JEE Main 27 January 2024 Shift 1 question paper with solutions

JEE Main 27 January 2024 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.

Maths

Question 1

Maths · Binomial Theorem · Single correct

${}^{n-1}C_r = (k^2 - 8)\, {}^nC_{r+1}$ if and only if:

$2\sqrt{2} < k \leq 3$
$2\sqrt{3} < k \leq 3\sqrt{2}$
$2\sqrt{3} < k < 3\sqrt{3}$
$2\sqrt{2} < k < 2\sqrt{3}$

Answer: (a)

Solution

${}^{n-1}C_r = (k^2 - 8)\, {}^nC_{r+1}$ $\underbrace{r + 1 \geq 0,\ r \geq 0}_{r \geq 0}$ $$\frac{{}^nC_r}{{}^nC_{r+1}} = k^2 - 8$$ $$\frac{r+1}{n} = k^2 - 8$$ $\Rightarrow k^2 - 8 > 0$ $(k - 2\sqrt{2})(k + 2\sqrt{2}) > 0$ $k \in (-\infty,\ -2\sqrt{2}) \cup (2\sqrt{2},\ \infty) \hfill \ldots\text{(I)}$ $\therefore\ n \geq r + 1,\ \dfrac{r+1}{n} \leq 1$ $\Rightarrow k^2 - 8 \leq 1$ $k^2 - 9 \leq 0$ $-3 \leq k \leq 3 \hfill \ldots\text{(II)}$ From equation (I) and (II) we get $k \in [-3,\ -2\sqrt{2}) \cup (2\sqrt{2},\ 3]$

Question 2

Maths · Three Dimensional Geometry · Single correct

The distance, of the point $(7, -2, 11)$ from the line $\frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$ along the line $\frac{x-5}{2} = \frac{y-1}{-3} = \frac{z-5}{6}$, is:

Answer: (b)

Solution

Given $B = (2\lambda + 7, -3\lambda - 2, 6\lambda + 11)$ and $(7, -2, 11)$. Point B lies on $$\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$$ $$\frac{2\lambda + 7 - 6}{1} = \frac{-3\lambda - 2 - 4}{0} = \frac{6\lambda + 11 - 8}{3}$$ $$-3\lambda - 6 = 0$$ $$\lambda = -2$$ $$AB = \sqrt{(7 - 3)^2 + (4 + 2)^2 + (11 + 1)^2}$$ $$= \sqrt{16 + 36 + 144}$$ $$= \sqrt{196} = 14$$

Question 3

Maths · Differential Equations · Single correct

Let $x = x(t)$ and $y = y(t)$ be solutions of the differential equations $\frac{dx}{dt} + ax = 0$ and $\frac{dy}{dt} + by = 0$ respectively, $a, b \in \mathbb{R}$. Given that $x(0) = 2; y(0) = 1$ and $3y(1) = 2x(1)$, the value of $t$, for which $x(t) = y(t)$, is :

$\log_{\frac{2}{3}} 2$
$\log_4 3$
$\log_3 4$
$\log_{\frac{4}{3}} 2$

Answer: (d)

Solution

Given $\dfrac{dx}{dt}+ax=0$ $\dfrac{dx}{x}=-a\,dt$ $\int \dfrac{dx}{x}=-a\int dt$ $\ln|x|=-at+c$ At $t=0,\ x=2$ $\ln 2=0+c$ $\ln x=-at+\ln 2$ $\dfrac{x}{2}=e^{-at}$ $x=2e^{-at}$ ...(i) $\dfrac{dy}{dt}+by=0$ $\dfrac{dy}{y}=-b\,dt$ $\ln|y|=-bt+\lambda$ At $t=0,\ y=1$ $0=0+\lambda$ $\lambda=0$ $y=e^{-bt}$ ...(ii) According to the question, $3y(1)=2x(1)$ $3e^{-b}=2(2e^{-a})$ $e^{a-b}=\dfrac{4}{3}$ For $x(t)=y(t)$, $\Rightarrow 2e^{-at}=e^{-bt}$ $2=e^{(a-b)t}$ $2=\left(\dfrac{4}{3}\right)^t$ $\therefore\ t=\log_{\frac{4}{3}}2$

Question 4

Maths · Properties of Triangles · Single correct

If $(a, b)$ be the orthocentre of the triangle whose vertices are $(1, 2)$, $(2, 3)$ and $(3, 1)$, and $$I_1 = \int_a^b x \sin(4x - x^2) \, dx, \ I_2 = \int_a^b \sin(4x - x^2) \, dx,$$ then $36 \frac{I_1}{I_2}$ is equal to:

Answer: (a)

Solution

Equation of CE $$y - 1 = -(x - 3)$$ $$x + y = 4$$ Orthocentre lies on the line $x + y = 4$ so, $a + b = 4$ $$I_1 = \int_a^b x \sin(x(4-x)) \, dx \ldots (i)$$ Using king rule $$I_1 = \int_a^b (4-x) \sin(x(4-x)) \, dx \ldots (ii)$$ (i) + (ii) $$2I_1 = \int_a^b 4 \sin(x(4-x)) \, dx$$ $$2I_1 = 4I_2$$ $$I_1 = 2I_2$$ $$\frac{I_1}{I_2} = 2$$ $$\frac{36I_1}{I_2} = 72$$

Question 5

Maths · Binomial Theorem · Single correct

If $A$ denotes the sum of all the coefficients in the expansion of $(1 - 3x + 10x^2)^n$ and $B$ denotes the sum of all the coefficients in the expansion of $(1 + x^2)^n$, then:

$A = B^3$
$3A = B$
$B = A^3$
$A = 3B$

Answer: (a)

Solution

Sum of coefficients in the expansion of $(1 - 3x + 10x^2)^n = A$ then $A = (1 - 3 + 10)^n = 8^n$ (put $x = 1$) and sum of coefficients in the expansion of $(1 + x^2)^n = B$ then $B = (1 + 1)^n = 2^n$ $A = B^3$

Question 6

Maths · Sequences and Series · Single correct

The number of common terms in the progressions $4, 9, 14, 19, \ldots$, up to $25^{\text{th}}$ term and $3, 6, 9, 12, \ldots$, up to $37^{\text{th}}$ term is:

Answer: (c)

Solution

4, 9, 14, 19, $\ldots$, up to 25^{th} term $$T_{25} = 4 + (25 - 1)5 = 4 + 120 = 124$$ 3, 6, 9, 12, $\ldots$, up to 37^{th} term $$T_{37} = 3 + (37 - 1)3 = 3 + 108 = 111$$ Common difference of I^{st} series d_1 = 5 Common difference of II^{nd} series d_2 = 3 First common term = 9, and their common difference = 15 (LCM of d_1 and d_2) then common terms are 9, 24, 39, 54, 69, 84, 99

Question 7

Maths · Conic Sections · Single correct

If the shortest distance of the parabola $y^2 = 4x$ from the centre of the circle $x^2 + y^2 - 4x - 16y + 64 = 0$ is $d$, then $d^2$ is equal to :

Answer: (c)

Solution

Equation of normal to parabola $$y = mx - 2m - m^3$$ this normal passing through center of circle $(2, 8)$ $$8 = 2m - 2m - m^3$$ $$m = -2$$ So point $P$ on parabola $\Rightarrow (am^2, -2am) = (4, 4)$ And $C = (2, 8)$ $$PC = \sqrt{4 + 16} = \sqrt{20}$$ $$d^2 = 20$$

Question 8

Maths · Three Dimensional Geometry · Single correct

If the shortest distance between the lines $\frac{x-4}{1} = \frac{y+1}{2} = \frac{z}{-3}$ and $\frac{x-\lambda}{2} = \frac{y+1}{4} = \frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$, then the sum of all possible values of $\lambda$ is:

Answer: (b)

Solution

The shortest distance between the lines is given by $$\frac{(\vec{a} - \vec{b}) \cdot (\vec{d_1} \times \vec{d_2})}{|\vec{d_1} \times \vec{d_2}|}$$ where $$\begin{vmatrix} \lambda - 4 & 0 & 2 \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix}$$ This is equal to $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix}$$ Calculating the determinant, we have $$\frac{(\lambda - 4)(-10 + 12) - 0 + 2(4 - 4)}{|2\hat{i} - 1\hat{j} + 0\hat{k}|}$$ Simplifying, we get $$\frac{6}{\sqrt{5}} = \frac{2(\lambda - 4)}{\sqrt{5}}$$ Thus, $$3 = |\lambda - 4|$$ This gives $$\lambda - 4 = \pm 3$$ So, $$\lambda = 7, 1$$ The sum of all possible values of $\lambda$ is 8.

Question 9

Maths · Integrals · Single correct

If $\displaystyle \int_{0}^{1} \frac{1}{\sqrt{3+x}+\sqrt{1+x}}\,dx = a+b\sqrt{2}+c\sqrt{3}$, where $a$, $b$, and $c$ are rational numbers, then $2a+3b-4c$ is equal to:

Answer: (d)

Solution

The integral is given by $$\int_0^1 \frac{1}{\sqrt{3+x} + \sqrt{1+x}} \, dx = \int_0^1 \frac{\sqrt{3+x} - \sqrt{1+x}}{(3+x) - (1+x)} \, dx$$ Simplifying, we have $$\frac{1}{2} \left[ \int_0^1 \sqrt{3+x} \, dx - \int_0^1 (\sqrt{1+x}) \, dx \right]$$ Evaluating the integrals, we get $$\frac{1}{2} \left[ 2 \frac{(3+x)^{\frac{3}{2}}}{3} - 2 \frac{(1+x)^{\frac{3}{2}}}{3} \right]_0^1$$ This simplifies to $$\frac{1}{2} \left[ \frac{2}{3} (8 - 3\sqrt{3}) - \frac{2}{3} \left( 2^{\frac{3}{2}} - 1 \right) \right]$$ Further simplification gives $$\frac{1}{3} [8 - 3\sqrt{3} - 2\sqrt{2} + 1]$$ This can be expressed as $$= 3 - \sqrt{3} - \frac{2}{3} \sqrt{2} = a + b\sqrt{2} + c\sqrt{3}$$ where $$a = 3, b = -\frac{2}{3}, c = -1$$ Finally, we calculate $$2a + 3b - 4c = 6 - 2 + 4 = 8$$

Question 10

Maths · Sets · Single correct

Let $S = \{1, 2, 3, \ldots, 10\}$. Suppose $M$ is the set of all the subsets of $S$, then the relation $R = \{(A, B): A \cap B \neq \emptyset; A, B \in M\}$ is:

symmetric and reflexive only
reflexive only
symmetric and transitive only
symmetric only

Answer: (d)

Solution

Question 11

Maths · Complex Numbers and Quadratic Equations · Single correct

If $S = \{ z \in \mathbb{C} : |z - i| = |z + i| = |z - 1| \}$, then, $n(S)$ is:

Answer: (a)

Solution

Given $|z - i| = |z + i| = |z - 1|$. ABC is a triangle. Hence its circum-centre will be the only point whose distance from A, B, C will be same. So $n(S) = 1$

Question 12

Maths · Conic Sections · Single correct

Four distinct points $(2k, 3k)$, $(1, 0)$, $(0, 1)$ and $(0, 0)$ lie on a circle for $k$ equal to:

$\frac{2}{13}$
$\frac{3}{13}$
$\frac{5}{13}$
$\frac{1}{13}$

Answer: (c)

Solution

(2k, 3k) will lie on circle whose diameter is AB. $$ (x - 1)(x) + (y - 1)(y) = 0 $$ $$ x^2 + y^2 - x - y = 0 \ldots (i) $$ Satisfy (2k, 3k) in (i) $$ (2k)^2 + (3k)^2 - 2k - 3k = 0 $$ $$ 13k^2 - 5k = 0 $$ $$ k = 0, k = \frac{5}{13} $$ hence k = $\frac{5}{13}$

Question 13

Maths · Continuity and Differentiability · Single correct

Consider the function. $$f(x) = \begin{cases} \frac{a(7x-12-x^2)}{b|x^2-7x+12|}, & x 3 \\ b, & x = 3 \end{cases}$$ Where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$. If $S$ denotes the set of all ordered pairs $(a, b)$ such that $f(x)$ is continuous at $x = 3$, then the number of elements in $S$ is :

2
Infinitely many
4
1

Answer: (d)

Solution

Given $f(3^-) = \frac{a}{b} \frac{(7x - 12 - x^2)}{|x^2 - 7x + 12|}$ (for $f(x)$ to be continuous). Therefore, $f(3^-) = \frac{-a}{b} \frac{(x-3)(x-4)}{(x-3)(x-4)}; x < 3$. Hence $f(3^-) = \frac{-a}{b}$. Then $f(3^+) = 2 \lim_{x \to 3^+} \left( \frac{\sin(x-3)}{x-3} \right) = 2a$ and $f(3) = b$. Hence $f(3) = f(3^+) = f(3^-)$. Therefore, $b = 2 = \frac{-a}{b}$. Solving gives $b = 2, a = -4$. Hence only 1 ordered pair $(-4, 2)$.

Question 14

Maths · Statistics · Single correct

Let $a_1, a_2, \ldots, a_{10}$ be 10 observations such that $\sum_{k=1}^{10} a_k = 50$ and $\sum_{k<j} a_k \cdot a_j = 1100$. Then the standard deviation of $a_1, a_2, \ldots, a_{10}$ is equal to:

5
$\sqrt{5}$
10
$\sqrt{115}$

Answer: (b)

Solution

Given $\sum_{k=1}^{10} a_k = 50$. $a_1 + a_2 + \ldots + a_{10} = 50 \ldots (i)$ $\sum_{\forall k < j} a_k a_j = 1100 \ldots (ii)$ If $a_1 + a_2 + \ldots + a_{10} = 50$. $(a_1 + a_2 + \ldots + a_{10})^2 = 2500$ $$\Rightarrow \sum_{i=1}^{10} a_i^2 + 2 \sum_{k < j} a_k a_j = 2500$$ $$\Rightarrow \sum_{i=1}^{10} a_i^2 = 2500 - 2(1100)$$ $\sum_{i=1}^{10} a_i^2 = 300$, Standard deviation $\sigma$ $$= \sqrt{\frac{\sum a_i^2}{10} - \left(\frac{\sum a_i}{10}\right)^2} = \sqrt{\frac{300}{10} - \left(\frac{50}{10}\right)^2}$$ $$= \sqrt{30 - 25} = \sqrt{5}$$

Question 15

Maths · Conic Sections · Single correct

The length of the chord of the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$, whose mid point is $\left(1, \frac{2}{5}\right)$, is equal to:

$\frac{\sqrt{1691}}{5}$
$\frac{\sqrt{2009}}{5}$
$\frac{\sqrt{1741}}{5}$
$\frac{\sqrt{1541}}{5}$

Answer: (a)

Solution

Equation of chord with given middle point. $T = S_1$ $$\frac{x}{25} + \frac{y}{40} = \frac{1}{25} + \frac{1}{100}$$ $$8x + 5y = \frac{8 + 2}{200}$$ $$y = \frac{10 - 8x}{5} ...(i)$$ $$\frac{x^2}{25} + \frac{(10 - 8x)^2}{400} = 1 (put in original equation)$$ $$\frac{16x^2 + 100 + 64x^2 - 160x}{400} = 1$$ $$4x^2 - 8x - 15 = 0$$ $$x = \frac{8 \pm \sqrt{304}}{8}$$ $$x_1 = \frac{8 + \sqrt{304}}{8} ; x_2 = \frac{8 - \sqrt{304}}{8}$$ Similarly, $$y = \frac{10 - 18 \pm \sqrt{304}}{5} = \frac{2 \pm \sqrt{304}}{5}$$ $$y_1 = \frac{2 - \sqrt{304}}{5} ; y_2 = \frac{2 + \sqrt{304}}{5}$$ Distance $$= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$$ $$= \sqrt{\frac{4 \times 304}{64} + \frac{4 \times 304}{25}} = \sqrt{\frac{1691}{5}}$$

Question 16

Maths · Straight Lines and Pair of Straight Lines · Single correct

The portion of the line $4x + 5y = 20$ in the first quadrant is trisected by the lines $L_1$ and $L_2$ passing through the origin. The tangent of an angle between the lines $L_1$ and $L_2$ is:

$\frac{8}{5}$
$\frac{25}{41}$
$\frac{2}{5}$
$\frac{30}{41}$

Answer: (d)

Solution

Co-ordinates of A are $\left( \frac{5}{3}, \frac{8}{3} \right)$. Co-ordinates of B are $\left( \frac{10}{3}, \frac{4}{3} \right)$. Slope of OA is $m_1 = \frac{8}{5}$. Slope of OB is $m_2 = \frac{2}{5}$. The formula for $\tan \theta$ is given by: $$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$ Substituting the values, we get: $$\tan \theta = \frac{\frac{6}{5}}{1 + \frac{16}{25}} = \frac{30}{41}$$ Therefore, $\tan \theta = \frac{30}{41}$.

Question 17

Maths · Vector Algebra · Single correct

Let $\vec{a}$ = $\hat{i}$ + 2$\hat{j}$ + $\hat{k}$, $\vec{b}$ = 3($\hat{i}$ - $\hat{j}$ + $\hat{k}$) . Let $\vec{c}$ be the vector such that $\vec{a}$ $\times$ $\vec{c}$ = $\vec{b}$ and $\vec{a}$ $\cdot$ $\vec{c}$ = 3 . Then $\vec{a}$ $\cdot$ (($\vec{c}$ $\times$ $\vec{b}$) - $\vec{b}$ - $\vec{c}$) is equal to:

Answer: (b)

Solution

Given $\vec{a} \cdot [(\vec{c} \times \vec{b}) - \vec{b} - \vec{c}]$. $\vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c} \ldots (i)$ Given $\vec{a} \times \vec{c} = \vec{b}$ $$\Rightarrow (\vec{a} \times \vec{c}) \cdot \vec{b} = \vec{b} \cdot \vec{b} = |\vec{b}|^2 = 27$$ $$\Rightarrow \vec{a} \cdot (\vec{c} \times \vec{b}) = [\vec{a} \ \vec{c} \ \vec{b}] = (\vec{a} \times \vec{c}) \cdot \vec{b} = 27 \ldots (ii)$$ Now $\vec{a} \cdot \vec{b} = 3 - 6 + 3 = 0 \ldots (iii)$ $\vec{a} \cdot \vec{c} = 3 \ldots (iv) (given)$ By $(i), (ii), (iii) \ \& \ (iv)$ $$27 - 0 - 3 = 24$$

Question 18

Maths · Limits and Derivatives · Single correct

If $a = \lim_{x \to 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$ and $b = \lim_{x \to 0} \frac{\sin^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$, then the value of $ab^3$ is:

Answer: (b)

Solution

Given $$a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^4} - \sqrt{2}}}{x^4}$$ This simplifies to $$= \lim_{x \to 0} \frac{\sqrt{1 + x^4} - 1}{x^4 \left(\sqrt{1 + \sqrt{1 + x^4} + \sqrt{2}}\right)}$$ Further simplifying gives $$= \lim_{x \to 0} \frac{x^4}{x^4 \left(\sqrt{1 + \sqrt{1 + x^4} + \sqrt{2}}\right) \left(\sqrt{1 + x^4 + 1}\right)}$$ Applying the limit, we find $$a = \frac{1}{4\sqrt{2}}$$ Next, consider $$b = \lim_{x \to 0} \frac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}}$$ This simplifies to $$= \lim_{x \to 0} \frac{(1 - \cos^2 x) \left(\sqrt{2} + \sqrt{1 + \cos x}\right)}{2 - (1 + \cos x)}$$ Further simplifying gives $$b = \lim_{x \to 0} (1 + \cos x)(\sqrt{2} + \sqrt{1 + \cos x})$$ Applying the limits, we find $$b = 2(\sqrt{2} + \sqrt{2}) = 4\sqrt{2}$$ Now, $$ab^3 = \frac{1}{4\sqrt{2}} \times (4\sqrt{2})^3 = 32$$

Question 19

Maths · Matrices · Single correct

Consider the matrix $f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$. Given below are two statements: Statement I: $f(-x)$ is the inverse of the matrix $f(x)$. Statement II: $f(x)f(y) = f(x + y)$. 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
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Both Statement I and Statement II are true

Answer: (d)

Solution

Given $$f(-x) = \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$f(x) \cdot f(-x) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$$ Hence statement-I is correct. Now, checking statement II $$f(y) = \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$f(x) \cdot f(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$\Rightarrow f(x) \cdot f(y) = f(x+y)$$ Hence statement-II is also correct.

Question 20

Maths · Relations and Functions · Single correct

The function $f : \mathbb{N} - \{1\} \to \mathbb{N}$; defined by $f(n) = the highest prime factor of n$, is :

both one-one and onto
one-one only
onto only
neither one-one nor onto

Answer: (d)

Solution

Given $f : \mathbb{N} - \{1\} \to \mathbb{N}$. $f(n) =$ The highest prime factor of $n$. $f(2) = 2$ $f(4) = 2$ This implies many one. 4 is not the image of any element, which implies into. Hence many one and into. Neither one-one nor onto.

Question 21

Maths · Vector Algebra · Numerical

The least positive integral value of $\alpha$, for which the angle between the vectors $\alpha \hat{i} - 2 \hat{j} + 2 \hat{k}$ and $\alpha \hat{i} + 2 \alpha \hat{j} - 2 \hat{k}$ is acute, is_____

Answer: 5

Solution

Given $$\cos \theta = \frac{(\alpha \hat{i} - 2 \hat{j} + 2 \hat{k}) \cdot (\alpha \hat{i} + 2 \hat{j} - 2 \hat{k})}{\sqrt{\alpha^2 + 4 + 4} \sqrt{\alpha^2 + 4}}$$ Simplifying, we have $$\cos \theta = \frac{\alpha^2 - 4\alpha - 4}{\sqrt{\alpha^2 + 8 \sqrt{5} \alpha^2 + 4}}$$ This implies $$\alpha^2 - 4\alpha - 4 > 0$$ Rearranging gives $$\alpha^2 - 4\alpha + 4 > 8 \Rightarrow (\alpha - 2)^2 > 8$$ Thus, $$\alpha - 2 > 2\sqrt{2} or \alpha - 2 2 + 2\sqrt{2} or \alpha < 2 - 2\sqrt{2}$$ Therefore, $$\alpha \in (-\infty, -0.82) \cup (4.82, \infty)$$ The least positive integral value of $\alpha$ is $5$.

Question 22

Maths · Continuity and Differentiability · Numerical

Let for a differentiable function $f : (0, \infty) \to \mathbb{R}$, $f(x) - f(y) \geq \log_e \left( \frac{x}{y} \right) + x - y$, $\forall x, y \in (0, \infty)$. Then $$\sum_{n=1}^{20} f' \left( \frac{1}{n^2} \right)$$ is equal to .

Answer: 2890

Solution

Given $f(x) - f(y) \geq \ln x - \ln y + x - y$. $$\frac{f(x) - f(y)}{x-y} \geq \frac{\ln x - \ln y}{x-y} + 1$$ Let $x > y$. $$\lim_{y \to x} f'(x^-) \geq \frac{1}{x} + 1 \ldots (1)$$ Let $x < y$. $$\lim_{y \to x} f'(x^+) \leq \frac{1}{x} + 1 \ldots (2)$$ $f^1(x^-) = f^1(x^+)$ $$f^1(x) = \frac{1}{x} + 1$$ $$f'\left(\frac{1}{x^2}\right) = x^2 + 1$$ $$\sum_{x=1}^{20} (x^2 + 1) = \sum_{x=1}^{20} x^2 + 20$$ $$= \frac{20 \times 21 \times 41}{6} + 20$$ $$= 2890$$

Question 23

Maths · Differential Equations · Fill in the blank

If the solution of the differential equation $(2x + 3y - 2)dx + (4x + 6y - 7)dy = 0, y(0) = 3$, is $\alpha x + \beta y + 3 \log_e |2x + 3y - \gamma| = 6$, then $\alpha + 2\beta + 3\gamma$ is equal to _____

Answer: 29

Solution

Given the equations $2x + 3y - 2 = t$ and $4x + 6y - 4 = 2t$. We have $2 + 3 \frac{dy}{dx} = \frac{dt}{dx}$ and $4x + 6y - 7 = 2t - 3$. The derivative $\frac{dy}{dx}$ is given by $$\frac{dy}{dx} = \frac{-(2x + 3y - 2)}{4x + 6y - 7}.$$ The derivative $\frac{dt}{dx}$ is given by $$\frac{dt}{dx} = \frac{-3t + 4t - 6}{2t - 3} = \frac{t - 6}{2t - 3}.$$ Integrating both sides, we have $$\int \frac{2t - 3}{t - 6} \, dt = \int \, dx.$$ This simplifies to $$\int \left( \frac{2t - 12}{t - 6} + \frac{9}{t - 6} \right) \cdot dt = x.$$ Thus, $$2t + 9 \ln(t - 6) = x + c.$$ Substituting $2(2x + 3y - 2) + 9 \ln(2x + 3y - 8) = x$, we find $$x = 0, y = 3$$ and $$c = 14.$$ Therefore, $$4x + 6y - 4 + 9 \ln(2x + 3y - 8) = x + 14.$$ Simplifying gives $$x + 2y + 3 \ln(2x + 3y - 8) = 6.$$ With $\alpha = 1, \beta = 2, \gamma = 8$, we find $$\alpha + 2\beta + 3\gamma = 1 + 4 + 24 = 29.$$

Question 24

Maths · Applications of Integrals · Numerical

Let the area of the region $\{(x, y) : x - 2y + 4 \geq 0, x + 2y^2 \geq 0, x + 4y^2 \leq 8, y \geq 0\}$ be $\frac{m}{n}$, where $m$ and $n$ are coprime numbers. Then $m + n$ is equal to

Answer: 119

Solution

The area $A$ is given by the integral: $$A = \int_0^1 \left[(8 - 4y^2) - (-2y^2)\right] \, dy + \int_1^{3/2} \left[(8 - 4y^2) - (2y - 4)\right] \, dy$$ Evaluating the integrals, we have: $$= \left[8y - \frac{2y^3}{3}\right]_0^1 + \left[12y - y^2 - \frac{4y^3}{3}\right]_1^{3/2}$$ This simplifies to: $$= \frac{107}{12} = \frac{m}{n}$$ Therefore, $m + n = 119$.

Question 25

Maths · Sequences and Series · Numerical

If $$8 = 3 + \frac{1}{4}(3 + p) + \frac{1}{4^2}(3 + 2p) + \frac{1}{4^3}(3 + 3p) + \cdots \infty,$$ then the value of $p$ is

Answer: 9

Solution

Given $$8 = \frac{3}{1 - \frac{1}{4}} + \frac{p \cdot \frac{1}{4}}{\left(1 - \frac{1}{4}\right)^2}$$ The sum of infinite terms of an A.G.P is $$\left(\frac{a}{1-r} + \frac{dr}{(1-r)^2}\right)$$ Thus, $$\frac{4p}{9} = 4 \Rightarrow p = 9$$

Question 26

Maths · Probability · Numerical

A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let $a = P(X = 3)$, $b = P(X \geq 3)$ and $c = P(X \geq 6 \mid X > 3)$. Then $\frac{b+c}{a}$ is equal to

Answer: 12

Solution

a = $\mathrm{P}$(X = 3) = $\frac{5}{6}$ $\times$ $\frac{5}{6}$ $\times$ $\frac{1}{6}$ = $\frac{25}{216}$ b = $\mathrm{P}$(X $\geq$ 3) = $\frac{5}{6}$ $\times$ $\frac{5}{6}$ $\times$ $\frac{1}{6}$ + $\left$($\frac{5}{6}$$\right$)^3 $\cdot$ $\frac{1}{6}$ + $\left$($\frac{5}{6}$$\right$)^4 $\cdot$ $\frac{1}{6}$ + $\ldots$ = $\frac{25}{216}$ $\frac{1}{1 - \frac{5}{6}}$ = $\frac{25}{216}$ $\times$ $\frac{6}{1}$ = $\frac{25}{36}$ $\mathrm{P}$(X $\geq$ 6) = $\left$($\frac{5}{6}$$\right$)^5 $\cdot$ $\frac{1}{6}$ + $\left$($\frac{5}{6}$$\right$)^6 $\cdot$ $\frac{1}{6}$ + $\ldots$ = $\frac{\left(\frac{5}{6}\right)^5 \cdot \frac{1}{6}}{1 - \frac{5}{6}}$ = $\left$($\frac{5}{6}$$\right$)^5 c = $\left$($\frac{5}{6}$$\right$)^5 = $\frac{25}{36}$ $\frac{b + c}{a}$ = $\frac{\left(\frac{5}{6}\right)^2 + \left(\frac{5}{6}\right)^2}{\left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}}$ = 12

Question 27

Maths · Trigonometric Functions · Numerical

Let the set of all $a \in \mathbb{R}$ such that the equation $\cos 2x + a \sin x = 2a - 7$ has a solution be $[p, q]$ and $$r = \tan 9^\circ - \tan 27^\circ - \frac{1}{\cot 63^\circ} + \tan 81^\circ$$, then $pqr$ is equal to

Answer: 48

Solution

Given $\cos 2x + a \cdot \sin x = 2a - 7$. $a(\sin x - 2) = 2(\sin x - 2)(\sin x + 2)$ $\sin x = 2$, $a = 2(\sin x + 2)$ Therefore, $a \in [2, 6]$. $p = 2$, $q = 6$ $r = \tan 9^\circ + \cot 9^\circ - \tan 27 - \cot 27$ $$r = \frac{1}{\sin 9 \cdot \cos 9} - \frac{1}{\sin 27 \cdot \cos 27}$$ $$= 2 \left[ \frac{4}{\sqrt{5} - 1} - \frac{4}{\sqrt{5} + 1} \right]$$ $r = 4$ Therefore, $p \cdot q \cdot r = 2 \times 6 \times 4 = 48$

Question 28

Maths · Continuity and Differentiability · Numerical

Let $f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3), x \in R.$ Then $f'(10)$ is equal to _____

Answer: 202

Solution

Given $$f(x) = x^3 + x^2 \cdot f'(1) + x \cdot f''(2) + f'''(3)$$ $$f'(x) = 3x^2 + 2x f'(1) + f''(2)$$ $$f''(x) = 6x + 2f'(1)$$ $$f'''(x) = 6$$ $$f'(1) = -5, f''(2) = 2, f'''(3) = 6$$ Substitute these values into the equation: $$f(x) = x^3 + x^2 \cdot (-5) + x \cdot (2) + 6$$ Simplify: $$f'(x) = 3x^2 - 10x + 2$$ Evaluate at $x = 10$: $$f'(10) = 300 - 100 + 2 = 202$$

Question 29

Maths · Determinants · Numerical

Let $A = \begin{bmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$, $B = [B_1, B_2, B_3]$, where $B_1, B_2, B_3$ are column matrices, and $A_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$, $AB_2 = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}$, $AB_3 = \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$ If $\alpha = |B|$ and $\beta$ is the sum of all the diagonal elements of $B$, then $\alpha^3 + \beta^3$ is equal to

Answer: 28

Solution

Given matrices: $$A = \begin{bmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}, B = [B_1, B_2, B_3]$$ Where: $$B_1 = \begin{bmatrix} x_1 \\ y_1 \\ z_1 \end{bmatrix}, B_2 = \begin{bmatrix} x_2 \\ y_2 \\ z_2 \end{bmatrix}, B_3 = \begin{bmatrix} x_3 \\ y_3 \\ z_3 \end{bmatrix}$$ Calculating $AB_1$: $$AB_1 = \begin{bmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ y_1 \\ z_1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$ Solving gives $x_1 = 1$, $y_1 = -1$, $z_1 = -1$. Calculating $AB_2$: $$AB_2 = \begin{bmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_2 \\ y_2 \\ z_2 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}$$ Solving gives $x_2 = 2$, $y_2 = 1$, $z_2 = -2$. Calculating $AB_3$: $$AB_3 = \begin{bmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_3 \\ y_3 \\ z_3 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$$ Solving gives $x_3 = 2$, $y_3 = 0$, $z_3 = -1$. Thus, matrix $B$ is: $$B = \begin{bmatrix} 1 & 2 & 2 \\ -1 & 1 & 0 \\ -1 & -2 & -1 \end{bmatrix}$$ Calculating $\alpha$ and $\beta$: $$\alpha = |B| = 3$$ $$\beta = 1$$ Finally, $\alpha^3 + \beta^3 = 27 + 1 = 28$.

Question 30

Maths · Complex Numbers and Quadratic Equations · Numerical

If $\alpha$ satisfies the equation $x^2 + x + 1 = 0$ and $(1 + \alpha)^7 = A + B\alpha + C^2$, $A$, $B$, $C \geq 0$, then $5(3A - 2B - C)$ is equal to

Answer: 5

Solution

Given $x^2 + x + 1 = 0 \Rightarrow x = \omega, \omega^2 = \alpha$. Let $\alpha = \omega$. Now $(1 + \alpha)^7 = -\omega^{14} = -\omega^2 = 1 + \omega$. $A = 1$, $B = 1$, $C = 0$. Therefore, $5(3A - 2B - C) = 5(3 - 2 - 0) = 5$.

Physics

Question 31

Physics · Moving Charges and Magnetism · Single correct

Position of an ant (S in metres) moving in Y - Z plane is given by $S = 2t^2 \hat{j} + 5 \hat{k}$ (where t is in second). The magnitude and direction of velocity of the ant at $t = 1 \, \mathrm{s}$ will be :

16 m/s in $y$-direction
4 m/s in $x$-direction
9 m/s in $z$-direction
4 m/s in $y$-direction

Answer: (d)

Solution

Given $\vec{v} = \frac{d\vec{s}}{dt} = 4t \hat{\jmath}$. At $t = 1 \, sec$, $\vec{v} = 4 \hat{\jmath}$.

Question 32

Physics · Mechanical Properties of Fluids · Single correct

Given below are two statements : Statement (I) : Viscosity of gases is greater than that of liquids. Statement (II) : Surface tension of a liquid decreases due to the presence of insoluble impurities. In the light of the above statements, choose the most appropriate answer from the options given below:

Statement I is correct but statement II is incorrect
Statement I is incorrect but Statement II is correct
Both Statement I and Statement II are incorrect
Both Statement I and Statement II are correct

Answer: (b)

Solution

Gases have less viscosity. Due to insoluble impurities like detergent, surface tension decreases.

Question 33

Physics · Ray Optics and Optical Instruments · Single correct

If the refractive index of the material of a prism is $\cot \left( \frac{A}{2} \right)$, where $A$ is the angle of prism then the angle of minimum deviation will be

$\pi - 2A$
$\frac{\pi}{2} - 2A$
$\pi - A$
$\frac{\pi}{2} - A$

Answer: (a)

Solution

Given $\cot\frac{A}{2}=\frac{\sin\left(\frac{A+\delta_{\min}}{2}\right)}{\sin\frac{A}{2}}$. This implies $\cos\frac{A}{2}=\sin\left(\frac{A+\delta_{\min}}{2}\right)$. $\frac{A+\delta_{\min}}{2}=\frac{\pi}{2}-\frac{A}{2}$ $\delta_{\min}=\pi-2A$

Question 34

Physics · Moving Charges and Magnetism · Multiple correct

A proton moving with a constant velocity passes through a region of space without any change in its velocity. If $\vec{E}$ and $\vec{B}$ represent the electric and magnetic fields respectively, then the region of space may have:

$E = 0$, $B = 0$
$E = 0$, $B \neq 0$
$E \neq 0$, $B = 0$
$E \neq 0$, $B \neq 0$
, (B) and (C) only
, (C) and (D) only
, (B) and (D) only
, (C) and (D) only

Answer: (c)

Solution

Net force on particle must be zero i.e. $q \vec{E} + q \vec{V} \times \vec{B} = 0$. Possible cases are (i) $\vec{E} \& \vec{B} = 0$ (ii) $\vec{V} \times \vec{B} = 0$, $\vec{E} = 0$. Solutions (iii) $q \vec{E} = -q \vec{V} \times \vec{B}$ $\vec{E} \neq 0 \& \vec{B} \neq 0$

Question 35

Physics · Gravitation · Single correct

The acceleration due to gravity on the surface of earth is $g$. If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be:

$g/4$
$2g$
$g/2$
$4g$

Answer: (d)

Solution

Given $g = \frac{GM}{R^2}$, it follows that $g \propto \frac{1}{R^2}$. $$\frac{g_2}{g_1} = \frac{R_1^2}{R_2^2}$$ $$g_2 = 4 \, g_1 \left( R_2 = \frac{R_1}{2} \right)$$

Question 36

Physics · Laws of Motion · Single correct

A train is moving with a speed of 12 m/s on rails which are 1.5 m apart. To negotiate a curve radius 400 m, the height by which the outer rail should be raised with respect to the inner rail is (Given, $g = 10 \, \mathrm{m/s^2}$):

6.0 cm
5.4 cm
4.8 cm
4.2 cm

Answer: (b)

Solution

Given $\tan \theta = \frac{v^2}{R g} = \frac{12 \times 12}{10 \times 400}$. $\tan \theta = \frac{h}{1.5}$ Therefore, $\frac{h}{1.5} = \frac{144}{4000}$ $h = 5.4 \, \mathrm{cm}$

Question 37

Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct

Which of the following circuits is reverse - biased?

$+2\vec{\nu}\longrightarrow \mathrm{Mu}^{+4}+\vec{\nu}$

Answer: (d)

Solution

The $P$ end should be at higher potential for forward biasing.

Question 38

Physics · Experimental Physics · Single correct

Identify the physical quantity that cannot be measured using spherometer:

Radius of curvature of concave surface
Specific rotation of liquids
Thickness of thin plates
Radius of curvature of convex surface

Answer: (b)

Solution

Spherometer can be used to measure curvature of surface.

Question 39

Physics · Work, Energy and Power · Single correct

Two bodies of mass $4 \, \mathrm{g}$ and $25 \, \mathrm{g}$ are moving with equal kinetic energies. The ratio of magnitude of their linear momentum is:

3 : 5
5 : 4
2 : 5
4 : 5

Answer: (c)

Solution

Given $$$ $\frac{P_1^2}{2 \, m_1}$ = $\frac{P_2^2}{2 \, m_2}$ $$$. Therefore, $$$ $\frac{P_1}{P_2}$ = $\sqrt{\frac{m_1}{m_2}}$ = $\frac{2}{5}$ $$$.

Question 40

Physics · Thermodynamics · Single correct

0.08 kg air is heated at constant volume through $5^\circ \mathrm{C}$. The specific heat of air at constant volume is $0.17 \mathrm{kcal/kg}^\circ \mathrm{C}$ and $J = 4.18 \mathrm{joule/cal}$. The change in its internal energy is approximately.

318 J
298 J
284 J
142 J

Answer: (c)

Solution

Q = $\Delta$ U as work done is zero [constant volume] $\Delta$ U = ms$\Delta$ T = 0.08 $\times$ (170 $\times$ 4.18) $\times$ 5 $\approx$ 284 $\mathrm{J}$

Question 41

Physics · Atoms · Single correct

The radius of third stationary orbit of electron for Bohr's atom is $R$. The radius of fourth stationary orbit will be:

$\frac{4}{3} R$
$\frac{16}{9} R$
$\frac{3}{4} R$
$\frac{9}{16} R$

Answer: (b)

Solution

Given $r \propto \frac{n^2}{Z}$. The ratio $\frac{r_4}{r_3} = \frac{4^2}{3^2}$. Therefore, $r_4 = \frac{16}{9} R$.

Question 42

Physics · Electromagnetic Induction · Single correct

A rectangular loop of length 2.5 m and width 2 m is placed at $60^\circ$ to a magnetic field of 4 $\mathrm{T}$. The loop is removed from the field in 10 $\mathrm{sec}$. The average emf induced in the loop during this time is

-2 $\mathrm{V}$
+2 $\mathrm{V}$
+1 $\mathrm{V}$
-1 $\mathrm{V}$

Answer: (c)

Solution

Average emf = $\frac{Change in flux}{Time}$ = -$\frac{\Delta \phi}{\Delta t}$ $$= -\frac{0 - (4 \times (2.5 \times 2) \cos 60^\circ)}{10}$$ $$= +1 \, \mathrm{V}$$

Question 43

Physics · Electric Charges and Fields · Single correct

An electric charge $10^{-6} \, \mu \mathrm{C}$ is placed at origin $(0, 0) \, \mathrm{m}$ of $X - Y$ co-ordinate system. Two points $P$ and $Q$ are situated at $(\sqrt{3}, \sqrt{3}) \, \mathrm{m}$ and $(\sqrt{6}, 0) \, \mathrm{m}$ respectively. The potential difference between the points $P$ and $Q$ will be:

$\sqrt{3} \, \mathrm{V}$
$\sqrt{6} \, \mathrm{V}$
$0 \, \mathrm{V}$
$3 \, \mathrm{V}$

Answer: (c)

Solution

Potential difference = $\frac{KQ}{r_1}$ - $\frac{KQ}{r_2}$ $\newline$ r_1 = $\sqrt{(\sqrt{3})^2 + (\sqrt{3})^2}$ $\newline$ r_2 = $\sqrt{(\sqrt{6})^2 + 0}$ $\newline$ As r_1 = r_2 = $\sqrt{6}$ $\,$ $\mathrm{m}$ $\newline$ So potential difference = 0

Question 44

Physics · Ray Optics and Optical Instruments · Single correct

A convex lens of focal length 40 cm forms an image of an extended source of light on a photoelectric cell. A current $I$ is produced. The lens is replaced by another convex lens having the same diameter but focal length 20 cm. The photoelectric current now is :

$\frac{I}{2}$
$4I$
$2I$
$I$

Answer: (d)

Solution

As amount of energy incident on cell is same so current will remain same.

Question 45

Physics · Work, Energy and Power · Single correct

A body of mass 1000 kg is moving horizontally with a velocity 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is:

Answer: (d)

Solution

Momentum will remain conserved. $$1000 \times 6 = 1200 \times v$$ $$v = 5 \, \mathrm{m/s}$$

Question 46

Physics · Electromagnetic Waves · Single correct

A plane electromagnetic wave propagating in x-direction is described by $$E_y = \left(200 \, \mathrm{Vm}^{-1}\right) \sin\left[1.5 \times 10^7 t - 0.05 x\right];$$ The intensity of the wave is : (Use $\epsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C}^2 \, \mathrm{N}^{-1} \, \mathrm{m}^{-2}$)

$35.4 \, \mathrm{Wm}^{-2}$
$53.1 \, \mathrm{Wm}^{-2}$
$26.6 \, \mathrm{Wm}^{-2}$
$106.2 \, \mathrm{Wm}^{-2}$

Answer: (b)

Solution

The intensity is given by the formula: $$I = \frac{1}{2} \varepsilon_0 E_0^2 \times c$$ Substituting the values: $$I = \frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^4 \times 3 \times 10^8$$ Calculating the result: $$I = 53.1 \, \mathrm{W/m^2}$$

Question 47

Physics · Physical World, Units and Measurements · Single correct

Given below are two statements : Statement (I) : Planck's constant and angular momentum have same dimensions. Statement (II) : Linear momentum and moment of force have same dimensions. 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: (a)

Solution

Question 48

Physics · Current Electricity · Single correct

A wire of length $10\,\mathrm{cm}$ and radius $\sqrt{7}\times10^{-4}\,\mathrm{m}$ is connected across the right gap of a meter bridge. When a resistance of $4.5\,\Omega$ is connected in the left gap using a resistance box, the balance length is found to be $60\,\mathrm{cm}$ from the left end. If the resistivity of the wire is $R\times10^{-7}\,\Omega\,\mathrm{m}$, then the value of $R$ is:

Answer: (c)

Solution

For null point, $$\frac{4.5}{60} = \frac{R}{40}$$ Also, $R = \frac{\rho \ell}{A} = \frac{\rho \ell}{\pi r^2}$ $$4.5 \times 40 = \rho \times \frac{0.1}{\pi \times 7 \times 10^{-8}} \times 60$$ $$\rho = 66 \times 10^{-7} \, \Omega \times \mathrm{m}$$

Question 49

Physics · Current Electricity · Single correct

A wire of resistance $R$ and length $L$ is cut into 5 equal parts. If these parts are joined parallely, then resultant resistance will be:

$\frac{1}{25} R$
$\frac{1}{5} R$
$25 R$
$5 R$

Answer: (a)

Solution

Resistance of each part = $\frac{R}{5}$ Total resistance = $\frac{1}{5} \times \frac{R}{5} = \frac{R}{25}$

Question 50

Physics · Kinetic Theory · Single correct

The average kinetic energy of a monatomic molecule is 0.414 eV at temperature: (Use $K_B = 1.38 \times 10^{-23} \, \mathrm{J/mol - K}$)

3000 K
3200 K
1600 K
1500 K

Answer: (b)

Solution

For monoatomic molecule degree of freedom = 3. Therefore, $K_{avg} = \frac{3}{2} K_B T$ $$T = \frac{0.414 \times 1.6 \times 10^{-19} \times 2}{3 \times 1.38 \times 10^{-23}}$$ $$= 3200 \, \mathrm{K}$$

Question 51

Physics · Motion in a Straight Line · Numerical

A particle starts from origin at $t = 0$ with a velocity $5\hat{i} \, \mathrm{m/s}$ and moves in $x-y$ plane under action of a force which produces a constant acceleration of $(3\hat{i} + 2\hat{j}) \, \mathrm{m/s^2}$. If the $x$-coordinate of the particle at that instant is $84 \, \mathrm{m}$, then the speed of the particle at this time is $\sqrt{\alpha} \, \mathrm{m/s}$. The value of $\alpha$ is

Answer: 673

Solution

Given $u_x = 5 \, \mathrm{m/s}$, $a_x = 3 \, \mathrm{m/s^2}$, $x = 84 \, \mathrm{m}$. $v_x^2 - u_x^2 = 2ax$. $v_x^2 - 25 = 2(3)(84)$. $V_x = 23 \, \mathrm{m/s}$. $v_x - u_x = a_x t$. $t = \frac{23 - 5}{3} = 6 \, \mathrm{s}$. $v_y = 0 + a_y t = 0 + 2 \times (6) = 12 \, \mathrm{m/s}$. $v^2 = v_x^2 + v_y^2 = 23^2 + 12^2 = 673$. $v = \sqrt{673} \, \mathrm{m/s}$.

Question 52

Physics · Electric Charges and Fields · Numerical

A thin metallic wire having cross sectional area of $10^{-4} \, \mathrm{m}^2$ is used to make a ring of radius $30 \, \mathrm{cm}$. A positive charge of $2\pi \, \mathrm{C}$ is uniformly distributed over the ring, while another positive charge of $30 \, \mathrm{pC}$ is kept at the centre of the ring. The tension in the ring is ______ N; provided that the ring does not get deformed (neglect the influence of gravity). (given, $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, \mathrm{SI \, units}$)

Answer: 3

Solution

Given the equation: $$2T \sin \frac{d\theta}{2} = \frac{kq_0}{R^2} \cdot \lambda R d\theta$$ Where $$\lambda = \frac{Q}{2\pi R}$$ We have: $$T = \frac{K q_0 Q}{(R^2) \times 2\pi}$$ Substituting the values: $$= \frac{(9 \times 10^9)(2\pi \times 30 \times 10^{-12})}{(0.30)^2 \times 2\pi}$$ Simplifying: $$= \frac{9 \times 10^{-3} \times 30}{9 \times 10^{-2}} = 3 \, \mathrm{N}$$

Question 53

Physics · Electromagnetic Induction · Numerical

Two coils have mutual inductance 0.002H. The current changes in the first coil according to the relation $i = i_0 \sin \omega t$, where $i_0 = 5 \, \mathrm{A}$ and $\omega = 50 \pi \, \mathrm{rad/s}$. The maximum value of emf in the second coil is $\frac{\pi}{\alpha} \, \mathrm{V}$. The value of $\alpha$ is

Answer: 2

Solution

Given $\phi = \mathrm{Mi} = \mathrm{Mi_0} \sin \omega t$. $$\mathrm{EMF} = -\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}} = -0.002 \left(i_0 \omega \cos \omega t \right)$$ $$\mathrm{EMF}_{\max} = i_0 \omega (0.002) = (5)(50\pi)(0.002)$$ $$\mathrm{EMF}_{\max} = \frac{\pi}{2} \, \mathrm{V}$$

Question 54

Physics · Ray Optics and Optical Instruments · Numerical

Two immiscible liquids of refractive indices $\frac{8}{5}$ and $\frac{3}{2}$ respectively are put in a beaker as shown in the figure. The height of each column is 6 cm. A coin is placed at the bottom of the beaker. For near normal vision, the apparent depth of the coin is $\frac{\alpha}{4}$ cm. The value of $\alpha$ is _____.

Answer: 31

Solution

Given $$$ h_{upp} = $\frac{h_1}{\mu_1}$ + $\frac{h_2}{\mu_2}$ = $\frac{6}{3/2}$ + $\frac{6}{8/5}$ = 4 + $\frac{15}{4}$ = $\frac{31}{4}$ $$$ cm.

Question 55

Physics · Nuclei · Numerical

In a nuclear fission process, a high mass nuclide $(A \approx 236)$ with binding energy $7.6 \mathrm{MeV/ Nucleon}$ dissociated into middle mass nuclides $(A \approx 118)$, having binding energy of $8.6 \mathrm{MeV/ Nucleon}$. The energy released in the process would be MeV.

Answer: 236

Solution

Given $Q=\mathrm{BE}_{\mathrm{Product}}-\mathrm{BE}_{\mathrm{Reactant}}$. $Q=2(118)(8.6)-236(7.6)$ $=236\times1=236\,\mathrm{MeV}$

Question 56

Physics · System of Particles and Rotational Motion · Numerical

Four particles each of mass 1 kg are placed at four corners of a square of side 2 m. Moment of inertia of system about an axis perpendicular to its plane and passing through one of its vertex is _____ kgm$^2$.

Answer: 16

Solution

The moment of inertia is calculated as follows: $$I = ma^2 + ma^2 + m(\sqrt{2}a)^2$$ Simplifying, we get: $$= 4ma^2$$ Substituting the given values: $$= 4 \times 1 \times (2)^2 = 16$$

Question 57

Physics · Oscillations · Numerical

A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 $\mathrm{cm/s}$. The distance of the particle from the mean position when its speed becomes 5 $\mathrm{cm/s}$ is $\sqrt{\alpha} \mathrm{cm}$, where $\alpha = \underline{ }$.

Answer: 12

Solution

At mean position, $V = A \omega \Rightarrow 10 = 4 \omega$. $$\omega = \frac{5}{2}$$ $$v = \omega \sqrt{A^2 - x^2}$$ $$5 = \frac{5}{2} \sqrt{4^2 - x^2} \Rightarrow x^2 = 16 - 4$$ $$x = \sqrt{12} \, cm$$

Question 58

Physics · Moving Charges and Magnetism · Numerical

Two long, straight wires carry equal currents in opposite directions as shown in figure. The separation between the wires is 5.0 cm. The magnitude of the magnetic field at a point $P$ midway between the wires is _____ $\mu \mathrm{T}$ (Given : $\mu_0 = 4\pi \times 10^{-7} \mathrm{TmA}^{-1}$)

Answer: 160

Solution

Given $$B = \left( \frac{\mu_0 i}{2 \pi a} \right) \times 2 = \frac{4 \pi \times 10^{-7} \times 10}{\pi \times \left( \frac{5}{2} \times 10^{-2} \right)}$$ $$= 16 \times 10^{-5} = 160 \mu \mathrm{T}$$

Question 59

Physics · Electrostatic Potential and Capacitance · Numerical

The charge accumulated on the capacitor connected in the following circuit is _____ $\mu \mathrm{C}$ ( Given $C = 150 \mu \mathrm{F}$)

Answer: 400

Solution

Given the circuit, we have the equation: $$V_A + \frac{10}{3}(1) - 6(1) = V_B$$ Solving for $V_A - V_B$: $$V_A - V_B = 6 - \frac{10}{3} = \frac{8}{3} volt$$ The charge $Q$ is given by: $$Q = C(V_A - V_B)$$ Substituting the values: $$= 150 \times \frac{8}{3} = 400 \mu C$$

Question 60

Physics · Mechanical Properties of Solids · Numerical

If average depth of an ocean is 4000 m and the bulk modulus of water is $2 \times 10^9 \, \mathrm{Nm}^{-2}$, then fractional compression $\frac{\Delta V}{V}$ of water at the bottom of ocean is $\alpha \times 10^{-2}$. The value of $\alpha$ is _____ (Given, $g = 10 \, \mathrm{ms}^{-2}$, $\rho = 1000 \, \mathrm{kg} \, \mathrm{m}^{-3}$)

Answer: 2

Solution

Given $$B = - \frac{\Delta P}{\left( \frac{\Delta V}{V} \right)}$$ $$- \left( \frac{\Delta V}{V} \right) = \frac{\rho gh}{B} = \frac{1000 \times 10 \times 4000}{2 \times 10^9}$$ $$= 2 \times 10^{-2} \quad \text{[-ve sign represent compression]}$$

Chemistry

Question 61

Chemistry · Biomolecules · Single correct

Two nucleotides are joined together by a linkage known as :

Phosphodiester linkage
Glycosidic linkage
Disulphide linkage
Peptide linkage

Answer: (a)

Solution

Phosphodiester linkage is shown in the diagram. It connects the 5' end of one sugar molecule to the 3' end of another sugar molecule in a nucleic acid chain. The linkage involves a phosphate group connecting two sugar molecules, forming the backbone of DNA or RNA.

Question 62

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Highest enol content will be shown by:

Answer: (b)

Solution

The given reaction shows the conversion of a cyclic diketone to an aromatic compound. The structure on the left is a cyclic diketone, and the structure on the right is a benzene ring with two hydroxyl groups, indicating an aromatic compound.

Question 63

Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct

Element not showing variable oxidation state is :

Bromine
Iodine
Chlorine
Fluorine

Answer: (d)

Solution

Fluorine does not show variable oxidation state.

Question 64

Chemistry · Equilibrium · Single correct

Which of the following is strongest Bronsted base?

Answer: (d)

Solution

The nitrogen atom has a localized lone pair and is $sp^3$ hybridized.

Question 65

Chemistry · Structure of Atom · Single correct

Which of the following electronic configuration would be associated with the highest magnetic moment?

[$\mathrm{Ar}$] $3d^7$
[$\mathrm{Ar}$] $3d^8$
[$\mathrm{Ar}$] $3d^3$
[$\mathrm{Ar}$] $3d^6$

Answer: (d)

Solution

\begin{tabular}{|l|l|l|l|l|} \hline & $3d^{7}$ & $3d^{8}$ & $3d^{3}$ & $3d^{6}$ \\ \hline No. of unpaired e & 3 & 2 & 3 & 4 \\ \hline Spin only Magnetic moment & $\sqrt{15}$ BM & $\sqrt{8}$ BM & $\sqrt{15}$ BM & $\sqrt{24}$BM\\ \hline \end{tabular}

Question 66

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

Which of the following has highly acidic hydrogen?

Answer: (d)

Solution

The conjugate base is more stable due to more resonance of negative charge.

Question 67

Chemistry · Solutions · Single correct

A solution of two miscible liquids showing negative deviation from Raoult's law will have:

increased vapour pressure, increased boiling point
increased vapour pressure, decreased boiling point
decreased vapour pressure, decreased boiling point
decreased vapour pressure, increased boiling point

Answer: (d)

Solution

Solution with negative deviation has $$P_T < P_A^0 X_A + P_B^0 X_B$$ $$P_A < P_A^0 X_A$$ $$P_B < P_B^0 X_B$$ If vapour pressure decreases so boiling point increases.

Question 68

Chemistry · Co-ordination Compounds · Single correct

Consider the following complex ions P = [ $\mathrm{FeF}_6$ $]^{3-}$ Q = $\left[\mathrm{V(H_2O)}_6\right]^{2+}$ R = $\left[\mathrm{Fe(H_2O)}_6\right]^{2+}$ The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :

R < Q < P
R < P < Q
Q < R < P
Q < P < R

Answer: (c)

Solution

\[ [\mathrm{FeF}_6]^{3-}:\ \mathrm{Fe}^{3+}:[\mathrm{Ar}]\,3d^5 \] \[ \mathrm{F^- :\ Weak\ field\ ligand} \] \[ \text{Number of unpaired electrons }=5 \] \[ \mu=\sqrt{5(5+2)} \] \[ \mu=\sqrt{35}\ \mathrm{BM} \] \[ [\mathrm{V(H_2O)}_6]^{2+}:\ \mathrm{V}^{2+}:3d^3 \] \[ \text{Number of unpaired electrons }=3 \] \[ \mu=\sqrt{3(3+2)} \] \[ \mu=\sqrt{15}\ \mathrm{BM} \] \[ [\mathrm{Fe(H_2O)}_6]^{2+}:\ \mathrm{Fe}^{2+}:3d^6 \] \[ \mathrm{H_2O:\ Weak\ field\ ligand} \] \[ \text{Number of unpaired electrons }=4 \] \[ \mu=\sqrt{4(4+2)} \] \[ \mu=\sqrt{24}\ \mathrm{BM} \]

Question 69

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Choose the polar molecule from the following:

$\mathrm{CCl}_4$
$\mathrm{CO}_2$
$\mathrm{CH}_2 = \mathrm{CH}_2$
$\mathrm{CHCl}_3$

Answer: (d)

Solution

The dipole moment $\mu \neq 0$. CHCl$_3$ is a polar molecule and the rest of the molecules are non-polar.

Question 70

Chemistry · The d-and f-Block Elements · Single correct

Given below are two statements : Statement (I) : The 4f and 5f - series of elements are placed separately in the Periodic table to preserve the principle of classification. Statement (II) :S-block elements can be found in pure form in nature. In the light of the above statements, choose the most appropriate answer from the options given below :

Statement I is false but Statement II is true
Both Statement I and Statement II are true
Statement I is true but Statement II is false
Both Statement I and Statement II are false

Answer: (c)

Solution

s-block elements are highly reactive and found in combined state.

Question 71

Chemistry · Alcohols, Phenols and Ethers · Single correct

Given below are two statements: Statement (I) : p-nitrophenol is more acidic than m-nitrophenol and o-nitrophenol. Statement (II) : Ethanol will give immediate turbidity with Lucas reagent. 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
Both Statement I and Statement II are false
Statement I is false but Statement II is true

Answer: (a)

Solution

Acidic strength Ethanol gives Lucas test after a long time. Statement (I) is correct. Statement (II) is incorrect.

Question 72

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

The ascending order of acidity of $-\mathrm{OH}$ group in the following compounds is : Choose the correct answer from the options given below:

< (D) < $(C)$ < (B) < (E)
$(C)$ < (A) < (D) < (B) < (E)
$(C)$ < (D) < (B) < (A) < (E)
< $(C)$ < (D) < (B) < (E)

Answer: (d)

Solution

The order of acidity is shown as follows: $$(E) > (B) > (D) > (C) > (A)$$. The nitro groups in (E) and (B) are electron-withdrawing groups, which increase acidity. The methoxy group in (C) is an electron-donating group, which decreases acidity.

Question 73

Chemistry · The p-Block Elements (Group-13 and 14) · Single correct

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason $(R)$. Assertion (A): Melting point of Boron (2453 K) is unusually high in group 13 elements. Reason $(R)$: Solid Boron has very strong crystalline lattice. In the light of the above statements, choose the most appropriate answer from the options given below;

Both (A) and $(R)$ are correct but $(R)$ is not the correct explanation of (A)
Both (A) and $(R)$ are correct and $(R)$ is the correct explanation of (A)
(A) is true but $(R)$ is false
(A) is false but $(R)$ is true

Answer: (b)

Solution

Solid Boron has a very strong crystalline lattice, so its melting point is unusually high in group 13 elements.

Question 74

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

Cyclohexene is ______ type of an organic compound.

Benzenoid aromatic
Benzenoid non-aromatic
Acyclic
Alicyclic

Answer: (d)

Solution

The structure shown is a cyclohexene. So, it is alicyclic.

Question 75

Chemistry · Co-ordination Compounds · Single correct

Yellow compound of lead chromate gets dissolved on treatment with hot NaOH solution. The product of lead formed is a :

Tetraanionic complex with coordination number six
Neutral complex with coordination number four
Dianionic complex with coordination number six
Dianionic complex with coordination number four

Answer: (d)

Solution

PbCrO_4 + $\mathrm{NaOH}$ (hot excess) $\rightarrow$ [$\mathrm{Pb(OH)_4}$]^{-2} + $\mathrm{Na_2CrO_4}$. Dianionic complex with coordination number four.

Question 76

Chemistry · Equilibrium · Single correct

Given below are two statements : Statement (I) : Aqueous solution of ammonium carbonate is basic. Statement (II) : Acidic/basic nature of salt solution of a salt of weak acid and weak base depends on $K_a$ and $K_b$ value of acid and the base forming it. In the light of the above statements, choose the most appropriate answer from the options given below :

Both Statement I and Statement II are correct
Statement I is correct but Statement II is incorrect
Both Statement I and Statement II are incorrect
Statement I is incorrect but Statement II is correct

Answer: (a)

Solution

Aqueous solution of $\mathrm{(NH_4)_2CO_3}$ is basic. pH of salt of weak acid and weak base depends on $K_a$ and $K_b$ value of acid and the base forming it.

Question 77

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

IUPAC name of following compound $(P)$ is:

1-Ethyl-5, 5-dimethylcyclohexane
3-Ethyl-1,1-dimethylcyclohexane
1-Ethyl-3, 3-dimethylcyclohexane
1,1-Dimethyl-3-ethylcyclohexane

Answer: (b)

Solution

The structure shown is 3-ethyl-1,1-dimethylcyclohexane.

Question 78

Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct

$\mathrm{NaCl}$ reacts with conc. $\mathrm{H_2SO_4}$ and $\mathrm{K_2Cr_2O_7}$ to give reddish fumes (B), which react with $\mathrm{NaOH}$ to give yellow solution (C). (B) and (C) respectively are:

CrO_2Cl_2, Na_2CrO_4
Na_2CrO_4, CrO_2Cl_2
CrO_2Cl_2, KHSO_4
CrO_2Cl_2, Na_2Cr_2O_7

Answer: (a)

Solution

NaCl + conc. $\mathrm{H_2SO_4}$ + $\mathrm{K_2Cr_2O_7}$ $\rightarrow$ $\mathrm{CrO_2Cl_2}$ + $\mathrm{KHSO_4}$ + $\mathrm{NaHSO_4}$ + $\mathrm{H_2O}$. (B) Reddish brown $\mathrm{CrO_2Cl_2}$ + $\mathrm{NaOH}$ $\rightarrow$ $\mathrm{Na_2CrO_4}$ + $\mathrm{NaCl}$ + $\mathrm{H_2O}$. (C) Yellow colour

Question 79

Chemistry · Haloalkanes and Haloarenes · Single correct

The correct statement regarding nucleophilic substitution reaction in a chiral alkyl halide is ;

Retention occurs in $S_{N}1$ reaction and inversion occurs in $S_{N}2$ reaction.
Racemisation occurs in $S_{N}1$ reaction and retention occurs in $S_{N}2$ reaction.
Racemisation occurs in both $S_{N}1$ and $S_{N}2$ reactions.
Racemisation occurs in $S_{N}1$ reaction and inversion occurs in $S_{N}2$ reaction.

Answer: (d)

Solution

For $\mathrm{S_N^1}$ reactions, racemisation occurs. For $\mathrm{S_N^2}$ reactions, inversion occurs.

Question 80

Chemistry · The d-and f-Block Elements · Single correct

The electronic configuration for Neodymium is: [Atomic Number for Neodymium 60]

$[Xe]4f^4 6s^2$
$[Xe]5f^4 7s^2$
$[Xe]4f^6 6s^2$
$[Xe]4f^1 5d^1 6s^2$

Answer: (a)

Solution

The electronic configuration of Nd (Z = 60) is $[\mathrm{Xe}] \, 4f^4 \, 6s^2$.

Question 81

Chemistry · Electrochemistry · Numerical

The mass of silver (Molar mass of Ag : $108 \, \mathrm{g/mol}^{-1}$) displaced by a quantity of electricity which displaces 5600 mL of $\mathrm{O}_2$ at S.T.P. will be _____ g.

Answer: 108

Solution

Eq. of Ag = Eq. of O_2 Let $x$ gm silver displaced, $$\frac{x \times 1}{108} = \frac{5.6}{22.7} \times 4$$ (Molar volume of gas at STP = 22.7 lit) $x = 106.57$ gm Ans. 107 OR, as per old STP data, molar volume = 22.4 lit $$\frac{x \times 1}{108} = \frac{5.6}{22.4} \times 4, x = 108 gm.$$ Ans. 108

Question 82

Chemistry · Chemical Kinetics and Nuclear Chemistry · Fill in the blank

Consider the following data for the given reaction: $2\mathrm{HI}_{(g)} \rightarrow \mathrm{H_2}_{(g)} + \mathrm{I_2}_{(g)}$ $\begin{array}{c|ccc} & 1 & 2 & 3 \\ \mathrm{HI}\,(\mathrm{mol\,L^{-1}}) & 0.005 & 0.01 & 0.02 \\ \mathrm{Rate}\,(\mathrm{mol\,L^{-1}\,s^{-1}}) & 7.5 \times 10^{-4} & 3.0 \times 10^{-3} & 1.2 \times 10^{-2} \end{array}$ The order of the reaction is $\underline{\hspace{2cm}}$.

Answer: 2

Solution

Let, $R = k[\mathrm{HI}]^n$ Using any two of given data, $$\frac{3 \times 10^{-3}}{7.5 \times 10^{-4}} = \left(\frac{0.01}{0.005}\right)^n$$ $n = 2$

Question 83

Chemistry · Some Basic Concepts of Chemistry · Numerical

Mass of methane required to produce 22 g of $CO_2$ after complete combustion is ______ g. (Given Molar mass in g $mol^{-1}$ C = 12.0 H = 1.0 O = 16.0)

Answer: 8

Solution

Given the reaction: $$$ $\mathrm{CH_4}$ + 2$\mathrm{O_2}$ $\rightarrow$ $\mathrm{CO_2}$ + 2$\mathrm{H_2O}$ $$$. Moles of $$$ $\mathrm{CO_2}$ = $\frac{22}{44}$ = 0.5 $$$. So, required moles of $$$ $\mathrm{CH_4}$ = 0.5 $$$. Mass = $$$ 0.5 $\times$ 16 = 8 $\mathrm{gm}$ $$$.

Question 84

Chemistry · Thermodynamics · Numerical

If three moles of an ideal gas at 300 \, $\mathrm{K}$ expand isothermally from 30 \, $\mathrm{dm}^3$ to 45 \, $\mathrm{dm}^3$ against a constant opposing pressure of 80 \, $\mathrm{kPa}$, then the amount of heat transferred is _______ \, $\mathrm{J}$.

Answer: 1200

Solution

Using, first law of thermodynamics, $$\Delta U = Q + W,$$ $$\Delta U = 0 : Process is isothermal$$ $$Q = -W$$ $$W = -P_{ext} \Delta V : Irreversible$$ $$= -80 \times 10^3 (45 - 30) \times 10^{-3}$$ $$= -1200 \, J$$

Question 85

Chemistry · Hydrocarbons · Numerical

3-Methylhex-2-ene on reaction with HBr in presence of peroxide forms an addition product (A). The number of possible stereoisomers for 'A' is

Answer: 4

Solution

The reaction of the given alkene with HBr in the presence of peroxide leads to the formation of a product with two chiral centers. The number of stereoisomers is calculated using the formula $2^n$, where $n$ is the number of chiral centers. Here, $n = 2$, so the number of stereoisomers is $2^2 = 4$.

Question 86

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

Among the given organic compounds, the total number of aromatic compounds is

Answer: (c)

Solution

B, C and D are Aromatic

Question 87

Chemistry · Hydrocarbons · Numerical

Among the following, total number of meta directing functional groups is (Integer based) $-\mathrm{OCH}_3$, $-\mathrm{NO}_2$, $-\mathrm{CN}$, $-\mathrm{CH}_3$ $-\mathrm{NHCOCH}_3$, $-\mathrm{COR}$, $-\mathrm{OH}$, $-\mathrm{COOH}$, $-\mathrm{Cl}$

Answer: 4

Solution

Q7 $-\mathrm{NO_2}$, $-\mathrm{C} \equiv \mathrm{N}$, $-\mathrm{COR}$, $-\mathrm{COOH}$ are meta directing.

Question 88

Chemistry · Structure of Atom · Numerical

The number of electrons present in all the completely filled subshells having $n = 4$ and $s = +\frac{1}{2}$ is (Where $n =$ principal quantum number and $s =$ spin quantum number)

Answer: 16

Solution

Given $n = 4$, the possible electron configurations are: \begin{tabular}{|l|l|l|l|l|} \hline & 4s & 4p & 4d & 4f \\ \hline Total e & 2 & 6 & 10 & 14 \\ \hline Total e- with S = +$\frac{1}{2}$ & 1 & 3 & 5 & 7 \\ \hline \end{tabular} So, the answer is 16.

Question 89

Chemistry · Chemical Bonding and Molecular Structure · Numerical

Sum of bond order of CO and $NO^+$ is

Answer: 6

Solution

CO $\Rightarrow \underline{\mathrm{C}} \equiv \mathrm{O}^{+}$ : BO = 3 NO$^{+}$ $\Rightarrow \mathrm{N} \equiv \mathrm{O}^{+}$ : BO = 3

Question 90

Chemistry · Redox Reactions · Numerical

From the given list, the number of compounds with $+4$ oxidation state of Sulphur: $\mathrm{SO_3}$, $\mathrm{H_2SO_3}$, $\mathrm{SOCl_2}$, $\mathrm{SF_4}$, $\mathrm{BaSO_4}$, $\mathrm{H_2S_2O_7}$

Answer: 3

Solution

The table shows the oxidation states of sulfur in various compounds. In $\mathrm{SO_3}$, the oxidation state is $+6$. In $\mathrm{H_2SO_3}$, it is $+4$. In $\mathrm{SOCl_2}$, it is $+4$. In $\mathrm{SF_4}$, it is $+4$. In $\mathrm{BaSO_4}$, it is $+6$. In $\mathrm{H_2S_2O_7}$, it is $+6$.