JEE Main 28 June 2022 Shift 1 question paper with solutions
JEE Main 28 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 · Binomial Theorem · Single correct
If $$\sum_{k=1}^{31} \binom{31}{k} \binom{31}{k-1} - \sum_{k=1}^{30} \binom{30}{k} \binom{30}{k-1} = \frac{\alpha (60!)}{(30!)(31!)},$$ Where $\alpha \in \mathbb{R}$, then the value of $16\alpha$ is equal to
Let a function $f : \mathbb{N} \to \mathbb{N}$ be defined by $$f(n) = \begin{cases} 2n, & n = 2, 4, 6, 8, \ldots \\ n - 1, & n = 3, 7, 11, 15, \ldots \\ \frac{n+1}{2}, & n = 1, 5, 9, 13, \ldots \end{cases}$$ then, $f$ is
one-one but not onto
onto but not one-one
neither one-one nor onto
one-one and onto
Answer: (d)
Solution
Given $$f(x) = \begin{cases} 4R & ; \ n = 2R \\ 4R - 2 & ; \ n = 4R - 1 \\ 2R - 1 & ; \ n = 4R - 3 \end{cases}$$ where $R \in \mathbb{N}$. Note that for any element, it will fall into exactly one of these sets. $$\{ y : y = 4R; \ y \in \mathbb{N} \}$$ $$\{ y : y = 4R - 2; \ y \in \mathbb{N} \}$$ $$\{ y : y = 2R - 1; \ y \in \mathbb{N} \}$$ Corresponding to that $y$, we will get exactly one value of $n$. Thus, $f$ is one-to-one and onto.
Question 3
Maths · Determinants · Single correct
If the system of linear equations $$2x + 3y - z = -2$$ $$x + y + z = 4$$ $$x - y + |\lambda| z = 4\lambda - 4$$ where $\lambda \in \mathbb{R}$, has no solution, then
$\lambda = 7$
$\lambda = -7$
$\lambda = 8$
$\lambda^2 = 1$
Answer: (b)
Solution
Given the determinant: $$\begin{vmatrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & \lambda \end{vmatrix} = 0$$ This implies $|\lambda| = 7 \Rightarrow \lambda = \pm 7$ $\hspace{0.5cm}$ ...(1) System: $$2x + 3y - z = -2 \hspace{0.5cm} ...(2)$$ $$x + y + z = 4 \hspace{0.5cm} ...(3)$$ $$x - y + |\lambda| z = 4\lambda - 4 \hspace{0.5cm} ...(4)$$ Eliminating $y$ from equations (2) and (3) we get: $$x + 4z = 14 \hspace{0.5cm} ...(5)$$ Adding equations (3) and (4) gives: $$x + \left( \frac{|\lambda| + 1}{2} \right) z = 2\lambda \hspace{0.5cm} ...(6)$$ Clearly for $\lambda = -7$, the system is inconsistent.
Question 4
Maths · Determinants · Single correct
Let A be a matrix of order 3 $\times$ 3 and $\det$(A) = 2. Then $\det$($\det$(A) $\operatorname{adj}$(5 $\operatorname{adj}(A^3)))$ is equal to
Maths · Permutations and Combinations · Single correct
The total number of 5-digit numbers, formed by using the digits 1, 2, 3, 5, 6, 7 without repetition, which are multiple of 6, is
36
48
60
72
Answer: (d)
Solution
To make a number divisible by 3 we can use the digits 1, 2, 5, 6, 7 or 1, 2, 3, 5, 7. Using 1, 2, 5, 6, 7, number of even numbers is $$= 4 \times 3 \times 2 \times 1 \times 2 = 48$$ Using 1, 2, 3, 5, 7, number of even numbers is $$= 4 \times 3 \times 2 \times 1 \times 1 = 24$$ Required answer is 72.
Question 6
Maths · Sequences and Series · Single correct
Let $A_1, A_2, A_3, \ldots$ be an increasing geometric progression of positive real numbers. If $A_1 A_3 A_5 A_7 = \frac{1}{1296}$ and $A_2 + A_4 = \frac{7}{36}$, then, the value of $A_6 + A_8 + A_{10}$ is equal to
Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral $$\int_{0}^{1} [-8x^2 + 6x - 1] \, dx$$ is equal to
-1
-$\frac{5}{4}$
$\frac{\sqrt{17} - 13}{8}$
$\frac{\sqrt{17} - 16}{8}$
Answer: (c)
Solution
Evaluate the integral from 0 to 1 of $[-8x^2 + 6x - 1] \, dx$. This can be split into three integrals: $$\int_0^{1/4} -1 \, dx + \int_{1/4}^{1/2} 0 \, dx + \int_{1/2}^{3/4} -1 \, dx$$ Using the graph, we add: $$\int_{3/4}^{\frac{3+\sqrt{17}}{8}} -2 \, dx + \int_{\frac{3+\sqrt{17}}{8}}^1 -3 \, dx$$ This simplifies to: $$= -\left[ x \right]_0^{1/4} + 0 - \left[ x \right]_{1/2}^{3/4} - 2 \left[ x \right]_{3/4}^{\frac{3+\sqrt{17}}{8}} - 3 \left[ x \right]_{\frac{3+\sqrt{17}}{8}}^1$$ Evaluating the integrals: $$= -\left( \frac{1}{4} - 0 \right) - \left( \frac{3}{4} - \frac{1}{2} \right) - 2 \left( \frac{3+\sqrt{17}}{8} - \frac{3}{4} \right) - 3 \left( 1 - \frac{3+\sqrt{17}}{8} \right)$$ Simplifying further: $$= -\frac{1}{4} - \frac{1}{4} + \frac{-6 - 2\sqrt{17}}{8} + \frac{3}{2} - 3 + \frac{9 + 3\sqrt{17}}{8}$$ Finally, $$= \frac{\sqrt{17} - 13}{8}$$
Question 8
Maths · Continuity and Differentiability · Single correct
Let $f : \mathbb{R} \to \mathbb{R}$ be defined as $$f(x) = \begin{cases} e^x, & x < 0 \\ ae^x + [x-1], & 0 \leq x < 1 \\ b + [\sin(\pi x)], & 1 \leq x < 2 \\ [e^{-x}] - c, & x \geq 2 \end{cases}$$ where $a, b, c \in \mathbb{R}$ and $[t]$ denotes greatest integer less than or equal to $t$. Then, which of the following statements is true?
There exists $a, b, c \in \mathbb{R}$ such that $f$ is continuous on $\mathbb{R}$.
If $f$ is discontinuous at exactly one point, then $a + b + c = 1$.
If $f$ is discontinuous at exactly one point, then $a + b + c \neq 1$.
$f$ is discontinuous at at least two points, for any values of $a, b$ and $c$.
Answer: (c)
Solution
Given $f(x)$ is discontinuous at $x = 1$. For continuous at $x = 0$; $a = 1$. For continuous at $x = 2$; $b + c = 1$. $a + b + c = 2$.
Question 9
Maths · Applications of Integrals · Single correct
The area of the region S = $\{$(x,y) : $y^2$ $\leq$ 8x, y $\geq$ $\sqrt{2}$x, x $\geq$ 1$\}$ is
$\frac{13\sqrt{2}}{6}$
$\frac{11\sqrt{2}}{6}$
$\frac{5\sqrt{2}}{6}$
$\frac{19\sqrt{2}}{6}$
Answer: (b)
Solution
Given $y^2 = 8x$ (1) $y = \sqrt{2x}$ (2) $y^2 = 2x^2$ Solving $8x = 2x^2$ gives $x = 0$ and $x = 4$. The area is given by: $$ Area := \int_{1}^{4} \left( 2\sqrt{2}\sqrt{x} - \sqrt{2}x \right) \, dx $$ Evaluating the integral: $$ = 2\sqrt{2} \left( \frac{x^{3/2}}{3/2} \right) \bigg|_1^4 - \sqrt{2} \left( \frac{x^2}{2} \right) \bigg|_1^4 $$ $$ = \frac{4\sqrt{2}}{3} (8 - 1) - \frac{\sqrt{2}}{3} (16 - 1) $$ $$ = \frac{28\sqrt{2}}{3} - \frac{15\sqrt{2}}{2} = \frac{11\sqrt{2}}{6} $$
Question 10
Maths · Differential Equations · Single correct
Let the solution curve $y = y(x)$ of the differential equation, $$\left[ \frac{x}{\sqrt{x^2 - y^2}} + e^x \right] x \frac{dy}{dx} = x + \left[ \frac{x}{\sqrt{x^2 - y^2}} + e^x \right] y$$ pass through the points $(1, 0)$ and $(2\alpha, \alpha),\ \alpha > 0$. Then $\alpha$ is equal to
Let $y = y(x)$ be the solution of the differential equation $x(1-x^2)\frac{dy}{dx} + (3x^2y - y - 4x^3) = 0, x > 1,$ with $y(2) = -2$. Then $y(3)$ is equal to
-18
-12
-6
-3
Answer: (a)
Solution
Given the equation $x(1-x^2) \frac{dy}{dx} + (3x^2y - y - 4x^3) = 0$. Simplifying, we have $x(1-x^2) \frac{dy}{dx} + (3x^2 - 1)y = 4x^3$. Using the integrating factor method, we find the integrating factor (IF) as follows: $$IF = e^{\int P \, dx} = e^{\int \frac{3x^2 - 1}{x - x^3} \, dx}$$ Let $x - x^3 = t$, then $IF = e^{\int \frac{-dt}{t}} = e^{-\ln t} = \frac{1}{t}$. Therefore, $$IF = \frac{1}{x - x^3}$$ The solution is given by $$y \times IF = \int Q \times IF \, dx$$ $$y \left( \frac{1}{x - x^3} \right) = \int \frac{4x^3}{x - x^3} \times \frac{1}{x - x^3} \, dx$$ $$= \int \frac{4x^3}{(x - x^3)^2} \, dx$$ $$= \int \frac{4x}{(1-x^2)^2} \, dx$$ Let $1-x^2 = K$, then $$-2x \, dx = dK$$ $$= 2 \int \frac{-dK}{K^2}$$ $$= -2 \left( -\frac{1}{K} \right) + c$$ $$\frac{y}{x - x^3} = \frac{2}{K} + c$$ $$\frac{y}{x - x^3} = \frac{2}{1-x^2} + c$$ At $x = 2$, $y = -2$, $$\frac{-2}{2-8} = \frac{2}{1-4} + c$$ $$\frac{1}{3} = \frac{-2}{3} + c$$ Solving, $$\frac{y}{3-27} = \frac{2}{1-9} + 1$$ $$\frac{y}{-24} = \frac{1}{4}$$ $$\frac{y}{-24} = \frac{3}{4}$$ $$y = \frac{3}{4}(-24) = -18$$
Question 12
Maths · Applications of Derivatives · Single correct
The number of real solutions of $x^7 + 5x^3 + 3x + 1 = 0$ is equal to _______.
0
1
3
5
Answer: (b)
Solution
Given $f(x) = x^7 + 5x^3 + 3x + 1$. The derivative is $f'(x) = 7x^6 + 15x^2 + 3 > 0$. Therefore, $f(x)$ is a strictly increasing function. As $x \to -\infty$, $y \to -\infty$. As $x \to \infty$, $y \to \infty$. Therefore, the number of real solutions is 1.
Question 13
Maths · Conic Sections · Single correct
Let the eccentricity of the hyperbola $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be $\sqrt{\frac{5}{2}}$ and length of its latus rectum be $6\sqrt{2}$. If $y = 2x + c$ is a tangent to the hyperbola $H$, then the value of $c^2$ is equal to
If the tangents drawn at the points $O(0,0)$ and $P(1+\sqrt{5},2)$ on the circle \[ x^2+y^2-2x-4y=0 \] intersect at the point $Q$, then the area of the triangle $OPQ$ is equal to
Maths · Three Dimensional Geometry · Single correct
If two distinct point Q, R lie on the line of intersection of the planes $-x + 2y - z = 0$ and $3x - 5y + 2z = 0$ and $PQ = PR = \sqrt{18}$ where the point P is $(1, -2, 3)$, then the area of the triangle PQR is equal to
Maths · Three Dimensional Geometry · Single correct
The acute angle between the planes $P_1$ and $P_2$, when $P_1$ and $P_2$ are the planes passing through the intersection of the planes $5x + 8y + 13z - 29 = 0$ and $8x - 7y + z - 20 = 0$ and the points $(2, 1, 3)$ and $(0, 1, 2)$, respectively, is
$\frac{\pi}{3}$
$\frac{\pi}{4}$
$\frac{\pi}{6}$
$\frac{\pi}{12}$
Answer: (a)
Solution
Equation of plane passing through the intersection of planes $5x + 8y + 13z - 29 = 0$ and $8x - 7y + z - 20 = 0$ is $$5x + 8y + 3z - 29 + \lambda (8x - 7y + z - 20) = 0$$ and if it is passing through $(2, 1, 3)$ then $\lambda = \frac{7}{2}$. $P_1$: Equation of plane through intersection of $5x + 8y + 13z - 29 = 0$ and $8x - 7y + z - 20 = 0$ and the point $(2, 1, 3)$ is $$5x + 8y + 3z - 29 + \frac{7}{2}(8x - 7y + z - 20) = 0$$ $$\Rightarrow 2x - y + z = 6$$ Similarly $P_2$: Equation of plane through intersection of $5x + 8y + 13z - 29 = 0$ and $8x - 7y + z - 20 = 0$ and the point $(0, 1, 2)$ is $$\Rightarrow x + y + 2z = 5$$ Angle between planes $\theta = \cos^{-1} \left( \frac{3}{\sqrt{6} \sqrt{6}} \right) = \frac{\pi}{3}$
Question 17
Maths · Three Dimensional Geometry · Single correct
Let the plane $P: \vec{r} \cdot \vec{a} = d$ contain the line of intersection of two planes $\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 6$ and $\vec{r} \cdot (-6\hat{i} + 5\hat{j} - \hat{k}) = 7$. If the plane $P$ passes through the point $\left(2, 3, \frac{1}{2}\right)$, then the value of $\frac{|13\vec{a}|^2}{d^2}$ is equal to
90
93
95
97
Answer: (b)
Solution
Equation of plane passing through line of intersection of planes $P_1 : \vec{r} \cdot \left( \hat{i} + 3 \hat{j} - \hat{k} \right) = 6$ and $P_2 : \vec{r} \cdot \left( -6 \hat{i} + 5 \hat{j} - \hat{k} \right) = 7$ is $P_1 + \lambda P_2 = 0$ $$\left( \vec{r} \cdot \left( \hat{i} + 3 \hat{j} - \hat{k} \right) - 6 \right) + \lambda \left( \vec{r} \cdot \left( -6 \hat{i} + 5 \hat{j} - \hat{k} \right) - 7 \right) = 0$$ and it passes through point $\left( 2, 3, \frac{1}{2} \right)$ $$\Rightarrow \left( 2 + 9 - \frac{1}{2} - 6 \right) + \lambda \left( -12 + 15 - \frac{1}{2} - 7 \right) = 0$$ $$\Rightarrow \lambda = 1$$ Equation of plane is $\vec{r} \cdot \left( -5 \hat{i} + 8 \hat{j} - 2 \hat{k} \right) = 13$ $$|\vec{a}|^2 = 25 + 64 + 4 = 93 ; \ d = 13$$ Value of $\frac{|13 \vec{a}|^2}{d^2} = 93$
Question 18
Maths · Probability · Single correct
The probability, that in a randomly selected 3-digit number at least two digits are odd, is
$\frac{19}{36}$
$\frac{15}{36}$
$\frac{13}{36}$
$\frac{23}{36}$
Answer: (d)
Solution
At least two digits are odd. This is equal to exactly two digits are odd plus exactly three digits are odd. For exactly three digits are odd: $$5 \times 5 \times 5 = 125$$ For exactly two digits odd: If 0 is used then: $$2 \times 5 \times 5 = 50$$ If 0 is not used then: $$^3C_1 \times 4 \times 5 \times 5 = 300$$ Required Probability: $$\frac{475}{900} = \frac{19}{36}$$
Question 19
Maths · Heights and Distances · Single correct
Let AB and PQ be two vertical poles, 160 $\,$ $\mathrm{m}$ apart from each other. Let C be the middle point of B and Q, which are feet of these two poles. Let $\frac{\pi}{8}$ and $\theta$ be the angles of elevation from C to P and A, respectively. If the height of pole PQ is twice the height of pole AB, then $\tan^2$ $\theta$ is equal to
Let $p,q,r$ be three logical statements. Consider the compound statements $S_1:\ ((\sim p)\vee q)\vee((\sim p)\vee r)$ and $S_2:\ p\rightarrow(q\vee r)$ Then, which of the following is NOT true?
\text{If } S_2 \text{ is True, then } S_1 \text{ is True}
\text{If } S_2 \text{ is False, then } S_1 \text{ is False}
\text{If } S_2 \text{ is False, then } S_1 \text{ is True}
\text{If } S_1 \text{ is False, then } S_2 \text{ is False}
Answer: (c)
Solution
Given $$s_1: (\sim p \lor q) \lor (\sim p \lor r)$$ $$\equiv \sim p \lor (q \lor r)$$ $$s_2: p \rightarrow (q \lor r)$$ $$\equiv \sim p \lor (q \lor r) \rightarrow$$ By conditional law $$s_1 \equiv s_2$$
Question 21
Maths · Relations and Functions · Fill in the blank
Let $R_1$ and $R_2$ be relations on the set $\{1, 2, \ldots, 50\}$ such that $R_1 = \{(p, p^n) : p$ is a prime and $n \geq 0$ is an integer$\}$ and $R_2 = \{(p, p^n) : p$ is a prime and $n = 0$ or $1\}$. Then, the number of elements in $R_1 - R_2$ is .
Answer: 8
Solution
Here, $p, p^n \in \{1, 2, \ldots, 50\}$. Now $p$ can take values $2, 3, 5, 7, 11, 13, 17, 23, 29, 31, 37, 41, 43$ and $47$. Therefore, we can calculate the number of elements in $R_1$ as $(2, 2^0), (2, 2^1), \ldots, (2, 2^5)$; $(3, 3^0), \ldots, (3, 3^3)$; $(5, 5^0), \ldots, (5, 5^2)$; $(7, 7^0), \ldots, (7, 7^2)$; $(11, 11^0), \ldots, (11, 11^1)$. And rest for all other two elements each. Therefore, $$n(R_1) = 6 + 4 + 3 + 3 + (2 \times 10) = 36$$ Similarly for $R_2$, $(2, 2^0), (2, 2^1)$; $(47, 47^0), (47, 47^1)$. Therefore, $$n(R_2) = 2 \times 14 = 28$$ Therefore, $$n(R_1) - n(R_2) = 36 - 28 = 8$$
Question 22
Maths · Complex Numbers and Quadratic Equations · Numerical
The number of real solutions of the equation $$e^{4x} + 4e^{3x} - 58e^{2x} + 4e^{x} + 1 = 0$$ is .
The mean and standard deviation of 15 observations are found to be 8 and 3 respectively. On rechecking it was found that, in the observations, 20 was misread as 5. Then, the correct variance is equal to .
Answer: 17
Solution
We have $$Variance = \frac{\sum_{r=1}^{15} x_r^2}{15} - \left( \frac{\sum_{r=1}^{15} x_r}{15} \right)^2$$ Now, as per information given in equation $$\frac{\sum x_r^2}{15} - 8^2 = 3^2 \implies \sum x_r^2 = \log 5$$ Now, the new $$\sum x_r^2 = \log 5 - 5^2 + 20^2 = 1470$$ And, new $$\sum x_r = (15 \times 8) - 5 + (20) = 135$$ Therefore, $$Variance = \frac{1470}{15} - \left( \frac{135}{15} \right)^2 = 98 - 81 = 17$$
Question 24
Maths · Vector Algebra · Numerical
If $\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k}$, $\vec{b} = 3\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$ are coplanar vectors and $\vec{a} \cdot \vec{c} = 5$, $\vec{b} \perp \vec{c}$, then $122 (c_1 + c_2 + c_3)$ is equal to
Maths · Straight Lines and Pair of Straight Lines · Numerical
A ray of light passing through the point P(2, 3) reflects on the x-axis at point A and the reflected ray passes through the point Q(5, 4). Let R be the point that divides the line segment AQ internally into the ratio 2 : 1. Let the co-ordinates of the foot of the perpendicular M from R on the bisector of the angle PAQ be $(\alpha, \beta)$. Then, the value of $7\alpha + 3\beta$ is equal to _______.
Answer: 31
Solution
By observation we see that $A(\alpha, 0)$. And $\beta = y-coordinate of R$ $$= \frac{2 \times 4 + 1 \times 0}{2 + 1} = \frac{8}{3} \ldots (1)$$ Now $P'$ is image of $P$ in $y = 0$ which will be $P'(2, -3)$. Therefore, equation of $P'Q$ is $(y + 3) = \frac{4 + 3}{5 - 2}(x - 2)$ i.e. $3y + 9 = 7x - 14$. $A \equiv \left(\frac{23}{7}, 0\right)$ by solving with $y = 0$. Therefore, $\alpha = \frac{23}{7} \ldots (2)$ By (1), (2) $$7\alpha + 3\beta = 23 + 8 = 31$$
Question 26
Maths · Applications of Derivatives · Numerical
Let $\ell$ be a line which is normal to the curve $y = 2x^2 + x + 2$ at a point $P$ on the curve. If the point $Q(6, 4)$ lies on the line $\ell$ and $O$ is origin, then the area of the triangle $OPQ$ is equal to .
Answer: 13
Solution
Given $y = 2x^2 + x + 2$. The derivative is $\frac{dy}{dx} = 4x + 1$. Let $P$ be $(h, k)$, then the normal at $P$ is $$y - k = -\frac{1}{4h + 1}(x - h)$$ This passes through $Q(6, 4)$. Therefore, $$4 - k = -\frac{1}{4h + 1}(6 - h)$$ $$\Rightarrow (4h + 1)(4 - k) + 6 - h = 0$$ Also $k = 2h^2 + h + 2$. Thus, $$(4h + 1)(4 - 2h^2 - h - 2) + 6 + h = 0$$ $$\Rightarrow 4h^3 - 3h^2 + 3h - 8 = 0$$ $$\Rightarrow h = 1, k = 5$$ Now the area of $\Delta OPQ$ will be $$\frac{1}{2}\begin{vmatrix} 1 & 0 & 0 \\ 1 & 1 & 5 \\ 1 & 6 & 4 \end{vmatrix} = 13$$
Question 27
Maths · Permutations and Combinations · Numerical
Let $A=\{1,a_1,a_2,\ldots,a_{18},77\}$ be a set of integers with $1<a_1<a_2<\cdots<a_{18}<77$. Let the set $A+A=\{x+y:x,y\in A\}$ contain exactly $39$ elements. Then, the value of $a_1+a_2+\cdots+a_{18}$ is equal to ______.
Answer: 702
Solution
$t_1+t_2+t_3+t_4=30$ Coefficient of $x^{30}$ in $(1+x+\cdots+x^{30})^2$ $=(x^4+x^5+x^6+x^7)$ $(x^2+x^3+x^4+x^5+x^6)$ $\left(\frac{1-x^{31}}{1-x}\right)^2$ $=x^6 \left(\frac{1-x^{31}}{1-x}\right)^2$ $(1+x+x^2+x^3) (1+x+x^2+x^3+x^4)$ $=x^6(1-x^3)(1-x^4)(1-x)^{-2}$ $=x^6(1-x^4-x^5+x^9)(1-x)^{-2}$ Coefficient of $x^n$ in $(1-x)^{-r}$ is ${}^{\,n+r-1}C_{r-1}$ $\Rightarrow {}^{27}C_3 -{}^{23}C_3 -{}^{22}C_3 +{}^{18}C_3$ $=2925-1771-1540+816$ $=430$ OR $x_2\in[4,7]$ $x_3\in[2,6]$ $\Rightarrow t_1+t_2+t_3+t_4=24$ total ways $= {}^{24+4-1}C_{4-1} -{}^{20+4-1}C_{4-1} -{}^{19+4-1}C_{4-1} +{}^{15+4-1}C_{4-1}$ $= {}^{27}C_3 -{}^{23}C_3 -{}^{22}C_3 +{}^{18}C_3$ $=430$
Question 28
Maths · Binomial Theorem · Numerical
The number of positive integers $k$ such that the constant term in the binomial expansion of $$\left(2x^3 + \frac{3}{x^k}\right)^{12}$$, $x \neq 0$ is $2^8 \cdot \ell$, where $\ell$ is an odd integer, is _______.
Answer: 3,6
Solution
Given $\($ $\left$( 2x^3 + $\frac{3}{x^k}$ $\right$)^{12} $\)$ $\($ t_{r+1} = $\binom{12}{r}$ $\left$( 2x^3 $\right$)^r $\left$( $\frac{3}{x^k}$ $\right$)^{12-r} $\)$ $\($ x^{3r - (12-r)k} $\rightarrow$ constant $\)$ Therefore, $\($ 3r - 12k + rk = 0 $\)$ $\($ $\Rightarrow$ k = $\frac{3r}{12 - r}$ $\)$ Therefore, possible values of $\($ r $\)$ are 3, 6, 8, 9, 10 and corresponding values of $\($ k $\)$ are 1, 3, 6, 9, 15. Now $\($ $\binom{12}{r}$ = 220, 924, 495, 220, 66 $\)$ Therefore, possible values of $\($ k $\)$ for which we will get $\($ 2^8 $\)$ are 3, 6.
Question 29
Maths · Complex Numbers and Quadratic Equations · Numerical
The number of elements in the set $\{$z = a + ib $\in$ $\mathbb{C}$ : a, b $\in$ $\mathbb{Z}$ and 1 < |z - 3 + 2i| < 4$\}$ is
Answer: 40
Solution
Given $1 < |Z - 3 + 2i| < 4$. This represents an annular region centered at $(3, -2)$ with inner radius $1$ and outer radius $4$. The inequality $1 < (a - 3)^2 + (b + 2)^2 < 16$ describes the same region in terms of $a$ and $b$. The possible integer solutions for $(a, b)$ are: $(0, \pm 2)$, $(\pm 2, 0)$, $(\pm 1, \pm 2)$, $(\pm 2, \pm 1)$, $(\pm 2, \pm 3)$, $(3 \pm 2, \pm 2)$, $(\pm 1, \pm 1)$, $(2 \pm, \pm 2)$, $(\pm 3, 0)$, $(0, \pm 3)$, $(\pm 3, \pm 1)$, $(\pm 1, \pm 3)$. Total 40 points.
Question 30
Maths · Conic Sections · Numerical
Let the lines $y + 2x = \sqrt{11} + 7\sqrt{7}$ and $2y + x = 2\sqrt{11} + 6\sqrt{7}$ be normal to a circle $C: (x-h)^2 + (y-k)^2 = r^2$. If the line $\sqrt{11}y - 3x = \frac{5\sqrt{77}}{3} + 11$ is tangent to the circle $C$, then the value of $(5h - 8k)^2 + 5r^2$ is equal to ____.
Answer: 816
Solution
Normal are $$y + 2x = \sqrt{11} + 7\sqrt{7},$$ $$2y + x = 2\sqrt{11} + 6\sqrt{7}$$ Center of the circle is point of intersection of normals i.e. $$\left( \frac{8\sqrt{7}}{3}, \sqrt{11} + \frac{5\sqrt{7}}{3} \right)$$ Tangent is $\sqrt{11}y - 3x = \frac{5\sqrt{77}}{3} + 11$ Radius will be perpendicular distance of tangent from center i.e. $4\sqrt{\frac{7}{5}}$ Now $(5h - 8k)^2 + 5r^2 = 816$
Physics
Question 31
Physics · Mechanical Properties of Fluids · Single correct
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Product of Pressure (P) and time (t) has the same dimension as that of coefficient of viscosity. Reason R: Coefficient of viscosity = $\frac{Force}{Velocity gradient}$ Question: Choose the correct answer from the options given below :
Both A and R are true, and R is correct explanation of A.
Both A and R are true but R is NOT the correct explanation of A.
A is true but R is false.
A is false but R is true.
Answer: (c)
Solution
Pressure and time $$P : \frac{N}{\mathrm{m}^2}, Time : Sec$$ $$Pt = \frac{N \cdot \mathrm{sec}}{\mathrm{m}^2}$$ $$\eta = \frac{F}{6 \pi r v} : \frac{N}{\mathrm{m} \cdot \mathrm{m/sec}} : \frac{N \cdot \mathrm{sec}}{\mathrm{m}^2}$$
Question 32
Physics · Work, Energy and Power · Single correct
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as $a = k^2 r t^2$. where k is a constant. The power delivered to the particle by the force acting on it is given as
zero
mk^2 r^2 t^2
mk^2 r t
mk^2 rt
Answer: (c)
Solution
Given $a = k^2 r t^2 = \frac{V^2}{r}$. $V = k r t$. $a_t = \frac{dv}{dt} = k r$. $F_t = m a_t = m k r$. $P = \mathbf{F} \cdot \mathbf{V}$. $= F \cos \theta V = F_t V = m k r (k r t)$. $P = m k^2 r^2 t$.
Question 33
Physics · Motion in a Plane · Single correct
Motion of a particle in x-y plane is described by a set of following equations $x = 4 \sin \left( \frac{\pi}{2} - \omega t \right)$ m and $y = 4 \sin (\omega t)$ m. The path of particle will be –
circular
helical
parabolic
elliptical
Answer: (a)
Solution
Given $$x = 4 \sin \left( \frac{\pi}{2} - \omega t \right)$$ $$y = 4 \cos(\omega t)$$ or $$x = 4 \cos(\omega t)$$ $$y = 4 \sin(\omega t)$$ Eliminate 't' to find relation between $x$ and $y$ $$x^2 + y^2 = y^2 \cos^2 \omega t + y^2 \sin^2 \omega t = 4^2$$ Thus, $$x^2 + y^2 = 4^2$$
Question 34
Physics · System of Particles and Rotational Motion · Single correct
Match List-I with List-II Question: Choose the correct answer from the options given below
Two planets A and B of equal mass are having their period of revolutions $T_A$ and $T_B$ such that $T_A = 2T_B$. These planets are revolving in the circular orbits of radii $r_A$ and $r_B$ respectively. Which out of the following would be the correct relationship of their orbits?
Given the formula for the period $T$: $$T = \frac{2\pi}{\sqrt{Gm_a}} r^{\frac{3}{2}}$$ We have $T^2 \propto r^3$. Using the ratio of periods and radii: $$\left( \frac{T_A}{T_B} \right)^2 = \left( \frac{r_A}{r_B} \right)^3$$ Substituting the given values: $$\left( \frac{2}{1} \right)^2 = \frac{r_A}{r_B} \Rightarrow r_A^3 = 4r_B^3$$
Question 36
Physics · Mechanical Properties of Fluids · Single correct
A water drop of diameter $\mathrm{cm}$ is broken into 64 equal droplets. The surface tension of water is $0.075\,\mathrm{N\,m^{-1}}$. In this process, the gain in surface energy will be:
Given below are two statement : Statement – I : What $\mu$ amount of an ideal gas undergoes adiabatic change from state $(P_1, V_1, T_1)$ to state $(P_2, V_2, T_2)$, the work done is $W = \frac{1R(T_2 - T_1)}{1 - \gamma}$, where $\gamma = \frac{C_P}{C_V}$ and $R$ = universal gas constant, Statement — II: In the above case, when work is done on the gas, the temperature of the gas would rise. Choose the correct answer from the options given below:
Both statement—I and statement-II are true.
Both statement—I and statement-II are false.
Statement-I is true but statement-II is false.
Statement-I is false but statement-II is true.
Answer: (a)
Solution
Given $$W_{adiabatic} = \frac{NR(T_f - T_i)}{1 - \gamma}$$ statement 1. $$Q = W + \Delta U$$ $$0 = W + \Delta U$$ $$\Delta U = -W$$ If work is done on the gas, i.e. work is negative, $$\therefore \Delta U$$ is positive. $$\therefore$$ Temperature will increase.
Question 38
Physics · Electric Charges and Fields · Single correct
Given below are two statements : Statement-I : A point charge is brought in an electric field. The value of electric field at a point near to the charge may increase if the charge is positive. Statement-II : An electric dipole is placed in a non-uniform electric field. The net electric force on the dipole will not be zero. Choose the correct answer from the options given below :
Both statement-I and statement-II are true.
Both statement-I and statement-I are false.
Statement-I is true but statement-II is false.
Statement-I is false but statement-II is true.
Answer: (a)
Solution
If the electric field is in the positive direction and the positive charge is to the left of that point then the electric field will increase. But to the left of the positive charge the electric field would decrease. If the dipole is kept at the point where the electric field is maximum then the force on it will be zero.
Question 39
Physics · Electric Charges and Fields · Single correct
The three charges $q/2$, $q$ and $q/2$ are placed at the corners $A$, $B$ and $C$ of a square of side $'a'$ as shown in figure. The magnitude of electric field $(E)$ at the corner $D$ of the square, is:
The electric field at point D due to the charges is calculated as follows. The net electric field at D is given by: $$ (E_{net})_D = \frac{kq}{2a^2} + \frac{\sqrt{2}kq}{2a^2} $$ This can be simplified to: $$ (E_{net})_D = \frac{kq}{a^2} \left( \frac{1}{2} + \frac{1}{\sqrt{2}} \right) $$ Further simplifying, we have: $$ (E_{net})_D = \frac{q}{4\pi \varepsilon_0 a^2} \left( \frac{1}{2} + \frac{1}{\sqrt{2}} \right) $$
Question 40
Physics · Moving Charges and Magnetism · Single correct
An infinitely long hollow conducting cylinder with radius $R$ carries a uniform current along its surface. Choose the correct representation of magnetic field $(B)$ as a function of radial distance $(r)$ from the axis of cylinder.
A radar sends an electromagnetic signal of electric field $(E_0) = 2.25 \, \mathrm{V/m}$ and magnetic field $(B_0) = 1.5 \times 10^{-8} \, \mathrm{T}$ which strikes a target on line of sight at a distance of $3 \, \mathrm{km}$ in a medium. After that, a pail of signal (echo) reflects back towards the radar with same velocity and by same path. If the signal was transmitted at time $t_0$ from radar, then after how much time echo will reach to the radar?
Physics · Ray Optics and Optical Instruments · Single correct
The refracting angle of a prism is $A$ and refractive index of the material of the prism is $\cot (A/2)$. Then the angle of minimum deviation will be -
The aperture of the objective is 24.4 $\,$ $\mathrm{cm}$. The resolving power of this telescope. If a light of wavelength 2440 $\,$ $\mathrm{\AA}$ is used to see the object will be
Physics · Dual Nature of Radiation and Matter · Single correct
The de Brogue wavelengths for an electron and a photon are $\lambda_e$ and $\lambda_p$ respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Brogue wavelengths of two?
$\lambda_p \propto \lambda_e^2$
$\lambda_p \propto \lambda_e$
$\lambda_p \propto \sqrt{\lambda_e}$
$\lambda_p \propto \sqrt{\frac{1}{\lambda_e}}$
Answer: (a)
Solution
Given $$\lambda_e = \frac{h}{\sqrt{2mk}}$$. Also for photon, $$k = \frac{hc}{\lambda_p}$$. Therefore, $$\lambda_e = \frac{h \sqrt{\lambda_p}}{\sqrt{2m hc}}$$. Hence, $$\lambda_p \propto \lambda e^2$$.
Question 45
Physics · Nuclei · Single correct
The Q-value of a nuclear reaction and kinetic energy of the projectile particle, $K_p$ are related as:
$Q = K_p$
$(K_p + Q) < 0$
$Q < K_p$
$(K_p + Q) > 0$
Answer: (d)
Solution
The reaction is given by $x + p \rightarrow \gamma + b$. The equation for $Q$ is $Q = k_\gamma + k_b - k_p$. Rearranging gives $Q + k_p = k_\gamma + k_b$. Therefore, $Q + k_p > 0$.
Question 46
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
In the following circuit, the correct relation between output (Y) and inputs A and B will be:
Y = AB
Y = A + B
Y = $\overline{AB}$
Y = $\overline{A + B}$
Answer: (c)
Solution
This is a NAND gate.
Question 47
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
For using a multimeter to identify diode from electrical components, choose the correct statement out of the following about the diode:
It is two terminal device which conducts current in both directions.
It is two terminal device which conducts current in one direction only.
It does not conduct current gives an initial deflection which decays to zero.
It is three terminal device which conducts current in one direction only between central terminal and either of the remaining two terminals.
Answer: (b)
Solution
In forward bias diode conducts. In reverse bias it does not conduct.
Question 48
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: n-p-n transistor permits more current than a p-n-p transistor. Reason R: Electrons have greater mobility as a charge carrier. Choose the correct answer from the options given below:
Both A and R true, and R is correct explanation of A.
Both A and R are true but R is NOT the correct explanation of A.
A is true but R is false.
A is false but R is true.
Answer: (a)
Solution
Theory
Question 49
Physics · Communication Systems · Single correct
Match List-I with List-II Choose the correct answer from the options given below:
A-I, B-II, C-III, D-IV
A-IV, B-III, C-I, D-II
A-IV, B-III, C-II, D-I
A-I, B-II, C-IV, D-III
Answer: (c)
Solution
Theory
Question 50
Physics · Waves · Single correct
The velocity of sound in a gas, in which two wavelengths 4.08 m and 4.16 m produce 40 beats in 12 s, will be:
A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at $60^\circ$ with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be ______ ms$^{-1}$. (if $g = 10$ m/s$^2$)
Answer: 5
Solution
The maximum speed is given by $V_{max} = \sqrt{2gh}$. The speed will be highest at the lowest position. $$h = (\ell - \ell \cos 60^\circ) = \frac{\ell}{2}$$ $$V_{max} = \sqrt{2 \times g \times \frac{\ell}{2}} = \sqrt{10 \times 2.5} = 5 \, \mathrm{m/s}$$
Question 52
Physics · Current Electricity · Numerical
A meter bridge setup is shown in the figure. It is used to determine an unknown resistance $R$ using a given resistor of $15 \, \Omega$. The galvanometer (G) shows null deflection when tapping key is at $43 \, \mathrm{cm}$ mark from end A. If the end correction for end A is $2 \, \mathrm{cm}$, then the determined value of $R$ will be _______ $\Omega$.
Answer: 19
Solution
Using the conditions of a balanced wheatstone bridge and adding the end correction. $$\frac{15}{(43+2)} = \frac{R}{(102-45)} \Rightarrow R = \frac{57}{45} \times 15$$ $$R = 19 \, \Omega$$
Question 53
Physics · Current Electricity · Numerical
Current measured by the ammeter \text{Ⓐ} in the reported circuit when no current flows through 10 $\Omega$ resistance, will be A.
Answer: 10
Solution
Using the condition of a balanced wheat stone bridge, $$\Rightarrow \frac{R}{3} = \frac{4}{6} \Rightarrow R = 2\, \Omega$$ So the effective resistance of the circuit is $$R_{eq} = \frac{6 \times 9}{6 + 9} = \frac{18}{5} \, \Omega$$ $$i = \frac{36}{R_{eq}} = 10\, A$$
Question 54
Physics · Alternating Current · Numerical
An AC source is connected to an inductance of $100\,\mathrm{mH}$, a capacitance of $100\,\mu\mathrm{F}$ and a resistance of $120\,\Omega$ as shown in figure. The time in which the resistance having a thermal capacity $2\,\mathrm{J/°C}$ will get heated by $16°\mathrm{C}$ is ________ s.
Physics · System of Particles and Rotational Motion · Numerical
The position vector of 1 kg object is $\vec{r} = (3\hat{i} - \hat{j}) \, \mathrm{m}$ and its velocity $\vec{v} = (3\hat{j} + \hat{k}) \, \mathrm{ms^{-1}}$. The magnitude of its angular momentum is $\sqrt{x} \, \mathrm{Nm}$ where $x$ is _______.
A man of 60 $\mathrm{\, kg}$ is running on the road and suddenly jumps into a stationary trolly car of mass 120 kg. Then, the trolly car starts moving with velocity 2 $ms^{-1}$. The velocity of the running man was $ms^{-1}$ when he jumps into the car.
Answer: 6
Solution
Taking the system as man and trolley and using conservation of linear momentum. $$60 \times v = (60 + 120) \times 2$$ $$\Rightarrow v = 6 \, \mathrm{m/s}$$
Question 57
Physics · Laws of Motion · Fill in the blank
A hanging mass M is connected to a four times bigger mass by using a string-pulley arrangement as shown in the figure. The bigger mass is placed on a horizontal ice slab and is being pulled by a force of 2Mg. In this situation, the tension in the string is $\frac{x}{5}Mg$ for $x = \_\_\_\_$. Neglect the mass of the string and the friction between the bigger mass and the ice slab. (Given $g$ = acceleration due to gravity)
Answer: 6
Solution
Using $\vec{F}_{net} = \mu \vec{a}$, $$2Mg - T = 4Ma$$ $$T - Mg = Ma$$ $$\Rightarrow a = \frac{g}{5}$$ $$T = Mg + Ma = Mg + \frac{Mg}{5} = \frac{6}{5}Mg$$
Question 58
Physics · Thermodynamics · Numerical
The total internal energy of two mole monoatomic ideal gas at temperature $T = 300 \, \mathrm{K}$ will be J. (Given $R = 8.31 \, \mathrm{J/mol.K}$)
Answer: 7479
Solution
Question 59
Physics · Moving Charges and Magnetism · Numerical
A singly ionized magnesium atom (A24) ion is accelerated to kinetic energy $5 \, \mathrm{keV}$ and is projected perpendicularly into a magnetic field $B$ of the magnitude $0.5 \, \mathrm{T}$. The radius of path formed will be __________ cm.
Answer: 10
Solution
The radius of the circular path is given by the equation $$R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}.$$
Question 60
Physics · Alternating Current · Numerical
A telegraph line of length 100 km has a capacity of $0.01 \, \mu\mathrm{F/km}$ and it carries an alternating current at $0.5$ kilo cycle per second. If minimum impedance is required, then the value of the inductance that needs to be introduced in series is ________ mH. (if $\pi = \sqrt{10}$)
Answer: 100
Solution
For minimum impedance $X_L = X_C$ $$\Rightarrow \omega L = \frac{1}{\omega C} \Rightarrow L = \frac{1}{\omega^2 C} = 10^{-1} \, \mathrm{H} = 100 \, \mathrm{mH}$$
Chemistry
Question 61
Chemistry · The Solid State · Single correct
The incorrect statement about the imperfections in solids is :
Schottky defect decreases the density of the substance.
Interstitial defect increases the density of the substance.
Frenkel defect does not alter the density of the substance.
Vacancy defect increases the density of the substance.
Answer: (d)
Solution
Due to vacancy defect density of the substance will decrease.
Question 62
Chemistry · Surface Chemistry · Single correct
The Zeta potential is related to which property of colloids?
Colour
Tyndall effect
Charge on the surface of colloidal particles
Brownian movement
Answer: (c)
Solution
The potential difference between the fixed and diffused layer of charges in a colloidal particle is called zeta potential.
Question 63
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Element "E" belongs to the period 4 and group 16 of the periodic table. The valence shell electron configuration of the element, which is just above ‘E’ in the group is
$3s^2$, $3p^4$
$3d^{10}$, $4s^2$, $4p^4$
$4d^{10}$, $5s^2$, $5p^4$
$2s^2$, $p^4$
Answer: (a)
Solution
E $\Rightarrow [\mathrm{Ar}]\,3d^{10}\,4s^2\,4p^4$ Element above E $\Rightarrow [\mathrm{Ne}]\,3s^2\,3p^4$
Question 64
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Given are two statements one is labelled as Assertion A and other is labelled as Reason R. Assertion A : Magnesium can reduce Al$_2$O$_3$ at a temperature below 1350$^\circ$C, while above 1350$^\circ$C aluminium can reduce MgO. Reason R : The melting and boiling points of magnesium are lower than those of aluminium. In light of the above statements, choose most appropriate answer from the options given below:
Both A and R are correct, and R is correct explanation of A.
Both A and R are correct, but R is NOT the correct explanation of A.
A is correct R is not correct.
A is not correct, R is correct.
Answer: (b)
Solution
From the Ellingham diagram given in NCERT, it can be seen that the $\mathrm{Mg,MgO}$ line crosses the $\mathrm{Al,Al_2O_3}$ line after $1350^\circ\mathrm{C}$. Hence, the Assertion is true. Yes, Mg has a lower melting point (MP) and boiling point (BP) than aluminium, but this does not explain the above fact.
Question 65
Chemistry · The d-and f-Block Elements · Single correct
Dihydrogen reacts with CuO to give
CuH_2
Cu
Cu_2O
Cu(OH)_2
Answer: (b)
Solution
The reaction is: $$\mathrm{CuO + H_2 \rightarrow Cu + H_2O}$$ under hot conditions.
Question 66
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Nitrogen gas is obtained by thermal decomposition of
Ba(NO$_3$)$_2$
Ba(N$_3$)$_2$
NaNO$_2$
NaNO$_3$
Answer: (b)
Solution
The reaction is given by the equation: $$\mathrm{Ba(N_3)_2} \rightarrow \mathrm{Ba} + 3\mathrm{N_2}$$
Question 67
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Given below are two statements : Statement -I :The pentavalent oxide of group- 15 element. $\mathrm{E}_2\mathrm{O}_5$. is less acidic than trivalent oxide. $\mathrm{E}_2\mathrm{O}_3$. of the same element. Statement -II :The acidic character of trivalent oxide of group 15 elements. $\mathrm{E}_2\mathrm{O}_3$. decreases down the group. In light of the above statements, choose most appropriate answer from the options given below:
Both Statement I and Statement II are true.
Both Statement I and Statement II are false.
Statement I true, but statement II is false.
Statement I is false but statement II is true.
Answer: (d)
Solution
As positive oxidation state increases, electronegativity of element increases hence acidic character increases. Down the group, non-metallic character decreases, acidic character decreases. Acidic character: $E_2O_5 > E_2O_3$. Down the group, acidic character of $E_2O_3$ decreases.
Question 68
Chemistry · The d-and f-Block Elements · Single correct
Which one of the lanthanoids given below is the most stable in divalent form?
Ce (Atomic Number 58)
Sm (Atomic Number 62)
Eu (Atomic Number 63)
Yb (Atomic Number 70)
Answer: (c)
Solution
Given $E^\circ_{\mathrm{M^{3+}/M^{2+}}} \Rightarrow \begin{array}{cc} \mathrm{Eu} & \mathrm{Yb} \\ -0.35 & -1.05 \end{array}$. Hence, due to more reduction potential in Eu as compared to Yb, it can be concluded that $\mathrm{Eu^{2+}}$ is more stable than $\mathrm{Yb^{2+}}$.
Question 69
Chemistry · Co-ordination Compounds · Single correct
Given below are two statements: Statement I: $[\mathrm{Ni(CN)}_4]^{2-}$ is square planar and diamagnetic complex, with $dsp^2$ hybridization for Ni but $[\mathrm{Ni(CO)}_4]$ is tetrahedral, paramagnetic and with $sp^3$-hybridization for Ni. Statement II: $[\mathrm{NiCl}_4]^{2-}$ and $[\mathrm{Ni(CO)}_4]$ both have same d-electron configuration, have same geometry and are paramagnetic. In light of the above statements, choose the correct answer from the options given below:
Both Statement I and Statement II are true.
Both Statement I and Statement II are false.
Statement I is correct but Statement II is false.
Statement I is incorrect but Statement II is true.
Chemistry · Environmental Chemistry · Single correct
Which amongst the following is not a pesticide?
DDT
Organophosphates
Dieldrin
Sodium arsenite
Answer: (d)
Solution
Sodium arsenite is a herbicide.
Question 71
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Which one of the following techniques is not used to spot components of a mixture separated on thin layer chromatographic plate?
$\mathrm{I}_2$ (Solid)
U.V. Light
Visualisation agent as a component of mobile phase
Spraying of an appropriate reagent
Answer: (c)
Solution
The function of mobile phase is to carry the components present on TLC.
Question 72
Chemistry · Hydrocarbons · Single correct
Which of the following structures are aromatic in nature?
A, B, C and D
Only A and B
Only A and C
Only B, C and D
Answer: (b)
Solution
A, B aromatic. C, D is nonaromatic.
Question 73
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The major product (P ) in the reaction is
Answer: (c)
Solution
The reaction involves the addition of HBr to the alkene. The bromine ion attacks the more substituted carbon, leading to the formation of the product shown.
Question 74
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The correct structure of product ‘A’ formed in the following reaction.
Answer: (a)
Solution
PhCH = O + PhCH $\xrightarrow{OD/D_2O}$ $PhCH_2OD + PhCO_2^-$
Question 75
Chemistry · Haloalkanes and Haloarenes · Single correct
Which one of the following compounds is inactive towards $\mathrm{S_N1}$ reaction?
Answer: (c)
Solution
Question 76
Chemistry · Amines · Single correct
Identify the major product formed in the following sequence of reactions:
Answer: (c)
Solution
The reaction starts with aniline ($\mathrm{C_6H_5NH_2}$) which undergoes bromination in the presence of $\mathrm{Br_2}$ and $\mathrm{H_2O}$ to form 2,4,6-tribromoaniline. This compound is then treated with $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ to form the diazonium salt. Finally, the diazonium salt is reduced using $\mathrm{H_3PO_2}$ to yield 1,3,5-tribromobenzene.
Question 77
Chemistry · Amines · Single correct
A primary aliphatic amine on reaction with nitrous acid in cold (273 $\mathrm{K}$) and there after raising temperature of reaction mixture to room temperature (298 $\mathrm{K}$). Gives a/an
Buna-S, PHBr and Butadiene-styrene are copolymer. Only neoprene is namopolymer.
Question 79
Chemistry · Biomolecules · Single correct
Stability of $\alpha$ - Helix structure of proteins depends upon
dipolar interaction
H-bonding interaction
van der Waals forces
$\pi$ -stacking interaction
Answer: (b)
Solution
Mostly H-bonding is responsible for the stability of $\alpha$-helix form.
Question 80
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The formula of the purple colour formed in Laissagne’s test for sulphur using sodium nitroprusside is
NaFe[Fe(CN)_6]
Na[Cr(NH_3)_2(NCS)_4]
Na_2[Fe(CN)_5(NO)]
Na_4[Fe(CN)_5(NOS)]
Answer: (d)
Solution
The reaction is given by: $$\mathrm{Na_2S + Na_2[Fe(CN)_5NO] \rightarrow Na_4[Fe(CN)_5NO_5]}$$
Question 81
Chemistry · Redox Reactions · Numerical
A $2.0\,\mathrm{g}$ sample containing $\mathrm{MnO_2}$ is treated with $\mathrm{HCl}$ liberating $\mathrm{Cl_2}$. The $\mathrm{Cl_2}$ gas is passed into a solution of $\mathrm{KI}$ and $60.0\,\mathrm{mL}$ of $0.1\,\mathrm{M}$ $\mathrm{Na_2S_2O_3}$ is required to titrate the liberated iodine. The percentage of $\mathrm{MnO_2}$ in the sample is \_\_\_\_. (Nearest integer) [Atomic masses (in u): $\mathrm{Mn} = 55$; $\mathrm{Cl} = 35.5$; $\mathrm{O} = 16$; $\mathrm{I} = 127$; $\mathrm{Na} = 23$; $\mathrm{K} = 39$; $\mathrm{S} = 32$]
Answer: 13
Solution
The reaction is given by: $$\mathrm{MnO_2 + HCl \rightarrow Cl_2 + Mn^{+2}}$$ with 6 meq of $\mathrm{Cl_2}$ produced, which equals 3 mmol. Next reaction: $$\mathrm{Cl_2 + KI \rightarrow Cl^- + I_2}$$ with 6 meq of $\mathrm{I_2}$ produced. Then: $$\mathrm{I_2 + Na_2S_2O_3 \rightarrow I^- + Na_2S_4O_6}$$ with 6 meq of $\mathrm{I_2}$ reacting, which equals 6 mmol. The percentage of $\mathrm{MnO_2}$ is calculated as: $$\%\mathrm{MnO_2} = \frac{3 \times 10^{-3} \times 87}{2} \times 100$$ which equals 13.05%. Answer: 13
Question 82
Chemistry · Structure of Atom · Numerical
If the work function of a metal is $6.63 \times 10^{-19} \, \mathrm{J}$, the maximum wavelength of the photon required to remove a photoelectron from the metal is ______ nm. (Nearest integer) [Given : $h = 6.63 \times 10^{-34} \, \mathrm{J \, s}$, and $c = 3 \times 10^{8} \, \mathrm{m \, s^{-1}}$]
Chemistry · Chemical Bonding and Molecular Structure · Fill in the blank
The hybridization of P exhibited in PF$_5$ is sp$^x$d$^y$. The value of $y$ is .
Answer: 1
Solution
PF$_5$ implies $sp^3d$ hybridisation. (5 sigma bonds, zero lone pair on central atom) Value of $y = 1$
Question 84
Chemistry · Thermodynamics · Numerical
4.0 $\mathrm{L}$ of an ideal gas is allowed to expand isothermally into vacuum until the total volume is 20 $\mathrm{L}$. The amount of heat absorbed in this expansion is ____ $\mathrm{L}$ \, $\mathrm{atm}$.
Answer: 0
Solution
For free expansion, $\($ P_{ext} = 0 $\)$, $\($ w = P_{ext} $\cdot$ 0 = 0 $\)$, $\($ w = 0 $\)$. Therefore, $\($ q = 0 $\)$, $\($ $\Delta$ U = q + w = 0 $\)$, $\($ $\Delta$ U = 0 $\)$. Ans. 0
Question 85
Chemistry · Solutions · Numerical
The vapour pressures of two volatile liquids A and B at 25°C are 50 Torr and 100 Torr, respectively. If the liquid mixture contains 0.3 mole fraction of A, then the mole fraction of liquid B in the vapour phase is $\frac{x}{17}$. The value of x is _______.
The solubility product of a sparingly soluble salt $\mathrm{A}_2\mathrm{X}_3$ is $1.1 \times 10^{-23}$. If specific conductance of the solution is $3 \times 10^{-5} \, \mathrm{S} \, \mathrm{m}^{-1}$, the limiting molar conductivity of the solution is $x \times 10^{-3} \, \mathrm{S} \, \mathrm{m}^2 \, \mathrm{mol}^{-1}$. The value of $x$ is .
Answer: 3
Solution
The reaction is given by $$\mathrm{A_2X_3(s) \rightleftharpoons 2A^{+3}_{(aq)} + 3X^{-2}_{(aq)}}$$ Solubility is $s \mathrm{M}$. Therefore, $2s$ and $3s$. $$(2s)^2 (3s)^3 = 1.1 \times 10^{-23}$$ $$108 \, s^5 = 1.1 \times 10^{-23}$$ $$s \approx 10^{-5} \, \mathrm{M} = 10^{-5} \, \mathrm{mol/L} = 0.01 \, \mathrm{mol/m^3}$$ Now, $\wedge_m \approx \wedge_m^\infty = \frac{k}{m} = \frac{k}{s}$. $$\Rightarrow \wedge_m^\infty = \frac{3 \times 10^{-5}}{0.01} = 3 \times 10^{-3} \, \mathrm{S \cdot m^2/mol}$$ Ans. 3
Question 87
Chemistry · Electrochemistry · Numerical
The quantity of electricity in Faraday needed to reduce 1 mol of $\mathrm{Cr_2O_7^{2-}}$ to $\mathrm{Cr^{3+}}$ is ________.
Answer: 6
Solution
Given the reaction: $$\mathrm{Cr_2O_7^{2-} + 6e^- \rightarrow 2Cr^{3+}}$$ 1 mol of $$\mathrm{Cr_2O_7^{2-}}$$ reacts with 6 mol of electrons. Therefore, the number of faradays equals the moles of electrons, which is 6.
Question 88
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
For a first order reaction $A \rightarrow B$, the rate constant, $k = 5.5 \times 10^{-14} \, \mathrm{s}^{-1}$. The time required for 67$\%$ completion of reaction is $x \times 10^{-1}$ times the half life of reaction. The value of $x$ is _______ (Nearest integer) (Given : $\log 3 = 0.4771$)
Answer: 16
Solution
Given $$t_{67\%} = \frac{1}{k} \ln \left( \frac{1}{1 - 0.67} \right) = \frac{t_{1/2}}{\ln 2} \times \ln \left( \frac{1}{1 - \frac{2}{3}} \right)$$ We have $$t_{67\%} = \frac{t_{1/2}}{\log 2} \times \log 3 = \frac{t_{1/2} \times 0.4771}{0.301}$$ Thus, $$\Rightarrow t_{67\%} = 1.585 \times t_{1/2}$$ Given $$X \times 10^{-1} = 1.585$$ Therefore, $$\Rightarrow X = 15.85$$ The answer is 16.
Question 89
Chemistry · Co-ordination Compounds · Numerical
Number of complexes which will exhibit synergic bonding amongst, $[Cr(CO)_6]$, $[Mn(CO)_5]$ and $[Mn_2(CO)_{10}]$ is .
Answer: 3
Solution
Carbonyl complex compounds have tendency to show synergic bonding.
Question 90
Chemistry · Analytical Chemistry · Numerical
In the estimation of bromine, 0.5 $\mathrm{g}$ of an organic compound gave 0.40 $\mathrm{g}$ of silver bromide. The percentage of bromine in the given compound is ________$\%$ (nearest integer) (Relative atomic masses of Ag and Br are 108u and 80u, respectively).
Answer: 34
Solution
The reaction is given as: $$\mathrm{O.C} \rightarrow \mathrm{AgBr}$$ with masses 0.5 g and 0.4 g respectively. The moles of Br are equal to the moles of AgBr: $$mol of Br = mol of AgBr = \frac{0.4}{188}$$ The percentage of Br is calculated as: $$\% Br = \frac{\frac{0.4}{188} \times 80}{0.5} \times 100$$ This results in: $$= 34.04\%$$