JEE Main 8 April 2024 Shift 2 question paper with solutions
JEE Main 8 April 2024 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 · Conic Sections · Single correct
If the image of the point $(-4, 5)$ in the line $x + 2y = 2$ lies on the circle $(x + 4)^2 + (y - 3)^2 = r^2$, then $r$ is equal to:
Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}$ and $\vec{c} = 3\hat{i} - \hat{j} + \lambda \hat{k}$ be three vectors. Let $\vec{r}$ be a unit vector along $\vec{b} + \vec{c}$. If $\vec{r} \cdot \vec{a} = 3$, then $3\lambda$ is equal to:
If $\alpha \neq a$, $\beta \neq b$, $\gamma \neq c$ and $$\begin{vmatrix} \alpha & b & c \\ a & \beta & c \\ a & b & \gamma \end{vmatrix} = 0$$, then $\frac{a}{\alpha-a} + \frac{b}{\beta-b} + \frac{\gamma}{\gamma-c}$ is equal to:
3
0
1
2
Answer: (b)
Solution
Perform the row operations $R_1 \rightarrow R_1 - R_2$, $R_2 \rightarrow R_2 - R_3$ on the matrix: $$\begin{vmatrix} \alpha - a & b - \beta & 0 \\ 0 & \beta - b & c - \gamma \\ a & b & \gamma \end{vmatrix} = 0$$ The expression becomes: $$(\alpha - a)(\gamma(\beta - b) - b(c - \gamma)) - (b - \beta)(-a(c - \gamma)) = 0$$ Simplifying further: $$\gamma(\alpha - a)(\beta - b) - b(\alpha - a)(c - \gamma) + a(b - \beta)(c - \gamma)$$ Finally, the equation: $$\frac{\gamma}{\gamma - c} + \frac{b}{\beta - b} + \frac{a}{\alpha - a} = 0$$
Question 4
Maths · Sequences and Series · Single correct
In an increasing geometric progression of positive terms, the sum of the second and sixth terms is $\frac{70}{3}$ and the product of the third and fifth terms is $49$. Then the sum of the $4^{th}$, $6^{th}$ and $8^{th}$ terms is equal to:
96
91
84
78
Answer: (b)
Solution
Given $T_2 + T_6 = \frac{70}{3}$, we have $ar + ar^5 = \frac{70}{3}$. Also, $T_3 \cdot T_5 = 49$, which gives $ar^2 \cdot ar^4 = 49$. Therefore, $a^2 r^6 = 49$. From $ar^3 = +7$, we find $a = \frac{7}{r^3}$. Now, $ar \left(1 + r^4\right) = \frac{70}{3}$. Solving $\frac{7}{r^2} \left(1 + r^4\right) = \frac{70}{3}$, we set $r^2 = t$. Then, $\frac{1}{t} \left(1 + t^2\right) = \frac{10}{3}$. Solving $3t^2 - 10t + 3 = 0$, we find $t = 3, \frac{1}{3}$. Since it is an increasing G.P., $r^2 = 3$, so $r = \sqrt{3}$. Now, $T_4 + T_6 + T_8 = ar^3 + ar^5 + ar^7 = ar^3 \left(1 + r^2 + r^4\right) = 7(1 + 3 + 9) = 91$.
Question 5
Maths · Permutations and Combinations · Single correct
The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to:
179
177
181
175
Answer: (a)
Solution
AA, MM, TT, H, I, C, S, E (1) All distinct $$^8C_5 \rightarrow 56$$ (2) 2 same, 3 different $$^3C_1 \times ^7C_3 \rightarrow 105$$ (3) 2 same 1st kind, 2 same 2nd kind, 1 different $$^3C_2 \times ^6C_1 \rightarrow 18$$ Total $\rightarrow 179$
Question 6
Maths · Complex Numbers and Quadratic Equations · Single correct
The sum of all possible values of $\theta \in [-\pi, 2\pi]$, for which $\frac{1+i \cos \theta}{1-2i \cos \theta}$ is purely imaginary, is equal
$3\pi$
$2\pi$
$5\pi$
$4\pi$
Answer: (a)
Solution
Given $Z = \frac{1 + i \cos \theta}{1 - 2i \cos \theta}$. Since $Z = -\overline{Z}$, we have $$\frac{1 + i \cos \theta}{1 - 2i \cos \theta} = -\left( \frac{1 + i \cos \theta}{1 - 2i \cos \theta} \right)$$ Multiplying both sides by the conjugate, $$(1 + i \cos \theta)(1 - 2i \cos \theta) = -(1 - 2i \cos \theta)(1 + i \cos \theta)$$ Simplifying, $$(1 + i \cos \theta)(1 + 2i \cos \theta) = -(1 - i \cos \theta)$$ This gives $$1 + 3i \cos \theta - 2 \cos^2 \theta = -(1 - 3i \cos \theta - 2 \cos^2 \theta)$$ Thus, $$2 - 4 \cos^2 \theta = 0$$ Solving for $\cos^2 \theta$, $$\Rightarrow \cos^2 \theta = \frac{1}{2} \Rightarrow \theta = -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$ The sum is $3\pi$.
Question 7
Maths · Determinants · Single correct
If the system of equations $x + 4y - z = \lambda$, $7x + 9y + \mu z = -3$, $5x + y + 2z = -1$ has infinitely many solutions, then $(2\mu + 3\lambda)$ is equal to:
Maths · Three Dimensional Geometry · Single correct
If the shortest distance between the lines $\frac{x-\lambda}{2} = \frac{y-4}{3} = \frac{z-3}{4}$ and $\frac{x-2}{4} = \frac{y-4}{6} = \frac{z-7}{8}$ is $\frac{13}{\sqrt{29}}$, then a value of $\lambda$ is :
If the value of $\($ $\frac{3 \cos 36^\circ + 5 \sin 18^\circ}{5 \cos 36^\circ - 3 \sin 18^\circ}$ $\)$ is $\($ $\frac{a \sqrt{5} - b}{c}$ $\)$, where $\($ a, b, c $\)$ are natural numbers and $\($ $\gcd$(a, c) = 1 $\)$, then $\($ a + b + c $\)$ is equal to:
40
52
50
54
Answer: (b)
Solution
Given $\($ $\frac{3(\sqrt{5}+1)}{4}$ + 5$\left$($\frac{\sqrt{5}-1}{4}$$\right$) $\)$ over $\($ 5$\left$($\frac{\sqrt{5}+1}{4}$$\right$) - 3$\left$($\frac{\sqrt{5}-1}{4}$$\right$) $\)$ equals $\($ $\frac{8\sqrt{5} - 2}{2\sqrt{5} + 8}$ $\)$. This simplifies to: $\[$ $\frac{4\sqrt{5} - 1}{\sqrt{5} + 4}$ $\times$ $\frac{\sqrt{5} - 4}{\sqrt{5} - 4}$ $\]$ $\[$ = $\frac{20 - 16\sqrt{5} - \sqrt{5} + 4}{-11}$ $\]$ $\[$ = $\frac{17\sqrt{5} - 24}{11}$ $\Rightarrow$ a = 17, b = 27, c = 11 $\]$ Thus, $\($ a + b + c = 52 $\)$.
Question 10
Maths · Differential Equations · Single correct
Let $y = y(x)$ be the solution curve of the differential equation $\sec y \frac{dy}{dx} + 2x \sin y = x^3 \cos y, y(1) = 0$. Then $y(\sqrt{3})$ is equal to:
$\frac{\pi}{3}$
$\frac{\pi}{6}$
$\frac{\pi}{12}$
$\frac{\pi}{4}$
Answer: (d)
Solution
Given $\sec^2 y \frac{dy}{dx} + 2x \sin y \sec y = x^3 \cos y \sec y$. This simplifies to $\sec^2 y \frac{dy}{dx} + 2x \tan y = x^3$. Let $\tan y = t \Rightarrow \sec^2 y \frac{dy}{dx} = \frac{dt}{dx}$. Thus, $\frac{dt}{dx} + 2xt = x^3$. If $t = e^{\int 2x \, dx} = e^{x^2}$, then $tx^2 = \int x^3 \cdot e^{x^2} \, dx + c$. Let $x^2 = Z \Rightarrow t \cdot e^Z = \frac{1}{2} \left[ e^Z \cdot Z - e^Z \right] + c$. Therefore, $2 \tan y = (x^2 - 1) + 2e^{-x^2}$. Given $y(1) = 0 \Rightarrow c = 0 \Rightarrow y(\sqrt{3}) = \frac{\pi}{4}$.
Question 11
Maths · Applications of Integrals · Single correct
The area of the region in the first quadrant inside the circle $x^2 + y^2 = 8$ and outside the parabola $y^2 = 2x$ is equal to:
Maths · Straight Lines and Pair of Straight Lines · Single correct
If the line segment joining the points $(5, 2)$ and $(2, a)$ subtends an angle $\frac{\pi}{4}$ at the origin, then the absolute value of the product of all possible values of $a$ is:
6
8
2
-4
Answer: (d)
Solution
Given the points $A(5, 2)$ and $B(2, a)$, and the angle $\pi/4$ at $O$, we have: $$m_{OA} = \frac{2}{5}$$ $$m_{OB} = \frac{a}{2}$$ The equation is: $$4 - 5a = \pm (10 + 2a)$$ Solving for $a$: Case 1: $$4 - 5a = 10 + 2a$$ $$\Rightarrow 7a + 6 = 0$$ $$\Rightarrow a = -\frac{6}{7}$$ Case 2: $$4 - 5a = -10 - 2a$$ $$3a = 14$$ $$a = \frac{14}{3}$$ The tangent of the angle is: $$\tan \frac{\pi}{4} = \left| \frac{\frac{2}{5} - \frac{a}{2}}{1 + \frac{2}{5} \cdot \frac{a}{2}} \right| = \left| \frac{4 - 5a}{10 + 2a} \right|$$ This gives: $$1 = \left| \frac{4 - 5a}{10 + 2a} \right|$$
Question 13
Maths · Vector Algebra · Single correct
Let $\vec{a}=4\hat{i}-\hat{j}+\hat{k}$, $\vec{b}=11\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $$(\vec{a}+\vec{b})\times\vec{c}=\vec{c}\times(-2\vec{a}+3\vec{b}).$$ If $$(2\vec{a}+3\vec{b})\cdot\vec{c}=1670,$$ then $$|\vec{c}|^2$$ is equal to:
1609
1618
1600
1627
Answer: (b)
Solution
Given $(\vec{a}+\vec{b})\times\vec{c}-\vec{c}\times(-2\vec{a}+3\vec{b})=0$. This implies $(\vec{a}+\vec{b})\times\vec{c}+(-2\vec{a}+3\vec{b})\times\vec{c}=0$. Therefore, $((\vec{a}+\vec{b})-2\vec{a}+3\vec{b})\times\vec{c}=0$. This implies $\vec{c}=\lambda(4\vec{b}-\vec{a})$. Substituting, $\lambda(44\hat{i}-4\hat{j}+4\hat{k}-4\hat{i}+\hat{j}-\hat{k}) =\lambda(40\hat{i}-3\hat{j}+3\hat{k})$. Now $(8\hat{i}-2\hat{j}+2\hat{k}+33\hat{i}-3\hat{j}+3\hat{k}) \cdot \lambda(40\hat{i}-3\hat{j}+3\hat{k}) =1670$. This implies $(41\hat{i}-5\hat{j}+5\hat{k}) \cdot (40\hat{i}-3\hat{j}+3\hat{k}) \times\lambda =1670$. Therefore, $(1640+15+15)\lambda=1670 \Rightarrow \lambda=1$. So $\vec{c}=40\hat{i}-3\hat{j}-3\hat{k}$. This implies $|\vec{c}|^2=1600+9+9=1618$.
Question 14
Maths · Applications of Derivatives · Single correct
If the function $f(x) = 2x^3 - 9ax^2 + 12a^2 x + 1, a > 0$ has a local maximum at $x = \alpha$ and a local minimum at $x = \alpha^2$, then $\alpha$ and $\alpha^2$ are the roots of the equation:
$x^2 - 6x + 8 = 0$
$x^2 + 6x + 8 = 0$
$8x^2 + 6x - 1 = 0$
$8x^2 - 6x + 1 = 0$
Answer: (a)
Solution
Given $f(x) = 6x^2 - 18ax + 12a^2 = 0$. Let $\alpha$ and $\alpha^2$ be the roots. Then $\alpha + \alpha^2 = 3a$ and $\alpha \times \alpha^2 = 2a^2$. Thus, $$(\alpha + \alpha^2)^3 = 27a^3$$ This implies $$2a^2 + 4a^4 + 3(3a)(2a^2) = 27a^3$$ Simplifying, we get $$2 + 4a^2 + 18a = 27a$$ Further simplifying, $$4a^2 - 9a + 2 = 0$$ This can be rewritten as $$4a^2 - 8a - a + 2 = 0$$ Factoring gives $$(4a - 1)(a - 2) = 0$$ Thus, $a = 2$. Substituting back, $$6x^2 - 36x + 48 = 0$$ This simplifies to $$x^2 - 6x + 8 = 0 \ldots (1)$$ If we take $a = \frac{1}{4}$ then $\alpha = \frac{1}{2}$ which is not possible.
Question 15
Maths · Probability · Single correct
There are three bags $X$, $Y$ and $Z$. Bag $X$ contains 5 one-rupee coins and 4 five-rupee coins; Bag $Y$ contains 4 one-rupee coins and 5 five-rupee coins and Bag $Z$ contains 3 one-rupee coins and 6 five-rupee coins. A bag is selected at random and a coin drawn from it at random is found to be a one-rupee coin. Then the probability, that it came from bag $Y$, is:
$\frac{1}{4}$
$\frac{1}{2}$
$\frac{5}{12}$
$\frac{1}{3}$
Answer: (d)
Solution
Given the values for X, Y, and Z: X: 5 one & 4 five Y: 4 one & 5 five Z: 3 one & 6 five The probability P is calculated as follows: $$P = \frac{4/9}{5/9 + 4/9 + 3/9} = \frac{4}{12} = \frac{1}{3}.$$
Question 16
Maths · Integrals · Single correct
Let $\int_{\alpha}^{\log_e 4}$ $\frac{dx}{\sqrt{e^x - 1}}$ = $\frac{\pi}{6}$. Then $e^\alpha$ and $e^{-\alpha}$ are the roots of the equation:
Let $f(x) = \begin{cases} -a & if -a \leq x \leq 0 \\ x + a & if 0 0$ and $g(x) = (f|x| - |f(x)|)/2$. Then the function $g : [-a, a] \to [-a, a]$ is
neither one-one nor onto.
onto.
both one-one and onto.
one-one.
Answer: (a)
Solution
The given graphs represent different transformations of the function $f(x)$. The first graph shows $y = f(x)$ with points at $(-a, -a)$, $(0, a)$, and $(a, 2a)$. The second graph shows $y = f(|x|)$, which reflects the graph of $f(x)$ for $x < 0$ onto the positive $x$-axis. The third graph shows $y = |f(x)|$, which reflects any negative parts of $f(x)$ above the $x$-axis. The fourth graph shows $g(x) = \frac{f(|x|) - |f(x)|}{2}$, which combines the transformations of $f(x)$ and $|f(x)|$ to create a new function.
Question 18
Maths · Relations and Functions · Single correct
Let $A = \{2, 3, 6, 8, 9, 11\}$ and $B = \{1, 4, 5, 10, 15\}$. Let $R$ be a relation on $A \times B$ defined by $(a, b)R(c, d)$ if and only if $3ad - 7bc$ is an even integer. Then the relation $R$ is
an equivalence relation.
reflexive and symmetric but not transitive.
transitive but not symmetric.
reflexive but not symmetric.
Answer: (b)
Solution
Given sets $A = \{2, 3, 6, 8, 9, 11\}$ and $B = \{1, 4, 5, 10, 15\}$, consider the relation $(a, b)R(c, d)$ defined by $3ad - 7bc$. Reflexive: $(a, b)R(a, b)$ $$3ab - 7ba = -4ab$$ This is always even, so it is reflexive. Symmetric: If $3ad - 7bc$ is even Case-I: odd odd Case-II: even even For $(c, d)R(a, b)$, we have $3bc - 3ab$. Case-I: odd odd Case-II: even even Thus, it is a symmetric relation. Transitive: Set $(3, 4)R(6, 4)$ satisfies the relation. Set $(6, 4)R(3, 1)$ satisfies the relation. But $(3, 4)R(3, 1)$ does not satisfy the relation, so it is not transitive.
Question 19
Maths · Continuity and Differentiability · Single correct
For $a, b > 0$, let $f(x) = \begin{cases} \frac{\tan((a+1)x) + b \tan x}{x}, & x 0 \end{cases}$ be a continuous function at $x = 0$. Then $\frac{b}{a}$ is equal to:
If the term independent of $x$ in the expansion of $\left( \sqrt{ax^2} + \frac{1}{2x^3} \right)^{10}$ is 105, then $a^2$ is equal to:
2
4
6
9
Answer: (b)
Solution
Given $\left(\sqrt{ax^2} + \dfrac{1}{2x^3}\right)^{10}$. The general term is ${}^{10}C_r \left(\sqrt{ax^2}\right)^{10-r} \left(\dfrac{1}{2x^3}\right)^r$. Solving $20 - 2r - 3r = 0$ gives $r = 4$. Then, ${}^{10}C_4\, a^3 \cdot \dfrac{1}{16} = 105$. Solving for $a^3 = 8$ gives $a^2 = 4$.
Question 21
Maths · Applications of Derivatives · Numerical
Let A be the region enclosed by the parabola $y^2 = 2x$ and the line $x = 24$. Then the maximum area of the rectangle inscribed in the region A is _______
Answer: 128
Solution
The area is given by the formula: $$A = 2 \left( 24 - \frac{b^2}{2} \right) \cdot b$$ To find the maximum area, we set the derivative equal to zero: $$\frac{dA}{db} = 0 \Rightarrow b = 4$$ Substituting back to find the area: $$A = 2(24 - 8)4$$ $$= 128$$
Question 22
Maths · Limits and Derivatives · Numerical
If $\alpha = \lim_{x \to 0^+} \left( \frac{e^{\sqrt{\tan x}} - e^{\sqrt{x}}}{\sqrt{\tan x} - \sqrt{x}} \right)$ and $\beta = \lim_{x \to 0} (1 + \sin x)^{\frac{1}{2} \cot x}$ are the roots of the quadratic equation $ax^2 + bx - \sqrt{e} = 0$, then $12 \log_e (a + b)$ is equal to ________
Let S be the focus of the hyperbola $\frac{x^2}{3} - \frac{y^2}{5} = 1$, on the positive $x$-axis. Let $C$ be the circle with its centre at $A(\sqrt{6}, \sqrt{5})$ and passing through the point $S$. If $O$ is the origin and $SAB$ is a diameter of $C$, then the square of the area of the triangle $OSB$ is equal to
Answer: 40
Solution
Area = $\frac{1}{2}$(OS)h = $\frac{1}{2}$ $\sqrt{8}$ 2 $\sqrt{5}$ = $\sqrt{40}$
Question 24
Maths · Three Dimensional Geometry · Numerical
Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the image of the point $\mathrm{Q}(1, 6, 4)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$. Then $2\alpha + \beta + \gamma$ is equal to
Answer: 11
Solution
Point A is given as $A(t, 2t + 1, 3t + 2)$. The vector $\overrightarrow{QA}$ is $\overrightarrow{QA} = (t - 1) \hat{i} + (2t - 5) \hat{j} + (3t - 2) \hat{k}$. The dot product $\overrightarrow{QA} \cdot \overrightarrow{b} = 0$ gives the equation $$(t - 1) + 2(2t - 5) + 3(3t - 2) = 0.$$ Solving this, we find $14t = 17$. Therefore, $$t = \frac{17}{14}.$$ The coordinates $\alpha$, $\beta$, and $\gamma$ are calculated as follows: $$\alpha = \frac{20}{14}, \beta = \frac{12}{14}, \gamma = \frac{102}{14}.$$ The equation $2\alpha + \beta + \gamma = \frac{154}{14} = 11$ is satisfied.
Question 25
Maths · Sequences and Series · Fill in the blank
An arithmetic progression is written in the following way The sum of all the terms of the 10th row is
Answer: 1505
Question 26
Maths · Complex Numbers and Quadratic Equations · Numerical
The number of distinct real roots of the equation $|x + 1||x + 3| - 4|x + 2| + 5 = 0$, is
Maths · Straight Lines and Pair of Straight Lines · Numerical
Let a ray of light passing through the point $(3, 10)$ reflects on the line $2x + y = 6$ and the reflected ray passes through the point $(7, 2)$. If the equation of the incident ray is $ax + by + 1 = 0$, then $a^2 + b^2 + 3ab$ is equal to
Let a, b, c $\in$ $\mathbb{N}$ and a < b < c. Let the mean, the mean deviation about the mean and the variance of the 5 observations 9, 25, a, b, c be 18, 4 and $\frac{136}{5}$, respectively. Then 2a + b - c is equal to
Answer: 33
Solution
Given $a, b, c \in \mathbb{N}$ and $a < b < c$. The mean $\bar{x} = mean = \frac{9 + 25 + a + b + c}{5} = 18$. Thus, $a + b + c = 56$. Mean deviation $= \frac{\Sigma |x_i - \bar{x}|}{n} = 4$. $$= 9 + 7 + |18 - a| + |18 - b| + |18 - c| = 20$$ $$= |18 - a| + |18 - b| + |18 - c| = 4$$ Variance $= \frac{\Sigma |x_i - \bar{x}|^2}{n} = \frac{136}{5}$. $$= 81 + 49 + |18 - a|^2 + |18 - b|^2 + |18 - c|^2 = 136$$ $$= (18 - a)^2 + (18 - b)^2 + (18 - c)^2 = 6$$ Possible values $(18 - a)^2 = 1$, $(18 - b)^2 = 1$, $(18 - c)^2 = 4$ with $a < b < c$. So $$18 - a = 1 18 - b = -1 18 - c = -2$$ $$a = 17 b = 19 c = 20$$ $a + b + c = 56$. $2a + b - c = 34 = 19 - 20 = 33$.
Question 29
Maths · Differential Equations · Fill in the blank
Let $\alpha |x| = |y| e^{xy - \beta}$, $\alpha, \beta \in \mathbb{N}$ be the solution of the differential equation $$x \, dy - y \, dx + xy(x \, dy + y \, dx) = 0, \, y(1) = 2.$$ Then $\alpha + \beta$ is equal to ______.
Answer: 4
Solution
Given $a|x| = |y|e^{yx - \beta}$, $a, b \in \mathbb{N}$. The equation is $xdy - ydx + xy(xdy + ydx) = 0$. Rewriting, we have $$\frac{dy}{y} - \frac{dx}{x} + (xdy + ydx) = 0.$$ Integrating, we get $$\ln|y| - \ln|x| + xy = c.$$ Given $y(1) = 2$, we find $$\ln|2| - 0 + 2 = c.$$ Thus, $c = 2 + \ln 2$. Substituting back, $$\ln|y| - \ln|x| + xy = 2 + \ln 2.$$ This implies $$\ln|x| = \ln\left|\frac{y}{2}\right| - 2 + xy.$$ Therefore, $$|x| = \left|\frac{y}{2}\right| e^{xy - 2}.$$ Thus, $2|x| = |y|e^{xy - 2}$. Finally, $\alpha = 2$, $\beta = 2$, and $\alpha + \beta = 4$.
Question 30
Maths · Integrals · Fill in the blank
If $\int \dfrac{1}{\sqrt[5]{(x-1)^4(x+3)^6}}\, dx = A\left(\dfrac{\alpha x-1}{\beta x+3}\right)^B + C$, where $C$ is the constant of integration, then the value of $\alpha + \beta + 20AB$ is _____
Answer: 7
Solution
The integral $$\int \frac{1}{\sqrt[5]{(x-1)^4}(x+3)^6} \, dx = A \left( \frac{\alpha x - 1}{\beta x + 3} \right)^B + C$$ is given. Let $$I = \int \frac{1}{(x-1)^{4/5}(x+3)^{6/5}} \, dx$$. Rewriting, we have $$I = \int \frac{1}{\left(\frac{x-1}{x+3}\right)^{4/5} (x+3)^2} \, dx$$. Let $$\left(\frac{x-1}{x+3}\right) = t \Rightarrow \frac{4}{(x+3)^2} \, dx = dt$$ and $$t^{-4/5+1}$$. Then, $$I = \frac{1}{4} \int \frac{1}{t^{4/5}} \, dt = \frac{1}{4} \frac{t^{1/5}}{1/5} + c$$. Thus, $$I = \frac{5}{4} \left(\frac{x-1}{x+3}\right)^{1/5} + C$$. From this, $$A = \frac{5}{4}$$, $$\alpha = \beta = 1$$, $$B = \frac{1}{5}$$. Calculating, $$\alpha + \beta + 20AB = 2 + 20 \times \frac{5}{4} \times \frac{1}{5} = 7$$.
Physics
Question 31
Physics · Work, Energy and Power · Single correct
A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is :
Physics · Ray Optics and Optical Instruments · Single correct
The position of the image formed by the combination of lenses is :
15 cm (right of second lens)
30 cm (left of third lens)
15 cm (left of second lens)
30 cm (right of third lens)
Answer: (d)
Solution
For lens 1: $f_1=10,\ u=-30,\ v=?$ $v=\frac{uf}{u+f}=\frac{-30\times10}{-30+10}=15$ For lens 2: $f_1=-10,\ u=10,\ v=?$ $v=\frac{uf}{u+f}=\frac{10\times(-10)}{10-10}=\infty$ For lens 3: $f=30,\ u=-\infty,\ v=?$ So $v$ will be $30$.
Question 35
Physics · Waves · Single correct
A plane progressive wave is given by $y = 2 \cos 2\pi(330t - x)\, \mathrm{m}$. The frequency of the wave is :
330 $\mathrm{Hz}$
660 $\mathrm{Hz}$
340 $\mathrm{Hz}$
165 $\mathrm{Hz}$
Answer: (a)
Solution
Given $$y = 2 \cos 2\pi (330t - x) \, \mathrm{m}$$ $$y = A \cos(\omega t - kx)$$ By comparing, $\($ $\omega$ = 2$\pi$ $\times$ 330 $\)$. $$2\pi f = 2\pi \times 330$$ $$f = 330$$
Question 36
Physics · System of Particles and Rotational Motion · Single correct
A thin circular disc of mass $M$ and radius $R$ is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with angular velocity $\omega$. If another disc of same dimensions but of mass $M/2$ is placed gently on the first disc co-axially, then the new angular velocity of the system is :
$\frac{3}{2} \omega$
$\frac{5}{4} \omega$
$\frac{2}{3} \omega$
$\frac{4}{5} \omega$
Answer: (c)
Solution
Given $I_1 \omega = I_2 \omega_2$. $$\frac{MR^2}{2} \omega = \frac{3}{2} \left( \frac{MR^2}{2} \right) \omega_2$$ Solving for $\omega_2$, we get: $$\omega_2 = \frac{2}{3} \omega$$
Question 37
Physics · Mechanical Properties of Fluids · Single correct
A cube of ice floats partly in water and partly in kerosene oil. The ratio of volume of ice immersed in water to that in kerosene oil (specific gravity of Kerosene oil = 0.8, specific gravity of ice = 0.9)
Given below are two statements : Statement (I) : The mean free path of gas molecules is inversely proportional to square of molecular diameter. Statement (II) : Average kinetic energy of gas molecules is directly proportional to absolute temperature of gas. In the light of the above statements, choose the correct answer from the options given below :
Statement I is true but Statement II is false
Both Statement I and Statement II are false
Both Statement I and Statement II are true
Statement I is false but Statement II is true
Answer: (c)
Solution
For question 9, the mean free path $\lambda$ is given by the equation: $$\lambda = \frac{RT}{\sqrt{2\pi} d^2 N_A P}$$ The kinetic energy $\mathrm{KE}$ is given by: $$\mathrm{KE} = \frac{f}{2} nRT$$
Question 39
Physics · Gravitation · Single correct
Two satellite $A$ and $B$ go round a planet in circular orbits having radii $4R$ and $R$ respectively. If the speed of $A$ is $3v$, the speed of $B$ will be:
Physics · Moving Charges and Magnetism · Single correct
A long straight wire of radius $a$ carries a steady current $I$. The current is uniformly distributed across its cross section. The ratio of the magnetic field at $\frac{a}{2}$ and $2a$ from axis of the wire is :
Water boils in an electric kettle in 20 minutes after being switched on. Using the same main supply, the length of the heating element should be _____ to _____ times of its initial length if the water is to be boiled in 15 minutes.
Physics · Electrostatic Potential and Capacitance · Single correct
A capacitor has air as dielectric medium and two conducting plates of area $12 \, \mathrm{cm}^2$ and they are $0.6 \, \mathrm{cm}$ apart. When a slab of dielectric having area $12 \, \mathrm{cm}^2$ and $0.6 \, \mathrm{cm}$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \, \mathrm{cm}$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $\varepsilon_0 = 8.834 \times 10^{-12} \, \mathrm{F/m}$)
1
1.33
0.66
1.50
Answer: (d)
Solution
Given $\($ $\frac{A \varepsilon_o}{d}$ = $\frac{A \varepsilon_o}{\left(0.2 + \frac{d}{k}\right)}$ $\)$. $\($ 0.6 = 0.2 + $\frac{0.6}{k}$ $\)$ Solving for $\($ k $\)$, we get $\($ k = $\frac{3}{2}$ $\)$
Question 44
Physics · Laws of Motion · Single correct
A given object takes n times the time to slide down $45^\circ$ rough inclined plane as it takes the time to slide down an identical perfectly smooth $45^\circ$ inclined plane. The coefficient of kinetic friction between the object and the surface of inclined plane is:
$\sqrt{1 - \frac{1}{n^2}}$
$1 - n^2$
$1 - \frac{1}{n^2}$
$\sqrt{1 - n^2}$
Answer: (c)
Solution
Case-1: No friction $a = g \sin \theta$ $\ell = \frac{1}{2} (g \sin \theta) t_1^2$ $$t_1 = \sqrt{\frac{2\ell}{g \sin \theta}}$$ Case-2: With friction $a = g \sin \theta - \mu g \cos \theta$ $\ell = \frac{1}{2} (g \sin \theta - \mu g \cos \theta) t_2^2$ $$\sqrt{\frac{2\ell}{g \sin \theta - \mu g \cos \theta}} = n \sqrt{\frac{2\ell}{g \sin \theta}}$$ $\mu = 1 - \frac{1}{n^2}$
Question 45
Physics · Alternating Current · Single correct
A coil of negligible resistance is connected in series with $90\,\Omega$ resistor across $120\,\mathrm{V}$, $60\,\mathrm{Hz}$ supply. A voltmeter reads $36\,\mathrm{V}$ across resistance. Inductance of the coil is:
0.286 H
0.76 H
2.86 H
0.91 H
Answer: (b)
Solution
Given $36 = I_{rms} R$ and $I_{rms} = \frac{120}{\sqrt{X_L^2 + R^2}}$. Therefore, $36 = \frac{120}{\sqrt{X_L^2 + R^2}} \times R$. Given $R = 90 \, \Omega$, we have $36 = \frac{120 \times 90}{\sqrt{X_L^2 + 90^2}}$. Solving for $\sqrt{X_L^2 + 90^2} = 300$, we find $X_L^2 = 81900$. Thus, $X_L = 286.18$. Since $\omega L = 286.18$, we have $L = \frac{286.18}{376.8} = 0.76 \, H$.
Question 46
Physics · Experimental Physics · Single correct
There are 100 divisions on the circular scale of a screw gauge of pitch 1 mm. With no measuring quantity in between the jaws, the zero of the circular scale lies 5 divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found that 4 linear scale divisions are clearly visible while 60 divisions on circular scale coincide with the reference line. The diameter of the wire is :
Physics · Dual Nature of Radiation and Matter · Single correct
A proton and an electron have the same de Broglie wavelength. If $K_p$ and $K_e$ be the kinetic energies of proton and electron respectively, then choose the correct relation:
$K_p > K_e$
$K_p < K_e$
$K_p = K_e$
$K_p = K_e^2$
Answer: (b)
Solution
De Broglie wavelength of proton and electron is $\lambda$. Therefore, $\lambda = \frac{h}{p}$. Thus, $p_{proton} = p_{electron}$. Therefore, $KE = \frac{p^2}{2m}$. Hence, $KE_{proton} < KE_{electron}$ which implies $[K_p < K_c]$.
Question 48
Physics · Experimental Physics · Single correct
Least count of a vernier caliper is $\frac{1}{20N}$ cm. The value of one division on the main scale is 1 mm. Then the number of divisions of main scale that coincide with $N$ divisions of vernier scale is :
$(2N - 1)$
$\left(\frac{2N - 1}{2N}\right)$
$\left(\frac{2N - 1}{2}\right)$
$\left(\frac{2N - 1}{20N}\right)$
Answer: (c)
Solution
Least count of vernier calipers is $\frac{1}{20N} \, \mathrm{cm}$. Therefore, least count is $1 \mathrm{MSD} - 1 \mathrm{VSD}$. Let $x$ be the number of divisions of the main scale that coincides with $N$ divisions of the vernier scale, then $$1 \mathrm{VSD} = \frac{x \times 1 \, \mathrm{mm}}{N}$$ Therefore, $$\frac{1}{20N} \, \mathrm{cm} = 1 \, \mathrm{mm} - \frac{x \times 1 \, \mathrm{mm}}{N}$$ $$\frac{1}{2N} \, \mathrm{mm} = 1 \, \mathrm{mm} - \frac{x}{N} \, \mathrm{mm}$$ $$x = \left(1 - \frac{1}{2N}\right)N$$ $$x = \frac{2N - 1}{2}$$
Question 49
Physics · Nuclei · Single correct
If $M_o$ is the mass of isotope $^{12}_5 B$, $M_P$ and $M_n$ are the masses of proton and neutron, then nuclear binding energy of isotope is :
$(M_o - 5M_P) C^2$
$(5M_P + 7M_n - M_o) C^2$
$(M_0 - 12M_n) C^2$
$(M_o - 5M_P - 7M_n) C^2$
Answer: (b)
Solution
B.E. = $\Delta m C^2$ $(5M_p + 7M_n - M_o) C^2$
Question 50
Physics · Thermodynamics · Single correct
A diatomic gas ($\gamma = 1.4$) does $100 \, \mathrm{J}$ of work in an isobaric expansion. The heat given to the gas is :
$250 \, \mathrm{J}$
$150 \, \mathrm{J}$
$350 \, \mathrm{J}$
$490 \, \mathrm{J}$
Answer: (c)
Solution
For Isobaric process $w = P \Delta v = nR \Delta T = 100 \, \mathrm{J}$ $Q = \Delta u + w$ $$\Delta Q = \frac{F}{2} nR \Delta T + nR \Delta T$$ $$\left( \frac{f}{2} + 1 \right) nR \Delta T$$ $$\left( \frac{5}{2} + 1 \right) 100 = 350 \, \mathrm{J}$$
Question 51
Physics · Magnetism and Matter · Numerical
The coercivity of a magnet is $5 \times 10^3 \, \mathrm{A/m}$. The amount of current required to be passed in a solenoid of length $30 \, \mathrm{cm}$ and the number of turns $150$, so that the magnet gets demagnetised when inside the solenoid is ____ A.
Physics · Mechanical Properties of Fluids · Numerical
Small water droplets of radius $0.01 \, \mathrm{mm}$ are formed in the upper atmosphere and falling with a terminal velocity of $10 \, \mathrm{cm/s}$. Due to condensation, if 8 such droplets are coalesced and formed a larger drop, the new terminal velocity will be ____ $\mathrm{cm/s}$.
Answer: 40
Solution
m = mass of small drop M = mass of bigger drop $$V_t = \frac{2}{9} \frac{R^2 (\rho - \sigma) g}{\eta}$$ $8 \propto m = M$ $$8r^3 = R^3 \Rightarrow R = 2R$$ As $V_t \times R^2$, therefore radius doubles so $V_t$ becomes 4 times. Thus, $4 \times 10 = 40 \, \mathrm{cm/s}$
Question 53
Physics · Electric Charges and Fields · Fill in the blank
If the net electric field at point P along Y axis is zero, then the ratio of $\left| \frac{q_2}{q_3} \right|$ is $\frac{8}{5\sqrt{x}}$, where $x =$.
Answer: 5
Solution
Given the diagram, we have the equation: $$\frac{K q_2}{20} \cos \beta = \frac{K q_3}{25} \cos \theta$$ Substituting the values, we get: $$\frac{K q_2}{20} \frac{4}{\sqrt{20}} = \frac{K q_3}{25} \frac{4}{\sqrt{25}}$$ Simplifying, we find: $$\frac{q_2}{q_3} = \frac{20}{25} \frac{\sqrt{20}}{\sqrt{25}} = \frac{8}{5 \sqrt{x}}$$ Solving for $\sqrt{x}$, we have: $$\Rightarrow \sqrt{x} = \frac{8 \times 25 \sqrt{25}}{5 \times 20 \sqrt{20}}$$ Therefore, $x = 5$.
Question 54
Physics · Current Electricity · Numerical
A heater is designed to operate with a power of 1000 $\mathrm{W}$ in a 100 $\mathrm{V}$ line. It is connected in combination with a resistance of 10 $\Omega$ and a resistance $R$, to a 100 $\mathrm{V}$ mains as shown in figure. For the heater to operate at 62.5 \, \mathrm{W}, the value of $R$ should be ___ $\Omega$.
Answer: 5
Solution
The resistance of the heater is calculated as follows: $$R_{heater} = \frac{V^2}{P} = \frac{(100)^2}{1000} = 10\, \Omega$$ For the heater, the power is given by $$P = \frac{V^2}{R} \implies V = \sqrt{PR}$$ Therefore, $$V = \sqrt{62.5 \times 10}$$ $$V = 25\, V$$ The current $i_1$ is calculated as $$i_1 = \frac{75}{10} = 7.5\, A, i_H = \frac{25}{10} = 2.5\, A.$$ The current $i_R$ is $$i_R = i_1 - i_H = 5$$ The voltage $V$ is given by $$V = IR$$ Therefore, the resistance $R$ is $$R = \frac{25}{5} = 5\, \Omega$$
Question 55
Physics · Alternating Current · Numerical
An alternating emf $E = 110\sqrt{2} \sin 100t$ volt is applied to a capacitor of $2\mu F$, the rms value of current in the circuit is _____ mA,
Answer: 22
Solution
Given $C = 2 \mu f$ and $E = 110 \sqrt{2} \sin(100t)$. The capacitive reactance $X_C$ is calculated as follows: $$X_C = \frac{1}{\omega C} = \frac{1}{100 \times 2 \times 10^{-6}}$$ $$= \frac{10000}{2} = 5000 \Omega$$ The peak current $i_o$ is given by: $$i_o = \frac{110 \sqrt{2}}{5000}$$ The RMS current $i_{rms}$ is: $$i_{rms} = \frac{110 \sqrt{2}}{5000 \sqrt{2}}$$ $$= \frac{110}{5} \, mA$$ $$= 22 \, mA$$
Question 56
Physics · Wave Optics · Numerical
Two slits are 1 mm apart and the screen is located 1 m away from the slits. A light of wavelength 500 nm is used. The width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern is _____ $\times 10^{-4}$ $\mathrm{m}$
An object of mass 0.2 kg executes simple harmonic motion along x axis with frequency of $\left( \frac{25}{\pi} \right)$ Hz. At the position $x = 0.04$ m the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation is _____ cm.
Answer: 6
Solution
Total energy = K.E. + P.E. At $x = 0.04 \, \mathrm{m}$, T.E. = 0.5 + 0.4 = 0.9 J. T.E = $\frac{1}{2}$ m $\omega$^2 A^2 = 0.9 $$= \frac{1}{2} \times 0.2 \left( 2\pi \times \frac{25}{\pi} \right)^2 \times A^2 = 0.9$$ $\Rightarrow$ A = 0.06 $\mathrm{m}$ $_$ A = 6 $\mathrm{cm}$
A potential divider circuit is connected with a dc source of 20 V, a light emitting diode of glow in voltage 1.8 V and a zener diode of breakdown voltage of 3.2 V. The length (PR) of the resistive wire is 20 cm. The minimum length of PQ to just glow the LED is _____ cm
A body of mass $M$ thrown horizontally with velocity $v$ from the top of the tower of height $H$ touches the ground at a distance of $100 \, \mathrm{m}$ from the foot of the tower. A body of mass $2M$ thrown at a velocity $\frac{v}{2}$ from the top of the tower of height $4H$ will touch the ground at a distance of _____ m.
Answer: 100
Solution
Given the equations for the horizontal distances: $$100 = v \sqrt{\frac{2H}{g}};$$ For the second scenario, $$x = \frac{v}{2} \sqrt{\frac{2(4H)}{g}} = v \sqrt{\frac{2H}{g}}.$$ Thus, $$\Rightarrow x = 100.$$
Question 60
Physics · Work, Energy and Power · Numerical
A circular table is rotating with an angular velocity of $\omega \, \mathrm{rad/s}$ about its axis (see figure). There is a smooth groove along a radial direction on the table. A steel ball is gently placed at a distance of $1 \, \mathrm{m}$ on the groove. All the surfaces are smooth. If the radius of the table is $3 \, \mathrm{m}$, the radial velocity of the ball w.r.t. the table at the time ball leaves the table is $x \sqrt{2} \omega \mathrm{m/s}$, where the value of $x$ is _____.
Answer: 2
Solution
Given $a_c = \omega^2 x$. We have $v \frac{dv}{dx} = \omega^2 x$. Integrating both sides, $$\int_0^v v \, dv = \int_1^3 \omega^2 x \, dx$$ $$\frac{v^2}{2} = \omega^2 \left[ \frac{x^2}{2} \right]$$ $$\frac{v^2}{2} = \frac{\omega^2}{2} \left[ 3^2 - 1^2 \right]$$ $$v = 2 \sqrt{2} \omega$$ Therefore, $x = 2$.
Chemistry
Question 61
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
In qualitative test for identification of presence of phosphorous, the compound is heated with an oxidising agent. Which is further treated with nitric acid and ammonium molybdate respectively. The yellow coloured precipitate obtained is :
The reaction is as follows: $$\mathrm{PO_4^{3-}} or \mathrm{HPO_4^{2-}} + (\mathrm{NH_4})_2\mathrm{MoO_4} \xrightarrow{\mathrm{H^+}} (\mathrm{NH_4})_3\mathrm{PO_4} \cdot 12\mathrm{MoO_3} \downarrow$$ This forms a canary yellow precipitate known as ammonium phospho molybdate.
Question 62
Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct
For a reaction $\mathrm{A} \xrightarrow{K_1} \mathrm{B} \xrightarrow{K_2} \mathrm{C}$ If the rate of formation of $\mathrm{B}$ is set to be zero then the concentration of $\mathrm{B}$ is given by:
$(K_1 + K_2) [\mathrm{A}]$
$(K_1/K_2) [\mathrm{A}]$
$(K_1 - K_2) [\mathrm{A}]$
$K_1 K_2 [\mathrm{A}]$
Answer: (b)
Solution
Rate of formation of $B$ is $$\frac{\mathrm{d}[B]}{\mathrm{d}t} = k_1 [A] - k_2 [B]$$ $$0 = k_1 [A] - k_2 [B]$$ $$\left( \frac{k_1}{k_2} \right) [A] = [B]$$
Question 63
Chemistry · Chemical Bonding and Molecular Structure · Single correct
When $\psi_A$ and $\psi_B$ are the wave functions of atomic orbitals, then $\sigma^*$ is represented by:
$\psi_A + 2\psi_B$
$\psi_A - \psi_B$
$\psi_A + \psi_B$
$\psi_A - 2\psi_B$
Answer: (b)
Solution
Antibonding molecular orbitals are formed by destructive interference of wave functions. $$(ABMO) \sigma^* = \psi_A - \psi_B$$
Question 64
Chemistry · Alcohols, Phenols and Ethers · Single correct
Which one the following compounds will readily react with dilute NaOH ?
C_2H_5OH
C_6H_5OH
C_6H_5CH_2OH
(CH_3)_3COH
Answer: (b)
Solution
The given reaction shows that the phenol is reacting with NaOH to form phenoxide ion and water. This indicates that phenol is a stronger acid than water.
Question 65
Chemistry · Chemical Bonding and Molecular Structure · Single correct
The shape of carbocation is:
diagonal pyramidal
trigonal planar
tetrahedral
diagonal
Answer: (b)
Solution
The image shows a carbocation with a trigonal planar geometry. The central carbon atom is bonded to three hydrogen atoms and carries a positive charge.
Question 66
Chemistry · Haloalkanes and Haloarenes · Single correct
Given below are two statements : Statement (I) : $S_{N}2$ reactions are 'stereospecific', indicating that they result in the formation of only one stereoisomer as the product. Statement (II) : $S_{N}1$ reactions generally result in formation of product as racemic mixtures. In the light of the above statements, choose the correct answer from the options given below :
Chemistry · Co-ordination Compounds · Single correct
Match List-I with List-II.
(A)-(IV), (B)-(III), \text(C)-(III), (D)-(I)
(A)-(I), (B)-(IV), \text(C)-(II), (D)-(III)
(A)-(III), (B)-(II), \text(C)-(I), (D)-(IV)
(A)-(II), (B)-(IV), \text(C)-(I), (D)-(III)
Answer: (d)
Solution
Question 68
Chemistry · Co-ordination Compounds · Single correct
Match List - I with List - II Choose the correct answer from the options given below:
(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Solution
(A) Bayer's test → Unsaturation (B) Ceric ammonium nitrate test → Alcoholic-OH group (C) Phthalein dye test → Phenol (D) Schiff's test → Aldehyde
Question 69
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Identify the incorrect statements about group 15 elements : (A) Dinitrogen is a diatomic gas which acts like an inert gas at room temperature. (B) The common oxidation states of these elements are $-3$, $+3$ and $+5$ . (C) Nitrogen has unique ability to form $p\pi - p\pi$ multiple bonds. (D) The stability of $+5$ oxidation states increases down the group. (E) Nitrogen shows a maximum covalency of 6. Choose the correct answer from the options given below :
(A), (C), (E) only
(A), (D), (E) only
(A) and (E) only
(A), (B), (D) only
Answer: (c)
Solution
(D) Due to inert pair effect lower oxidation state is more stable. (E) Nitrogen belongs to $2^{nd}$ period and cannot expand its octet.
Question 70
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
IUPAC name of the following hydrocarbon (X) is :
\quad 2\text{-Ethyl-}3,6\text{-dimethylheptane}
\quad 2,5,6\text{-Trimethyloctane}
\quad 3,4,7\text{-Trimethyloctane}
\quad 2\text{-Ethyl-}2,6\text{-dimethylheptane}
Answer: (b)
Question 71
Chemistry · Equilibrium · Single correct
The equilibrium $Cr_2O_7^{2-}⇌2CrO_4^{2-}$ is shifted to the right in :
Given below are two statements : Statement (I) : A Buffer solution is the mixture of a salt and an acid or a base mixed in any particular quantities. Statement (II) : Blood is naturally occurring buffer solution whose pH is maintained by $\mathrm{H_2CO_3/HCO_3^-}$ concentrations. In the light of the above statements, choose the correct 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: (c)
Solution
Buffer solution is a mixture of either weak acid / weak base and its respective conjugate. Blood is a buffer solution of carbonic acid $\mathrm{H_2CO_3}$ and bicarbonate $\mathrm{HCO_3^-}$. Statement 1 is false but Statement II is true.
Question 73
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The correct sequence of acidic strength of the following aliphatic acids in their decreasing order is: $CH_3CH_2COOH,\ CH_3COOH,$ $CH_3CH_2CH_2COOH,\ HCOOH$
The correct order is: $$\mathrm{HCOOH} > \mathrm{CH_3COOH} > \mathrm{CH_3CH_2COOH} > \mathrm{CH_3CH_2CH_2COOH}$$
Question 74
Chemistry · Amines · Single correct
Given below are two statements: Statement (I): All the following compounds react with p-toluenesulfonyl chloride. $\mathrm{C_6H_5NH_2}$ $(\mathrm{C_6H_5})_2\mathrm{NH}$ $(\mathrm{C_6H_5})_3\mathrm{N}$ Statement (II): Their products in the above reaction are soluble is aqueous NaOH. In the light of the above statements, choose the correct answer from the options given below
Statement I is false but Statement II is true
Statement I is true but Statement II is false
Both Statement I and Statement II are true
Both Statement I and Statement II are false
Answer: (d)
Solution
Hinsberg test given by $1^\circ$ amine only.
Question 75
Chemistry · Electrochemistry · Single correct
The emf of cell $\mathrm{Tl}\ |\ \mathrm{Tl}^{+}_{(0.001M)}\ ||\ \mathrm{Cu}^{2+}_{(0.01M)}\ |\ \mathrm{Cu}$ is $0.83\ \mathrm{V}$ at $298\ \mathrm{K}$. It could be increased by :
decreasing concentration of both $\mathrm{Tl}^{+}$ and $\mathrm{Cu}^{2+}$ ions
increasing concentration of $\mathrm{Cu}^{2+}$ ions
increasing concentration of $\mathrm{Tl}^{+}$ ions
increasing concentration of both $\mathrm{Tl}^{+}$ and $\mathrm{Cu}^{2+}$ ions
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Identify the correct statements about p-block elements and their compounds. (A) Non metals have higher electronegativity than metals. (B) Non metals have lower ionisation enthalpy than metals. (C) Compounds formed between highly reactive nonmetals and highly reactive metals are generally ionic. (D) The non-metal oxides are generally basic in nature. (E) The metal oxides are generally acidic or neutral in nature. Choose the correct answer from the options given below :
(1) (B) and (D) only
(2) (A) and (C) only
(3) (D) and (E) only
(4) (B) and (E) only
Answer: (b)
Solution
As electronegativity increases non-metallic nature increases. Along the period ionisation energy increases. High electronegativity difference results in ionic bond formation. Oxides of metals are generally basic and that of non-metals are acidic in nature.
Question 77
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Given below are two statements : Statement (I) : Kjeldahl method is applicable to estimate nitrogen in pyridine. Statement (II) : The nitrogen present in pyridine can easily be converted into ammonium sulphate in Kjeldahl method. 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 false but Statement II is true
Statement I is true but Statement II is false
Answer: (b)
Solution
Nitrogen present in pyridine can not be estimated by Kjeldahl method as the nitrogen present in pyridine can not be easily converted into ammonium sulphate.
Question 78
Chemistry · Electrochemistry · Single correct
The reaction; $$\frac{1}{2} \mathrm{H}_2 (g) + \mathrm{AgCl} (s) \rightarrow \mathrm{H}^+ (aq) + \mathrm{Cl}^- (aq) + \mathrm{Ag} (s)$$ occurs in which of the following galvanic cell:
Chemistry · The d-and f-Block Elements · Single correct
Given below are two statements : Statement (I) : Fusion of $\mathrm{MnO_2}$ with $\mathrm{KOH}$ and an oxidising agent gives dark green $\mathrm{K_2MnO_4}$. Statement (II) : Manganate ion on electrolytic oxidation in alkaline medium gives permanganate ion. In the light of the above statements, choose the correct answer from the options given below :
Statement I is true but Statement II is false
Both Statement I and Statement II are false
Statement I is false but Statement II is true
Both Statement I and Statement II are true
Answer: (d)
Solution
Given the reaction: $$\mathrm{MnO_2 + 4KOH + O_2 \xrightarrow{fused} 2\, K_2MnO_4 + 2H_2O}$$ The color is dark green. Electrolytic oxidation in alkaline medium: At anode: $$\mathrm{MnO_4^{2-} \rightarrow MnO_4^- + e^-}$$
Question 80
Chemistry · Co-ordination Compounds · Single correct
Match List - I with List - II. \[ \begin{array}{ll} \textbf{List-I} & \textbf{List-II} \\ \text{(Complex ion)} & \text{(Spin only magnetic moment in B.M.)} \end{array} \] \[ \begin{array}{ll} \text{(A)}\ [\mathrm{Cr(NH_3)_6}]^{3+} & \text{(I)}\ 4.90 \\[6pt] \text{(B)}\ [\mathrm{NiCl_4}]^{2-} & \text{(II)}\ 3.87 \\[6pt] \text{(C)}\ [\mathrm{CoF_6}]^{3-} & \text{(III)}\ 0.0 \\[6pt] \text{(D)}\ [\mathrm{Ni(CN)_4}]^{2-} & \text{(IV)}\ 2.83 \end{array} \]
(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
(A)-(I), (B)-(IV), (C)-(II), (D)-(III)
(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Answer: (b)
Solution
For option (A) $[\mathrm{Cr(NH_3)_6}]^{3+}$, $\mathrm{Cr^{3+}}$ has configuration $3d^3$. The number of unpaired electrons $n = 3$. The magnetic moment $\mu \simeq 3.87 \, \mathrm{B.M.}$ (II). For option (B) $[\mathrm{NiCl_4}]^{2-}$, $\mathrm{Ni^{2+}}$ has configuration $3d^8$. The number of unpaired electrons $n = 2$. The magnetic moment $\mu \simeq 2.83 \, \mathrm{B.M.}$ (IV). For option (C) $[\mathrm{CoF_6}]^{3-}$, $\mathrm{Co^{3+}}$ has configuration $3d^6$. The number of unpaired electrons $n = 4$. The magnetic moment $\mu \simeq 4.90 \, \mathrm{B.M.}$ (I). For option (D) $[\mathrm{Ni(CN)_4}]^{2-}$, $\mathrm{Ni^{2+}}$ has configuration $3d^8$. The number of unpaired electrons $n = 0$. The magnetic moment $\mu = 0 \, \mathrm{B.M.}$ (III).
Question 81
Chemistry · Thermodynamics · Numerical
$\Delta_{vap} H^\circ$ for water is $+40.79 \, \mathrm{kJ \, mol^{-1}}$ at 1 bar and $100^\circ \mathrm{C}$. Change in internal energy for this vapourisation under same condition is $\_\_\_\_ \, \mathrm{kJ \, mol^{-1}}$. (Integer answer) (Given R = 8.3J $K^{-1}$ $mol^{-1}$)
Total number of optically active compounds from the following is
Answer: 1
Solution
Question 84
Chemistry · Biomolecules · Numerical
The total number of carbon atoms present in tyrosine, an amino acid, is
Answer: 9
Solution
Tyrosine. Number of carbon atoms = 9.
Question 85
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical
Two moles of benzaldehyde and one mole of acetone under alkaline conditions using aqueous NaOH after heating gives $x$ as the major product. The number of $\pi$ bonds in the product $x$ is ______
Total number of aromatic compounds among the following compounds is .
Answer: 1
Solution
Question 87
Chemistry · Solutions · Numerical
Molality of an aqueous solution of urea is 4.44 $\mathrm{m}$. Mole fraction of urea in solution is $x \times 10^{-3}$. Value of $x$ is
Answer: 74
Solution
Molality of urea is 4.44 m, that means 4.44 moles of urea present in 1000 gm of water. Therefore, $X_{urea} = \frac{4.44}{4.44 + \frac{1000}{18}} = 0.0740$ OR $74 \times 10^{-3}$ $X = 74$
Question 88
Chemistry · Co-ordination Compounds · Numerical
Total number of unpaired electrons in the complex ions $[Co(NH_3)_6]^{3+}$ and $[NiCl_4]^{2-}$ is
Answer: 2
Solution
For $\mathrm{Co^{+3}}$: $3d^6 \ t_{2g}^{2,2,2} \ e_g^{0,0}$. Unpaired $e^-$ is $0$. For $\mathrm{Ni^{+2}}$: $3d^8 \ e_g^{2,2} \ t_{2g}^{2,1,1}$. Unpaired $e^-$ is $2$.
Question 89
Chemistry · Some Basic Concepts of Chemistry · Numerical
Wavenumber for a radiation having $5800\,\mathring{A}$ wavelength is x $\times 10$ $cm^{-1}$. The value of x is (Integer answer)
Answer: 1724
Solution
The wave number $\bar{\nu}$ is given by $$\bar{\nu} (wave no.) = \frac{1}{\lambda} = \frac{1}{5800 \times 10^{-8} \, cm} = 17241.$$ OR $$1724 \times 10 \, cm^{-1} \Rightarrow x = 1724.$$
Question 90
Chemistry · Solutions · Numerical
A solution is prepared by adding 1 mole ethyl alcohol in 9 mole water. The mass percent of solute in the solution is ________ (Integer answer) (Given: Molar mass in g $mol^{-1}$ Ethyl alcohol: 46 water: 18)
Answer: 22
Solution
Mass percent of Alcohol $=\frac{\mathrm{Mass\ of\ ethyl\ alcohol}}{\mathrm{Total\ mass\ of\ solution}} \times 100$ $=\frac{1 \times 46}{1 \times 46 + 9 \times 18} \times 100 = \frac{4600}{208}$ $=22.11$ Or $22$