JEE Main 27 July 2021 Shift 1 question paper with solutions
JEE Main 27 July 2021 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 · Statistics · Single correct
If the mean and variance of the following data: 6, 10, 7, 13, a, 12, b, 12 are 9 and $\frac{37}{4}$ respectively, then $(a - b)^2$ is equal to:
The value of $\lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{(2j-1)+8n}{(2j-1)+4n}$ is equal to:
5 + $\\log_e$ ($\frac{3}{2}$)
2 - $\\log_e$ $(\frac{2}{3})$
3 + 2 $\\log_e$ $(\\frac{2}{3})$
1 + 2 $\\log_e$ $(\\frac{3}{2})$
Answer: (d)
Solution
Given $$\lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{\left( \frac{2j}{n} - \frac{1}{n} + 8 \right)}{\left( \frac{2j}{n} - \frac{1}{n} + 4 \right)}$$ We have $$\int_{0}^{1} \frac{2x+8}{2x+4} \, dx = \int_{0}^{1} \, dx + \int_{0}^{1} \frac{4}{2x+4} \, dx$$ This simplifies to $$= 1 + 4 \frac{1}{2} \left( \ln |2x + 4| \right) \bigg|_{0}^{1}$$ Evaluating the integral, we get $$= 1 + 2 \ln \left( \frac{3}{2} \right)$$
Question 3
Maths · Vector Algebra · Single correct
Let $\vec{a}=\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b}=-\hat{i}+2\hat{j}+3\hat{k}$. Then the vector product $(\vec{a}+\vec{b})\times\left(\left(\vec{a}\times\left((\vec{a}-\vec{b})\times\vec{b}\right)\right)\times\vec{b}\right)$ is equal to $\underline{\hspace{2cm}}$.
The value of the definite integral $$\int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{dx}{(1+e^x \cos x)(\sin^4 x + \cos^4 x)}$$ is equal to:
$-\frac{\pi}{2}$
$\frac{\pi}{2\sqrt{2}}$
$-\frac{\pi}{4}$
$\frac{\pi}{\sqrt{2}}$
Answer: (b)
Solution
Given $$I = \int_{\pi/4}^{\pi/4} \frac{dx}{(1+e^{x \cos x})(\sin^4 x + \cos^4 x)} \ldots (1)$$ Using $$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$$ $$I = \int_{-\pi/4}^{\pi/4} \frac{dx}{(1+e^{-x \cos x})(\sin^4 x + \cos^4 x)}$$ Add (1) and (2) $$2I = \int_{\pi/4}^{x/4} \frac{dx}{\sin^4 x + \cos^4 x}$$ $$2I = 2 \int_0^{\pi/4} \frac{dx}{\sin^4 x + \cos^4 x}$$ $$I = \int_0^{\pi/4} \frac{(1+\tan^4 x) \sec^2 x}{\tan^4 x + 1} dx$$ $$I = \int_0^{\pi/4} (1+\frac{1}{\tan^2 x}) \sec^2 x \left( \tan x - \frac{1}{\tan x} \right)^2 + 2 dx$$ Let $\tan x - \frac{1}{\tan x} = t$ $$\left(1 + \frac{1}{\tan^2 x}\right) \sec^2 x dx = dt$$ $$I = \int_{-\infty}^0 \frac{dt}{t^2 + 2} = \left[ \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{t}{\sqrt{2}} \right) \right]_0^{-\infty}$$
Question 5
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $\mathbb{C}$ be the set of all complex numbers. Let $S_1=\{\,z\in\mathbb{C}\mid |z-3-2i|^2=8\,\}$ $S_2=\{\,z\in\mathbb{C}\mid \mathrm{Re}(z)\ge 5\,\}$ and $S_3=\{\,z\in\mathbb{C}\mid |z-\bar z|\ge 8\,\}$. Then the number of elements in $S_1\cap S_2\cap S_3$ is equal to:
Maths · Applications of Integrals · Single correct
If the area of the bounded region $$R = \left\{ (x, y) : \max \{0, \log_e x\} \leq y \leq 2^x, \frac{1}{2} \leq x \leq 2 \right\}$$ is, $\alpha (\log_e 2)^{-1} + \beta (\log_e 2) + \gamma$, then the value of $$(\alpha + \beta - 2\gamma)^2$$ is equal to :
8
2
4
1
Answer: (b)
Solution
Given the region $R = \{ (x, y) \cdot \max (0, \log_e x) \leq y \leq 2^x, \frac{1}{2} \leq x \leq 2 \}$. The integral is evaluated as follows: $$\int_{\frac{1}{2}}^2 2^x \, dx - \int_1^2 \ln x \, dx$$ This simplifies to: $$\left[ \frac{2^x}{\ln 2} \right]_{\frac{1}{2}}^2 - \left[ x \ln x - x \right]_1^2$$ Evaluating the first integral: $$\frac{(2^2) - 2^{1/2}}{\log_2 2} - (2 \ln 2 - 1)$$ Simplifying further: $$\frac{(2^2 - \sqrt{2})}{\log_2 2} - 2 \ln 2 + 1$$ Thus, $\alpha = 2^2 - \sqrt{2}$, $\beta = -2$, $\gamma = 1$. Now, evaluating: $$(\alpha + \beta + 2 \gamma)^2$$ This becomes: $$\left(2^2 - \sqrt{2} - 2 - 2 \right)^2$$ Finally: $$(\sqrt{2})^2 = 2$$
Question 7
Maths · Conic Sections · Single correct
A ray of light through $(2, 1)$ is reflected at a point $P$ on the $y$-axis and then passes through the point $(5, 3)$. If this reflected ray is the directrix of an ellipse with eccentricity $\frac{1}{3}$ and the distance of the nearer focus from this directrix is $\frac{8}{\sqrt{53}}$, then the equation of the other directrix can be:
$11x + 7y + 8 = 0$ or $11x + 7y - 15 = 0$
$11x - 7y - 8 = 0$ or $11x + 7y + 15 = 0$
$2x - 7y + 29 = 0$ or $2x - 7y - 7 = 0$
$2x - 7y - 39 = 0$ or $2x - 7y - 7 = 0$
Answer: (c)
Solution
Equation of reflected Ray $y - 1 = \frac{2}{7}(x + 2)$ $7y - 7 = 2x + 4$ $2x - 7y + 11 = 0$ Let the equation of other directrix is $2x - 7y + \lambda$ Distance of directrix from Focus $\frac{a}{e} - ae = \frac{8}{\sqrt{53}}$ $3a - \frac{a}{3} = \frac{8}{\sqrt{53}}$ or $a = \frac{3}{\sqrt{53}}$ Distance from other focus $\frac{a}{e} + ae$ $3a + \frac{a}{3} = \frac{10a}{3} = \frac{10}{3} \times \frac{3}{\sqrt{53}} = \frac{10}{\sqrt{53}}$ Distance between two directrix $= \frac{2a}{e}$ $= 2 \times 3 \times \frac{3}{\sqrt{53}} = \frac{18}{\sqrt{53}}$ $\frac{|\lambda - 11|}{\sqrt{53}} = \frac{18}{\sqrt{53}}$ $\lambda - 11 = 18$ or $-18$ $\lambda = 29$ or $-7$ $2x - 7y - 7 = 0$ or $2x - 7y + 29 = 0$
Question 8
Maths · Binomial Theorem · Single correct
If the coefficients of $x^7$ in $\left(x^2 + \frac{1}{bx}\right)^{11}$ and $x^{-7}$ in $\left(x - \frac{1}{bx^2}\right)^{11}$, $b \neq 0$, are equal, then the value of $b$ is equal to:
2
-1
1
-2
Answer: (c)
Solution
Coefficient of $x^7$ in $\left(x^2 + \frac{1}{bx}\right)^{11}$ is given by $$\binom{11}{r}(x^2)^{11-t} \cdot \left(\frac{1}{bx}\right)^t$$ $$= \binom{11}{r} x^{22-3t} \cdot \frac{1}{b^t}$$ Solving $22 - 3r = 7$ gives $r = 5$. Thus, $\binom{11}{5} \cdot \frac{1}{b^5} x^7$. Coefficient of $x^{-7}$ in $\left(x - \frac{b}{bx^2}\right)^{11}$ is given by $$\binom{11}{r}(x)^{11-t} \cdot \left(-\frac{1}{bx^2}\right)^t$$ $$= \binom{11}{r} x^{11-3t} \cdot \frac{(-1)^t}{b^t}$$ Solving $11 - 3r = -7$ gives $r = 6$. Thus, $\binom{11}{6} \cdot \frac{1}{b^6} x^{-7}$. Equating $\binom{11}{5} \cdot \frac{1}{b^5} = \binom{11}{6} \cdot \frac{1}{b^6}$ gives $b \neq 0$, therefore $b = 1$.
Question 9
Maths · Mathematical Reasoning · Single correct
The compound statement $(P \lor Q) \land (\sim P) \Rightarrow Q$ is equivalent to:
$P \lor Q$
$P \land \sim Q$
$\sim (P \Rightarrow Q)$
$\sim (P \Rightarrow Q) \iff P \land \sim Q$
Answer: (d)
Solution
Using Truth Table For the first table:
Question 10
Maths · Trigonometric Functions · Single correct
If $\sin \theta + \cos \theta = \frac{1}{2}$, then $16(\sin(2\theta) + \cos(4\theta) + \sin(6\theta))$ is equal to:
Let $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$. If $A^{-1} = \alpha I + \beta A$, $\alpha, \beta \in \mathbb{R}$, $I$ is a $2 \times 2$ identity matrix, then $4(\alpha - \beta)$ is equal to:
5
$\frac{8}{3}$
2
4
Answer: (d)
Solution
Given the matrix $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$ with $|A| = 6$. The inverse of $A$ is given by $$A^{-1} = \frac{adj A}{|A|} = \frac{1}{6} \begin{bmatrix} 4 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix}.$$ We equate this to $$\begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix} = \begin{bmatrix} \alpha & 0 \\ 0 & \alpha \end{bmatrix} + \begin{bmatrix} \beta & 2\beta \\ -\beta & 4\beta \end{bmatrix}.$$ Solving the equations, we have $$\alpha + \beta = \frac{2}{3}$$ $$\beta = -\frac{1}{6}$$ which implies $$\alpha = \frac{2}{3} + \frac{1}{6} = \frac{5}{6}.$$ Therefore, $$4(\alpha - \beta) = 4(1) = 4.$$
Question 12
Maths · Continuity and Differentiability · Single correct
Let $f : \left( -\frac{\pi}{4}, \frac{\pi}{4} \right) \to \mathbb{R}$ be defined as $$f(x) = \begin{cases} (1 + |\sin x|)^{\frac{3a}{|\sin x|}}, & -\frac{\pi}{4} < x < 0 \\ b, & x = 0 \\ e^{\cot 4x / \cot 2x}, & 0 < x < \frac{\pi}{4} \end{cases}$$ If $f$ is continuous at $x = 0$, then the value of $6a + b^2$ is equal to:
Let y = y(x) be solution of the differential equation $$ \log_e \left( \frac{dy}{dx} \right) = 3x + 4y, with y(0) = 0 $$ If $$ y \left( -\frac{2}{3} \log_e 2 \right) = \alpha \log_e 2, $$ then the value of $\($ $\alpha$ $\)$ is equal to:
Maths · Three Dimensional Geometry · Single correct
Let the plane passing through the point $(-1, 0, -2)$ and perpendicular to each of the planes $2x + y - z = 2$ and $x - y - z = 3$ be $ax + by + cz + 8 = 0$. Then the value of $a + b + c$ is equal to:
3
8
5
4
Answer: (d)
Solution
Normal of required plane $(2\hat{i}+\hat{j}-\hat{k})\times(\hat{i}-\hat{j}-\hat{k})$ $=-2\hat{i}+\hat{j}-3\hat{k}$ Equation of plane $-2(x+1)+1(y-0)-3(z+2)=0$ $-2x+y-3z-8=0$ $\Rightarrow 2x-y+3z+8=0$ $\therefore\ a+b+c=4$
Question 15
Maths · Conic Sections · Single correct
Two tangents are drawn from the point $P(-1, 1)$ to the circle $x^2 + y^2 - 2x - 6y + 6 = 0$. If these tangents touch the circle at points $A$ and $B$, and if $D$ is a point on the circle such that length of the segments $AB$ and $AD$ are equal, then the area of the triangle $ABD$ is equal to:
2
(3$\sqrt{2}$ + 2)
4
3($\sqrt{2}$ - 1)
Answer: (c)
Solution
The area of triangle $\Delta ABD$ is calculated as follows: $$\Delta ABD = \frac{1}{2} \times 2 \times 4$$ which simplifies to $$= 4$$
Question 16
Maths · Limits and Derivatives · Single correct
Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that $f(2) = 4$ and $f'(2) = 1$. Then, the value of $\lim_{x \to 2} \frac{x^2 f(2) - 4f(x)}{x - 2}$ is equal to:
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let P and Q be two distinct points on a circle which has center at C(2, 3) and which passes through origin O. If OC is perpendicular to both the line segments CP and CQ, then the set $\{$P, Q$\}$ is equal to
Given $\tan \theta = -\frac{2}{3}$. Using symmetric form of line $P, Q$: $(2 \pm \sqrt{13} \cos \theta, 3 \pm \sqrt{13} \sin \theta)$, we have: $$\left(2 \pm \sqrt{13} \cdot \left(-\frac{3}{\sqrt{13}}\right), 3 \pm \sqrt{13} \left(\frac{2}{\sqrt{13}}\right)\right)$$ $(-1, 5)$ and $(5, 1)$.
Question 18
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $\alpha, \beta$ be two roots of the equation $x^2 + (20)^{1/4}x + (5)^{1/2} = 0$. Then $\alpha^8 + \beta^8$ is equal to
10
100
50
160
Answer: (c)
Solution
Given the equation $$\left( x^2 + \sqrt{5} \right)^2 = \sqrt{20} x^2$$. We have $$x^4 = -5 \Rightarrow x^8 = 25$$. Also, $$\alpha^8 + \beta^8 = 50$$.
Question 19
Maths · Probability · Single correct
The probability that a randomly selected 2-digit number belongs to the set $\{ n \in N : (2^n - 2) \}$ is a multiple of 3 is equal to
$\frac{1}{6}$
$\frac{2}{3}$
$\frac{1}{2}$
$\frac{1}{3}$
Answer: (c)
Solution
Total number of cases $={}^{90}C_{1}=90$ Now, $n^2-2=(3-1)^n$ $={}^{n}C_{0}3^n-{}^{n}C_{1}3^{\,n-1}+\cdots+(-1)^{n-1}\,{}^{n}C_{n-1}3+(-1)^n\,{}^{n}C_{n}$ $=3\left(3^{\,n-1}-{}^{n}C_{1}3^{\,n-2}+\cdots+(-1)^{n-1}n\right)+(-1)^n-2$ $(2^n-2)$ is divisible by $3$ only when $n$ is odd. Required Probability $=\dfrac{45}{90}=\dfrac{1}{2}$
Question 20
Maths · Conic Sections · Single correct
Let \[ A=\left\{(x,y)\in\mathbb{R}\times\mathbb{R}\mid 2x^2+2y^2-2x-2y=1\right\}, \] \[ B=\left\{(x,y)\in\mathbb{R}\times\mathbb{R}\mid 4x^2+4y^2-16y+7=0\right\}, \] and \[ C=\left\{(x,y)\in\mathbb{R}\times\mathbb{R}\mid x^2+y^2-4x-2y+5\le r^2\right\}. \] Then the minimum value of $|r|$ such that \[ A\cup B\subseteq C \] is equal to
For real numbers $\alpha$ and $\beta$, consider the following system of linear equations: $$x + y - z = 2, x + 2y + \alpha z = 1, 2x - y + z = \beta$$ If the system has infinite solutions, then $\alpha + \beta$ is equal to
Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b}$ and $\vec{c} = \hat{j} - \hat{k}$ be three vectors such that $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 1$. If the length of projection vector of the vector $\vec{b}$ on the vector $\vec{a} \times \vec{c}$ is $l$, then the value of $3l^2$ is equal to
Answer: 2
Solution
Given $\vec{a} \times \vec{b} = \vec{c}$. Take dot with $\vec{c}$. $$(\vec{a} \times \vec{b}) \cdot \vec{c} = |\vec{c}|^2 = 2$$ Projection of $\vec{b}$ or $\vec{a} \times \vec{c} = \ell$. $$\frac{|\vec{b} \cdot (\vec{a} \times \vec{c})|}{|\vec{a} \times \vec{c}|} = \ell$$ Therefore, $$\ell = \frac{2}{\sqrt{6}} \implies \ell^2 = \frac{4}{6}$$ $$3\ell^2 = 2$$
Question 23
Maths · Sequences and Series · Numerical
If $\log_3 2$, $\log_3 \left(2^x - 5\right)$, $\log_3 \left(2^x - \frac{7}{2}\right)$ are in an arithmetic progression, then the value of $x$ is equal to
Let the domain of the function $$f(x) = \log_4 \left( \log_5 \left( \log_3 \left( 18x - x^2 - 77 \right) \right) \right)$$ be (a, b) Then the value of the integral $$\int_a^b \frac{\sin^3 x}{\sin^3 x + \sin^3 (a + b - x)} \, dx$$ is equal to
Let $f(x) = \begin{vmatrix} \sin^2 x & -2 + \cos^2 x & \cos 2x \\ 2 + \sin^2 x & \cos^2 x & \cos 2x \\ \sin^2 x & \cos^2 x & 1 + \cos 2x \end{vmatrix}, x \in [0, \pi]$ Then the maximum value of $f(x)$ is equal to
Answer: 6
Solution
Given the matrix: $$\begin{vmatrix} -2 & -2 & 0 \\ 2 & 0 & -1 \\ \sin^2 x & \cos^2 x & 1 + \cos 2x \end{vmatrix}$$ Perform the row operations: $$(R_1 \rightarrow R_1 - R_2)$$ $$(R_2 \rightarrow R_2 - R_3)$$ Calculate: $$-2 (\cos^2 x) + 2 \left(2 + 2 \cos 2x + \sin^2 x\right)$$ Simplify to: $$4 + 4 \cos 2x - 2 \left(\cos^2 x - \sin^2 x\right)$$ Thus, $$f(x) = 4 + \underbrace{2 \cos 2x}_{max = 1}$$ The maximum value of $f(x)$ is: $$f(x)_{max} = 4 + 2 = 6$$
Question 26
Maths · Integrals · Numerical
Let $F: [3,5] \to \mathbb{R}$ be a twice differentiable function on $(3,5)$ such that \[ F(x) = e^{-x}\int_3^x \left(3t^2 + 2t + 4F'(t)\right)dt \] If $F'(4) = \dfrac{\alpha e^{\beta} - 224}{\left(e^{\beta}-4\right)^2}$, then $\alpha + \beta$ is equal to
Let a plane P pass through the point $(3, 7, -7)$ and contain the line, $\frac{x-2}{-3} = \frac{y-3}{2} = \frac{z+2}{1}$. If distance of the plane P from the origin is $d$, then $d^2$ is equal to
Let $S = \{1, 2, 3, 4, 5, 6, 7\}$. Then the number of possible functions $f : S \to S$ such that $f(m \cdot n) = f(m) \cdot f(n)$ for every $m, n \in S$ and $m \cdot n \in S$ is equal to
Answer: 490
Solution
Given $\(F(mn)=f(m)\cdot f(n)\). Put \(m=1\)$. \[ f(n)=f(1)\cdot f(n)\Rightarrow f(1)=1. \] Put \(m=n=2\). \[ f(4)=f(2)\cdot f(2) \] \[ f(2)=1 \Rightarrow f(4)=1 \qquad \text{or} \qquad f(2)=2 \Rightarrow f(4)=4 \] Put \(m=2,\; n=3\). \[ f(6)=f(2)\cdot f(3) \] When \(f(2)=1\), \[ f(3)=1 \text{ to } 7 \] When \(f(2)=2\), \[ f(3)=1 \text{ or } 2 \text{ or } 3 \] \(f(5)\) and \(f(7)\) can take any value. \[ |x|\times 7\times 1\times 7\times 1\times 7 \] \[ +\,|x|\times 3\times 1\times 7\times 1\times 7 \] \[ =490 \]
Question 29
Maths · Differential Equations · Numerical
If $y = y(x)$, $y \in \left[0, \frac{\pi}{2}\right]$ is the solution of the differential equation $$\sec y \frac{dy}{dx} - \sin(x + y) - \sin(x - y) = 0, with y(0) = 0$$ then $5y'\left(\frac{\pi}{2}\right)$ is equal to
Answer: 2
Solution
Given $\sec y \frac{dy}{dx} = 2 \sin x \cos y$. Multiply both sides by $\sec y$ to get $\sec^2 y \, dy = 2 \sin x \, dx$. Integrating both sides, we have $\tan y = -2 \cos x + c$. Given $c = 2$, we substitute to get $\tan y = -2 \cos x + 2$. At $x = \frac{\pi}{2}$, $\tan y = 2$. Differentiating again, $\sec^2 y \frac{dy}{dx} = 2 \sin x$. Substitute $\tan y = 2$ to get $5 \frac{dy}{dx} = 2$.
Question 30
Maths · Continuity and Differentiability · Numerical
Let $f : [0, 3] \to \mathbb{R}$ be defined by $$f(x) = \min\{x - \lfloor x \rfloor, 1 + \lfloor x \rfloor - x\}$$ where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. Let $P$ denote the set containing all $x \in [0, 3]$ where $f$ is discontinuous, and $Q$ denote the set containing all $x \in (0, 3)$ where $f$ is not differentiable. Then the sum of number of elements in $P$ and $Q$ is equal to
Answer: 5
Solution
The function is given by $1 - \{x\} = 1 - x$ for $0 \leq x < 1$. The function is non-differentiable at $x = \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}$.
Physics
Question 31
Physics · Current Electricity · Single correct
In the given figure, a battery of emf $E$ is connected across a conductor $PQ$ of length $l$ and different area of cross-sections having radii $r_1$ and $r_2$ ($r_2 < r_1$). Choose the correct option as one moves from $P$ to $Q$:
Drift velocity of electron increases.
Electric field decreases.
Electron current decreases.
All of these
Answer: (a)
Solution
Current is constant in conductor, $i = constant$. Resistance of element $dR = \frac{\rho \, dx}{\pi r^2}$. $$dV = i \, dR = \frac{i \rho \, dx}{\pi r^2}$$ $$E = \frac{dV}{dx} = \frac{i \rho}{\pi r^2}$$ $$V_d = \frac{e E \tau}{m}$$ Thus, $V_d \propto E$. Therefore, $E \propto \frac{1}{r^2}$. If $r$ decreases, $E$ will increase. Therefore, $V_d$ will increase.
Question 32
Physics · Kinetic Theory · Single correct
The number of molecules in one litre of an ideal gas at 300 K and 2 atmospheric pressure with mean kinetic energy $2 \times 10^{-9}$ J per molecules is :
0.75 $\times$ 10^{11}
3 $\times$ 10^{11}
1.5 $\times$ 10^{11}
6 $\times$ 10^{11}
Answer: (c)
Solution
KE = $\frac{3}{2}$ k T PV = $\frac{N}{N_A}$ RT N = $\frac{PV}{k T}$ = N = 1.5 $\times$ 10^{11}
Question 33
Physics · Electric Charges and Fields · Single correct
The relative permittivity of distilled water is 81. The velocity of light in it will be: ( Given $\mu_r = 1$)
$4.33 \times 10^7 \, \mathrm{m/s}$
$2.33 \times 10^7 \, \mathrm{m/s}$
$3.33 \times 10^7 \, \mathrm{m/s}$
$5.33 \times 10^7 \, \mathrm{m/s}$
Answer: (c)
Solution
Given $$V = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$$ The velocity is $$= 3.33 \times 10^7 \, \mathrm{m/sec}$$
Question 34
Physics · Ray Optics and Optical Instruments · Single correct
Choose the correct answer from the options given below :
Physics · System of Particles and Rotational Motion · Single correct
Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. The masses of A, B and C are $m$, $2m$ and $2m$ respectively. A moves towards B with a speed of $9 \, \mathrm{m/s}$ and makes an elastic collision with it. Thereafter B makes a completely inelastic collision with C. All motions occur along same straight line. The final speed of C is:
$6 \, \mathrm{m/s}$
$9 \, \mathrm{m/s}$
$4 \, \mathrm{m/s}$
$3 \, \mathrm{m/s}$
Answer: (d)
Solution
Collision between A and B $$m \times 9 = mv_1 + 2 \, mv_2 (from momentum conservation)$$ $$e = 1 = \frac{v_2 - v_1}{9}$$ $$\Rightarrow v_2 = 6 \, \mathrm{m/s}, \ v_1 = -3 \, \mathrm{m/s}$$ Collision between B and C $$2 \, m \times 6 = 4 \, mv (from momentum conservation)$$ $$v = 3 \, \mathrm{m/s}$$
Question 36
Physics · Electrostatic Potential and Capacitance · Single correct
A capacitor of capacitance $C = 1 \, \mu \mathrm{F}$ is suddenly connected to a battery of $100 \, \mathrm{volt}$ through a resistance $R = 100 \, \Omega$. The time taken for the capacitor to be charged to get $50 \, \mathrm{V}$ is: [ Take $\ln 2 = 0.69$ ]
$1.44 \times 10^{-4} \, \mathrm{s}$
$3.33 \times 10^{-4} \, \mathrm{s}$
$0.69 \times 10^{-4} \, \mathrm{s}$
$0.30 \times 10^{-4} \, \mathrm{s}$
Answer: (c)
Solution
Given the equation for voltage: $$V = V_0 \left(1 - e^{-\frac{t}{RC}}\right)$$ Substitute the values: $$50 = 100 \left(1 - e^{-\frac{t}{RC}}\right)$$ Solving for $t$: $$t = 0.69 \times 10^{-4} sec.$$
Question 37
Physics · Electrostatic Potential and Capacitance · Single correct
In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be: (Given area of plate = A)
$\frac{15}{34} \frac{K \varepsilon_0 A}{d}$
$\frac{15}{6} \frac{K \varepsilon_0 A}{d}$
$\frac{25}{6} \frac{K \varepsilon_0 A}{d}$
$\frac{9}{6} \frac{K \varepsilon_0 A}{d}$
Answer: (a)
Solution
Given the equation for the effective capacitance: $$\frac{1}{C_{eff}} = \frac{d}{K \epsilon_0 A} + \frac{2d}{3K \epsilon_0 A} + \frac{3d}{5K \epsilon_0 A}$$ Solving for $C_{eff}$, we have: $$C_{eff} = \frac{15K \epsilon_0 A}{34d}$$
Question 38
Physics · System of Particles and Rotational Motion · Single correct
The figure shows two solid discs with radii $R$ and $r$, respectively. If mass per unit area is same for both, what is the ratio of MI of bigger disc about axis $AB$ (which is $\perp$ to the plane of the disc and passing through its centre) of MI of smaller disc around one of its diameters lying on its plane? Given '$M$' is the mass of the larger disc. (MI stands for moment of inertia.)
$R^2 : r^2$
$2r^4 : R^4$
$2R^2 : r^2$
$2R^4 : r^4$
Answer: (d)
Solution
Ratio of moment of inertia = $\($ $\frac{\frac{1}{2} MR^2}{\frac{1}{4} mr^2}$ $\)$ \[ = \frac{2 \sigma \pi R^2 R^2}{\sigma \pi r^2 r^2} = \frac{2R^4}{r^4} \]
Question 39
Physics · Wave Optics · Single correct
In Young's double slit experiment, if the source of light changes from orange to blue then:
the central bright fringe will become a dark fringe.
the distance between consecutive fringes will decrease.
the distance between consecutive fringes will increase.
the intensity of the minima will increase.
Answer: (b)
Solution
Fringe width = $\frac{\lambda D}{d}$ as $\lambda$ decreases, fringe width also decreases
Question 40
Physics · Thermodynamics · Single correct
In the reported figure, there is a cyclic process ABCDA on a sample of $1 \, \mathrm{mol}$ of a diatomic gas. The temperature of the gas during the process $A \rightarrow B$ and $C \rightarrow D$ are $T_1$ and $T_2$ $(T_1 > T_2)$ respectively. Choose the correct option out of the following for work done if processes BC and DA are adiabatic.
$W_{AB} = W_{DC}$
$W_{AD} = W_{BC}$
$W_{BC} + W_{DA} > 0$
$W_{AB} < W_{CD}$
Answer: (b)
Solution
Work done in adiabatic process = $$\frac{-nR}{\gamma - 1} \left( T_f - T_i \right)$$ Therefore, $$W_{AD} = \frac{-nR}{\gamma - 1} \left( T_2 - T_1 \right)$$ and $$W_{BC} = \frac{-nR}{\gamma - 1} \left( T_2 - T_1 \right)$$ Therefore, $$W_{AD} = W_{BC}$$
Question 41
Physics · Mathematics in Physics · Single correct
Assertion $A$: If $A, B, C, D$ are four points on a semi-circular arc with centre at 'O' such that $|\overrightarrow{AB}| = |\overrightarrow{BC}| = |\overrightarrow{CD}|$, then $\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} = 4\overrightarrow{AO} + \overrightarrow{OB} + \overrightarrow{OC}$ Reason $R$: Polygon law of vector addition yields $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{AD} = 2\overrightarrow{AO}$ In the light of the above statements, choose the most appropriate answer from the options given below:
$A$ is correct but $R$ is not correct.
$A$ is not correct but $R$ is correct.
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 $A$.
Answer: (d)
Solution
Polygon law is applicable in both but the equation given in the reason is not useful in explaining the assertion.
Question 42
Physics · Mechanical Properties of Fluids · Single correct
A light cylindrical vessel is kept on a horizontal surface. Area of base is $A$. A hole of cross-sectional area $a$ is made just at its bottom side. The minimum coefficient of friction necessary to prevent sliding the vessel due to the impact force of the emerging liquid is (a<<A)
$\frac{A}{2a}$
None of these
$\frac{2a}{A}$
$\frac{a}{A}$
Answer: (c)
Solution
For no sliding $$f \geq \rho a v^2$$ $$\mu mg \geq \rho a v^2$$ $$\mu \rho A h g \geq \rho a 2 g h$$ $$\mu \geq \frac{2a}{A}$$ Option (3)
Question 43
Physics · Oscillations · Single correct
A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kinetic energy is $\frac{3E}{4}$ then its displacement 'y' is given by:
$y = a$
$y = \frac{a}{\sqrt{2}}$
$y = \frac{a\sqrt{3}}{2}$
$y = \frac{a}{2}$
Answer: (d)
Solution
Given $E = \frac{1}{2} K a^2$. $$\frac{3E}{4} = \frac{1}{2} K \left(a^2 - y^2\right)$$ $$\frac{3}{4} \times \frac{1}{2} K a^2 = \frac{1}{2} K \left(a^2 - y^2\right)$$ $$y^2 = a^2 - \frac{3a^2}{4}$$ $$y = \frac{a}{2}$$
Question 44
Physics · Nuclei · Single correct
If 'f denotes the ratio of the number of nuclei decayed ($N_d$) to the number of nuclei at $t = 0$ ($N_0$) then for a collection of radioactive nuclei, the rate of change of 'f with respect to time is given as: $[\lambda$ is the radioactive decay constant $]$
Physics · Electrostatic Potential and Capacitance · Single correct
Two capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be:
Physics · Motion in a Straight Line · Single correct
A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching $\frac{h}{3}$ in both the directions.
$\frac{\sqrt{2} - 1}{\sqrt{2} + 1}$
$\frac{1}{3}$
$\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$
$\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$
Answer: (c)
Solution
Given $u = \sqrt{2gh}$. Now, $S = \frac{h}{3}$ and $a = -g$. The equation for $S$ is $S = ut + \frac{1}{2}at^2$. Substituting the values, we have $$\frac{h}{3} = \sqrt{2gh}t + \frac{1}{2}(-g)t^2$$ which simplifies to $$t^2 \left( \frac{g}{2} \right) - \sqrt{2gh}t + \frac{h}{3} = 0.$$ From the quadratic equation, $$t_1, t_2 = \frac{\sqrt{2gh} \pm \sqrt{2gh - \frac{4g}{2} \frac{h}{3}}}{g}.$$ Simplifying further, $$\frac{t_1}{t_2} = \frac{\sqrt{2gh} - \sqrt{\frac{4gh}{3}}}{\sqrt{2gh} + \sqrt{\frac{4gh}{3}}} = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}.$$
Question 47
Physics · Alternating Current · Single correct
A 0.07$\mathrm{H}$ inductor and a 12$\Omega$ resistor are connected in series to a 220$\mathrm{V}$, 50$\mathrm{Hz}$ ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take $\pi$ as $\frac{22}{7}$]
8.8$\mathrm{A}$ and $\tan^{-1}\left(\frac{11}{6}\right)$
88$\mathrm{A}$ and $\tan^{-1}\left(\frac{11}{6}\right)$
0.88$\mathrm{A}$ and $\\tan^{-1}\left(\frac{11}{6}\right)$
8.8$\mathrm{A}$ and $\tan^{-1}\left(\frac{6}{11}\right)$
Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'l'. What is the equilibrium separation when each thread makes a small angle 'θ' with the vertical?
$x = \left( \frac{q^2 l}{2 \pi \varepsilon_0 m g} \right)^{\frac{1}{2}}$
$x = \left( \frac{q^2 l}{2 \pi \varepsilon_0 m g} \right)^{\frac{1}{3}}$
Given the forces, we have: $$T \cos \theta = mg$$ $$T \sin \theta = \frac{kq^2}{x^2}$$ Thus, $$\tan \theta = \frac{kq^2}{x^2 mg}$$ As $\tan \theta \approx \sin \theta \approx \frac{x}{2L}$, we have: $$\frac{x}{2L} = \frac{kq^2}{x^2 mg}$$ Solving for $x$, we get: $$x = \left( \frac{q^2 L}{2 \pi \varepsilon_0 mg} \right)^{1/3}$$
Question 49
Physics · Experimental Physics · Single correct
Assertion A: If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is $5 \, \mathrm{mm}$ and there are $50$ total divisions on circular scale, then least count is $0.001 \, \mathrm{cm}$. Reason R: Least Count $= \frac{\text{Pitch}}{\text{Total divisions on circular scale}}$ In the light of the above statements, choose the most appropriate answer from the options given below:
A is not correct but R is correct.
Both A and R are correct and R is the correct explanation of A.
A is correct but R is not correct.
Both A and R are correct and R is NOT the correct explanation of A.
Answer: (a)
Solution
Least count = $\($ $\frac{Pitch}{total division on circular scale}$ $\)$ In 5 revolutions, distance traveled is 5 mm. In 1 revolution, it will travel 1 mm. So least count = $\($ $\frac{1}{50}$ = 0.02 $\)$
Question 50
Physics · Thermal Properties of Matter · Single correct
A body takes 4 min. to cool from $61^\circ \mathrm{C}$ to $59^\circ \mathrm{C}$. If the temperature of the surroundings is $30^\circ \mathrm{C}$, the time taken by the body to cool from $51^\circ \mathrm{C}$ to $49^\circ \mathrm{C}$ is:
Consider an electrical circuit containing a two way switch 'S'. Initially S is open and then $T_1$ is connected to $T_2$. As the current in $R = 6\Omega$ attains a maximum value of steady state level, $T_1$ is disconnected from $T_2$ and immediately connected to $T_3$. Potential drop across $r = 3\Omega$ resistor immediately after $T_1$ is connected to $T_3$ is ___ V. (Round off to the Nearest Integer)
Answer: 3
Solution
When $T_1$ and $T_2$ are connected, then the steady state current in the inductor $I = \frac{6}{6} = 1 \, \mathrm{A}$. When $T_1$ and $T_3$ are connected then current through inductor remains same. So potential difference across $3 \, \Omega$ $$V = Ir = 1 \times 3 = 3 \, volt$$
Question 52
Physics · Gravitation · Numerical
Suppose two planets (spherical in shape) of radii $R$ and $2R$, but mass $M$ and $9M$ respectively have a centre to centre separation $8R$ as shown in the figure. A satellite of mass 'm' is projected from the surface of the planet of mass 'M' directly towards the centre of the second planet. The minimum speed 'v' required for the satellite to reach the surface of the second planet is $\sqrt{\frac{a}{7} \frac{GM}{R}}$ then the value of 'a' is ____ [Given : The two planets are fixed in their position]
Answer: 4
Solution
Acceleration due to gravity will be zero at P therefore, $$\frac{GM}{x^2} = \frac{G9M}{(8R-x)^2}$$ $$8R - x = 3x$$ $$x = 2R$$ Apply conservation of energy and consider velocity at P is zero. $$\frac{1}{2} mv^2 - \frac{GMm}{R} - \frac{G9Mm}{7R} = 0 - \frac{GMm}{2R} - \frac{G9Mm}{6R}$$ $$\therefore \; V = \sqrt{\frac{4}{7} \frac{GM}{R}}$$
Question 53
Physics · Atoms · Numerical
In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius 0.5$\AA$. If the speed of electron is $2.2 \times 10^6 \, \mathrm{m/s}$, then the current associated with the electron will be ___ $\times 10^{-2}$ mA. [ Take $\pi$ as $\frac{22}{7}$ ]
Answer: 112
Solution
Given $$I = \frac{e}{T} = \frac{e \omega}{2 \pi} = \frac{eV}{2 \pi}$$ Calculate $$I = \frac{1.6 \times 10^{-19} \times 2.2 \times 10^6 \times 7}{2 \times 22 \times 0.5 \times 10^{-10}}$$ This simplifies to $$= 1.12 \, \mathrm{mA}$$ Which is equivalent to $$112 \times 10^{-2} \, \mathrm{mA}$$
Question 54
Physics · Nuclei · Numerical
A radioactive sample has an average life of 30 ms and is decaying. A capacitor of capacitance 200 $\mu$F is first charged and later connected with resistor 'R'. If the ratio of charge on capacitor to the activity of radioactive sample is fixed with respect to time then the value of 'R' should be ___ $\Omega$.
Answer: 150
Solution
Given $T_m = 30 \, \mathrm{ms}$ and $C = 200 \, \mu \mathrm{F}$. The equation is $$\frac{q}{N} = \frac{Q_0 e^{-t/RC}}{N_0 e^{-\lambda t}} = \frac{Q_0}{N_0} e^{\left( \lambda - \frac{1}{RC} \right)}.$$ Since $q/N$ is constant, hence $$\lambda = \frac{1}{RC}.$$ Therefore, $$R = \frac{1}{\lambda C} = \frac{T_m}{C} = \frac{30 \times 10^{-3}}{200 \times 10^{-6}} = 150 \, \Omega.$$
Question 55
Physics · Dual Nature of Radiation and Matter · Numerical
A particle of mass $9.1 \times 10^{-31} \, \mathrm{kg}$ travels in a medium with a speed of $10^6 \, \mathrm{m/s}$ and a photon of a radiation of linear momentum $10^{-27} \, \mathrm{kg \, m/s}$ travels in vacuum. The wavelength of photon is ___ times the wavelength of the particle.
Physics · Ray Optics and Optical Instruments · Numerical
A prism of refractive index $n_1$ and another prism of refractive index $n_2$ are stuck together (as shown in the figure). $n_1$ and $n_2$ depend on $\lambda$, the wavelength of light, according to the relation $$n_1 = 1.2 + \frac{10.8 \times 10^{-14}}{\lambda^2} and n_2 = 1.45 + \frac{1.8 \times 10^{-14}}{\lambda^2}$$ The wavelength for which rays incident at any angle on the interface BC pass through without bending at that interface will be ___ nm
Physics · Mechanical Properties of Solids · Numerical
A stone of mass 20 g is projected from a rubber catapult of length 0.1 m and area of cross section $10^{-6} \, \mathrm{m}^2$ stretched by an amount 0.04 m. The velocity of the projected stone is ___ m/s (Young's modulus of rubber = $0.5 \times 10^9 \, \mathrm{N/m}^2$)
A transistor is connected in common emitter circuit configuration, the collector supply voltage is 10 V and the voltage drop across a resistor of 1000 $\Omega$ in the collector circuit is 0.6 V. If the current gain factor ($\beta$) is 24, then the base current is ___ $\mu$A. (Round off to the Nearest Integer)
The amplitude of upper and lower side bands of A.M. wave where a carrier signal with frequency 11.21 MHz, peak voltage 15 V is amplitude modulated by a 7.7 kHz sine wave of 5 V amplitude are $\frac{a}{10}$ V and $\frac{b}{10}$ V respectively. Then the value of $\frac{a}{b}$ is ____
Answer: 1
Solution
Given the amplitude diagram, we have the following equation: $$\frac{a}{10} = \frac{b}{10} = \frac{\mu A_c}{2}$$ which implies $$\frac{a}{b} = 1$$
Question 60
Physics · Magnetism and Matter · Numerical
In a uniform magnetic field, the magnetic needle has a magnetic moment $9.85 \times 10^{-2} \, \mathrm{A/m^2}$ and moment of inertia $5 \times 10^{-6} \, \mathrm{kg \, m^2}$. If it performs 10 complete oscillations in 5 seconds then the magnitude of the magnetic field is ___ mT [Take $\pi^2$ as 9.85]
Answer: 8
Solution
Given the formula for the period, $$T = 2\pi \sqrt{\frac{I}{MB}}$$. The magnetic field is calculated as $$B = 80 \times 10^{-4} = 8 \, \mathrm{mT}$$.
Chemistry
Question 61
Chemistry · Alcohols, Phenols and Ethers · Single correct
Which one of the following compounds will give orange precipitate when treated with 2,4-dinitrophenyl hydrazine?
Answer: (d)
Solution
Explanation: 2,4-D.N.P test is used for carbonyl compound (aldehyde and ketone).
Question 62
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
The product obtained from the electrolytic oxidation of acidified sulphate solutions, is :
$\mathrm{HSO}_4^-$
$\mathrm{HO}_3\mathrm{SOOSO}_3\mathrm{H}$
$\mathrm{HO}_2\mathrm{SOSO}_2\mathrm{H}$
$\mathrm{HO}_3\mathrm{SOSO}_3\mathrm{H}$
Answer: (b)
Solution
Electrolysis of concentrated solution of acidified sulphate solution yields $\mathrm{H_2S_2O_8}$.
Question 63
Chemistry · The Solid State · Single correct
The parameters of the unit cell of a substance are $a = 2.5, \ b = 3.0, \ c = 4.0, \ \alpha = 90^\circ, \ \beta = 120^\circ, \ \gamma = 90^\circ$ The crystal system of the substance is :
Hexagonal
Orthorhombic
Monoclinic
Triclinic
Answer: (c)
Solution
Given $a \neq b \neq c$ and $\alpha = \gamma = 90^\circ \neq \beta$ are parameters of monoclinic unit cell.
Question 64
Chemistry · Redox Reactions · Single correct
The oxidation states of 'P' in $\mathrm{H_4P_2O_7}$, $\mathrm{H_4P_2O_5}$ and $\mathrm{H_4P_2O_6}$, respectively, are:
7, 5 and 6
5, 4 and 3
5, 3 and 4
6, 4 and 5
Answer: (c)
Solution
Oxidation state of P in $\mathrm{H_4P_2O_7}$, $\mathrm{H_4P_2O_5}$ and $\mathrm{H_4P_2O_6}$ is 5, 3, and 4 respectively. $\mathrm{H_4P_2O_7}$ $$2x + 4(+1) + 7(-2) = 0$$ $$x = +5$$ $\mathrm{H_4P_2O_5}$ $$2x + 4(+1) + 5(-2) = 0$$ $$x = +3$$ $\mathrm{H_4P_2O_6}$ $$2x + 4(+1) + 6(-2) = 0$$ $$x = +4$$
Question 65
Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct
For a reaction of order $n$, the unit of the rate constant is:
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Given below are two statements: Statement I: Aniline is less basic than acetamide. Statement II: In aniline, the lone pair of electrons on nitrogen atom is delocalised over benzene ring due to resonance and hence less available to a proton. Choose the most appropriate option:
Statement I is true but statement II is false.
Statement I is false but statement II is true.
Both statement I and statement II are true.
Both statement I and statement II are false.
Answer: (b)
Solution
Explanation: aniline is more basic than acetamide because in acetamide, lone pair of nitrogen is delocalised to more electronegative element oxygen. In aniline lone pair of nitrogen delocalised over benzene ring.
Question 67
Chemistry · Co-ordination Compounds · Single correct
The type of hybridisation and magnetic property of the complex $[MnCl_6]^{3-}$, respectively, are:
$sp^3 d^2$ and diamagnetic
$d^2 sp^3$ and diamagnetic
$d^2 sp^3$ and paramagnetic
$sp^3 d^2$ and paramagnetic
Answer: (d)
Solution
The complex $[\mathrm{MnCl}_6]^{3-}$ has the electron configuration for $\mathrm{Mn}^{3+}$ as $[\mathrm{Ar}]\,3d^4$. The $3d$ orbitals have 4 unpaired electrons. This makes the complex paramagnetic and having 4 unpaired electrons.
Question 68
Chemistry · Co-ordination Compounds · Single correct
The number of geometrical isomers found in the metal complexes $[PtCl_2(NH_3)_2]$, $[Ni(CO)_4]$, $[Ru(H_2O)_3Cl_3]$ and $[CoCl_2(NH_3)_4]^+$ respectively, are:
1, 1, 1, 1
2, 1, 2, 2
2, 0, 2, 2
2, 1, 2, 1
Answer: (b)
Solution
[$\mathrm{PtCl_2(NH_3)_2}$] 2 Geometrical isomers [$\mathrm{Ni(CO)_4}$] $\rightarrow$ All ligands are same Zero Geometrical isomers [$\mathrm{Ru(H_2O)_3Cl_3}$] 2 Geometrical isomers [$\mathrm{CoCl_3(NH_3)_3}$]^+ 2 Geometrical isomers
Question 69
Chemistry · Environmental Chemistry · Single correct
Which one of the following statements is NOT correct?
Eutrophication indicates that water body is polluted?
The dissolved oxygen concentration below 6 ppm inhibits fish growth
Eutrophication leads to increase in the oxygen level in water
Eutrophication leads to anaerobic conditions
Answer: (c)
Solution
Eutrophication leads to decrease in oxygen level of water. 3rd statement is incorrect.
Question 70
Chemistry · Structure of Atom · Single correct
Given below are two statements: Statement I : Rutherford's gold foil experiment cannot explain the line spectrum of hydrogen atom. Statement II : Bohr's model of hydrogen atom contradicts Heisenberg's uncertainty principle. In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I is false but statement II is true.
Statement I is true but statement II is false.
Both statement I and statement II are false.
Both statement I and statement II are true.
Answer: (d)
Solution
Rutherford's gold foil experiment only proved that electrons are held towards the nucleus by electrostatic forces of attraction and move in circular orbits with very high speeds. Bohr's model gave an exact formula for simultaneous calculation of speed and distance of electron from the nucleus, something which was deemed impossible according to Heisenberg.
Question 71
Chemistry · Hydrocarbons · Single correct
Presence of which reagent will affect the reversibility of the following reaction, and change it to a irreversible reaction:
$HOCl$
dilute $HNO_2$
Liquid $NH_3$
Concentrated $HIO_3$
Answer: (d)
Solution
Iodination of alkane is a reversible reaction. It can be irreversible in the presence of a strong oxidising agent like conc. $\mathrm{HNO_3}$ or conc. $\mathrm{HIO_3}$.
Question 72
Chemistry · Biomolecules · Single correct
Which one among the following chemical tests is used to distinguish monosaccharide from disaccharide?
Seliwanoff's test
Iodine test
Barfoed test
Tollen's test
Answer: (c)
Solution
Barford test is used for distinguish monosaccharide from disaccharide.
Question 73
Chemistry · Chemistry in Everyday Life · Single correct
Match List-I with List-II : \begin{tabular}{|l|l|} \hline List-I & List-II \\ (Drug) & (Class of Drug) \\ \hline (a) Furacin & (i) Antibiotic \\ \hline (b) Arsphenamine & (ii) Tranquilizers \\ \hline (c) Dimetone & (iii) Antiseptic \\ \hline (d) Valium & (iv) Synthetic antihistamines \\ \hline \end{tabular} Choose the most appropriate match:
(a) - (i), (b) - (iii), $(c)$ - (iv), (d) - (ii)
(a) -(iii), (b)-(iv), $(c)$-(ii), (d)-(i)
(a) -(ii), (b)-(i), $(c)$-(iii), (d)-(iv)
(a) -(iii), (b)-(i), $(c)$-(iv), (d)-(ii)
Answer: (d)
Solution
Furacine acts as antiseptic. Arsphenamine, also known as salvarsan, acts as antibiotic. Dimetone is synthetic histamine. Valium is a tranquilizer.
Question 74
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
The statement that is INCORRECT about Ellingham diagram is
provides idea about the reaction rate.
provides idea about free energy change.
provides idea about changes in the phases during the reaction.
provides idea about reduction of metal oxide.
Answer: (a)
Solution
Ellingham diagram is a plot between $\Delta G^\circ$ and $T$ and does not give any information regarding rate of reaction.
Question 75
Chemistry · Alcohols, Phenols and Ethers · Single correct
Consider the above reaction and identify the Product P:
Answer: (d)
Solution
Dehydration of alcohol by $E^1$ reaction. H-attached at more hindered site while OH attached at less hindered site in HBO reaction.
Question 76
Chemistry · Biomolecules · Single correct
The compound 'A' is a complementary base of in DNA stands.
Uracil
Guanine
Adenine
Cytosine
Answer: (c)
Solution
Given structure is Thymine and Thymine being paired with adenine
Question 77
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Staggered and eclipsed conformers of ethane are:
Polymers
Rotamers
Enantiomers
Mirror images
Answer: (b)
Solution
Staggered and eclipsed conformers of ethane are also known as rotamers.
Question 78
Chemistry · The s-Block Elements · Single correct
\text{Match List-I with List-II :} \begin{tabular}{|l|l|} \hline $\textbf{List-I}$ & $\textbf{List-II}$ \\[4pt] \hline \text{(a) } $\mathrm{NaOH}$ & \text{(i) Acidic} \\ \hline \text{(b) } $\mathrm{Be(OH)_2}$ & \text{(ii) Basic} \\ \hline \text{(c) } $\mathrm{Ca(OH)_2}$ & \text{(iii) Amphoteric} \\ \hline \text{(d) } $\mathrm{B(OH)_3}$ & \\ \hline \text{(e) } $\mathrm{Al(OH)_3}$ & \\ \hline \end{tabular} Choose the most appropriate answer from the options given below
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The correct order of stability of given carbocation is:
A > C > B > D
D > B > C > A
D > B > A > C
C > A > D > B
Answer: (a)
Solution
The stability order is given as follows: $Ph-CH_2^+ > CH_3-CH_2^+ > CH_2=CH^+ > CH \equiv C^+$. The phenyl group is stable due to resonance. The positive charge is more stable on less electronegative elements.
Question 80
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Given below are two statements: One is labelled as Assertion $A$ and the other labelled as Reason $R$. Assertion $A$: Lithium halides are somewhat covalent in nature. Reason $R$: Lithium possess high polarisation capability. In the light of the above statements, choose the most appropriate answer from the options given below:
$A$ is true but $R$ is false
$A$ is false but $R$ is true
Both $A$ and $R$ are true but $R$ is NOT the correct explanation of $A$
Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
Answer: (d)
Solution
Lithium due to small size has very high polarization capability and thus increases covalent nature in Halides.
Question 81
Chemistry · Solutions · Numerical
The density of NaOH solution is $1.2 \, \mathrm{g \, cm^{-3}}$. The molality of this solution is ___ $\mathrm{m}$. (Round off to the Nearest Integer) [Use : Atomic masses: Na : $23.0\mathrm{u}$ O : $16.0\mathrm{u}$ H : $1.0\mathrm{u}$ Density of $\mathrm{H_2O}$ : $1.0 \, \mathrm{g \, cm^{-3}}$]
Answer: 5
Solution
Consider $1 \, \ell$ solution. Mass of solution $= (1.2 \times 1000) \, \mathrm{g}$ $= 1200 \, \mathrm{gm}$ Neglecting volume of NaOH, mass of water $= 1000 \, \mathrm{gm}$ $\Rightarrow$ Mass of NaOH $= (1200 - 1000) \, \mathrm{gm}$ $= 200 \, \mathrm{gm}$ $\Rightarrow$ Moles of NaOH $= \frac{200 \, \mathrm{g}}{50 \, \mathrm{g/mol}} = 5 \, \mathrm{mol}$ $\Rightarrow$ molality $= \frac{5 \, \mathrm{mol}}{1 \, \mathrm{kg}} = 5 \, \mathrm{m}$
Question 82
Chemistry · Surface Chemistry · Numerical
$\mathrm{CO_2}$ gas adsorbs on charcoal following Freundlich adsorption isotherm. For a given amount of charcoal, the mass of $\mathrm{CO_2}$ adsorbed becomes $64$ times when the pressure of $\mathrm{CO_2}$ is doubled. The value of $n$ in the Freundlich isotherm equation is ____ $\times 10^{-2}$. (Round off to the Nearest Integer)
Answer: 17
Question 83
Chemistry · Electrochemistry · Numerical
The conductivity of a weak acid HA of concentration $0.001 \, \mathrm{mol} \, \mathrm{L}^{-1}$ is $2.0 \times 10^{-5} \, \mathrm{S} \, \mathrm{cm}^{-1}$. If $\Lambda_m^\circ(\mathrm{HA}) = 190 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$, the ionization constant $(K_a)$ of HA is equal to $\times 10^{-6}$ (Round off to the Nearest Integer)
1.46 $\mathrm{g}$ of a biopolymer dissolved in a 100 $\mathrm{mL}$ water at 300 $\mathrm{K}$ exerted an osmotic pressure of 2.42 $\times 10^{-3}$ $\mathrm{bar}$. The molar mass of the biopolymer is $-10^4$ $\mathrm{g \, mol^{-1}}$. (Round off to the Nearest Integer) [Use: R = 0.083 $\mathrm{L}$ \, $\mathrm{bar}$ \, $\mathrm{mol^{-1}}$ \, $\mathrm{K^{-1}}$]
Answer: 15
Solution
Given $\pi = CRT$; $\pi =$ osmotic pressure. $C =$ molarity. $T =$ Temperature of solution. Let the molar mass be $M \mathrm{gm/mol}$. $$2.42 \times 10^{-3} \, \mathrm{bar} = \left( \frac{1.46 \, \mathrm{g}}{M \mathrm{gm/mol}} \right) \times \left( \frac{0.083 \, \mathrm{L} - \mathrm{bar}}{\mathrm{mol} \cdot \mathrm{K}} \right) \times (300 \, \mathrm{K})$$ $$\Rightarrow M = 15.02 \times 10^4 \, \mathrm{g/mol}$$
An organic compound is subjected to chlorination to get compound A using 5.0 $\mathrm{g}$ of chlorine. When 0.5 $\mathrm{g}$ of compound A is reacted with $\mathrm{AgNO_3}$ [Carius Method], the percentage of chlorine in compound A is when it forms 0.3849 $\mathrm{g}$ of $\mathrm{AgCl}$. (Round off to the Nearest Integer) (Atomic masses of Ag and Cl are 107.87 and 35.5 respectively)
Answer: 19
Solution
Mass of organic compound = 0.5 $\,$ $\mathrm{gm}$. mass of formed $\mathrm{AgCl}$ = 0.3849 $\,$ $\mathrm{gm}$. $\%$ of Cl = $\frac{atomic mass of Cl \times mass formed \mathrm{AgCl}}{molecular mass of \mathrm{AgCl} \times mass of organic compound}$ $\times$ 100 = $\frac{35.5 \times 0.3849}{143.37 \times 0.5}$ $\times$ 100 = 19.06 $\approx$ 19
Question 86
Chemistry · Co-ordination Compounds · Numerical
The number of geometrical isomers possible in triamminetrinitrocobalt (III) is $X$ and in trioxalatochromate (III) is $Y$. Then the value of $X + Y$ is
Answer: 2
Solution
Triamminetrinitrocobalt(III) goes to $[\mathrm{Co(NO_2)_3(NH_3)_3}]$. Ion goes to $[\mathrm{Cr(C_2O_4)_3}]^{3-}$. $[\mathrm{Co(NO_2)_3(NH_3)_3}]$ has two geometrical isomers (X). $[\mathrm{Cr(C_2O_4)_3}]^{3-}$ has zero geometrical isomer (Y). $X + Y = 2 + 0 = 2.0$
Question 87
Chemistry · Chemical Bonding and Molecular Structure · Numerical
In gaseous triethyl amine the "-C-N-C-" bond angle is ______ degree.
Answer: 108
Solution
In gaseous triethyl amine the "-C-N-C-" bond angle is 108 degree.
Question 88
Chemistry · Thermodynamics · Numerical
For water at $100^\circ\mathrm{C}$ and $1\,\mathrm{bar}$, $\Delta_{\mathrm{vap}}H-\Delta_{\mathrm{vap}}U=$ _______ $\times10^2\,\mathrm{J\,mol^{-1}}$ (Round off to the Nearest Integer) $\left[K_b=0.52\,\mathrm{K\,kg\,mol^{-1}}\right]$ [Use: $R=8.31\,\mathrm{J\,mol^{-1}\,K^{-1}}$] Assume volume of $\mathrm{H_2O(l)}$ is much smaller than volume of $\mathrm{H_2O(g)}$. Assume $\mathrm{H_2O(g)}$ is treated as an ideal gas.
Answer: 31
Solution
Question 89
Chemistry · Equilibrium · Numerical
$\mathrm{PCl_5 \rightleftharpoons PCl_3+Cl_2}$, $K_c=1.844$. $3.0$ moles of $\mathrm{PCl_5}$ are introduced into a $1\,\mathrm{L}$ closed reaction vessel at $380\,\mathrm{K}$. The number of moles of $\mathrm{PCl_5}$ at equilibrium is $\underline{\hspace{1cm}}\times10^{-3}$. (Round off to the nearest integer.)
Answer: 1400
Solution
Given the reaction $\mathrm{PCl_5}_{(g)} \rightleftharpoons \mathrm{PCl_3}_{(g)} + \mathrm{Cl_2}_{(g)}$, with $K_2 = 1.844$. At $t = 0$, there are 3 moles. At $t = \infty$, let the change be $x$ for both products. $$\frac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]} = \frac{x^2}{3-x} = 1.844$$ This leads to the equation: $$x^2 + 1.844x - 5.532 = 0$$ Solving for $x$ using the quadratic formula: $$x = \frac{-1.844 + \sqrt{(1.844)^2 + 4 \times 5.532}}{2}$$ This gives $x \approx 1.604$. Therefore, the moles of $\mathrm{PCl_5}$ remaining are $3 - 1.604 \approx 1.396$.
Question 90
Chemistry · Chemical Bonding and Molecular Structure · Fill in the blank
The difference between the bond orders of $\mathrm{CO}$ and $\mathrm{NO^+}$ is $\frac{x}{2}$, where $x=\underline{\hspace{1cm}}$. (Round off to the nearest integer.)
Answer: 0
Solution
Bond order of CO = 3. Bond order of NO^+ = 3. Difference = 0 = $\frac{x}{2}$. x = 0.