JEE Main 26 August 2021 Shift 2 question paper with solutions
JEE Main 26 August 2021 Shift 2: 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 · Continuity and Differentiability · Single correct
Let [t] denote the greatest integer less than or equal to t. Let f(x) = x - [x], g(x) = 1 - x + [x], and h(x) = $\min$$\{$f(x), g(x)$\}$, x $\in$ [-2, 2]. Then h is :
continuous in [-2, 2] but not differentiable at more than four points in (-2, 2)
not continuous at exactly three points in [-2, 2]
continuous in [-2, 2] but not differentiable at exactly three points in (-2, 2)
not continuous at exactly four points in [-2, 2]
Answer: (a)
Solution
Given $\min\{x - \lfloor x \rfloor, 1 - x + \lfloor x \rfloor\}$. Define $h(x) = \min\{x - \lfloor x \rfloor, 1 - \lfloor x \rfloor\}$. The function is always continuous in $[-2, 2]$ but non-differentiable at 7 points.
Question 2
Maths · Matrices · Single correct
Let $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{pmatrix}$. Then $A^{2025} - A^{2020}$ is equal to:
Maths · Applications of Derivatives · Single correct
The local maximum value of the function $$f(x) = \left( \frac{2}{x} \right)^{x^2}, \; x > 0,$$ is
$\left( 2 \sqrt{e} \right)^{\frac{1}{e}}$
$\left( \frac{4}{\sqrt{e}} \right)^{\frac{e}{4}}$
$(e)^{\frac{2}{e}}$
$1$
Answer: (c)
Solution
Given $f(x) = \left( \frac{2}{x} \right) x^2$ for $x > 0$. Then $\ln f(x) = x^2 (\ln 2 - \ln x)$. Differentiating, we have $$f'(x) = f(x) \{-x + (\ln 2 - \ln x) 2x \}$$ $$f'(x) = f(x) \underbrace{x (2 \ln 2 - 2 \ln x - 1)}_{g(x)}$$ where $g(x) = 2 \ln 2 - 2 \ln x - 1$. Solving $\ln \frac{4}{x^2} - 1 = 0$, we find $x = \frac{2}{\sqrt{e}}$. The function is increasing for $x \frac{2}{\sqrt{e}}$. The local maximum (LM) is at $x = \frac{2}{\sqrt{e}}$. The local maximum value is $$\left( \frac{2}{2/\sqrt{e}} \right)^{\frac{4}{e}} \Rightarrow e^{\frac{2}{e}}.$$
Question 4
Maths · Integrals · Single correct
If the value of the integral $\int_{0}^{5} \frac{x + \lfloor x \rfloor}{e^{x - \lfloor x \rfloor}} \, dx = \alpha e^{-1} + \beta$, where $\alpha, \beta \in \mathbb{R}$, $5\alpha + 6\beta = 0$, and $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$; then the value of $(\alpha + \beta)^2$ is equal to:
100
25
16
36
Answer: (b)
Solution
Given $$I = \int_0^5 \frac{x + \lfloor x \rfloor}{e^{x - \lfloor x \rfloor}} \, dx$$ We can split the integral as follows: $$\int_0^1 \frac{x}{e^x} \, dx + \int_1^2 \frac{x + 1}{e^{x-1}} \, dx + \int_2^3 \frac{x + 2}{e^{x-2}} \, dx + \ldots + \int_4^5 \frac{x + 4}{e^{x-4}} \, dx$$ Substituting, we have: For $x = t + 1$, $x = z + 2$, $x = y + 4$: $$\int_0^1 \frac{t + 2}{e^t} \, dt + \int_0^1 \frac{z + 4}{e^z} \, dz + \ldots + \int_0^1 \frac{y + 8}{e^y} \, dy$$ This implies: $$\Rightarrow \int_0^5 \frac{5x + 20}{e^x} \, dt = 5 \int_0^1 \frac{x + 4}{e^x} \, dx$$ $$\Rightarrow 5 \int_0^1 (x + 4) e^{-x} \, dx$$ $$\Rightarrow 5e^{-x}(-x - 5) \bigg|_0^1 \Rightarrow -\frac{30}{e} + 25$$ Let $\alpha = -30$ and $\beta = 25$. Then $5\alpha + 6\beta = 0$. $$(\alpha + \beta)^2 = 5^2 = 25$$
Question 5
Maths · Conic Sections · Single correct
The point $\mathrm{P}(-2\sqrt{6}, \sqrt{3})$ lies on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ having eccentricity $\frac{\sqrt{5}}{2}$. If the tangent and normal at $\mathrm{P}$ to the hyperbola intersect its conjugate axis at the point $\mathrm{Q}$ and $\mathrm{R}$ respectively, then $\mathrm{QR}$ is equal to:
$4\sqrt{3}$
$6$
$6\sqrt{3}$
$3\sqrt{6}$
Answer: (c)
Solution
P $(-2\sqrt{6}, \sqrt{3})$ lies on hyperbola $$\Rightarrow \frac{24}{a^2} - \frac{3}{b^2} = 1 \ldots (i)$$ $$e = \frac{\sqrt{5}}{2} \Rightarrow b^2 = a^2 \left(\frac{5}{4} - 1\right) \Rightarrow 4b^2 = a^2$$ Put in (i) $$\Rightarrow \frac{6}{b^2} - \frac{3}{b^2} = 1 \Rightarrow b = \sqrt{3}$$ $$\Rightarrow a = \sqrt{12}$$ $$\frac{x^2}{12} - \frac{y^2}{3} = 1$$ Tangent at $P$: $$-\frac{x}{\sqrt{6}} - \frac{y}{\sqrt{3}} = 1 \Rightarrow Q(0, \sqrt{3})$$ Slope of $T = -\frac{1}{\sqrt{2}}$ Normal at $P$: $$y - \sqrt{3} = \sqrt{2}(x + 2\sqrt{6})$$ $$\Rightarrow R = (0, 5\sqrt{3})$$ $$QR = 6\sqrt{3}$$
Question 6
Maths · Differential Equations · Single correct
Let $y(x)$ be the solution of the differential equation $$2x^2 \, dy + (e^y - 2x) \, dx = 0, x > 0.$$ If $y(e) = 1$, then $y(1)$ is equal to:
0
2
$\log$_e 2
$\log$_e (2e)
Answer: (c)
Solution
Given the equation $2x^2 dy + (e^y - 2x) dx = 0$. We have $$\frac{dy}{dx} + \frac{e^y - 2x}{2x^2} = 0 \implies \frac{dy}{dx} + \frac{e^y}{2x^2} - \frac{1}{x} = 0$$ Multiply by $e^{-y}$: $$e^{-y} \frac{dy}{dx} - \frac{e^{-y}}{x} = -\frac{1}{2x^2} \implies Put e^{-y} = z$$ Then, $$-\frac{dz}{dx} - \frac{z}{x} = -\frac{1}{2x^2} \implies xdz + zdx = \frac{dx}{2x}$$ Integrate: $$d(xz) = \frac{dx}{2x} \implies xz = \frac{1}{2} \log_e x + c$$ Thus, $$xe^{-y} = \frac{1}{2} \log_e x + c, passes through (e, 1)$$ Therefore, $$C = \frac{1}{2}$$ So, $$xe^{-y} = \frac{\log_e x}{2}$$ Finally, $$e^{-y} = \frac{1}{2} \implies y = \log_e 2$$
Question 7
Maths · Mathematical Reasoning · Single correct
Consider the two statements: (S1) : (p $\rightarrow$ q) $\lor$ ($\sim$ q $\rightarrow$ p) is a tautology. (S2) : (p $\land$ $\sim$ q) $\land$ ($\sim$ p $\lor$ q) is a fallacy. Then :
only (S1) is true.
both (S1) and (S2) are false.
both (S1) and (S2) are true.
only (S2) is true.
Answer: (c)
Solution
$S_1$: $(\sim p \lor q) \lor (q \lor p) = (q \lor \sim p) \lor (q \lor p)$ $S_1 = q \lor (\sim p \lor p) = q \lor t = t = \text{tautology}$ $S_2$: $(p \land \sim q) \land (\sim p \lor q) = (p \land \sim q) \land \sim (p \land \sim q) = c = \text{fallacy}$
Question 8
Maths · Inverse Trigonometric Functions · Single correct
The domain of the function $\operatorname{cosec}^{-1}\!\left(\dfrac{1+x}{x}\right)$ is:
(-1, -$\frac{1}{2}$] $\cup$ (0, $\infty$)
[-$\frac{1}{2}$, 0) $\cup$ [1, $\infty$)
(-$\frac{1}{2}$, $\infty$) - {0}
[-$\frac{1}{2}$, $\infty$) - {0}
Answer: (d)
Solution
The solution involves determining the correct interval for the given expressions. For the expression $\frac{1+x}{x}$, the interval is $(-\infty, -1] \cup [1, \infty)$. For the expression $\frac{1}{x}$, the interval is $(-\infty, -2] \cup [0, \infty)$. For $x$, the interval is $\left[-\frac{1}{2}, 0\right) \cup (0, \infty)$. For $x$, the interval is $\left[-\frac{1}{2}, \infty\right) - \{0\}$.
Question 9
Maths · Probability · Single correct
A fair die is tossed until six is obtained on it. Let $X$ be the number of required tosses, then the conditional probability $\mathbb{P}(X \geq 5 \mid X > 2)$ is:
Maths · Inverse Trigonometric Functions · Single correct
If $\sum_{r=1}^{50}$ $\tan^{-1}$ $\frac{1}{2r^2}$ = p, then the value of $\tan$ p is :
$\frac{101}{102}$
$\frac{50}{51}$
100
$\frac{51}{50}$
Answer: (b)
Solution
The given expression is $$\sum_{r=1}^{50} \tan^{-1}\left(\frac{2}{4r^2}\right) = \sum_{r=1}^{50} \tan^{-1}\left(\frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}\right)$$ This simplifies to $$\sum_{r=1}^{50} \tan^{-1}(2r+1) - \tan^{-1}(2r-1)$$ Evaluating the series, we get $$\tan^{-1}(101) - \tan^{-1} 1 \Rightarrow \tan^{-1} \frac{50}{51}$$
Question 11
Maths · Determinants · Single correct
Two fair dice are thrown. The numbers on them are taken as $\lambda$ and $\mu$, and a system of linear equations $$x + y + z = 5$$ $$x + 2y + 3z = \mu$$ $$x + 3y + \lambda z = 1$$ is constructed. If $p$ is the probability that the system has a unique solution and $q$ is the probability that the system has no solution, then :
The locus of the mid points of the chords of the hyperbola $x^2 - y^2 = 4$, which touch the parabola $y^2 = 8x$, is :
$y^3(x - 2) = x^2$
$x^3(x - 2) = y^2$
$y^2(x - 2) = x^3$
$x^2(x - 2) = y^3$
Answer: (c)
Solution
Given $T = S_1$. The equation is $xh - yk = h^2 - k^2$. Solving for $y$, we have: $$y = \frac{x h}{k} - \frac{(h^2 - k^2)}{k}$$ This touches $y^2 = 8x$ then $c = \frac{a}{m}$. We have: $$\left( \frac{k^2 - h^2}{k} \right) = \frac{2k}{h}$$ Thus: $$2y^2 = x \left( y^2 - x^2 \right)$$ Finally: $$y^2 (x - 2) = x^3$$
Question 13
Maths · Trigonometric Functions · Single correct
The value of $$2 \sin\left(\frac{\pi}{8}\right) \sin\left(\frac{2\pi}{8}\right) \sin\left(\frac{3\pi}{8}\right) \sin\left(\frac{5\pi}{8}\right) \sin\left(\frac{6\pi}{8}\right) \sin\left(\frac{7\pi}{8}\right)$$ is:
$\frac{1}{4\sqrt{2}}$
$\frac{1}{4}$
$\frac{1}{8}$
$\frac{1}{8\sqrt{2}}$
Answer: (c)
Solution
Given the expression: $$2 \sin \left( \frac{\pi}{8} \right) \sin \left( \frac{2\pi}{8} \right) \sin \left( \frac{3\pi}{8} \right) \sin \left( \frac{5\pi}{8} \right) \sin \left( \frac{6\pi}{8} \right) \sin \left( \frac{7\pi}{8} \right)$$ Simplify it step by step: $$2 \sin^2 \frac{\pi}{8} \sin^2 \frac{2\pi}{8} \sin^2 \frac{3\pi}{8}$$ This can be rewritten as: $$\sin^2 \frac{\pi}{8} \sin^2 \frac{3\pi}{8}$$ Using the identity $\sin^2 x = \cos^2 x$ for complementary angles: $$\sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}$$ This simplifies to: $$\frac{1}{4} \sin^2 \left( \frac{\pi}{4} \right) = \frac{1}{8}$$
Question 14
Maths · Complex Numbers and Quadratic Equations · Single correct
If $\left( \sqrt{3} + i \right)^{100} = 2^{99}(p + iq)$, then $p$ and $q$ are roots of the equation:
$x^2 - (\sqrt{3} - 1)x - \sqrt{3} = 0$
$x^2 + (\sqrt{3} + 1)x + \sqrt{3} = 0$
$x^2 + (\sqrt{3} - 1)x - \sqrt{3} = 0$
$x^2 - (\sqrt{3} + 1)x + \sqrt{3} = 0$
Answer: (a)
Solution
Given $\left(2 e^{i \pi / 6}\right)^{100} = 2^{99} (p + iq)$. We have: $$2^{100} \left( \cos \frac{50 \pi}{3} + i \sin \frac{50 \pi}{3} \right) = 2^{99} (p + iq)$$ This implies: $$p + iq = 2 \left( \cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3} \right)$$ Therefore, $p = -1$, $q = \sqrt{3}$. The equation is: $$x^2 - (\sqrt{3} - 1)x - \sqrt{3} = 0.$$
Question 15
Maths · Three Dimensional Geometry · Single correct
A hall has a square floor of dimension $10 \, \mathrm{m} \times 10 \, \mathrm{m}$ (see the figure) and vertical walls. If the angle GPH between the diagonals AG and BH is $\cos^{-1} \frac{1}{5}$, then the height of the hall (in meters) is:
5
2$\sqrt{10}$
5$\sqrt{3}$
5$\sqrt{2}$
Answer: (d)
Solution
Given points: A $(\hat{j})$, B $(10\hat{i})$ H $(h\hat{i} + 10\hat{k})$ G $(10\hat{i} + h\hat{j} + 10\hat{k})$ Vectors: $$\overrightarrow{AG} = 10\hat{i} + h\hat{j} + 10\hat{k}$$ $$\overrightarrow{BH} = -10\hat{i} + h\hat{j} + 10\hat{k}$$ The cosine of the angle $\theta$ is given by: $$\cos \theta = \frac{\overrightarrow{AG} \cdot \overrightarrow{BH}}{|\overrightarrow{AG}||\overrightarrow{BH}|}$$ Solving for $h$: $$\frac{1}{5} = \frac{h^2}{h^2 + 200}$$ $$4h^2 = 200 \Rightarrow h = 5\sqrt{2}$$
Question 16
Maths · Three Dimensional Geometry · Single correct
Let P be the plane passing through the point (1, 2, 3) and the line of intersection of the planes $$\vec{r} \cdot (\hat{i} + \hat{j} + 4\hat{k}) = 16$$ and $$\vec{r} \cdot (-\hat{i} + \hat{j} + \hat{k}) = 6$$. Then which of the following points does NOT lie on P?
(3, 3, 2)
(6, -6, 2)
(4, 2, 2)
(-8, 8, 6)
Answer: (c)
Solution
Given $$(x + y + 4z - 16) + \lambda (-x + y + z - 6) = 0$$ Passes through $(1, 2, 3)$ $$-1 + \lambda (-2) \Rightarrow \lambda = -\frac{1}{2}$$ $$2(x + y + 4z - 16) - (-x + y + z - 6) = 0$$ $$3x + y + 7z - 26 = 0$$
Question 17
Maths · Inverse Trigonometric Functions · Single correct
A 10 inches long pencil AB with mid point C and a small eraser P are placed on the horizontal top of a table such that PC = $\sqrt{5}$ inches and $\angle$ PCB = $\tan^{-1}(2)$. The acute angle through which the pencil must be rotated about C so that the perpendicular distance between eraser and pencil becomes exactly 1 inch is:
The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{1+\sin^2 x}{1+\pi^{sin x}} \right) dx$ is
$\frac{\pi}{2}$
$\frac{5\pi}{4}$
$\frac{3\pi}{4}$
$\frac{3\pi}{2}$
Answer: (c)
Solution
Given $$I = \int_0^{\pi/2} \frac{(1+\sin^2 x)}{(1+\pi^{\sin x})} + \frac{\pi^{\sin x} (1+\sin^2 x)}{(1+\pi^{\sin x})} \, dx$$ Simplifying, we have $$I = \int_0^{\pi/2} (1 + \sin^2 x) \, dx$$ Evaluating the integral, $$I = \frac{\pi}{2} + \frac{\pi}{2} \cdot \frac{1}{2} = \frac{3\pi}{4}$$
Question 19
Maths · Conic Sections · Single correct
A circle $C$ touches the line $x = 2y$ at the point $(2, 1)$ and intersects the circle $C_1 : x^2 + y^2 + 2y - 5 = 0$ at two points $P$ and $Q$ such that $PQ$ is a diameter of $C_1$. Then the diameter of $C$ is :
The sum of all 3-digit numbers less than or equal to 500, that are formed without using the digit "1" and they all are multiple of 11, is.
Answer: 7744
Solution
Given the sequence 209, 220, 231, $\ldots$, 495. The sum is calculated as: $$Sum = \frac{27}{2} (209 + 495) = 9504$$ Number containing 1 at unit place: $$231, 341, 451$$ Number containing 1 at 10^{th} place: $$319, 418$$ Required = 9501 - (231 + 341 + 451 + 319 + 418) The result is 7744.
Question 22
Maths · Applications of Integrals · Numerical
Let a and b respectively be the points of local maximum and local minimum of the function $f(x) = 2x^3 - 3x^2 - 12x$. If $A$ is the total area of the region bounded by $y = f(x)$, the x-axis and the lines $x = a$ and $x = b$, then $4A$ is equal to .
Answer: 114
Solution
Given $f'(x) = 6x^2 - 6x - 12 = 6(x - 2)(x + 1)$. The points are $(2, -20)$ and $(-1, 7)$. The area $A$ is calculated as follows: $$A = \int_{-1}^{0} (2x^3 - 3x^2 - 12x) \, dx + \int_{0}^{2} (12x + 3x^2 - 2x^3) \, dx$$ Evaluating the integrals, we have: $$A = \left( \frac{x^4}{2} - x^3 - 6x^2 \right)_{-1}^{0} + \left( 6x^2 + x^3 - \frac{x^4}{2} \right)_{0}^{2}$$ Finally, $4A = 114$.
Question 23
Maths · Vector Algebra · Numerical
If the projection of the vector $\hat{i} + 2\hat{j} + \hat{k}$ on the sum of the two vectors $2\hat{i} + 4\hat{j} - 5\hat{k}$ and $-\lambda \hat{i} + 2\hat{j} + 3\hat{k}$ is 1, then $\lambda$ is equal to .
Let $a_1, a_2, \ldots, a_{10}$ be an $AP$ with common difference $-3$ and $b_1, b_2, \ldots, b_{10}$ be a $GP$ with common ratio $2$. Let $c_k = a_k + b_k, k = 1, 2, \ldots, 10$. If $c_2 = 12$ and $c_3 = 13$, then $\sum_{k=1}^{10} c_k$ is equal to.
Let Q be the foot of the perpendicular from the point P(7, -2, 13) on the plane containing the lines $\frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8}$ and $\frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7}$. Then $(PQ)^2$, is equal to .
Maths · Complex Numbers and Quadratic Equations · Numerical
Let $\lambda \neq 0$ be in $\mathbb{R}$. If $\alpha$ and $\beta$ are the roots of the equation $x^2 - x + 2\lambda = 0$, and $\alpha$ and $\gamma$ are the roots of equation $3x^2 - 10x + 27\lambda = 0$, then $\frac{\beta \gamma}{\lambda}$ is equal to .
Let the mean and variance of four numbers $3, 7, x$ and $y (x > y)$ be $5$ and $10$ respectively. Then the mean of four numbers $3 + 2x, 7 + 2y, x + y$ and $x - y$ is.
Answer: 12
Solution
Given $5 = \frac{3+7+x+y}{4}$, we have $x + y = 10$. The variance is given by $Var(x) = 10 = \frac{3^2 + 7^2 + x^2 + y^2}{4} - 25$. This simplifies to $140 = 49 + 9 + x^2 + y^2$. Thus, $x^2 + y^2 = 82$. We have $x + y = 10$. Solving these equations gives $\Rightarrow (x, y) = (9, 1)$. The four numbers are 21, 9, 10, 8. The mean is $\frac{48}{4} = 12$.
Question 29
Maths · Matrices · Numerical
Let A be a $3 \times 3$ real matrix. If $\det(2 \ \mathrm{Adj}(2 \ \mathrm{Adj}(\mathrm{Adj}(2 \ A)))) = 2^{41}$, then the value of $\det(A^2)$ equal .
Answer: 4
Solution
Given $adj(2\, A) = 2^2 adj A$. Therefore, $adj(adj(2\, A)) = adj(4 adj A) = 16 adj (adj A) = 16 \mid A \mid A$. Thus, $adj(32 \mid A \mid A) = (32 \mid A \mid)^2 adj A$. Therefore, $12(32 \mid A \mid)^2 \mid adj A \mid = 2^3 (32 \mid A \mid)^6 \mid adj A \mid$. Hence, $2^3 \cdot 2^{30} \mid A \mid^6 \cdot \mid A \mid^2 = 2^{41}$. Therefore, $\mid A \mid^8 = 2^8 \Rightarrow \mid A \mid = \pm 2$. Finally, $\mid A \mid^2 = \mid A \mid^2 = 4$.
Question 30
Maths · Complex Numbers and Quadratic Equations · Numerical
The least positive integer $n$ such that $\frac{{(2i)^n}}{{(1-i)^{n-2}}}, \ i = \sqrt{-1}$ is a positive integer, is.
Answer: 6
Solution
Given $\frac{(2i)^n}{(1-i)^{\,n-2}}$ $= \frac{(2i)^n}{(-2i)^{\frac{n-2}{2}}}$ $= \frac{(2i)^{\frac{n+2}{2}}}{(-1)^{\frac{n-2}{2}}}$ $= \frac{2^{\frac{n+2}{2}}\,i^{\frac{n+2}{2}}}{(-1)^{\frac{n-2}{2}}}$ This is a positive integer for $n=6$.
Physics
Question 31
Physics · Thermal Properties of Matter · Single correct
The temperature of equal masses of three different liquids $x$, $y$ and $z$ are $10^\circ \mathrm{C}$, $20^\circ \mathrm{C}$ and $30^\circ \mathrm{C}$ respectively. The temperature of mixture when $x$ is mixed with $y$ is $16^\circ \mathrm{C}$ and that when $y$ is mixed with $z$ is $26^\circ \mathrm{C}$. The temperature of mixture when $x$ and $z$ are mixed will be:
$28.32^\circ \mathrm{C}$
$25.62^\circ \mathrm{C}$
$23.84^\circ \mathrm{C}$
$20.28^\circ \mathrm{C}$
Answer: (c)
Solution
Given: For X: $m_1 = m$ $T_1 = 10^\circ \mathrm{C}$ For Y: $m_2 = m$ $T_2 = 20^\circ \mathrm{C}$ For Z: $m_3 = m$ $T_3 = 30^\circ \mathrm{C}$ When X and Y are mixed, $T_{f_1} = 16^\circ \mathrm{C}$ $$m_1 \, s_1 \, T + m_2 \, s_2 \, T_2 = (m_1 \, s_1 + m_2 \, s_2) \, T_{f_1}$$ $$s_1 \times 10 + s_2 \times 20 = (s_1 + s_2) \times 16$$ $$s_1 = \frac{2}{3} s_2 \ldots (i)$$ When Y and Z are mixed, $T_{f_2} = 26^\circ \mathrm{C}$ $$m_2 \, s_2 \, T + m_3 \, s_3 \, T_3 = (m_3 \, s_3 + m_3 \, s_3) \, T_{f_2}$$ $$s_2 \times 20 + s_3 \times 30 = (s_2 + s_3) \times 26$$ $$s_3 = \frac{3}{2} s_2$$ When X and Z are mixed: $$m_1 \, s_1 \, T_1 + m_3 \, s_3 \, T_3 = (m_1 \, s_1 + m_3 \, s_3) \, T_f$$ $$\frac{2}{3} s_2 \times 10 + \frac{3}{2} s_2 \times 20 = \left( \frac{2}{3} s_2 + \frac{3}{2} s_2 \right) \, T_f$$ $$T_f = 23.84^\circ \mathrm{C}$$
Question 32
Physics · Dual Nature of Radiation and Matter · Single correct
The de-Broglie wavelength of a particle having kinetic energy $E$ is $\lambda$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75$\%$ of the initial value?
$\frac{1}{9}E$
$\frac{7}{9}E$
$E$
$\frac{16}{9}E$
Answer: (b)
Solution
Given $$\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mE}}$$, $$mv = \sqrt{2mE}$$. $$\lambda \propto \frac{1}{\sqrt{E}}$$, $$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{E_1}{E_2}} = \frac{3}{4}$$, $$\lambda_2 = 0.75 \lambda_1$$. $$\frac{E_1}{E_2} = \left(\frac{3}{4}\right)^2$$. $$E_2 = \frac{16}{9} E_1 = \frac{16}{9} E$$. ($\($E_1 = E$\)$) Extra energy given $$= \frac{16}{9} E - E = \frac{7}{9} E$$
Question 33
Physics · Motion in a Plane · Single correct
A particle of mass m is suspended from a ceiling through a string of length L. The particle moves in a horizontal circle of radius r such that $r = \frac{L}{\sqrt{2}}$. The speed of particle will be :
$\sqrt{rg}$
$\sqrt{2rg}$
2$\sqrt{rg}$
$\sqrt{\frac{rg}{2}}$
Answer: (a)
Solution
For a conical pendulum, we have: $$r = \frac{\ell}{\sqrt{2}}$$ The sine of the angle $\theta$ is given by: $$\sin \theta = \frac{r}{\ell} = \frac{1}{\sqrt{2}}$$ Thus, $\theta = 45^\circ$. The tension components are: $$T \sin \theta = \frac{mv^2}{r}$$ $$T \cos \theta = mg$$ Using the tangent of the angle $\theta$: $$\tan \theta = \frac{v^2}{rg} \implies v = \sqrt{rg}$$
Question 34
Physics · Kinetic Theory · Single correct
A cylindrical container of volume $4.0 \times 10^{-3} \, \mathrm{m}^3$ contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is $400 \, \mathrm{K}$. The pressure of the mixture of gases is: [Take gas constant as $8.3 \, \mathrm{J} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}$]
$249 \times 10^1 \, \mathrm{Pa}$
$24.9 \times 10^3 \, \mathrm{Pa}$
$24.9 \times 10^5 \, \mathrm{Pa}$
$24.9 \, \mathrm{Pa}$
Answer: (c)
Solution
Given $V = 4 \times 10^{-3} \, \mathrm{m^3}$, $n = 3 \, \mathrm{moles}$, $T = 400 \, \mathrm{K}$. Using the ideal gas law $PV = nRT$, we have $$P = \frac{nRT}{V}$$ $$P = \frac{3 \times 8.3 \times 400}{4 \times 10^{-3}}$$ $$= 24.9 \times 10^5 \, \mathrm{Pa}$$
Question 35
Physics · Motion in a Plane · Single correct
The angle between vector $\vec{A}$ and $\vec{A} - \vec{B}$ is:
$\tan^{-1}\left(\frac{-B}{A - B \frac{\sqrt{3}}{2}}\right)$
$\tan^{-1}\left(\frac{A}{0.7B}\right)$
$\tan^{-1}\left(\frac{\sqrt{3}B}{2A - B}\right)$
$\tan^{-1}\left(\frac{B \cos \theta}{A - B \sin \theta}\right)$
Answer: (c)
Solution
Angle between $\vec{A}$ and $\vec{B}$, $\theta = 60^\circ$. Angle between $\vec{A}$ and $\vec{A} - \vec{B}$. $$\tan \alpha = \frac{B \sin \theta}{A - B \cos \theta}$$ $$= \frac{B \sqrt{\frac{3}{2}}}{A - B \times \frac{1}{2}}$$ $$\tan \alpha = \frac{\sqrt{3}B}{2A - B}$$
Question 36
Physics · Electromagnetic Waves · Single correct
A light beam is described by $E = 800 \sin \omega \left( t - \frac{x}{c} \right)$ . An electron is allowed to move normal to the propagation of light beam with a speed of $3 \times 10^7 \, \mathrm{ms}^{-1}$. What is the maximum magnetic force exerted on the electron?
Physics · Electric Charges and Fields · Single correct
The two thin coaxial rings, each of radius 'a' and having charges +Q and -Q respectively are separated by a distance of 's'. The potential difference between the centres of the two rings is :
The potential at point A is given by $$V_A = \frac{KQ}{a} - \frac{KQ}{\sqrt{a^2 + s^2}}.$$ The potential at point B is given by $$V_B = -\frac{KQ}{a} + \frac{KQ}{\sqrt{a^2 + s^2}}.$$ The potential difference $V_A - V_B$ is $$V_A - V_B = \frac{2KQ}{a} - \frac{2KQ}{\sqrt{a^2 + s^2}}.$$ This can be simplified to $$= \frac{Q}{2\pi \varepsilon_0} \left( \frac{1}{a} - \frac{1}{s^2 + a^2} \right).$$
Question 38
Physics · Current Electricity · Single correct
If you are provided a set of resistances $2\,\Omega$, $4\,\Omega$, $6\,\Omega$ and $8\,\Omega$. Connect these resistances so as to obtain an equivalent resistance of $\frac{46}{3}\,\Omega$
$4\,\Omega$ and $6\,\Omega$ are in parallel with $2\,\Omega$ and $8\,\Omega$ in series
$6\,\Omega$ and $8\,\Omega$ are in parallel with $2\,\Omega$ and $4\,\Omega$ in series
$2\,\Omega$ and $6\,\Omega$ are in parallel with $4\,\Omega$ and $8\,\Omega$ in series
$2\,\Omega$ and $4\,\Omega$ are in parallel with $6\,\Omega$ and $8\,\Omega$ in series
Answer: (d)
Solution
The resistors of 2 $\($ $\Omega$ $\)$ and 4 $\($ $\Omega$ $\)$ are in parallel. The equivalent resistance $\($ R_1 $\)$ is given by: $$ \frac{1}{R_1} = \frac{1}{2} + \frac{1}{4} $$ $$ \frac{1}{R_1} = \frac{2 + 1}{4} = \frac{3}{4} $$ $$ R_1 = \frac{4}{3} \Omega $$ This equivalent resistance $\($ R_1 $\)$ is in series with the 6 $\($ $\Omega$ $\)$ resistor. The total resistance $\($ R_2 $\)$ is: $$ R_2 = R_1 + 6 = \frac{4}{3} + 6 = \frac{4}{3} + \frac{18}{3} = \frac{22}{3} \Omega $$ Finally, $\($ R_2 $\)$ is in series with the 8 $\($ $\Omega$ $\)$ resistor. The total resistance $\($ R_T $\)$ is: $$ R_T = R_2 + 8 = \frac{22}{3} + 8 = \frac{22}{3} + \frac{24}{3} = \frac{46}{3} \Omega $$
Question 39
Physics · System of Particles and Rotational Motion · Single correct
The solid cylinder of length 80 cm and mass M has a radius of 20 cm. Calculate the density of the material used if the moment of inertia of the cylinder about an axis CD parallel to AB as shown in figure is 2.7 kg m^2.
Physics · Electrostatic Potential and Capacitance · Single correct
A parallel-plate capacitor with plate area $A$ has separation $d$ between the plates. Two dielectric slabs of dielectric constant $K_1$ and $K_2$ of same area $A/2$ and thickness $d/2$ are inserted in the space between the plates. The capacitance of the capacitor will be given by:
The equivalent capacitance is given by $$C_{eq} = \frac{A}{2} \frac{\varepsilon_0}{d} + \frac{A \varepsilon_0}{d} \frac{K_1 K_2}{K_1 + K_2}$$ This simplifies to $$= \frac{A \varepsilon_0}{d} \left( \frac{1}{2} + \frac{K_1 K_2}{K_1 + K_2} \right)$$
Question 41
Physics · Motion in a Plane · Single correct
A bomb is dropped by fighter plane flying horizontally. To an observer sitting in the plane, the trajectory of the bomb is a :
hyperbola
parabola in the direction of motion of plane
straight line vertically down the plane
parabola in a direction opposite to the motion of plane
Answer: (c)
Solution
The velocity at point B is given by $$\mathbf{v}_B = u_0 \hat{\mathbf{i}} - gt \hat{\mathbf{j}}.$$ The relative velocity of B with respect to P is $$\overrightarrow{\mathbf{V}_{B/P}} = \overrightarrow{\mathbf{V}_B} - \overrightarrow{\mathbf{V}_p}.$$ The relative velocity of R with respect to P is $$\overrightarrow{\mathbf{V}_{R/P}} = -8t \hat{\mathbf{j}}.$$ The path is a straight line vertically down.
Question 42
Physics · Nuclei · Single correct
At time $t = 0$, a material is composed of two radioactive atoms A and B, where $N_A(0) = 2 \, N_B(0)$. The decay constant of both kind of radioactive atoms is $\lambda$. However, A disintegrates to B and B disintegrates to C. Which of the following figures represents the evolution of $N_B(t)/N_B(0)$ with respect to time $t$ ? $\begin{aligned} N_A(0) &= No. of A atoms at t = 0 \\ N_B(0) &= No. of B atoms at t = 0 \end{aligned}$
A transmitting antenna at top of a tower has a height of 50 m and the height of receiving antenna is 80 m. What is range of communication for Line of Sight (LoS) mode ? [use radius of earth = 6400 km ]
45.5 km
80.2 km
144.1 km
57.28 km
Answer: (d)
Solution
Given the diagram, we have the equation for $d_1$ as follows: $$d_1 = \sqrt{2Rh_1} + \sqrt{2Rh_2}$$ This can be rewritten as: $$= \sqrt{2R} \left( \sqrt{h_1} + \sqrt{h_2} \right)$$ Substituting the given values: $$= \left( 2 \times 6400 \times 10^3 \right)^{1/2} \left( \sqrt{50} + \sqrt{80} \right)$$ Calculating further: $$= 3578(7.07 + 8.94)$$ Finally, we get: $$= 57.28 \, \mathrm{Km}$$
Question 44
Physics · Thermodynamics · Single correct
A refrigerator consumes an average $35 \, \mathrm{W}$ power to operate between temperature $-10^\circ \mathrm{C}$ to $25^\circ \mathrm{C}$. If there is no loss of energy then how much average heat per second does it transfer?
$263 \, \mathrm{J/s}$
$298 \, \mathrm{J/s}$
$350 \, \mathrm{J/s}$
$35 \, \mathrm{J/s}$
Answer: (a)
Solution
Question 45
Physics · Current Electricity · Single correct
An electric bulb of 500 watt at 100 volt is used in a circuit having a 200 V supply. Calculate the resistance R to be connected in series with the bulb so that the power delivered by the bulb is 500 W.
20$\Omega$
30$\Omega$
5$\Omega$
10$\Omega$
Answer: (a)
Solution
Given the power equation $P = Vi$, we have $500 = Vi$. Solving for current $i$, we get $i = 5 \, \mathrm{Amp}$. Using Ohm's law $V = i \times R$, we find $R = 20$.
Question 46
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
Four NOR gates are connected as shown in figure. The truth table for the given figure is :
Answer: (d)
Solution
The expression for the circuit is given by: $$y = \overline{(A + \overline{A + B}) + (B + \overline{A + B})}$$ This simplifies to: $$y = \overline{(A + \overline{A + B})} \cdot \overline{(B + \overline{A + B})}$$ The truth table is as follows: $\begin{array}{|c|c|c|}$ $\hline$ A & B & y $\\$ $\hline$ 0 & 0 & 1 $\\$ 0 & 1 & 0 $\\$ 1 & 0 & 0 $\\$ 1 & 1 & 1 $\\$ $\hline$ $\end{array}$
Question 47
Physics · Current Electricity · Single correct
Match List-I with List-II. Choose the most appropriate answer from the options given below:
$(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)$
$(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)$
$(a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)$
$(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)$
Answer: (d)
Solution
(a) Magnetic Induction = $M T^{-A} A^{-}$ (b) Magnetic Flux = $M L^2 T^{-2} A^{-1}$ (c) Magnetic Permeability = $M L T^{-2} A^{-2}$ (d) Magnetization = $M^{0} L^{-1} A$ Ans. (4)
Question 48
Physics · Alternating Current · Single correct
In the given circuit the AC source has $\omega = 100 \, \mathrm{rads}^{-1}$. Considering the inductor and capacitor to be ideal, what will be the current $I$ flowing through the circuit?
If the length of the pendulum in pendulum clock increases by 0.1$\%$, then the error in time per day is
$86.4 \, \mathrm{s}$
$4.32 \, \mathrm{s}$
$43.2 \, \mathrm{s}$
$8.64 \, \mathrm{s}$
Answer: (c)
Solution
Given $T = 2\pi \sqrt{\frac{\ell}{g}}$. $$\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta \ell}{\ell}$$ $$\Delta T = \frac{1}{2} \times \frac{0.1}{100} \times 24 \times 3600$$ $$\Delta T = 43.2$$
Question 50
Physics · Mechanical Properties of Solids · Single correct
Two blocks of masses 3 kg and 5 kg are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is $\frac{24}{\pi} \times 10^2 \, \mathrm{Nm}^{-2}$. What is the minimum radius of the wire? (Take $g = 10 \, \mathrm{ms}^{-2}$)
125 cm
1250 cm
12.5 cm
1.25 cm
Answer: (c)
Solution
The tension $T$ is given by $$T = \frac{2 \, m_1 \, m_2 \, g}{m_1 + m_2} = \frac{2 \times 3 \times 5 \times 10}{8}$$ which simplifies to $$= \frac{75}{2}$$ The stress is given by $$Stress = \frac{T}{A}$$ Substituting the values, we have $$\frac{24}{\pi} \times 10^2 = \frac{75}{2 \times \pi R^2}$$ Solving for $R^2$, we get $$R^2 = \frac{75}{2 \times 24 \times 100} = \frac{3}{8 \times 24}$$ Thus, $$\Rightarrow R = 0.125 \, m$$ Therefore, $R = 12.5 \, cm$.
Question 51
Physics · Waves · Numerical
Two waves are simultaneously passing through a string and their equations are : $$y_1 = A_1 \sin k(x - vt), y_2 = A_2 \sin k(x - vt + x_0)$$. Given amplitudes $A_1 = 12 \, \mathrm{mm}$ and $A_2 = 5 \, \mathrm{mm}$ $x_0 = 3.5 \, \mathrm{cm}$ and wave number $k = 6.28 \, \mathrm{cm}^{-1}$. The amplitude of resulting wave will be mm
Answer: 7
Solution
Given $$y_1 = A_1 \sin k(x - vt)$$ $$y_1 = 12 \sin 6.28(x - vt)$$ $$y_2 = 5 \sin 6.28(x - vt + 3.5)$$ The phase difference is $$\Delta \phi = \frac{2\pi}{\lambda}(\Delta x)$$ $$= K(\Delta x)$$ $$= 6.28 \times 3.5 = \frac{7}{2} \times 2\pi = 7\pi$$ The net amplitude is given by $$A_{nd} = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$$ Substituting the values, we get $$A_{net} = \sqrt{(12)^2 + (5)^2 + 2(12)(5) \cos(7\pi)}$$ $$= \sqrt{144 + 25 - 120}$$
Question 52
Physics · Wave Optics · Numerical
A source of light is placed in front of a screen. Intensity of light on the screen is I. Two Polaroids $P_1$ and $P_2$ are so placed in between the source of light and screen that the intensity of light on screen is $I/2$. $P_2$ should be rotated by an angle of (degrees) so that the intensity of light on the screen becomes $\frac{3I}{8}$.
Physics · Moving Charges and Magnetism · Numerical
If the maximum value of accelerating potential provided by a radio frequency oscillator is $12\ \mathrm{kV}$, the number of revolutions made by a proton in a cyclotron to achieve one sixth of the speed of light is $m_{p}=1.67\times10^{-27}\ \mathrm{kg}$ $e=1.6\times10^{-19}\ \mathrm{C}$ Speed of light $=3\times10^{8}\ \mathrm{m/s}$
The acceleration due to gravity is found upto an accuracy of 4$\%$ on a planet. The energy supplied to a simple pendulum to known mass 'm' to undertake oscillations of time period $T$ is being estimated. If time period is measured to an accuracy of 3$\%$, the accuracy to which $E$ is known as ..........%
A circular coil of radius $8.0\,\mathrm{cm}$ and $20$ turns is rotated about its vertical diameter with an angular speed of $50\,\mathrm{rad\,s}^{-1}$ in a uniform horizontal magnetic field of $3.0 \times 10^{-2}\,\mathrm{T}$. The maximum emf induced in the coil will be $\ldots \times 10^{-2}\,\mathrm{volt}$ (rounded off to the nearest integer).
Answer: 60
Solution
Maximum emf $\varepsilon = N \omega AB$ Given $N = 20$, $\omega = 50$, $B = 3 \times 10^{-2} \, \mathrm{T}$ $$\varepsilon = 20 \times 50 \times \pi \times (0.08)^2 \times 3 \times 10^{-2} = 60.28 \times 10^{-2}$$ Rounded off to nearest integer $= 60$ Ans. 60
Question 56
Physics · Oscillations · Numerical
Two simple harmonic motions are represented by the equations $$x_1 = 5 \sin \left( 2\pi t + \frac{\pi}{4} \right)$$ and $$x_2 = 5\sqrt{2}(\sin 2\pi t - \cos 2\pi t)$$ The amplitude of second motion is times the amplitude in first motion.
Answer: 2
Solution
Given $$x_2 = 5 \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin 2\pi t + \frac{1}{\sqrt{2}} \cos 2\pi t \right) \sqrt{2}$$ This simplifies to $$= 10 \sin \left( 2\pi t + \frac{\pi}{4} \right)$$ Therefore, $$\frac{A_2}{A_1} = \frac{10}{5} = 2$$
Question 57
Physics · Moving Charges and Magnetism · Numerical
A coil in the shape of an equilateral triangle of side 10 cm lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field 20 $\mathrm{mT}$. The torque acting on the coil when a current of 0.2 $\mathrm{A}$ is passed through it and its plane becomes parallel to the magnetic field will be $\sqrt{x} \times 10^{-5}$ $\mathrm{Nm}$. The value of $x$ is.
Answer: 3
Solution
The torque $\vec{\tau}$ is given by $\vec{M} \times \vec{B} = MB \sin 90^\circ$. Therefore, $$MB = \frac{i \sqrt{3} \ell^2}{4} - B$$ which simplifies to $$= \sqrt{3} \times 10^{-5} \, \mathrm{N \cdot m}.$$
Physics · Ray Optics and Optical Instruments · Numerical
An object is placed at a distance of 12 cm from a convex lens. A convex mirror of focal length 15 cm is placed on other side of lens at 8 cm as shown in the figure. Image of object coincides with the object. When the convex mirror is removed, a real and inverted image is formed at a position. The distance of the image from the object will be (cm)
Answer: 50
Solution
For the object to coincide with the image, the light must fall perpendicularly to the mirror. Which means that the light will have to converge at C of the mirror. Without the mirror also, the light would converge at C. So the distance is: $12 + 8 + 30 = 50 \, \mathrm{cm}$
Question 60
Physics · Laws of Motion · Numerical
The coefficient of static friction between two blocks is 0.5 and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is $\ldots$$\ldots$ N. (take $g = 10 \, \mathrm{ms}^{-2}$)
Answer: 15
Solution
Given $F = 3a$ for the system. The maximum static friction $f_{s_{max}} = 1a$ for the 1 kg block. Using $\mu \times 1 \times g = a$, we have $\mu = 0.5$ and $g = 10 \, \mathrm{m/s^2}$. Therefore, $a = 5$. Thus, $F = 15 \, \mathrm{N}$.
Chemistry
Question 61
Chemistry · Alcohols, Phenols and Ethers · Single correct
Which one of the following phenols does not give colour when condensed with phthalic anhydride in presence of conc. $\mathrm{H_2SO_4}$ ?
Answer: (b)
Solution
Only p-methyl, phenol does not give any colour with phthalic anhydroxide with cons. $\mathrm{H_2SO_4}$.
Question 62
Chemistry · Environmental Chemistry · Single correct
Given below are two statements : one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$. Assertion $(A)$ : Photochemical smog causes cracking of rubber. Reason $(R)$ : Presence of ozone, nitric oxide, acrolein, formaldehyde and peroxyacetyl nitrate in photochemical smog makes it oxidizing. Choose the most appropriate answer from the options given below :
Both $(A)$ and $(R)$ are true but $(R)$ is not the true explanation of $(A)$
$(A)$ is false but $(R)$ is true.
$(A)$ is true but $(R)$ is false
Both $(A)$ and $(R)$ are true and $(R)$ is the true explanation of $(A)$
Answer: (d)
Solution
Photochemical smog causes cracking of rubber, the common component of photochemical smog are ozone, nitric oxide, acrolein, formaldehyde and peroxyacetyle nitrate (PAN).
Question 63
Chemistry · Chemical Bonding and Molecular Structure · Single correct
The interaction energy of London forces between two particles is proportional to $r^x$, where $r$ is the distance between the particles. The value of $x$ is :
3
-3
-6
6
Answer: (c)
Solution
For london dispersion forces. $$E \propto \frac{1}{r^6}$$ Hence $x = -6$
Question 64
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
The number of non-ionisable hydrogen atoms present in the final product obtained from the hydrolysis of $\mathrm{PCl}_5$ is :
0
2
1
3
Answer: (a)
Solution
The reaction is as follows: $$\mathrm{PCl_5 + H_2O \rightarrow POCl_3 + 2HCl}$$ Adding more water: $$+ 3\mathrm{H_2O}$$ Results in: $$\mathrm{H_3PO_4 + 3HCl}$$ In the structure of $\mathrm{H_3PO_4}$, all hydrogens are ionisable. Therefore, the answer is zero.
Question 65
Chemistry · Chemical Bonding and Molecular Structure · Single correct
The bond order and magnetic behaviour of $\mathrm{O}_2^{-}$ ion are, respectively:
1.5 and paramagnetic
1.5 and diamagnetic
2 and diamagnetic
1 and paramagnetic
Answer: (a)
Solution
Given $\mathrm{O}_2^- = (\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2$. $\left( \pi_{2p_x}^2 = \pi_{2p_y}^2 \right) \left( \pi_{2p_x}^{*2} = \pi_{2p_y}^{*1} \right)$. Bond order $= \frac{10 - 7}{2} = 1.5$ and paramagnetic.
Question 66
Chemistry · Biomolecules · Single correct
Given below are two statements: one is labelled as Assertion $(A)$ and other is labelled as Reason $(R)$. Assertion $(A)$: Sucrose is a disaccharide and a non-reducing sugar. Reason $(R)$: Sucrose involves glycosidic linkage between $\mathrm{C_1}$ of $\beta$-glucose and $\mathrm{C_2}$ of $\alpha$-fructose. Choose the most appropriate answer from the options given below:
Both $(A)$ and $(R)$ are true but $(R)$ is not the true explanation of $(A)$
$(A)$ is false but $(R)$ is true.
$(A)$ is true but $(R)$ is false
Both $(A)$ and $(R)$ are true and $(R)$ is the true explanation of $(A)$
Answer: (c)
Solution
Sucrose is an example of disaccharide and non-reducing sugar. Assertion: correct Sucrose involves glycosidic linkage between $C_1$ of $\alpha$-D-glucose and $C_2$ of $\beta$-D-fructose. Reason: Incorrect
Question 67
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Match List I with List II: List-I (Chemical Reaction) List-II (Reagent used) Choose the most appropriate match:
a - ii, b - iv, c - iii, d - i
a - iv, b - ii, c - iii, d - i
a - ii, b - iii, c - iv, d - i
a - iii, b - ii, c - i, d - iv
Answer: (c)
Solution
The first reaction involves the hydrolysis of an ester. The ester $CH_3COOCH_2CH_3$ reacts with $H_3O^+$ to form acetic acid $CH_3CO_2H$ and ethanol $CH_3CH_2OH$. The second reaction is the reduction of an ester using DIBAL-H. The ester $CH_3COOCH_3$ is reduced by DIBAL-H in the presence of water at $-78^\circC$ to form acetaldehyde $CH_3CHO$. The third reaction involves the reduction of a nitrile. The nitrile $CH_3CN$ is reduced by $SnCl_2$ and $HCl$ in the presence of water to form acetaldehyde $CH_3CH=O$. The fourth reaction is the reaction of a nitrile with a Grignard reagent. The nitrile $CH_3C\equivN$ reacts with $CH_3MgBr$ (1 equivalent) in the presence of $H_3O^+$ to form a ketone.
Question 68
Chemistry · The s-Block Elements · Single correct
Given below are two statements : one is labelled as Assertion $(\mathbf{A})$ and the other is labelled as Reason $(\mathbf{R})$. Assertion $(\mathbf{A})$: Barium carbonate is insoluble in water and is highly stable. Reason $(\mathbf{R})$: The thermal stability of the carbonates increases with increasing cationic size.
Both $(\mathbf{A})$ and $(\mathbf{R})$ are true but $(\mathbf{R})$ is the true explanation of $(\mathbf{A})$
$(\mathbf{A})$ is true but $(\mathbf{R})$ is false
Both $(\mathbf{A})$ and $(\mathbf{R})$ are true and $(\mathbf{R})$ is not the true explanation of $(\mathbf{A})$
$(\mathbf{A})$ is false but $(\mathbf{R})$ is true
Answer: (a)
Solution
In IIA group on moving down the group size of cation increases and show thermal stability of carbonate increases.
Question 69
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The major product in the above reaction is:
Answer: (d)
Solution
Question 70
Chemistry · Co-ordination Compounds · Single correct
Indicate the complex/complex ion which did not show any geometrical isomerism:
$[CoCl_2(en)_2]$
$[Co(CN)_5(NC)]^{3-}$
$[Co(NH_3)_3(NO_2)_3]$
$[Co(NH_3)_4Cl_2]^+$
Answer: (b)
Solution
Q11 (2) (1) $[\mathrm{CoCl_2(en)_2}]$ show cis-trans isomerism. (2) $[\mathrm{Co(CN)_5(NC)]^{-3}}$ can't show G.I. (3) $[\mathrm{Co(NH_3)_3(NO_2)_3}]$ show fac and mer isomerism. (4) $[\mathrm{Co(NH_3)_4Cl_2}]^{\oplus}$ show cis and trans isomerism.
Question 71
Chemistry · Surface Chemistry · Single correct
The sol given below with negatively charged colloidal particles is :
$\mathrm{FeCl}_3$ added to hot water
$\mathrm{KI}$ added to $\mathrm{AgNO}_3$ solution
$\mathrm{AgNO}_3$ added to $\mathrm{KI}$ solution
$\mathrm{Al}_2\mathrm{O}_3 \cdot x\mathrm{H}_2\mathrm{O}$ in water
Answer: (c)
Solution
$AgNO_3$ added to KI solution
Question 72
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Given below are two statements : Statement I : Sphalerite is a sulphide ore of zinc and copper glance is a sulphide ore of copper. Statement II : It is possible to separate two sulphide ores by adjusting proportion of oil to water or by using 'depressants' in a froth flotation method. Choose the most appropriate answer from the options given below :
Statement I is true but Statement II is false.
Both Statement I and Statement II are true.
Statement I is false but Statement II is true.
Both Statement I and Statement II are false.
Answer: (b)
Solution
Sphalerite $-- ZnS$, copper glance $-- Cu_2S$ two sulphide ores can be separated by adjusting proportions of oil to water or by using 'Depressants'.
Question 73
Chemistry · Hydrogen · Single correct
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Heavy water is used for the study of reaction mechanism. Reason (R): The rate of reaction for the cleavage of O - H bond is slower than that of O - D bond. Choose the most appropriate answer from the options given below :
Both (A) and (R) are true but (R) is not the true explanation of (A).
Both (A) and (R) are true and (R) is the true explanation of (A).
is false but (R) is true.
is true but (R) is false.
Answer: (d)
Solution
D$_2$O is used for the study of reaction mechanism. Rate of reaction for the cleavage of O-H bond is greater than O-D bond.
Question 74
Chemistry · Co-ordination Compounds · Single correct
Arrange the following Cobalt complexes in the order of increasing Crystal Field Stabilization Energy (CFSE) value. Complexes : \[ \begin{aligned} A &= [\mathrm{CoF}_6]^{3-} \\ B &= [\mathrm{Co(H_2O)}_6]^{2+} \\ C &= [\mathrm{Co(NH_3)}_6]^{3+} \\ D &= [\mathrm{Co(en)}_3]^{3+} \end{aligned} \] Therefore, the correct order is:
A < B < C < D
B < A < C < D
B < C < D < A
C < D < B < A
Answer: (b)
Solution
(i) CFSE is proportional to charge or oxidation number of central metal ion. (ii) CFSE is proportional to strength of ligand. en $>$ $\mathrm{NH_3}$ $>$ $\mathrm{H_2O}$ $>$ $\mathrm{F^-}$ Therefore, order of CFSE $$[\mathrm{Co(en)_3}]^{3+} > [\mathrm{Co(NH_3)_6}]^{3+} > [\mathrm{CoF_6}]^{3-} > [\mathrm{Co(H_2O)_6}]^{2+}$$
Question 75
Chemistry · Chemistry in Everyday Life · Single correct
The class of drug to which chlordiazepoxide with above structure belongs is :
Antacid
Analgesic
Tranquilizer
Antibiotic
Answer: (c)
Solution
The drug named chlordiate poxide is an example of a tranquilizer.
Question 76
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Chalcogen group elements are :
Se, Tb and Pu.
Se, TeTe and Po.
S, Te and Pm.
O, Ti and Po.
Answer: (b)
Solution
Group 16/oxygen family is known as Chalcogens. The members are O, S, Se, Te, Po.
Question 77
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Which one of the following compounds is not aromatic ?
Answer: (c)
Solution
Question 78
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The number of stereoisomers possible for 1,2 dimethyl cyclopropane is:
One
Four
Two
Three
Answer: (d)
Solution
The image shows three diagrams labeled A, B, and C. Each diagram represents a different orientation or configuration of a triangular shape with a shaded section. The task is to determine the correct combination or sequence of these diagrams.
Question 79
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Consider the given reaction, Identify 'X' and 'Y':
Answer: (c)
Solution
The reaction starts with the compound undergoing a reaction with NaOH and HCN in the presence of water. This results in the formation of a cyanohydrin. The cyanohydrin is then reduced using LiAlH4 to give the major product, which is an amino alcohol.
Question 80
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Consider the given reaction, the product A is:
Answer: (c)
Solution
The given compound is an acyl benzene with a propyl group, which is meta-directing. When it reacts with $\mathrm{Br_2}$ and $\mathrm{AlBr_3}$ in $\mathrm{Et_2O}$, bromination occurs at the meta position relative to the acyl group. The product is a brominated acyl benzene with the bromine at the meta position.
In the sulphur estimation, 0.471 g of an organic compound gave 1.44 g of barium sulphate. The percentage of sulphur in the compound is ____ $\%$. (Nearest integer) (Atomic Mass of Ba = 137 u)
The equilibrium constant $K_c$ at 298 K for the reaction $\mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}$ is 100. Starting with an equimolar solution with concentrations of $\mathrm{A}$, $\mathrm{B}$, $\mathrm{C}$ and $\mathrm{D}$ all equal to 1 M, the equilibrium concentration of $\mathrm{D}$ is ___ $\times 10^{-2}$ M. (Nearest integer)
Answer: 182
Solution
First check direction of reversible reaction. Since $Q_c = \frac{[C][D]}{[A][B]} = 1 < K_{eq}$. Therefore, reaction will move in forward direction to attain equilibrium state. $$A + B \rightleftharpoons C + D : K_{eq} = 100$$ $$to 1 1 1 1$$ $$teq. 1-x 1-x 1+x 1+x$$ Now: $K_{eq} = 100 = \frac{(1+x)(1+x)}{(1-x)(1-x)}$ $$\Rightarrow 100 = \left(\frac{1+x}{1-x}\right)^2$$ (i) $10 = \left(\frac{1+x}{1-x}\right)$ $$\Rightarrow 10 - 10x = 1 + x$$ $$\Rightarrow 11x = 9$$ $$\Rightarrow x = \frac{9}{11}$$ (ii) $-10 = \frac{1+x}{1-x}$ $$\Rightarrow -10 + 10x = 1 + x$$ $$\Rightarrow -9x = -11$$ $$\Rightarrow x = \frac{11}{9}$$ '$x$' cannot be more than one, therefore not valid. Therefore, equation concentration of $(D) = 1 + x$ $$= 1 + \frac{9}{11} = \frac{20}{11}$$ $$= 1.8181 = 181.81 \times 10^{-2}$$ $$\approx 182 \times 10^{-2}$$
Question 83
Chemistry · Thermodynamics · Numerical
For water $\Delta_{vap} H = 41 \, \mathrm{kJ \, mol^{-1}}$ at $373 \, \mathrm{K}$ and $1 \, \mathrm{bar}$ pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ____ $\mathrm{kJ \, mol^{-1}}$ [ Use : $R = 8.3 \, \mathrm{J \, mol^{-1} \, K^{-1}}$ ]
Answer: 38
Solution
Given equation is $$\mathrm{H_2O(\ell)} \rightarrow \mathrm{H_2O(g)} : \Delta H = 41 \, \mathrm{\frac{kJ}{mol}}$$ From the relation: $\Delta H = \Delta U + \Delta n_g RT$ $$41 \, \mathrm{\frac{kJ}{mol}} = \Delta U + (1) \times \frac{8.3}{1000} \times 373$$ $$\Delta U = 41 - 3.0959$$ $$= 38 \, \mathrm{\frac{kJ}{mol}}$$
Question 84
Chemistry · Structure of Atom · Numerical
A metal surface is exposed to $500\,\mathrm{nm}$ radiation. The threshold frequency of the metal for photoelectric current is $4.3 \times 10^{14}\,\mathrm{Hz}$. The velocity of ejected electron is ____ $\times 10^5\,\mathrm{ms}^{-1}$. (Nearest integer) (Given: $h = 6.63 \times 10^{-34}\,\mathrm{Js}$, $m_e = 9.0 \times 10^{-31}\,\mathrm{kg}$)
Answer: 5
Solution
v : speed of electron having max. K. E. From Einstein equation: $E = \phi + K. E._{max}$ $$\frac{hc}{\lambda} = h\nu_0 + \frac{1}{2}mv^2$$ $$\Rightarrow \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} = 6.63 \times 10^{-34} \times 4.3 \times 10^{14} + \frac{1}{2}mv^2$$ $$\Rightarrow \frac{6.63 \times 30 \times 10^{-20}}{5} = 6.63 \times 4.3 \times 10^{-20} + \frac{1}{2}mv^2$$ $$\Rightarrow 11.271 \times 10^{-20} \, \mathrm{J} = \frac{1}{2} \times 9 \times 10^{-31} \times v^2$$ $$\Rightarrow v = 5 \times 10^5 \, \mathrm{m/s}$$
Chemistry · Some Basic Concepts of Chemistry · Numerical
100 $\mathrm{mL}$ of $\mathrm{Na_3PO_4}$ solution contains 3.45 $\mathrm{g}$ of sodium. The molarity of the solution is ____ $\times$ $10^{-2}$ $\mathrm{mol \, L^{-1}}$. (Nearest integer) [Atomic Masses - Na : 23.0 $\mathrm{u}$, O : 16.0 $\mathrm{u}$, P : 31.0 $\mathrm{u}$]
Answer: 50
Solution
Therefore, molarity of $\mathrm{Na_3PO_4}$ solution is given by $$\frac{n_{\mathrm{Na_3PO_4}}}{volume of solution in L}$$ $$= \frac{\frac{1}{3} \times \frac{3.45}{23} \, \mathrm{mol}}{0.1 \, \mathrm{L}}$$ $$= 0.5 = 50 \times 10^{-2}$$
Question 87
Chemistry · Co-ordination Compounds · Numerical
The overall stability constant of the complex ion $$[\mathrm{Cu(NH_3)_4}]^{2+}$$ is $2.1 \times 10^{13}$. The overall dissociation constant is $y \times 10^{-14}$. Then $y$ is ____. (Nearest integer)
83 g of ethylene glycol is dissolved in 625 g of water. The freezing point of the solution is ____ K. (Nearest integer) [Use: Molal freezing point depression constant of water = $1.86\ \mathrm{K\,kg\,mol^{-1}}$] Freezing point of water = $273\ \mathrm{K}$ Atomic masses: $\mathrm{C}=12.0\,\mathrm{u}$, $\mathrm{O}=16.0\,\mathrm{u}$, $\mathrm{H}=1.0\,\mathrm{u}$
The reaction rate for the reaction $\mathrm{[PtCl_4]^{2-} + H_2O \rightleftharpoons [Pt(H_2O)Cl_3]^- + Cl^-}$ was measured as a function of the concentrations of different species. It was observed that $-\dfrac{d[\mathrm{PtCl_4}^{2-}]}{dt}=4.8\times10^{-5}[\mathrm{PtCl_4}^{2-}]-2.4\times10^{-3}[\mathrm{Pt(H_2O)Cl_3}^{-}][\mathrm{Cl^-}]$ where square brackets denote molar concentrations. The equilibrium constant, $K_c=$ _____ (Round off to the nearest integer.)
Answer: 50
Solution
The reaction is given by: $$[\mathrm{PtCl_4}]^{-2} + \mathrm{H_2O} \rightleftharpoons [\mathrm{Pt(H_2O)Cl_3}]^- + \mathrm{Cl}^-$$ The rate of change is: $$-\frac{d[\mathrm{PtCl_4}]^{-2}}{dt} = 4.8 \times 10^{-5} [\mathrm{PtCl_4}]^{-2} - 2.4 \times 10^3$$ The equilibrium expression is: $$[\mathrm{Pt(H_2O)Cl_3}]^0 [\mathrm{u}]$$ Therefore, the equilibrium constant is: $$\Rightarrow K_{eq} = \frac{k_f}{k_b} = \frac{4.8 \times 10^{-5}}{2.4 \times 10^{-3}} = 0.02$$
A chloro compound "A". (i) forms aldehydes on ozonolysis followed by the hydrolysis. (ii) when vaporized completely 1.53 g of A, gives 448 $\mathrm{mL}$ of vapour at STP. The number of carbon atoms in a molecule of compound A is ____.
Answer: 3
Solution
448 $\mathrm{ml}$ of A $\Rightarrow$ 1.53 $\mathrm{gm}$ A 22400 $\mathrm{ml}$ of A $\Rightarrow$ $\frac{1.53}{448}$ $\times$ 22400 gm A = 7650 $\mathrm{H_3CHC=CHCl}$ $\xrightarrow{\mathrm{O_3} Zn/ \mathrm{H_2O}}$ $\mathrm{CH_3CH=O}$ It has 3 carbon atoms and mm is 36 + 5 + 35.5 = 76.5