JEE Main 29 June 2022 Shift 1 question paper with solutions
JEE Main 29 June 2022 Shift 1: 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
Question 1
Maths · Matrices · Single correct
The probability that a randomly chosen $2 \times 2$ matrix with all the entries from the set of first 10 primes, is singular, is equal to :
$\frac{133}{10^4}$
$\frac{18}{10^3}$
$\frac{19}{10^3}$
$\frac{271}{10^4}$
Answer: (c)
Solution
Let matrix $A$ be singular then $|A| = 0$. Number of singular matrices = All entries are the same + only two prime numbers are used in the matrix $$= 10 + 10 \times 9 \times 2$$ $$= 190$$ Required probability $$= \frac{190}{10^4} = \frac{19}{10^3}$$
Question 2
Maths · Differential Equations · Single correct
Let the solution curve of the differential equation $$x \frac{dy}{dx} - y = \sqrt{y^2 + 16x^2}$$, $$y(1) = 3$$ be $$y = y(x)$$. Then $$y(2)$$ is equal to :
15
11
13
17
Answer: (a)
Solution
Given $y = vx$, we have $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Therefore, $$x \frac{dv}{dx} = \sqrt{v^2 + 16}$$ Integrating both sides, $$\int \frac{dv}{\sqrt{v^2 + 16}} = \int \frac{dx}{x}$$ This gives $$\ln |v + \sqrt{v^2 + 16}| = \ln x + \ln C$$ Therefore, $$y + \sqrt{y^2 + 16x^2} = Cx^2$$ Given $y(1) = 3$, we find $C = 8$. Thus, $$y(2) = 15$$
Question 3
Maths · Three Dimensional Geometry · Single correct
If the mirror image of the point (2, 4, 7) in the plane $3x - y + 4z = 2$ is $(a, b, c)$, the $2a + b + 2c$ is equal to:
Maths · Continuity and Differentiability · Single correct
Let $f : \mathbb{R} \to \mathbb{R}$ be a function defined by : $$f(x) = \begin{cases} \max\{t^3 - 3t\} ; x \leq 2 \\ t \leq x \\ x^2 + 2x - 6 ; 2 5 \end{cases}$$ Where [t] is the greatest integer less than or equal to t. Let m be the number of points where f is not differentiable and $I = \int_{-2}^{2} f(x) \, dx$. Then the ordered pair (m, I) is equal to :
(3, $\frac{27}{4}$)
(3, $\frac{23}{4}$)
(4, $\frac{27}{4}$)
(4, $\frac{23}{4}$)
Answer: (c)
Solution
Given $$f(x) = \begin{cases} x^3 - 3x, & x \leq -1 \\ 2, & -1 5 \end{cases}$$ Clearly $f(x)$ is not differentiable at $x = 2, 3, 4, 5 \Rightarrow m = 4$ $$I = \int_{-2}^{-1} (x^3 - 3x) \, dx + \int_{-1}^{2} 2 \, dx = \frac{27}{4}$$
Question 5
Maths · Vector Algebra · Single correct
Let $\vec{a} = \alpha \hat{i} + 3 \hat{j} - \hat{k}$, $\vec{b} = 3 \hat{i} - \beta \hat{j} + 4 \hat{k}$ and $\vec{c} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ where $\alpha, \beta \in \mathbb{R}$, be three vectors. If the projection of $\vec{a}$ on $\vec{c}$ is $\frac{10}{3}$ and $\vec{b} \times \vec{c} = -6 \hat{i} + 10 \hat{j} + 7 \hat{k}$, then the value of $\alpha + \beta$ equal to:
Maths · Applications of Integrals · Single correct
The area enclosed by $y^2 = 8x$ and $y = \sqrt{2x}$ that lies outside the triangle formed by $y = \sqrt{2x}, x = 1, y = 2\sqrt{2}$, is equal to:
$\frac{16\sqrt{2}}{6}$
$\frac{11\sqrt{2}}{6}$
$\frac{13\sqrt{2}}{6}$
$\frac{5\sqrt{2}}{6}$
Answer: (c)
Solution
Area of $\($ $\triangle$ ABC = $\frac{1}{2}$($\sqrt{2}$) $\cdot$ 1 = $\frac{\sqrt{2}}{2}$ $\)$. So required Area = $\($ $\int$_{0}^{4} ($\sqrt{8x}$ - $\sqrt{2x}$) $\,$ dx - $\frac{\sqrt{2}}{2}$ $\)$ = $\($ $\frac{32\sqrt{2}}{3}$ - 8$\sqrt{2}$ - $\frac{\sqrt{2}}{2}$ = $\frac{13\sqrt{2}}{6}$ $\)$
Question 7
Maths · Determinants · Single correct
If the system of linear equations $$ \begin{aligned} 2x+y-z&=7,\\ x-3y+2z&=1,\\ x+4y+\delta z&=k, \end{aligned} $$ where $\delta,k\in\mathbb{R}$, has infinitely many solutions, then $(\delta+k)$ is equal to:
-3
3
6
9
Answer: (b)
Solution
\[ \begin{aligned} 2x+y-z &= 7 \qquad\qquad\qquad\cdots(1)\\ x-3y+2z &= 1 \qquad\qquad\qquad\cdots(2)\\ x+4y+\delta z &= k \qquad\qquad\cdots(3) \end{aligned} \] $_$ Equation (2) + (3) \[ \text{We get } 2x+y+(2+\delta)z=1+k \qquad\cdots(4) \] For infinitely solution \[ \text{From equation (1) and (4)} \] \[ 2+\delta=-1 \Rightarrow \boxed{\delta=-3} \] $_$
Question 8
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + (2i - 1) = 0$. Then, the value of $|\alpha^8 + \beta^8|$ is equal to:
50
250
1250
1500
Answer: (a)
Solution
Given $X^2 = 1 - 2i$, we have $\alpha^2 = 1 - 2i$ and $\beta^2 = 1 - 2i$. Hence $\alpha^8 = \beta^8$. The expression $|\alpha^8 + \beta^8| = |2\alpha^8| = 2|\alpha^2|^4$ simplifies to $2 \sqrt{5}^4 = 50$.
Question 9
Maths · Mathematical Reasoning · Single correct
Let $\Delta \in \{\land, \lor, \Rightarrow, \Leftrightarrow\}$ be such that $(p \land q) \Delta ((p \lor q) \Rightarrow q)$ is a tautology. Then $\Delta$ is equal to:
$\land$
$\lor$
$\Rightarrow$
$\Leftrightarrow$
Solution
Given $p \lor q \Rightarrow q$. Therefore, $$\Rightarrow \sim (p \lor q) \lor q$$ $$\Rightarrow (\sim p \land \sim q) \lor q$$ $$\Rightarrow (\sim p \lor q) \land (\sim q \lor q)$$ $$\Rightarrow (\sim p \lor q) \land t = \sim p \lor q$$ Now by taking option C $$(p \land q) \Rightarrow \sim p \lor q$$ $$\Rightarrow \sim p \lor \sim q \lor \sim p \lor q$$ $$\Rightarrow t$$ Hence C
Question 10
Maths · Matrices · Single correct
Let $A=[a_{ij}]$ be a square matrix of order $3$ such that $a_{ij}=2^{\,j-i},\ \text{for all }i,j=1,2,3.$ Then, the matrix $A^2+A^3+\cdots+A^{10}$ is equal to
$(\\frac{3^{10} - 3}{2})$ A
$(\\frac{3^{10} - 1}{2})$ A
$(\\frac{3^{10} + 1}{2})$ A
$(\\frac{3^{10} + 3}{2})$ A
Answer: (a)
Solution
Given the matrix $$A = \begin{pmatrix} 1 & 2 & 2^2 \\ 1/2 & 1 & 2 \\ 1/2^2 & 1/2 & 1 \end{pmatrix}$$ we have the following equations: $$A^2 = 3A$$ $$A^3 = 3^2 A$$ The expression $$A^2 + A^3 + \ldots + A^{10}$$ can be simplified as follows: $$= 3A + 3^2 A + \ldots + 3^9 A = \frac{3(3^9 - 1)}{3 - 1} A$$ Finally, $$= \frac{3^{10} - 3}{2} A$$
Question 11
Maths · Relations and Functions · Single correct
Let a set $A = A_1 \cup A_2 \cup \ldots \cup A_k$, where $A_i \cap A_j = \phi$ for $i \neq j$, $1 \leq i, j \leq k$. Define the relation $R$ from $A$ to $A$ by $R = \{(x,y): y \in A_i$ if and only if $x \in A_i, 1 \leq i \leq k\}$. Then, $R$ is:
reflexive, symmetric but not transitive
reflexive, transitive but not symmetric
reflexive but not symmetric and transitive
an equivalence relation
Answer: (d)
Solution
Given the set $A = \{1, 2, 3\}$ and the relation $R = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\}$.
Question 12
Maths · Sequences and Series · Single correct
Let $\{a_n\}_{n=0}^{\infty}$ be a sequence such that $a_0 = a_1 = 0$ and $$a_{n+2} = 2a_{n+1} - a_n + 1$$ for all $n \geq 0$. Then, $$\sum_{n=2}^{\infty} \frac{a_n}{7^n}$$ is equal to
Maths · Straight Lines and Pair of Straight Lines · Single correct
The distance between the two points A and A' which lie on $y = 2$ such that both the line segments AB and A'B (where B is the point (2, 3)) subtend angle $\frac{\pi}{4}$ at the origin, is equal to:
10
$\frac{48}{5}$
$\frac{52}{5}$
3
Answer: (c)
Solution
Given $M_1 = \frac{3}{2}$ and $M_2 = \frac{2}{x}$. The equation for $\tan \frac{\pi}{4}$ is given by $$\left| \frac{\frac{3}{2} - \frac{2}{x}}{1 + \frac{6}{2x}} \right| = 1.$$ Solving this equation gives $x_1 = 10$ and $x_2 = -\frac{2}{5}$. Therefore, $AA' = \frac{52}{5}$.
Question 14
Maths · Applications of Derivatives · Single correct
A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is:
$\frac{22}{9 + 4\sqrt{3}}$
$\frac{66}{9 + 4\sqrt{3}}$
$\frac{22}{4 + 9\sqrt{3}}$
$\frac{66}{4 + 9\sqrt{3}}$
Answer: (b)
Solution
Given $3a = x$ and $4b = 22 - x$. We have $a = 2/13$. The area $A_T$ is given by: $$A_T = \frac{\sqrt{3}}{4} a^2 + b^2$$ Substituting the values, we get: $$= \frac{\sqrt{3}}{4} \frac{x^2}{9} + \frac{(22-x)^2}{16}$$ To find the derivative, we set: $$\frac{dA}{dx} = 0 \implies x \left( \frac{\sqrt{3}}{2 \times 9} + \frac{1}{8} \right) - \frac{22}{8} = 0$$ Solving for $x$, we have: $$\implies x \left( \frac{4\sqrt{3} + 9}{36} \right) = \frac{11}{2}$$ Thus, $a = x/3$. Finally, we calculate $a$: $$a = \left( \frac{11/2}{4\sqrt{3} + 9} \right) \left( \frac{1}{3} \right) = \frac{66}{4\sqrt{3} + 9}$$
Question 15
Maths · Inverse Trigonometric Functions · Single correct
The domain of the function $\cos^{-1}\left(\frac{2\sin^{-1}\left(\frac{1}{4x^2-1}\right)}{\pi}\right)$ is :
Given the inequality $$-1 \leq \frac{2 \sin^{-1} \left( \frac{1}{4x^2 - 1} \right)}{\pi} \leq 1$$ we have $$-\pi/2 \leq \sin^{-1} \frac{1}{4x^2 - 1} \leq \pi/2$$ which implies $$-1 \leq \frac{1}{4x^2 - 1} \leq 1$$ is always true. Therefore, $$x \in \left( -\infty, -\frac{1}{\sqrt{2}} \right) \cup \left[ \frac{1}{\sqrt{2}}, \infty \right).$$
Question 16
Maths · Binomial Theorem · Single correct
If the constant term in the expansion of $$\left(3x^3 - 2x^2 + \frac{5}{x^5}\right)^{10}$$ is $2^k \cdot l$, where $l$ is an odd integer, then the value of $k$ is equal to:
$\int_{0}^{5} \cos \left( \pi \left( x - \left\lfloor \frac{x}{2} \right\rfloor \right) \right) \, dx,$ Where [t] denotes greatest integer less than or equal to t, is equal to :
$-3$
$-2$
$2$
$0$
Answer: (d)
Solution
Given $$I = \int_0^5 \cos \left( \pi x - \pi \left[ \frac{x}{2} \right] \right) \, dx$$ This can be split into three integrals: $$\Rightarrow I = \int_0^2 \cos(\pi x) \, dx + \int_2^4 \cos(\pi x - \pi) \, dx + \int_4^5 \cos(\pi x - 2\pi) \, dx$$ Evaluating each integral: $$\Rightarrow I = \left[ \frac{\sin \pi x}{\pi} \right]_0^2 + \left[ \frac{\sin(\pi x - \pi)}{\pi} \right]_2^4 + \left[ \frac{\sin(\pi x - 2\pi)}{\pi} \right]_4^5$$ Simplifying gives: $$\Rightarrow I = 0$$
Question 18
Maths · Conic Sections · Single correct
Let PQ be a focal chord of the parabola $y^2 = 4x$ such that it subtends an angle of $\frac{\pi}{2}$ at the point $(3, 0)$. Let the line segment PQ be also a focal chord of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a^2 > b^2$. If $e$ is the eccentricity of the ellipse $E$, then the value of $\frac{1}{e^2}$ is equal to:
$1 + \sqrt{2}$
$3 + 2\sqrt{2}$
$1 + 2\sqrt{3}$
$4 + 5\sqrt{3}$
Answer: (b)
Solution
PQ is a focal chord. The slope condition is given by: $$m_{PR} \cdot m_{PQ} = -1$$ Calculating the slopes: $$\frac{2t}{t^2 - 3} \times \frac{-2/t}{1} = -1$$ Simplifying: $$(t^2 - 1)^2 = 0$$ This implies: $$t = 1$$ Therefore, P and Q must be the endpoints of the latus rectum: $$P(1, 2) and Q(1, -2)$$ Given: $$\frac{2b^2}{a} = 4 and ae = 1$$ We know that: $$b^2 = a^2 (1 - e^2)$$ Thus: $$a = 1 + \sqrt{2}$$ And: $$e^2 = 1 - \frac{b^2}{a^2}$$ Therefore: $$e^2 = 3 - 2\sqrt{2}$$ Finally: $$\frac{1}{e^2} = 3 + 2\sqrt{2}$$
Question 19
Maths · Conic Sections · Single correct
Let the tangent to the circle $C_1 : x^2 + y^2 = 2$ at the point $M(-1,1)$ intersect the circle $C_2 : (x-3)^2 + (y-2)^2 = 5$, at two distinct points $A$ and $B$. If the tangents to $C_2$ at the points $A$ and $B$ intersect at $N$, then the area of the triangle $ANB$ is equal to:
$\frac{1}{2}$
$\frac{2}{3}$
$\frac{1}{6}$
$\frac{5}{3}$
Answer: (c)
Solution
OP = $\left$| $\frac{2 - 3 + 2}{\sqrt{2}}$ $\right$| OP = $\frac{3}{\sqrt{2}}$ AP = $\sqrt{OA^2 - OP^2}$ = $\frac{1}{\sqrt{2}}$ tan $\theta$ = 3 $\therefore$ $\sin$ $\theta$ = $\frac{3}{\sqrt{10}}$ = $\frac{AP}{AN}$ $\Rightarrow$ AN = $\frac{\sqrt{5}}{3}$ = BN Area of $\triangle$ ANB = $\frac{1}{2}$ $\cdot$ (AN^2) $\sin$ 2$\theta$ = $\frac{1}{6}$
Question 20
Maths · Statistics · Single correct
Let the mean and the variance of 5 observations $x_1, x_2, x_3, x_4, x_5$ be $\frac{24}{5}$ and $\frac{194}{25}$ respectively. If the mean and variance of the first 4 observation are $\frac{7}{2}$ and $a$ respectively, then $(4a + x_5)$ is equal to:
Maths · Complex Numbers and Quadratic Equations · Numerical
Let $S = \{ z \in \mathbb{C} : |z - 2| \leq 1, z(1+i) + \overline{z}(1-i) \leq 2 \}$. Let $|z - 4i|$ attains minimum and maximum values, respectively, at $z_1 \in S$ and $z_2 \in S$. If $5(|z_1|^2 + |z_2|^2) = \alpha + \beta \sqrt{5}$, where $\alpha$ and $\beta$ are integers, then the value of $\alpha + \beta$ is equal to _____.
Let $y = y(x)$ be the solution of the differential equation $$\frac{dy}{dx} + \frac{\sqrt{2}\,y}{2\cos^4 x - \cos 2x} = x e^{\tan^{-1}(\sqrt{2}\cot 2x)}, \quad 0 < x < \frac{\pi}{2}$$ with $y\left(\dfrac{\pi}{4}\right) = \dfrac{\pi^2}{32}$. If $y\left(\dfrac{\pi}{3}\right) = \dfrac{\pi^2}{18}\, e^{-\tan^{-1}(\alpha)}$, then the value of $3\alpha^2$ is equal to _____.
Answer: 2
Solution
Given $\dfrac{dy}{dx} + \dfrac{\sqrt{2}}{2\cos^4 x - \cos 2x}\,y = x e^{\tan^{-1}(\sqrt{2}\cot 2x)}$. $$\int \frac{dx}{2\cos^4 x - \cos 2x}$$ $$= \int \frac{dx}{\cos^4 x + \sin^4 x} = \int \frac{\csc^4 x\,dx}{1 + \cot^4 x}$$ $$= \int \frac{t^2+1}{t^4+1}\,dt = -\int \left(\frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2+2}\right)dt = -\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)$$ Let $\cot x = t$ $$= -\frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\cot 2x)$$ Therefore, if $E = e^{-\tan^{-1}(\sqrt{2}\cot 2x)}$ $$y\, e^{-\tan^{-1}(\sqrt{2}\cot 2x)} = \int x\,dx$$ $$y\, e^{-\tan^{-1}(\sqrt{2}\cot 2x)} = \frac{x^2}{2} + c$$ $$y\left(\frac{\pi}{4}\right) = \frac{\pi^2}{32} + c \Rightarrow c = 0$$ $$y = \frac{x^2}{2}\,e^{\tan^{-1}(\sqrt{2}\cot 2x)}$$ $$y\left(\frac{\pi}{3}\right) = \frac{\pi^2}{18}\,e^{\tan^{-1}\left(\sqrt{2}\cot\frac{2\pi}{3}\right)}$$ $$= \frac{\pi^2}{18}\,e^{-\tan^{-1}\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}$$ $\alpha = \dfrac{\sqrt{2}}{\sqrt{3}} \Rightarrow 3\alpha^2 = 2$
Question 23
Maths · Three Dimensional Geometry · Numerical
Let $d$ be the distance between the foot of perpendiculars of the points $P(1, 2 - 1)$ and $Q(2, -1, 3)$ on the plane $- x + y + z = 1$. Then $d^2$ is equal to _______.
Answer: 26
Solution
Points $P(1, 2, -1)$ and $Q(2, -1, 3)$ lie on the same side of the plane. Perpendicular distance of point $P$ from plane is $$\frac{| -1 + 2 - 1 - 1 |}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{1}{\sqrt{3}}$$ Perpendicular distance of point $Q$ from plane is $$\frac{| -2 - 1 + 3 - 1 |}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{1}{\sqrt{3}}$$ Therefore, $\overline{PQ}$ is parallel to the given plane. So, distance between $P$ and $Q$ = distance between their foot of perpendiculars. $$\Rightarrow |\overline{PQ}| = \sqrt{(1 - 2)^2 + (2 + 1)^2 + (-1 - 3)^2}$$ $$= \sqrt{26}$$ $$|\overline{PQ}|^2 = 26 = d^2$$ Alternate $$-x + y + z - 1 = 0$$ $P(1, 2, -1)$ $Q(2, -1, 3)$ $$M(x_1, y_1, z_1)$$ $$\frac{x_1 - 1}{-1} = \frac{y_1 - 2}{1} = \frac{z_1 + 1}{1} = \frac{1}{3}$$ $$x_1 = \frac{2}{3}, y_1 = \frac{7}{3}, z_1 = \frac{-2}{3}$$ $$M\left(\frac{2}{3}, \frac{7}{3}, \frac{-2}{3}\right)$$ $$N(x_2, y_2, z_2)$$ $$\frac{x_2 - 2}{-1} = \frac{y_2 + 1}{1} = \frac{z_2 - 3}{1} = \frac{1}{3}$$ $$x_2 = \frac{5}{3}, y_2 = \frac{-2}{3}, z_2 = \frac{10}{3}$$ $$N = \left(\frac{5}{3}, \frac{-2}{3}, \frac{10}{3}\right)$$
Question 24
Maths · Trigonometric Functions · Numerical
The number of elements in the set $S = \{ \theta \in [-4\pi, 4\pi] : 3 \cos^2 2\theta + 6 \cos 2\theta - 10 \cos^2 \theta + 5 = 0 \}$ is
Maths · Relations and Functions · Fill in the blank
Let c, k $\in$ $\mathbb{R}$. If f(x) = $\frac{(c+1) x^2 + (1-c^2) x + 2k}{\overline{}}$ and f(x+y) = f(x) + f(y) - xy, for all x, y $\in$ $\mathbb{R}$, then the value of |2(f(1) + f(2) + f(3) + $\ldots$ + f(20))| is equal to _______.
Let $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, $a > 0$, $b > 0$, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4\left(2\sqrt{2} + \sqrt{14}\right)$. If the eccentricity $H$ is $\frac{\sqrt{11}}{2}$, then value of $a^2 + b^2$ is equal to ______.
Let $P_1 : \vec{r} \cdot (2\hat{i} + \hat{j} - 3\hat{k}) = 4$ be a plane. Let $P_2$ be another plane which passes through the points $(2, -3, 2)$ $(2, -2, -3)$ and $(1, -4, 2)$. If the direction ratios of the line of intersection of $P_1$ and $P_2$ be $16, \alpha, \beta$, then the value of $\alpha + \beta$ is equal to ______.
Answer: 28
Solution
Given $\($ P_1 : $\mathbf{r}$ $\cdot$ (2$\hat{i}$ + $\hat{j}$ - 3$\hat{k}$) = 4 $\)$ and $\($ P_1 : 2x + y - 3z = 4 $\)$. For $\($ P_2 $\)$, $\[$ $\begin{vmatrix}$ x - 2 & y + 3 & z - 2 $\\$ 0 & 1 & -5 $\\$ -1 & -1 & 0 $\end{vmatrix}$ = 0 $\]$ This simplifies to $\($-5x + 5y + z + 23 = 0$\)$. Let $\($ a, b, c $\)$ be the direction ratios of the line of intersection. Then $\($ a = $\frac{16\lambda}{15}$ $\)$; $\($ b = $\frac{13\lambda}{15}$ $\)$; $\($ c = $\frac{15\lambda}{15}$ $\)$. Therefore, $\($ $\alpha$ = 13 $\)$ and $\($ $\beta$ = 15 $\)$.
Question 30
Maths · Permutations and Combinations · Numerical
Let $b_1b_2b_3b_4$ be a 4-element permutation with $b_i \in \{1, 2, 3, \ldots, 100\}$ for $1 \leq i \leq 4$ and $b_i \neq b_j$ for $i \neq j$, such that either $b_1, b_2, b_3$ are consecutive integers or $b_2, b_3, b_4$ are consecutive integers. Then the number of such permutations $b_1b_2b_3b_4$ is equal to .
Answer: 18915
Solution
Let $A$ be the set when $b_1, b_2, b_3$ are consecutive. $$n(A) = \frac{97 + 97 + \ldots + 97}{98 times} = 97 \times 98$$ Similarly, when $b_2, b_3, b_4$ are consecutive, $$N(A) = 97 \times 98$$ $$n(A \cap B) = \frac{97 + 97 + \ldots + 97}{98 times} = 97 \times 98$$ Similarly, when $b_2, b_3, b_4$ are consecutive, $$n(B) = 97 \times 98$$ $$n(A \cap B) = 97$$ $$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$ Number of permutations = 18915
Physics
Question 31
Physics · Motion in a Straight Line · Single correct
Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at $t = 0$ s. Ball B is thrown vertically down with an initial velocity 'u' at $t = 2$ s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ms$^{-1}$. [use $g = 10$ ms$^{-2}$]:
10
15
20
30
Answer: (d)
Solution
Let they meet at time t. $$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 80}{10}}$$ $$= 4 sec$$ Time taken by ball B to meet A = 2 sec Using $S = ut + \frac{1}{2}at^2$ $$-80 = -u \times 2 + \frac{1}{2}(-10)(2)^2$$ $$u = 30$$
Question 32
Physics · System of Particles and Rotational Motion · Single correct
A body of mass $M$ at rest explodes into three pieces, in the ratio of masses $1 : 1 : 2$. Two smaller pieces fly off perpendicular to each other with velocities of $30 \, \mathrm{ms^{-1}}$ and $40 \, \mathrm{ms^{-1}}$ respectively. The velocity of the third piece will be:
$15 \, \mathrm{ms^{-1}}$
$25 \, \mathrm{ms^{-1}}$
$35 \, \mathrm{ms^{-1}}$
$50 \, \mathrm{ms^{-1}}$
Answer: (b)
Solution
Mass of pieces by $\frac{M}{4}$, $\frac{M}{4}$, $\frac{M}{2}$. Conserving momentum $\vec{P}_1 + \vec{P}_2 + \vec{P}_3 = 0$. $\vec{P}_3 = - (\vec{P}_1 + \vec{P}_2)$. As $\vec{P}_1$ and $\vec{P}_2$ are perpendicular, so $P_3 = \sqrt{P_1^2 + P_2^2}$. $P_3 = (50) \frac{M}{4}$. $\& \ P_3 = \frac{M}{2} v$. So $v = 25$.
Question 33
Physics · Nuclei · Single correct
The activity of a radioactive material is $2.56 \times 10^{-3}$ Ci. If the half life of the material is 5 days, after how many days the activity will become $2 \times 10^{-5}$ Ci?
30 days
35 days
40 days
25 days
Answer: (b)
Solution
Given $\($ $\frac{A}{A_0}$ = $\frac{N}{N_0}$ $\)$. \[ \frac{2 \times 10^{-5}}{2.56 \times 10^{-3}} = \frac{N}{N_0} \] \[ \frac{N}{N_0} = \frac{1}{128} \Rightarrow N = \frac{N_0}{128} \] After 7 half life activity comes down to given value. $\($ T = 7 $\times$ 5 $\)$ $\($ = 35 $\)$ days
Question 34
Physics · System of Particles and Rotational Motion · Single correct
A spherical shell of 1 kg mass and radius R is rolling with angular speed $\omega$ on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is $\frac{a}{3} R^2 \omega$. The value of $a$ will be:
2
3
5
4
Answer: (c)
Solution
Let $L_0$ be the angular momentum of the shell about $O$. As the shell is rolling, we have $V_{cm} = \omega R$. Therefore, $L_0 = m V_{cm} R + I \omega$. This becomes $$L_0 = 1 \times \omega R \times R + \frac{2}{3} R^2 \omega$$ $$= \frac{5}{3} R^2 \omega$$ so $a = 5$.
Question 35
Physics · Thermodynamics · Single correct
A cylinder of fixed capacity of 44.8 litres contains helium gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by $20.0^\circ\mathrm{C}$ will be: (Given gas constant $R = 8.3\,\mathrm{J\,K^{-1}\,mol^{-1}}$)
249 \, $\mathrm{J}$
415 \, $\mathrm{J}$
498 \, $\mathrm{J}$
830 \, $\mathrm{J}$
Answer: (c)
Solution
Question 36
Physics · Laws of Motion · Single correct
A wire of length L is hanging from a fixed support. The length changes to $L_1$ and $L_2$ when masses $1\,\mathrm{kg}$ and $2\,\mathrm{kg}$ are suspended respectively from its free end. Then the value of L is equal to:
Physics · Dual Nature of Radiation and Matter · Single correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : The photoelectric effect does not take place, if the energy of the incident radiation is less than the work function of a metal. Reason R : Kinetic energy of the photoelectrons is zero, if the energy of the incident radiation is equal to the work function of a metal. In the light of the above statements, choose the most appropriate answer from the options given below.
Both A and R are correct and R is the correct explanation of A
Both A and R are correct but R is not the correct explanation of A
A is correct but R is not correct
A is not correct but R is correct
Answer: (b)
Solution
To free the electron from metal surface minimum energy required, is equal to the work function of that metal. So Assertion A, is correct. $$h\nu = w_0 + K. E._{\max}$$ if $$h\nu = w_0$$ $$\Rightarrow K.E._{\max} = 0$$ Hence reason R, is correct, But R is not the correct explanation of A.
Question 38
Physics · Work, Energy and Power · Single correct
A particle of mass 500 gm is moving in a straight line with velocity $\upsilon = b \, x^{5/2}$. The work done by the net force during its displacement from $x = 0$ to $x = 4 \, \mathrm{m}$ is : (Take $b = 0.25 \, \mathrm{m}^{-3/2} \, \mathrm{s}^{-1}$).
2 J
4 J
8 J
16 J
Answer: (d)
Solution
By work energy theorem, work done by net force equals the change in kinetic energy. $$w = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$$ $$w = \frac{1}{2} \times 0.5 \times (0.25)^2 \times (4)^5$$ $$w = 16 \, \mathrm{J}$$
Question 39
Physics · Moving Charges and Magnetism · Single correct
A charged particle moves along circular path in a uniform magnetic field in a cyclotron. The kinetic energy of the charged particle increases to 4 times its initial value. What will be the ratio of new radius to the original radius of circular path of the charged particle:
1 : 1
1 : 2
2 : 1
1 : 4
Answer: (c)
Solution
Radius of particle in cyclotron $$r = \frac{\sqrt{2m \mathrm{K.E.}}}{qB}$$ So ratio of new radius to original $$\frac{r_n}{r_0} = \sqrt{\frac{(\mathrm{K.E.})_n}{(\mathrm{K.E.})_0}} = \sqrt{4} \Rightarrow 2:1 \ (\mathrm{C})$$
Question 40
Physics · Alternating Current · Multiple correct
For a series LCR circuit, $I$ vs $\omega$ curve is shown: \begin{enumerate} \item[(a)] To the left of $\omega_r$, the circuit is mainly capacitive. \item[(b)] To the left of $\omega_r$, the circuit is mainly inductive. \item[(c)] At $\omega_r$, the impedance of the circuit is equal to the resistance of the circuit. \item[(d)] At $\omega_r$, impedance of the circuit is $0$. \end{enumerate}
$(a)$ and $(d)$ only
$(b)$ and $(d)$ only
$(a)$ and $(c)$ only
$(b)$ and $(c)$ only
Answer: (c)
Solution
At $\omega_r$, $X_C = X_L$. $$\frac{1}{\omega_r C} = \omega_r L$$ So if $\omega < \omega_r$ then $x_C$ will increase and $X_L$ will decrease. Hence to the left of $\omega_r$, the circuit is capacitive. $$Z = \sqrt{R^2 + (X_C - X_L)^2}$$ At $\omega_r$, $Z = \sqrt{R^2 + 0^2} = R$ (C)
Question 41
Physics · Laws of Motion · Single correct
A block of metal weighing 2 kg is resting on a frictionless plane (as shown in figure). It is struck by a jet releasing water at a rate of $1 \, \mathrm{kg \, s^{-1}}$ and at a speed of $10 \, \mathrm{m \, s^{-1}}$. Then, the initial acceleration of the block, in $\mathrm{m \, s^{-2}}$, will be:
Physics · Physical World, Units and Measurements · Single correct
In Vander Waals equation $$ P + \frac{a}{V^2} \left[ V - b \right] = RT; $$ P is pressure, V is volume, R is universal gas constant and T is temperature. The ratio of constants $\frac{a}{b}$ is dimensionally equal to:
$\frac{P}{V}$
$\frac{V}{P}$
PV
$PV^3$
Answer: (c)
Solution
By principle of homogeneity $$[P] = \left[ \frac{a}{v^2} \right] and [b] = [v]$$ $$\Rightarrow \left[ \frac{a}{b} \right] = [PV] (C)$$
Question 43
Physics · Mathematics in Physics · Single correct
Two vectors $\vec{A}$ and $\vec{B}$ have equal magnitudes. If magnitude of $\vec{A} + \vec{B}$ is equal to two times the magnitude of $\vec{A} - \vec{B}$, then the angle between $\vec{A}$ and $\vec{B}$ will be:
The escape velocity of a body on a planet 'A' is 12 $\mathrm{km\ s^{-1}}$. The escape velocity of the body on another planet 'B', whose density is four times and radius is half of the planet 'A', is :
12 $\mathrm{km\ s^{-1}}$
24 $\mathrm{km\ s^{-1}}$
36 $\mathrm{km\ s^{-1}}$
6 $\mathrm{km\ s^{-1}}$
Answer: (a)
Solution
Given $V_{escape} = \sqrt{\frac{2Gm}{R}} \implies \sqrt{\frac{2G \rho \times \frac{4}{3} \pi R^3}{R}}$. $V_{escape} \propto \sqrt{\rho R^2}$. Therefore, if $\rho$ is 4 times and Radius is halved, $\implies V_{escape}$ will remain same. Therefore, Ans (A)
Question 45
Physics · Magnetism and Matter · Single correct
At a certain place the angle of dip is $30^\circ$ and the horizontal component of earth's magnetic field is $0.5 \, \mathrm{G}$. The earth's total magnetic field (in G), at that certain place, is:
$\frac{1}{\sqrt{3}}$
$\frac{1}{2}$
$\sqrt{3}$
1
Answer: (a)
Solution
Given $B_H = B \cos \theta$. Therefore, $$B = \frac{B_H}{\cos \theta} = \frac{0.5G}{\cos 30^\circ} \Rightarrow \frac{G}{\sqrt{3}}$$
Question 46
Physics · Waves · Single correct
A longitudinal wave is represented by $$x = 10 \sin 2\pi \left( nt - \frac{x}{\lambda} \right) cm.$$ The maximum particle velocity will be four times the wave velocity if the determined value of wavelength is equal to:
$2\pi$
$5\pi$
$\pi$
$\frac{5\pi}{2}$
Answer: (b)
Solution
Given $V_{\mathrm{p \ max}} = 4V_{\mathrm{wave}}$. We have $\omega A = 4 \left( \frac{\omega}{k} \right) \Rightarrow A = \frac{4 \lambda}{2 \pi}$. Therefore, $\lambda = \frac{2 \pi A}{4} \Rightarrow \frac{20 \pi}{4} \Rightarrow 5 \pi$.
Question 47
Physics · Electrostatic Potential and Capacitance · Single correct
A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will:
increase by 50%
decrease by 15%
increase by 25%
increase by 33%
Answer: (a)
Solution
E is given by $$\frac{1}{2}(KC)v^2$$. Therefore, the percentage change is given by $$\frac{\frac{1}{2}K_2CV^2 - \frac{1}{2}K_1CV^2}{\frac{1}{2}K_1CV^2} = \frac{K_2 - K_1}{K_1} \times 100$$. This simplifies to $$\frac{15 - 10}{10} \times 100 = 50\%$$.
Question 48
Physics · Electric Charges and Fields · Single correct
A positive charge particle of $100 \, \mathrm{mg}$ is thrown in opposite direction to a uniform electric field of strength $1 \times 10^5 \, \mathrm{NC^{-1}}$. If the charge on the particle is $40 \, \mu \mathrm{C}$ and the initial velocity is $200 \, \mathrm{ms^{-1}}$, how much distance it will travel before coming to the rest momentarily:
$1\,\mathrm{m}$
$5\,\mathrm{m}$
$10\,\mathrm{m}$
$0.5\,\mathrm{m}$
Answer: (d)
Solution
Distance travelled by particle before stopping $$\frac{V^2}{2a} = S \Rightarrow \frac{v^2 m}{2qE} \Rightarrow \frac{(200)^2 \times 100 \times 10^{-6}}{2 \times 40 \times 10^{-6} \times 10^5} = 0.5 \, \mathrm{m}$$
Question 49
Physics · Wave Optics · Single correct
Using Young's double slit experiment, a monochromatic light of wavelength 5000 Å produces fringes of fringe width 0.5 mm. If another monochromatic light of wavelength 6000 Å is used and the separation between the slits is doubled, then the new fringe width will be:
Only 2$\%$ of the optical source frequency is the available channel bandwidth for an optical communicating system operating at 1000 $\mathrm{nm}$. If an audio signal requires a bandwidth of 8 $\mathrm{kHz}$, how many channels can be accommodated for transmission:
375 $\times$ 10^7
75 $\times$ 10^7
375 $\times$ 10^8
75 $\times$ 10^9
Answer: (b)
Solution
Frequency at 1000 nm = $\frac{3 \times 10^8}{1000 \times 10^{-9}} \Rightarrow 3 \times 10^{14} \, \mathrm{Hz}$ available for channel bandwidth $$= \frac{2}{100} \times 3 \times 10^{14} \Rightarrow 6 \times 10^{12} \, \mathrm{Hz}$$ Bandwidth for 1 channel = $8000 \, \mathrm{Hz}$ Therefore, number of channels $$= \frac{6 \times 10^{12}}{8 \times 10^3} \Rightarrow \frac{600}{8} \times 10^7 = 75 \times 10^7$$
Question 51
Physics · Current Electricity · Fill in the blank
Two coils require 20 minutes and 60 minutes respectively to produce same amount of heat energy when connected separately to the same source. If they are connected in parallel arrangement to the same source; the time required to produce same amount of heat by the combination of coils, will $\dots$ min.
Answer: 15
Solution
We know $$\frac{dQ}{dt} = i^2 R = \frac{V^2}{R}$$. In 't' time, $$\Delta Q = \left( \frac{V^2}{R} \right) t$$. Given that, for the same source, $$v = same$$, $$Q_0 = \frac{V^2}{R_1} \times 20 = \frac{V^2}{R_2} \times 60 \ldots (1)$$. Thus, $$R_2 = 3R_1 \ldots (ii)$$. If they are connected in parallel, then $$Req = \frac{R_2 R_1}{R_1 + R_2} = \frac{3R_1 \cdot R_1}{3R_1 + R_1} = \frac{3R_1}{4}$$. To produce the same heat, using equation (1), $$Q_0 = \frac{V^2}{R_1} \times 20 = \frac{V^2}{\frac{3R_1}{4}} \times t$$. Therefore, $$t = \frac{3 \times 20}{4} = 15 min$$.
Question 52
Physics · Electromagnetic Waves · Numerical
The intensity of the light from a bulb incident on a surface is $0.22 \, \mathrm{W/m^2}$. The amplitude of the magnetic field in this light-wave is _____ $\times 10^{-9} \, \mathrm{T}$. (Given: Permittivity of vacuum $\epsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C^2 N^{-1} m^{-2}}$, speed of light in vacuum $c = 3 \times 10^8 \, \mathrm{ms^{-1}}$)
Physics · Thermal Properties of Matter · Numerical
As per the given figure, two plates A and B of thermal conductivity K and 2 K are joined together to form a compound plate. The thickness of plates are 4.0 $\mathrm{cm}$ and 2.5 $\mathrm{cm}$ respectively and the area of cross-section is 120 $\mathrm{cm}^2$ for each plate. The equivalent thermal conductivity of the compound plate is $\left(1 + \frac{5}{\alpha}\right) K$, then the value of $\alpha$ will be ________.
Answer: 21
Solution
The heat transfer rate is given by $$\frac{\Delta Q}{\Delta t} = \left( \frac{1}{R} \right) \Delta T$$ where $R$ is the thermal resistivity. Therefore, $$R_1 = \frac{L_1}{K_1 A} = \frac{L_1}{K(120)}$$ with $L_1 = 4 \, \mathrm{cm}$ and $A = 120 \, \mathrm{cm^2}$. For $R_2$, $$R_2 = \frac{2.5}{(2K)(120)}.$$ Now, the equivalent resistance $R_{eq}$ of this series combination is $$R_{eq} = R_1 + R_2$$ where $$L_{eq} = 4 + 2.5 = 6.5.$$ Thus, $$\frac{L_{eq}}{K_{eq}(A)} = \frac{4}{K(120)} + \frac{5}{2(2K)(120)}.$$ Simplifying, $$\frac{6.5}{K_{eq}(120)} = \frac{4}{K(120)} + \frac{5}{4K(120)}.$$ Therefore, $$\frac{6.5}{K_{eq}} = \frac{21}{4K}$$ and $$K_{eq} = \frac{26}{21} K = \left( 1 + \frac{5}{21} \right) K.$$ Thus, the answer is $a = 21.$
Question 54
Physics · Oscillations · Numerical
A body is performing simple harmonic with an amplitude of 10 $\mathrm{\ cm}$. The velocity of the body was tripled by air Jet when it is at 5 $\mathrm{\ cm}$ from its mean position. The new amplitude of vibration is $\sqrt{x}$ $\mathrm{\ cm}$. The value of x is__________.
Answer: 700
Solution
Given $A = 10 \, \mathrm{cm}$. Therefore, Total Energy $= \frac{1}{2} KA^2$. By energy conservation we can find $v$ at $x = 5$: $$\frac{1}{2} K(10)^2 = \frac{1}{2} K(5)^2 + \frac{1}{2} mv^2$$ $$V = \sqrt{\frac{75K}{m}}$$ Now, velocity is tripled through external means so the amplitude of SHM will change and so the total energy, (but potential energy at this moment will remain same). $$\therefore \frac{1}{2} K(5)^2 + \frac{1}{2} m \left(3 \sqrt{\frac{75K}{m}}\right)^2 = \frac{1}{2} KA^2$$ $$\Rightarrow 25 \, K + 675 \, K = KA^2$$ $$\therefore A = \sqrt{700}$$ Therefore, $x = 700$
Question 55
Physics · Current Electricity · Numerical
The variation of applied potential and current flowing through a given wire is shown in figure. The length of wire is 31.4 cm. The diameter of wire is measured as 2.4 cm. The resistivity of the given wire is measured as $x \times 10^{-3} \, \Omega \, \mathrm{cm}$. The value of $x$ is _________. [Take $\pi = 3.14$]
Answer: 144
Solution
Given the equation $1 = \rho \frac{\ell}{A}$. Substituting the values, we have: $$1 = \frac{\rho \times 31.4}{\frac{\pi (2.4)^2}{4}}$$ Simplifying, we get: $$\frac{\pi (2.4)^2}{4} = \rho \times 314$$ Further simplification gives: $$\frac{2.4 \times 2.4}{4} = \rho \times 10$$ Continuing, we have: $$\frac{0.6 \times 2.4}{10} = \rho$$ Finally, we find: $$\frac{1.44}{10} = \rho$$ Thus, $0.144 = \rho$.
Question 56
Physics · Thermodynamics · Fill in the blank
300 cal. of heat is given to a heat engine and it rejects 225 cal. of heat. If source temperature is $227^\circ \mathrm{C}$, then the temperature of sink will be __ $^\circ \mathrm{C}$.
Answer: 102
Solution
Question 57
Physics · Atoms · Fill in the blank
$\sqrt{d_1}$ and $\sqrt{d_2}$ are the impact parameters corresponding to scattering angles $60°$ and $90°$ respectively, when an $\alpha$ particle is approaching a gold nucleus. For $d_1 = x\,d_2$, the value of $x$ will be _______.
Answer: 3
Solution
Given $\sqrt{d} \propto \cot \frac{\theta}{2}$. $$\cot^2 30^\circ = x \cot^2 45^\circ$$ $$3 = x$$
A transistor is used in an amplifier circuit in common emitter mode. If the base current changes by $100 \, \mu \mathrm{A}$, it brings a change of $10 \, \mathrm{mA}$ in collector current. If the load resistance is $2 \, \mathrm{k}\Omega$ and input resistance is $1 \, \mathrm{k}\Omega$, the value of power gain is $x \times 10^4$. The value of $x$ is _______.
Answer: 2
Solution
Given $\Delta i_B = 100 \, \mu\mathrm{A}$ and $\Delta i_C = 10 \, \mathrm{mA}$. The current gain $\beta$ is given by $\beta = \frac{\Delta i_C}{\Delta i_B}$. The power is calculated as $power = \beta^2 \times \frac{R_0}{R_{in}}$. Substituting the values, we have $$Power = \left(\frac{10}{0.1}\right)^2 \times \frac{2}{1}$$ $$Power = 100 \times 100 \times 2$$ Therefore, the gain is $2 \times 10^4$.
Question 59
Physics · Ray Optics and Optical Instruments · Numerical
A parallel beam of light is allowed to fall on a transparent spherical globe of diameter 30 cm and refractive index 1.5. The distance from the centre of the globe at which the beam of light can converge is ________ mm.
Answer: 225
Solution
Given $\mu = \frac{3}{2}$ and the lens setup with $15 \, \mathrm{cm}$ radii. $$\frac{3}{2V} - \frac{1}{\infty} = \frac{3}{2} - \frac{1}{15}$$ $$\frac{3}{2V} = \frac{1}{30}$$ Solving for $V$, we get $V = 45 \, \mathrm{cm}$. Next, using the lens formula: $$\frac{1}{V} - \frac{3}{2} \frac{1}{15} = \frac{1 - \frac{3}{2}}{-15}$$ $$\frac{1}{V} - \frac{1}{10} = \frac{1}{30}$$ $$\frac{1}{V} = \frac{1}{10} + \frac{1}{30} = \frac{4}{30}$$ Solving for $V$, we find $V = 7.5$. Then, $V = 22.5$. Finally, $v = 225 \, \mathrm{mm}$.
Question 60
Physics · Current Electricity · Numerical
For the network shown below, the value $V_B - V_A$ is _______ V.
Answer: 10
Solution
The current $i$ is calculated as $$i = \frac{15}{3} = 5 \, \mathrm{A}$$ The voltage is calculated as $$15 - 5 \times 1 = 10 \, \mathrm{Volt}$$
Chemistry
Question 61
Chemistry · Some Basic Concepts of Chemistry · Single correct
Production of iron in blast furnace follows the following equation $$\mathrm{Fe_3O_4(s) + 4CO(g) \rightarrow 3Fe(l) + 4CO_2(g)}$$ when 4.640 kg of $\mathrm{Fe_3O_4}$ and 2.520 kg of CO are allowed to react then the amount of iron (in g) produced is : [Given : Molar Atomic mass (g mol$^{-1}$): Fe = 56 Molar Atomic mass (g mol$^{-1}$) : O = 16 Molar Atomic mass (g mol$^{-1}$): C = 12
1400
2200
3360
4200
Answer: (c)
Solution
Moles of $\mathrm{Fe_3O_4} = \frac{4.640 \times 10^3}{232} = 20$ Moles of CO $= \frac{2.52 \times 10^3}{28} = 90$ So limiting Reagent $= \mathrm{Fe_3O_4}$ So moles of Fe formed $= 60$ Weight of Fe $= 60 \times 56 = 3360$ gms
Question 62
Chemistry · Structure of Atom · Single correct
Which of the following statements are correct ? (A) The electronic configuration of Cr is $[\mathrm{Ar}] \, 3d^5 \, 4s^1$. (B) The magnetic quantum number may have a negative value. $(C)$ In the ground state of an atom, the orbitals are filled in order of their increasing energies. (D) The total number of nodes are given by $n - 2$. Choose the most appropriate answer from the options given below :
$(A), (C)$ and $(D)$ only
$(A)$ and $(B)$ only
$(A)$ and $(C)$ only
$(A), (B)$ and $(C)$ only
Answer: (d)
Solution
(A) Cr = $[\mathrm{Ar}]3d^5 \, 4s^1$ (B) $m = -\ell$ to $+\ell$ (C) According to Aufbau principle, orbitals are filled in order of their increasing energies. (D) Total nodes = $n - 1$
Question 63
Chemistry · Solutions · Numerical
1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is $0.0198^\circ\mathrm{C}$. The percentage of dissociation of the acid is ________. (Nearest integer) [Given : Density of acetic acid is $1.02 \, \mathrm{g} \, \mathrm{mL}^{-1}$ Molar mass of acetic acid is $60 \, \mathrm{g} \, \mathrm{mol}^{-1}$ $K_f (\mathrm{H_2O}) = 1.85 \, \mathrm{K} \, \mathrm{kg} \, \mathrm{mol}^{-1}$]
Answer: 5
Solution
$M = d \times V = 1.02 \times 1.2 = 1.224\,\mathrm{g}$ Moles of acetic acid $= 0.0204\,\mathrm{mol}$ in $2\,\mathrm{L}$ So molality $= 0.0102\,\mathrm{mol/kg}$ Now $\Delta T_f = i \times K_f \times m$ $i = 1 + \alpha$ for acetic acid $0.0198 = (1 + \alpha) \times 1.85 \times 0.0102$ $\alpha = 0.04928 \cong 5\%$
Question 64
Chemistry · Equilibrium · Single correct
The solubility of AgCl will be maximum in which of the following?
0.01 M KCl
0.01 M HCl
0.01 M AgNO_3
Deionised water
Answer: (d)
Solution
In deionized water no common ion effect will take place so maximum solubility
Question 65
Chemistry · Surface Chemistry · Single correct
Which of the following is a correct statement?
Brownian motion destabilises sols.
Any amount of dispersed phase can be added to emulsion without destabilising it.
Mixing two oppositely charged sols in equal amount neutralises charges and stabilises colloids.
Presence of equal and similar charges on colloidal particles provides stability to the colloidal solution.
Answer: (d)
Solution
As equal and similar charge particles will repel each other, hence will never precipitate.
Question 66
Chemistry · Structure of Atom · Single correct
The electronic configuration of Pt (atomic number 78) is:
[Xe] $4f^{14}$ $5d^9$ $6s^1$
[Kr] $4f^{14}$ $5d^{10}$
[Xe] $4f^{14}$ $5d^{10}$
[Xe] $4f^{14}$ $5d^8$ $6s^2$
Answer: (a)
Solution
The electronic configuration of $^{78}\mathrm{Pt}$ is $[\mathrm{Xe}] \, 4f^{14} \, 5d^9 \, 6s^1$. This is an exceptional electronic configuration.
Question 67
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
In isolation of which one of the following metals from their ores, the use of cyanide salt is not commonly involved?
Zinc
Gold
Silver
Copper
Answer: (d)
Solution
For $\mathrm{ZnS}$, $\mathrm{KCN}$ is used as depressant. For Gold and silver $\Rightarrow$ leaching [Cyanide process]
Question 68
Chemistry · Hydrogen · Single correct
Which one of the following reactions indicates the reducing ability of hydrogen peroxide in basic medium?
In option (A) and (C) reducing action of hydrogen peroxide is shown. In option (A) it is in acidic medium, in option (B) it is in basic medium. For reducing ability $\mathrm{H_2O_2}$ changes to $\mathrm{O_2}$, i.e. oxidize, so in option 'A' & 'C' $\mathrm{O_2}$ is formed but 'A' is in acidic medium so option - C correct.
Question 69
Chemistry · Amines · Single correct
Mathch List (I) with List ( II ) choose the answer:
(A)-(I), (B)-(II), (C )-(III), (D)-(IV)
(A)-(III), (B)-(I), (C )-(I), (D)-(IV)
(A)-(III), (B)-(I), (C )-(II), (D)-(IV)
(A)-(IV), (B)-(II), (C )-(I), (D)-(III)
Answer: (a)
Solution
Question 70
Chemistry · The s-Block Elements · Single correct
Match the List-I with List-II. \begin{tabular}{|c|c|c|l|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{(Metal)} & \multicolumn{2}{c|}{Application} \\ \hline (A) & Cs & (I) & High temperature thermometer \\ \hline (B) & Ga & (II) & Water repellent sprays \\ \hline (C) & B & (III) & Photoelectric cells \\ \hline (D) & Si & (IV) & Bullet proof vest \\ \hline \end{tabular} Choose the most appropriate answer from the option given below:
(A) -(III), (B)-(I), $(C)$-(IV), (D)-(II)
(A) -(IV), (B)-(III), $(C)$-(II), (D)-(I)
(A) -(II), (B)-(III), $(C)$-(IV), (D)-(I)
(A) -(I), (B)-(IV), $(C)$-(II), (D)-(III)
Answer: (a)
Solution
Caesium is used in devising photoelectric cells. Boron fibres are used in making bullet-proof vest. Silicones being surrounded by non-polar alkyl groups are water repelling in nature. Gallium is less toxic and has a very high boiling point, so it is used in high temperature thermometers.
Question 71
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
The oxoacid of phosphorus that is easily obtained from a reaction of alkali and white phosphorus and has two P-H bonds, is:
Phosphonic acid
Phosphinic acid
Pyrophosphorus acid
Hypophosphoric acid
Answer: (b)
Solution
The reaction is given by: $$\mathrm{P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2}$$ The oxoacid is $\mathrm{H_3PO_2}$ (hypophosphorous acid) or (phosphinic acid).
Question 72
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
The acid that is believed to be mainly responsible for the damage of Taj Mahal is
Sulfuric acid
Hydrofluoric acid
Phosphoric acid
Hydrochloric acid
Answer: (a)
Solution
The reaction is given by: $$\mathrm{CaCO_3 + H_2SO_4 \rightarrow CaSO_4 + H_2O + CO_2}$$
Question 73
Chemistry · Hydrocarbons · Single correct
Two isomers 'A' and 'B' with molecular formula $\mathrm{C_4H_8}$ give different products on oxidation with $\mathrm{KMnO_4}$ in acidic medium. Isomer 'A' on reaction with $\mathrm{KMnO_4/H^+}$ results in effervescence of a gas and gives ketone. The compound 'A' is
But-1-ene
cis-But-2-ene
trans-But-2-ene
2-methyl propene
Answer: (d)
Solution
Question 74
Chemistry · Alcohols, Phenols and Ethers · Single correct
In the given conversion the compound A is:
Answer: (B)
Solution
The reaction starts with the brominated aromatic compound. It reacts with RLi to form a lithium intermediate. This intermediate then reacts with carbon dioxide to form a lithium carboxylate. Finally, the addition of $\mathrm{H_3O^+}$ leads to the formation of the carboxylic acid.
Question 75
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Given below are two statements : Statement I : The esterification of carboxylic acid with an alcohol is a nucleophilic acyl substitution. Statement II : Electron withdrawing groups in the carboxylic acid will increase the rate of esterification reaction. Choose the most appropriate option :
Both Statement I and Statement II are correct.
Both Statement I and Statement II are incorrect.
Statement I is correct but Statement II is incorrect.
Statement I is incorrect but Statement II is correct.
Answer: (a)
Solution
The reaction is: $$R--OH + R--C--OH \rightarrow R--O--C--R$$ This is a nucleophilic acyl substitution. An electron withdrawing group on carboxylic acid will increase the rate of esterification.
Question 76
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Consider the above reaction, the product A and product B respectively are
Answer: (c)
Solution
Question 77
Chemistry · Polymers · Single correct
The polymer, which can be stretched and retains its original status on releasing the force is
Bakelite
Nylon 6,6
Buna-N
Terylene
Answer: (c)
Solution
Buna-N is synthetic rubber which can be stretched and retains its original status on releasing the force.
Question 78
Chemistry · Biomolecules · Single correct
Sugar moiety in DNA and RNA molecules respectively are
$\beta$-D-2-deoxyribose, $\beta$-D-deoxyribose
$\beta$-D-2-deoxyribose, $\beta$-D-ribose
$\beta$-D-ribose, $\beta$-D-2-deoxyribose
$\beta$-D-deoxyribose, $\beta$-D-2-deoxyribose
Answer: (b)
Solution
DNA contains $\Rightarrow \beta - D - 2 - deoxyribose$ RNA contains $\Rightarrow \beta - D - ribose$
Question 79
Chemistry · Chemistry in Everyday Life · Single correct
Which of the following compound does not contain sulfur atom?
Cimetidine
Ranitidine
Histamine
Saccharin
Answer: (c)
Solution
Histamine is a nitrogenous compound. It does not contain sulfur.
Question 80
Chemistry · Alcohols, Phenols and Ethers · Single correct
Given below are two statements. Statement I : Phenols are weakly acidic. Statement II : Therefore they are freely soluble in NaOH solution and are weaker acids than alcohols and water. Choose the most appropriate option:
Both Statement I and Statement II are correct.
Both Statement I and Statement II are incorrect.
Statement I is correct but Statement II is incorrect.
Statement I is incorrect but Statement II is correct.
Answer: (c)
Solution
Phenol are weakly acidic. Phenol is more acidic than alcohol and $\mathrm{H_2O}$. Statement (I) is correct. (II) is incorrect.
Question 81
Chemistry · Some Basic Concepts of Chemistry · Numerical
Geraniol, a volatile organic compound, is a component of rose oil. The density of the vapour is $0.46 \, \mathrm{gL^{-1}}$ at $257^\circ\mathrm{C}$ and $100 \, \mathrm{mm}$ \mathrm{Hg} . The molar mass of geraniol is _________ (Nearest Integer) [Given R = 0.082 \, \mathrm{L} \, \mathrm{atm} \, $\mathrm{K^{-1}}$ \, $\mathrm{mol^{-1}}$]
Answer: 152
Solution
Assuming ideal behaviour, $P=\frac{dRT}{M}$ $P=\frac{100}{760}\,\mathrm{atm},\quad T=257+273=530\,\mathrm{K}$ $d=0.46\,\mathrm{gm/L}$ So, $M=\frac{0.46\times0.082\times530}{100}\times760$ $=151.93\approx152$
Question 82
Chemistry · Thermodynamics · Numerical
17.0 $\mathrm{g}$ of $\mathrm{NH}_3$ completely vapourises at $-33.42^\circ$ $\mathrm{C}$ and 1 bar pressure and the enthalpy change in the process is 23.4 $\mathrm{kJ}$ $\mathrm{mol}^{-1}$. The enthalpy change for the vapourisation of 85 $\mathrm{g}$ of $\mathrm{NH}_3$ under the same conditions is _____ $\mathrm{kJ}$.
Answer: 117
Solution
Given data is for 1 mole and asked for 5 moles so value is $23.4 \times 5 = 117 \, \mathrm{kJ}$
Question 83
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Arrange the following in the decreasing order of their covalent character: (A) LiCl (B) NaCl $(C)$ KCl (D) CsCl Question: Choose the most appropriate answer from the options given below:
$(A)> (C) > (B) > (D)$
$(B) > (A) > (C) > (D)$
$(A)> (B) > (C) > (D)$
$(A) > (B) > (D) > (C)$
Answer: (c)
Solution
LiCl > NaCl > KCl > CsCl (Covalent character)
Question 84
Chemistry · Electrochemistry · Numerical
A dilute solution of sulphuric acid is electrolysed using a current of $0.10 \, \mathrm{A}$ for 2 hours to produce hydrogen and oxygen gas. The total volume of gases produced at STP is ______ cm$^3$. (Nearest integer) [Given : Faraday constant $F = 96500 \, \mathrm{C} \, \mathrm{mol}^{-1}$ at STP, molar volume of an ideal gas is $22.7 \, \mathrm{L} \, \mathrm{mol}^{-1}$]
Answer: 127
Solution
At anode $$2\mathrm{H_2O} \rightarrow \mathrm{O_2}(g) + 4\mathrm{H^+} + 4e^-$$ At cathode $$2\mathrm{H^+} + 2e^- \rightarrow \mathrm{H_2}(g)$$ Now number of gm eq. = $\($ $\frac{i \times t}{96500}$ $\)$ $$= \frac{0.1 \times 2 \times 60 \times 60}{96500}$$ $$= 0.00746$$ $$V_{\mathrm{O_2}} = \frac{0.00746}{4} \times 22.7 = 0.0423$$ $$V_{\mathrm{H_2}} = \frac{0.00746}{2} \times 22.7 = 0.0846$$ $$V_{Total} \approx 127 ml or cc$$
Question 85
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
The activation energy of one of the reactions in a biochemical process is $53261 \, \mathrm{J \, mol^{-1}}$. When the temperature falls from $310 \, \mathrm{K}$ to $300 \, \mathrm{K}$, the change in rate constant observed is $k_{300} = x \times 10^{-3} k_{310}$. The value of $x$ is ___. [Given: $\ln 10 = 2.3$ $\newline$ $R=8.3 \, \mathrm{J \, K^{-1} \, mol^{-1}}$]
Answer: 1
Solution
Given $$\ln \left( \frac{K_2}{K_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$$ Substituting the values, $$\ln \left( \frac{K_2}{K_1} \right) = \frac{532611}{8.3} \times \left( \frac{10}{310 \times 300} \right)$$ where $K_2$ is at $310 \, \mathrm{K}$ and $K_1$ is at $300 \, \mathrm{K}$. This simplifies to $$\ln \left( \frac{K_2}{K_1} \right) = 6.9$$ which is equal to $$3 \times \ln 10$$ Thus, $$\ln \frac{K_2}{K_1} = \ln 10^3$$ Therefore, $$K_2 = K_1 \times 10^3$$ and $$K_1 = K_2 \times 10^3$$ So $K = 1$.
Question 86
Chemistry · The d-and f-Block Elements · Fill in the blank
The number of terminal oxygen atoms present in the product B obtained from the following reaction is ________: $\mathrm{FeCr_2O_4 + Na_2CO_3 + O_2 \rightarrow A + Fe_2O_3 + CO_2}$ $\mathrm{A + H^+ \rightarrow B + H_2O + Na^+}$
Answer: 6
Solution
The reaction is given as follows: $$4 \mathrm{FeCr_2O_4} + 8 \mathrm{Na_2CO_3} + 7 \mathrm{O_2} \rightarrow 8 \mathrm{Na_2CrO_4} +$$ $$2 \mathrm{Fe_2O_3} + 8 \mathrm{CO_2}$$ The next step involves the reaction: $$2 \mathrm{Na_2CrO_4} + 2 \mathrm{H}^+ \rightarrow \underset{B}{\mathrm{Na_2Cr_2O_7}} + 2 \mathrm{Na}^+ + \mathrm{H_2O}$$ The structure of the dichromate ion is shown as: $$2 \mathrm{Na}^+ \left[ \begin{array}{c} \mathrm{O} \mathrm{O} \\ \backslash / \\ \mathrm{Cr} \mathrm{Cr} \\ / \backslash \\ \mathrm{O} \mathrm{O} \end{array} \right]^{2-}$$
Question 87
Chemistry · The d-and f-Block Elements · Numerical
An acidified manganate solution undergoes disproportionation reaction. The spin-only magnetic moment value of the product having manganese in higher oxidation state is ________ B.M. (Nearest integer)
Answer: 0
Solution
The reaction is given by: $$3\mathrm{MnO_4^{2-}} + 4\mathrm{H^+} \longrightarrow 2\mathrm{MnO_4^-} + \mathrm{MnO_2} + 2\mathrm{H_2O}$$ For $\mathrm{Mn}^{+7}$, the number of unpaired electrons is '0'. The magnetic moment $\mu = 0$ B.M.
Kjeldahl's method was used for the estimation of nitrogen in an organic compound. The ammonia evolved from $0.55 \, \mathrm{g}$ of the compound neutralised $12.5 \, \mathrm{mL}$ of $1 \, \mathrm{M} \, \mathrm{H_2SO_4}$ solution. The percentage of nitrogen in the compound is ________. (Nearest integer)
Answer: 64
Solution
Meq of $\mathrm{H_2SO_4}$ used by $\mathrm{NH_3} = 12.5 \times 1 \times 2 = 25$ % of N in the compound = $$\frac{25 \times 10^{-3} \times 14 \times 100}{0.55} = 63.6$$ or Meq. of $\mathrm{H_2SO_4} = Meq. of \mathrm{NH_3}$ $12.5 \times 1 \times 2 = 25$ meq. of $\mathrm{NH_3}$ $= 25$ millimoles of $\mathrm{NH_3}$ So Millimoles of 'N' = 25 Moles of 'N' = $25 \times 10^{-3}$ wt. of N = $14 \times 25 \times 10^{-3}$ %N = $$\frac{14 \times 25 \times 10^{-3}}{0.55} \times 100$$ $= 63.66$ $\approx 64\%$