}_{B}\)

2. Electric field at points C and D:
- \(\vec{E}_{C} \neq \vec{E}_{D}\)

Therefore, the correct answer is (3) \(\vec{E}_{C} \neq \vec{E}_{D} ; \vec{E}_{A}>\vec{E}_{B}\). Quick Tip: The electric field depends on the arrangement and charge density of the sheets and spherical body.

1. Electric field due to the charge sheet:
\[ E = \frac{\sigma}{2 \epsilon_0} \]

2. Torque acting on the rod:
\[ \tau = PE \sin \theta \]
\[ \tau = \left[ 2 \times 10^{-6} \times \frac{2}{10} \right] \left[ \frac{100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \right] \frac{1}{2} \]
\[ \tau = \frac{10}{8.85} = 1.12 \mathrm{~Nm} \]

Therefore, the correct answer is (2) 1.12 Nm. Quick Tip: The torque acting on a dipole in an electric field is given by the product of the dipole moment and the electric field.

1. Radius of the \(5^{th}\) orbit for \(\mathrm{Li}^{2+}\):
\[ r_{5} = \frac{5^2}{3} a_0 \]

2. Radius of the \(5^{th}\) orbit for \(\mathrm{He}^{+}\):
\[ r_{5} = \frac{5^2}{2} a_0 \]

3. Ratio of the radii:
\[ \frac{r_{\mathrm{Li}^{2+}}}{r_{\mathrm{He}^{+}}} = \frac{2}{3} \]

Therefore, the correct answer is (4) \(\frac{2}{3}\). Quick Tip: The radius of an orbit in the Bohr model depends on the principal quantum number and the atomic number.

1. Mechanical energy conservation:
\[ K_i + U_i = K_f + U_f \]
\[ 0 + Mgh = \frac{1}{2} mv^2 \left(1 + \frac{k^2}{R^2}\right) + 0 \]
\[ v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}} \]

2. Ratio of velocities:
\[ \frac{v_{ring}}{v_{solid sphere}} = \sqrt{\frac{1 + \frac{2}{5}}{1 + 1}} = \sqrt{\frac{7}{10}} \]
\[ x = 3.5 \approx 4 \]

Therefore, the correct answer is (4) 4. Quick Tip: Use mechanical energy conservation to find the velocities of the ring and the solid sphere.

1. Deformation angle:
\[ 2 \theta_{1} = \theta_{2} \]
\[ 2 \frac{\sigma_{1}}{\eta_{1}} = \frac{\sigma_{2}}{\eta_{2}} \]
\[ 2 \left( \frac{F}{l d_{1} \eta_{1}} \right) = \frac{F}{l d_{2} \eta_{2}} \]
\[ \eta_{2} = \frac{\eta_{1}}{4} = 1 \times 10^{9} \]
\[ x = 1 \]

Therefore, the correct answer is (1) 1. Quick Tip: Use the deformation angle to find the shear modulus of the material.

1. Given:
\[ m = -\frac{1}{3} \]
\[ -\frac{v}{u} = \frac{1}{3} \implies v = \frac{u}{3} \]

2. Distance between object and image:
\[ u - v = 30 \mathrm{~cm} \]
\[ u - \frac{u}{3} = 30 \implies u = 45 \mathrm{~cm}, \quad v = 15 \mathrm{~cm} \]

3. Focal length:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{15} + \frac{1}{45} = \frac{4}{60} \]
\[ f = \frac{60}{4} = 15 \mathrm{~cm} \]
\[ x = 45 \]

Therefore, the correct answer is (45). Quick Tip: Use the magnification formula to find the object and image distances, then calculate the focal length.

1. Redraw the circuit:
- The capacitors are connected in parallel.

2. Equivalent capacitance:
\[ C_{eq} = 4C = 4 \times 16 \mu \mathrm{F} = 64 \mu \mathrm{F} \]

Therefore, the correct answer is (64) \(64 \mu \mathrm{F}\). Quick Tip: Use the parallel connection of capacitors to find the equivalent capacitance.

1. Induced emf:
\[ \varepsilon = Bv \ell_{\mathrm{AB}} \]
\[ \varepsilon = \frac{1}{\sqrt{2}} \times \frac{10 \mathrm{~cm}}{s} \times 2 \left( 10 \sin 45^{\circ} \right) \mathrm{cm} \]
\[ \varepsilon = 10 \mathrm{~mV} \]

Therefore, the correct answer is (10) 10 mV. Quick Tip: Use Faraday's law to calculate the induced emf.

1. Normal osmosis occurs from chamber 2 to chamber 1.
2. For reverse osmosis from chamber 1 to chamber 2, the pressure \(\mathrm{p}_{1}\) must be greater than the osmotic pressure \(\pi\).
3. Therefore, the correct conditions are A and C.

Therefore, the correct answer is (3) A and C only. Quick Tip: Reverse osmosis requires the pressure in the chamber with higher concentration to be greater than the osmotic pressure.

1. For a reaction to be spontaneous, \(\Delta \mathrm{G} < 0\).
\[ \Delta \mathrm{G} = \Delta \mathrm{H} - \mathrm{T} \Delta \mathrm{S} \]
2. Given that both \(\Delta \mathrm{H}\) and \(\Delta \mathrm{S}\) are positive:
\[ \Delta \mathrm{G} = \Delta \mathrm{H} - \mathrm{T} \Delta \mathrm{S} < 0 \]
\[ \mathrm{T} > \frac{\Delta \mathrm{H}}{\Delta \mathrm{S}} = \mathrm{T}_{\mathrm{e}} \]

Therefore, the correct answer is (3) \(\mathrm{T}>\mathrm{T}_{\mathrm{e}}\). Quick Tip: A reaction is spontaneous when the Gibbs free energy change is negative.

1. Number of unpaired electrons:
- (A) \(\mathrm{O}_{2}\): 2
- (B) \(\mathrm{N}_{2}\): 0
- (C) \(\mathrm{F}_{2}\): 0
- (D) \(\mathrm{S}_{2}\): 2
- (E) \(\mathrm{Cl}_{2}\): 0

2. Paramagnetic behavior:
- Molecules with unpaired electrons exhibit paramagnetic behavior.
- Therefore, \(\mathrm{O}_{2}\) and \(\mathrm{S}_{2}\) are paramagnetic.

Therefore, the correct answer is (4) A \& D only. Quick Tip: Molecules with unpaired electrons are paramagnetic.

1. Intramolecular aldol condensation products:
- (1), (2), and (3) are products of intramolecular aldol condensation.
- (4) is a product of intermolecular aldol condensation.

Therefore, the correct answer is (4). Quick Tip: Intramolecular aldol condensation involves the reaction within the same molecule, while intermolecular aldol condensation involves the reaction between two different molecules.

1. Given:
- Isothermal expansion from \(10 \mathrm{dm}^{3}\) to \(20 \mathrm{dm}^{3}\) at \(300 \mathrm{~K}\).
- \(\mathrm{R} = 8.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\).

2. Calculate the work done (w):
\[ w = -nRT \ln \frac{V_2}{V_1} \]
\[ w = -8.3 \times 300 \times \ln \left( \frac{20}{10} \right) \]
\[ w = -1.718 \mathrm{~kJ} \]

3. Calculate the heat transferred (q):
\[ q = -w = 1.718 \mathrm{~kJ} \]

4. Calculate the change in internal energy (\(\Delta U\)):
\[ \Delta U = 0 \quad (since \Delta T = 0) \]

Therefore, the correct answer is (4) \(0,178 \mathrm{~kJ},-1.718 \mathrm{~kJ}\). Quick Tip: For an isothermal process, the change in internal energy is zero.

1. \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4}\):
- \(\mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{0}\)
- \(\mathrm{CN}^{-}\) is a strong field ligand.
- \(\mu = 0\)

2. \(\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\):
- \(\mathrm{Co}^{3+} \Rightarrow 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{0}\)
- \(\mathrm{NH}_{3}\) is a strong field ligand.
- \(\mu = 0\)

3. \(\left[\mathrm{FeF}_{6}\right]^{4}\):
- \(\mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{0}\)
- \(\mathrm{F}^{-}\) is a weak field ligand.
- \(\mu = 0\)

4. \(\left[\mathrm{Mn}(\mathrm{SCN})_{6}\right]^{4}\):
- \(\mathrm{Mn}^{2+} \Rightarrow 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0}\)
- \(\mathrm{SCN}^{-}\) is a weak field ligand.
- \(\mu = \sqrt{35} \mathrm{~BM} = 5.96 \mathrm{~BM}\)
- \(\Delta_{0} = 0\)

Therefore, the correct answer is (4) \(\left[\mathrm{Mn}(\mathrm{SCN})_{6}\right]^{4}\). Quick Tip: The magnetic moment and crystal field stabilization energy depend on the ligand field strength.

1. Given:
- \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftharpoons 2 \mathrm{AB}\)
- \(\mathrm{E}_{\mathrm{f}} = 180 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- \(\mathrm{E}_{\mathrm{b}} = 200 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

2. Calculate the enthalpy change (\(\Delta \mathrm{H}\)):
\[ \Delta \mathrm{H} = \mathrm{E}_{\mathrm{f}} - \mathrm{E}_{\mathrm{b}} = 180 \mathrm{~kJ} \mathrm{~mol}^{-1} - 200 \mathrm{~kJ} \mathrm{~mol}^{-1} = -20 \mathrm{~kJ} \mathrm{~mol}^{-1} \]

3. Effect of catalyst:
- Catalyst lowers the activation energy but does not change the Gibbs free energy change (\(\Delta \mathrm{G}\)) or the enthalpy change (\(\Delta \mathrm{H}\)) of the reaction.

Therefore, the correct answer is (1) Catalyst does not alter the Gibbs energy change of a reaction. Quick Tip: Catalysts lower the activation energy but do not change the thermodynamic properties of the reaction.

1. Initial rate law:
\[ \mathrm{r}_{1} = \mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}} \]

2. New concentrations:
- Concentration of A is doubled: \(2[\mathrm{A}]\)
- Concentration of B is halved: \(\frac{[\mathrm{B}]}{2}\)

3. New rate law:
\[ \mathrm{r}_{2} = \mathrm{k}(2[\mathrm{A}])^{\mathrm{n}} \left( \frac{[\mathrm{B}]}{2} \right)^{\mathrm{m}} \]
\[ \mathrm{r}_{2} = \mathrm{k} \cdot 2^{\mathrm{n}} [\mathrm{A}]^{\mathrm{n}} \cdot \frac{[\mathrm{B}]^{\mathrm{m}}}{2^{\mathrm{m}}} \]

4. Ratio of new rate to initial rate:
\[ \frac{\mathrm{r}_{2}}{\mathrm{r}_{1}} = \frac{\mathrm{k} \cdot 2^{\mathrm{n}} [\mathrm{A}]^{\mathrm{n}} \cdot \frac{[\mathrm{B}]^{\mathrm{m}}}{2^{\mathrm{m}}}}{\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}} = 2^{\mathrm{n}} \cdot \frac{1}{2^{\mathrm{m}}} = 2^{(\mathrm{n}-\mathrm{m})} \]

Therefore, the correct answer is (1) \(2^{(\mathrm{n}-\mathrm{m})}\). Quick Tip: The rate law depends on the concentrations of the reactants raised to their respective orders.

1. \(\left[\mathrm{CrCl}_{3}(\mathrm{py})_{3}\right]\):
- Facial and meridional isomers are possible.
- Total stereoisomers = 2.

2. \(\left[\mathrm{CrCl}_{2}(\mathrm{ox})_{2}\right]^{3-}\):
- Geometrical isomers: cis and trans.
- Optical isomers for cis: 2.
- Optical isomers for trans: 1.
- Total stereoisomers = 3.

Therefore, the correct answer is (3) \(2 \& 3\). Quick Tip: Stereoisomers include geometrical and optical isomers.

1. Reaction sequence:
- The major product formed is (2).

Therefore, the correct answer is (2). Quick Tip: Follow the reaction sequence to determine the major product.

1. Anode reaction:
- \(\mathrm{PbSO}_{4}\) is reduced to \(\mathrm{Pb}\).
- \(\mathrm{Pb}^{2+} \rightarrow \mathrm{Pb}^{0}\)
- \(\mathrm{x}_{1} = +2\), \(\mathrm{y}_{1} = 0\)

2. Cathode reaction:
- \(\mathrm{PbSO}_{4}\) is oxidized to \(\mathrm{PbO}_{2}\).
- \(\mathrm{Pb}^{2+} \rightarrow \mathrm{Pb}^{4+}\)
- \(\mathrm{x}_{2} = +2\), \(\mathrm{y}_{2} = +4\)

Therefore, the correct answer is (2) \(+2,0,+2,+4\). Quick Tip: The oxidation states change during the charging of a lead-acid battery.

1. Statement I:
- Nitrogen can form oxides with oxidation states from +1 to +5 due to the formation of \(\mathrm{p} \pi-\mathrm{p} \pi\) bonds with oxygen.
- This statement is true.

2. Statement II:
- Nitrogen does not form halides with a +5 oxidation state due to the absence of d-orbitals.
- This statement is true.

Therefore, the correct answer is (4) Both Statement I and Statement II are true. Quick Tip: Nitrogen's ability to form oxides and halides depends on its electronic configuration and the availability of d-orbitals.

1. Reaction sequence:
- Benzene treated with oleum produces benzene sulfonic acid (X).
- Heating with molten sodium hydroxide followed by acidification produces phenol (Y).
- Treatment with zinc metal reduces phenol to cyclohexanol (Z).

Therefore, the correct answer is (2). Quick Tip: Follow the reaction sequence to identify the final product.

1. Reactants:
- \(\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{COCl}\) (Glycine chloride)
- \(\mathrm{NH}_{2}-\mathrm{CH}-\mathrm{COOH}\) (Alanine)

2. Reaction:
- The reaction between these reactants with the elimination of HCl will produce the dipeptide Gly-Ala.

Therefore, the correct answer is (1) \(\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{COCl}\) and \(\mathrm{NH}_{2}-\mathrm{CH}-\mathrm{COOH}\). Quick Tip: The formation of a dipeptide involves the reaction between an amino acid and its chloride derivative with the elimination of HCl.

1. Incorrect pair:
- \(\mathrm{Al}<\mathrm{Ga}\)

2. Ionic radius comparison:
- \(\mathrm{Al}^{3+}<\mathrm{Ga}^{3+}\)
- The atomic number of \(\mathrm{Ga}\) is 31.

Therefore, the correct answer is (1) 31. Quick Tip: The atomic radius and ionic radius depend on the atomic number and the periodic trends.

1. Compound (X):
- \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) (Acetone)

2. Reduction to (Y):
- \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}\) (Isopropyl alcohol)

3. Reaction with HBr to form (Z):
- \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{3}\) (2-Bromopropane)

4. Grignard reagent and reaction with (X):
- The Grignard reagent formed from (Z) reacts with acetone to form 2,3-dimethylbutan-2-ol after hydrolysis.

Therefore, the correct answer is (2) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}, \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{3}\). Quick Tip: Follow the reaction sequence to identify the compounds involved.

% Option
(A) Step 1: Bromination (Br\(_2\)/hv)
\[\ce{CH3 ->[Br2/hv] CH2Br}\]

% Option
(B) Step 2: Elimination (Alcoholic KOH)
\[\ce{CH2Br ->[Alc. KOH][\Delta] CH2=}\]

% Option
(C) Step 3: Anti-Markovnikov addition (HBr/ROOR)
\[\ce{CH2= ->[HBr/ROOR][hv] Br-CH3}\]


Mechanistic Explanation:

Free radical bromination converts methyl to bromomethyl
Elimination forms methylene intermediate
Peroxide effect gives anti-Markovnikov product


Therefore, the correct answer is \boxed{(2) \ce{Br-CH3. Quick Tip: Key Points: Radical bromination prefers allylic position Alcoholic KOH causes elimination ROOR reverses normal addition orientation

1. Analysis of Statement I:

For \chemfig{CH_3-CH=CH-CH=O: \(\mu = q \times d\)
Conjugated system creates greater charge separation
More distance between charges than in saturated compound
Therefore, greater dipole moment
Statement I is TRUE


2. Analysis of Statement II:

In \chemfig{CH_3-CH=CH-CH=O, \(C_1-C_2\) has partial double bond character
Double bond character means shorter bond length
Compared to pure single bond in \chemfig{CH_3-CH_2-CH_2-CH=O
Statement II is FALSE (actual bond length is shorter) Quick Tip: Key concepts: Dipole moment depends on charge magnitude and separation distance Conjugation affects both electronic distribution and bond lengths Partial double bond character decreases bond length

1. \(\mathrm{V}^{2+}\):
- \(\mathrm{V}^{2+} \Rightarrow 3 \mathrm{~d}^{3} 4 \mathrm{~s}^{0}\)
- Number of unpaired electrons = 3

2. \(\mathrm{Co}^{2+}\):
- \(\mathrm{Co}^{2+} \Rightarrow 3 \mathrm{~d}^{7} 4 \mathrm{~s}^{0}\)
- Number of unpaired electrons = 3

3. \(\mathrm{Ti}^{2+}\):
- \(\mathrm{Ti}^{2+} \Rightarrow 3 \mathrm{~d}^{2} 4 \mathrm{~s}^{0}\)
- Number of unpaired electrons = 2

4. \(\mathrm{Fe}^{3+}\):
- \(\mathrm{Fe}^{3+} \Rightarrow 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0}\)
- Number of unpaired electrons = 5

5. \(\mathrm{Cr}^{2+}\):
- \(\mathrm{Cr}^{2+} \Rightarrow 3 \mathrm{~d}^{4} 4 \mathrm{~s}^{0}\)
- Number of unpaired electrons = 4

6. \(\mathrm{Ti}^{3+}\):
- \(\mathrm{Ti}^{3+} \Rightarrow 3 \mathrm{~d}^{1} 4 \mathrm{~s}^{0}\)
- Number of unpaired electrons = 1

7. \(\mathrm{Mn}^{2+}\):
- \(\mathrm{Mn}^{2+} \Rightarrow 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0}\)
- Number of unpaired electrons = 5

Therefore, the correct answer is (1) \(\mathrm{V}^{2+}, \mathrm{Co}^{2+}\). Quick Tip: The number of unpaired electrons in transition metal ions depends on their electronic configuration.

1. Probability density:
- The probability density of finding the electron is maximum at the nucleus.

2. Distance from the nucleus:
- The electron can be found at a distance \(2 \mathrm{a}_{0}\) from the nucleus.

3. Spherical symmetry:
- The 1s orbital is spherically symmetrical.

4. Total energy:
- The total energy of the electron is maximum when it is at a distance \(\mathrm{a}_{0}\) from the nucleus. This statement is incorrect.

Therefore, the correct answer is (4). Quick Tip: The probability density, distance from the nucleus, spherical symmetry, and total energy of an electron in the 1s orbital are important properties to consider.

1. Calculate the partial pressure of \(\mathrm{N}_{2}\):
\[ \mathrm{p}_{\mathrm{N}_2} = 900 \mathrm{~mm~Hg} - 15 \mathrm{~mm~Hg} = 885 \mathrm{~mm~Hg} \]

2. Calculate the moles of \(\mathrm{N}_{2}\):
\[ Moles of \mathrm{N}_2 = \frac{885 \mathrm{~mm~Hg} \times 0.15 \mathrm{~L}}{0.0821 \mathrm{~L~atm/mol} \times 300 \mathrm{~K}} = 0.0071 \mathrm{~mol} \]

3. Calculate the percentage of nitrogen:
\[ Percentage of nitrogen = \frac{0.0071 \mathrm{~mol} \times 28 \mathrm{~g/mol}}{1 \mathrm{~g}} \times 100 = 19.85% \approx 20% \]

Therefore, the correct answer is (20). Quick Tip: Use the ideal gas law to calculate the moles of nitrogen and then determine the percentage composition.

1. Oxidation states of Mn:
- Reactant: \(\mathrm{Mn}^{7+}\)
- Product: \(\mathrm{Mn}^{2+}\)
- Difference in oxidation states: \(X = 7 - 2 = 5\)

2. Brown red precipitate:
- The brown red precipitate is \(\mathrm{Fe}(\mathrm{OH})_2(\mathrm{CH}_3\mathrm{COO})_n\).
- \(\mathrm{Fe}^{3+}\) has 5 d-electrons.
- Therefore, \(Y = 5\).

3. Calculate \(\mathrm{X}+\mathrm{Y}\):
\[ \mathrm{X} + \mathrm{Y} = 5 + 5 = 10 \]

Therefore, the correct answer is (10). Quick Tip: Determine the oxidation states and the number of d-electrons to find the values of X and Y.

1. Calculate the mass of iron required:
\[ Mass of iron = \frac{12 \mathrm{~ppm} \times 150 \mathrm{~kg}}{10^6} = 1.8 \mathrm{~g} \]

2. Calculate the moles of iron:
\[ Moles of iron = \frac{1.8 \mathrm{~g}}{56 \mathrm{~g/mol}} = 0.0321 \mathrm{~mol} \]

3. Calculate the moles of \(\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}\):
\[ Moles of \mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O} = 0.0321 \mathrm{~mol} \]

4. Calculate the mass of \(\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}\):
\[ Molar mass of \mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O} = 56 + 32 + 7 \times 18 = 277 \mathrm{~g/mol} \]
\[ Mass of \mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O} = 0.0321 \mathrm{~mol} \times 277 \mathrm{~g/mol} = 8.8935 \mathrm{~g} \approx 9 \mathrm{~g} \]

Therefore, the correct answer is (9). Quick Tip: Use the molar mass and stoichiometry to calculate the mass of the compound required.

1. Initial pH calculation:
\[ \mathrm{HX}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{X}_{(\mathrm{aq})}^{-} \quad \mathrm{K}_{\mathrm{a}}=4 \times 10^{-10} \]
\[ 0.01(1-\alpha) \quad 0.01 \alpha \quad 0.01 \alpha \quad Not justified \]
\[ \Rightarrow 0.01 \alpha=10^{-5} \Rightarrow \alpha=10^{-3} \]

2. Calculate \(\mathrm{K}_{\mathrm{a}}\):
\[ \mathrm{K}_{\mathrm{a}}=0.01 \alpha^{2}=10^{-8} \]

3. Data given is inconsistent \& contradictory. This should be bonus. Quick Tip: Check the consistency of the given data and ensure the calculations are justified.

1. Hydrogen bonding in DNA:
- Adenine (A) forms two hydrogen bonds with Thymine (T).
- Guanine (G) forms three hydrogen bonds with Cytosine (C).

2. Count the hydrogen bonds:
- Number of G-C pairs: 4
- Number of A-T pairs: 4

3. Total number of hydrogen bonds:
\[ Total hydrogen bonds = 4 \times 3 + 4 \times 2 = 12 + 8 = 20 \]

Therefore, the correct answer is (33). Quick Tip: Count the number of hydrogen bonds formed by each base pair in the DNA strand.