JEE Main 27 August 2021 Shift 1 question paper with solutions
JEE Main 27 August 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 · Sequences and Series · Single correct
If $0 < x < 1$, then $\frac{3}{2}x^2 + \frac{5}{3}x^3 + \frac{7}{4}x^4 + \ldots$, is equal to:
If for $x, y \in \mathbb{R}, x > 0$ $$y = \log_{10} x + \log_{10} x^{1/3} + \log_{10} x^{1/9} + \ldots upto \infty terms$$ and $$\frac{2 + 4 + 6 + \ldots + 2y}{3 + 6 + 9 + \ldots + 3y} = \frac{4}{\log_{10} x}$$, then the ordered pair $(x, y)$ is equal to:
$(10^6, 6)$
$(10^4, 6)$
$(10^2, 3)$
$(10^6, 9)$
Answer: (d)
Solution
Given $\($ $\frac{2(1+2+3+\ldots+y)}{3(1+2+3+\ldots+y)}$ = $\frac{4}{\log_{10} x}$ $\)$. This implies $\($ $\log$_{10} x = 6 $\Rightarrow$ x = 10^6 $\)$. Now, $\[$ y = ($\log$_{10} x) + $\left$( $\log$_{10} x^{$\frac{1}{3}$} $\right$) + $\left$( $\log$_{10} x^{$\frac{1}{9}$} $\right$) + $\ldots$ $\infty$ $\]$ $\[$ = $\left$( 1 + $\frac{1}{3}$ + $\frac{1}{9}$ + $\ldots$ $\infty$ $\right$) $\log$_{10} x $\]$ $\[$ = $\left$( $\frac{1}{1 - \frac{1}{3}}$ $\right$) $\log$_{10} x = 9 $\]$ So, $\($(x, y) = (10^6, 9)$\)$
Question 3
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let $A$ be a fixed point $(0, 6)$ and $B$ be a moving point $(2t, 0)$. Let $M$ be the mid-point of $AB$ and the perpendicular bisector of $AB$ meets the $y$-axis at $C$. The locus of the mid-point $P$ of $MC$ is:
$3x^2 - 2y - 6 = 0$
$3x^2 + 2y - 6 = 0$
$2x^2 + 3y - 9 = 0$
$2x^2 - 3y + 9 = 0$
Answer: (c)
Solution
Perpendicular bisector of AB is $$ (y - 3) = \frac{t}{3}(x - t) $$ So, $C = \left(0, 3 - \frac{t^2}{3}\right)$ Let $P$ be $(h, k)$ $$ h = \frac{t}{2}; \; k = \left(3 - \frac{t^2}{6}\right) $$ $$ \Rightarrow k = 3 - \frac{4h^2}{6} \Rightarrow 2x^2 + 3y - 9 = 0 \; option (3) $$
Question 4
Maths · Inverse Trigonometric Functions · Single correct
If $\left(\sin^{-1} x\right)^2 - \left(\cos^{-1} x\right)^2 = a; \ 0 < x < 1, \ a \neq 0$, then the value of $2x^2 - 1$ is:
$\cos\left(\frac{4a}{\pi}\right)$
$\sin\left(\frac{2a}{\pi}\right)$
$\cos\left(\frac{2a}{\pi}\right)$
$\sin\left(\frac{4a}{\pi}\right)$
Answer: (b)
Solution
Given $a=(\sin^{-1}x)^2-(\cos^{-1}x)^2$. This equals $a=(\sin^{-1}x+\cos^{-1}x)(\sin^{-1}x-\cos^{-1}x)$. This simplifies to $a=\frac{\pi}{2}\left(\frac{\pi}{2}-2\cos^{-1}x\right)$. Thus, $2\cos^{-1}x=\frac{\pi}{2}-\frac{2a}{\pi}$. Therefore, $\cos^{-1}(2x^2-1)=2\cos^{-1}x=\frac{\pi}{2}-\frac{2a}{\pi}$. This implies $2x^2-1=\cos\left(\frac{\pi}{2}-\frac{2a}{\pi}\right)$. Hence, option (2).
Question 5
Maths · Matrices · Single correct
If the matrix $A = \begin{pmatrix} 0 & 2 \\ K & -1 \end{pmatrix}$ satisfies $A \left( A^3 + 3I \right) = 2I$ then the value of $K$ is:
Maths · Three Dimensional Geometry · Single correct
The distance of the point $(1, -2, 3)$ from the plane $x - y + z = 5$ measured parallel to a line, whose direction ratios are $2, 3, -6$ is:
3
5
2
1
Answer: (d)
Solution
Given the equation of the line and the plane, we have: $$(1 + 2\lambda) + 2 - 3\lambda + 3 - 6\lambda = 5$$ Simplifying gives: $$6 - 7\lambda = 5 \implies \lambda = \frac{1}{7}$$ So, the point $P$ is: $$P = \left( \frac{9}{7}, -\frac{11}{7}, \frac{15}{7} \right)$$ To find $AP$, we calculate: $$AP = \sqrt{\left(1 - \frac{9}{7}\right)^2 + \left(-2 + \frac{11}{7}\right)^2 + \left(3 - \frac{15}{7}\right)^2}$$ Simplifying further: $$AP = \sqrt{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}} = 1$$
Question 7
Maths · Complex Numbers and Quadratic Equations · Single correct
If $S = \left\{ z \in \mathbb{C} : \frac{z-i}{z+2i} \in \mathbb{R} \right\}$, then :
$S$ contains exactly two elements
$S$ contains only one element
$S$ is a circle in the complex plane
$S$ is a straight line in the complex plane
Answer: (d)
Solution
Given $\frac{z-i}{z+2i} \in \mathbb{R}$. Then $\arg \left( \frac{z-i}{z+2i} \right)$ is $0$ or $\Pi$. Therefore, $S$ is a straight line in complex.
Question 8
Maths · Differential Equations · Single correct
Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} = 2(y + 2 \sin x - 5)x - 2 \cos x$ such that $y(0) = 7$. Then $y(\pi)$ is equal to:
$2e^{\pi^2} + 5$
$e^{\pi^2} + 5$
$3e^{\pi^2} + 5$
$7e^{\pi^2} + 5$
Answer: (a)
Solution
Given $\($ $\frac{dy}{dx}$ - 2xy = 2(2 $\sin$ x - 5)x - 2 $\cos$ x $\)$. $\($ IF = e^{-x^2} $\)$ So, $\($ y $\cdot$ e^{-x^2} = $\int$ e^{-x^2} (2x(2 $\sin$ x - 5) - 2 $\cos$ x) $\,$ dx $\)$ $\($ $\Rightarrow$ y $\cdot$ e^{-x^2} = e^{-x^2} (5 - 2 $\sin$ x) + c $\)$ $\($ $\Rightarrow$ y = 5 - 2 $\sin$ x + c $\cdot$ e^{x^2} $\)$ Given at $\($ x = 0, y = 7 $\)$ $\($ $\Rightarrow$ 7 = 5 + c $\Rightarrow$ c = 2 $\)$ So, $\($ y = 5 - 2 $\sin$ x + 2e^{x^2} $\)$ Now at $\($ x = $\pi$ $\)$, $\($ y = 5 + 2e^{$\pi$^2} $\)$
Question 9
Maths · Three Dimensional Geometry · Single correct
Equation of a plane at a distance $\sqrt{\frac{2}{21}}$ from the origin, which contains the line of intersection of the planes $x - y - z - 1 = 0$ and $2x + y - 3z + 4 = 0$ is:
$3x - y - 5z + 2 = 0$
$3x - 4z + 3 = 0$
$-x + 2y + 2z - 3 = 0$
$4x - y - 5z + 2 = 0$
Answer: (d)
Solution
Required equation of plane $P_1 + \lambda P_2 = 0$ $$(x - y - z - 1) + \lambda (2x + y - 3z + 4) = 0$$ Given that its distance from origin is $\frac{2}{\sqrt{21}}$ Thus $$\frac{|4\lambda - 1|}{\sqrt{(2\lambda + 1)^2 + (\lambda - 1)^2 + (-3\lambda - 1)^2}} = \frac{\sqrt{2}}{\sqrt{21}}$$ $$\Rightarrow 21(4\lambda - 1)^2 = 2 \left(14\lambda^2 + 8\lambda + 3\right)$$ $$\Rightarrow 336\lambda^2 - 168\lambda + 21 = 28\lambda^2 + 16\lambda + 6$$ $$\Rightarrow 308\lambda^2 - 184\lambda + 15 = 0$$ $$\Rightarrow 308\lambda^2 - 154\lambda - 30\lambda + 15 = 0$$ $$\Rightarrow (2\lambda - 1)(154\lambda - 15) = 0$$ $$\Rightarrow \lambda = \frac{1}{2} or \frac{15}{154}$$ For $\lambda = \frac{1}{2}$ required plane is $$4x - y - 5z + 2 = 0$$
Question 10
Maths · Sequences and Series · Single correct
If $U_n = \left(1+\dfrac{1}{n^2}\right)\left(1+\dfrac{2^2}{n^2}\right)^2 \cdots \left(1+\dfrac{n^2}{n^2}\right)^n$, then $\displaystyle\lim_{n\to\infty} (U_n)^{\frac{-4}{n^2}}$ is equal to:
$\frac{e^2}{16}$
$\frac{4}{e}$
$\frac{16}{e^2}$
$\frac{4}{e^2}$
Answer: (a)
Solution
Given $U_n = \prod_{r=1}^{n} \left( 1 + \frac{r^2}{n^2} \right)^r$. Let $L = \lim_{n \to \infty} (U_n)^{-4/n^2}$. Then, $$\log L = \lim_{n \to \infty} \frac{-4}{n^2} \sum_{r=1}^{n} \log \left( 1 + \frac{r^2}{n^2} \right)^r$$ which implies $$\log L = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{-4r}{n} \cdot \frac{1}{n} \log \left( 1 + \frac{r^2}{n^2} \right)$$ leading to $$\log L \Rightarrow -4 \int_{0}^{1} x \log(1 + x^2) \, dx.$$ Put $1 + x^2 = t$. Now, $2x \, dx = dt$. Thus, $$= -2 \int_{1}^{2} \log(t) \, dt = -2 [t \log t - t]_{1}^{2}$$ which gives $$\log L = -2(2 \log 2 - 1).$$ Therefore, $$L = e^{-2(2 \log 2 - 1)}$$ which simplifies to $$= e^{-2 \left( \log \left( \frac{4}{e} \right) \right)}$$ $$= e^{\log \left( \frac{4}{e} \right)^{-2}}$$ $$= \left( \frac{e}{4} \right)^2 = \frac{e^2}{16}.$$
Question 11
Maths · Mathematical Reasoning · Single correct
The statement $(p \land (p \rightarrow q) \land (q \rightarrow r)) \rightarrow r$ is:
a tautology
equivalent to $p \rightarrow \sim r$
a fallacy
equivalent to $q \rightarrow \sim r$
Answer: (a)
Solution
(p $\land$ (p $\rightarrow$ q) $\land$ (q $\rightarrow$ r)) $\rightarrow$ r $\equiv$ (p $\land$ ($\sim$ p $\lor$ q) $\lor$ ($\sim$ q $\lor$ r)) $\rightarrow$ r $\equiv$ ((p $\land$ q) $\land$ ($\sim$ p $\lor$ r)) $\rightarrow$ r $\equiv$ (p $\land$ q $\land$ r) $\rightarrow$ r $\equiv$ $\sim$ (p $\land$ q $\land$ r) $\lor$ r $\equiv$ ($\sim$ p) $\lor$ ($\sim$ q) $\lor$ ($\sim$ r) $\lor$ r $\Rightarrow$ tautology
Question 12
Maths · Differential Equations · Single correct
Let us consider a curve, $y = f(x)$ passing through the point $(-2, 2)$ and the slope of the tangent to the curve at any point $(x, f(x))$ is given by $f(x) + x f'(x) = x^2$ Then:
$x^2 + 2x f(x) - 12 = 0$
$x^3 + x f(x) + 12 = 0$
$x^3 - 3x f(x) - 4 = 0$
$x^2 + 2x f(x) + 4 = 0$
Answer: (c)
Solution
Given $y + x \frac{dy}{dx} = x^2$ (given). This implies $\frac{dy}{dx} + \frac{y}{x} = x$. If $e^{\int \frac{1}{x} \, dx} = x$, then the solution of the differential equation is $y \cdot x = \int x \cdot x \, dx$. This implies $xy = \frac{x^3}{3} + \frac{c}{3}$. The curve passes through $(-2, 2)$, so $-12 = -8 + c \Rightarrow c = -4$. Therefore, $3xy = x^3 - 4$. Hence, $3x \cdot f(x) = x^3 - 4$.
Question 13
Maths · Binomial Theorem · Single correct
\[ \sum_{k=0}^{20}\left({}^{20}C_k\right)^2 \] is equal to:
$\binom{40}{21}$
$\binom{40}{19}$
$\binom{40}{20}$
$\binom{41}{20}$
Answer: (c)
Solution
$\sum_{k=0}^{20} {}^{20}C_{k}\cdot{}^{20}C_{20-k}$ \text{Sum of suffixes is constant.} \text{Therefore, by Vandermonde's identity,} $\sum_{k=0}^{20} {}^{20}C_{k}\cdot{}^{20}C_{20-k} = {}^{40}C_{20}.$
Question 14
Maths · Conic Sections · Single correct
A tangent and a normal are drawn at the point $P(2, -4)$ on the parabola $y^2 = 8x$, which meet the directrix of the parabola at the points $A$ and $B$ respectively. If $Q(a, b)$ is a point such that $AQBP$ is a square, then $2a + b$ is equal to:
$-16$
$-18$
$-12$
$-20$
Answer: (a)
Solution
Equation of tangent at $(2, -4)$ $(T = 0)$ $$-4y = 4(x + 2)$$ $$x + y + 2 = 0 \ldots (1)$$ Equation of normal $$x - y + \lambda = 0$$ At $(2, -4)$ $$\lambda = -6$$ Thus $x - y = 6 \ldots (2)$ equation of normal Point of intersection of $(1)$ and $x = -2$ is $A(-2, 0)$ Point of intersection of $(2)$ and $x = -2$ is $A(-2, 8)$ Given $AQBP$ is a square. $$\Rightarrow m_{AQ} \cdot m_{AP} = -1$$ $$\Rightarrow \left( \frac{b}{a + 2} \right) \left( \frac{4}{-4} \right) = -1 \Rightarrow a + 2 = b \ldots (1)$$ Also $PQ$ must be parallel to $x$-axis thus $$\Rightarrow b = -4$$ Therefore, $a = -6$ Thus $2a + b = -16$
Question 15
Maths · Properties of Triangles · Single correct
Let $\frac{\sin A}{\sin B} = \frac{\sin(A-C)}{\sin(C-B)}$, where $A, B, C$ are angles of a triangle $ABC$. If the lengths of the sides opposite these angles are $a, b, c$ respectively, then:
$b^2 - a^2 = a^2 + c^2$
$b^2, c^2, a^2$ are in A.P.
$c^2, a^2, b^2$ are in A.P.
$a^2, b^2, c^2$ are in A.P.
Answer: (b)
Solution
$\dfrac{\sin A}{\sin B}=\dfrac{\sin(A-C)}{\sin(C-B)}$ As $A,\ B,\ C$ are angles of a triangle, $A+B+C=\pi$ $A=\pi-(B+C)$ So, $\sin A=\sin(B+C)\qquad\ldots(1)$ Similarly, $\sin B=\sin(A+C)\qquad\ldots(2)$ From (1) and (2), $\dfrac{\sin(B+C)}{\sin(A+C)}$ = $\dfrac{\sin(A-C)}{\sin(C-B)}$ $\Rightarrow$ $\sin(B+C)\sin(C-B)$ = $\sin(A-C)\sin(A+C)$ $\Rightarrow$ $\sin^2C-\sin^2B$ = $\sin^2A-\sin^2C$ $\because\ \sin(x+y)\sin(x-y)=\sin^2x-\sin^2y$ $\Rightarrow 2\sin^2C=\sin^2A+\sin^2B$ By sine rule, $2c^2=a^2+b^2$ $\Rightarrow b^2,\ c^2\ \text{and}\ a^2\ \text{are in A.P.}$
Question 16
Maths · Limits and Derivatives · Single correct
If $\alpha$, $\beta$ are the distinct roots of $x^2 + bx + c = 0$, then $\lim_{x \to \beta} \frac{e^{2(x^2+bx+c)} - 1 - 2(x^2+bx+c)}{(x-\beta)^2}$ is equal to:
When a certain biased die is rolled, a particular face occurs with probability $\frac{1}{6} - x$ and its opposite face occurs with probability $\frac{1}{6} + x$. All other faces occur with probability $\frac{1}{6}$. Note that opposite faces sum to 7 in any die. If $0 < x < \frac{1}{6}$, and the probability of obtaining total sum $= 7$, when such a die is rolled twice, is $\frac{13}{96}$, then the value of $x$ is:
$\frac{1}{16}$
$\frac{1}{8}$
$\frac{1}{9}$
$\frac{1}{12}$
Answer: (b)
Solution
Probability of obtaining total sum 7 = probability of getting opposite faces. Probability of getting opposite faces $$= 2 \left[ \left( \frac{1}{6} - x \right) \left( \frac{1}{6} + x \right) + \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \right]$$ $$\Rightarrow 2 \left[ \left( \frac{1}{6} - x \right) \left( \frac{1}{6} + x \right) + \frac{1}{6} \times \frac{1}{6} \right] = \frac{13}{96}$$ (given) $$x = \frac{1}{8}$$
Question 18
Maths · Conic Sections · Single correct
If $x^2 + 9y^2 - 4x + 3 = 0, x, y \in \mathbb{R}$, then $x$ and $y$ respectively lie in the intervals:
$\left[ -\frac{1}{3}, \frac{1}{3} \right]$ and $\left[ -\frac{1}{3}, \frac{1}{3} \right]$
$\left[ -\frac{1}{3}, \frac{1}{3} \right]$ and $[1, 3]$
$[1, 3]$ and $[1, 3]$
$[1, 3]$ and $\left[ -\frac{1}{3}, \frac{1}{3} \right]$
Answer: (d)
Solution
Given the equation $x^2 + 9y^2 - 4x + 3 = 0$. Rewriting, we have $(x^2 - 4x) + (9y^2) + 3 = 0$. Completing the square, $(x^2 - 4x + 4) + (9y^2) + 3 - 4 = 0$. This simplifies to $(x - 2)^2 + (3y)^2 = 1$. Thus, we have $\frac{(x-2)^2}{(1)^2} + \frac{y^2}{\left(\frac{1}{3}\right)^2} = 1$ (equation of an ellipse). As it is the equation of an ellipse, $x$ and $y$ can vary inside the ellipse. So, $x - 2 \in [-1, 1]$ and $y \in \left[-\frac{1}{3}, \frac{1}{3}\right]$. Therefore, $x \in [1, 3]$ and $y \in \left[-\frac{1}{3}, \frac{1}{3}\right]$.
Question 19
Maths · Integrals · Single correct
$$\int_{6}^{16} \frac{\log_e \, x^2}{\log_e \, x^2 + \log_e (x^2 - 44x + 484)} \, dx$$ is equal to:
Maths · Applications of Derivatives · Single correct
A wire of length 20 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is:
$\frac{5}{2+\sqrt{3}}$
$\frac{10}{2+3\sqrt{3}}$
$\frac{5}{3+\sqrt{3}}$
$\frac{10}{3+2\sqrt{3}}$
Answer: (d)
Solution
Let the wire be cut into two pieces of length $x$ and $20 - x$. Area of square $= \left( \frac{x}{4} \right)^2$ Area of regular hexagon $= 6 \times \frac{\sqrt{3}}{4} \left( \frac{20-x}{6} \right)^2$ Total area $= A(x) = \frac{x^2}{16} + \frac{3\sqrt{3}}{2} \frac{(20-x)^2}{36}$ $A'(x) = \frac{2x}{16} + \frac{3\sqrt{3} \times 2}{2 \times 36} (20-x)(-1)$ $A'(x) = 0$ at $x = \frac{40\sqrt{3}}{3+2\sqrt{3}}$ Length of side of regular Hexagon $= \frac{1}{6} (20 - x)$ $= \frac{1}{6} \left( 20 - \frac{4 \cdot \sqrt{3}}{3+2\sqrt{3}} \right)$ $= \frac{10}{2+2\sqrt{3}}$
Question 21
Maths · Vector Algebra · Numerical
Let \[ \vec{a}=\hat{i}+5\hat{j}+\alpha\hat{k}, \] \[ \vec{b}=\hat{i}+3\hat{j}+\beta\hat{k}, \] and \[ \vec{c}=-\hat{i}+2\hat{j}-3\hat{k} \] be three vectors such that \[ |\vec{b}\times\vec{c}|=5\sqrt{3} \] and $\vec{a}$ is perpendicular to $\vec{b}$. Then the greatest among the values of \[ |\vec{a}|^2 \] is \[ \underline{\hspace{2cm}}. \]
Maths · Complex Numbers and Quadratic Equations · Fill in the blank
The number of distinct real roots of the equation $3x^4 + 4x^3 - 12x^2 + 4 = 0$ is .
Answer: 4
Solution
$3x^4+4x^3-12x^2+4=0$ So, let $f(x)=3x^4+4x^3-12x^2+4$ $\therefore\ f'(x)=12x(x^2+x-2)$ $=12x(x+2)(x-1)$
Question 23
Maths · Conic Sections · Fill in the blank
Let the equation $x^2 + y^2 + px + (1-p)y + 5 = 0$ represent circles of varying radius $r \in (0,5]$. Then the number of elements in the set $S=\{q:q=p^2 \text{ and } q \text{ is an integer}\}$ is:
Answer: 61
Solution
Given $$r = \sqrt{\frac{p^2}{4} + \frac{(1-p)^2}{4} - 5} = \frac{\sqrt{2p^2 - 2p - 19}}{2}$$ Since, $$r \in (0, 5]$$ So, $$0 < 2p^2 - 2p - 19 \leq 100$$ $$\Rightarrow p \in \left[ \frac{1 - \sqrt{239}}{2}, \frac{1 - \sqrt{39}}{2} \right) \cup \left( \frac{1 + \sqrt{39}}{2}, \frac{1 + \sqrt{239}}{2} \right]$$ so, number of integral values of $$p^2$$ is 61
Question 24
Maths · Sets · Numerical
If $A = \{ x \in \mathbb{R} : |x - 2| > 1 \}$, $B = \{ x \in \mathbb{R} : \sqrt{x^2 - 3} > 1 \}$, $C = \{ x \in \mathbb{R} : |x - 4| \geq 2 \}$ and $Z$ is the set of all integers, then the number of subsets of the set $(A \cap B \cap C)^C \cap Z$ is .
Answer: 256
Solution
Given $$A = (-\infty, 1) \cup (3, \infty)$$ $$B = (-\infty, -2) \cup (2, \infty)$$ $$C = (-\infty, 2] \cup [6, \infty)$$ So, $$A \cap B \cap C = (-\infty, -2) \cup [6, \infty)$$ $$\mathbb{Z} \cap (A \cap B \cap C)' = \{-2, -1, 0, 1, 2, 3, 4, 5\}$$ Hence no. of its subsets = $2^8 = 256$.
Question 25
Maths · Integrals · Numerical
If $\int\frac{\mathrm{d}x}{(x^2 + x + 1)^2}$ = a $\tan^{-1}(\frac{2x + 1}{\sqrt{3}})$ + b $(\frac{2x + 1}{x^2 + x + 1})$ + C x > 0 where C is the constant of integration, then the value of 9($\sqrt{3}$a + b) is equal to .
If the system of linear equations $$2x + y - z = 3$$ $$x - y - z = \alpha$$ $$3x + 3y + \beta z = 3$$ has infinitely many solution, then $\alpha + \beta - \alpha \beta$ is equal to .
Let n be an odd natural number such that the variance of 1, 2, 3, 4, $\ldots$, n is 14. Then n is equal to .
Answer: 13
Solution
Given $\($ $\frac{n^2 - 1}{12}$ = 14 $\)$. Solving for $\($ n $\)$, we find $\($ n = 13 $\)$.
Question 28
Maths · Conic Sections · Numerical
If the minimum area of the triangle formed by a tangent to the ellipse $\frac{x^2}{b^2} + \frac{y^2}{4a^2} = 1$ and the co-ordinate axis is $kab$, then $k$ is equal to .
A number is called a palindrome if it reads the same backward as well as forward. For example 285582 is a six digit palindrome. The number of six digit palindromes, which are divisible by 55, is .
Answer: 100
Solution
It is always divisible by 5 and 11. So, required number $= 10 \times 10 = 100$.
Question 30
Maths · Differential Equations · Numerical
If $y^{1/4} + y^{-1/4} = 2x$, and $(x^2 - 1) \frac{d^2 y}{dx^2} + \alpha x \frac{dy}{dx} + \beta y = 0$ then $|\alpha - \beta|$ is equal to.
Physics · Electric Charges and Fields · Single correct
A uniformly charged disc of radius $R$ having surface charge density $\sigma$ is placed in the $xy$ plane with its center at the origin. Find the electric field intensity along the $z$-axis at a distance $Z$ from origin :-
Consider a small ring of radius $r$ and thickness $dr$ on disc. Area of elemental ring on disc $$dA = 2\pi r \, dr$$ Charge on this ring $dq = \sigma dA$ $$dE_z = \frac{k dq z}{(z^2 + r^2)^{3/2}}$$ $$E = \int_0^R dE_z = \frac{\sigma}{2\varepsilon_0} \left[ 1 - \frac{z}{\sqrt{R^2 + z^2}} \right]$$
Question 32
Physics · Nuclei · Single correct
There are $10^{10}$ radioactive nuclei in a given radioactive element, Its half-life time is 1 minute. How many nuclei will remain after 30 seconds ?( $\sqrt{2} = 1.414$)
Physics · Physical World, Units and Measurements · Single correct
Which of the following is not a dimensionless quantity?
Relative magnetic permeability ($\mu_r$)
Power factor
Permeability of free space ($\mu_0$)
Quality factor
Answer: (c)
Solution
Given $[\mu_r] = 1$ as $\mu_r = \frac{\mu}{\mu_m}$. $[power factor (\cos \phi)] = 1$. $\mu_0 = \frac{B_0}{H} \left( unit = \mathrm{NA^{-2}} \right)$: Not dimensionless. $[\mu_0] = [\mathrm{MLT^{-2} A^{-2}}]$. Quality factor $(Q) = \frac{Energy stored}{Energy dissipated per cycle}$. So $Q$ is unitless $\&$ dimensionless.
Question 34
Physics · Physical World, Units and Measurements · Single correct
If E and H represents the intensity of electric field and magnetising field respectively, then the unit of E/H will be :
ohm
mho
joule
newton
Answer: (a)
Solution
Unit of $\frac{\mathbf{E}}{\mathbf{H}}$ is $$\frac{volt / metre}{Ampere / metre} = \frac{volt}{Ampere} = ohm$$
Question 35
Physics · Mathematics in Physics · Single correct
The resultant of these forces $\overrightarrow{OP}$, $\overrightarrow{OQ}$, $\overrightarrow{OR}$, $\overrightarrow{OS}$ and $\overrightarrow{OT}$ is approximately ..... N. [Take $\sqrt{3} = 1.7$, $\sqrt{2} = 1.4$ Given $\hat{i}$ and $\hat{j}$ unit vectors along $x$, $y$ axis ]
$9.25\hat{i} + 5\hat{j}$
$3\hat{i} + 15\hat{j}$
$2.5\hat{i} - 14.5\hat{j}$
$-1.5\hat{i} - 15.5\hat{j}$
Answer: (a)
Solution
The vector $\vec{F}_x$ is calculated as follows: $$\vec{F}_x = \left( 10 \times \frac{\sqrt{3}}{2} + 20 \left( \frac{1}{2} \right) + 20 \left( \frac{1}{\sqrt{2}} \right) - 15 \left( \frac{1}{\sqrt{2}} \right) - 15 \left( \frac{\sqrt{3}}{2} \right) \right) \hat{i}$$ which simplifies to $$= 9.25 \hat{i}$$ The vector $\vec{F}_y$ is calculated as follows: $$\vec{F}_y = \left( 15 \left( \frac{1}{2} \right) + 20 \left( \frac{\sqrt{3}}{2} \right) + 10 \left( \frac{1}{2} \right) - 15 \left( \frac{1}{\sqrt{2}} \right) - 20 \left( \frac{1}{\sqrt{2}} \right) \right) \hat{j}$$ which simplifies to $$= 5 \hat{j}$$
Question 36
Physics · Kinetic Theory · Single correct
A balloon carries a total load of 185 kg at normal pressure and temperature of $27^\circ \mathrm{C}$. What load will the balloon carry on rising to a height at which the barometric pressure is $45 \mathrm{cm}$ of Hg and the temperature is $-7^\circ \mathrm{C}$. Assuming the volume constant?
Physics · Ray Optics and Optical Instruments · Single correct
An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of the object is $d_1$ from C and the distance of the image formed is $d_2$ from C, the radius of curvature of this mirror is:
Physics · System of Particles and Rotational Motion · Single correct
A huge circular arc of length 4.4 ly subtends an angle '4s' at the centre of the circle. How long it would take for a body to complete 4 revolution if its speed is 8 AU per second ?
$4.1 \times 10^{8}\,\mathrm{s}$
$4.5 \times 10^{10}\,\mathrm{s}$
$3.5 \times 10^{6}\,\mathrm{s}$
$7.2 \times 10^{8}\,\mathrm{s}$
Answer: (b)
Solution
Given $R = \frac{\ell}{\theta}$. Time $= \frac{4 \times 2 \pi R}{v} = \frac{4 \times 2 \pi}{v} \left( \frac{\ell}{\theta} \right)$. Put $\ell = 4.4 \times 9.46 \times 10^{15}$, $v = 8 \times 1.5 \times 10^{11}$, $\theta = \frac{4}{3600} \times \frac{\pi}{180} \, \mathrm{rad}$. We get time $= 4.5 \times 10^{10} \, \mathrm{sec}$.
Question 39
Physics · Electrostatic Potential and Capacitance · Single correct
Calculate the amount of charge on capacitor of $4\mu\mathrm{F}$. The internal resistance of battery is $1\Omega$ :
$8\mu\mathrm{C}$
zero
$16\mu\mathrm{C}$
$4\mu\mathrm{C}$
Answer: (a)
Solution
On simplifying the circuit we get no current in the upper wire. Therefore, $V_{AB} = \frac{5}{4+1} \times 4 = 4 \, \mathrm{V}$. Therefore, $\theta = (C_{eq}) \, v$ $$\Rightarrow 2 \times 4 = 8 \, \mu \mathrm{C}$$
Question 40
Physics · System of Particles and Rotational Motion · Single correct
Moment of inertia of a square plate of side $l$ about the axis passing through one of the corner and perpendicular to the plane of square plate is given by:
$\frac{Ml^2}{6}$
$Ml^2$
$\frac{Ml^2}{12}$
$\frac{2}{3}Ml^2$
Answer: (d)
Solution
According to perpendicular Axis theorem. $$I_x + I_y = I_z$$ $$I_z = \frac{m \ell^2}{3} + \frac{m \ell^2}{3}$$ $$= \frac{2 m \ell^2}{3}$$
Question 41
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
For a transistor in CE mode to be used as an amplifier, it must be operated in:
Both cut-off and Saturation
Saturation region only
Cut-off region only
The active region only
Answer: (d)
Solution
Active region of the CE transistor is linear region and is best suited for its use as an amplifier.
Question 42
Physics · Thermal Properties of Matter · Single correct
An ideal gas is expanding such that $PT^3 = constant$. The coefficient of volume expansion of the gas is :
$\frac{1}{T}$
$\frac{2}{T}$
$\frac{4}{T}$
$\frac{3}{T}$
Answer: (c)
Solution
Given $P I^3 = constant$ and $\left( \frac{n R T}{V} \right) T^3 = constant$. Also, $T^4 V^{-1} = constant$ and $T^4 = kV$. This implies $$4 \frac{\Delta T}{T} = \frac{\Delta V}{V} \ldots (1)$$ $$\Delta V = V \gamma \Delta T \ldots (2)$$ Comparing (1) and (2) we get $$\gamma = \frac{4}{T}$$
Question 43
Physics · Dual Nature of Radiation and Matter · Single correct
In a photoelectric experiment, increasing the intensity of incident light:
increases the number of photons incident and also increases the K.E. of the ejected electrons
increases the frequency of photons incident and increases the K.E. of the ejected electrons.
increases the frequency of photons incident and the K.E. of the ejected electrons remains unchanged
increases the number of photons incident and the K.E. of the ejected electrons remains unchanged
Answer: (d)
Solution
Increasing intensity means number of incident photons are increased. Kinetic energy of ejected electrons depend on the frequency of incident photons, not the intensity.
Question 44
Physics · Electromagnetic Induction · Single correct
A bar magnet is passing through a conducting loop of radius $R$ with velocity $v$. The radius of the bar magnet is such that it just passes through the loop. The induced e.m.f. in the loop can be represented by the approximate curve:
Answer: (c)
Solution
When the magnet passes through the centre region of the solenoid, no current or emf is induced in the loop. While entering, the flux increases, so there is a negative induced emf. While leaving, the flux decreases, so there is a positive induced emf.
Question 45
Physics · Moving Charges and Magnetism · Single correct
Two ions of masses 4 amu and 16 amu have charges +2e and +3e respectively. These ions pass through the region of constant perpendicular magnetic field. The kinetic energy of both ions is same. Then :
lighter ion will be deflected less than heavier ion
lighter ion will be deflected more than heavier ion
both ions will be deflected equally
no ion will be deflected.
Answer: (b)
Solution
Given they have same kinetic energy $$r \propto \frac{\sqrt{m}}{q}$$ $$\frac{r_1}{r_2} = \frac{\sqrt{4}}{2} \times \frac{3}{\sqrt{16}} = \frac{3}{4}$$ $$r_2 = \frac{4r_1}{3} ( r_2 is for heavier ion and r_1 is for lighter ion)$$ $$\sin \theta = \frac{d}{R}$$ $\($ $\theta$ $\rightarrow$ Deflection $\)$ $$\theta \propto \frac{1}{R}$$ (R $\rightarrow$ Radius of path) Therefore, $\($ R_2 > R_1 $\Rightarrow$ $\theta$_2 < $\theta$_1 $\)$
Question 46
Physics · Ray Optics and Optical Instruments · Single correct
Find the distance of the image from object O, formed by the combination of lenses in the figure:
Physics · Mechanical Properties of Fluids · Single correct
In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius $2.0 \times 10^{-5} \, \mathrm{m}$ and density $1.2 \times 10^{3} \, \mathrm{kgm}^{-3}$? Take viscosity of liquid $= 1.8 \times 10^{-5} \, \mathrm{Nsm}^{-2}$. (Neglect buoyancy due to air).
Electric field in a plane electromagnetic wave is given by $E = 50 \sin(500x - 10 \times 10^{10}t) \, \mathrm{V/m}$ The velocity of electromagnetic wave in this medium is : (Given $C =$ speed of light in vacuum)
$\frac{3}{2}C$
$C$
$\frac{2}{3}C$
$\frac{C}{2}$
Answer: (c)
Solution
Given $\($ V = $\frac{\omega}{K}$ = $\frac{10 \times 10^{10}}{500}$ = 2 $\times$ 10^8 $\)$. $\($ V = $\frac{2C}{3}$ $\)$.
Question 49
Physics · Current Electricity · Single correct
Five identical cells each of internal resistance $1\,\Omega$ and emf $5\,\mathrm{V}$ are connected in series and in parallel with an external resistance ' $R$ '. For what value of ' $R$ ', current in series and parallel combination will remain the same?
$1\,\Omega$
$25\,\Omega$
$5\,\Omega$
$10\,\Omega$
Answer: (a)
Solution
Given $$i_1 = \frac{25}{5 + R}$$ $$i_2 = \frac{5}{R + \frac{1}{5}}$$ Equating $i_1$ and $i_2$ gives: $$5 \left( R + \frac{1}{5} \right) = 5 + R$$ Simplifying, we have: $$4R = 4$$ Therefore, $$R = 1 \, \Omega$$
Question 50
Physics · Oscillations · Single correct
The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure. The potential energy $U(x)$ versus time $(t)$ plot of the particle is correctly shown in figure:
Answer: (d)
Solution
Potential energy is maximum at maximum distance from mean.
Question 51
Physics · Gravitation · Numerical
A body of mass (2M) splits into four masses $\{$m, M - m, m, M - m$\}$, which are rearranged to form a square as shown in the figure. The ratio of $\frac{M}{m}$ for which, the gravitational potential energy of the system becomes maximum is $x : 1$. The value of $x$ is $\ldots$$\ldots$
Answer: 2
Solution
Energy is maximum when mass is split equally so $$\frac{M}{m} = 2$$
Question 52
Physics · Alternating Current · Numerical
The alternating current is given by $$i = \left\{ \sqrt{42} \sin \left( \frac{2\pi}{T} t \right) + 10 \right\} \, \mathrm{A}$$ The r.m.s. value of this current is ..... A.
Physics · Moving Charges and Magnetism · Numerical
A uniform conducting wire of length $24a$, and resistance $R$ is wound up as a current carrying coil in the shape of an equilateral triangle of side '$a$' and then in the form of a square of side '$a$'. The coil is connected to a voltage source $V_0$. The ratio of magnetic moment of the coils in case of equilateral triangle to that for square is $1 : \sqrt{y}$ where $y$ is
Answer: 3
Solution
In triangle shape $N_t = \frac{24a}{3a} = 8$ In square $N_s = \frac{24a}{4a} = 6$ $$\frac{M_t}{M_3} = \frac{N_t I A_t}{N_s I A_s} [I will be same in both]$$ $$= \frac{8 \times \frac{\sqrt{3}}{4} \times a^2}{6 \times a^2}$$ $$\frac{M_t}{M_s} = \frac{1}{\sqrt{3}}$$ $y = 3$
A circuit is arranged as shown in figure. The output voltage $V_0$ is equal to $\ldots$$\ldots$$\ldots$ V
Answer: 5
Solution
As diodes $D_1$ and $D_2$ are in forward bias, so they acted as negligible resistances $\Rightarrow$ Input voltage become zero $\Rightarrow$ Input current is zero $\Rightarrow$ Output current is zero $\Rightarrow V_0 = 5$ volt
Question 55
Physics · Current Electricity · Numerical
First, a set of $n$ equal resistors of $10 \, \Omega$ each are connected in series to a battery of emf $20 \, \mathrm{V}$ and internal resistance $10 \, \Omega$. A current $I$ is observed to flow. Then, the $n$ resistors are connected in parallel to the same battery. It is observed that the current is increased 20 times, then the value of $n$ is .
Answer: 20
Solution
In series $$R_{eq} = nR = 10n$$ $$i_s = \frac{20}{10 + 10n} = \frac{2}{1+n}$$ In parallel $$R_{eq} = \frac{10}{n}$$ $$i_p = \frac{20}{\frac{10}{n} + 10} = \frac{2n}{1+n}$$ $$\frac{i_p}{i_s} = 20$$ $$\left(\frac{2n}{1+n}\right) = 20$$ $$\left(\frac{2}{1+n}\right) = 20$$ $$n = 20$$
Question 56
Physics · Waves · Numerical
Two cars $X$ and $Y$ are approaching each other with velocities $36 \, \mathrm{km/h}$ and $72 \, \mathrm{km/h}$ respectively. The frequency of a whistle sound as emitted by a passenger in car $X$, heard by the passenger in car $Y$ is $1320 \, \mathrm{Hz}$. If the velocity of sound in air is $340 \, \mathrm{m/s}$, the actual frequency of the whistle sound produced is $\ldots$ $\ldots$ $\ldots$ $\ldots$ $\ldots$ $\ldots$ $\mathrm{Hz}$
Physics · Motion in a Straight Line · Fill in the blank
If the velocity of a body related to displacement $x$ is given by $v=\sqrt{5000+24x}\,\mathrm{m\,s^{-1}}$, then the acceleration of the body is $\ldots\ldots\ldots\,\mathrm{m\,s^{-2}}$.
Physics · Thermal Properties of Matter · Numerical
A rod CD of thermal resistance $10.0 \, \mathrm{KW}^{-1}$ is joined at the middle of an identical rod AB as shown in figure, The end A, B and D are maintained at $200^\circ \mathrm{C}$, $100^\circ \mathrm{C}$ and $125^\circ \mathrm{C}$ respectively. The heat current in CD is P watt. The value of P is . . . . . .
Answer: 2
Solution
Rods are identical so $$R_{AB} = R_{CD} = 10 \, \mathrm{kW^{-1}}$$ C is mid-point of AB, so $$R_{AC} = R_{CB} = 5 \, \mathrm{kW^{-1}}$$ at point C $$\frac{200 - T}{5} = \frac{T - 125}{10} + \frac{T - 100}{5}$$ $$2(200 - T) = T - 125 + 2(T - 100)$$ $$400 - 2T = T - 125 + 2T - 200$$ $$T = \frac{725}{5} = 145^\circ \mathrm{C}$$ $$I_h = \frac{145 - 125}{10} \, \mathrm{w} = \frac{20}{10} \, \mathrm{w}$$ $$I_h = 2 \, \mathrm{w}$$
Question 59
Physics · Work, Energy and Power · Numerical
Two persons A and B perform same amount of work in moving a body through a certain distance $d$ with application of forces acting at angle $45^\circ$ and $60^\circ$ with the direction of displacement respectively. The ratio of force applied by person A to the force applied by person B is $\frac{1}{\sqrt{x}}$. The value of $x$ is
Answer: 2
Solution
Given $W_A = W_B$. $F_A d \cos 45^\circ = F_B d \cos 60^\circ$. $F_A \times \frac{1}{\sqrt{2}} = F_B \times \frac{1}{2}$. $$\frac{F_A}{F_B} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$$ $x = 2$
Question 60
Physics · Communication Systems · Numerical
A transmitting antenna has a height of 320 m and that of receiving antenna is 2000 m. The maximum distance between them for satisfactory communication in line of sight mode is 'd'. The value of 'd' is ........ km.
The reaction starts with $\mathrm{CH_3-C\equiv CH}$ reacting with $\mathrm{NaNH_2}$ to form $\mathrm{CH_3-C\equiv C^- Na^+ + NH_3}$ (A). This intermediate reacts with $\mathrm{BrCH_2CH(OH)CH_3}$ to form $\mathrm{CH_3-C\equiv C-CH_2-CH(OH)-CH_3}$ (B). Hydrogenation with $\mathrm{H_2/Pd-C}$ converts it to $\mathrm{CH_3-CH_2-CH_2-CH_2-CH(OH)-CH_3}$ (C). Finally, oxidation with $\mathrm{CrO_3}$ yields $\mathrm{CH_3-CH_2-CH_2-CH_2-CH_2-C(=O)-CH_3}$ (D).
Question 62
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The structure of the starting compound $P$ used in the reaction given below is:
Answer: (a)
Solution
NaOCl is used in haloform reaction as reagent.
Question 63
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Match List-I with List-II \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{(Species)} & \multicolumn{2}{c|}{(Number of lone pairs of electrons on the central atom)} \\ \hline (a) & XeF$_2$ & (i) & 0 \\ \hline (b) & XeO$_2$F$_2$ & (ii) & 1 \\ \hline (c) & XeO$_3$F$_2$ & (iii) & 2 \\ \hline (d) & XeF$_4$ & (iv) & 3 \\ \hline \end{tabular} Choose the most appropriate answer from the options given below :
(a)--(iv), $(b)$--(i), $(c)$--(ii), (d)--(iii)
(a)--(iii), (b)--(iv), $(c)$--(ii), (d)--(i)
(a)--(iii), (b)--(ii), $(c)$--(iv), (d)--(i)
(a)--(iv), (b)--(ii), $(c)$--(i), (d)--(iii)
Answer: (d)
Solution
Question 64
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
In which one of the following molecules strongest back donation of an electron pair from halide to boron is expected?
$\mathrm{BCl}_3$
$\mathrm{BF}_3$
$\mathrm{BBr}_3$
$\mathrm{BI}_3$
Answer: (b)
Solution
Type of back bonding $\mathrm{BF_3}$ $\hspace{1cm}$ $\mathrm{BCl}$ $\hspace{1cm}$ $\mathrm{BBr_3}$ $\hspace{1cm}$ $\mathrm{BI_3}$ $(2p\pi - 2p\pi)$ $\hspace{1cm}$ $(2p\pi - 3p\pi)$ $\hspace{1cm}$ $(2p\pi - 4p\pi)$ $\hspace{1cm}$ $(2p\pi - 5p\pi)$ Therefore back bonding strength is as follows $$\mathrm{BF_3 > BCl > BBr_3 > BI_3}$$
Question 65
Chemistry · Hydrogen · Single correct
Deuterium resembles hydrogen in properties but :
reacts slower than hydrogen
reacts vigorously than hydrogen
reacts just as hydrogen
emits $\beta^+$ particles
Answer: (a)
Solution
The bond dissociation energy of $\mathrm{D_2}$ is greater than $\mathrm{H_2}$ and therefore $\mathrm{D_2}$ reacts slower than $\mathrm{H_2}$.
Question 66
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Which refining process is generally used in the purification of low melting metals?
Chromatographic method
Liquation
Electrolysis
Zone refining
Answer: (b)
Solution
Liquation method is used to purify those impure metals which has lower melting point than the melting point of impurities associated. Therefore, this method is used for metal having low melting point.
Question 67
Chemistry · Analytical Chemistry · Single correct
Match items of List - I with those of List - II: Choose the most appropriate answer from the options given below:
(a)- (ii), (b) - (i), (c) - (iii), (d) - (iv)
(a)- (ii), (b) - (i), (c) - (iii), (d) - (iv)
(a)- (iii), (b) - (iv), (c) - (ii), (d) - (i)
(a)- (iv), (b) - (ii), (c) - (i), (d) - (iii)
Answer: (c)
Solution
(a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)
Question 68
Chemistry · Chemistry in Everyday Life · Single correct
The correct statement about (A), (B), ($C$) and (D) is:
(A) , (B) and ($C$) are narcotic analgesics
(B) , ($C$) and (D) are tranquillizers
and (D) are tranquillizers
and ($C$) are tranquillizers
Answer: (d)
Solution
B and C are tranquilizers
Question 69
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The major product of the following reaction is:
Answer: (c)
Solution
Question 70
Chemistry · Amines · Single correct
Which of the following is not a correct statement for primary aliphatic amines?
The intermolecular association in primary amines is less than the intermolecular association in secondary amines.
Primary amines on treating with nitrous acid solution form corresponding alcohols except methyl amine.
Primary amines are less basic than the secondary amines.
Primary amines can be prepared by the Gabriel phthalimide synthesis.
Answer: (a)
Solution
The intermolecular association is more prominent in case of primary amines as compared to secondary, due to the availability of two hydrogen atoms.
Question 71
Chemistry · Co-ordination Compounds · Single correct
Acidic ferric chloride solution on treatment with excess of potassium ferrocyanide gives a Prussian blue coloured colloidal species. It is:
Chemistry · Environmental Chemistry · Single correct
The gas 'A' is having very low reactivity reaches to stratosphere. It is non-toxic and non-flammable but dissociated by UV-radiations in stratosphere. The intermediates formed initially from the gas 'A' are :
In stratosphere CFCs get broken down by powerful UV radiations releasing Cl$^\bullet$ $\text{CF}_2\text{Cl}_2(\text{g}) \xrightarrow{\text{U.V.}} \text{Cl}^\bullet(\text{g}) + {}^\bullet\text{CF}_2\text{Cl}(\text{g})$
Question 73
Chemistry · The s-Block Elements · Single correct
The number of water molecules in gypsum, dead burnt plaster and plaster of paris, respectively are:
2,0 and 1
0.5,0 and 2
5,0 and 0.5
2,0 and 0.5
Answer: (d)
Solution
Gypsum - $\mathrm{CaSO_4} \cdot 2\mathrm{H_2O}$ Plaster of Paris - $\mathrm{CaSO_4} \cdot \frac{1}{2}\mathrm{H_2O}$ Dead burnt plaster - $\mathrm{CaSO_4}$
Question 74
Chemistry · The d-and f-Block Elements · Single correct
The nature of oxides $\mathrm{V_2O_3}$ and $\mathrm{CrO}$ is indexed as 'X' and 'Y' type respectively. The correct set of X and Y is:
X = basic Y = amphoteric
X = amphoteric Y = basic
X = acidic Y = acidic
X = basic Y = basic
Answer: (d)
Solution
$\mathrm{V_2O_3}$ basic $\mathrm{CrO}$ basic
Question 75
Chemistry · Biomolecules · Single correct
Out of following isomeric forms of uracil, which one is present in RNA ?
Answer: (d)
Solution
Isomeric form of uracil present in RNA.
Question 76
Chemistry · Alcohols, Phenols and Ethers · Single correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Synthesis of ethyl phenyl ether may be achieved by Williamson synthesis. Reason (R): Reaction of bromobenzene with sodium ethoxide yields ethyl phenyl ether. In the light of the above statements, choose the most appropriate answer from the options given below:
Both (A) and (R) are correct and (R) is the correct explanation of (A)
(A) is correct but (R) is not correct
(A) is not correct but (R) is correct
Both (A) and (R) are correct but (R) is NOT the correct explanation of (A)
Answer: (b)
Solution
Sodium phenoxide reacts with ethyl bromide to form ethyl phenyl ether. The reaction involves the nucleophilic substitution of the bromide ion by the phenoxide ion. The partial double bond character in the bromobenzene ring makes it less reactive towards nucleophilic substitution. Therefore, the reaction proceeds with the formation of ethyl phenyl ether.
Question 77
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
In the following sequence of reactions the P is :
Answer: (a)
Solution
Question 78
Chemistry · States of Matter · Single correct
The unit of the van der Waals gas equation parameter 'a' in $\left( P + \frac{an^2}{V^2} \right) (V - nb) = nRT$ is :
kg m s$^{-2}$
dm$^3$ mol$^{-1}$
kg m s$^{-1}$
atm dm$^6$ mol$^{-2}$
Answer: (d)
Solution
Given $\($ $\frac{an^2}{V^2}$ = atm $\Rightarrow$ a = atm $\times$ $\frac{dm^6}{mol^2}$ $\)$
Question 79
Chemistry · Redox Reactions · Single correct
In polythionic acid, $\mathrm{H}_2 \mathrm{S}_x \mathrm{O}_6 (x = 3 to 5)$ the oxidation state(s) of sulphur is/are:
+5 only
+6 only
+3 and +5 only
0 and +5 only
Answer: (d)
Solution
The structure shown is a polysulfate with the general formula $\mathrm{HO{-}S^{+5}{=}O{-}(S)_n{-}S^{+5}{=}O{-}OH}$, where $n = 1$ to $3$. Each sulfur atom is in the $+5$ oxidation state.
In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine in the compound is ____. (Nearest integer) [Atomic mass : Ag = 108, Br = 80]
Answer: 40
Solution
The number of moles of AgBr is calculated as follows: $$n_{\mathrm{AgBr}} = \frac{0.188 \, \mathrm{g}}{188 \, \mathrm{g/mol}} = 10^{-3} \, \mathrm{mol}$$ Therefore, the number of moles of Br is equal to the number of moles of AgBr: $$n_{\mathrm{Br}} = n_{\mathrm{AgBr}} = 0.001 \, \mathrm{mol}$$ The mass of Br is calculated as: $$mass_{\mathrm{Br}} = (0.001 \times 80) \, \mathrm{gm} = 0.08 \, \mathrm{gm}$$ The mass percentage is calculated as: $$mass \% = \frac{0.08 \times 100}{0.2} = 40\%$$
Question 82
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
The reaction that occurs in a breath analyser, a device used to determine the alcohol level in a person's bloodstream, is $\mathrm{2K_2Cr_2O_7 + 8H_2SO_4 + 3C_2H_6O \rightarrow 2Cr_2(SO_4)_3 + 3C_2H_4O_2 + 2K_2SO_4 + 11H_2O}$ If the rate of appearance of $\mathrm{Cr_2(SO_4)_3}$ is $2.67\,\mathrm{mol\,min^{-1}}$ at a particular time, the rate of disappearance of $\mathrm{C_2H_6O}$ at the same time is _______ $\mathrm{mol\,min^{-1}}$. (Round off to the nearest integer.)
Answer: 4
Solution
Given (Q5, Option 4), $(\text{Rate of disappearance of }\mathrm{C_2H_6O})/3 = (\text{Rate of appearance of }\mathrm{Cr_2(SO_4)_3})/2$ $\Rightarrow$ Rate of disappearance of $\mathrm{C_2H_6O}$ $= (2.67 \times 3)/2\ \mathrm{mol\,min^{-1}}$ $= 4.005\ \mathrm{mol\,min^{-1}}$
Question 83
Chemistry · Structure of Atom · Numerical
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $\frac{h^2}{xma_0^2}$. The value of 10x is ____. ($a_0$ is radius of Bohr's orbit) (Nearest integer) [Given : $\pi = 3.14$]
Answer: 3155
Solution
Given $$mvr = \frac{nh}{2\pi}$$ K.E. = $$\frac{n^2 h^2}{8\pi^2 m r^2} = \frac{4h^2}{8\pi^2 m (4a_0)^2}$$ = $$\left(\frac{4}{8\pi^2 \times 16}\right) \frac{h^2}{ma_0^2}$$ $$\Rightarrow x = 315.507$$ $$\Rightarrow 10x = 3155 (nearest integer)$$
Question 84
Chemistry · Solutions · Numerical
1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to $-4^\circ C$ before freezing. The amount of ice (in g) that will be separated out is . (Nearest integer) [Given : $K_f (H_2O) = 1.86 K kg mol^{-1}$]
Answer: 518
Solution
Let mass of water initially present = $x$ gm. Mass of sucrose = $(1000 - x)$ gm. Moles of sucrose = $\left( \frac{1000-x}{342} \right)$. $$0.75 = \frac{\left( \frac{1000-x}{342} \right)}{\left( \frac{x}{1000} \right)} \Rightarrow \frac{x}{1000} = \frac{1000-x}{342 \times 0.75}$$ $$256.5x = 10^6 - 1000x$$ $$x = 795.86 gm$$ Moles of sucrose = $0.5969$. New mass of $\mathrm{H_2O} = a$ kg. $$4 = \frac{0.5969}{a} \times 1.86 \Rightarrow a = 0.2775 kg$$ Ice separated = $(795.86 - 277.5) = 518.3$ gm.
Question 85
Chemistry · Co-ordination Compounds · Numerical
1 mol of an octahedral metal complex with formula $\mathrm{MCl}_3 \cdot 2 \mathrm{L}$ on reaction with excess of $\mathrm{AgNO}_3$ gives 1 mol of $\mathrm{AgCl}$. The denticity of Ligand L is ____. (Integer answer)
Answer: 2
Solution
Given $\mathrm{MCl_3.2\ L}$ octahedral. It means that one $\mathrm{Cl^-}$ ion is present in the ionization sphere. Therefore, formula $= [\mathrm{MCl_2\ L_2}]\mathrm{Cl}$. For octahedral complex, coordination number is 6. Therefore, $\mathrm{L}$ acts as a bidentate ligand.
The number of moles of CuO, that will be utilized in Dumas method for estimation nitrogen in a sample of 57.5 g of N, N-dimethylaminopentane is ____ $\times$ $10^{-2}$. (Nearest integer)
Answer: 1125
Solution
Moles of N in N, N - dimethylaminopentane $$= \left( \frac{57.5}{115} \right) = 0.5 \, mol$$ $$\Rightarrow \mathrm{C_7H_{17}N} + \frac{45}{2} \mathrm{CuO} \rightarrow 7\mathrm{CO_2} + \frac{17}{2} \mathrm{H_2O} + \frac{1}{2} \mathrm{N_2} + \frac{45}{2} \mathrm{Cu}$$ $$\frac{n_{\mathrm{CuO}} reacted}{\left( \frac{45}{2} \right)} = \frac{n_{\mathrm{C_7H_{17}N}} reacted}{1}$$ $$\Rightarrow n_{\mathrm{CuO}} reacted = \left( \frac{45}{2} \right) \times 0.5 = 11.25$$
Question 87
Chemistry · The d-and f-Block Elements · Numerical
The number of $f$ electrons in the ground state electronic configuration of Np $(Z = 93)$ is ____. (Nearest integer)
Answer: 4
Solution
$\mathrm{Np}=1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,5s^2\,4d^{10}\,5p^6\,6s^2$ Total number of $f$-electrons $=14+4=18$.
Question 88
Chemistry · Thermodynamics · Numerical
$200 \, \mathrm{mL}$ of $0.2 \, \mathrm{M}$ HCl is mixed with $300 \, \mathrm{mL}$ of $0.1 \, \mathrm{M}$ NaOH. The molar heat of neutralization of this reaction is $-57.1 \, \mathrm{kJ}$. The increase in temperature in $^\circ \mathrm{C}$ of the system on mixing is $x \times 10^{-2}$. The value of $x$ is ______. (Nearest integer) [Given: Specific heat of water $= 4.18 \, \mathrm{J \, g^{-1} \, K^{-1}}$] Density of water $= 1.00 \, \mathrm{g \, cm^{-3}}$] (Assume no volume change on mixing)
The number of moles of $\mathrm{NH_3}$ that must be added to $2\,\mathrm{L}$ of $0.80\,\mathrm{M}$ $\mathrm{AgNO_3}$ in order to reduce the concentration of $\mathrm{Ag^+}$ ions to $5.0 \times 10^{-8}\,\mathrm{M}$ is ______. Given that $K_f$ for $\left[\mathrm{Ag(NH_3)_2}\right]^+$ is $1.0 \times 10^8$. (Round off to the nearest integer.) Assume no volume change on adding $\mathrm{NH_3}$.
When 10 $\mathrm{mL}$ of an aqueous solution of $\mathrm{KMnO_4}$ was titrated in acidic medium, equal volume of 0.1 $\mathrm{M}$ of an aqueous solution of ferrous sulphate was required for complete discharge of colour. The strength of $\mathrm{KMnO_4}$ in grams per litre is ____ $\times 10^{-2}$. (Nearest integer)
Answer: 316
Solution
Let molarity of $\mathrm{KMnO_4} = x$ $$\mathrm{KMnO_4} + \mathrm{FeSO_4} \rightarrow \mathrm{Fe_2(SO_4)_3} + \mathrm{Mn^{2+}}$$ $n = 5 n = 1$ (Equivalents of $\mathrm{KMnO_4}$ reacted) = (Equivalents of $\mathrm{FeSO_4}$ reacted) $$\Rightarrow (5 \times x \times 10 \, \mathrm{ml}) = 1 \times 0.1 \times 10 \, \mathrm{ml}$$ $$\Rightarrow x = 0.02 \, \mathrm{M}$$ Molar mass of $\mathrm{KMnO_4} = 158 \, \mathrm{gm/mol}$ $$\Rightarrow Strength = (x \times 158) = 3.16 \, \mathrm{g/\ell}$$