JEE Main 24 February 2021 Shift 2: all 90 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 · Mathematical Reasoning · Single correct
For the statements $p$ and $q$, consider the following compound statements: (a) $\left( \sim q \land (p \to q) \right) \to \sim p$ (b) $\left( (p \lor q) \land \sim P \right) \to P$ Then which of the following statements is correct?
Answer: (c)
a $\&$ b are both tautologies.
Maths · Three Dimensional Geometry · Single correct
Let a, b $\in$ $\mathbb{R}$. If the mirror image of the point P(a, 6, 9) with respect to the line \[ \frac{x-3}{7} = \frac{y-2}{5} = \frac{z-1}{-9} \] is (20, b, -a - 9), then |a + b| is equal to:
Answer: (b)
Given P(a, 6, 9), Q(20, b, -a - 9). The midpoint of PQ is $$\left( \frac{a+20}{2}, \frac{b+6}{2}, -\frac{a}{2} \right)$$. This point lies on the line $$\frac{\frac{a+20}{2} - 3}{7} = \frac{\frac{b+6}{2} - 2}{5} = \frac{-\frac{a}{2} - 1}{-9}$$. Simplifying, $$\frac{a+20 - 6}{14} = \frac{b+6 - 4}{10} = \frac{-a - 2}{-18}$$. Further simplification gives $$\frac{a+14}{14} = \frac{a+2}{18}$$. Solving, $$18a + 252 = 14a + 28$$. Therefore, $$4a = -224$$, which gives $$a = -56$$. Now, $$\frac{b+2}{10} = \frac{a+2}{18}$$. Substituting $$a = -56$$, $$\frac{b+2}{10} = \frac{-54}{18}$$. Solving, $$\frac{b+2}{10} = -3 \Rightarrow b = -32$$. Finally, $$|a + b| = |-56 - 32| = 88$$.
Maths · Three Dimensional Geometry · Single correct
The vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $\vec{r} \cdot (\hat{i} - 2\hat{j}) = -2$, and the point $(1,0,2)$ is:
Answer: (b)
Plane passing through intersection of plane is $\{ \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 1 \} + \lambda \{ \vec{r} \cdot (\hat{i} - 2\hat{j}) + 2 \} = 0$. Passes through $\hat{i} + 2\hat{k}$, we get $$(3 - 1) + \lambda (1 + 2) = 0 \implies \lambda = -\frac{2}{3}$$ Hence, equation of plane is $$3 \{ \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 1 \} - 2 \{ \vec{r} \cdot (\hat{i} - 2\hat{j}) + 2 \} = 0$$ $$\Rightarrow \vec{r} \cdot (\hat{i} + 7\hat{j} + 3\hat{k}) = 7$$
Maths · Conic Sections · Single correct
If $P$ is a point on the parabola $y = x^2 + 4$ which is closest to the straight line $y = 4x - 1$, then the co-ordinates of $P$ are:
Answer: (d)
Given $\($ $\frac{dy}{dx}$ $\bigg$|_P = 4 $\)$. Therefore, $\($ 2x_1 = 4 $\)$. This implies $\($ x_1 = 2 $\)$. Therefore, the point will be $\($(2, 8)$\)$.
Maths · Heights and Distances · Single correct
The angle of elevation of a jet plane from a point A on the ground is $60^\circ$. After a flight of 20 seconds at the speed of $432 \, \mathrm{km/hour}$, the angle of elevation changes to $30^\circ$. If the jet plane is flying at a constant height, then its height is:
Answer: (a)
Given $v = 432 \times \frac{1000}{60 \times 60} \, \mathrm{m/sec} = 120 \, \mathrm{m/sec}$. Distance $AB = v \times 20 = 2400 \, \mathrm{meter}$. In $\triangle PAC$, $$\tan 60^\circ = \frac{h}{PC} \Rightarrow PC = \frac{h}{\sqrt{3}}$$ In $\triangle PBD$, $$\tan 30^\circ = \frac{h}{PD} \Rightarrow PD = \sqrt{3} \, h$$ $PD = PC + CD$ $$\sqrt{3}h = \frac{h}{\sqrt{3}} + 2400 \Rightarrow \frac{2h}{\sqrt{3}} = 2400$$ $$h = 1200 \sqrt{3} \, \mathrm{meter}$$
Maths · Sequences and Series · Single correct
If $n \geq 2$ is a positive integer, then the sum of the series $$\binom{n+1}{2} + 2 \left( \binom{2}{2} + \binom{3}{2} + \binom{4}{2} + \ldots + \binom{n}{2} \right)$$ is:
Answer: (c)
Given $^2C_2 = ^3C_3$. $S = ^3C_3 + ^3C_2 + \ldots + ^nC_2 = ^{n+1}C_3$. Therefore, $^nC_r + ^nC_{r-1} = ^{n+1}C_r$. Thus, $^{n+1}C_2 + ^{n+1}C_3 + ^{n+1}C_3 = ^{n+2}C_3$. $$= \frac{(n+1)!}{3!(n-1)!} + \frac{(n+1)!}{3!(n-2)!}$$ $$= \frac{(n+2)(n+1)n}{6} + \frac{(n+1)(n)(n-1)}{6}$$ $$= \frac{n(n+1)(2n+1)}{6}$$
Maths · Applications of Derivatives · Single correct
Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be defined as $$ f(x) = \begin{cases} -55x & \text{if } x 4 \end{cases} $$ Let $A = \{x \in \mathbb{R} : f \text{ is increasing}\}$. Then $A$ is equal to:
Answer: (a)
Given $$f(x) = \begin{cases} -55 & ; x 4 \end{cases}$$ and $$f(x) = \begin{cases} -55 & ; x 4 \end{cases}$$ Hence, $f(x)$ is monotonically increasing in interval $(-5,-4) \cup (4, \infty)$.
Maths · Differential Equations · Single correct
Let $f$ be a twice differentiable function defined on $\mathbb{R}$ such that $f(0) = 1$, $f'(0) = 2$ and $f'(x) \neq 0$ for all $x \in \mathbb{R}$. If $$\begin{vmatrix} f(x) & f'(x) \\ f'(x) & f''(x) \end{vmatrix} = 0$$, for all $x \in \mathbb{R}$ then the value of $f(1)$ lies in the interval:
Answer: (b)
Given $f(x)f''(X) - (f'(x))^2 = 0$. Let $h(x) = \frac{f(x)}{f'(x)}$. Therefore, $h'(x) = 0$ implies $h(x) = k$. This gives $\frac{f(x)}{f'(x)} = k$ which implies $f(x) = kf'(x)$. Thus, $f(0) = kf'(0)$ implies $1 = k(2)$, leading to $k = \frac{1}{2}$. Now $f(x) = \frac{1}{2} f'(x)$ implies $\int 2 \, dx = \int \frac{f'(x)}{f(x)} \, dx$. Therefore, $2x = \ln |f(x)| + C$. As $f(0) = 1$, it follows that $C = 0$. Hence, $2x = \ln |f(X)|$ implies $f(x) = \pm e^{2x}$. As $f(0) = 1$, it follows that $f(x) = e^{2x}$, leading to $f(1) = e^2$.
Maths · Applications of Derivatives · Single correct
For which of the following curves, the line $x + \sqrt{3}y = 2\sqrt{3}$ is the tangent at the point $\left( \frac{3\sqrt{3}}{2}, \frac{1}{2} \right)$?
Answer: (a)
Tangent to $x^2 + 9y^2 = 9$ at point $\left( \frac{3\sqrt{3}}{2}, \frac{1}{2} \right)$ is $\frac{3\sqrt{3}}{2} x + 9y \left( \frac{1}{2} \right) = 9$. $$3\sqrt{3}x + 9y = 18 \Rightarrow x + \sqrt{3}y = 2\sqrt{3}$$ Therefore, option (1) is true.
Maths · Integrals · Single correct
The value of the integral, $$\int_{1}^{3} \left[ x^2 - 2x - 2 \right] \, dx$$, where $[x]$ denotes the greatest integer less than or equal to $x$, is:
Answer: (c)
Given $$I = \int_1^3 3 \, dx + \int_1^3 [(x-1)^2] \, dx$$ Put $x - 1 = t$; $dx = dt$ $$I = (-6) + \int_0^2 [t^2] \, dt$$ $$I = -6 + \int_0^1 0 \, dt + \int_1^{\sqrt{2}} 1 \, dt + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dt + \int_{\sqrt{3}}^2 3 \, dt$$ $$I = -6 + (\sqrt{2} - 1) + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3}$$ $$I = -1 - \sqrt{2} - \sqrt{3}$$
Maths · Inverse Trigonometric Functions · Single correct
A possible value of $\tan \left( \frac{1}{4} \sin^{-1} \frac{\sqrt{63}}{8} \right)$ is:
Answer: (b)
Given $\tan \left( \frac{1}{4} \sin^{-1} \frac{\sqrt{63}}{8} \right)$. Let $\sin^{-1} \left( \frac{\sqrt{63}}{8} \right) = \theta$ so $\sin \theta = \frac{\sqrt{63}}{8}$. From the triangle, $\cos \theta = \frac{1}{8}$. Using the double angle formula, $2 \cos^2 \frac{\theta}{2} - 1 = \frac{1}{8}$. Solving for $\cos^2 \frac{\theta}{2}$, we have: $$\cos^2 \frac{\theta}{2} = \frac{9}{16}$$ Thus, $\cos \frac{\theta}{2} = \frac{3}{4}$. Using the identity: $$\frac{1 - \tan^2 \frac{\theta}{4}}{1 + \tan^2 \frac{\theta}{4}} = \frac{3}{4}$$ Solving for $\tan \frac{\theta}{4}$, we find: $$\tan \frac{\theta}{4} = \frac{1}{\sqrt{7}}$$
Maths · Mathematical Reasoning · Single correct
The negation of the statement $\sim p \land (p \lor q)$ is:
Answer: (d)
The truth table is constructed as follows: Therefore, $\sim p \land (p \lor q) \equiv p \lor \sim q$.
Maths · Applications of Derivatives · Single correct
If the curve $y = ax^2 + bx + c, x \in \mathbb{R}$, passes through the point (1,2) and the tangent line to this curve at origin is $y = x$, then the possible values of $a, b, c$ are:
Answer: (a)
Given the equation $2 = a + b + c \ldots (i)$. The derivative is $\frac{dy}{dx} = 2ax + b$. At the point $(0,0)$, $\frac{dy}{dx}\bigg|_{(0,0)} = 1$. This implies $b = 1$ and therefore $a + c = 1$. Since $(0,0)$ lies on the curve, it follows that $c = 0$ and $a = 1$.
Maths · Applications of Integrals · Single correct
The area of the region : $R = \{(x, y) : 5x^2 \leq y \leq 2x^2 + 9\}$ is:
Answer: (b)
Required area $$= 2 \int_0^{\sqrt{3}} (2x^2 + 9 - 5x^2) \, dx$$ $$= 2 \int_0^{\sqrt{3}} (9 - 3x^2) \, dx$$ $$= 2 \left[ 9x - x^3 \right]_0^{\sqrt{3}} = 12\sqrt{3}$$
Maths · Differential Equations · Single correct
If a curve $y = f(x)$ passes through the point $(1,2)$ and satisfies $x \frac{dy}{dx} + y = bx^4$, then for what value of $b$, $\int_1^2 f(x) \, dx = \frac{62}{5}$?
Answer: (d)
\[ \frac{dy}{dx}+\frac{y}{x}=bx^3 \] \[ \text{I.F.}=e^{\int \frac{dx}{x}}=x \] \[ \therefore\quad yx=\int bx^4\,dx=\frac{bx^5}{5}+c \] Passes through \(1,2\), we get \[ 2=\frac{b}{5}+c \qquad (1) \] Also, \[ \int_{1}^{2}\left(\frac{bx^4}{5}+\frac{c}{x}\right)\,dx=\frac{62}{5} \] \[ \Rightarrow \frac{b}{25}\times 32 +c\ln 2 -\frac{b}{25} =\frac{62}{5} \] \[ \Rightarrow \frac{31b}{25}+c\ln 2=\frac{62}{5} \] \[ \Rightarrow c=0,\quad b=10. \]
Maths · Integrals · Single correct
Let $f(x)$ be a differentiable function defined on $[0,2]$ such that $f'(x) = f'(2-x)$ for all $x \in (0,2)$, $f(0) = 1$ and $f(2) = e^2$. Then the value of $\int_0^2 f(x) \, dx$ is:
Answer: (a)
Given $f'(x) = f'(2-x)$. On integrating both sides, $f(x) = -f(2-x) + c$. Put $x = 0$, $f(0) + f(2) = c \Rightarrow c = 1 + e^2$. Therefore, $f(x) + f(2-x) = 1 + e^2 \ldots (i)$. $I = \int_0^2 f(x) \, dx = \int_0^1 \{ f(x) + f(2-x) \} \, dx = (1 + e^2)$
Maths · Matrices · Single correct
Let A and B be $3 \times 3$ real matrices such that A is symmetric matrix and B is skew-symmetric matrix. Then the system of linear equations $$\left( A^2 B^2 - B^2 A^2 \right) X = O$$, where X is a $3 \times 1$ column matrix of unknown variables and O is a $3 \times 1$ null matrix, has:
Answer: (c)
Given $A^\top = A$, $B^\top = -B$. Let $A^2B^2 - B^2A^2 = P$. Then, $$P^\top = \left( A^2 B^2 - B^2 A^2 \right)^\top = \left( A^2 B^2 \right)^\top - \left( B^2 A^2 \right)^\top$$ $$= \left( B^2 \right)^\top \left( A^2 \right)^\top - \left( A^2 \right)^\top \left( B^2 \right)^\top$$ $$= B^2 A^2 - A^2 B^2$$ Therefore, $P$ is a skew-symmetric matrix. $$\begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ Thus, $$ay + bz = 0$$ $$-ax + cz = 0$$ $$-bx - cy = 0$$ From equations 1, 2, and 3, $$\Delta = 0 \& \Delta_1 = \Delta_2 = \Delta_3 = 0$$ Therefore, the equation has an infinite number of solutions.
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $a, b, c$ be in arithmetic progression. Let the centroid of the triangle with vertices $(a, c)$, $(2, b)$ and $(a, b)$ be $\left(\dfrac{10}{3}, \dfrac{7}{3}\right)$. If $\alpha, \beta$ are the roots of the equation $ax^2 + bx + 1 = 0$, then the value of $\alpha^2 + \beta^2 - \alpha\beta$ is:
Answer: (d)
$2b = a + c$ $\dfrac{2a + 2}{3} = \dfrac{10}{3}$ and $\dfrac{2b + c}{3} = \dfrac{7}{3}$ $a = 4,\quad \begin{cases} 2b + c = 7\\ 2b - c = 4 \end{cases}$, solving $b = \dfrac{11}{4}$ $c = \dfrac{3}{2}$ $\therefore$ Quadratic Equation is $4x^2 + \dfrac{11}{4}x + 1 = 0$ $\therefore$ The value of $(\alpha + \beta)^2 - 3\alpha\beta = \dfrac{121}{256} - \dfrac{3}{4} = -\dfrac{71}{256}$
Maths · Determinants · Single correct
For the system of linear equations: $$x - 2y = 1, \; x - y + kz = -2, \; ky + 4z = 6, \; k \in \mathbb{R}$$ consider the following statements: (A) The system has unique solution if $k \neq 2, \; k \neq -2$. (B) The system has unique solution if $k = -2$. (C) The system has unique solution if $k = 2$. (D) The system has no-solution if $k = 2$. (E) The system has infinite number of solutions if $k \neq -2$. Which of the following statements are correct?
Answer: (c)
Given the system of equations: $$x - 2y + 0 \cdot z = 1$$ $$x - y + kz = -2$$ $$0 \cdot x + ky + 4z = 6$$ The determinant is: $$\Delta = \begin{vmatrix} 1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4 \end{vmatrix} = 4 - k^2$$ For a unique solution, we require $4 - k^2 \neq 0$. Thus, $k \neq \pm 2$. For $k = 2$: $$x - 2y + 0 \cdot z = 1$$ $$x - y + 2z = -2$$ $$0 \cdot x + 2y + 4z = 6$$ The determinant $\Delta x$ is: $$\Delta x = \begin{vmatrix} 1 & -2 & 0 \\ -2 & -1 & 2 \\ 6 & 2 & 4 \end{vmatrix} = (-8) + 2[-20]$$ $$\Delta x = -48 \neq 0$$ For $k = 2$, $\Delta x \neq 0$. For $k = 2$, the system has no solution.
Maths · Probability · Single correct
The probability that two randomly selected subsets of the set {1,2,3,4,5} have exactly two elements in their intersection, is:
Answer: (b)
Required probability $$= \frac{{^5C_2 \times 3^3}}{{4^5}}$$ $$= \frac{{10 \times 27}}{{2^{10}}} = \frac{{135}}{{2^9}}$$
Maths · Binomial Theorem · Numerical
For integers $n$ and $r$, let $\binom{n}{r} = {}^nC_r$, if $n \geq r \geq 0$, and $\binom{n}{r} = 0$, otherwise. The maximum value of $k$ for which the sum $\sum_{i=0}^{k}\binom{10}{i}\binom{15}{k-i}$ $+\sum_{i=0}^{k+1}\binom{12}{i}\binom{13}{k+1-i}$ exists, is equal to ________.
Answer: 12
Given $$(1 + x)^{10} = \binom{10}{0} + \binom{10}{1}x + \binom{10}{2}x^2 + \ldots + \binom{10}{10}x^{10}$$ $$(1 + x)^{15} = \binom{15}{0} + \binom{15}{1}x + \ldots \binom{15}{k-1}x^{k-1} + \binom{15}{k}x^k + \binom{15}{k+1}x^{k+1} + \ldots + \binom{15}{15}x^{15}$$ $$\sum_{i=0}^{k} \binom{10}{i} \left(15C_{k-i}\right) = \binom{10}{0} \cdot 15C_k + \binom{10}{1} \cdot 15C_{k-1} + \ldots + \binom{10}{k} \cdot 15C_0$$ Coefficient of $x^k$ in $(1 + x)^{25}$ $$= \binom{25}{C_k}$$ $$\sum_{i=0}^{k+1} \binom{12}{i} \left(13C_{k+1-i}\right) = \binom{12}{0} \cdot 13C_{k+1} + \binom{12}{1} \cdot 13C_k + \ldots + \binom{12}{k+1} \cdot 13C_0$$ Coefficient of $x^{k+1}$ in $(1 + x)^{25}$ $$= \binom{25}{C_{k+1}}$$ $$25C_k + 25C_{k+1} = 26C_{k+1}$$ For maximum value $$k + 1 = 13$$ $$K = 12$$
Maths · Three Dimensional Geometry · Numerical
Let $\lambda$ be an integer. If the shortest distance between the lines $x - \lambda = 2y - 1 = -2z$ and $x = y + 2\lambda = z - \lambda$ is $\frac{\sqrt{7}}{2\sqrt{2}}$, then the value of $|\lambda|$ is
Answer: 1
Given the equations $\frac{x-\lambda}{1}=\frac{y-\frac{1}{2}}{1}=\frac{z}{-\frac{1}{2}}$ and $\frac{x-\lambda}{2}=\frac{y-\frac{1}{2}}{1}=\frac{z}{-1}$ ...(1) The point on the line is $(\lambda,\frac{1}{2},0)$. Another point on the line is $(0,-2\lambda,\lambda)$. The distance between skew lines is given by $\frac{\left[\overrightarrow{a_2-a_1}\ \overrightarrow{b_1}\ \overrightarrow{b_2}\right]}{\left|\overrightarrow{b_1\times b_2}\right|}$ Substituting the values, $\left|\begin{matrix} \lambda & \frac{1}{2}+2\lambda & -\lambda \\ 2 & 1 & -1 \\ 1 & 1 & 1 \end{matrix}\right|$ and $\left|\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 1 & 1 \end{matrix}\right|$ This results in $\frac{\left|-5\lambda-\frac{3}{2}\right|}{\sqrt{14}}=\frac{\sqrt{7}}{2\sqrt{2}}$ (given) Solving, $|10\lambda+3|=7$ $\Rightarrow \lambda=-1$ Thus, $|\lambda|=1$.
Maths · Relations and Functions · Numerical
If $a + \alpha = 1$, $b + \beta = 2$ and $af(x) + \alpha f\left(\frac{1}{x}\right) = bx + \frac{\beta}{x}$, $x \neq 0$, then the value of the expression $$\frac{f(x) + f\left(\frac{1}{x}\right)}{x + \frac{1}{x}}$$ is
Answer: 2
Given $af(x) + \alpha f\left(\frac{1}{x}\right) = bx + \frac{\beta}{x}$. As $x \to \frac{1}{x}$, $af\left(\frac{1}{x}\right) + af(x) = \frac{b}{x} + \beta x$. Combining (i) and (ii), $$(a + \alpha) \left[f(x) + f\left(\frac{1}{x}\right)\right] = \left(x + \frac{1}{x}\right)(b + \beta)$$ $$\frac{f(x) + f\left(\frac{1}{x}\right)}{x + \frac{1}{x}} = \frac{2}{1} = 2$$
Maths · Straight Lines and Pair of Straight Lines · Numerical
Let a point P be such that its distance from the point (5,0) is thrice the distance of P from the point (-5,0). If the locus of the point P is a circle of radius r, then $4r^2$ is equal to (Round off to the nearest integer)
Answer: 56
Let $P(h,k)$ Given $PA=3\,PB$ $PA^2=9\,PB^2$ $\Rightarrow (h-5)^2+k^2=9\left[(h+5)^2+k^2\right]$ $\Rightarrow 8h^2+8k^2+100h+200=0$ $\therefore$ Locus $x^2+y^2+\left(\frac{25}{2}\right)x+25=0$ $\therefore\ c=\left(-\frac{25}{4},0\right)$ $\therefore\ r^2=\left(-\frac{25}{4}\right)^2-25$ $\therefore\ r^2=\frac{625}{16}-25$ $\therefore\ r^2=\frac{225}{16}$ $\therefore\ 4r^2=4\times\frac{225}{16}=\frac{225}{4}=56.25$
Maths · Applications of Derivatives · Numerical
If the area of the triangle formed by the positive $x$-axis, the normal and the tangent to the circle $(x-2)^2 + (y-3)^2 = 25$ at the point $(5,7)$ is $A$, then $24A$ is equal to
Answer: 1225
Equation of normal at P $$(y - 7) = \left( \frac{7 - 3}{5 - 2} \right)(x - 5)$$ $$3y - 21 = 4x - 20$$ $$\Rightarrow 4x - 3y + 1 = 0 \ldots (1)$$ $$\Rightarrow M \left( -\frac{1}{4}, 0 \right)$$ Equation of tangent at P $$(y - 7) = -\frac{3}{4}(x - 5)$$ $$4y - 28 = -3x + 15$$ $$\Rightarrow 3x + 4y = 43 \ldots \ldots (ii)$$ $$\Rightarrow N \left( \frac{43}{3}, 0 \right)$$ Hence area of $\triangle PMN = \frac{1}{2} \times \mathrm{MN} \times 7$ $$\lambda = \frac{1}{2} \times \frac{175}{12} \times 7$$ $$\Rightarrow 24 \lambda = 1225$$
Maths · Statistics · Numerical
If the variance of 10 natural numbers 1, 1, 1, $\ldots$, 1, $k$ is less than 10, then the maximum possible value of $k$ is
Answer: 11
Given $$\sigma^2 = \frac{\Sigma X^2}{n} - \left(\frac{\Sigma x}{n}\right)^2$$ We have $$\sigma^2 = \frac{(9+k^2)}{10} - \left(\frac{9+k}{10}\right)^2 < 10$$ Expanding and simplifying, we get: $$(90 + k^2) \cdot 10 - (81 + k^2 + 8k) < 1000$$ Simplifying further: $$90 + 10k^2 - k^2 - 18k - 81 < 1000$$ This reduces to: $$9k^2 - 18k + 9 < 1000$$ Rewriting, we have: $$(k - 1)^2 < \frac{1000}{9} \implies k - 1 < \frac{10\sqrt{10}}{3}$$ Thus, $$k < \frac{10\sqrt{10}}{3} + 1$$ The maximum integral value of $k$ is 11.
Maths · Sequences and Series · Numerical
The sum of first four terms of a geometric progression (G.P.) is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18}$. If the product of first three terms of the G.P. is 1, and the third term is $\alpha$, then $2\alpha$ is
Answer: 3
Given the sequence $a, ar, ar^2, ar^3$. The sum is given by: $$a + ar + ar^2 + ar^3 = \frac{65}{12} \ldots (1)$$ The reciprocal sum is: $$\frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} = \frac{65}{18}$$ This can be rewritten as: $$\frac{1}{a} \left( \frac{r^3 + r^2 + r + 1}{r^3} \right) = \frac{65}{18} \ldots (2)$$ From equations $(i)$ and $(ii)$, we have: $$a^2 r^3 = \frac{18}{12} = \frac{3}{2}$$ Solving for $a$: $$a^3 r^3 = 1 \Rightarrow a \left( \frac{3}{2} \right) = 1 \Rightarrow a = \frac{2}{3}$$ Solving for $r$: $$\frac{4}{9} r^3 = \frac{3}{2} \Rightarrow r^3 = \frac{3^3}{2^3} \Rightarrow r = \frac{3}{2}$$ Now, calculate $\alpha$: $$\alpha = ar^2 = \frac{2}{3} \cdot \left( \frac{3}{2} \right)^2 = \frac{3}{2}$$ Finally, we have: $$2\alpha = 3$$
Maths · Permutations and Combinations · Numerical
The students $S_1, S_2, \ldots, S_{10}$ are to be divided into 3 groups $A, B$ and $C$ such that each group has at least one student and the group $C$ has at most 3 students. Then the total number of possibilities of forming such groups is
Answer: 31650
The solution is as follows: $$= \binom{10}{1} [2^9 - 2] + \binom{10}{2} [2^8 - 2] + \binom{10}{3} [2^7 - 2]$$ $$= 2^7 \left[ \binom{10}{1} \times 4 + \binom{10}{2} \times 2 + \binom{10}{3} \right] - 20 - 90 - 240$$ $$= 128[40 + 90 + 120] - 350$$ $$= (128 \times 250) - 350$$ $$= 10[3165] = 31650$$
Maths · Complex Numbers and Quadratic Equations · Numerical
Let $i=\sqrt{-1}$. If $k=\frac{(-1+i\sqrt{3})^{21}}{(1-i)^{24}} +\frac{(1+i\sqrt{3})^{21}}{(1+i)^{24}}$, and $n=[|k|]$ be the greatest integral part of $|k|$. Then $\sum_{j=0}^{n+5}(j+5)^2-\sum_{j=0}^{n+5}(j+5)$ is equal to
Answer: 310
Given $$\left(2e^{\frac{2\pi}{3}i}\right)^{21} + \left(2e^{\frac{\pi}{3}i}\right)^{21}$$ over $$\left(\sqrt{2}e^{-i\frac{\pi}{4}}\right)^{24} + \left(\sqrt{2}e^{i\frac{\pi}{4}}\right)^{24}$$. This simplifies to $$\frac{2^{21}e^{i14\pi}}{2^{12}e^{-i6\pi}} + \frac{2^{21}(e^{i7\pi})}{2^{12}(e^{i6\pi})}$$ which further simplifies to $$2^9 e^{i(20\pi)} + 2^9 e^{i\pi}$$. This results in $$2^9 + 2^9(-1) = 0$$. Therefore, $$n = 0$$. Next, consider $$\sum_{j=0}^{5}(j+5)^2 - \sum_{j=0}^{5}(j+5)$$. This expands to $$[5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2] - [5 + 6 + 7 + 8 + 9 + 10]$$. Simplifying further, we have $$[(1^2 + 2^2 + \ldots + 10^2) - (1^2 + 2^2 + 3^2 + 4^2)] - [(1 + 2 + 3 + \ldots + 10) - (1 + 2 + 3 + 4)]$$. This results in $$(385 - 30) - [55 - 10]$$ which simplifies to $$355 - 45 \Rightarrow 310$$ ans.
Maths · Complex Numbers and Quadratic Equations · Numerical
The number of the real roots of the equation $(x + 1)^2 + |x - 5| = \frac{27}{4}$ is
Answer: 2
$x \ge 5$ $(x + 1)^2 + (x - 5) = \dfrac{27}{4}$ $\Rightarrow x^2 + 3x - 4 = \dfrac{27}{4}$ $\Rightarrow x^2 + 3x - \dfrac{43}{4} = 0$ $\Rightarrow 4x^2 + 12x - 43 = 0$ $x = \dfrac{-12 \pm \sqrt{144 + 688}}{8}$ $x = \dfrac{-12 \pm \sqrt{832}}{8} = \dfrac{-12 \pm 28.8}{8}$ $= \dfrac{-3 \pm 7.2}{2}$ $= \dfrac{-3 + 7.2}{2},\ \dfrac{-3 - 7.2}{2}$ (Therefore no solution) For $x \le 5$ $(x + 1)^2 - (x - 5) = \dfrac{27}{4}$ $x^2 + x + 6 - \dfrac{27}{4} = 0$ $4x^2 + 4x - 3 = 0$ $x = \dfrac{-4 \pm \sqrt{16 + 48}}{8}$ $x = \dfrac{-4 \pm 8}{8} \Rightarrow x = -\dfrac{12}{8},\ \dfrac{4}{8}$ $\therefore$ 2 Real Root's
Physics · Oscillations · Single correct
When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is :
Answer: (a)
We know that is SHM: $$V = \omega \sqrt{A^2 - x^2}$$ elliptical
Physics · Electric Charges and Fields · Single correct
Two electrons each are fixed at a distance '2d'. A third charge proton placed at the midpoint is displaced slightly by a distance $x (x << d)$ perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency: (m = mass of charged particle)
Answer: (a)
Restoring force on proton: $$F_r = \frac{2Kq^2y}{(d^2+y^2)^{\frac{3}{2}}}$$ For $y \ll d$ $$F_r = \frac{2kq^2y}{d^3} = \frac{q^2y}{2\pi \varepsilon_0 d^3} = ky$$ $$K = \frac{q^2}{2\pi \varepsilon_0 d^3}$$ Angular Frequency $\omega = \sqrt{\frac{k}{m}}$ $$\omega = \sqrt{\frac{q^2}{2\pi \varepsilon_0 md^3}}$$
Physics · Kinetic Theory · Single correct
On the basis of kinetic theory of gases, the gas exerts pressure because its molecules:
Answer: (a)
On the basis of kinetic theory of gases, the gas pressure is due to the molecules suffering change in momentum when impinge on the walls of container.
Physics · Magnetism and Matter · Single correct
A soft ferromagnetic material is placed in an external magnetic field. The magnetic domains:
Answer: (b)
Atoms of ferromagnetic material in unmagnetised state form domains inside the ferromagnetic material. These domains have large magnetic moment of atoms. In the absence of magnetic field, these domains have magnetic moment in different directions. But when the magnetic field is applied, domains aligned in the direction of the field grow in size and those aligned in the direction opposite to the field reduce in size and also its orientation changes.
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
The logic circuit shown above is equivalent to:
Answer: (b)
The circuit diagram shows an OR gate with inputs A and B. Input B is inverted before entering the OR gate. The output of the OR gate is then inverted to produce the final output C. The expression for the output C is given by: $$C = \overline{\overline{A} + \overline{B}}$$ This simplifies to: $$C = \overline{A} \cdot B$$
Physics · Mathematics in Physics · Single correct
The period of oscillation of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$. Measured value of ' $L$ ' is $1.0 \, \mathrm{m}$ from meter scale having a minimum division of $1 \, \mathrm{mm}$ and time of one complete oscillation is $1.95 \, \mathrm{s}$ measured from stopwatch of $0.01 \, \mathrm{s}$ resolution. The percentage error in the determination of ' $g$ ' will be :
Answer: (c)
Given the formula for the period $T$ of a pendulum: $$T = 2\pi \sqrt{\frac{\ell}{g}}$$ Squaring both sides, we have: $$T^2 = 4\pi^2 \left[ \frac{\ell}{g} \right]$$ Solving for $g$, we get: $$g = 4\pi^2 \left[ \frac{\ell}{T^2} \right]$$ The relative error in $g$ is given by: $$\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + \frac{2 \Delta T}{T}$$ Substituting the given values, we find: $$= \left[ \frac{1 \, \mathrm{mm}}{1 \, \mathrm{m}} + \frac{2(10 \times 10^{-3})}{1.95} \right] \times 100$$ This results in: $$= 1.13\%$$
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
Given below are two statements: Statement I: PN junction diodes can be used to function as transistor, simply by connecting two diodes, back to back, which acts as the base terminal. Statement II: In the study of transistor, the amplification factor $\beta$ indicates ratio of the collector current to the base current. In the light of the above statements, choose the correct answer from the options given below.
Answer: (a)
Statement 1 is false because in case of two discrete back to back connected diodes, there are four doped regions instead of three and there is nothing that resembles a thin base region between an emitter and a collector. Statement-2 is true, as $\beta = \frac{I_C}{I_B}$
Physics · Oscillations · Single correct
In the given figure, a body of mass $M$ is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant $k$, the frequency of oscillation of given body is :
Answer: (a)
Equivalent $K = K + K = 2K$ Now, $T = 2\pi \sqrt{\frac{m}{K_{eq}}}$ $$\Rightarrow T = 2\pi \sqrt{\frac{m}{2k}}$$ $$\therefore f = \frac{1}{2\pi} \sqrt{\frac{2k}{m}}$$
Physics · Electromagnetic Induction · Single correct
Figure shows a circuit that contains four identical resistors with resistance $R = 2.0\,\Omega$. Two identical inductors with inductance $L = 2.0\,\mathrm{mH}$ and an ideal battery with emf $E = 9\,\mathrm{V}$. The current 'i' just after the switch 's' is closed will be:
Answer: (c)
When switch $S$ is closed, given $v = 9 \, \mathrm{V}$. From $V = IR$, $$I = \frac{V}{R}$$ The equivalent resistance is $$R_{eq.} = 2 + 2 = 4 \, \Omega$$ Therefore, $$I = \frac{9}{4} = 2.25 \, \mathrm{A}$$
Physics · Atoms · Single correct
The de Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities is:
Answer: (b)
From De-broglie's wavelength: $\lambda = \frac{h}{mv}$ Given $\lambda_p = \lambda_\alpha$ $$v\alpha \frac{1}{m}$$ $$\frac{v_p}{v_\alpha} = \frac{m_\alpha}{m_p} = \frac{4 \, m_p}{m_p} = \frac{4}{1}$$
Physics · Kinetic Theory · Single correct
If one mole of an ideal gas at $(P_1, V_1)$ is allowed to expand reversibly and isothermally (A to B) its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value (B $\rightarrow$ C). Then it is restored to its initial state by a reversible adiabatic compression (C to A ). The net workdone by the gas is equal to :
Answer: (c)
AB is an isothermal process, so $W_{AB} \rightarrow nRT \ln 2 = RT \ln 2$. BC is an isochoric process, so $W_{BC} = 0$. CA is an adiabatic process, so $W_{CA} = \frac{P_1 V_1 - \frac{P_1}{4} \times 2V_1}{1 - \gamma} = \frac{P_1 V_1}{2(1 - \gamma)} = \frac{RT}{2(1 - \gamma)}$. Therefore, $W_{ABCA} = \frac{RT}{n2} + \frac{RT}{2(1 - \gamma)} = RT \left[ \ln 2 - \frac{1}{2(\gamma - 1)} \right]$.
Physics · Nuclei · Single correct
An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be :
Answer: (b)
\[ \lambda_{\min} = \frac{hc}{eV} \] \[ \lambda_{\min} = \frac{1240\ \mathrm{nm\cdot eV}}{1.24\times10^{6}\ \mathrm{eV}} \] \[ \lambda_{\min} = 10^{-3}\ \mathrm{nm} \]
Physics · Waves · Single correct
Which of the following equations represents a travelling wave?
Answer: (b)
$Y=F(x,t)$ For travelling wave $Y$ should be linear function of $x$ and $t$ and they must exist as $(x \pm vt)$. $Y=A\sin(15x-2t)$ $\rightarrow$ linear function in $x$ and $t$
Physics · Atoms · Single correct
According to Bohr atom model, in which of the following transitions will the frequency be maximum?
Answer: (a)
Given $\Delta E = hf$. The frequency $f$ is more for transition from $n = 2$ to $n = 1$.
Physics · Wave Optics · Single correct
If the source of light used in a Young's double slit experiment is changed from red to violet:
Answer: (b)
Given $\beta = \frac{\lambda D}{d}$. As $\lambda_V < \lambda_R$, it implies $\beta_V < \beta_R$. Therefore, consecutive fringe lines will come closer. Thus, the answer is (2).
Physics · System of Particles and Rotational Motion · Single correct
A circular hole of radius $\frac{a}{2}$ is cut out of a circular disc of radius 'a' shown in figure. The centroid of the remaining circular portion with respect to point 'O' will be:
Answer: (d)
Let $\sigma$ be the surface mass density of the disc. $$X_{com} = \frac{(\sigma \pi a^2 \cdot a) - \left(\sigma \frac{\pi a^2}{4} \cdot \frac{3a}{2}\right)}{\sigma \pi a^2 - \frac{\sigma \pi a^2}{4}}$$ $$X_{com} = \frac{a - \frac{3a}{8}}{1 - \frac{1}{4}}$$ $$X_{com} = \frac{5a}{8} \cdot \frac{4}{3}$$ $$X_{com} = \frac{5a}{6}$$
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
Zener breakdown occurs in a p-n junction having p and n both:
Answer: (b)
The zener breakdown occurs in the heavily doped $p-n$ junction diode. Heavily doped $p-n$ junction diodes have narrow depletion region.
Physics · Electromagnetic Waves · Single correct
Match List - I with List - II. List - I
Answer: (a)
Q2 (1) (a) Source of microwave frequency - (ii) Magnetron (b) Source of infra red frequency - (iv) Vibration of atom and molecules (c) Source of gamma ray - (i) Radio active decay of nucleus (d) Source of X-ray - (iii) inner shell electron
Physics · Laws of Motion · Single correct
A particle is projected with velocity $v_0$ along $x$-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. $ma = -\alpha X^2$. The distance at which the particle stops:
Answer: (c)
Given $a = \frac{v dv}{dx}$. $$\int_{v_i}^{v_f} V dV = \int_{x_i}^{x_f} a dx$$ Given: $-v_i = v_0$ $V_f = 0$ $X_i = 0$ $X_f = x$ From Damping Force: $a = -\frac{\alpha x^2}{m}$ $$\int_{0}^{V_0} V dV = -\int_{0}^{x} \frac{\alpha x^2}{m} dx$$ $$-\frac{v_0^2}{2} = -\frac{\alpha}{m} \left[ \frac{x^3}{3} \right]$$ $$x = \left[ \frac{3m v_0^2}{2\alpha} \right]^{\frac{1}{3}}$$
Physics · Gravitation · Single correct
A body weighs 49 N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator? [Use $g = \frac{GM}{R^2} = 9.8 \, \mathrm{ms^{-2}}$ and radius of earth, $R = 6400 \, \mathrm{km}$.]
Answer: (d)
At north pole, weight $Mg = 49$. Now, at equator $g' = g - \omega^2 R$. $$Mg' = M \left( g - \omega^2 R \right)$$ Weight will be less than $Mg$ at equator.
Physics · Mechanical Properties of Solids · Numerical
A uniform metallic wire is elongated by $0.04 \, \mathrm{m}$ when subjected to a linear force $F$. The elongation, if its length and diameter is doubled and subjected to the same force will be ___ cm.
Answer: 2
Given $$y = \frac{F/A}{\Delta \ell / \ell}$$ Therefore, $$\frac{F}{A} = y \frac{\Delta \ell}{\ell}$$ Thus, $$\frac{F}{A} = y \times \frac{0.04}{\ell} \cdots (1)$$ When length and diameter is doubled. Therefore, $$\frac{F}{4A} = y \times \frac{\Delta \ell}{2\ell} \cdots (2)$$ Dividing (1) by (2) $$\frac{F/A}{F/4A} = \frac{y \times \frac{0.04}{\ell}}{y \times \frac{\Delta \ell}{2\ell}}$$ This simplifies to $$4 = \frac{0.04 \times 2}{\Delta \ell}$$ Solving for $\Delta \ell$ $$\Delta \ell = 0.02$$ Thus, $$\Delta \ell = 2 \times 10^{-2}$$ Therefore, $$\therefore x = 2$$
Physics · Current Electricity · Numerical
A cylindrical wire of radius 0.5 mm and conductivity $5 \times 10^7 \, \mathrm{S/m}$ is subjected to an electric field of $10 \, \mathrm{mV/m}$. The expected value of current in the wire will be $x^3 \pi \, \mathrm{mA}$. The value of $x$ is
Answer: 5
We know that $J = \sigma E$ $$\Rightarrow J = 5 \times 10^7 \times 10 \times 10^{-3}$$ $$\Rightarrow J = 50 \times 10^4 \, \mathrm{A/m^2}$$ Current flowing $I = J \times \pi R^2$ $$I = 50 \times 10^4 \times \pi (0.5 \times 10^{-3})^2$$ $$I = 5 \times 10^4 \times \pi \times 0.25 \times 10^{-6}$$ $$I = 125 \times 10^{-3} \pi$$ $X = 5$
Physics · System of Particles and Rotational Motion · Numerical
A uniform thin bar of mass 6 kg and length 2.4 meter is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is ____ $\times 10^{-1} \, \mathrm{kg \, m^2}$
Answer: 8
MOI of AB about P: $I_{AB_p} = \frac{M}{6} \left( \frac{\ell}{6} \right)^2 \frac{1}{12}$ MOI of AB about O, $I_{AB_o} = \left[ \frac{M}{6} \left( \frac{\ell}{6} \right)^2 \frac{1}{12} + \frac{M}{6} \left( \frac{\ell}{6} \frac{\sqrt{3}}{2} \right)^2 \right]$ $I_{Hexagon} = 6 I_{AB_o} = M \left[ \frac{\ell^2}{12 \times 36} + \frac{\ell^2}{36} \times \frac{3}{4} \right]$ $= \frac{6}{100} \left[ \frac{24 \times 24}{12 \times 36} + \frac{24 \times 24}{36} \times \frac{3}{4} \right]$ $= 0.8 kgm^2$ $= 8 \times 10^{-2} kg/m^2$
Physics · Work, Energy and Power · Numerical
Two solids A and B of mass 1 kg and 2 kg respectively are moving with equal linear momentum. The ratio of their kinetic energies $(\mathrm{K. E.})_A : (\mathrm{K. E.})_B$ will be $\frac{A}{1}$. So the value of $A$ will be
Answer: 2
Given that, $\frac{M_1}{M_2} = \frac{1}{2}$. Also, $p_1 = p_2 = p$. Therefore, $M_1 \, V_1 = M_2 \, V_2 = p$. Also, we know that $$K = \frac{p^2}{2M} \implies K_1 = \frac{p^2}{2M_1} \& \implies K_2 = \frac{p^2}{2M_2}$$ $$\implies \frac{K_1}{K_2} = \frac{p^2}{2M_1} \times \frac{2M_2}{p^2} \implies \frac{K_1}{K_2} = \frac{M_2}{M_1} = \frac{2}{1}$$ $$\implies \frac{A}{1} = \frac{2}{1} \therefore A = 2$$
Physics · Kinetic Theory · Numerical
The root mean square speed of molecules of a given mass of a gas at $27^{\circ} \mathrm{C}$ and 1 atmosphere pressure is $200 \, \mathrm{ms}^{-1}$. The root mean square speed of molecules of the gas at $127^{\circ} \mathrm{C}$ and 2 atmosphere pressure is $\frac{x}{\sqrt{3}} \, \mathrm{ms}^{-1}$. The value of $x$ will be
Answer: 400
Given $V_{rms} = \sqrt{\frac{3RT_1}{M_0}}$. $$200 = \sqrt{\frac{3R \times 300}{M_0}}$$ Also, $$\frac{x}{\sqrt{3}} = \sqrt{\frac{3R \times 400}{M_0}}$$ Dividing (1) by (2): $$\frac{200}{x/\sqrt{3}} = \sqrt{\frac{300}{400}} = \sqrt{\frac{3}{4}}$$ Therefore, $x = 400 \, \mathrm{m/s}$.
Physics · Electric Charges and Fields · Numerical
A point charge of $+12\,\mu\mathrm{C}$ is at a distance $6\,\mathrm{cm}$ vertically above the centre of a square of side $12\,\mathrm{cm}$ as shown in figure. The magnitude of the electric flux through the square will be ____ $\times 10^3\,\mathrm{Nm}^2/\mathrm{C}$
Answer: 226
Using Gauss law, it is a part of cube of side 12 cm and charge at centre so; $$\phi = \frac{Q}{6\varepsilon_0} = \frac{12 \mu c}{6\varepsilon_0} = 2 \times 4\pi \times 9 \times 10^9 \times 10^{-6}$$ $$= 226 \times 10^3 \mathrm{Nm^2/C}$$
Physics · Communication Systems · Numerical
A signal of 0.1 $\mathrm{kW}$ is transmitted in a cable. The attenuation of cable is $-5 \, \mathrm{dB}$ per km and cable length is $20 \, \mathrm{km}$. the power received at receiver is $10^{-X} \, \mathrm{W}$. The value of $X$ is ____ $\left[ Gain in dB = 10 \log_{10} \left( \frac{P_0}{P_i} \right) \right]$
Answer: 8
Power of signal transmitted: $P_i = 0.1 \, \mathrm{KW} = 100 \, \mathrm{w}$ Rate of attenuation $= -5 \, \mathrm{dB/Km}$ Total length of path $= 20 \, \mathrm{km}$ Total loss suffered $= -5 \times 20 = -100 \, \mathrm{dB}$ Gain in dB $= 10 \log_{10} \frac{P_0}{P_i}$ $$-100 = 10 \log_{10} \frac{P_0}{P_i}$$ $$\Rightarrow \log_{10} \frac{P_i}{P_0} = 10$$ $$\Rightarrow \log_{10} \frac{P_i}{P_0} = \log_{10} 10^{10}$$ $$\Rightarrow \frac{100}{P_0} = 10^{10}$$ $$\Rightarrow P_0 = \frac{1}{10^8} = 10^{-8}$$ Therefore, $x = 8$
Physics · Alternating Current · Numerical
A series LCR circuit is designed to resonate at an angular frequency $\omega_0 = 10^5 \, \mathrm{rad/s}$. The circuit draws $16 \, \mathrm{W}$ power from $120 \, \mathrm{V}$ source at resonance. The value of resistance ' $R'$ in the circuit is ____ $\Omega$
Answer: 900
Given the formula for power, $P = \frac{V^2}{R}$. We have $16 = \frac{120^2}{R}$, which implies $R = \frac{14400}{16}$. Therefore, $R = 900 \, \Omega$.
Physics · Waves · Numerical
Two cars are approaching each other at an equal speed of $7.2 \, \mathrm{km/hr}$. When they see each other, both blow horns having frequency of $676 \, \mathrm{Hz}$. The beat frequency heard by each driver will be _____ Hz. [Velocity of sound in air is $340 \, \mathrm{m/s}$.]
Answer: 8
Speed = 7.2 km/h = 2 m/s. Frequency as heard by A $f'_A = f_B \left( \frac{V + V_0}{V - V_s} \right)$. $$f'_A = 676 \left( \frac{340 + 2}{340 - 2} \right)$$ $$f'_A = 684 \, \mathrm{Hz}$$ Therefore, $f_{Beat} = f'_A - f_B$. $$= 684 - 676$$ $$= 8 \, \mathrm{Hz}$$
Physics · Electromagnetic Waves · Numerical
An electromagnetic wave of frequency $3 \, \mathrm{GHz}$ enters a dielectric medium of relative electric permittivity $2.25$ from vacuum. The wavelength of this wave in that medium will be ____ $\times 10^{-2} \, \mathrm{cm}$
Answer: 667
Given $f = 3 \, \mathrm{GHz}$, $\varepsilon_r = 2.25$. $V = \lambda f \Rightarrow \lambda = \frac{v}{f}$ $C = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ $V = \frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}} \Rightarrow \lambda = \frac{C}{f \cdot \sqrt{\mu_0 \varepsilon_0} \cdot \sqrt{\mu_r \varepsilon_r} \cdot f}$ $$\Rightarrow \lambda = \frac{3 \times 10^8}{f \cdot \sqrt{\mu_r} \cdot \sqrt{\varepsilon_r}} \Rightarrow \lambda = \frac{1}{3 \times 10^9 \times \sqrt{1} \times \sqrt{2.25}}$$ $$\Rightarrow \lambda = 667 \times 10^{-2} \, \mathrm{cm}$$
Chemistry · Amines · Single correct
What is the correct sequence of reagents used for converting nitrobenzene into $m-$ dibromobenzene?
Answer: (d)
The reaction sequence begins with the nitration of benzene to form nitrobenzene. The first step involves bromination using $\mathrm{Br_2/Fe}$ to form bromonitrobenzene. Next, reduction with $\mathrm{Sn/HCl}$ converts the nitro group to an amino group, resulting in bromobenzeneamine. Diazotization with $\mathrm{NaNO_2/HCl}$ forms a diazonium salt. Finally, the Sandmeyer reaction with $\mathrm{CuBr/HBr}$ replaces the diazonium group with a bromine atom, yielding dibromobenzene.
Chemistry · Chemistry in Everyday Life · Single correct
Most suitable salt which can be used for efficient clotting of blood will be:
Answer: (d)
Blood is a negative sol. According to hardy-Schulz's rule, the cation with high charge has high coagulation power. Hence, $\mathrm{FeCl_3}$ can be used for clotting blood.
Chemistry · Haloalkanes and Haloarenes · Single correct
The correct order of the following compounds showing increasing tendency towards nucleophilic substitution reaction is :
Answer: (c)
Reactivity is proportional to the $-m$ group present at the $O/P$ position.
Chemistry · Structure of Atom · Single correct
According to Bohr's atomic theory: $(A)$ Kinetic energy of electron is $\propto \frac{Z^2}{n^2}$ $(B)$ The product of velocity $(v)$ of electron and principal quantum number $(n)$. '$vn$' $\propto Z^2$. $(C)$ Frequency of revolution of electron in an orbit is $\propto \frac{Z^3}{n^3}$. $(D)$ Coulombic force of attraction on the electron is $\propto \frac{Z^3}{n^4}$. Choose the most appropriate answer from the options given below:
Answer: (b)
(A) KE = -TE = 13.6 $\times$ $\frac{Z^2}{n^2}$ eV KE $\propto$ $\frac{Z^2}{n^2}$ (B) V = 2.188 $\times$ 10^6 $\times$ $\frac{Z}{n}$ m/sec So, Vn $\propto$ Z (C) Frequency = $\frac{V}{2 \pi r}$ So, F $\propto$ $\frac{Z^2}{n^3}$ $\left$[ $\therefore$ r $\propto$ $\frac{n^2}{Z}$ and V $\propto$ $\frac{Z}{n}$ $\right$] (D) Force $\propto$ $\frac{Z}{r^2}$ So, F $\propto$ $\frac{Z^3}{n^4}$ So, only statement (A) is correct
Chemistry · Co-ordination Compounds · Single correct
Match List-I with List-II Choose the correct answer from the options given below:
Answer: (d)
Chemistry · Co-ordination Compounds · Single correct
The calculated magnetic moments (spin only value) for species $[FeCl_4]^{2-}$, $[Co(C_2O_4)_3]^{3-}$ and $MnO_4^{2-}$ respectively are:
Answer: (c)
$[\text{FeCl}_4]^{2-}$ $\text{Fe}^{2+}$ 3 d$^6$ $\rightarrow$ 4 unpaired electron. as $\text{Cl}^-$ in a weak field liquid. $$\mu_{\text{spin}} = \sqrt{248} \text{ M}$$ $$= 4.9 \text{ BM}$$ $[\text{Co}(\text{C}_2\text{O}_4)_3]^{3-}$ $\text{Co}^{3+}$ 3 d$^6$ $\rightarrow$ for $\text{Co}^{3+}$ with coordination no. $6\text{C}_2\text{O}_4^{2-}$ is strong field ligand \& causes pairing \& hence no. unpaired electron $$\mu_{\text{spin}} = 0$$ $[\text{MnO}_4]^{2-}$ $\text{Mn}^{+6}$ it has one unpaired electron. $$\mu_{\text{spin}} = \sqrt{3} \text{ BM}$$
Chemistry · The s-Block Elements · Single correct
Match List-I with List-II \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{List - I (Salt)} & \multicolumn{2}{c|}{List - II (Flame colour wavelength)} \\ \hline (a) & $\mathrm{LiCl}$ & (i) & $455.5\,\mathrm{nm}$ \\ \hline (b) & $\mathrm{NaCl}$ & (ii) & $970.8\,\mathrm{nm}$ \\ \hline (c) & $\mathrm{RbCl}$ & (iii) & $780.0\,\mathrm{nm}$ \\ \hline (d) & $\mathrm{CsCl}$ & (iv) & $589.2\,\mathrm{nm}$ \\ \hline \end{tabular}
Answer: (b)
Range of visible region: $390 \, \mathrm{nm} - 760 \, \mathrm{nm}$ VIBGYOR Violet Red LiCl Crimson Red NaCl Golden yellow RbCl Violet CsCl Blue So LiCl which is crimson have wavelength close to red in the spectrum of visible region which is as per given data is.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Wich one of the following carbonyl compounds cannot be prepared by addition of wate on an alkyne in the presence of $HgSO_4$ and $H_2SO_4$ ?
Answer: (a)
Reaction of Alkyne with $\mathrm{HgSO_4}$ and $\mathrm{H_2SO_4}$ follow as $$\mathrm{CH \equiv CH \xrightarrow{HgSO_4, H_2SO_4 \atop H_2O} CH_3CHO}$$ $$\mathrm{CH_3 - C \equiv CH \xrightarrow{HgSO_4, H_2SO_4 \atop H_2O} CH_3 - C - CH_3}$$ Hence, by this process preparation of $\mathrm{CH_3CH_2CHO}$ cannot possible.
Chemistry · Polymers · Single correct
In polymer Buna-S; 'S' stands for :
Answer: (a)
Buna-S is the co-polymer of buta-1,3 diene & styrene.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Which of the following reagent is suitable for the preparation of the product in the above reaction.
Answer: (b)
It is wolf-kishner reduction of carbonyl compounds.
Chemistry · Chemistry in Everyday Life · Single correct
Match List-I and List-II. \begin{tabular}{|c|c|c|c|} \hline \multicolumn{2}{|c|}{\textbf{List-I}} & \multicolumn{2}{c|}{\textbf{List-II}} \\ \hline (a) & Valium & (i) & Antifertility drug \\ \hline (b) & Morphine & (ii) & Pernicious anaemia \\ \hline (c) & Norethindrone & (iii) & Analgesic \\ \hline (d) & Vitamin B$_{12}$ & (iv) & Tranquilizer \\ \hline \end{tabular}
Answer: (b)
(a) Valium with (iv) Tranquilizer (b) Morphine with (iii) Analgesic (c) Norethindrone with (i) Antifertility drug (d) Vitamin $B_{12}$ with (ii) Pernicious anaemia
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Match List-I with List-II \begin{tabular}{|l|l|} \hline \textbf{List-I} & \textbf{List-II} \\ (Metal) & (Ores) \\ \hline (a) Aluminium $\hspace{1cm}$ & (i) Siderite \\ \hline (b) Iron $\hspace{1cm}$ & (ii) Calamine \\ \hline (c) Copper $\hspace{1cm}$ & (iii) Kaolinite \\ \hline (d) Zinc $\hspace{1cm}$ & (iv) Malachite \\ \hline \end{tabular} Choose the correct answer from the options given below:
Answer: (c)
Siderite $\mathrm{FeCO_3}$ Calamine $\mathrm{ZnCO_3}$ Kaolinite $\mathrm{Si_2Al_2O_5(OH)_4}$ or $\mathrm{Al_2O_3 \cdot 2SiO_2 \cdot 2H_2O}$ Malachite $\mathrm{CuCO_3 \cdot Cu(OH)_2}$
Chemistry · Hydrocarbons · Single correct
Which one of the following compounds is non-aromatic?
Answer: (b)
The structure contains an $sp^3$ carbon atom, which is not planar. Hence it is non-aromatic.
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
What is the correct order of the following elements with respect to their density?
Answer: (d)
Fact Based Density depend on many factors like atomic mass, atomic radius and packing efficiency.
Chemistry · Environmental Chemistry · Single correct
Given below are two statements: Statement I: The value of the parameter "Biochemical Oxygen Demand (BOD)" is important for survival of aquatic life. Statement II: The optimum value of BOD is 6.5 ppm. In the light of the above statements, choose the most appropriate answer from the options given below.
Answer: (c)
For survival of aquatic life dissolved oxygen is responsible. Its optimum limit is $6.5 \, \mathrm{ppm}$ and the optimum limit of BOD ranges from $10 - 20 \, \mathrm{ppm}$. BOD stands for biochemical oxygen demand.
Chemistry · The d-and f-Block Elements · Single correct
The incorrect statement among the following is :
Answer: (b)
Red colour of ruby is due to presence of $\mathrm{CrO_3}$ or $\mathrm{Cr^{+6}}$ not $\mathrm{CO_3^{+}}$
Chemistry · Chemical Bonding and Molecular Structure · Single correct
The correct shape and I-I-I bond angles respectively in $\mathrm{I}_3^-$ ion are:
Answer: (c)
For $\mathrm{I_3^-}$, $\mathrm{sp^3d}$ hybridisation (2BP + 3L.P.) results in linear geometry.
Chemistry · Hydrogen · Single correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Hydrogen is the most abundant element in the Universe, but it is not the most abundant gas in the troposphere. Reason R : Hydrogen is the lightest element. In the light of the above statements, choose the correct answer from the given below
Answer: (b)
Hydrogen is the most abundant element in the universe because all luminous bodies of the universe, i.e., stars and nebulae, are made up of hydrogen, which acts as nuclear fuel, and fusion reaction is responsible for their light.
Chemistry · Amines · Single correct
The diazonium salt of which of the following compounds will form a coloured dye on reaction with $\beta$-Naphthol in NaOH?
Answer: (c)
The reaction of aniline with $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ followed by $\beta$-Naphthol produces an orange bright dye.
Chemistry · The Solid State · Single correct
The Correct set from the following in which both pairs are in correct order of melting point is
Answer: (b)
Generally M.P. $\propto$ Lattice energy = $\frac{KQ_1Q_2}{r^+ + r^-}$ $\propto$ (packing efficiency)
Chemistry · Amines · Single correct
The total number of amines among the following which can be synthesized by Gabriel synthesis is
Answer: (d)
Only aliphatic amines can be prepared by Gabriel synthesis. 1.86 g of aniline completely reacts to form acetanilide. 10$\%$ of the product is lost during purification. Amount of acetanilide obtained after purification (in g) is $\times 10^{-2}$.
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Among the following allotropic forms of sulphur, the number of allotropic forms, which will show paramagnetism is (1) $\alpha$-sulphur (2)$\beta$-sulphur (3)$\mathrm{S}_2$-form
Answer: a
$\mathrm{S_2}$ is like $\mathrm{O_2}$; e paramagnetic as per molecular orbital theory.
Chemistry · Hydrocarbons · Numerical
The formula of a gaseous hydrocarbon which requires 6 times of its own volume of $O_2$ for complete oxidation and produces 4 times its own volume of $CO_2$ is $C_xH_y$. The value of $y$ is
Answer: 8
Given the reaction $\mathrm{C_xH_y + 6O_2 \rightarrow 4CO_2 + \frac{y}{2}H_2O}$. Applying POAC on 'O' atoms $6 \times 2 = 4 \times 2 + \frac{y}{2} \times 1$. $$\frac{y}{2} = 4 \Rightarrow y = 8$$
Chemistry · States of Matter · Numerical
The volume occupied by 4.75 g of acetylene gas at $50^\circ$C and 740mmHg pressure is ___ L. (Rounded off to the nearest integer) $\left(\text{Given } R = 0.0826 \text{ L atm K}^{-1} \text{ mol}^{-1}\right)$
Answer: 5
Given: $T = 50^\circ \mathrm{C} = 323.15 \, \mathrm{K}$ $P = 740 \, \mathrm{mm \, of \, Hg} = \frac{740}{760} \, \mathrm{atm}$ $V = ?$ Moles $(n) = \frac{4.75}{26}$ $$V = \frac{4.75}{26} \times \frac{0.0821 \times 323.15}{740} \times 760$$ $$V = 4.97 \approx 5 \, \mathrm{Lit}$$
Chemistry · Solutions · Fill in the blank
$C_6H_6$ freezes at $5.5^\circ$ C. The temperature at which a solution of 10 g of $C_4H_{10}$ in 200 g of $C_6H_6$ freeze is ____ $^\circ$ C. (The molal freezing point depression constant of $C_6H_6$ is $5.12^\circ$ C/m)
Answer: 1
Given $\Delta T_f = i \times K_f \times m$. $$= (1) \times 5.12 \times \frac{10/58}{200} \times 1000 \implies \Delta T_f = \frac{5.12 \times 50}{58} = 4.414$$ $T_{f(solution)} = T_{K(solvent)} - \Delta T_f$ $$= 5.5 - 4.414$$ $$= 1.086^\circ C$$ $$\approx 1.09^\circ C = 1 (nearest integer)$$
Chemistry · Electrochemistry · Numerical
The magnitude of the change in oxidising power of the $\mathrm{MnO_4^-}/\mathrm{Mn^{2+}}$ couple is $x \times 10^{-4} \, \mathrm{V}$, if the $\mathrm{H^+}$ concentration is decreased from $1 \, \mathrm{M}$ to $10^{-4} \, \mathrm{M}$ at $25^\circ \mathrm{C}$. (Assume concentration of $\mathrm{MnO_4^-}$ and $\mathrm{Mn^{2+}}$ to be same on change in $\mathrm{H^+}$ concentration). The value of $x$ is ____ (Rounded off to the nearest integer) [Given $\frac{2303RT}{F} = 0.059$]
Answer: 3776
The reaction is given by $$\mathrm{Se^{-}} + MnO_4^{-} + 8H^{+} \rightarrow Mn^{+2} + 4H_2O$$ The reaction quotient $Q$ is $$Q = \frac{[\mathrm{Mn^{+2}}]}{[\mathrm{H^{+}}]^8 [\mathrm{MnO_4^{-}}]}$$ This implies $$\mathrm{E_1 = E^\circ - \frac{0.059}{5} \log(Q_1)}$$ $$\mathrm{E_2 = E^\circ - \frac{0.059}{5} \log(Q_2)} \implies \mathrm{E_2 - E_1 = \frac{0.059}{5} \log \left( \frac{Q_1}{Q_2} \right)}$$ $$= \frac{0.059}{5} \log \left\{ \left( \frac{[\mathrm{H^{+}}]_\mathrm{II}}{[\mathrm{H^{+}}]_\mathrm{I}} \right)^8 \right\} \implies \frac{0.059}{5} \log \left( \frac{10^{-4}}{1} \right)^8$$ $$\mathrm{(E_2 - E_1) = \frac{0.059}{5} \times (-32)} \implies |\mathrm{(E_2 - E_1)}| = 32 \times \frac{0.059}{5} = \mathrm{x} \times 10^{-4}$$ $$= \frac{32 \times 590}{5} \times 10^{-4} = 3776 \times 10^{-4} \mathrm{x} = 3776$$
Chemistry · Equilibrium · Numerical
The solubility product of $\mathrm{PbI_2}$ is $8.0 \times 10^{-9}$. The solubility of lead iodide in $0.1$ molar solution of lead nitrate is $x \times 10^{-6} \, \mathrm{mol/L}$. The value of $x$ is ____. (Rounded off to the nearest integer) [Given $\sqrt{2} = 1.41$]
Answer: 141
$K_{SP} (\mathrm{PbI_2}) = 8 \times 10^{-9}$ $\mathrm{PbI_2(s) \rightleftharpoons Pb^{+2}(aq) + 2I^-(aq)}$ $S + 0.1 \quad 2S$ $K_{SP} = [\mathrm{Pb}^{+2}] [\mathrm{I}^-]^2$ $8 \times 10^{-9} = (S + 0.1)(2S)^2 \Rightarrow 8 \times 10^{-9} \simeq 0.1 \times 4S^2$ $\Rightarrow S^2 = 2 \times 10^{-8}$ $S = 1.414 \times 10^{-4} \, \mathrm{mol/Lit}$ $= x \times 10^{-4} \, \mathrm{mol/Lit} \quad \therefore x = 141.4 \simeq 141$
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of $3.33 \, \mathrm{h}$ at $25^\circ \mathrm{C}$. After $9 \, \mathrm{h}$, the fraction of sucrose remaining is $f$. The value of $\log_{10} \left( \frac{1}{f} \right)$ is ____ $\times 10^{-2}$ (Rounded off to the nearest integer) [Assume: $\ln 10 = 2.303, \ln 2 = 0.693$]
Answer: 81
Sucrose undergoes hydrolysis to form glucose and fructose. The half-life $t_{1/2} = 3.33 \, \mathrm{h} = \frac{10}{3} \, \mathrm{h}$. Therefore, $$C_t = \frac{C_0}{2^{t/t_{1/2}}}$$ The fraction of sucrose remaining is given by $$f = \frac{C_t}{C_0} = \frac{1}{2^{t/t_{1/2}}}$$ Thus, $$\frac{1}{f} = 2^{t/t_{1/2}}$$ Taking logarithms, $$\log(1/f) = \log\left(2^{t/t_{1/2}}\right) = \frac{t}{t_{1/2}} \log(2)$$ Substituting the values, $$= \frac{9}{10/3} \times 0.3 = \frac{8.1}{10} = 0.81$$ Therefore, $$= x \times 10^{-2}$$ Hence, $$x = 81$$
Chemistry · Amines · Numerical
$1.86\,\mathrm{g}$ of aniline completely reacts to form acetanilide. $10\%$ of the product is lost during purification. The amount of acetanilide obtained after purification (in $\mathrm{g}$) is $x \times 10^{-2}$. The value of $x$ is.
Answer: 243
Molar mass $= 93$ Molar mass $= 135$ $93 \, \mathrm{g}$ Aniline produce $135 \, \mathrm{g}$ acetanilide $1.86 \, \mathrm{g}$ produce $$\frac{135 \times 1.86}{93} = 2.70 \, \mathrm{g}$$ At $10\%$ loss, $90\%$ product will be formed after purification. Amount of product obtained $$= \frac{2.70 \times 90}{100} = 2.43 \, \mathrm{g} = 243 \times 10^{-2} \, \mathrm{g}$$
Chemistry · Equilibrium · Fill in the blank
Assuming ideal behaviour, the magnitude of $\log K$ for the following reaction at $25^\circ\mathrm{C}$ is $x\times10^{-1}$. The value of $x$ is ____. (Nearest integer) $3\mathrm{HC\equiv CH(g)} \rightleftharpoons \mathrm{C_6H_6(l)}$ [Given : $\Delta_fG^\circ(\mathrm{HC\equiv CH})=-2.04\times10^5\ \mathrm{J\,mol^{-1}},\ \Delta_fG^\circ(\mathrm{C_6H_6})=-1.24\times10^5\ \mathrm{J\,mol^{-1}}$ $R=8.314\ J\,K^{-1}\,mol^{-1}$]
Answer: 855
The reaction is given by $3\mathrm{HC} \equiv \mathrm{CH}(g) \rightleftharpoons \mathrm{C_6H_6}(\ell)$. The change in Gibbs free energy is calculated as follows: $$\Delta G_r^\circ = \Delta G_f^\circ [\mathrm{C_6H_6}(\ell)] - 3 \times \Delta G_f^\circ [\mathrm{HC} \equiv \mathrm{CH}]$$ $$= \left[ -1.24 \times 10^5 - 3 \times (-2.04 \times 10^5) \right]$$ $$= 4.88 \times 10^5 \, \mathrm{J/mol}$$ The relationship between Gibbs free energy and equilibrium constant is given by: $$\Delta G_r^\circ = -RT \ln(K_{eq})$$ Therefore, $$\log(K_{eq}) = \frac{-\Delta G^\circ}{2.303RT}$$ $$= \frac{-4.88 \times 10^5}{2.303 \times 8.314 \times 298}$$ $$= -8.55 \times 10^1 = 855 \times 10^{-1}$$