JEE Main 27 June 2022 Shift 1 question paper with solutions
JEE Main 27 June 2022 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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Maths
Question 1
Maths · Complex Numbers and Quadratic Equations · Single correct
The area of the polygon, whose vertices are the non-real roots of the equation $\overline{z} = i z^2$ is:
$\frac{3\sqrt{3}}{4}$
$\frac{3\sqrt{3}}{2}$
$\frac{3}{2}$
$\frac{3}{4}$
Answer: (a)
Solution
Let $z = x + iy$, $x, y \in \mathbb{R}$. Now $\overline{z} = iz^2$. Then $x - iy = i(x^2 - y^2 + 2xyi)$. $x - iy = i(x^2 - y^2) - 2xy$. Therefore, $x = -2xy$ and $-y = x^2 - y^2$. This implies $x(1 + 2y) = 0$. So, $x = 0$ or $y = -\frac{1}{2}$. Put $x = 0$ in $-y = x^2 - y^2$. We get $y = y^2$. Thus, $y = 0, 1$. Similarly, put $y = -\frac{1}{2}$ in $-y = x^2 - y^2$. Then $\frac{1}{2} = x^2 - \frac{1}{4}$. This implies $x^2 = \frac{3}{4}$. Therefore, $x = \pm \frac{\sqrt{3}}{2}$. So, $z = \left(0, i, \frac{\sqrt{3}}{2} - \frac{1}{2}i, -\frac{\sqrt{3}}{2} - \frac{1}{2}i\right)$. The area is $\frac{1}{2} \cdot (\sqrt{3}) \left(\frac{3}{2}\right)$.
Question 2
Maths · Determinants · Single correct
Let the system of linear equations $x + 2y + z = 2$, $\alpha x + 3y - z = \alpha$, $-\alpha x + y + 2z = -\alpha$ be inconsistent. Then $\alpha$ is equal to:
$\frac{5}{2}$
$-\frac{5}{2}$
$\frac{7}{2}$
$-\frac{7}{2}$
Answer: (d)
Solution
Given $$\Delta = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 3 & -1 \\ -2 & 1 & 2 \end{vmatrix}$$ This equals $$(6 + y) - 2 \left( (2\alpha - \alpha) + 1(\alpha + 3\alpha) \right)$$ Simplifying, we get $$= 7 - 2\alpha + 4\alpha$$ Which simplifies to $$= 7 + 2\alpha$$ Setting $\Delta = 0$, we have $$\alpha = -\frac{7}{2}$$ Now consider $$\Delta_1 = \begin{vmatrix} 2 & 2 & 1 \\ \alpha & 3 & -1 \\ -\alpha & 1 & 2 \end{vmatrix}$$ This equals $$14 + 2\alpha$$ Given $$\alpha = -x_2 = 7$$ We find $$\Delta_1 \neq 0$$
Question 3
Maths · Sequences and Series · Single correct
If $x = \sum_{n=0}^{\infty} a^n$, $y = \sum_{n=0}^{\infty} b^n$, $z = \sum_{n=0}^{\infty} c^n$, where $a$, $b$, $c$ are in A.P. and $|a| < 1$, $|b| < 1$, $|c| < 1$, $abc \neq 0$, then
$x, y, z$ are in A.P.
$x, y, z$ are in G.P.
$\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in A.P.
$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 - (a + b + c)$
Answer: (c)
Solution
Given $x = 1 + a + a^2 = \ldots$ $$x = \frac{1}{1-a} \implies a = 1 - \frac{1}{x}$$ $$y = \frac{1}{1-b} \implies b = 1 - \frac{1}{y}$$ $$z = \frac{1}{1-c} \implies c = 1 - \frac{1}{z}$$ a, b, c are in A.P. $$\implies 1 - \frac{1}{x}, 1 - \frac{1}{y}, 1 - \frac{1}{z} are in A.P.$$ $$\implies -\frac{1}{x}, -\frac{1}{y}, -\frac{1}{z} are in A.P.$$ $$\implies \frac{1}{x}, \frac{1}{y}, \frac{1}{z} are in A.P.$$
Question 4
Maths · Differential Equations · Single correct
Let $\frac{dy}{dx} = \frac{ax - by + a}{bx + cy + a}$, where $a, b, c$ are constants, represent a circle passing through the point $(2, 5)$. Then the shortest distance of the point $(11, 6)$ from this circle is:
10
8
7
5
Answer: (b)
Solution
Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. $$\Rightarrow \frac{dy}{dx} = \frac{-(2x + 2g)}{(2y + 2f)}$$ Comparing with $\frac{dy}{dx} = \frac{ax - by + a}{bx + cy + a}$ $$\Rightarrow b = 0, \ a = -2, \ c = 2$$ $$\Rightarrow -2g = -2 \Rightarrow g = 1 2f = -2$$ $$f = -1$$ Now the circle will be $x^2 + y^2 + 2x - 2y + c = 0$ It passes through $(2, 5)$ which will give $c = -23$ So the circle will be $x^2 + y^2 + 2x - 2y - 23 = 0$ Centre $C = (-1, 1)$ and radius $5$ Now $P$ is $(11, 6)$ So the minimum distance of $P$ from the circle will be $$= \sqrt{(11 + 1)^2 + (6 - 1)^2} - 5$$ $$= 13 - 5$$ $$= 8$$
Question 5
Maths · Limits and Derivatives · Single correct
Let a be an integer such that $\lim_{x \to 7} \frac{18 - [1-x]}{[x - 3a]}$ exists, where $[t]$ is greatest integer $\leq t$. Then a is equal to:
Maths · Applications of Derivatives · Single correct
The number of distinct real roots of $x^4 - 4x + 1 = 0$ is :
4
2
1
0
Answer: (b)
Solution
Let $f(x) = x^4 - 4x + 1$. $f'(x) = 4x^3 - 4$. $f'(x) = 0 \implies x = 1$. $x = 1$ is point of minima. $f(1) = -2$. $f(0) = 1$. Hence 2 solutions.
Question 7
Maths · Applications of Derivatives · Single correct
The lengths of the sides of a triangle are $10 + x^2$, $10 + x^2$ and $20 - 2x^2$. If for $x = k$, the area of the triangle is maximum, then $3k^2$ is equal to:
5
8
10
12
Answer: (c)
Solution
Given $a = 20 - 2x^2$, $b = 10 + x^2$, $c = 10 + x^2$. The semi-perimeter $s$ is given by $$s = \frac{a + b + c}{2} = 20.$$ The area $\Delta$ is calculated as $$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{20(2x^2)(10-x^2)(10-x^2)} = 2\sqrt{10} \sqrt{x^2(10-x^2)^2} = 2\sqrt{10} \left| x(10-x^2) \right| = 2\sqrt{10} \left| 10x - x^3 \right|.$$ The expression for $S$ is $S = 10x - x^3$. Differentiating with respect to $x$, we have $$\frac{dS}{dx} = 10 - 3x^2.$$ Setting $\frac{dS}{dx} = 0$ gives $x^2 = \frac{10}{3}$. Solving $3x^2 = 10$ gives the same result.
Question 8
Maths · Continuity and Differentiability · Single correct
If $\cos^{-1}\left(\frac{y}{2}\right) = \log_e\left(\frac{x}{5}\right)^5$, $|y| < 2$, then:
The value of the integral $$\int_{-2}^{2} \frac{|x^3 + x|}{(e^{x|x|} + 1)} \, dx$$ is equal to:
5e^2
3e^{-2}
4
6
Answer: (d)
Solution
Given $f(x) = \frac{|x^3 + x|}{e^{|x|} + 1}$. The integral from $-2$ to $2$ of $f(x) \, dx$ is equal to the integral from $0$ to $2$ of $(f(x) + f(-x)) \, dx$. $$\int_{-2}^{2} f(x) \, dx = \int_{0}^{2} \left( f(x) + f(-x) \right) \, dx$$ This becomes: $$= \int_{0}^{2} \left( \frac{x^3 + x}{e^{|x|} + 1} + \frac{-x^3 - x}{e^{-|x|} + 1} \right) \, dx$$ Simplifying further: $$= \int_{0}^{2} \left( \frac{x^3 + x}{e^{|x|} + 1} + \frac{x^3 + x}{e^{-|x|} + 1} \right) \, dx$$ This can be rewritten as: $$= \int_{0}^{2} \left( \frac{x^3 + x}{e^{x^2} + 1} + \frac{x^3 + x}{e^{-x^2} + 1} \right) \, dx$$ Let $I = \int_{0}^{2} \left( \frac{x^3 + x}{1 + e^{x^2}} + \frac{e^{-x^2}(x^3 + x)}{1 + e^{x^2}} \right) \, dx$. This simplifies to: $$= \int_{0}^{2} (x^3 + x) \, dx$$ Evaluating the integral: $$= \left[ \frac{x^4}{4} + \frac{x^2}{2} \right]_{0}^{2}$$ Calculating the result: $$= 4 + 2 = 6$$
Question 11
Maths · Differential Equations · Single correct
If $\frac{dy}{dx} + \frac{2^{x-y} \left(2^y - 1\right)}{2^x - 1} = 0$, $x, y > 0$, $y(1) = 1$, then y(2) is equal to:
2 + $\log$_2 3
2 + $\log$_2 2
2 - $\log$_2 3
2 - $\log$_2 3
Answer: (d)
Solution
Given $\left( \frac{dy}{dx} + \frac{2^{x-y}(2^x-1)}{2^y-1} = 0 \right)$, $\left( x, y > 0,\ y(1) = 1,\ y(2) = ? \right)$. $$\frac{dy}{dx} = -\frac{2^x(2^x-1)}{2^y(2^y-1)}$$ $$\int \frac{2^y}{2^y-1}\,dy = -\int \frac{2^x}{2^x-1}\,dx$$ $$\frac{1}{\ln 2}\int \frac{2^y \ln 2}{2^y-1}\,dy = -\frac{1}{\ln 2}\int \frac{2^x \ln 2}{2^x-1}\,dx$$ $$\frac{1}{\ln 2}\ln|2^y-1| = -\frac{1}{\ln 2}\ln|2^x-1| + C$$ At $\left( x = 1,\ y = 1 \right)$ Putting this values in above relation we get $\left( C = 0 \right)$ $$\ln|2^y-1| + \ln|2^x-1| = 0$$ $$(2^x-1)(2^y-1) = 1$$ $$2^y - 1 = \frac{1}{2^x-1}$$ At $\left( x = 2 \right)$ $$2^y = \frac{1}{3} + 1 = \frac{4}{3}$$ $$y = \log_2 \frac{4}{3} = \log_2 4 - \log_2 3 = 2 - \log_2 3$$
Question 12
Maths · Straight Lines and Pair of Straight Lines · Single correct
In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x + y = 4. Let the point B lie on the line x + 3y = 7. If $(\alpha, \beta)$ is the centroid $\triangle ABC$, then $15(\alpha + \beta)$ is equal to:
Let the eccentricity of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \ a > b,$ be $\frac{1}{4}$. If this ellipse passes through the point $\left( -4 \sqrt{\frac{2}{5}}, 3 \right)$, then $a^2 + b^2$ is equal to:
Maths · Three Dimensional Geometry · Single correct
If two straight lines whose direction cosines are given by the relations $l + m - n = 0$, $3l^2 + m^2 + cnl = 0$ are parallel, then the positive value of $c$ is:
6
4
3
2
Answer: (a)
Solution
Given $$l + m - n = 0$$ $$3l^2 + m^2 + cl (l + m) = 0$$ $$n = l + m$$ $$3l^2 + m^2 + cl^2 + clm = 0$$ $$(3 + c) l^2 + clm + m^2 = 0$$ $$\left(3 + c\right) \left(\frac{l}{m}\right)^2 + c \left(\frac{l}{m}\right) + 1 = 0 \ldots (1)$$ Therefore, lines are parallel. Roots of (1) must be equal $$\Rightarrow D = 0$$ $$c^2 - 4 \left(3 + c\right) = 0$$ $$c^2 - 4c - 12 = 0$$ $$(c - 6)(c + 2) = 0$$ $$c = 6 or c = -2$$ Positive value of $c = 6$
Question 15
Maths · Vector Algebra · Single correct
Let $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} - 3\hat{j} + 2\hat{k}$. Then the number of vectors $\vec{b}$ such that $\vec{b} \times \vec{c} = \vec{a}$ and $|\vec{b}| \in \{1, 2, \ldots, 10\}$ is:
0
1
2
3
Answer: (a)
Solution
Given $\vec{a} = i + j - k$ and $\vec{c} = 2i - 3j + 2k$. We have $\vec{b} \times \vec{c} = \vec{a}$. The magnitude $|\vec{b}| \in \{1, 2, \ldots, 10\}$. Therefore, $\vec{b} \times \vec{c} = \vec{a}$. This implies $\vec{a}$ is perpendicular to $\vec{b}$ as well as $\vec{a}$ is perpendicular to $\vec{c}$. Now $\vec{a} \cdot \vec{c} = 2 - 3 - 2 = -3 \neq 0$. This $\vec{b} \times \vec{c} = \vec{a}$ is not possible. Number of vectors $\vec{b} = 0$.
Question 16
Maths · Probability · Single correct
Five numbers $x_1, x_2, x_3, x_4, x_5$ are randomly selected from the numbers 1, 2, 3, $\ldots$, 18 and are arranged in the increasing order ($x_1 < x_2 < x_3 < x_4 < x_5$). The probability that $x_2 = 7$ and $x_4 = 11$ is:
$\frac{1}{136}$
$\frac{1}{72}$
$\frac{1}{68}$
$\frac{1}{34}$
Answer: (c)
Solution
No. of ways to select and arrange $x_1,x_2,x_3,x_4,x_5$ from $1,2,3,\ldots,18$ $n(S)$ $={}^{18}C_5$ $\begin{array}{ccccc} x_1 & x_2 & x_3 & x_4 & x_5\\ 7 & & & 11 & \end{array}$ $n(E)$ $={}^{6}C_1\times{}^{3}C_1\times{}^{7}C_1$ $P(E)$ $=\frac{6\times3\times7}{{}^{18}C_5}$ $=\frac{1}{17\times4}$ $=\frac{1}{68}$
Question 17
Maths · Probability · Single correct
Let X be a random variable having binomial distribution B(7, p). If P(X = 3) = 5P(X = 4), then the sum of the mean and the variance of X is :
The Boolean expression $\left( \sim (p \land q) \right) \lor q$ is equivalent to:
$q \rightarrow (p \land q)$
$p \rightarrow q$
$p \rightarrow (p \rightarrow q)$
$p \rightarrow (p \lor q)$
Answer: (d)
Solution
$\sim(p\land q)\lor q$ $=(\sim p\lor\sim q)\lor q$ $=\sim p\lor(\sim q\lor q)$ $=\sim p\lor T$ $=T$ This statement is a tautology. Option D $\sim p\Rightarrow(p\lor q)$ is also a tautology. OR
Question 21
Maths · Relations and Functions · Fill in the blank
Let $f : \mathbb{R} \to \mathbb{R}$ be a function defined $f(x) = \frac{2e^{2x}}{e^{2x} + e}$. Then $f\left(\frac{1}{100}\right) + f\left(\frac{2}{100}\right) + f\left(\frac{3}{100}\right) + \ldots + f\left(\frac{99}{100}\right)$ is equal to ________.
Maths · Complex Numbers and Quadratic Equations · Fill in the blank
If the sum of all the roots of the equation $$e^{2x} - 11e^x - 45e^{-x} + \frac{81}{2} = 0$$ is $\log_e P$, then $p$ is equal to ______.
Answer: 45
Solution
Given the equation $e^{2x} - 11e^x - 45e^{-x} + \frac{81}{2} = 0$. Rewriting it as $\left(e^x\right)^3 - 11\left(e^x\right)^2 - 45 + \frac{81e^x}{2} = 0$. Let $e^x = t$. Then the equation becomes $2t^3 - 22t^2 + 81t - 90 = 0$. The product of the roots is $t_1 t_2 t_3 = 45$. Therefore, $e^{x_1} e^{x_2} e^{x_3} = 45$. This implies $e^{x_1 + x_2 + x_3} = 45$. Taking the logarithm, $\log_e e^{x_1 + x_2 + x_3} = \log_e 45$. Thus, $x_1 + x_2 + x_3 = \log_e 45$. Let $\log_e P = \log_e 45$. Therefore, $P = 45$.
Question 23
Maths · Matrices · Numerical
The positive value of the determinant of the matrix $$\begin{pmatrix} 14 & 28 & -14 \\ -14 & 14 & 28 \\ 28 & -14 & 14 \end{pmatrix}$$, A, whose $\mathrm{Adj} (\mathrm{Adj} (A)) = A$, is _________.
The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is
Answer: 56
Solution
Given 16 cubes, 11 are blue and 5 are red. $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 11$$ where $x_1, x_6 \geq 0$ and $x_2, x_3, x_4, x_5 \geq 2$. Let $x_2 = t_1 + 2$, $x_3 = t_3 + 2$, $x_4 = t_4 + 2$, $x_5 = t_5 + 2$. Then $x_1, t_2, t_3, t_4, t_5, x_6 \geq 0$. The number of solutions is given by: $$\binom{6+3-1}{3} = \binom{8}{3} = 56$$
Question 25
Maths · Binomial Theorem · Numerical
If the coefficient of $x^{10}$ in the binomial expansion of $$\left( \frac{\sqrt{x}}{5^4} + \frac{\sqrt{5}}{x^3} \right)^{60}$$ is $5^k l$, $k \in \mathbb{N}$ and $l$ is co-prime to 5, then $k$ is equal to .
Let $$A_1 = \left\{ (x, y) : |x| \leq y^2, |x| + 2y \leq 8 \right\}$$ and $$A_2 = \left\{ (x, y) : |x| + |y| \leq k \right\}$$. If 27 (Area $A_1$) = 5 (Area $A_2$), then $k$ is equal to :
Answer: 8
Solution
Given $A_1 = \{(x, y) : |x| \leq y^2, |x| + 2y \leq 8\}$ and $A_2 = \{(x, y) : |x| + |y| \leq k\}$. The area of $A_1$ is calculated as follows: $$area(A_1) = 2 \left[ \int_0^2 y^2 \, dy + \int_2^4 (8 - 2y) \, dy \right]$$ Evaluating the integrals: $$= 2 \left[ \left. \frac{y^3}{3} \right|_0^2 + \left. (8y - y^2) \right|_2^4 \right]$$ The area of $A_1$ is: $$area(A_1) = 2 \times \frac{20}{3} = \frac{40}{3}$$ For $A_2$, the area is calculated as: $$Area (A_2) = 4 \times \frac{1}{2} k^2$$ Thus, $$Area (A_2) = 2k^2$$ Now, $$27 \times (Area A_1) = 5 \times (Area A_2)$$
Question 27
Maths · Sequences and Series · Numerical
If the sum of the first ten terms of the series $$\frac{1}{5} + \frac{2}{65} + \frac{3}{325} + \frac{4}{1025} + \frac{5}{2501} + \ldots$$ is $\frac{m}{n}$, where $m$ and $n$ are co-prime numbers, then $m + n$ is equal to
Answer: 276
Solution
Given the series: $$\frac{1}{5} + \frac{2}{65} + \frac{3}{325} + \frac{4}{1025} + \frac{5}{2501} + \ldots$$ The general term $T_n$ is given by: $$T_n = \frac{n}{4n^4 + 1}$$ This can be rewritten as: $$= \frac{n}{(2n^2 + 1)^2 - (2n)^2} = \frac{n}{(2n^2 + 2n + 1)(2n^2 - 2n + 1)}$$ Simplifying further: $$= \frac{1}{4} \left[ \frac{1}{2n^2 - 2n + 1} - \frac{1}{2n^2 + 2n + 1} \right]$$ The sum $S_{10}$ is: $$S_{10} = \sum_{n=1}^{10} T_n = \frac{1}{4} \left[ 1 - \frac{1}{5} + \frac{1}{5} - \frac{1}{13} + \ldots + \frac{1}{200 + 20 + 1} \right]$$ Simplifying the sum: $$= \frac{1}{4} \left[ 1 - \frac{1}{221} \right] = \frac{1}{4} \times \frac{220}{221} = \frac{m}{n}$$ Where $m + n = 55 + 221 = 276$
Question 28
Maths · Conic Sections · Fill in the blank
A rectangle R with end points of the one of its dies as (1, 2) and (3, 6) is inscribed in a circle. If the equation of a diameter of the circle is $2x - y + 4 = 0$, then the area of R is _______.
Answer: 16
Solution
Equation of line AB is given by $$y = 2x$$ The slope of AB is 2. The slope of the given diameter is also 2. So the diameter is parallel to AB. The distance between the diameter and line AB is $$= \left( \frac{4}{\sqrt{2^2 + 12}} \right) = \frac{4}{\sqrt{5}}$$ Thus, BC is $$2 \times \frac{4}{\sqrt{5}} = \frac{8}{\sqrt{5}}$$ The length of AB is $$\sqrt{(1-3)^2 + (2-6)^2} = \sqrt{20} = 2\sqrt{5}$$ The area is given by $$Area = AB \times BC = \frac{8}{\sqrt{5}} \times 2\sqrt{5} = 16$$ Answer.
Question 29
Maths · Conic Sections · Numerical
A circle of radius 2 unit passes through the vertex and the focus of the parabola $y^2 = 2x$ and touches the parabola $y = \left( x - \frac{1}{4} \right)^2 + \alpha$, where $\alpha > 0$. Then $(4\alpha - 8)^2$ is equal to ________.
Answer: 63
Solution
Vertex and focus of parabola $y^2 = 2x$ are $V \,(0, 0)$ and $S \left( \frac{1}{2}, 0 \right)$ respectively. Let equation of circle be $(x - h)^2 + (y - k)^2 = 4$. Therefore, circle passes through $(0, 0)$ $$\Rightarrow h^2 + k^2 = 4 \ldots (1)$$ Therefore, circle passes through $\left( \frac{1}{2}, 0 \right)$ $$\left( \frac{1}{2} - h \right)^2 + k^2 = 4$$ $$\Rightarrow h^2 + k^2 - h = \frac{15}{4} \ldots (2)$$ On solving (1) and (2) $$4 - h = \frac{15}{4}$$ $$h = 4 - \frac{15}{4} = \frac{1}{4}$$ $$k = + \frac{\sqrt{63}}{4}$$ $$k = - \frac{\sqrt{63}}{4}$$ is rejected as circle with centre $\left( \frac{1}{4}, -\frac{\sqrt{63}}{4} \right)$ can't touch given parabola. Equation of circle is $$\left( x - \frac{1}{4} \right)^2 + \left( k - \frac{\sqrt{63}}{4} \right)^2 = 4$$ From figure $$\alpha = 2 + \frac{\sqrt{63}}{4} = \frac{8 + \sqrt{63}}{4}$$ $$4\alpha - 8 = \sqrt{63}$$ $$(4\alpha - 8)^2 = 63$$
Question 30
Maths · Three Dimensional Geometry · Numerical
Let the mirror image of the point $(a, b, c)$ with respect to the plane $3x - 4y + 12z + 19 = 0$ be $(a-6, \beta, \gamma)$. If $a + b + c = 5$, then $7 \beta - 9 \gamma$ is equal to ___________.
Answer: 137
Solution
Given $\mathbf{P}(a, b, c)$ and $\mathbf{P'}(a - 6, \beta, \gamma)$, the midpoint $\mathbf{M}$ is given by: $$\mathbf{M} = \left( a - 3, \frac{\beta + b}{2}, \frac{\gamma + c}{2} \right)$$ Since $\mathbf{M}$ lies on $3x + 4y + 12z + 19 = 0$, we have: $$6a - 4b + 12c - 4\beta + 12\gamma + 20 = 0 ...(1)$$ Since $\mathbf{PP'}$ is parallel to the normal of the plane, then: $$\frac{6}{3} = \frac{b - \beta}{-4} = \frac{c - \gamma}{12}$$ This implies $\beta = b + 8$ and $\gamma = c - 24$. Given $a + b + c = 5$, we have $a + \beta - 8 + \gamma + 24 = 5$. Thus, $a = -\beta - \gamma - 11$. Now, putting these values in equation (1), we get: $$6(-\beta - \gamma - 11) - 4(\beta - 8) + 12(\gamma + 24) - 4\beta + 12\gamma + 20 = 0$$ Simplifying gives: $$7\beta - 9\gamma = 170 - 33$$ Therefore, $\boxed{137}$.
Physics
Question 31
Physics · Motion in a Plane · Single correct
A projectile is launched at an angle $\alpha$ with the horizontal with a velocity $20 \, \mathrm{ms^{-1}}$. After $10 \, \mathrm{s}$, its inclination with horizontal is $\beta$. The value of $\tan \beta$ will be: $(g = 10 \, \mathrm{ms^{-2}})$
A girl standing on road holds her umbrella at 45^$\circ$ with the vertical to keep the rain away. If she starts running without umbrella with a speed of $15\sqrt{2} \, \mathrm{kmh}^{-1}$, the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is:
$30 \, \mathrm{kmh}^{-1}$
$\frac{25}{\sqrt{2}} \, \mathrm{kmh}^{-1}$
$\frac{30}{\sqrt{2}} \, \mathrm{kmh}^{-1}$
$25 \, \mathrm{kmh}^{-1}$
Answer: (c)
Solution
Given the triangle, we have $V = \tan \theta = \frac{V_G}{V_{RG}}$. Since $\theta = 45^\circ$, we have: $$1 = \frac{V_G}{V_{RG}} \implies 15\sqrt{2} = V_{RG}$$
Question 33
Physics · Mathematics in Physics · Single correct
A sliver wire has mass $(0.6 \pm 0.006) \, \mathrm{g}$, radius $(0.5 \pm 0.005) \, \mathrm{mm}$ and length $(4 \pm 0.04) \, \mathrm{cm}$. The maximum percentage error in the measurement of its density will be:
4%
3%
6%
7%
Answer: (a)
Solution
Given $M = (0.6 \pm 0.006) \, \mathrm{g}$, $r = (0.5 \pm 0.005) \, \mathrm{mm}$, $l = (4 \pm 0.04) \, \mathrm{cm}$. The density $\rho$ is given by $\rho = \frac{m}{V}$. Therefore, $$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{2 \Delta r}{r} + \frac{\Delta l}{l}$$ (Volume of cylinder $= \pi r^2 l$). Substituting the values, $$= \frac{0.006}{0.6} + \frac{2 \times 0.005}{0.5} + \frac{0.04}{4}$$ $$100 \times \frac{\Delta \rho}{\rho} = 4 \times 10^{-2} \times 100$$ $$\frac{\Delta \rho}{\rho} \times 100 = 4\%$$
Question 34
Physics · Laws of Motion · Single correct
A system of two blocks of masses $m = 2 \, \mathrm{kg}$ and $M = 8 \, \mathrm{kg}$ is placed on a smooth table as shown in figure. The coefficient of static friction between two blocks is $0.5$. The maximum horizontal force $F$ that can be applied to the block of mass $M$ so that the blocks move together will be:
$9.8 \, \mathrm{N}$
$39.2 \, \mathrm{N}$
$49 \, \mathrm{N}$
$78.4 \, \mathrm{N}$
Answer: (c)
Solution
The maximum acceleration $\left(a_A\right)_{\max} = 0.5g = 4.9 \, \mathrm{m/s^2}$. For moving together, the maximum force $F_{\max} = m_T a_A$. Therefore, $$F_{\max} = 10 \times 4.9 = 49 \, \mathrm{N}.$$
Question 35
Physics · System of Particles and Rotational Motion · Single correct
Two blocks of masses 10 kg and 30 kg are placed on the same straight line with coordinates (0, 0) cm and (x, 0) cm respectively. The block of 10 kg is moved on the same line through a distance of 6 cm towards the other block. The distance through which the block of 30 kg must be moved to keep the position of centre of mass of the system unchanged is :
4 cm towards the 10 kg block
2 cm away from the 10 kg block
2 cm towards the 10 kg block
4 cm away from the 10 kg block
Answer: (c)
Solution
Given the equation for the center of mass displacement: $$\Delta x_G = \frac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2}$$ Substituting the given values: $$0 = \frac{10 \times 6 + 30(\Delta x_2)}{40}$$ Solving for $\Delta x_2$: $$\Delta x_2 = -2 \, \mathrm{cm}$$ The block of mass 30 kg will move towards 10 kg.
Question 36
Physics · Current Electricity · Single correct
A 72 $\Omega$ galvanometer is shunted by a resistance of 8 $\Omega$. The percentage of the total current which passes through the galvanometer is:
0.1$\%$
10$\%$
25$\%$
0.25$\%$
Answer: (b)
Solution
Given $$S = \frac{R_G}{\frac{I}{I_g} - 1}$$ Substituting the values, we have $$8 = \frac{72}{\frac{I}{I_g} - 1}$$ Solving for $\($ $\frac{I}{I_g}$ $\)$, we get $$\frac{I}{I_g} - 1 = 9$$ Therefore, $$\frac{I}{I_g} = 10 \Rightarrow \frac{I_g}{I} = \frac{1}{10}$$ The percentage $\($ I $\)$ is $$\% I = \frac{I_g}{I} \times 100 = 10\%$$
Question 37
Physics · Gravitation · Single correct
Given below are two statements : Statement I : The law of gravitation holds good for any pair of bodies in the universe. Statement II : The weight of any person becomes zero when the person is at the centre of the earth. In the light of the above statements, choose the correct answer from the options given below.
Both statement I and Statement II are true
Both statement I and Statement II are false
Statement I is true but Statement II are false
Statement I is false but Statement II is true
Answer: (a)
Solution
Since it is universal law so it hold good for any pair of bodies. The value of $g$ at centre is zero. So statement I and Statement II are true.
Question 38
Physics · System of Particles and Rotational Motion · Single correct
What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 5 times its mass? (Assume the collision to be head-on elastic collision)
Physics · Mechanical Properties of Fluids · Single correct
The velocity of a small ball of mass 'm' and density $d_1$, when dropped in a container filled with glycerine, becomes constant after some time. If the density of glycerine is $d_2$, then the viscous force acting on the ball, will be:
$mg \left( 1 - \frac{d_1}{d_2} \right)$
$mg \left( 1 - \frac{d_2}{d_1} \right)$
$mg \left( \frac{d_1}{d_2} - 1 \right)$
$mg \left( \frac{d_2}{d_1} - 1 \right)$
Answer: (b)
Solution
The force $F_V$ is given by the equation: $$F_V = mg - F_B$$ Substituting the expression for $F_B$: $$= mg - \left( \frac{m}{d_1} \times d_2 \right) g$$ Simplifying further: $$= mg \left( 1 - \frac{d_2}{d_1} \right)$$
Question 40
Physics · Magnetism and Matter · Single correct
The susceptibility of a paramagnetic material is 99. The permeability of the material in Wb/A-m is : [Permeability of free space $\mu_0 = 4\pi \times 10^{-7} \, \mathrm{Wb/A-m}$]
The current flowing through an ac circuit is given by $$I = 5 \sin(120 \pi t) \, \mathrm{A}$$ How long will the current take to reach the peak value starting from zero?
$\frac{1}{60} \, \mathrm{s}$
$60 \, \mathrm{s}$
$\frac{1}{120} \, \mathrm{s}$
$\frac{1}{240} \, \mathrm{s}$
Answer: (d)
Solution
Given $\omega = 120\pi = \frac{2\pi}{T} \Rightarrow T = \frac{1}{60} sec$. Time taken to reach peak value $= \frac{T}{4} = \frac{1}{240} s$.
Question 42
Physics · Electromagnetic Waves · Single correct
Match List-I with List-II: Choose the correct answer from the options given below:
$(A)$–(iii), $(B)$–(iv), $(C)$–(ii), $(D)$–(i)
$(A)$–(iii), $(B)$–(i), $(C)$–(ii), $(D)$–(iv)
$(A)$–(iv), $(B)$–(iii), $(C)$–(ii), $(D)$–(i)
$(A)$–(iv), $(B)$–(iii), $(C)$–(ii), $(D)$–(i)
Answer: (a)
Solution
Given $k = \frac{P^2}{2m} \implies P \alpha \sqrt{m}$. Now $\lambda = \frac{h}{p}$. So, $\lambda \alpha \frac{1}{p} \implies \lambda \alpha \frac{1}{\sqrt{m}}$. $$\frac{\lambda_\alpha}{\lambda_{\mathrm{C_{12}}}} = \frac{\sqrt{3}}{1}$$
Question 43
Physics · Dual Nature of Radiation and Matter · Single correct
An $\alpha$ particle and a carbon 12 atom has same kinetic energy $K$. The ratio of their de-Broglie wavelength $\left( \lambda_a : \lambda_{C_{12}} \right)$ is:
1 : $\sqrt{3}$
$\sqrt{3}$ : 1
3 : 1
2 : $\sqrt{3}$
Answer: (b)
Solution
Given $F = qE = q \left( \frac{Q}{A \varepsilon_0} \right) = \frac{qQ}{A \varepsilon_0} = 10 \, \mathrm{N}$. Now, when one plate is removed. $$E' = \frac{Q}{2A \varepsilon_0}$$ $$F = qE' = \frac{Qq}{2A \varepsilon_0} = 5 \, \mathrm{N}$$
Question 44
Physics · Electrostatic Potential and Capacitance · Single correct
A force of $10\,\mathrm{N}$ acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be:
5 N
10 N
20 N
Zero
Answer: (a)
Solution
Given $X = A \sin \omega t$ with $t = 3$ and $X = \frac{A}{2}$. Therefore, $$\frac{A}{2} = A \sin 3\omega$$ This implies $$\sin 3\omega = \frac{1}{2}$$ Thus, $$3\omega = \frac{\pi}{6}$$ So, $$\omega = \frac{\pi}{18} = \frac{2\pi}{T}$$ Therefore, $$T = 36 \, \mathrm{s}$$
Question 45
Physics · Oscillations · Single correct
The displacement of simple harmonic oscillator after 3 seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is :
6 $\mathrm{s}$
8 $\mathrm{s}$
12 $\mathrm{s}$
36 $\mathrm{s}$
Answer: (d)
Solution
Given $$f_0 = \left( \frac{v + v_0}{v} \right) f_s$$ Substituting $$f_0 = \left( \frac{v + \frac{v}{5}}{v} \right) f_s$$ Simplifying $$f_0 = \frac{6}{5} f_s$$ The percentage change is given by $$\% change = \frac{f_0 - f_s}{f_s} \times 100$$ Calculating $$= \frac{1}{5} \times 100 = 20\%$$
Question 46
Physics · Waves · Single correct
An observer moves towards a stationary source of sound with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be:
Physics · Ray Optics and Optical Instruments · Single correct
Consider a light ray travelling in air is incident into a medium of refractive index $\sqrt{2n}$. The incident angle is twice that of refracting angle. Then, the angle of incidence will be:
$\sin^{-1}(\sqrt{n})$
$\cos^{-1}\left(\sqrt{\frac{n}{2}}\right)$
$\sin^{-1}(\sqrt{2n})$
$2\cos^{-1}\left(\sqrt{\frac{n}{2}}\right)$
Answer: (d)
Solution
Given the equation: $$13.6 \left( \frac{1}{1^2} - \frac{1}{n^2} \right) = 10.2$$ Solving for $n$ gives: $$n = 2$$ The initial angular momentum $L_i$ is: $$L_i = \frac{h}{2\pi} \times 1$$ The final angular momentum $L_F$ is: $$L_F = \frac{2h}{2\pi}$$ The change in angular momentum $\Delta L$ is: $$\Delta L = L_F - L_i = \frac{h}{2\pi} = \frac{6.6 \times 10^{-34}}{2 \times \frac{22}{7}}$$ This simplifies to: $$= 1.05 \times 10^{-34} \, \mathrm{J \cdot s}$$
Question 48
Physics · Atoms · Single correct
A hydrogen atom in its ground state absorbs $10.2\,\mathrm{eV}$ of energy. The angular momentum of electron of the hydrogen atom will increase by the value of: (Given, Planck's constant $= 6.6 \times 10^{-34}\,\mathrm{Js}$)
2.10 $\times$ 10^{-34} \, $\mathrm{Js}$
1.05 $\times$ 10^{-34} \, $\mathrm{Js}$
3.15 $\times$ 10^{-34} \, $\mathrm{Js}$
4.2 $\times$ 10^{-34} \, $\mathrm{Js}$
Answer: (b)
Solution
Given the equation: $$13.6 \left( \frac{1}{1^2} - \frac{1}{n^2} \right) = 10.2$$ We find that $$n = 2$$ The initial angular momentum is given by $$L_i = \frac{h}{2\pi} \times 1$$ The final angular momentum is $$L_F = \frac{2h}{2\pi}$$ The change in angular momentum is $$\Delta L = L_F - L_i = \frac{h}{2\pi} = \frac{6.6 \times 10^{-34}}{2 \times \frac{22}{7}}$$ This simplifies to $$= 1.05 \times 10^{-34} \, \mathrm{J \cdot s}$$
Question 49
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
Identify the correct Logic Gate for the following output (Y) of two inputs A and B.
Answer: (b)
Solution
The truth table for the NAND gate is shown. The output $Y$ is $0$ only when both inputs $A$ and $B$ are $1$. Otherwise, the output is $1$. The expression for the NAND gate is given by: $$Y = \overline{A \cdot B}$$
Question 50
Physics · Kinetic Theory · Single correct
A mixture of hydrogen and oxygen has volume $2000 \, \mathrm{cm}^3$, temperature $300 \, \mathrm{K}$, pressure $100 \, \mathrm{kPa}$ and mass $0.76 \, \mathrm{g}$. The ratio of number of moles of hydrogen to number of moles of oxygen in the mixture will be:
In a carnot engine, the temperature of reservoir is $527^\circ \mathrm{C}$ and that of sink is $200 \, \mathrm{K}$. If the workdone by the engine when it transfers heat from reservoir to sink is $12000 \, \mathrm{kJ}$, the quantity of heat absorbed by the engine from reservoir is ______ $\times 10^6 \, \mathrm{J}$.
Answer: 16
Solution
Question 52
Physics · Alternating Current · Numerical
A 220 $\mathrm{V}$, 50 $\mathrm{Hz}$ AC source is connected to a 25 $\mathrm{V}$, 5 $\mathrm{W}$ lamp and an additional resistance R in series (as shown in figure) to run the lamp at its peak brightness, then the value of R (in ohm) will be
Answer: 975
Solution
Given $P = V i$. $5 = 25 i$. Solving for $i$, we have $i = \frac{1}{5}$. The voltage across the resistor $V_R = i R$. Substituting the values, $(220 - 25) = \frac{1}{5} R$. Solving for $R$, we get $R = 195 \times 5 = 975 \, \Omega$.
Question 53
Physics · Wave Optics · Numerical
In Young’s double slit experiment the two slits are $0.6 \, \mathrm{mm}$ distance apart. Interference pattern is observed on a screen at a distance $80 \, \mathrm{cm}$ from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be _____ nm.
Answer: 450
Solution
Given $d = 0.6 \times 10^{-3}$ and $D = 80 \times 10^{-2}$. For the 1st dark fringe, $$\frac{D \lambda}{2d} = \frac{d}{2},$$ $$\lambda = \frac{d^2}{D}$$ $$= 450 \times 10^{-9} \, \mathrm{m}$$
Question 54
Physics · Atoms · Numerical
A beam of monochromatic light is used to excite the electron in $\mathrm{Li}^{++}$ from the first orbit to the third orbit. The wavelength of monochromatic light is found to be $x \times 10^{-10} \, \mathrm{m}$. The value of $x$ is ______. [Given $hc = 1242 \, \mathrm{eV} \, \mathrm{nm}$]
Answer: 114
Solution
Given $Z = 3$. The formula for the wavelength is: $$\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$ Given $n_1 = 1$, $n_2 = 3$, Substitute the values: $$\frac{1}{\lambda} = R(9) \left( \frac{1}{1} - \frac{1}{9} \right) = 8R$$ Therefore, the wavelength is: $$\lambda = \frac{1}{8R} = 114 \times 10^{-10} \, \mathrm{m}$$
Question 55
Physics · Current Electricity · Numerical
A cell, shunted by a 8 $\Omega$ resistance, is balanced across a potentiometer wire of length $3\, \mathrm{m}$. The balancing length is $2\, \mathrm{m}$ when the cell is shunted by $4\Omega$ resistance. The value of internal resistance of the cell will be _____ $\Omega$.
The current density in a cylindrical wire of radius 4 mm is $4 \times 10^6 \, \mathrm{Am}^{-2}$. The current through the outer portion of the wire between radial distance $\frac{R}{2}$ and $R$ is _____ $\pi \, \mathrm{A}$.
Physics · Electrostatic Potential and Capacitance · Numerical
A capacitor of capacitance 50 $\mathrm{pF}$ is charged by 100 $\mathrm{V}$ source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is ____ $\mathrm{nJ}$.
The height of a transmitting antenna at the top of a tower is 25 m and that of receiving antenna is, 49 m. The maximum distance between them, for satisfactory communication in LOS (Line-Of-Sight) is $K \sqrt{5} \times 10^2 \, \mathrm{m}$. The value of $K$ is . [Assume radius of Earth is $64 \times 10^5 \, \mathrm{m}$] (Calculate upto nearest integer value)
Answer: 192
Solution
Given $$LOS = \sqrt{2R h_T} + \sqrt{2R h_R}$$ Simplifying, we have: $$= \sqrt{2R} \left( \sqrt{h_T} + \sqrt{h_R} \right)$$ Substituting the values: $$= \sqrt{2 \times 64 \times 10^5} \left( \sqrt{25} + \sqrt{49} \right)$$ This simplifies to: $$= 192 \sqrt{5} \times 10^2 \, \mathrm{m}.$$ Therefore, $$K = 192$$
Question 59
Physics · Mechanical Properties of Fluids · Numerical
The area of cross-section of a large tank is $0.5 \, \mathrm{m}^2$. It has a narrow opening near the bottom having area of cross-section $1 \, \mathrm{cm}^2$. A load of $25 \, \mathrm{kg}$ is applied on the water at the top in the tank. Neglecting the speed of water in the tank, the velocity of the water, coming out of the opening at the time when the height of water level in the tank is $40 \, \mathrm{cm}$ above the bottom, will be _______ $\mathrm{cms}^{-1}$. [Take $g = 10 \, \mathrm{ms}^{-2}$]
Physics · System of Particles and Rotational Motion · Numerical
A pendulum of length 2 m consists of a wooden bob of mass 50 g. A bullet of mass 75 g is fired towards the stationary bob with a speed $v$. The bullet emerges out of the bob with a speed $\frac{v}{3}$ and the bob just completes the vertical circle. The value of $v$ is _________ $\mathrm{ms^{-1}}$. (if $g = 10 \, \mathrm{m/s^2}$)
Answer: 10
Solution
Considering only horizontal direction. Initial momentum $P_i$ is equal to final momentum $P_f$. $$(75v) + 0 = 50(\sqrt{5gR}) + 75 \frac{v}{3}$$ $$75 \left( v - \frac{v}{3} \right) = 50 \sqrt{100}$$ $$v = 10 \, \mathrm{m/s}$$
Chemistry
Question 61
Chemistry · Solutions · Single correct
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R ) Assertion (A) : At $10^\circ\mathrm{C}$, the density of a $5\,\mathrm{M}$ solution of $\mathrm{KCl}$ [atomic masses of K and Cl are $39$ $\&$ $35.5\,\mathrm{g\,mol^{-1}}$]. The solution is cooled to $-21^\circ\mathrm{C}$. The molality of the solution will remain unchanged. Reason (R ) : The molality of a solution does not change with temperature as mass remains unaffected with temperature. In the light of the above statements, choose the correct answer from the options given below:
Both (A) and ( R) are true and (R ) is the correct explanation of (A)
Both (A) and (R ) are true but (R ) is not the correct explanation of (A)
(A) is true but (R ) is false
(A) is false but ( R) is true
Answer: (a)
Solution
Molality is independent of temperature and hence both assertion and reason are true.
Question 62
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Based upon VSEPR theory, match the shape (geometry) of the molecules in List-I with the molecules in List-II and select the most appropriate option \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{(Shape)} & \multicolumn{2}{c|}{(Molecules)} \\ \hline (A) & T-shaped & (I) & XeF$_4$ \\ \hline (B) & Trigonal planar & (II) & SF$_4$ \\ \hline (C) & Square planar & (III) & ClF$_3$ \\ \hline (D) & See-saw & (IV) & BF$_3$ \\ \hline \end{tabular}
Chemistry · Co-ordination Compounds · Single correct
Match List-I with List-II Choose the correct answer from the options given below:
$(A)- (III), (B) - (II), (C) - (IV), (D) - (I)$
$(A)- (II), (B) - (III), (C) - (IV), (D) - (I)$
$(A)- (II), (B) - (III), (C) - (I), (D) - (IV)$
$(A)- (II), (B) - (I), (C) - (III), (D) - (IV)$
Answer: (b)
Solution
(A) For a spontaneous process $\Delta G_{T,P} < 0$. (B) $\Delta P = 0 \rightarrow$ Isobaric process. $\Delta T = 0 \rightarrow$ Isothermal process. $(C)$ $\Delta H_{reaction}$ = $(\Sigma Bond energies of reactants)$ - $(\Sigma bond energies of products)$. (D) $\Delta H < 0$ is for exothermic reaction.
Question 64
Chemistry · Co-ordination Compounds · Single correct
Match List-I with List-II Choose the correct answer from the options given below:
$(A)- (II), (B) - (I), (C) - (IV), (D) - (III)$
$(A)- (III), (B) - (I), (C) - (IV), (D) - (II)$
$(A)- (II), (B) - (I), (C) - (III), (D) - (IV)$
$(A)- (III), (B) - (II), (C) - (I), (D) - (IV)$
Answer: (a)
Solution
Q5 (A) (A) Protective colloids are lyophilic colloids (B) Emulsions are liquid in liquid colloidal solutions (C) $\mathrm{FeCl_3}$ + hot water forms positively charged colloidal solution of hydrated ferric oxide. (D) $\mathrm{FeCl_3}$ + $\mathrm{NaOH}$ forms negatively charged colloidal solution due to preferential adsorption of $\mathrm{OH^-}$ ions
Question 65
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason(R) Assertion (A): The ionic radii of $\mathrm{O}^{2-}$ and $\mathrm{Mg}^{2+}$ are same. Reason (R) : Both $\mathrm{O}^{2-}$ and $\mathrm{Mg}^{2+}$ are isoelectronic species In the light of the above statements, choose the correct answer from the options given below
Both (A) and (R) are true and (R) is the correct explanation of (A)
Both (A) and (R) are true but (R) is not the correct explanation of (A)
(A) is true but (R) is false
(A) is false but (R) is true
Answer: (d)
Solution
Ionic radius of $\mathrm{O^{2-}}$ is more than that of $\mathrm{Mg^{2+}}$. Both $\mathrm{O^{2-}}$ and $\mathrm{Mg^{2+}}$ are isoelectronic with 10 electrons.
Question 66
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Match List-I with List-II \begin{tabular}{ll \qquad ll} \textbf{List-I} & \textbf{List-II} \\ (A) Concentration of gold ore & (I) $\mathrm{Aniline}$ \\ (B) Leaching of alumina & (II) $\mathrm{NaOH}$ \\ (C) Froth stabiliser & (III) $\mathrm{SO_2}$ \\ (D) Blister copper & (IV) $\mathrm{NaCN}$ \end{tabular} Choose the correct answer from the options given below.
(A)- (IV), (B) - (III), (C ) - (II), (D) - (I)
(A)- (IV), (B) - (II), (C ) - (I), (D) - (III)
(A)- (III), (B) - (II), (C ) - (I), (D) - (IV)
(A)- (II), (B) - (IV), (C ) - (III), (D) - (I)
Answer: (b)
Solution
Gold is concentrated by cyanidation. Leaching of alumina is done by NaOH. Froth stabiliser is aniline. Blister copper has condensed $\mathrm{SO_2}$ on the surface.
Question 67
Chemistry · The s-Block Elements · Single correct
Addition of $\mathrm{H_2SO_4}$ to $\mathrm{BaO_2}$ produces:
$\mathrm{BaO}$, $\mathrm{SO_2}$ and $\mathrm{H_2O}$
$\mathrm{BaHSO_4}$ and $\mathrm{O_2}$
$\mathrm{BaSO_4}$, $\mathrm{H_2}$ and $\mathrm{O_2}$
$\mathrm{BaSO_4}$ and $\mathrm{H_2O_2}$
Answer: (d)
Solution
The reaction is $\mathrm{BaO_2} + \mathrm{H_2SO_4} \rightarrow \mathrm{BaSO_4} + \mathrm{H_2O_2}$. This is a common method to prepare hydrogen peroxide.
Question 68
Chemistry · The s-Block Elements · Single correct
$\mathrm{BeCl_2}$ reacts with $\mathrm{LiAlH_4}$ to give
$2\mathrm{BeCl_2} + \mathrm{LiAlH_4} \rightarrow 2\mathrm{BeH_2} + \mathrm{LiCl} + \mathrm{AlCl_3}$ This is the method to prepare $\mathrm{BeH_2}$.
Question 69
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Match List-I with List-II \begin{tabular}{|c|c|c|l|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{(Si-Compounds)} & \multicolumn{2}{c|}{(Si-Polymer/other products)} \\ \hline (A) & (CH$_3$)$_4$Si & (I) & Chain silicone \\ \hline (B) & (CH$_3$)Si(OH)$_3$ & (II) & Dimeric silicone \\ \hline (C) & (CH$_3$)$_2$Si(OH)$_2$ & (III) & Silane \\ \hline (D) & (CH$_3$)$_3$Si(OH) & (IV) & 2D-Silicone \\ \hline \end{tabular} Choose the correct answer from the options given below:
(A) – (III), (B) – (II), $(C)$ – (I), (D) – (IV)
(A) – (IV), (B) – (I), $(C)$ – (II), (D) – (III)
(A) – (II), (B) – (I), $(C)$ – (IV), (D) – (III)
(A) – (III), (B) – (IV), $(C)$ – (I), (D) – (II)
Answer: (d)
Solution
($\mathrm{CH_3}$)_4$\mathrm{Si}$ is a silane. ($\mathrm{CH_3}$)$\mathrm{Si(OH)_3}$ polymerise to form 2D silicone. ($\mathrm{CH_3}$)_2$\mathrm{Si(OH)_2}$ polymerise to form chain silicone. ($\mathrm{CH_3}$)_3$\mathrm{Si(OH)}$ form dimer ($\mathrm{CH_3}$)_3$\mathrm{Si-O-Si(CH_3)_3}$.
Question 70
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Heating white phosphorus with conc. NaOH solution gives mainly
$Na_3P$ and $H_2O$
$H_3PO$ and $NaH$
$P(OH)_3$ and $NaH_2PO_4$
$PH_3$ and $NaH_2PO_2$
Answer: (d)
Solution
The reaction is given by the equation: $$\mathrm{P_4 + 3NaOH + 3H_2O \rightarrow 3NaH_2PO_2 + PH_3}$$
Question 71
Chemistry · Co-ordination Compounds · Single correct
Which of the following will have maximum stabilization due to crystal field?
[$\mathrm{Ti(H_2O)_6}$]^{3+}
[$\mathrm{Co(H_2O)_6}$]^{2+}
[$\mathrm{Co(CN)_6}$]^{3-}
[$\mathrm{Cu(NH_3)_4}$]^{2+}
Answer: (c)
Solution
$\mathrm{Co^{3+}}$ has maximum effective nuclear charge and $\mathrm{CN^-}$ is the strongest ligand in the given options.
Question 72
Chemistry · Environmental Chemistry · Single correct
Given below are two statements: Statement I: Classical smog occurs in cool humid climate. It is a reducing mixture of smoke, fog and sulphur dioxide Statement II: Photochemical smog has components, ozone, nitric oxide, acrolein, formaldehyde, PAN etc. In the light of above statements, choose the most appropriate answer from the options give below
Both Statement I and Statement II are correct
Both Statement I and Statement II are incorrect
Statement I is correct but statement II is incorrect
Statement I is incorrect but Statement II is correct
Answer: (a)
Solution
Classical smog occurs in cool humid climate. It is a reducing mixture of smoke, fog and sulphur dioxide. Photochemical smog has components, ozone, nitric oxide, acrolein, formaldehyde, PAN etc. $$\mathrm{CH_4 + O_3 \rightarrow HCHO + H_2O + CH_2 = CH - CHO +}$$ $$\begin{array}{c} \mathrm{H_3C} \\ \mathrm{|} \\ \mathrm{O} \\ \mathrm{|} \\ \mathrm{O-ONO_2} \end{array}$$ (PAN - peroxyacetyl nitrate)
Question 73
Chemistry · Analytical Chemistry · Single correct
Which of the following is structure of a separating funnel?
Answer: (a)
Solution
It is used to separate liquid-liquid mixture which is immiscible with different densities.
Question 74
Chemistry · Hydrocarbons · Single correct
'A' and 'B' respectively are: A $\xrightarrow[\mathrm{(2)\ Zn-H_2O}]{\mathrm{(1)\ O_3}}$ Ethane-1,2-dicarbaldehyde $+$ Glyoxal/Oxaldehyde B $\xrightarrow[\mathrm{(2)\ Zn-H_2O}]{\mathrm{(1)\ O_3}}$ 5-oxohexanal
Chemistry · Haloalkanes and Haloarenes · Single correct
The major product of the following reaction is:
Answer: (a)
Solution
It is bimolecular nucleophilic substitution (SN$^2$) which occur at benzylic carbon by inversion in configuration. This reaction cannot undergo substitution at benzene ring.
Question 76
Chemistry · Alcohols, Phenols and Ethers · Multiple correct
Which of the following reactions will yield benzaldehyde as a product?
(B) and $(C)$
$(C)$ and (D)
(A) and (D)
(A) and $(C)$
Answer: (c)
Solution
The reaction starts with benzoic acid, which is converted to benzoyl chloride using $\mathrm{SOCl_2}$ and quinoline. Then, benzoyl chloride undergoes Rosenmund reduction with $\mathrm{H_2/Pd/BaSO_4}$ to form benzaldehyde. In the second reaction, benzyl alcohol is oxidized to benzoic acid using $\mathrm{CrO_3/H_2SO_4}$. In the third reaction, methyl benzoate is treated with $\mathrm{NaBH_4}$, but no reduction occurs. In the fourth reaction, toluene is oxidized using $\mathrm{CrO_3.(CH_3CO)_2O}$ to form an intermediate, which upon hydrolysis with $\mathrm{H_3O^+}$ gives benzaldehyde.
Question 77
Chemistry · Amines · Single correct
Given below are two statements: Statements-I : In Hofmann degradation reaction, the migration of only an alkyl group takes place from carbonyl carbon of the amide to the nitrogen atom. Statement-II : The group is migrated in Hofmann degradation reaction to electron deficient atom. In the light of the above statement, choose the most appropriate answer from the options given below:
Both Statement-I and Statement-II are correct
Both Statement-I and Statement-II are incorrect
Statement-I is correct but Statement-II is incorrect
Statement-I is incorrect but Statement-II is correct
Answer: (d)
Solution
R - $\mathrm{CO}$ - $\mathrm{NH_2}$ + $\mathrm{Br_2}$ + $\mathrm{NaOH}$ $\rightarrow$ R - $\mathrm{NH_2}$ + $\mathrm{Na_2CO_3}$ + $\mathrm{NaBr}$ + $\mathrm{H_2O}$ R - $\mathrm{CO}$ - $\mathrm{NH_2}$ + $\mathrm{OH^-}$ $\rightarrow$ R - $\mathrm{CO}$ - $\mathrm{NH}$ $\xrightarrow{\mathrm{Br_2}}$ R - $\mathrm{CO}$ - $\mathrm{NH}$ - $\mathrm{Br}$ $\xrightarrow{\mathrm{OH^-}}$ R - $\mathrm{CO}$ - $\mathrm{N^-Br}$ $\xrightarrow{migration of \overset{-}{R}}$ R - $\mathrm{NCO}$ $\xrightarrow{2\mathrm{OH^-}}$ $\mathrm{RNH_2}$ + $\mathrm{CO_3^{2-}}$ In this reaction of alkyl as well as aryl group can migrate to electron deficient nitrogen atom.
Question 78
Chemistry · Polymers · Single correct
Match List-I with List-II \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{\textbf{List-I (Polymer)}} & \multicolumn{2}{c|}{\textbf{List-II (Used in)}} \\ \hline (A) & Bakelite & (I) & Radio and television Cabinets \\ \hline (B) & Glyptal & (II) & Electrical switches \\ \hline (C) & PVC & (III) & Paints and Lacquers \\ \hline (D) & Polystyrene & (IV) & Water pipes \\ \hline \end{tabular} Choose the correct answer from the options given below:
Bakelite - It is a thermosetting polymer used for making electrical switches. Glyptal – manufacture of paints and lacquers. PVC – manufacture of water pipes, rain coats, hand bags. Polystyrene – manufacture of radio and television cabinets.
Question 79
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
L-isomer of a compound ‘A’ ($C_4H_8O_4$) gives a positive test with $[Ag(NH_3)_2]^+$. Treatment of ‘A’ with acetic anhydride yield triacetate derivative. Compound ‘A’ produces an optically active compound (B) and an optically inactive compound $(C)$ on treatment with bromine water and $HNO_3$ respectively, compound (A) is:
Answer: (a)
Solution
The L-isomer reacts with $\mathrm{Br_2|H_2O}$ to form an optically active compound. The same L-isomer reacts with concentrated $\mathrm{HNO_3}$ to form an optically inactive compound.
Question 80
Chemistry · Co-ordination Compounds · Single correct
Match List I with List II
(A)– (III), (B) – (II), $(C)$ – (IV), (D) – (I)
(A)– (IV), (B) – (II), $(C)$ – (III), (D) – (I)
(A)– (IV), (B) – (III), $(C)$ – (II), (D) – (I)
(A)– (III), (B) – (IV), $(C)$ – (I), (D) – (II)
Answer: (b)
Solution
Question 81
Chemistry · The Solid State · Numerical
Metal deficiency defect is shown by $\mathrm{Fe}_{0.93}\mathrm{O}$. In the crystal, some $\mathrm{Fe}^{2+}$ cations are missing and loss of positive charge is compensated by the presence of $\mathrm{Fe}^{3+}$ ions. The percentage of $\mathrm{Fe}^{2+}$ ions in the $\mathrm{Fe}_{0.93}\mathrm{O}$ crystals is . (Nearest integer)
Answer: 85
Solution
In $\mathrm{Fe}_{0.93}\mathrm{O}$ for every 93 Fe ions, 14 are $\mathrm{Fe}^{+3}$ and $(93 - 14) = 79$ are $\mathrm{Fe}^{+2}$ ions. Therefore, $\% \mathrm{Fe}^{+2} = \frac{79}{93} \times 100 = 84.9\%$. Thus, nearest integer = 85$\%$.
Question 82
Chemistry · Structure of Atom · Numerical
If the uncertainty in velocity and position of a minute particle in space are, $2.4 \times 10^{-26} \, (\mathrm{ms}^{-1})$ and $10^{-7} \, (\mathrm{m})$ respectively. The mass of the particle of g is _________ (Nearest integer) (Given : $h = 6.626 \times 10^{-34} \, \mathrm{Js}$)
Answer: 22
Solution
Given $\Delta V = 2.4 \times 10^{-26} \, \mathrm{ms^{-1}}$ and $\Delta x = 10^{-7} \, \mathrm{m}$. Therefore, $\Delta p \cdot \Delta x = \frac{h}{4\pi}$. Thus, $m \Delta V \cdot \Delta x = \frac{h}{4\pi}$. $$\Rightarrow m \times 2.4 \times 10^{-26} \times 10^{-7} = \frac{6.626 \times 10^{-34}}{4 \times \pi}$$ $$m = \frac{6.626}{9.6 \times \pi} \times 10^{-1}$$ $$m = 0.02198 \, \mathrm{kg}$$ $$m = 21.98 \, \mathrm{gm}$$ nearest integer = 22
Question 83
Chemistry · Solutions · Numerical
$2\,\mathrm{g}$ of a non-volatile non-electrolyte solute is dissolved in $200\,\mathrm{g}$ of two different solvents A and B whose ebullioscopic constants are in the ratio of $1:8$. The elevation in boiling points of A and B are in the ratio $\dfrac{x}{y}$ $(x:y)$. The value of $y$ is
$2\mathrm{NOCl(g)} \rightleftharpoons 2\mathrm{NO(g)} + \mathrm{Cl_2(g)}$ In an experiment, $2.0$ moles of $\mathrm{NOCl}$ was placed in a one-litre flask and the concentration of $\mathrm{NO}$ after equilibrium established, was found to be $0.4\ \mathrm{mol/L}$. The equilibrium constant at $30°\mathrm{C}$ is \_\_\_\_\_ $\times 10^{-4}$.
Answer: 125
Solution
The reaction is given by $2\mathrm{NOCl}(g) \rightleftharpoons 2\mathrm{NO}(g) + \mathrm{Cl_2}(g)$. Initially, at $t=0$, the concentrations are $2\,\mathrm{M}$ for $\mathrm{NOCl}$ and $0$ for both $\mathrm{NO}$ and $\mathrm{Cl_2}$. At equilibrium, $t=eq$, the concentrations are $(2-x)\,\mathrm{M}$ for $\mathrm{NOCl}$, $x\,\mathrm{M}$ for $\mathrm{NO}$, and $\frac{x}{2}\,\mathrm{M}$ for $\mathrm{Cl_2}$. Therefore, $x = 0.4\,\mathrm{M}$. Thus, $[\mathrm{NOCl}]_{eq} = 1.6\,\mathrm{M}$, $[\mathrm{NO}]_{eq} = 0.4\,\mathrm{M}$, and $[\mathrm{Cl_2}]_{eq} = 0.2\,\mathrm{M}$. The equilibrium constant $K_c$ is given by: $$K_c = \frac{[\mathrm{NO}]^2[\mathrm{Cl_2}]}{[\mathrm{NOCl}]^2} = \frac{[0.4]^2[0.2]}{[1.6]^2}$$ Calculating this gives: $$K_c = \frac{32}{2.56} \times 10^{-3}$$ $$K_c = 12.5 \times 10^{-3}$$ $$K_c = 125 \times 10^{-4}$$ The integer answer is 125.
Question 85
Chemistry · Electrochemistry · Fill in the blank
The limiting molar conductivities of $\mathrm{NaI}$, $\mathrm{NaNO_3}$ and $\mathrm{AgNO_3}$ are $12.7$, $12.0$ and $13.3\,\mathrm{mS\,m^2\,mol^{-1}}$, respectively (all at $25^\circ\mathrm{C}$). The limiting molar conductivity of $\mathrm{AgI}$ at this temperature is $\underline{\hspace{1cm}}$.
Answer: 14
Solution
Given $\lambda_m^\infty(\mathrm{NaI})=12.7\,\mathrm{mS\,m^2\,mol^{-1}}$ $\lambda_m^\infty(\mathrm{NaNO_3})=12.0\,\mathrm{mS\,m^2\,mol^{-1}}$ $\lambda_m^\infty(\mathrm{AgNO_3})=13.3\,\mathrm{mS\,m^2\,mol^{-1}}$ $\lambda_m^\infty(\mathrm{AgI})=12.7+13.3-12.0=14.0\,\mathrm{mS\,m^2\,mol^{-1}}$
Question 86
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
The rate constant for a first order reaction is given by the following equation: $$\ln k = 33.24 - \frac{2.0 \times 10^4 \, \mathrm{K}}{T}$$ The Activation energy for the reaction is given by _____ kJ mol$^{-1}$. (In Nearest integer) (Given: R = 8.3 J K$^{-1}$ mol$^{-1}$)
Answer: 166
Solution
Given the equation $\ln k = \ln A - \frac{E_A}{RT}$. Given: $\ln k = 33.24 - \frac{2.0 \times 10^4}{T}$. Therefore, on comparing $\frac{E_A}{R} = 2.0 \times 10^4$. Thus, $E_A = 2.0 \times 10^4 \times R$. Therefore, $E_A = 2.0 \times 10^4 \times 8.3 \, \mathrm{J}$. Hence, $E_A = 16.6 \times 10^4 \, \mathrm{J} = 166 \, \mathrm{kJ}$.
Question 87
Chemistry · The d-and f-Block Elements · Numerical
The number of statement(s) correct from the following for copper (at no. 29) is/are _______ $(A)$ Cu(II) complexes are always paramagnetic $(B)$ Cu(I) complexes are generally colourless $(C)$ Cu(I) is easily oxidized $(D)$ In Fehling solution, the active reagent has Cu(I)
Cu(II) complexes are always paramagnetic
Cu(I) complexes are generally colourless
Cu(I) is easily oxidized
In Fehling solution, the active reagent has Cu(I)
Answer: (c)
Solution
A, B, C are correct and D is incorrect because Fehling solution has Cu(II).
Question 88
Chemistry · The d-and f-Block Elements · Numerical
Acidified potassium permanganate solution oxidises oxalic acid. The spin-only magnetic moment of the manganese product formed from the above reaction is _____ B.M. (Nearest Integer)
Answer: 6
Solution
Given the reaction: $$2\mathrm{KMnO_4} + 5\mathrm{H_2C_2O_4} + 3\mathrm{H_2SO_4} \rightarrow \mathrm{K_2SO_4} + 2\mathrm{MnSO_4} + 10\mathrm{CO_2} + 8\mathrm{H_2O}$$ $\mathrm{Mn^{2+}}$ has $5$ unpaired electrons therefore the magnetic moment is $\sqrt{35}$ BM.
Question 89
Chemistry · The Solid State · Numerical
Two elements A and B which form 0.15 moles of $\mathrm{A}_2\mathrm{B}$ and $\mathrm{AB}_3$ type compounds. If both $\mathrm{A}_2\mathrm{B}$ and $\mathrm{AB}_3$ weigh equally, then the atomic weight of A is _____ times of atomic weight of B.
Answer: 2
Solution
Given: Molar mass of $\mathrm{A_2B} = \mathrm{AB_3}$. Therefore, $(2A + B) = (A + 3B)$ where $$\begin{align*} A & \rightarrow Atomic wt. of A \\ B & \rightarrow Atomic wt. of B \end{align*}$$ This implies $A = 2B$. Therefore, the atomic weight of A is $2$ times the atomic weight of B. Integer answer is 2.