JEE Advanced 26 May 2024 Paper 1 question paper with solutions
JEE Advanced 26 May 2024 Paper 1: all 51 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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Maths
Question 1
Maths · Continuity and Differentiability · Single correct
Let f(x) be a continuously differentiable function on the interval (0, $\infty$) such that f(1) = 2 and $$\lim_{t \to x} \frac{t^{10}f(x) - x^{10}f(t)}{t^9 - x^9} = 1$$ for each x > 0. Then, for all x > 0, f(x) is equal to
A student appears for a quiz consisting of only true-false type questions and answers all the questions. The student knows the answers of some questions and guesses the answers for the remaining questions. Whenever the student knows the answer of a question, he gives the correct answer. Assume that the probability of the student giving the correct answer for a question, given that he has guessed it, is $\frac{1}{2}$. Also assume that the probability of the answer for a question being guessed, given that the student's answer is correct, is $\frac{1}{6}$. Then the probability that the student knows the answer of a randomly chosen question is
Let $\frac{\pi}{2} < x < \pi$ be such that $\cot x = \frac{-5}{\sqrt{11}}$. Then $\left( \sin \frac{11x}{2} \right) (\sin 6x - \cos 6x) + \left( \cos \frac{11x}{2} \right) (\sin 6x + \cos 6x)$ is equal to
$\frac{\sqrt{11} - 1}{2\sqrt{3}}$
$\frac{\sqrt{11} + 1}{2\sqrt{3}}$
$\frac{\sqrt{11} + 1}{3\sqrt{2}}$
$\frac{\sqrt{11} - 1}{3\sqrt{2}}$
Answer: (b)
Question 4
Maths · Conic Sections · Single correct
Consider the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Let $S(p, q)$ be a point in the first quadrant such that $\frac{p^2}{9} + \frac{q^2}{4} > 1$. Two tangents are drawn from $S$ to the ellipse, of which one meets the ellipse at one end point of the minor axis and the other meets the ellipse at a point $T$ in the fourth quadrant. Let $R$ be the vertex of the ellipse with positive x-coordinate and $O$ be the center of the ellipse. If the area of the triangle $\triangle ORT$ is $\frac{3}{2}$, then which of the following options is correct?
$q = 2, p = 3\sqrt{3}$
$q = 2, p = 4\sqrt{3}$
$q = 1, p = 5\sqrt{3}$
$q = 1, p = 6\sqrt{3}$
Answer: (a)
Solution
Given $q = 2$. The area of $\triangle ORT = \frac{3}{2}$. Therefore, $$\frac{1}{2} \times OR \times QT = \frac{3}{2}$$ $$\frac{1}{2} \times 3 \times \beta = \frac{3}{2}$$ This implies $\beta = -1$. Thus, $$\frac{\alpha^2}{9} + \frac{\beta^2}{4} = 1$$ $$\frac{\alpha^2}{9} = 1 - \frac{1}{4} = \frac{3}{4}$$ $$\alpha^2 = \frac{27}{4} \implies \alpha = \frac{3\sqrt{3}}{2}$$ The tangent at $T$ gives $T = 0$. Therefore, $$x \cdot \frac{3\sqrt{3}}{2} \cdot \frac{2}{9} + y(-1) = 1 \bigg|_{(p, 2)}$$ $$p \sqrt{3} \cdot \frac{1}{6} = 1 \implies p \sqrt{3} = \frac{3}{2} \implies p = 3\sqrt{3}$$ Thus, $p = 3\sqrt{3}, q = 2$.
Question 5
Maths · Complex Numbers and Quadratic Equations · Multiple correct
Let $S = \{ a + b \sqrt{2} : a, b \in \mathbb{Z} \}$, $T_1 = \{ (-1 + \sqrt{2})^n : n \in \mathbb{N} \}$ and $T_2 = \{ (1 + \sqrt{2})^n : n \in \mathbb{N} \}$. Then which of the following statements is (are) TRUE?
$\mathbb{Z} \cup T_1 \cup T_2 \subset S$
$T_1 \cap \left( 0, \frac{1}{2024} \right) = \phi$, where $\phi$ denotes the empty set
$T_2 \cap (2024, \infty) \neq \phi$
For any given $a, b \in \mathbb{Z}$, $\cos \left( \pi (a + b \sqrt{2}) \right) + i \sin \left( \pi (a + b \sqrt{2}) \right) \in \mathbb{Z}$ if and only if $b = 0$, where $i = \sqrt{-1}$
Answer: (a), (c), (d)
Solution
Given $S = \{ a + b \sqrt{2} : a, b \in \mathbb{Z} \}$. For $b = 0$; $\mathbb{Z} \subset S$. $$T_1 = \left\{ (-1 + \sqrt{2})^n : n \in \mathbb{N} \right\} and T_2 = \left\{ (1 + \sqrt{2})^n : n \in \mathbb{N} \right\}$$ For $n \in \mathbb{N}$, elements of $T_1$ and $T_2$ are of the form $a + b \sqrt{2}$. Hence $\mathbb{Z} \cup T_1 \cup T_2 \subset S$. Now, $-1 + \sqrt{2} 1$ and its higher power increases. $$\Rightarrow (1 + \sqrt{2})^n can be made in (2024, \infty) for some higher n.$$ $\cos \pi (a + b \sqrt{2}) + i \sin \pi (a + b \sqrt{2}) \in \mathbb{Z}$ if $a + b \sqrt{2}$ is an integer $\Rightarrow b = 0$
Question 6
Maths · Determinants · Multiple correct
Let $\mathbb{R}^2$ denote $\mathbb{R} \times \mathbb{R}$. Let $S = \{(a, b, c) : a, b, c \in \mathbb{R}$ and $ax^2 + 2bxy + cy^2 > 0$ for all $(x, y) \in \mathbb{R}^2 - \{(0, 0)\}\}$. Then which of the following statements is (are) TRUE?
$\left(2, \frac{7}{2}, 6\right) \in S$
If $\left(3, b, \frac{1}{12}\right) \in S$, then $|2b| < 1$
For any given $(a, b, c) \in S$, the system of linear equations $ax + by = 1$, $bx + cy = -1$ has a unique solution.
For any given $(a, b, c) \in S$, the system of linear equations $(a+1)x + by = 0$, $bx + (c+1)y = 0$ has a unique solution.
Answer: (b), (c), (d)
Solution
Given $ax^2 + 2bxy + cy^2 > 0$ for $y, x \in \mathbb{R} - \{(0, 0)\}$. This implies $c \left( \frac{y}{x} \right)^2 + 2b \left( \frac{y}{x} \right) + a > 0$. Thus, $c > 0$, $D 2 \times 6$$ Therefore, option A is incorrect. (B) If $\left( 3, b, \frac{1}{12} \right) \in S$ This implies $b^2 0$. Therefore, unique solution. Option D is correct.
Question 7
Maths · Three Dimensional Geometry · Multiple correct
Let $\mathbb{R}^3$ denote the three-dimensional space. Take two points $P = (1, 2, 3)$ and $Q = (4, 2, 7)$. Let $dist(X, Y)$ denote the distance between two points $X$ and $Y$ in $\mathbb{R}^3$. Let $$S = \left\{ X \in \mathbb{R}^3 : (dist(X, P))^2 - (dist(X, Q))^2 = 50 \right\}$$ and $$T = \left\{ Y \in \mathbb{R}^3 : (dist(Y, Q))^2 - (dist(Y, P))^2 = 50 \right\}.$$ Then which of the following statements is (are) TRUE?
There is a triangle whose area is 1 and all of whose vertices are from $S$.
There are two distinct points $L$ and $M$ in $T$ such that each point on the line segment $LM$ is also in $T$.
There are infinitely many rectangles of perimeter 48, two of whose vertices are from $S$ and the other two vertices are from $T$.
There is a square of perimeter 48, two of whose vertices are from $S$ and the other two vertices are from $T$.
Answer: (a), (b), (c), (d)
Solution
Given $S : \{((x - 1)^2 + (y - 2)^2 + (z - 3)^2) - ((x - 4)^2 + (y - 2)^2 + (z - 7)^2) = 50\}$. Therefore, $S : \{6x + 8z - 105 = 0\}$. Similarly, $T = \{6x + 8z - 5 = 0\}$. $S$ represents a plane. So it will contain a triangle of area 1. So (A) is correct. $T$ represents a plane. So (B) is correct. $S$ and $T$ are two parallel planes at a distance of 10 units from each other. Therefore, (C) is correct and (D) is incorrect.
Question 8
Maths · Determinants · Fill in the blank
Let $a = 3\sqrt{2}$ and $b = \frac{1}{5^{\frac{1}{6}} \sqrt{6}}$. If $x, y \in \mathbb{R}$ are such that $$3x + 2y = \log_a \left(18^{\frac{5}{4}}\right)$$ and $$2x - y = \log_b \left(\sqrt{1080}\right),$$ then $4x + 5y$ is equal to ________.
Maths · Complex Numbers and Quadratic Equations · Numerical
Let $f(x) = x^4 + ax^3 + bx^2 + c$ be a polynomial with real coefficients such that $f(1) = -9$. Suppose that $i\sqrt{3}$ is a root of the equation $4x^3 + 3ax^2 + 2bx = 0$, where $i = \sqrt{-1}$. If $\alpha_1, \alpha_2, \alpha_3$, and $\alpha_4$ are all the roots of the equation $f(x) = 0$, then $|\alpha_1|^2 + |\alpha_2|^2 + |\alpha_3|^2 + |\alpha_4|^2$ is equal to .
Answer: 20
Solution
Given $f(1) = -9$, we have $1 + a + b + c = -9$. $$4x^3 + 3ax^2 + 2bx = 0$$ This implies $x = 0$, and $4x^2 + 3ax + 2b = 0$. The roots of equation (2) are $\sqrt{3}i$ and $-\sqrt{3}i$. Thus, $\sqrt{3}i - (-\sqrt{3}i) = \frac{-3a}{4}$ and $\sqrt{3}i(-\sqrt{3}i) = \frac{2b}{4}$. Solving gives $a = 0$, $b = 6$, $c = -16$. For $f(x) = 0$, we have $x^4 + 6x^2 - 16 = 0$. Solving for $x^2$, we get $$x^2 = \frac{-6 \pm \sqrt{36 + 64}}{2} = -3 \pm 5 = 2, -8.$$ Thus, $x = -\sqrt{2}, \sqrt{2}, -2\sqrt{2}i, 2\sqrt{2}i$. Finally, $|\alpha_1|^2 + |\alpha_2|^2 + |\alpha_3|^2 + |\alpha_4|^2 = 20$.
Question 10
Maths · Determinants · Numerical
Let $S = \left\{ A = \begin{pmatrix} 1 & a & d \\ 1 & b & e \end{pmatrix} : a, b, c, d, e \in \{0, 1\} and |A| \in \{-1, 1\} \right\}$, where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is .
Answer: 16
Solution
Given $|A| = -(e - d) + c(b - a) = \pm 1$. Case (i): $c = 0 \Rightarrow (e, d) = (1, 0), (0, 1) \rightarrow 2$ ways. $b$ and $a$ can be each 2 ways. $\Rightarrow$ Total $= 8$ ways. Case (ii): $c \neq 0 \Rightarrow c = 1$ $\Rightarrow d - e + b - a = \pm 1$ $$\begin{bmatrix} 1 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \rightarrow 4 \times 2 = 8 ways$$ Total $= 16$ ways
Question 11
Maths · Permutations and Combinations · Numerical
A group of 9 students, $s_1, s_2, \ldots, s_9$, is to be divided to form three teams $X$, $Y$ and $Z$ of sizes 2, 3, and 4, respectively. Suppose that $s_1$ cannot be selected for the team $X$, and $s_2$ cannot be selected for the team $Y$. Then the number of ways to form such teams, is______.
Answer: 665
Solution
Number of required ways $$= \frac{9!}{2!3!4!} - (n(s_1 \in X) + n(s_2 \in Y) - n(s_1 \in X and s_2 \in Y))$$ $$= \frac{9!}{2!3!4!} - \left( \frac{8!}{1!3!4!} + \frac{8!}{2!2!4!} - \frac{7!}{1!2!4!} \right)$$ $$= 665$$
Question 12
Maths · Vector Algebra · Numerical
Let $\overrightarrow{OP}$ = $\frac{\alpha - 1}{\alpha}$ $\hat{i}$ + $\hat{j}$ + $\hat{k}$, $\overrightarrow{OQ}$ = $\frac{\beta - 1}{\beta}$ $\hat{j}$ + $\hat{k}$ and $\overrightarrow{OR}$ = $\hat{i}$ + $\hat{j}$ + $\frac{1}{2}$ $\hat{k}$ be three vectors, where $\alpha$, $\beta$ $\in$ $\mathbb{R}$ - $\{$0$\}$ and O denotes the origin. If ($\overrightarrow{OP}$ $\times$ $\overrightarrow{OQ}$) $\cdot$ $\overrightarrow{OR}$ = 0 and the point ($\alpha$, $\beta$, 2) lies on the plane 3x + 3y - z + l = 0, then the value of l is
Let X be a random variable, and let P(X = x) denote the probability that X takes the value x. Suppose that the points (x, P(X = x)), x = 0, 1, 2, 3, 4, lie on a fixed straight line in the xy-plane, and P(X = x) = 0 for all x $\in$ $\mathbb{R}$ - $\{$0, 1, 2, 3, 4$\}$. If the mean of X is $\frac{5}{2}$, and the variance of X is $\alpha$, then the value of 24$\alpha$ is .
Let $\alpha$ and $\beta$ be the distinct roots of the equation $x^2 + x - 1 = 0$. Consider the set $T = \{1, \alpha, \beta\}$. For a $3 \times 3$ matrix $M = (a_{ij})_{3 \times 3}$, define $R_i = a_{i1} + a_{i2} + a_{i3}$ and $C_j = a_{1j} + a_{2j} + a_{3j}$ for $i = 1, 2, 3$ and $j = 1, 2, 3$. Match each entry in List-I to the correct entry in List-II.
→ (4) (Q) → (2) (R) → (5) (S) → (1)
→ (2) (Q) → (4) (R) → (1) (S) → (5)
→ (2) (Q) → (4) (R) → (3) (S) → (5)
→ (1) (Q) → (5) (R) → (3) (S) → (4)
Answer: (c)
Solution
Given $x^2 + x - 1 = 0$, the roots are $\alpha$ and $\beta$. $\alpha + \beta = -1$ and $\alpha \beta = -1$. Set $T = \{1, \alpha, \beta\}$, $M = (a_{ij})_{3 \times 3}$. $R_i = a_{i1} + a_{i2} + a_{i3}$, $C_j = a_{1j} + a_{2j} + a_{3j}$. (P) $R_i = C_j = 0$ for all $i, j$. $\alpha + \beta = -1$, $T = \{1, \alpha, \beta\}$. Number of matrices $= |3| \times 2 \times 1 = 12$. Number of ways to arrange $1, \alpha, \beta$ in $R$, and number of ways to arrange $1, \alpha, \beta$ in $R_2$. (Q) Number of symmetric matrices $= ?$ $C_j = 0 \forall \, j$. Number of symmetric matrices $$= \begin{bmatrix} 1 & \alpha & \beta \\ \alpha & \beta & 1 \\ \beta & 1 & \alpha \end{bmatrix}$$ $= |3| \times 1 = 6$. (R) $M \rightarrow$ skew symmetric of $3 \times 3$. $|M| = 0$, $a_{ij} \in T$ for $i > j$. $$M = \begin{bmatrix} x & a_{12} & 0 \\ y & 0 & -a_{23} \\ z & a_{31} & a_{32} \end{bmatrix}$$ $$= \begin{bmatrix} 0 & -a_{21} & -a_{31} \\ a_{21} & 0 & -a_{32} \\ a_{31} & a_{32} & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} a_{12} \\ 0 \\ -a_{23} \end{bmatrix}$$ As $x, y, z \in R$ and $a_{12} \& a_{23} \in R$ $\therefore$ System has infinite solutions. (S) $R_i = 0 \forall \, i$. $$M = \begin{bmatrix} 1 & \alpha & \beta \\ \alpha & \beta & 1 \\ \beta & 1 & \alpha \end{bmatrix}$$ $$= \begin{bmatrix} 1 + \alpha + \beta & \alpha & \beta \\ 1 + \alpha + \beta & \beta & 1 \\ 1 + \alpha + \beta & 1 & \alpha \end{bmatrix} = 0$$ (P) $\rightarrow$ (2) (Q) $\rightarrow$ (4) (R) $\rightarrow$ (3) (S) $\rightarrow$ (5)
Question 15
Maths · Conic Sections · Single correct
Let the straight line $y = 2x$ touch a circle with center $(0, \alpha)$, $\alpha > 0$, and radius $r$ at a point $A_1$. Let $B_1$ be the point on the circle such that the line segment $A_1B_1$ is a diameter of the circle. Let $\alpha + r = 5 + \sqrt{5}$. Match each entry in List-I to the correct entry in List-II.
→ (4) (Q) → (2) (R) → (1) (S) → (3)
→ (2) (Q) → (4) (R) → (1) (S) → (3)
→ (4) (Q) → (2) (R) → (5) (S) → (3)
→ (2) (Q) → (4) (R) → (3) (S) → (5)
Answer: (c)
Solution
The slope of the line is 2, which implies $\tan \theta = 2$. The point $C(0, \alpha)$ with $\alpha > 0$ is given. From the equation $\alpha + r = 5 + \sqrt{5}$, we have equation (1). The line $y = 2x$ is tangent to the circle, so $$\left| \frac{0 - \alpha}{\sqrt{4 + 1}} \right| = r$$ which implies $| -\alpha | = r \sqrt{5}$. Therefore, $\alpha = r \sqrt{5}$ as $\alpha > 0$. From equation (1), $r \sqrt{5} + r = 5 + \sqrt{5}$, which simplifies to $$r (\sqrt{5} + 1) = \sqrt{5} (\sqrt{5} + 1)$$ leading to $r = \sqrt{5}$. Thus, $\alpha = r \sqrt{5} = \sqrt{5} \times \sqrt{5} = 5$. The center is $C(0, 5)$. The distance $OC = 5$ and $A_1C = \sqrt{5}$. Therefore, $$OA_1 = \sqrt{25 - 5} = \sqrt{20} = 2 \sqrt{5}$$ with $\tan \theta = 2$ (from the figure). We have $\cos \theta = \frac{1}{\sqrt{5}}$ and $\sin \theta = \frac{2}{\sqrt{5}}$. The point $A_1(0 + OA_1 \cos \theta, 0 + OA_1 \sin \theta)$ gives $$A_1 \left( 2 \sqrt{5} \times \frac{1}{\sqrt{5}}, 2 \sqrt{5} \times \frac{2}{\sqrt{5}} \right)$$ which simplifies to $A_1(2, 4)$. Let $B_1(x_1, y_1)$. Then, $$\frac{x_1 + 2}{2} = 0 and \frac{y_1 + 4}{2} = 5$$ solving gives $x_1 = -2$ and $y_1 = 6$. Thus, $B_1(-2, 6)$. Therefore, $\alpha = 5$, $r = \sqrt{5}$, $A_1(2, 4)$, $B_1(-2, 6)$.
Question 16
Maths · Three Dimensional Geometry · Single correct
Let $\gamma \in \mathbb{R}$ be such that the lines $L_1 : \frac{x+11}{1} = \frac{y+21}{2} = \frac{z+29}{3}$ and $L_2 : \frac{x+16}{3} = \frac{y+11}{2} = \frac{z+4}{\gamma}$ intersect. Let $R_1$ be the point of intersection of $L_1$ and $L_2$. Let $O = (0, 0, 0)$, and $\hat{n}$ denote a unit normal vector to the plane containing both the lines $L_1$ and $L_2$. Match each entry in List-I to the correct entry in List-II.
→ (3) (Q) → (4) (R) → (1) (S) → (2)
→ (5) (Q) → (4) (R) → (1) (S) → (2)
→ (3) (Q) → (4) (R) → (1) (S) → (5)
→ (3) (Q) → (1) (R) → (4) (S) → (5)
Answer: (c)
Solution
Vector parallel to the line $L_1$ (say $\vec{b}_1$) is $\vec{i} + 2\vec{j} + 3\vec{k}$. Normal vector of plane ($\vec{n}$) containing $L_1$ and $L_2$ will be perpendicular to both $\vec{b}_1$ and $\overrightarrow{AB}$. $$\vec{n} = p(\overrightarrow{AB} \times \vec{n}) = p(5\vec{i} - 10\vec{j} - 25\vec{k}) \times (\vec{i} + 2\vec{j} + 3\vec{k})$$ $$= p(20\vec{i} - 40\vec{j} + 20\vec{k})$$ $$\Rightarrow \hat{n} = \frac{1}{\sqrt{6}} \vec{i} - \frac{2}{\sqrt{6}} \vec{j} + \frac{1}{\sqrt{6}} \vec{k}$$ Now, vector parallel to $L_2$ (say $\vec{b}_2$) is perpendicular to $\vec{n} \Rightarrow \vec{b}_2 \cdot \vec{n} = 0$. $$(3\vec{i} + 2\vec{j} + \gamma \vec{k}) \cdot p(20\vec{i} - 40\vec{j} + 20\vec{k}) = 0$$ $$\Rightarrow \gamma = 1$$ Now, for point of intersection (POI) $$L_1 : \frac{x + 11}{1} = \frac{y + 21}{2} = \frac{z + 29}{3} = \lambda$$ and $$L_2 : \frac{x + 16}{3} = \frac{y + 11}{2} = \frac{z + 4}{\gamma} = u$$ Comparing $x$ and $y$ coordinates, $-11 + \lambda = -16 + 3u$ and $-21 + 2\lambda = -11 + 2u$. $$\Rightarrow \lambda = 10, \ u = 5$$ POI i.e., $\overline{OR_1} : (-\vec{i} - \vec{j} + \vec{k})$ and $\overline{OR} \cdot \hat{n} = \frac{\sqrt{2}}{3}$
Question 17
Maths · Relations and Functions · Single correct
Let $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ be functions defined by $$f(x) = \begin{cases} x \left| x \right| \sin \left( \frac{1}{x} \right), & x \neq 0, \\ 0, & x = 0, \end{cases}$$ and $$g(x) = \begin{cases} 1 - 2x, & 0 \leq x \leq \frac{1}{2}, \\ 0, & otherwise \end{cases}$$ Let $a, b, c, d \in \mathbb{R}$. Define the function $h : \mathbb{R} \to \mathbb{R}$ by $$h(x) = af(x) + b \left( g(x) + g \left( \frac{1}{2} - x \right) \right) + c(x - g(x)) + d \ g(x), \ x \in \mathbb{R}$$ Match each entry in List-I to the correct entry in List-II.
→ (4) (Q) → (3) (R) → (1) (S) → (2)
→ (5) (Q) → (2) (R) → (4) (S) → (3)
→ (5) (Q) → (3) (R) → (2) (S) → (4)
→ (4) (Q) → (2) (R) → (1) (S) → (3)
Answer: (c)
Solution
Given $$g(x) = \begin{cases} 1 - 2x, & 0 \leq x \leq \frac{1}{2} \\ 0, & otherwise \end{cases}$$ $$= \begin{cases} 2x, & 0 \leq x \leq \frac{1}{2} \\ 0, & otherwise \end{cases}$$ $$g(x) + g\left(\frac{1}{2} - x\right) = \begin{cases} 1, & 0 \leq x \leq \frac{1}{2} \\ 0, & otherwise \end{cases}$$ Now, option (P), at $b = 1$ $$h(x) = g(x) + g\left(\frac{1}{2} - x\right)$$ has range $\{$0, 1$\}$ (P) $\rightarrow$ (5) Option (Q) at $a = 1$, $h(x) = f(x)$ $$f'(0^+) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin \frac{1}{h}}{h}$$ $$f'(0^-) = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{-h} = \lim_{h \to 0} h \sin h = 0$$ $$\Rightarrow f'(0^+) = f'(0^-)$$ as $f'(0^+) = f'(0^-)$ $\Rightarrow h(x) = f(x)$ is differentiable at $x = 0$ and all other points $f(x) = h(x)$ is differentiable as product of two differentiable functions (Q) $\rightarrow$ (3) Option (R) $$h(x) = x - g(x) = \begin{cases} 3x - 1, & 0 \leq x \leq \frac{1}{2} \\ x, & otherwise \end{cases}$$ by graph $h(x)$ has range $R$ and onto (R) $\rightarrow$ (2) (S) at $d = 1$ $h(x) = g(x)$ has range $\{$0, 1$\}$ (S) $\rightarrow$ (4)
Physics
Question 18
Physics · Physical World, Units and Measurements · Single correct
A dimensionless quantity is constructed in terms of electronic charge $e$, permittivity of free space $\varepsilon_0$, Planck's constant $h$, and speed of light $c$. If the dimensionless quantity is written as $e^\alpha \varepsilon_0^\beta h^\gamma c^\delta$ and $n$ is a non-zero integer, then $(\alpha, \beta, \gamma, \delta)$ is given by
$(2n, -n, -n, -n)$
$(n, -n, -2n, -n)$
$(n, -n, -n, -2n)$
$(2n, -n, -2n, -2n)$
Answer: (a)
Solution
Given $[AT]^\alpha [M^{-1}L^{-3}T^4A^2]^\beta [ML^2T^{-1}]^\gamma [LT^{-1}]^\delta = 0$. This implies $$\alpha + 2\beta = 0$$ $$-\beta + \gamma = 0$$ $$-3\beta + 2\gamma + \delta = 0$$ $$\alpha + 4\beta - \gamma - \delta = 0$$ Solving these equations, we find $$\alpha = -2\beta$$ $$\gamma = \beta$$ $$\delta = \beta$$ The solution is $(-2\beta, \beta, \beta, \beta)$.
Question 19
Physics · Moving Charges and Magnetism · Single correct
An infinitely long wire, located on the z-axis, carries a current $I$ along the $+z$-direction and produces the magnetic field $\vec{B}$. The magnitude of the line integral $\int \vec{B} \cdot d\vec{l}$ along a straight line from the point $(-\sqrt{3}a, a, 0)$ to $(a, a, 0)$ is given by [$\mu$_0 is the magnetic permeability of free space.]
7$\mu$_0 I / 24
7$\mu$_0 I / 12
$\mu$_0 I / 8
$\mu$_0 I / 6
Answer: (a)
Solution
Given $\theta = \pi - \frac{\pi}{4} - \frac{\pi}{6}$. Therefore, $$\theta = \frac{12\pi - 3\pi - 2\pi}{12} = \frac{7\pi}{12}$$ So, $\int \vec{B} \cdot d\vec{l}$ along the line is $$\int \vec{B} \cdot d\vec{l} = -\frac{\mu_0 (I)}{2\pi} \cdot \theta = \frac{\mu_0 I}{2\pi} \cdot \frac{7\pi}{12}$$ $$\Rightarrow \left| \int \vec{B} \cdot d\vec{l} \right| = \frac{7 \mu_0 I}{24}$$ Option (A) is correct Answer.
Question 20
Physics · Oscillations · Single correct
Two beads, each with charge $q$ and mass $m$, are on a horizontal, frictionless, non-conducting, circular hoop of radius $R$. One of the beads is glued to the hoop at some point, while the other one performs small oscillations about its equilibrium position along the hoop. The square of the angular frequency of the small oscillations is given by $[\varepsilon_0$ is the permittivity of free space.]
$q^2 / (4\pi\varepsilon_0 R^3 m)$
$q^2 / (32\pi\varepsilon_0 R^3 m)$
$q^2 / (8\pi\varepsilon_0 R^3 m)$
$q^2 / (16\pi\varepsilon_0 R^3 m)$
Answer: (b)
Solution
As the hoop mass is not given so it must not move or else its inertia must have some effect. Here $r = 2R \cos \phi$ Also $\theta = 2\phi$ Thus $\theta = \frac{\phi}{2}$ And $\theta = \frac{x}{R}$ If $\theta$ is the small angular displacement of free charge, then $F(\phi) = \frac{KqQ^2}{r^2}$ So, restoring force towards mean position is $F_{(R)} = \frac{KqQ^2}{r^2} \sin \phi$ $$a_R = \frac{F_{(R)}}{m} = \frac{-KqQ^2}{mr^2} \cdot \sin \phi = \frac{-KqQ^2}{m \cdot 4R^2 \cos^2 \phi}$$ $$\Rightarrow a_R = \frac{-KqQ^2}{4mR^2 \cos^2 \frac{\phi}{2}} \cdot \sin \left( \frac{\theta}{2} \right) \approx \frac{-KqQ^2}{4mR^2} \cdot \frac{1}{2} \cdot \frac{x}{R} = \omega^2 \cdot x$$ So, $$\omega^2 = \frac{q^2}{32 \pi \varepsilon_0 mR^3}$$ Option (B) is correct answer.
Question 21
Physics · Laws of Motion · Single correct
A block of mass 5 kg moves along the x-direction subject to the force $F = (-20x + 10) \mathrm{\, N}$, with the value of $x$ in metre. At time $t = 0 \mathrm{\, s}$, it is at rest at position $x = 1 \mathrm{\, m}$. The position and momentum of the block at $t = (\pi/4) \mathrm{\, s}$ are
A particle of mass $m$ is moving in a circular orbit under the influence of the central force $F(r) = -kr$, corresponding to the potential energy $V(r) = kr^2/2$, where $k$ is a positive force constant and $r$ is the radial distance from the origin. According to the Bohr's quantization rule, the angular momentum of the particle is given by $L = nh$, where $h = h/(2\pi)$, $h$ is the Planck's constant, and $n$ a positive integer. If $v$ and $E$ are the speed and total energy of the particle, respectively, then which of the following expression(s) is(are) correct?
Two uniform strings of mass per unit length $\mu$ and $4\mu$, and length $L$ and $2L$, respectively, are joined at point $O$, and tied at two fixed ends $P$ and $Q$, as shown in the figure. The strings are under a uniform tension $T$. If we define the frequency $v_0 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, which of the following statement(s) is(are) correct?
With a node at $O$, the minimum frequency of vibration of the composite string is $v_0$
With an antinode at $O$, the minimum frequency of vibration of the composite string is $2v_0$
When the composite string vibrates at the minimum frequency with a node at $O$, it has 6 nodes, including the end nodes
No vibrational mode with an antinode at $O$ is possible for the composite string
Answer: (a), (c), (d)
Solution
With node at $O$ $$\nu = \sqrt{\frac{T}{\mu}}, \nu' = \sqrt{\frac{T}{\mu'}} = \frac{1}{2} \nu$$ $$\Rightarrow \frac{m}{n} = \frac{1}{4}$$ Therefore, $m = 1$, $n = 4$. With antinode at $O$ $$m \frac{1}{4l} \sqrt{\frac{T}{\mu}} = n \frac{1}{4(2l)} \sqrt{\frac{T}{4\mu}}$$ $$\Rightarrow \frac{m}{n} = \frac{1}{2} \times \frac{1}{2}$$ $$\Rightarrow \frac{m}{n} = \frac{1}{4}$$ Therefore, $m = 1$, $f_{\min} = 1 \frac{1}{4l} \sqrt{\frac{T}{\mu}} = \frac{v_0}{2}$. (B is wrong) Also, when node at $O$. Total nodes = 6 (C is correct) A, C, D are correct
Question 24
Physics · Ray Optics and Optical Instruments · Multiple correct
A glass beaker has a solid, plano-convex base of refractive index 1.60, as shown in the figure. The radius of curvature of the convex surface (SPU) is 9 cm, while the planar surface (STU) acts as a mirror. This beaker is filled with a liquid of refractive index $n$ up to the level $QPR$. If the image of a point object $O$ at a height of $h$ ($OT$ in the figure) is formed onto itself, then, which of the following option(s) is (are) correct?
For $n = 1.42$, $h = 50 \, \mathrm{cm}$
For $n = 1.35$, $h = 36 \, \mathrm{cm}$
For $n = 1.45$, $h = 65 \, \mathrm{cm}$
For $n = 1.48$, $h = 85 \, \mathrm{cm}$
Answer: (a), (b)
Solution
For image to coincide with object $$-h = 2(f_{net})$$ $$\Rightarrow \frac{-1}{f_{net}} = 2 \left( \frac{1}{f_{liq}} \right) + 2 \left( \frac{1}{f_{lens}} \right) + \left( \frac{-1}{f_{mirror}} \right) (i)$$ $$\frac{-1}{f_{net}} = 2 \left( \frac{n-1}{-9} \right) + 2 \left( \frac{0.6}{9} \right) + \left( \frac{-1}{\infty} \right) (ii)$$ From (i) and (ii) $$h = \frac{9}{(1.6-n)}$$ For $n = 1.42$, $h = 50 \, cm$ (A is correct) For $n = 1.35$, $h = 36 \, cm$ (B is correct) For $n = 1.45$, $h = 60 \, cm$ (C is incorrect) For $n = 1.48$, $h = 75 \, cm$ (D is incorrect)
Question 25
Physics · Thermal Properties of Matter · Numerical
The specific heat capacity of a substance is temperature dependent and is given by the formula $C = kT$, where $k$ is a constant of suitable dimensions in SI units, and $T$ is the absolute temperature. If the heat required to raise the temperature of 1 kg of the substance from $-73^\circ \mathrm{C}$ to $27^\circ \mathrm{C}$ is $nk$, the value of $n$ is . [Given: $0 \, \mathrm{K} = -273^\circ \mathrm{C}$.]
Answer: 25000
Solution
Given $C = \frac{dQ/m}{dT}$. Therefore, $dQ = m \cdot C \cdot dT$. This implies $dQ = 1 \cdot kT dT$. The heat $Q$ is given by $$Q = \int_{200}^{300} kT dT = \frac{k}{2} \left[ 300^2 - 200^2 \right]$$ which simplifies to $$= \frac{10^4}{2} \cdot k \cdot 5$$ $$= 25000k$$ Therefore, $n = 25000$.
Question 26
Physics · System of Particles and Rotational Motion · Fill in the blank
A disc of mass $M$ and radius $R$ is free to rotate about its vertical axis as shown in the figure. A battery operated motor of negligible mass is fixed to this disc at a point on its circumference. Another disc of the same mass $M$ and radius $R/2$ is fixed to the motor's thin shaft. Initially, both the discs are at rest. The motor is switched on so that the smaller disc rotates at a uniform angular speed $\omega$. If the angular speed at which the large disc rotates is $\omega/n$, then the value of $n$ is _____.
Answer: 12
Solution
Conserving angular momentum of the system (bigger disc + smaller disc) about the symmetric axis of bigger disc: $$\frac{MR^2}{2} \omega' + M \cdot R \omega' \cdot R + \frac{M(R/2)^2}{2} \omega = 0$$ Where $\omega'$: required angular speed. $$\Rightarrow \frac{3}{2} MR^2 \omega' = -\frac{MR^2 \omega}{8}$$ $$\Rightarrow \omega' = -\frac{\omega}{12}$$ $$\Rightarrow n = 12$$
Question 27
Physics · Wave Optics · Numerical
A point source $S$ emits unpolarized light uniformly in all directions. At two points $A$ and $B$, the ratio $r = I_A/I_B$ of the intensities of light is 2. If a set of two polaroids having $45^\circ$ angle between their pass-axes is placed just before point $B$, then the new value of $r$ will be
Answer: 8
Solution
Given $I \propto \frac{1}{l^2}$ where $l$ is the distance from the point source. Therefore, $$\frac{I_A}{I_B} = \frac{l_B^2}{l_A^2} = 2$$ This implies $$I_B = \sqrt{2} \, I_A ...(1)$$ Also, due to polaroids: $$I_B' = \frac{I_B}{2} \cos^2 45^\circ = \frac{I_B}{4}$$ Thus, $$I_B' = \frac{I_B}{4} ...(2)$$ Therefore, the ratio becomes 4 times. Hence, $r_{new} = 8$
Question 28
Physics · Waves · Numerical
A source (S) of sound has frequency 240 $\,$ $\mathrm{Hz}$. When the observer (O) and the source move towards each other at a speed $v$ with respect to the ground (as shown in Case 1 in the figure), the observer measures the frequency of the sound to be 288 $\,$ $\mathrm{Hz}$. However, when the observer and the source move away from each other at the same speed $v$ with respect to the ground (as shown in Case 2 in the figure), the observer measures the frequency of sound to be $n$ $\,$ $\mathrm{Hz}$. The value of $n$ is .
Answer: 200
Solution
For Case 1: $$f_{app} = f \left( \frac{c+v}{c-v} \right) \Rightarrow 288 = 240 \left( \frac{c+v}{c-v} \right) ...(i)$$ For Case 2: $$f_{app} = f \left( \frac{c-v}{c+v} \right)$$ $$= 240 \left( \frac{c-v}{c+v} \right) ...(ii)$$ From (i) and (ii) $$288 \times f_{app} = 240 \left( \frac{c+v}{c-v} \right) \times 240 \left( \frac{c-v}{c+v} \right)$$ $$288 \times f_{app} = 240 \times 240$$ $$f_{app} = 200 \, Hz$$
Question 29
Physics · Mechanical Properties of Fluids · Numerical
Two large, identical water tanks, 1 and 2, kept on the top of a building of height $H$, are filled with water up to height $h$ in each tank. Both the tanks contain an identical hole of small radius on their sides, close to their bottom. A pipe of the same internal radius as that of the hole is connected to tank 2, and the pipe ends at the ground level. When the water flows the tanks 1 and 2 through the holes, the times taken to empty the tanks are $t_1$ and $t_2$, respectively. If $H = \left( \frac{16}{9} \right) h$, then the ratio $t_1/t_2$ is _______.
Answer: 3
Solution
In general case $$a \sqrt{2gh} = -A \frac{dh}{dt}$$ $$dt = -\frac{A}{a \sqrt{2gh}} \, dh$$ $$T = \int dt = \frac{-2A}{a \sqrt{2g}} \left( \sqrt{h_f} - \sqrt{h_i} \right)$$ $$T = \frac{2A}{a \sqrt{2g}} \left( \sqrt{h_i} - \sqrt{h_f} \right)$$ For tank 1: $$h_i = h, \; h_f = 0$$ $$T_1 = \frac{2A}{a \sqrt{2g}} \left( \sqrt{h} \right)$$ For tank 2: $$h_f = \frac{16h}{9}, \; h_i = h + H = \frac{25h}{9}$$ $$T_2 = \frac{2A \sqrt{h_f}}{a \sqrt{2g}} \left( \frac{5}{3} - \frac{4}{3} \right) = \frac{2A \sqrt{h}}{a \sqrt{2g}} \times \frac{1}{3}$$ $$\frac{T_1}{T_2} = 3$$
Question 30
Physics · System of Particles and Rotational Motion · Fill in the blank
A thin uniform rod of length $L$ and certain mass is kept on a frictionless horizontal table with a massless string of length $L$ fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point $O$. If a horizontal impulse $P$ is imparted to the rod at a distance $x = L/n$ from the mid-point of the rod (see figure), then the rod and string revolve together around the point $O$, with the rod remaining aligned with the string. In such a case, the value of $n$ is ______.
Answer: 18
Solution
M.I. of rod about centre (O) is $\frac{mL^2}{12} + m \left( L + \frac{L}{2} \right)^2$. $$I_0 = \frac{7mL^2}{3}$$ Since rod is in pure rotation about 'O'. So, angular impulse $= P \left( x + \frac{3L}{2} \right) = I_0 \omega_0$ and linear impulse $= mv_c$ where $v_c = \frac{3L}{2} \omega_0$ (20 m (i)) $m \left( \frac{3L}{2} \omega_0 \right) \left( x + \frac{3L}{2} \right) = \frac{7}{3} mL^2 \cdot \omega_0$ $$x + \frac{3L}{2} = \frac{14}{9} L$$ $$x = \frac{L}{18}$$
Question 31
Physics · Thermodynamics · Single correct
One mole of a monatomic ideal gas undergoes the cyclic process $J \rightarrow K \rightarrow L \rightarrow M \rightarrow J$, as shown in the $P-T$ diagram. Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [R is the gas constant.]
P $\rightarrow$ 1; Q $\rightarrow$ 3; R $\rightarrow$ 5; S $\rightarrow$ 4
P $\rightarrow$ 4; Q $\rightarrow$ 3; R $\rightarrow$ 5; S $\rightarrow$ 2
P $\rightarrow$ 4; Q $\rightarrow$ 1; R $\rightarrow$ 2; S $\rightarrow$ 2
P $\rightarrow$ 2; Q $\rightarrow$ 5; R $\rightarrow$ 3; S $\rightarrow$ 4
Answer: (b)
Solution
From $M \rightarrow J$, isothermal $$W = nRT \ln 2$$ $$= RT_0 \ln 2$$ $$\Delta U = 0$$ From $J \rightarrow K$, isobaric $$W = nR(3T_0 - T_0)$$ $$W = 2RT_0$$ $$\Delta U = nC_V \Delta T$$ $$= \frac{3R}{2} \times 2T_0 = 3RT_0$$ From $K \rightarrow L$, isothermal $$W = -nR(3T_0) \ln 2$$ $$W = -3RT_0 \ln 2$$ $$\Delta U = 0$$ $$Q = -3RT_0 \ln 2$$ From $L \rightarrow M$, isobaric $$W = nR(T_0 - 3T_0)$$ $$W = -2RT_0$$ $$\Delta U = nC_V \Delta T$$ $$= n \times \frac{3R}{2} \times (-2T_0)$$ $$= -3RT_0$$ For $P$: $$W_{net} = RT_0 \ln 2 + 2RT_0 - 3RT_0 \ln 2 - 2RT_0$$ $$= -2RT_0 \ln 2$$ $P \rightarrow 4$ for $Q \rightarrow 3$ for $R \rightarrow 5$ for $S \rightarrow 2$
Question 32
Physics · Electrostatic Potential and Capacitance · Single correct
Four identical thin, square metal sheets, $S_1$, $S_2$, $S_3$, and $S_4$, each of side $a$ are kept parallel to each other with equal distance $d(<<a)$ between them, as shown in the figure. Let $C_0 = \varepsilon_0 \frac{a^2}{d}$, where $\varepsilon_0$ is the permittivity of free space. Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
P $\rightarrow$ 3; Q $\rightarrow$ 2; R $\rightarrow$ 4; S $\rightarrow$ 5
P $\rightarrow$ 2; Q $\rightarrow$ 3; R $\rightarrow$ 2; S $\rightarrow$ 1
P $\rightarrow$ 3; Q $\rightarrow$ 2; R $\rightarrow$ 4; S $\rightarrow$ 1
P $\rightarrow$ 3; Q $\rightarrow$ 2; R $\rightarrow$ 2; S $\rightarrow$ 5
Answer: (c)
Solution
For $P$, all are in series. $$C_{eq} = \frac{C_0}{3}$$ For $P \to (3)$, $$C_{eq} = \frac{C_0}{2}$$ For $Q \to (2)$, for $R$, $$C_{eq} = \frac{2C}{3}$$ For $R \to (4)$, for $S$, $$C_{eq} = 3C_0$$ $S \to (1)$
Question 33
Physics · Ray Optics and Optical Instruments · Single correct
A light ray is incident on the surface of a sphere of refractive index $n$ at an angle of incidence $\theta_0$. The ray partially refracts into the sphere with angle of refraction $\phi_0$ and then partly reflects from the back surface. The reflected ray then emerges out of the sphere after a partial refraction. The total angle of deviation of the emergent ray with respect to the incident ray is $\alpha$. Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
The circuit shown in the figure contains an inductor $L$, a capacitor $C_0$, a resistor $R_0$ and an ideal battery. The circuit also contains two keys $K_1$ and $K_2$. Initially, both the keys are open and there is no charge on the capacitor. At an instant, key $K_1$ is closed and immediately after this the current in $R_0$ is found to be $I_1$. After a long time, the current attains a steady state value $I_2$. Thereafter, $K_2$ is closed and simultaneously $K_1$ is opened and the voltage across $C_0$ oscillates with amplitude $V_0$ and angular frequency $\omega_0$. Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
P $\rightarrow$ 1; Q $\rightarrow$ 3; R $\rightarrow$ 2; S $\rightarrow$ 5
P $\rightarrow$ 1; Q $\rightarrow$ 2; R $\rightarrow$ 3; S $\rightarrow$ 5
P $\rightarrow$ 1; Q $\rightarrow$ 3; R $\rightarrow$ 2; S $\rightarrow$ 4
P $\rightarrow$ 2; Q $\rightarrow$ 5; R $\rightarrow$ 3; S $\rightarrow$ 4
Answer: (a)
Solution
Just after closing $K_1$, $i_1 = 0$. After a long time, $I_2 = \frac{20}{5} = 4 \, \mathrm{A}$. After opening $K_1$ and closing $K_2$, $$\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{25 \times 10^{-3} \times 10 \times 10^{-6}}}$$ $$\omega = 2000 \, \mathrm{rad/s}$$ $$\omega = 2 \, \mathrm{krad/s}$$ $i_0 = 4$, $$Q_0 = \frac{i_0}{\omega} = \frac{4}{2 \times 10^3} = 2 \, \mathrm{mC}$$ $$\therefore V_0 = \frac{Q_0}{C} = \frac{2 \times 10^3}{10} = 200$$
Chemistry
Question 35
Chemistry · States of Matter · Single correct
A closed vessel contains 10 g of an ideal gas X at 300 K, which exerts 2 atm pressure. At the same temperature, 80 g of another ideal gas Y is added to it and the pressure becomes 6 atm. The ratio of root mean square velocities of X and Y at 300 K is
2$\sqrt{2}$ : $\sqrt{3}$
2$\sqrt{2}$ : 1
1 : 2
2 : 1
Answer: (d)
Solution
Given, $W_X = 10 \, \mathrm{g}$ $P_X = 2 \, \mathrm{atm}$ $W_Y = 80 \, \mathrm{g}$ $P_Y = P_{total} - P_X$ $$\Rightarrow 6 - 2 = 4 \, \mathrm{atm}$$ As $V_{rms} = \sqrt{\frac{3RT}{M}}$, $$\frac{(V_{rms})_X}{(V_{rms})_Y} = \sqrt{\frac{M_Y}{M_X}}$$ As we know, $$PV = nRT$$ Volume and temperature remains same. $$P_X V = \frac{W_X}{M_X} RT$$ $$P_Y V = \frac{W_Y}{M_Y} RT$$ $$M_X \propto \frac{W_X}{P_X}$$ $$M_Y \propto \frac{W_Y}{P_Y}$$
Question 36
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
At room temperature, disproportionation of an aqueous solution of ${in}$ ${situ}$ generated nitrous acid $(HNO_2)$ gives the species
$\mathrm{H_3O^+}$, $\mathrm{NO_3^-}$ and NO
$\mathrm{H_3O^+}$, $\mathrm{NO_3^-}$ and NO_2
$\mathrm{H_3O^+}$, $\mathrm{NO^-}$ and NO_2
$\mathrm{H_3O^+}$, $\mathrm{NO_3^-}$ and N_2O
Answer: (a)
Solution
The reaction is given by: $$3\mathrm{HNO_2} \rightleftharpoons \mathrm{HNO_3} + 2\mathrm{NO} + \mathrm{H_2O}$$ The oxidation states are indicated as follows: $$+3 +5 +1$$ The products formed are $\mathrm{H_3O^+}$, $\mathrm{NO_3^-}$, and $\mathrm{NO}$.
Question 37
Chemistry · Biomolecules · Single correct
Aspartame, an artificial sweetener, is a dipeptide aspartyl phenylalanine methyl ester. The structure of aspartame is
Answer: (b)
Solution
Aspartame is
Question 38
Chemistry · Co-ordination Compounds · Single correct
Among the following options, select the option in which each complex in Set-I shows geometrical isomerism and the two complexes in Set-II are ionization isomers of each other. $[\mathrm{en} = \mathrm{H_2NCH_2CH_2NH_2}]$
\textbf{Set-I:} $[\mathrm{Ni(CO)_4}]$ and $[\mathrm{PdCl_2(PPh_3)_2}]$ \hspace{1.5cm}\textbf{Set-II:} $[\mathrm{Co(NH_3)_5Cl}]\mathrm{SO_4}$ and $[\mathrm{Co(NH_3)_5(SO_4)}]\mathrm{Cl}$
\textbf{Set-I:} $[\mathrm{Co(en)(NH_3)_2Cl_2}]$ and $[\mathrm{PdCl_2(PPh_3)_2}]$ \hspace{1.5cm}\textbf{Set-II:} $[\mathrm{Co(NH_3)_6}][\mathrm{Cr(CN)_6}]$ and $[\mathrm{Cr(NH_3)_6}][\mathrm{Co(CN)_6}]$
\textbf{Set-I:} $[\mathrm{Co(NH_3)_3(NO_2)_3}]$ and $[\mathrm{Co(en)_2Cl_2}]$ \hspace{1.5cm}\textbf{Set-II:} $[\mathrm{Co(NH_3)_5Cl}]\mathrm{SO_4}$ and $[\mathrm{Co(NH_3)_5(SO_4)}]\mathrm{Cl}$
\textbf{Set-I:} $[\mathrm{Cr(NH_3)_5Cl}]\mathrm{Cl_2}$ and $[\mathrm{Co(en)(NH_3)_2Cl_2}]$ \hspace{1.5cm}\textbf{Set-II:} $[\mathrm{Cr(H_2O)_6}]\mathrm{Cl_3}$ and $[\mathrm{Cr(H_2O)_5Cl}]\mathrm{Cl_2}\cdot\mathrm{H_2O}$
Answer: (c)
Solution
Sol. Set-I: (Facial) and (Meridional) isomers of $\mathrm{Co(NH_3)_3(NO_2)_3}$. (Cis) and (Trans) isomers of $\mathrm{Co(en)_2Cl_2}$. Set-II: $\mathrm{[Co(NH_3)_5Cl]SO_4}$ and $\mathrm{[Co(NH_3)_5SO_4]Cl}$ are ionisation isomers.
Question 39
Chemistry · Structure of Atom · Multiple correct
Among the following the correct statement(s) for electrons in an atom is(are)
Uncertainty principle rules out the existence of definite paths for electrons.
The energy of an electron in 2s orbital of an atom is lower than the energy of an electron that is infinitely far away from the nucleus.
According to Bohr's model, the most negative energy value for an electron is given by $n = 1$, which corresponds to the most stable orbit.
According to Bohr's model, the magnitude of velocity of electrons increases with increase in values of $n$.
Answer: (a), (b), (c)
Solution
(A) Uncertainty principle rules out existence of definite paths or trajectories of electron and other similar particles. So, option (A) is correct. (B) Shell or orbit more near to nucleus has less energy than faraway. So, option (B) is also correct. (C) $E = -13.6 \frac{Z^2}{n^2} eV/atom$ So, $n = 1$ has most negative energy. So, option (C) is also correct. (D) $V = V_0 \times \frac{Z}{n}$ when $n$ increases velocity decreases. So, option (D) is incorrect.
Question 40
Chemistry · Haloalkanes and Haloarenes · Multiple correct
Reaction of $\textit{iso}$-propylbenzene with $\mathrm{O}_2$ followed by the treatment with $\mathrm{H}_3\mathrm{O}^+$ forms phenol and a by-product $P$. Reaction of $P$ with 3 equivalents of $\mathrm{Cl}_2$ gives compound $Q$. Treatment of $Q$ with $\mathrm{Ca(OH)}_2$ produces compound $R$ and calcium salt $S$. The correct statement(s) regarding $P$, $Q$, $R$ and $S$ is(are)
Reaction of $P$ with $R$ in the presence of KOH followed by acidification gives
Reaction of $R$ with $\mathrm{O}_2$ in the presence of light gives phosgene gas
$Q$ reacts with aqueous NaOH to produce $\mathrm{Cl}_3\mathrm{CCH}_2\mathrm{OH}$ and $\mathrm{Cl}_3\mathrm{CCOONa}$
$S$ on heating gives $P$
Answer: (a), (b), (d)
Solution
Iso-propyl benzene reacts with $\mathrm{O_2}$ to form a hydroperoxide intermediate, which upon acid catalysis ($\mathrm{H_3O^+}$) gives phenol and acetone (P). Acetone (P) reacts with 3 equivalents of $\mathrm{Cl_2}$ to form trichloroacetone (Q). Trichloroacetone (Q) reacts with chloroform (R) and calcium acetate (S) to form the final product. Option (A) shows the reaction of acetone (P) with chloroform (R) in the presence of $\mathrm{KOH}$ and $\mathrm{H^+}$ to form a different product. Option (B) shows the reaction of chloroform (R) with $\mathrm{O_2}$ to form phosgene, which is not the desired reaction. Option (C) shows the reaction of trichloroacetone (Q) with aqueous $\mathrm{NaOH}$ to form a carboxylate and chloroform, which is not the desired reaction. Option (D) shows the reaction of calcium acetate (S) with heat to form acetone, which is not the desired reaction.
Question 41
Chemistry · Chemical Bonding and Molecular Structure · Multiple correct
The option(s) in which at least three molecules follow Octet Rule is(are)
CO_2, C_2H_4, NO and HCl
NO_2, O_3, HCl and H_2SO_4
BCl_3, NO, NO_2 and H_2SO_4
CO_2, BCl_3, O_3 and C_2H_4
Answer: (a), (d)
Solution
Sol. (A) $\mathrm{CO_2}$, $\mathrm{C_2H_4}$ and HCl follow octet rule. (B) $\mathrm{O_3}$ and HCl and follow octet rule. (C) None of them follow octet rule. (D) $\mathrm{CO_2}$, $\mathrm{O_3}$ and $\mathrm{C_2H_4}$ follow octet rule. Correct answer is (A) and (D)
Question 42
Chemistry · Thermodynamics · Numerical
Consider the following volume–temperature (V–T) diagram for the expansion of 5 moles of an ideal monoatomic gas. Considering only P-V work is involved, the total change in enthalpy (in Joule) for the transformation of state in the sequence X $\rightarrow$ Y $\rightarrow$ Z is . [Use the given data: Molar heat capacity of the gas for the given temperature range, $C_{v,m} = 12 \, \mathrm{J \, K^{-1} \, mol^{-1}}$ and gas constant, $R = 8.3 \, \mathrm{J \, K^{-1} \, mol^{-1}}$]
Answer: 8120
Solution
X $\rightarrow$ Y is an isothermal process an ideal gas: $\Delta H = 0$ Y $\rightarrow$ Z is an isochoric process $\therefore \, w = 0$ $\Delta U = nC_{V,m} (T_2 - T_1)$ $= 5 \times 12 \,(415 - 335)$ $= 4800 \, \mathrm{J}$ $\Delta H = \Delta U + \Delta (PV)$ $= \Delta U + nR\Delta T$ $= 4800 + 5 \times 8.3 \times (415 - 335)$ $= 8120 \, \mathrm{J}$
Question 43
Chemistry · Chemical Kinetics and Nuclear Chemistry · Fill in the blank
Consider the following reaction, $2\mathrm{H}_2(g) + 2\mathrm{NO}(g) \rightarrow \mathrm{N}_2(g) + 2\mathrm{H}_2\mathrm{O}(g)$ which follows the mechanism given below: $$2\mathrm{NO}(g) \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} \mathrm{N}_2\mathrm{O}_2(g)$$ (fast equilibrium) $$\mathrm{N}_2\mathrm{O}_2(g) + \mathrm{H}_2(g) \xrightarrow{k_2} \mathrm{N}_2\mathrm{O}(g) + \mathrm{H}_2\mathrm{O}(g)$$ (slow reaction) $$\mathrm{N}_2\mathrm{O}(g) + \mathrm{H}_2(g) \xrightarrow{k_3} \mathrm{N}_2(g) + \mathrm{H}_2\mathrm{O}(g)$$ (fast reaction) The order of the reaction is_____?
Answer: 3
Solution
Rate of reaction (according to slowest step) $$r = k_2 [\mathrm{N_2O_2}] [\mathrm{H_2}] \ldots (1)$$ Now for intermediate $[\mathrm{N_2O_2}]$, $$\frac{k_1}{k_{-1}} = \frac{[\mathrm{N_2O_2}]}{[\mathrm{NO}]^2}$$ $$\Rightarrow [\mathrm{N_2O_2}] = \frac{k_1}{k_{-1}} [\mathrm{NO}]^2 \ldots (2)$$ From equation (1) and (2) $$r = \frac{k_2 k_1}{k_{-1}} [\mathrm{NO}]^2 [\mathrm{H_2}]$$ Overall order of reaction = 2 + 1 = 3
Question 44
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical
Complete reaction of acetaldehyde with excess formaldehyde, upon heating with conc. NaOH solution, gives $\textbf{P}$ and $\textbf{Q}$. Compound $\textbf{P}$ does not give Tollens' test, whereas $\textbf{Q}$ on acidification gives positive Tollens' test. Treatment of $\textbf{P}$ with excess cyclohexanone in the presence of catalytic amount of p-toluenesulfonic acid (PTSA) gives product $\textbf{R}$. Sum of the number of methylene groups (–CH$_2$–) and oxygen atoms in $\textbf{R}$ is____.
Answer: 18
Solution
The reaction involves acetaldehyde and formaldehyde in excess with NaOH, resulting in compound (P) and formate ion. Compound (P) does not give Tollen's Test. The formate ion (Q) is protonated to form formic acid, which gives a positive Tollen's Test. Compound (P) reacts with an excess of cyclohexanone in the presence of PTSA to form compound (R). The number of $\mathrm{CH_2}$ groups in $R$ is 14. The number of oxygen atoms is 4. The required answer is $14 + 4 = 18$.
Question 45
Chemistry · Co-ordination Compounds · Numerical
Among $\mathrm{V(CO)_6}$, $\mathrm{Cr(CO)_5}$, $\mathrm{Cu(CO)_3}$, $\mathrm{Mn(CO)_5}$, $\mathrm{Fe(CO)_5}$, $[\mathrm{Co(CO)_3}]^{3-}$, $[\mathrm{Cr(CO)_4}]^{4-}$, and $\mathrm{Ir(CO)_3}$, the total number of species isoelectronic with $\mathrm{Ni(CO)_4}$ is ______. [Given atomic number : $\mathrm{V} = 23$, $\mathrm{Cr} = 24$, $\mathrm{Mn} = 25$, $\mathrm{Fe} = 26$, $\mathrm{Co} = 27$, $\mathrm{Ni} = 28$, $\mathrm{Cu} = 29$, $\mathrm{Ir} = 77$]
Answer: 1
Solution
Total number of electrons in $\mathrm{Ni(CO)_4} = 84$. Species and their total electrons: $\mathrm{V(CO)_6}$: 107 $\mathrm{Cr(CO)_5}$: 94 $\mathrm{Cu(CO)_3}$: 71 $\mathrm{Mn(CO)_5}$: 95 $\mathrm{Fe(CO)_5}$: 96 $[\mathrm{Co(CO)_3}]^{3-}$: 72 $[\mathrm{Cr(CO)_4}]^{4-}$: 84 $\mathrm{Ir(CO)_3}$: 119
Question 46
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical
In the following reaction sequence, the major product $P$ is formed. Glycerol reacts completely with excess $P$ in the presence of an acid catalyst to form $Q$. Reaction of $Q$ with excess NaOH followed by the treatment with CaCl$_2$ yields Ca-soap $R$, quantitatively. Starting with one mole of $Q$, the amount of $R$ produced in gram is _____. [Given, atomic weight: H = 1, C = 12, N = 14, O = 16, Na = 23, Cl = 35, Ca = 40]
Answer: 909
Solution
The reaction sequence starts with the compound $\mathrm{H-C\equiv C-(CH_2)_{15}-COO-Et}$. In the presence of $\mathrm{Hg^{2+}}$ and $\mathrm{H_3O^+}$, it is converted to $\mathrm{CH_3-C-(CH_2)_{15}-C-OEt}$, which upon treatment with $\mathrm{Zn-Hg/HCl}$ gives $\mathrm{CH_3-CH_2-(CH_2)_{15}-C-OEt}$. This compound is then hydrolyzed with $\mathrm{H_3O^+}$ to form $\mathrm{CH_3-(CH_2)_{16}-COOH}$, labeled as (P). Compound (P) reacts with excess $\mathrm{CH_2-OH}$ in the presence of $\mathrm{H^+}$ to form the ester $\mathrm{CH_2-O-C-(CH_2)_{16}-CH_3}$, labeled as (Q). The ester (Q) is then treated with $\mathrm{NaOH}$ in excess to form $\mathrm{3CH_3-(CH_2)_{16}-COO^-Na^+}$, which upon reaction with $\mathrm{CaCl_2}$ gives $\frac{3}{2}(\mathrm{CH_3-(CH_2)_{16}-COO})_2\mathrm{Ca}$, labeled as (R). 1 mole of Q will give 1.5 mole of R. So, mass of R produced = $606 \, \mathrm{g} \times 1.5$ = $909 \, \mathrm{g}$
Question 47
Chemistry · Co-ordination Compounds · Numerical
Among the following complexes, the total number of diamagnetic species is . $[\mathrm{Mn(NH_3)_6}]^{3+}$, $[\mathrm{MnCl_6}]^{3-}$, $[\mathrm{FeF_6}]^{3-}$, $[\mathrm{CoF_6}]^{3-}$, $[\mathrm{Fe(NH_3)_6}]^{3+}$ and $[\mathrm{Co(en)_3}]^{3+}$ [Given, atomic number: Mn = 25, Fe = 26, Co = 27; en = $\mathrm{H_2NCH_2CH_2NH_2}$]
In a conductometric titration, small volume of titrant of higher concentration is added stepwise to a larger volume of titrate of much lower concentration, and the conductance is measured after each addition. The limiting ionic conductivity $(\Lambda_0)$ values (in $\mathrm{mS\ m^2\ mol^{-1}}$) for different ions in aqueous solutions are given below: \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|} \hline Ions & $\mathrm{Ag^+}$ & $\mathrm{K^+}$ & $\mathrm{Na^+}$ & $\mathrm{H^+}$ & $\mathrm{NO_3^-}$ & $\mathrm{Cl^-}$ & $\mathrm{SO_4^{2-}}$ & $\mathrm{OH^-}$ & $\mathrm{CH_3COO^-}$ \\ \hline $\Lambda_0$ & 6.2 & 7.4 & 5.0 & 35.0 & 7.2 & 7.6 & 16.0 & 19.9 & 4.1 \\ \hline \end{tabular} For different combinations of titrates and titrants given in \textbf{List-I}, the graphs of `conductance' versus `volume of titrant' are given in \textbf{List-II}. Match each entry in \textbf{List-I} with the appropriate entry in \textbf{List-II} and choose the correct option.
P-4, Q-3, R-2, S-5
P-2, Q-4, R-3, S-1
P-3, Q-4, R-2, S-5
P-4, Q-3, R-2, S-1
Answer: (c)
Solution
(P) $\mathrm{KCl} + \mathrm{AgNO_3} \longrightarrow \mathrm{AgCl} \downarrow + \mathrm{KNO_3}$ $\mathrm{Cl^-}$ is replaced by $\mathrm{NO_3^-}$. Conductance will first decrease and then after equivalence point, it will increase. $P \longrightarrow 3$ (Q) $\mathrm{AgNO_3} + \mathrm{KCl} \longrightarrow \mathrm{AgCl} + \mathrm{KNO_3}$ $\mathrm{Ag^+}$ is replaced by $\mathrm{K^+}$. Conductance will first increase slightly and then will increase further. (R) $\mathrm{NaOH} + \mathrm{HCl} \longrightarrow \mathrm{NaCl} + \mathrm{H_2O}$ $\mathrm{OH^-}$ is replaced by $\mathrm{Cl^-}$. (S) $\mathrm{NaOH} + \mathrm{CH_3COOH} \longrightarrow \mathrm{CH_3COONa} + \mathrm{H_2O}$ $\mathrm{OH^-}$ is replaced by $\mathrm{CH_3COO^-}$. Conductance will first decrease and then become almost constant due to buffer formation.
Question 49
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Based on VSEPR model, match the xenon compounds given in List-I with the corresponding geometries and the number of lone pairs on xenon given in List-II and choose the correct option.
P-5, Q-2, R-3, S-1
P-5, Q-3, R-2, S-4
P-4, Q-3, R-2, S-1
P-4, Q-2, R-5, S-3
Answer: (b)
Solution
XeF_2: Hybridisation – $sp^3d$. Geometry: Trigonal bipyramidal and three lone pair of electrons. XeF_4: Hybridisation – $sp^3d^2$. Geometry: Octahedral and two lone pair of electrons. XeO_3: Hybridisation - $sp^3$. Geometry: Tetrahedral & one lone pair of electrons. XeO_3F_2: Hybridisation - $sp^3d$. Geometry: Trigonal bipyramidal and no lone pair of electrons. Correct match: P $\rightarrow$ 5; Q $\rightarrow$ 3; R $\rightarrow$ 2; S $\rightarrow$ 4
Question 50
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
List-I contains various reaction sequences and List-II contains the possible products. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.
P-3, Q-5, R-4, S-1
P-3, Q-2, R-4, S-1
P-3, Q-5, R-1, S-4
P-5, Q-2, R-4, S-1
Answer: (a)
Solution
The solution involves several steps of organic reactions. For compound (P): 1. The starting compound undergoes ozonolysis with $\mathrm{O_3}$ and Zn to form an aldehyde. 2. This aldehyde reacts with aqueous $\mathrm{NaOH}$ under heat to form a cyclic compound. 3. The cyclic compound is treated with PTSA and ethylene glycol to form an acetal. 4. Hydroboration-oxidation with $\mathrm{BH_3}$, $\mathrm{H_2O_2}$, and $\mathrm{NaOH}$ converts the acetal to a diol. 5. Acidic hydrolysis with $\mathrm{H_3O^+}$ and reduction with $\mathrm{NaBH_4}$ yields the final product. For compound (Q): 1. The starting compound undergoes ozonolysis with $\mathrm{O_3}$ and Zn to form a diketone. 2. This diketone reacts with aqueous $\mathrm{NaOH}$ under heat to form a cyclic compound. 3. The cyclic compound is treated with PTSA and ethylene glycol to form an acetal. 4. Hydroboration-oxidation with $\mathrm{BH_3}$, $\mathrm{H_2O_2}$, and $\mathrm{NaOH}$ converts the acetal to a diol. 5. Acidic hydrolysis with $\mathrm{H_3O^+}$ and reduction with $\mathrm{NaBH_4}$ yields the final product. For compound (R): 1. The starting compound reacts with ethylene glycol and PTSA to form an acetal. 2. Oxymercuration-demercuration with $\mathrm{Hg(OAc)_2}$ and $\mathrm{H_2O}$, followed by reduction with $\mathrm{NaBH_4}$, yields a diol. 3. Acidic hydrolysis with $\mathrm{H_3O^+}$ yields the final product. For compound (S): 1. The starting compound reacts with ethylene glycol and PTSA to form an acetal. 2. Hydroboration-oxidation with $\mathrm{BH_3}$, $\mathrm{H_2O_2}$, and $\mathrm{NaOH}$ converts the acetal to a diol. 3. Acidic hydrolysis with $\mathrm{H_3O^+}$ and reduction with $\mathrm{NaBH_4}$ yields the final product.
Question 51
Chemistry · Alcohols, Phenols and Ethers · Single correct
List-I contains various reaction sequences and List-II contains different phenolic compounds. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.
P-2, Q-3, R-4, S-5
P-2, Q-3, R-5, S-1
P-3, Q-2, R-4, S-1
P-3, Q-2, R-5, S-4
Answer: (c)
Solution
The solution involves multiple steps of chemical reactions starting from different compounds (P), (Q), (R), and (S). For compound (P): 1. React with molten NaOH and $H_3O^+$. 2. Followed by reaction with conc. $HNO_3$ to form compound (3). For compound (Q): 1. React with conc. $HNO_3$ and conc. $H_2SO_4$. 2. Reduction with Sn/HCl to form an amine. 3. Diazotization with $NaNO_2$/HCl at $0-5^\circC$, followed by hydrolysis to form compound (5). For compound (R): 1. React with conc. $H_2SO_4$. 2. Followed by reaction with conc. $HNO_3$ to form a sulfonated nitro compound. 3. Hydrolysis with $H_3O^+$ and heat to form compound (4). For compound (S): 1. Oxidation with $KMnO_4$/KOH and heat, followed by hydrolysis. 2. Nitration with conc. $HNO_3$, conc. $H_2SO_4$, and heat. 3. Reaction with $SOCl_2$ and $NH_3$. 4. Bromination with $Br_2$ and NaOH. 5. Diazotization with $NaNO_2$/HCl at $0-5^\circC$, followed by hydrolysis (not forming the expected product).