JEE Main 5 April 2024 Shift 2 question paper with solutions

JEE Main 5 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.

Register free to attempt this shift as a timed 180-minute test with instant scoring and chapter-wise analysis.

Maths

Question 1

Maths · Continuity and Differentiability · Single correct

Let $f : [-1, 2] \to \mathbb{R}$ be given by $f(x) = 2x^2 + x + \lfloor x^2 \rfloor - \lfloor x \rfloor$, where $\lfloor t \rfloor$ denotes the greatest integer less than or equal to $t$. The number of points, where $f$ is not continuous, is :

  1. 5
  2. 6
  3. 3
  4. 4

Answer: (d)

Solution

Doubtful points: $-1, 0, 1, \sqrt{2}, \sqrt{3}, 2$ at $x = \sqrt{2}, \sqrt{3}$ $$f(x) = (2x^2 + x - [x]) + [x^2] = Discount$$ at $x = -1$: RHL $\Rightarrow f(x) = (2 - 1 - (-1)) + 0 = 2$ $$f(-1) = 2 - 1 - (-1) + 1 = 3 Dis.$$ at $x = 2$: LHL $\Rightarrow f(x) = 8 + 2 - 1 + 3 = 12$ $$f(2) = 8 + 2 - 2 + 4 = 12 Cont.$$ at $x = 0$: LHL $\Rightarrow 0 + 0 - (-1) + 0 = 1$ $$f(0) = 0 Dis.$$ at $x = 1$: LHL $\Rightarrow 2 + 1 - 0 + 0 = 3$ $$f(1) = 3 - 1 + 1 = 3 Cont.$$ RHL $\Rightarrow 2 + 1 - 1 + 1 = 3$

Question 2

Maths · Differential Equations · Single correct

The differential equation of the family of circles passing through the origin and having centre at the line $y = x$ is :

  1. $(x^2 - y^2 + 2xy) \, dx = (x^2 - y^2 - 2xy) \, dy$
  2. $(x^2 + y^2 + 2xy) \, dx = (x^2 + y^2 - 2xy) \, dy$
  3. $(x^2 + y^2 - 2xy) \, dx = (x^2 + y^2 + 2xy) \, dy$
  4. $(x^2 - y^2 + 2xy) \, dx = (x^2 - y^2 + 2xy) \, dy$

Answer: (a)

Solution

Given $C \equiv x^2 + y^2 + gx + gy = 0 \ldots (1)$. $2x + 2yy' + g + gy' = 0$ $g = -\left( \frac{2x + 2yy'}{1 + y'} \right)$ Put in (1) $$x^2 + y^2 - \left( \frac{2x + 2yy'}{1 + y'} \right)(x + y) = 0$$ $$(x^2 - y^2 - 2xy)y' = x^2 - y^2 + 2xy$$

Question 3

Maths · Complex Numbers and Quadratic Equations · Single correct

Let $S_1 = \{ z \in \mathbb{C} : |z| \leq 5 \}, S_2 = \{ z \in \mathbb{C} : \mathrm{Im} \left( \frac{z+1-\sqrt{3}i}{1-\sqrt{3}i} \right) \geq 0 \}$ and $S_3 = \{ z \in \mathbb{C} : \mathrm{Re}(z) \geq 0 \}$. Then the area of the region $S_1 \cap S_2 \cap S_3$ is:

  1. $\frac{125\pi}{12}$
  2. $\frac{125\pi}{4}$
  3. $\frac{125\pi}{24}$
  4. $\frac{125\pi}{6}$

Answer: (a)

Solution

Given $S_1: x^2 + y^2 \leq 25 \ldots (1)$ $S_2$: Im of $\frac{z + (1 - \sqrt{3}i)}{(1 - \sqrt{3}i)} \geq 0$ Im of $\left( \frac{x + iy}{1 - \sqrt{3}i} + 1 \right) \geq 0$ Im of $\left( \frac{(x + iy)(1 + \sqrt{3}i)}{4} \right) \geq 0$ $\Rightarrow \sqrt{3}x + y \geq 0 \ldots (2)$ $S_3: x \geq 0 \ldots (3)$ Area $= \frac{5}{12} \left( \pi (5)^2 \right)$

Question 4

Maths · Applications of Integrals · Single correct

The area enclosed between the curves $y = x|x|$ and $y = x - |x|$ is :

  1. $\frac{4}{3}$
  2. 1
  3. $\frac{2}{3}$
  4. $\frac{8}{3}$

Answer: (a)

Solution

The area is given by the integral from $-2$ to $0$ of $-x^2 - 2x$. Therefore, $$A = \int_{-2}^{0} -x^2 - 2x = \frac{4}{3}.$$

Question 5

Maths · Permutations and Combinations · Single correct

60 words can be made using all the letters of the word BHBJO, with or without meaning. If these words are written as in a dictionary, then the $50^{\text{th}}$ word is

  1. JBBOH
  2. OBBJH
  3. OBBHJ
  4. HBBJO

Answer: (b)

Solution

BBHJO B ____ $4! = 24$ H ____ $\frac{4!}{2!} = 12$ J ____ $\frac{4!}{2!} = 12$ O B B H J O B B J H $\rightarrow 50^{th}$ rank

Question 6

Maths · Vector Algebra · Single correct

Let $\vec{a}$ = 2$\hat{i}$ + 5$\hat{j}$ - $\hat{k}$, $\vec{b}$ = 2$\hat{i}$ - 2$\hat{j}$ + 2$\hat{k}$ and $\vec{c}$ be three vectors such that ($\vec{c}$ + $\hat{i}$) $\times$ ($\vec{a}$ + $\vec{b}$ + $\hat{i}$) = $\vec{a}$ $\times$ ($\vec{c}$ + $\hat{i}$). If $\($ $\vec{a}$ $\cdot$ $\vec{c}$ = -29 $\)$, then $\vec{c}$ $\cdot$ (-2$\hat{i}$ + $\hat{j}$ + $\hat{k}$) is equal to:

  1. 15
  2. 12
  3. 10
  4. 5

Answer: (d)

Solution

Let's assume $\vec{v} = \vec{a} + \vec{b} + \hat{i}$ which is equal to $5\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{c} + \hat{i} = \vec{p}$. So, $\vec{p} \times \vec{v} = \vec{a} \times \vec{p}$. This implies $\vec{p} \times \vec{v} + \vec{p} \times \vec{a} = \vec{0}$. Therefore, $\vec{p} \times (\vec{v} + \vec{a}) = \vec{0}$. This implies $\vec{p} = \lambda(\vec{v} + \vec{a})$. Therefore, $\vec{c} + i = \lambda(7\hat{i} + 8\hat{j})$. Also, $\vec{a} \cdot \vec{c} + \vec{a} \cdot \hat{i} = \lambda \vec{a} \cdot (7\hat{i} + 8\hat{j})$. Therefore, $-29 + 2 = \lambda(14 + 40)$. Solving for $\lambda$, we get $\lambda = -\frac{1}{2}$. Therefore, $\vec{c} \cdot (-2\hat{i} + \hat{j} + \hat{k}) + \hat{i} \cdot (-2\hat{i} + \hat{j} + \hat{k}) = \lambda(7\hat{i} + 8\hat{j}) \cdot (-2\hat{i} + \hat{j} + \hat{k})$. This simplifies to $= -\frac{1}{2}(-14 + 8) + 2 = 5$.

Question 7

Maths · Vector Algebra · Single correct

Consider three vectors $\vec{a}, \vec{b}, \vec{c}$. Let $|\vec{a}| = 2$, $|\vec{b}| = 3$ and $\vec{a} = \vec{b} \times \vec{c}$. If $\alpha \in \left[0, \frac{\pi}{3}\right]$ is the angle between the vectors $\vec{b}$ and $\vec{c}$, then the minimum value of $27|\vec{c} - \vec{a}|^2$ is equal to:

  1. 110
  2. 124
  3. 121
  4. 105

Answer: (b)

Solution

Given $|\vec{c} - \vec{a}| = |\vec{c}|^2 + |\vec{a}|^2 - 2\vec{a} \cdot \vec{c}$. This simplifies to $|\vec{c}|^2 + 4 - 0$. Therefore, $\vec{a} = \vec{b} \times \vec{c}$. $|\vec{a}| = |\vec{b} \times \vec{c}|$. $2 = 3|\vec{c}| \sin \alpha$. $|\vec{c}| = \frac{2}{3} \csc \alpha$, where $\alpha \in \left[0, \frac{\pi}{3}\right]$. $|\vec{c}|_{\min} = \frac{2}{3} \times \frac{2}{\sqrt{3}}$, with $\csc \alpha \in \left[\frac{2}{\sqrt{3}}, \infty\right)$. Therefore, $27|\vec{c} - \vec{a}|_{\min}^2 = 27 \left(\frac{16}{27} + 4\right) = 124$.

Question 8

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

Let $A(-1, 1)$ and $B(2, 3)$ be two points and $P$ be a variable point above the line $AB$ such that the area of $\triangle PAB$ is $10$. If the locus of $P$ is $ax + by = 15$, then $5a + 2b$ is:

  1. 6
  2. -$\frac{6}{5}$
  3. 4
  4. -$\frac{12}{5}$

Answer: (d)

Solution

The determinant of the matrix is calculated as follows: $$\frac{1}{2} \begin{vmatrix} h & k & 1 \\ -1 & 1 & 1 \\ 2 & 3 & 1 \end{vmatrix} = 10$$ Expanding the determinant, we have: $$-2x + 3y = 25$$ Simplifying, we get: $$-\frac{6}{5}x + \frac{9}{5}y = 15$$ Solving for $a$ and $b$, we find: $$a = -\frac{6}{5}, b = \frac{9}{5}$$ Therefore, $$5a = -6, 2b = \frac{18}{5}$$

Question 9

Maths · Three Dimensional Geometry · Single correct

Let $(\alpha, \beta, \gamma)$ be the image of the point $(8, 5, 7)$ in the line $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{5}$. Then $\alpha + \beta + \gamma$ is equal to :

  1. 16
  2. 20
  3. 14
  4. 18

Answer: (c)

Solution

The vector $\($ $\overrightarrow{AM}$ $\cdot$ (2$\hat{i}$ + 3$\hat{j}$ + 5$\hat{k}$) = 0 $\)$. Expanding this, we have: $$(2\lambda - 7)(2) + (3\lambda - 6)(3) + (5\lambda - 5)(5) = 0$$ Simplifying, we get: $$38\lambda = 57$$ Solving for $\($ $\lambda$ $\)$, we find: $$\lambda = \frac{3}{2}$$ Substituting $\($ $\lambda$ = $\frac{3}{2}$ $\)$ into the coordinates of $\($ M $\)$, we get: $$M \left( 4, \frac{7}{2}, \frac{19}{2} \right)$$ Finally, the coordinates of $\($ A' $\)$ are $\($ (0, 2, 12) $\)$.

Question 10

Maths · Binomial Theorem · Single correct

If the constant term in the expansion of $\left(\dfrac{\sqrt[5]{3}}{x} + \dfrac{2x}{\sqrt[3]{5}}\right)^{12}$, $x \neq 0$, is $\alpha \times 2^8 \times \sqrt[5]{3}$, then $25\alpha$ is equal to:

  1. 724
  2. 742
  3. 639
  4. 693

Answer: (d)

Solution

Given $$T_{r+1} = 12 C_r \left( \frac{3^{1/5}}{x} \right)^{12-r} \left( \frac{2x}{5^{1/3}} \right)^r$$ We have $$T_{r+1} = \frac{12 C_r (3)^{\frac{12-r}{5}} (2)^r (x)^{2r-12}}{(5)^{r/3}}$$ Let $r = 6$. Then $$T_7 = \frac{12 C_6 (3)^{6/5} (2)^6}{5^2} = \left( \frac{9 \times 11 \times 7}{25} \right) 2^8 \cdot 3^{1/5}$$ Finally, $$25 \alpha = 693$$

Question 11

Maths · Relations and Functions · Single correct

Let $f, g : \mathbb{R} \to \mathbb{R}$ be defined as : $$f(x) = |x - 1| and g(x) = \begin{cases} e^x, & x \geq 0 \\ x + 1, & x \leq 0 \end{cases}$$ Then the function $f(g(x))$ is

  1. neither one-one nor onto.
  2. one-one but not onto.
  3. onto but not one-one.
  4. both one-one and onto.

Answer: (a)

Solution

Given $f(g(x)) = |g(x) - 1|$. For $fog$, we have: $$fog = \begin{cases} |e^x - 1| & x \geq 0 \\ |x + 1 - 1| & x \leq 0 \end{cases}$$ Simplifying further: $$fog = \begin{cases} e^x - 1 & x \geq 0 \\ -x & x \leq 0 \end{cases}$$

Question 12

Maths · Conic Sections · Single correct

Let the circle $C_1 : x^2 + y^2 - 2(x + y) + 1 = 0$ and $C_2$ be a circle having centre at $(-1, 0)$ and radius $2$. If the line of the common chord of $C_1$ and $C_2$ intersects the $y$-axis at the point $P$, then the square of the distance of $P$ from the centre of $C_1$ is:

  1. 2
  2. 1
  3. 4
  4. 6

Answer: (a)

Solution

Given $S_1: x^2 + y^2 - 2x - 2y + 1 = 0$ and $S_2: x^2 + y^2 + 2x - 3 = 0$. The common chord is $S_1 - S_2 = 0$. This simplifies to $-4x - 2y + 4 = 0$. Solving for $y$, we get $2x + y = 2$, which implies the point $P(0, 2)$. The squared distance $d_{(c,p)}^2 = (1 - 0)^2 + (2 - 1)^2 = 2$.

Question 13

Maths · Sets · Single correct

Let the set $S = \{2, 4, 8, 16, \ldots, 512\}$ be partitioned into 3 sets $A, B, C$ with equal number of elements such that $A \cup B \cup C = S$ and $A \cap B = B \cap C = A \cap C = \phi$. The maximum number of such possible partitions of $S$ is equal to:

  1. 1680
  2. 1640
  3. 1520
  4. 1710

Answer: (a)

Solution

The solution is calculated using the formula for permutations of multiset. The number of ways to arrange 9 items where there are three groups of 3 identical items each is given by: $$\frac{9!}{(3!3!3!)} \times 3!$$

Question 14

Maths · Determinants · Single correct

The values of $m, n$, for which the system of equations $$x + y + z = 4,$$ $$2x + 5y + 5z = 17,$$ $$x + 2y + mz = n$$ has infinitely many solutions, satisfy the equation:

  1. $m^2 + n^2 - mn = 39$
  2. $m^2 + n^2 - m - n = 46$
  3. $m^2 + n^2 + m + n = 64$
  4. $m^2 + n^2 + mn = 68$

Answer: (a)

Solution

Given the determinant equation: $$D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & 5 \\ 1 & 2 & m \end{vmatrix} = 0 \Rightarrow m = 2$$ And for the second determinant: $$D_3 = \begin{vmatrix} 1 & 1 & 4 \\ 2 & 5 & 17 \\ 1 & 2 & n \end{vmatrix} = 0 \Rightarrow n = 7$$

Question 15

Maths · Probability · Single correct

The coefficients $a, b, c$ in the quadratic equation $ax^2 + bx + c = 0$ are from the set {$1, 2, 3, 4, 5, 6$\}. If the probability of this equation having one real root bigger than the other is $p$, then $216 \, p$ equals:

  1. 57
  2. 76
  3. 38
  4. 19

Answer: (c)

Solution

$D>0$ $b^2>4ac$ $b=3$ $(a,c)=(1,1),(1,2),(2,1)$ $b=4$ $(a,c)=(1,1),(1,2),(2,1),(1,3),(3,1)$ $b=5$ $(a,c)=(1,1),(1,2),(2,1),(1,3),(3,1)$ $(1,4),(4,1),(1,5),(5,1)$ $(1,6),(6,1),(2,3),(3,2)$ $b=6$ $(a,c)=(1,1),(1,2),(2,1),(1,3),(3,1)$ $(1,4),(4,1),(1,5),(5,1),(1,6),(6,1)$ $(2,3),(3,2),(2,4),(4,2),(2,2)$ fav. cases $=38$ Prob. $=\frac{38}{6\times6\times6}$ $=\frac{19}{108}$

Question 16

Maths · Conic Sections · Single correct

Let ABCD and AEFG be squares of side 4 and 2 units, respectively. The point E is on the line segment AB and the point F is on the diagonal AC. Then the radius r of the circle passing through the point F and touching the line segments BC and CD satisfies:

  1. r = 0
  2. 2r^2 - 4r + 1 = 0
  3. 2r^2 - 8r + 7 = 0
  4. r^2 - 8r + 8 = 0

Answer: (d)

Solution

Given the coordinates of the points, we have $\mathrm{OF}^2 = r^2$. Using the coordinates of $F(2,2)$ and the center $(4-r, 4-r)$, we find: $$(2-r)^2 + (2-r)^2 = r^2.$$ Simplifying, we get: $$r^2 - 8r + 8 = 0.$$

Question 17

Maths · Integrals · Single correct

Let $\beta(m, n) = \int_0^1 x^{m-1} (1-x)^{n-1} \, dx$, $m, n > 0$. If $\int_0^1 (1-x^{10})^{20} \, dx = a \times \beta(b, c)$, then $100(a+b+c)$ equals

  1. 1021
  2. 2120
  3. 2012
  4. 1120

Answer: (b)

Solution

Given $$I = \int_0^1 1 \cdot (1 - x^{10})^{20} \, dx$$ Let $x^{10} = t$. Then $x = t^{1/10}$ and $$dx = \frac{1}{10} (t)^{-9/10} \, dt$$ Substitute to get $$I = \int_0^1 (1 - t)^{20} \frac{1}{10} (t)^{-9/10} \, dt$$ Simplifying, $$I = \frac{1}{10} \int_0^1 t^{-9/10} (1 - t)^{20} \, dt$$ Thus, $$a = \frac{1}{10}, b = \frac{1}{10}, c = 21$$

Question 18

Maths · Determinants · Single correct

Let $\alpha \beta \neq 0$ and $A = \begin{bmatrix} \beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2 \alpha \end{bmatrix}$. If $B = \begin{bmatrix} 3 \alpha & -9 & 3 \alpha \\ -\alpha & 7 & -2 \alpha \\ -2 \alpha & 5 & -2 \beta \end{bmatrix}$ is the matrix of cofactors of the elements of $A$, then $\det(AB)$ is equal to:

  1. 64
  2. 216
  3. 343
  4. 125

Answer: (b)

Solution

Equating co-factor for $A_{21}$ $$(2\alpha^2 - 3\alpha) = \alpha$$ $$\alpha = 0, 2 (accept)$$ Now, $2\alpha^2 - \alpha\beta = 3\alpha$ $$\alpha = 2 \beta = 1$$ $$|AB| = |A cof(A)| = |A|^3$$ $$A = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 2 & 1 \\ -1 & 2 & 4 \end{vmatrix} = 6 - 2(9) + 3(6) = 6$$

Question 19

Maths · Continuity and Differentiability · Single correct

If $y(\theta) = \frac{2 \cos \theta + \cos 2\theta}{\cos 3\theta + 4 \cos 2\theta + 5 \cos \theta + 2}$, then at $\theta = \frac{\pi}{2}$, $y'' + y' + y$ is equal to:

  1. $\frac{1}{2}$
  2. $1$
  3. $2$
  4. $\frac{3}{2}$

Answer: (c)

Solution

Given $$y = \frac{2 \cos \theta + 2 \cos^2 \theta - 1}{4 \cos^3 \theta - 3 \cos \theta + 8 \cos^2 \theta - 4 + 5 \cos \theta + 2}$$ Simplifying, $$y = \frac{(2 \cos^2 \theta + 2 \cos \theta - 1)}{(2 \cos^2 \theta + 2 \cos \theta - 1)(2 \cos \theta + 2)}$$ This reduces to $$y = \frac{1}{2} \left( \frac{1}{1 + \cos \theta} \right)$$ Thus, $$\Rightarrow \theta = \frac{\pi}{2}, y = \frac{1}{2}$$ Differentiating, $$y' = \frac{1}{2} \left( \frac{-1}{(1 + \cos \theta)^2} \times (- \sin \theta) \right)$$ Again, $$\Rightarrow \theta = \frac{\pi}{2}, y = \frac{1}{2}$$ Second derivative, $$y'' = \frac{1}{2} \left[ \frac{\cos \theta (1 + \cos \theta)^2 - \sin \theta (2)(1 + \cos \theta)(- \sin \theta)}{(1 + \cos \theta)^4} \right]$$ Finally, $$\Rightarrow \theta = \frac{\pi}{2}, y = 1$$

Question 20

Maths · Sequences and Series · Single correct

For $x \geq 0$, the least value of $K$, for which $4^{1+x} + 4^{1-x}, \frac{K}{2}, 16^x + 16^{-x}$ are three consecutive terms of an A.P., is equal to:

  1. 8
  2. 4
  3. 10
  4. 16

Answer: (c)

Solution

Given $$k = 4 \left( 4^x + \frac{1}{4^x} \right) + \left( 4^{2x} + \frac{1}{4^{2x}} \right)$$ which is greater than or equal to 2. Also, $$k \geq 10$$

Question 21

Maths · Probability · Numerical

Let the mean and the standard deviation of the probability distribution \[ \begin{tabular}{|c|c|c|c|c|} \hline $X$ & $\alpha$ & $1$ & $0$ & $-3$ \\ \hline $P(X)$ & $\dfrac13$ & $k$ & $\dfrac16$ & $\dfrac14$ \\ \hline \end{tabular} be $\mu$ and $\sigma$, respectively. If $\sigma - \mu = 2$, then $\sigma + \mu$ is equal to _______

Answer: 5

Solution

Given $\($ $\frac{1}{3}$ + k + $\frac{1}{6}$ + $\frac{1}{4}$ = 1 $\)$, it implies $\($ k = $\frac{1}{4}$ $\)$. $\($ $\mu$ = $\frac{\alpha}{3}$ + $\frac{1}{4}$ - $\frac{3}{4}$ $\)$ $\($ $\mu$ = $\frac{\alpha}{3}$ - $\frac{1}{2}$ $\)$ $\($ $\sigma$ = $\sqrt{\left( \alpha^2 \frac{1}{3} + \frac{1}{4} + 9 \frac{1}{4} \right) - \left( \frac{\alpha}{3} - \frac{1}{2} \right)^2}$ $\)$ $\($ $\sigma$ = $\sqrt{\frac{2\alpha^2}{9} + \frac{\alpha}{3} + \frac{9}{4}}$ $\)$ $\($ $\sigma$ = $\mu$ + 2 $\)$ $\($ $\sigma$^2 = ($\mu$ + 2)^2 $\Rightarrow$ $\frac{2\alpha^2}{9}$ + $\frac{\alpha}{3}$ + $\frac{9}{4}$ = $\frac{\alpha^2}{9}$ + $\frac{9}{4}$ + $\alpha$ $\)$ $\($ $\frac{\alpha^2}{9}$ - $\frac{2\alpha}{3}$ = 0 $\)$ $\($ $\alpha$ = 0, (reject) or $\alpha$ = 6 $\)$ ($\($ $\therefore$ x = 0 is already given $\)$) $\($ $\Rightarrow$ $\sigma$ + $\mu$ = 2$\mu$ + 2 $\)$ $\($ = 5 $\)$

Question 22

Maths · Differential Equations · Numerical

Let $y = y(x)$ be the solution of the differential equation $$\frac{dy}{dx} + \frac{2x}{(1+x^2)^2}y = xe^{\frac{1}{(1+x^2)}}; \; y(0) = 0.$$ Then the area enclosed by the curve $f(x) = y(x)e^{-\frac{1}{(1+x^2)}}$ and the line $y - x = 4$ is

Answer: 18

Solution

Given the integrating factor $IF = e^{\int \frac{2x}{(1+x^2)^2} \, dx} = e^{-\frac{1}{1+x^2}}$. Then $y \cdot e^{-\frac{1}{1+x^2}} = \int x \cdot e^{\frac{1}{1+x^2}} \cdot e^{-\frac{1}{1+x^2}} \, dx$. This simplifies to $y \cdot e^{-\frac{1}{1+x^2}} = \frac{x^2}{2} + c$. Using the initial condition $(0, 0)$ implies $c = 0$. Therefore, $y(x) = \frac{x^2}{2} e^{\frac{1}{1+x^2}}$. The function $f(x) = \frac{x^2}{2}$. The area $A = \int_{-2}^{4} (x + 4) - \frac{x^2}{2} \, dx = 18$.

Question 23

Maths · Trigonometric Functions · Fill in the blank

The number of solutions of $\sin^2 x + (2 + 2x - x^2) \sin x - 3(x - 1)^2 = 0$, where $-\pi \leq x \leq \pi$, is ________

Answer: 2

Solution

Given the equation $\sin^2 x - (x^2 - 2x - 2) \sin x - 3(x-1)^2 = 0$. Rewriting, we have $\sin^2 x - (x-1)^2 \sin x - 3(x-1)^2 = 0$. The roots are $-3$ and $(x-1)^2$. Therefore, $\sin x = -3$ (reject) or $\sin x = (x-1)^2$. Thus, $\sin x = (x-1)^2$.

Question 24

Maths · Three Dimensional Geometry · Numerical

Let the point $(-1, \alpha, \beta)$ lie on the line of the shortest distance between the lines $\frac{x+2}{-3} = \frac{y-2}{4} = \frac{z-5}{2}$ and $\frac{x+2}{-1} = \frac{y+6}{2} = \frac{z-1}{0}$. Then $(\alpha - \beta)^2$ is equal to

Answer: 25

Solution

P $(-3\lambda - 2, 4\lambda + 2, 2\lambda + 5)$ Q $(-\mu - 2, 2\mu - 6, 1)$ DRS of $PQ = (3\lambda - \mu, 2\mu - 4\lambda - 8, -2\lambda - 4)$ $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 0 \\ -3 & 4 & 2 \end{vmatrix}$$ DRS of $PQ = (4\hat{i} + 2\hat{j} + 2\hat{k})$ OR $(2, 1, 1)$ $$\frac{3\lambda - \mu}{2} = \frac{2\mu - 4\lambda - 8}{1} = \frac{-2\lambda - 4}{1}$$ $$\Rightarrow \mu = \lambda + 2 \& 7\lambda = \mu - 8$$ $$\lambda = -1 \mu = 1$$ $Q : (-3, -4, 1)$ $$L_{PQ} = \frac{x + 3}{2} = \frac{y + 4}{1} = \frac{z - 1}{1}$$ $(-1, \alpha, \beta) \Rightarrow 1 = \frac{\alpha + 4}{1} = \frac{\beta - 1}{1}$ $$\Rightarrow \alpha = -3, \beta = 2$$ $$(\alpha - \beta)^2 = 25$$

Question 25

Maths · Sequences and Series · Numerical

If $1 + \frac{\sqrt{3} - \sqrt{2}}{2\sqrt{3}} + \frac{5 - 2\sqrt{6}}{18} + \frac{9\sqrt{3} - 11\sqrt{2}}{36\sqrt{3}} + \frac{49 - 20\sqrt{6}}{180} + \dots \text{ upto } \infty = 2 + \left( \sqrt{\frac{b}{a}} + 1 \right) \log_e \left( \frac{a}{b} \right)$, where $a$ and $b$ are integers with $\gcd(a, b) = 1$, then $11a + 18b$ is equal to \text{_____}.

Answer: 76

Solution

Given $$S = 1 + \frac{x}{2\sqrt{3}} + \frac{x^2}{18} + \frac{x^3}{36\sqrt{3}} + \frac{x^4}{180} + \ldots \infty$$ Put $\frac{x}{\sqrt{3}} = t$, where $x = \sqrt{3} - \sqrt{2}$. Then $$S = 1 + \frac{t}{2} + \frac{t^2}{6} + \frac{t^3}{12} + \frac{t^4}{20} + \ldots$$ $$S = 1 + t \left(1 - \frac{1}{2}\right) + t^2 \left(\frac{1}{2} - \frac{1}{3}\right) + t^3 \left(\frac{1}{3} - \frac{1}{4}\right) + t^4 \left(\frac{1}{4} - \frac{1}{5}\right)$$ $$S = \left(1 + t + \frac{t^2}{2} + \frac{t^3}{3} + \frac{t^3}{4} + \ldots \right) - \left(\frac{t}{2} + \frac{t^2}{3} + \frac{t^3}{4} + \frac{t^4}{5} + \ldots \right)$$ $$S = \left(t + \frac{t^2}{2} + \ldots \right) - \frac{1}{t} \left(t + \frac{t^2}{2} + \frac{t^3}{3} + \ldots \right) + 2$$ $$S = 2 + \left(1 - \frac{1}{t}\right)(-\log(1-t)) = \left(\frac{1}{t} - 1\right) \log(1-t) + 2$$ $$S = 2 + \left(\frac{\sqrt{3}}{\sqrt{3} - \sqrt{2}} - 1\right) \log \left(1 - \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3}}\right)$$ $$S = 2 + \left(\frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}\right) \log e \frac{\sqrt{2}}{\sqrt{3}}$$ $$S = 2 + \frac{(\sqrt{6} + 2)}{2} \log \frac{2}{3} = 2 + \left(\sqrt{\frac{3}{2}} + 1\right) \log e \frac{2}{3}$$ $a = 2, b = 3$ $11a + 18b = 11 \times 2 + 18 \times 3 = 76$

Question 26

Maths · Limits and Derivatives · Numerical

Let $a > 0$ be a root of the equation $2x^2 + x - 2 = 0$. If $\lim_{x \to \frac{1}{a}} \frac{16(1 - \cos(2 + x - 2x^2))}{(1 - ax)^2} = \alpha + \beta \sqrt{17}$, where $\alpha, \beta \in \mathbb{Z}$, then $\alpha + \beta$ is equal to _____.

Answer: 170

Solution

Given the equations $2x^2 + x - 2 = 0$ and $2x^2 - x - 2 = 0$, we have roots $\frac{1}{a}$ and $\frac{1}{b}$. The limit is given by: $$ \lim_{x \to \frac{1}{a}} 16 \cdot \frac{\left(1 - \cos 2 \left(x - \frac{1}{a}\right) \left(x - \frac{1}{b}\right)\right) \cdot 4\left(x - \frac{1}{b}\right)^2}{4\left(x - \frac{1}{b}\right)^2 \cdot a^2\left(x - \frac{1}{a}\right)^2} $$ Simplifying, we get: $$ = 16 \times \frac{2}{a^2} \left(\frac{1}{a} - \frac{1}{b}\right)^2 $$ Further simplification gives: $$ = \frac{32}{a^2} \left(\frac{17}{4}\right) = \frac{17.8}{a^2} = \frac{17 \times 8 \times 16}{(-1 + \sqrt{117})^2} $$ Calculating further: $$ = \frac{136.16}{18.2\sqrt{7}} \times \frac{18 + 2\sqrt{7}}{18 + 2\sqrt{7}} $$ This simplifies to: $$ = \frac{136}{256} (18 + 2\sqrt{7}) \cdot 16 $$ Finally, we have: $$ = 153 + 17\sqrt{17} = \alpha + \beta\sqrt{17} $$ Thus, $\alpha + \beta = 153 + 17 = 170$.

Question 27

Maths · Integrals · Numerical

If $f(t) = \int_{0}^{\pi} \frac{2x \, dx}{1 - \cos^2 t \sin^2 x}$, $0 < t < \pi$, then the value of $\int_{0}^{\frac{\pi}{2}} \frac{\pi^2 \, dt}{f(t)}$ equals

Answer: 1

Solution

Given $$f(t) = \int_0^\pi \frac{2x}{1 - \cos^2 t \sin^2 x} \, dx ....(1)$$ $$= 2 \int_0^\pi \frac{(\pi - x) \, dx}{1 - \cos^2 \sin^2 x} .....(2)$$ $$2f(t) = 2 \int_0^\pi \frac{\pi}{1 - \cos^2 \sin^2 x} \, dx$$ $$f(t) = \int_0^\pi \frac{\pi}{1 - \cos^2 t \sin^2 x} \, dx$$ Divide $\&$ by $\cos^2 x$ $$f(t) = \pi \int_0^\pi \frac{\sec^2 x \, dx}{\sec^2 x - \cos^2 t x}$$ $$f(t) = 2\pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{\sec^2 x - \cos^2 t^2 \tan^2 x}$$ Let $\tan x = z$ $\sec^2 x \, dx = dz$ $$f(t) = 2\pi \int_0^\infty \frac{dz}{1 + \sin^2 t \cdot z^2}$$ $$= \frac{\pi^2}{\sin t}$$ Then $$\int_0^{\pi/2} \frac{\pi^2}{f(t)} \, dt$$ $$= \int_0^{\pi/2} \sin t \, dt$$ $$= 1$$

Question 28

Maths · Applications of Derivatives · Numerical

Let the maximum and minimum values of $\left( \sqrt{8x - x^2 - 12 - 4} \right)^2 + (x - 7)^2, x \in \mathbb{R}$ be M and m, respectively. Then $M^2 - m^2$ is equal to

Answer: 1600

Solution

$(x-4)^2+(y-4)^2=16$ $y=\sqrt{8x-x^2}-12$ $y^2=(x-4)^2+16-12$ $(x-4)^2+y^2=4$ Circle: $(x-4)^2+y^2=2^2$ $C(4,0)$ and $r=2$ $P(7,4)$ $PC=\sqrt{(7-4)^2+(4-0)^2}$ $=\sqrt{3^2+4^2}$ $=5$ $m=PC+r$ $=5+4$ $=9$ $M=PC^2+r^2$ $=25+16$ $=41$ $M^2-m^2$ $=41^2-9^2$ $=1600$

Question 29

Maths · Conic Sections · Numerical

Let a line perpendicular to the line $2x - y = 10$ touch the parabola $y^2 = 4(x - 9)$ at the point $P$. The distance of the point $P$ from the centre of the circle $x^2 + y^2 - 14x - 8y + 56 = 0$ is

Answer: 10

Solution

Given $y^2 = 4(x - 9)$. The slope of the tangent is $-\frac{1}{2}$. The point of contact $P$ is given by $$\left(9 + \frac{1}{\left(-\frac{1}{2}\right)^2}, \frac{2 \times 1}{-\frac{1}{2}}\right).$$ This simplifies to $P(13, -4)$. The center of the circle $C$ is $(7, 4)$. The distance $CP$ is $$\sqrt{(13 - 7)^2 + (-4 - 4)^2} = 10.$$

Question 30

Maths · Complex Numbers and Quadratic Equations · Numerical

The number of real solutions of the equation $x|x + 5| + 2|x + 7| - 2 = 0$ is ________

Answer: 3

Solution

Case I: $x \geq -5$ $$x^2 + 5x + 2x + 12 = 0$$ $$x^2 + 7x + 12 = 0$$ $$x = -3, -4$$ Case II: $-7 < x < -5$ $$-x^2 - 5x + 2x + 14 - 2 = 0$$ $$-x^2 - 3x + 12 = 0$$ $$x = \frac{-3 \pm \sqrt{9 + 48}}{2}$$ $$= \frac{-3 \pm \sqrt{57}}{2}$$ $$x = \frac{-3 - \sqrt{57}}{2}, \frac{-3 + \sqrt{57}}{2} (rejected)$$ Case III: $x \leq -7$ $$-x^2 - 5x - 2x - 14 - 2 = 0$$ $$x^2 + 7x + 16 = 0$$ $$D = 49 - 64 < 0$$ No solutions No. of solutions = 3

Physics

Question 31

Physics · Ray Optics and Optical Instruments · Single correct

Given below are two statements: Statement I: When the white light passed through a prism, the red light bends lesser than yellow and violet. Statement II: The refractive indices are different for different wavelengths in dispersive medium. In the light of the above statements, chose the correct answer from the options given below:

  1. Statement I is false but Statement II is true
  2. Statement I is true but Statement II is false
  3. Both Statement I and Statement II are true
  4. Both Statement I and Statement II are false

Answer: (c)

Solution

As $\lambda_{red} > \lambda_{yellow} > \lambda_{violet}$. Light ray with longer wavelength bends less.

Question 32

Physics · Dual Nature of Radiation and Matter · Single correct

Which of the following statement is not true about stopping potential $(V_0)$?

  1. It is $1/e$ times the maximum kinetic energy of electrons emitted.
  2. It increases with increase in intensity of the incident light.
  3. It depends on the nature of emitter material.
  4. It depends upon frequency of the incident light.

Answer: (b)

Solution

The maximum kinetic energy is given by the equation $$\mathrm{KE_{max}} = h\nu - \phi_0 = eV$$

Question 33

Physics · Atoms · Single correct

The angular momentum of an electron in a hydrogen atom is proportional to: (Where $r$ is the radius of orbit of electron)

  1. $r$
  2. $\sqrt{r}$
  3. $\frac{1}{\sqrt{r}}$
  4. $\frac{1}{r}$

Answer: (b)

Solution

Given $F_C = \frac{mv^2}{r}$. $$\frac{K q_1 q_2}{r^2} = \frac{mv^2}{r}$$ $$mv^2 r^2 = K K_1 q_1 q_2 r$$ $$\frac{L^2}{m} = KK q_1 q_2 r$$ $L \propto \sqrt{r}$

Question 34

Physics · Current Electricity · Single correct

A galvanometer of resistance $100\,\Omega$ when connected in series with $400\,\Omega$ measures a voltage of upto $10\,\mathrm{V}$. The value of resistance required to convert the galvanometer into ammeter to read upto $10\,\mathrm{A}$ is $x \times 10^{-2}\,\Omega$. The value of $x$ is :

  1. 2
  2. 800
  3. 20
  4. 200

Answer: (c)

Solution

Given $i_g = \frac{10}{400 + 100} = 20 \times 10^{-3} \, \mathrm{A}$. For ammeter, let shunt resistance $= S$. $i_g R = (i - i_g) S$. $20 \times 10^{-3} \times 100 = 10 \, S$. $S = 20 \times 10^{-2} \, \Omega$.

Question 35

Physics · Electric Charges and Fields · Single correct

The vehicles carrying inflammable fluids usually have metallic chains touching the ground:

  1. To protect tyres from catching dirt from ground
  2. To alert other vehicles
  3. It is a custom
  4. To conduct excess charge due to air friction to ground and prevent sparking

Answer: (d)

Solution

Static charge is developed due to air friction. This can result in combustion. So, metallic chains are used to discharge excess charge.

Question 36

Physics · Kinetic Theory · Single correct

If n is the number density and d is the diameter of the molecule, then the average distance covered by a molecule between two successive collisions (i.e. mean free path) is represented by:

  1. $\sqrt{2n\pi d^2}$
  2. $\frac{1}{\sqrt{2n\pi d^2}}$
  3. $\frac{1}{\sqrt{2n\pi d^2}}$
  4. $\frac{1}{\sqrt{2n^2\pi^2 d^2}}$

Answer: (c)

Solution

n = number of molecule per unit volume d = diameter of the molecule $$\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$$ (By Theory)

Question 37

Physics · Mathematics in Physics · Single correct

A particle moves in $x-y$ plane under the influence of a force $\vec{F}$ such that its linear momentum is $\vec{p}(t) = \hat{i} \cos(kt) - \hat{j} \sin(kt)$. If $k$ is constant, the angle between $\vec{F}$ and $\vec{p}$ will be:

  1. $\frac{\pi}{4}$
  2. $\frac{\pi}{6}$
  3. $\frac{\pi}{2}$
  4. $\frac{\pi}{3}$

Answer: (c)

Solution

Given $\vec{P} = \cos(kt) \hat{i} - \sin(kt) \hat{j}$ and $|\vec{P}| = 1$. Therefore, $\vec{P} = m \vec{v}$. Thus, $\hat{P} = \hat{v}$. Hence, $\hat{v} = \cos(kt) \hat{i} - \sin(kt) \hat{j}$. The acceleration $\hat{a}$ is given by $$\hat{a} = \frac{-k \sin(kt) \hat{i} - k \cos(kt) \hat{j}}{k}$$ which simplifies to $$\hat{a} = -\sin kt \hat{i} - \cos kt \hat{j}.$$ Therefore, $\hat{F} = \hat{a} = -\sin kt \hat{i} - \cos kt \hat{j}$. The cosine of the angle $\theta$ is given by $$\cos \theta = \frac{\hat{F} \cdot \hat{P}}{|\hat{F}||\hat{P}|} = \frac{-\sin kt \cos t + \sin kt \cos t}{1 \times 1} = 0.$$ Thus, $\theta = \frac{\pi}{2}$.

Question 38

Physics · Moving Charges and Magnetism · Single correct

The electrostatic force $\left( \vec{F}_1 \right)$ and magnetic force $\left( \vec{F}_2 \right)$ acting on a charge $q$ moving with velocity $v$ can be written:

  1. $\vec{F}_1 = q\vec{E}, \vec{F}_2 = q(\vec{V} \times \vec{B})$
  2. $\vec{F}_1 = q\vec{B}, \vec{F}_2 = q(\vec{B} \times \vec{V})$
  3. $\vec{F}_1 = q\vec{E}, \vec{F}_2 = q(\vec{B} \times \vec{V})$
  4. $\vec{F}_1 = q\vec{V} \cdot \vec{E}, \vec{F}_2 = q(\vec{B} \cdot \vec{V})$

Answer: (a)

Solution

The forces are given by the equations: $$\vec{F}_1 = q \vec{E}$$ $$\vec{F}_2 = q (\vec{V} \times \vec{B})$$ (Theory)

Question 39

Physics · Laws of Motion · Single correct

A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius $9 \, \mathrm{m}$ and completes $120$ revolutions in $3$ minutes. The magnitude of centripetal acceleration of the monkey is \text{_____} $\mathrm{m/s^2}$.

  1. 57600$\pi^2$ $\mathrm{ms}^{-2}$
  2. Zero
  3. 4$\pi^2$ $\mathrm{ms}^{-2}$
  4. 16$\pi^2$ $\mathrm{ms}^{-2}$

Answer: (d)

Solution

Given: $R = 9 \, \mathrm{m}$, 120 revolutions in 3 minutes. $$\omega = \frac{120 \, \mathrm{Rev.}}{3 \, \mathrm{min.}} = \frac{120 \times 2\pi \, \mathrm{rad}}{3 \times 60 \, \mathrm{sec}} = \frac{4\pi}{3} \, \mathrm{rad/s}$$ $$a_{centripetal} = \omega^2 R = \left(\frac{4\pi}{3}\right)^2 \times 9 = 16\pi^2 \, \mathrm{m/s^2}$$

Question 40

Physics · Alternating Current · Single correct

A series LCR circuit is subjected to an ac signal of $200 \, \mathrm{V}$, $50 \, \mathrm{Hz}$. If the voltage across the inductor ($L = 10 \, \mathrm{mH}$) is $31.4 \, \mathrm{V}$, then the current in this circuit is _____.

  1. $68\,\mathrm{A}$
  2. $63\,\mathrm{A}$
  3. $10\,\mathrm{A}$
  4. $10\,\mathrm{mA}$

Answer: (c)

Solution

Voltage across inductor $V_L = I X_L$ $$31.4 = I [L \omega]$$ $$31.4 = I [L (2 \pi f)]$$ $$31.4 = I \left[ 10 \times 10^{-3} (2 \times 3.14) \times 50 \right]$$ $$\Rightarrow I = 10 \, \mathrm{A}$$

Question 41

Physics · Physical World, Units and Measurements · Single correct

What is the dimensional formula of $ab^{-1}$ in the equation $$\left( P + \frac{a}{V^2} \right) (V - b) = RT$$, where letters have their usual meaning.

  1. $[M^{-1}L^5T^{3}]$
  2. $[M^{5}L^7T^{4}]$
  3. $[ML^2T^{-2}]$
  4. $[M^0L^3T^{-2}]$

Answer: (c)

Solution

Since $[V] = [b]$, the dimension of $b$ is $[L^3]$. Therefore, $[P] = \left[ \frac{a}{V^2} \right]$. So, $[a] = [PV^2] = [ML^{-1}T^{-2}][L^6]$. The dimension of $a$ is $[ML^5T^{-2}]$. Therefore, $ab^{-1} = \frac{[ML^5T^{-2}]}{[L^3]} = [ML^2T^{-2}]$.

Question 42

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

The output ($Y$) of logic circuit given below is 0 only when :

  1. $A = 1$, $B = 0$
  2. $A = 0$, $B = 1$
  3. $A = 0$, $B = 0$
  4. $A = 1$, $B = 1$

Answer: (c)

Solution

The circuit consists of two OR gates and one AND gate. The inputs to the first OR gate are 0 and 0, resulting in an output of 0. The inputs to the AND gate are 0 and 1, resulting in an output of 0. The second OR gate receives inputs 0 and 0, resulting in an output of 0. Therefore, the final output Y0 is 0.

Question 43

Physics · Work, Energy and Power · Single correct

A body is moving unidirectionally under the influence of a constant power source. Its displacement in time $t$ is proportional to:

  1. $t$
  2. $t^{3/2}$
  3. $t^2$
  4. $t^{2/3}$

Answer: (b)

Solution

Given $P = constant$, it implies $FV = constant$. Therefore, $m \frac{dV}{dt} V = constant$. Integrating, we have $$\int_0^V V dV = (C) \int_0^t dt$$ $$\left( \frac{V^2}{2} \right) = Ct$$ Thus, $V \propto t^{1/2}$. Therefore, $$\frac{ds}{dt} \propto t^{1/2}$$ Integrating again, $$\int_0^s ds = K \int_0^t t^{1/2} dt$$ $$S = K \times \frac{2}{3} t^{3/2}$$ Therefore, $S \propto t^{3/2}$. Thus, displacement is proportional to $(t)^{3/2}$.

Question 44

Physics · Current Electricity · Single correct

Match List-I with List-II : Choose the correct answer from the options given below:

  1. $(A)-(III), (B)-(II), (C)-(IV), (D)-(I)$
  2. $(A)-(II), (B)-(I), (C)-(IV), (D)-(III)$
  3. $(A)-(IV), (B)-(III), (C)-(II), (D)-(I)$
  4. $(A)-(I), (B)-(III), (C)-(II), (D)-(IV)$

Answer: (a)

Solution

Infrared is the least energetic thus having the biggest wavelength ($\lambda$) and gamma rays are most energetic thus having the smallest wavelength ($\lambda$).

Question 45

Physics · Kinetic Theory · Single correct

During an adiabatic process, if the pressure of a gas is found to be proportional to the cube of its absolute temperature, then the ratio of $\frac{C_P}{C_V}$ for the gas is :

  1. $\frac{5}{3}$
  2. $\frac{9}{7}$
  3. $\frac{3}{2}$
  4. $\frac{7}{5}$

Answer: (c)

Solution

Given $P \propto T^3$. Therefore, $PT^{-3} = constant$. Thus, $\frac{PV}{T} = nR = constant from ideal gas equation$. $P(PV)^{-3} = constant$. $P^{-2} V^{-3} = constant \ldots (i)$ Therefore, the process equation for adiabatic process is $PV^\gamma = constant \ldots (ii)$ Comparing equation (1) and (2), $$\frac{C_P}{C_V} = \gamma = \frac{3}{2}$$

Question 46

Physics · Current Electricity · Single correct

Match List-I with List-II : Choose the correct answer from the options given below :

  1. $(A)$–(IV), $(B)$–(II), $(C)$–(III), $(D)$–(I)
  2. $(A)$–(III), $(B)$–(IV), $(C)$–(I), $(D)$–(II)
  3. $(A)$–(II), $(B)$–(IV), $(C)$–(I), $(D)$–(III)
  4. $(A)$–(III), $(B)$–(I), $(C)$–(II), $(D)$–(IV)

Answer: (b)

Solution

Given the equation for stress: $$stress = \frac{F_{restoring}}{A}$$ If $A = 1$, then $$Stress = F_{restoring}$$ (A)-(III) corresponds to shear stress. (B)-(IV) corresponds to volumetric stress. (C)-(I) corresponds to longitudinal stress. (D)-(II) corresponds to shear stress.

Question 47

Physics · Physical World, Units and Measurements · Single correct

A vernier callipers has 20 divisions on the vernier scale, which coincide with the $19^{\text{th}}$ division on the main scale. The least count of the instrument is $0.1\,\mathrm{mm}$. One main scale division is equal to \text{_____} $\mathrm{mm}$.

  1. 0.5
  2. 2
  3. 5
  4. 1

Answer: (b)

Solution

Given: $$20 VSD = 19 MSD$$ Therefore, $$1 VSD = \frac{19}{20} MSD$$ Least count (L.C.) is given by: $$L.C. = 1 MSD - 1 VSD$$ Substituting the values: $$0.1 \, mm = 1 MSD - \frac{19}{20} MSD$$ Simplifying: $$0.1 = \frac{1}{20} MSD$$ Thus, $$1 MSD = 2 \, mm$$

Question 48

Physics · Laws of Motion · Single correct

A heavy box of mass $50 \, \mathrm{kg}$ is moving on a horizontal surface. If co-efficient of kinetic friction between the box and horizontal surface is $0.3$ then force of kinetic friction is :

  1. 1.47 N
  2. 147 N
  3. 14.7 N
  4. 1470 N

Answer: (b)

Solution

The kinetic friction force is given by the formula $F_k = \mu_k N$. Substituting the given values, we have $$F_k = 0.3 \times 50 \times 9.8 = 147 \, \mathrm{N}.$$

Question 49

Physics · Gravitation · Single correct

A satellite revolving around a planet in stationary orbit has time period 6 hours. The mass of planet is one-fourth the mass of earth. The radius orbit of planet is : ( Given = Radius of geo-stationary orbit for earth is $4.2 \times 10^4$ km )

  1. $1.4 \times 10^4$ km
  2. $1.05 \times 10^4$ km
  3. $8.4 \times 10^4$ km
  4. $1.68 \times 10^5$ km

Answer: (b)

Solution

Given $$T = \frac{2\pi r^{3/2}}{\sqrt{GM}}$$ $$\frac{T_1}{T_2} = \left(\frac{r_1}{r_2}\right)^{3/2} \left(\frac{M_2}{M_1}\right)^{1/2}$$ $$\frac{6}{24} = \frac{(r_1)^{3/2}}{(4.2 \times 10^4)^{3/2}} \left(\frac{M}{M/4}\right)^{1/2}$$ $$r_1 = 1.05 \times 10^4 \, \mathrm{km}$$

Question 50

Physics · Current Electricity · Single correct

The ratio of heat dissipated per second through the resistance $5\,\Omega$ and $10\,\Omega$ in the circuit given below is :

  1. 1 : 2
  2. 2 : 1
  3. 4 : 1
  4. 1 : 1

Answer: (b)

Solution

The ratio of currents is given by $\($ $\frac{i_1}{i_2}$ = $\frac{10}{5}$ = $\frac{2}{1}$ $\)$. The ratio of powers is $\($ $\frac{P_1}{P_2}$ = $\frac{i_1^2 R_1}{i_2^2 R_2}$ = $\left$( $\frac{2}{1}$ $\right$)^2 $\times$ $\frac{5}{10}$ = $\frac{2}{1}$ $\)$.

Question 51

Physics · Moving Charges and Magnetism · Numerical

A solenoid of length 0.5 m has a radius of 1 cm and is made up of ' m ' number of turns. It carries a current of 5 A. If the magnitude of the magnetic field inside the solenoid is $6.28 \times 10^{-3} \, \mathrm{T}$ then the value of $m$ is _____.

Answer: 500

Solution

Given $\mu_0 n i = B$ where $n$ is the number of turns per unit length. $$\mu_0 \left( \frac{m}{\ell} \right) i = B$$ The expression for $m$ is given by: $$m = \frac{B \cdot \ell}{\mu_0 i} = \frac{6.28 \times 10^{-3} \times 0.5}{12.56 \times 10^{-7} \times 5}$$ Thus, $m = 500$.

Question 52

Physics · Atoms · Numerical

The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is $915\,\mathrm{\mathring{A}}$. The longest wavelength of spectral lines in the Balmer series will be \text{_____} $\mathrm{\mathring{A}}$.

Answer: 6588

Solution

Lyman Series Shortest, $\frac{hc}{\lambda} = -13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ $\lambda \downarrow E \uparrow$; $\frac{hc}{\lambda_0} = -13.6(1)$ Balmer Series: $$\frac{hc}{\lambda_1} = -13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$$ $$\frac{hc}{\lambda_1} = -13.6 \left( \frac{1}{4} - \frac{1}{9} \right)$$ $$\frac{hc}{\lambda_1} = -13.6 \times \left( \frac{5}{36} \right)$$ $$\Rightarrow \frac{-13.6 \lambda_0}{\lambda_1} = -13.6 \times \frac{5}{36}$$ $$\lambda_1 = \frac{\lambda_0 \times 36}{5} = \frac{915 \times 36}{5} = 6588$$

Question 53

Physics · Wave Optics · Numerical

In a single slit experiment, a parallel beam of green light of wavelength 550 nm passes through a slit of width 0.20 mm. The transmitted light is collected on a screen 100 cm away. The distance of first order minima from the central maximum will be $x \times 10^{-5}$ m. The value of $x$ is :

Answer: 275

Solution

Given $\lambda = 550 \, \mathrm{nm}$ and $d = 0.2 \, \mathrm{mm}$, with a distance of $100 \, \mathrm{cm}$, we calculate $y$ using the formula: $$y = \frac{\lambda D}{d} = \frac{550 \times 10^{-9} \times 100 \times 10^{-2}}{0.2 \times 10^{-3}} = 275.$$

Question 54

Physics · Waves · Numerical

A sonometer wire of resonating length 90 $\mathrm{\ cm}$ has a fundamental frequency of 400 $\mathrm{\ Hz}$ when kept under some tension. The resonating length of the wire with fundamental frequency of 600 $\mathrm{\ Hz}$ under same tension _____ cm.

Answer: 60

Solution

Given $f_0 = 400 \, \mathrm{Hz}$; $v = \sqrt{\frac{T}{\mu}} = constant$. $\frac{\lambda}{2} = L$; $v = f_0 \lambda$. $\frac{v}{2f_0} = L \Rightarrow v = 2Lf_0$. $L' = \frac{v}{2f'} = \frac{2Lf_0}{2f'}$. $\frac{L f_0}{f'} = \frac{90 \times 400}{600} = 60$.

Question 55

Physics · System of Particles and Rotational Motion · Numerical

A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is $\frac{x}{5}$. The value of $x$ is .

Answer: 2

Solution

Given $\($ $\frac{1}{2}$ I $\omega$^2 $\)$ over $\($ $\frac{1}{2}$ I $\omega$^2 + $\frac{1}{2}$ m v^2 $\)$ equals $\($ $\frac{\left( \frac{1}{2} \right) \left( \frac{2}{3} m R^2 \right) \omega^2}{\left( \frac{1}{2} \right) \left( \frac{2}{3} m R^2 \right) \omega^2 + \frac{1}{2} m (R \omega)^2}$ $\)$. This simplifies to $\($ $\frac{\frac{2}{3}}{\frac{2}{3} + 1}$ = $\frac{2}{5}$ $\)$. Thus, $\($ x = 2 $\)$.

Question 56

Physics · Mechanical Properties of Fluids · Numerical

A hydraulic press containing water has two arms with diameters as mentioned in the figure. A force of $10 \, \mathrm{N}$ is applied on the surface of water in the thinner arm. The force required to be applied on the surface of water in the thicker arm to maintain equilibrium of water is _____N.

Answer: 1000

Solution

$\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}$ $\dfrac{F_1}{\pi(7)^2} = \dfrac{10}{\pi \times (0.7)^2}$ $F_1 = 1000\,\text{N}$

Question 57

Physics · Electric Charges and Fields · Numerical

The electric field at point p due to an electric dipole is E. The electric field at point R on equitorial line will be $\frac{E}{x}$. The value of $x$:

Answer: 16

Solution

Given $$E_P = \frac{2KP}{r^3} = E$$ $$E_R = \frac{KP}{(2r)^3} = \frac{E}{16}$$ Therefore, $$x = 16$$

Question 58

Physics · Motion in a Plane · Fill in the blank

The maximum height reached by a projectile is 64 m. If the initial velocity is halved, the new maximum height of the projectile is _____ m.

Answer: 16

Solution

Given $$H_{max} = \frac{u^2 \sin^2 \theta}{2g}$$ $$\frac{H_{1 max}}{H_{2 max}} = \frac{u_1^2}{u_2^2}$$ $$\frac{64}{H_{2 max}} = \frac{u^2}{(u/2)^2}$$ $$H_{2 max} = 16 \, m$$

Question 59

Physics · Current Electricity · Numerical

A wire of resistance $20\,\Omega$ is divided into 10 equal parts, resulting pairs. A combination of two parts are connected in parallel and so on. Now resulting pairs of parallel combination are connected in series. The equivalent resistance of final combination is _____ $\Omega$.

Answer: 5

Solution

Each part has resistance $= 2\,\Omega$. 2 parts are connected in parallel so, $R = 1\,\Omega$. Now, there will be 5 parts each of resistance $1\,\Omega$, they are connected in series. $R_{eq} = 5R$, $R_{eq} = 5\,\Omega$.

Question 60

Physics · Electromagnetic Induction · Numerical

The current in an inductor is given by $I = (3t + 8)$ where $t$ is in second. The magnitude of induced emf produced in the inductor is $12\,\mathrm{mV}$. The self-inductance of the inductor ____ mH.

Answer: 4

Solution

Given $I = 3t + 8$ and $\varepsilon = 12 \, \mathrm{mV}$. The equation $|\varepsilon| = L \left| \frac{dI}{dt} \right|$ is used. Differentiating $I$ with respect to $t$, we get $\frac{dI}{dt} = 3$. Substituting, $12 = L \times 3$. Solving for $L$, we find $L = 4 \, \mathrm{mH}$.

Chemistry

Question 61

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Match List I with List II List - I \begin{tabular}{|c|c|c|l|} \hline \multicolumn{2}{|c|}{List - I} & \multicolumn{2}{c|}{List - II} \\ \hline (A) & ICl & (I) & T-shape \\ \hline (B) & ICl$_3$ & (II) & pyramidal \\ \hline (C) & ClF$_5$ & (III) & Pentagonal bipyramidal \\ \hline (D) & XeF$_2$ & (IV) & Linear \\ \hline \end{tabular} Choose the correct answer from the options given below :

  1. (A)-(IV), (B)-(I), $(C)$-(II), (D)-(III)
  2. (A)-(I), (B)-(IV), $(C)$-(III), (D)-(II)
  3. (A) -(IV), (B)-(III), $(C)$-(II), (D)-(I)
  4. (A) -(I), (B)-(III), $(C)$-(II), (D)-(IV)

Answer: (a)

Solution

A. I - Cl (iv) linear B. (I) T-shape C. (II) Square pyramidal D. (III) Pentagonal bipyramidal

Question 62

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

While preparing crystals of Mohr's salt, dil $\mathrm{H_2SO_4}$ is added to a mixture of ferrous sulphate and ammonium sulphate, before dissolving this mixture in water, dil $\mathrm{H_2SO_4}$ is added here to:

  1. prevent the hydrolysis of ferrous sulphate
  2. prevent the hydrolysis of ammonium sulphate
  3. make the medium strongly acidic
  4. increase the rate of formation of crystals

Answer: (a)

Solution

Fe^{+2} ions undergoes hydrolysis, therefore while preparing aqueous solution of ferrous sulphate and ammonium sulphate in water dilute sulphuric acid is added to prevent hydrolysis of ferrous sulphate.

Question 63

Chemistry · Haloalkanes and Haloarenes · Single correct

Identify the major product in the following reaction.

Answer: (c)

Solution

The reaction involves an elimination process where the bromine (Br) is removed, and a double bond is formed between the adjacent carbon atoms. The base, $\mathrm{OH}^-$, in ethanol ($\mathrm{EtOH}$) facilitates this elimination by abstracting a proton ($\mathrm{H}^+$) from the carbon adjacent to the one bearing the bromine. This results in the formation of a double bond, yielding the final product, a cyclopentene with a methyl group ($\mathrm{CH_3}$) attached.

Question 64

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

The correct nomenclature for the following compound is :

  1. 2-formyl-4-hydroxyhept-7-enoic acid
  2. 2-formyl-4-hydroxyhept-6-enoic acid
  3. 2-carboxy-4-hydroxyhept-7-enal
  4. 2-carboxy-4-hydroxyhept-6-enal

Answer: (b)

Solution

The compound shown is named 2-formyl-4-hydroxyhept-6-enoic acid.

Question 65

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason $(R)$. Assertion (A) : $\mathrm{NH}_3$ and $\mathrm{NF}_3$ molecule have pyramidal shape with a lone pair of electrons on nitrogen atom. The resultant dipole moment of $\mathrm{NH}_3$ is greater than that of $\mathrm{NF}_3$. Reason $(R)$ : In $\mathrm{NH}_3$, the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the $\mathrm{N} - \mathrm{H}$ bonds. $\mathrm{F}$ is the most electronegative element. In the light of the above statements, choose the correct answer from the options given below :

  1. Both (A) and $(R)$ are true and $(R)$ is the correct explanation of (A)
  2. is false but $(R)$ is true
  3. Both (A) and $(R)$ are true but $(R)$ is NOT the correct explanation of (A)
  4. is true but $(R)$ is false

Answer: (a)

Solution

The first molecule has a resultant dipole moment of $0.80 \times 10^{-30} \, \mathrm{Cm}$. The second molecule has a resultant dipole moment of $4.90 \times 10^{-30} \, \mathrm{cm}$.

Question 66

Chemistry · Equilibrium · Single correct

Given below are two statements : Statement I : On passing $\mathrm{HCl}_{(g)}$ through a saturated solution of $\mathrm{BaCl}_2$, at room temperature white turbidity appears. Statement II : When $\mathrm{HCl}$ gas is passed through a saturated solution of $\mathrm{NaCl}$, sodium chloride is precipitated due to common ion effect. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Both Statement I and Statement II are incorrect
  4. Statement I is incorrect but Statement II is correct

Answer: (b)

Solution

$BaCl_2, NaCl$ are soluble but on adding $HCl(g) to BaCl_2, NaCl$ solutions, Sodium or Barium chlorides may precipitate out, as a consequence of the law of mass action.

Question 67

Chemistry · Co-ordination Compounds · Single correct

The metal atom present in the complex MABXL (where A, B, X and L are unidentate ligands and M is metal) involves $sp^3$ hybridization. The number of geometrical isomers exhibited by the complex is:

  1. 2
  2. 0
  3. 4
  4. 3

Answer: (b)

Solution

Tetrahedral complex does not show geometrical isomerism.

Question 68

Chemistry · Co-ordination Compounds · Single correct

Match List I with List II Choose the correct answer from the options given below :

  1. (A)-(II), (B)-(I), $(C)$-(IV), (D)-(III)
  2. (A)-(III), (B)-(I), $(C)$-(II), (D)-(IV)
  3. (A)-(I), (B)-(III), $(C)$-(IV), (D)-(II)
  4. (A)-(III), (B)-(I), $(C)$-(IV), (D)-(II)

Answer: (d)

Solution

The given compounds are classified as follows: Position isomers: The compounds with OH groups in different positions are position isomers. Metamers: The compounds with different alkyl groups attached to the same functional group are metamers. Functional isomers: The compounds with different functional groups are functional isomers. Chain isomers: Neopentane and isopentane are chain isomers.

Question 69

Chemistry · Electrochemistry · Single correct

The quantity of silver deposited when one coulomb charge is passed through $\mathrm{AgNO_3}$ solution:

  1. 1 g of silver
  2. 1 electrochemical equivalent of silver
  3. 1 chemical equivalent of silver
  4. 0.1 g atom of silver

Answer: (b)

Solution

Given $$W = ZIt$$ $$W = ZQ$$ $$Q = \frac{W}{Z}$$ $$W = ZQ =$$ (electrochemical equivalent)

Question 70

Chemistry · Alcohols, Phenols and Ethers · Single correct

Which one of the following reactions is NOT possible?

Answer: (c)

Solution

The reaction of phenol with HCl is not possible because the hydroxyl group is attached to an $sp^2$ hybridized carbon, which does not allow the substitution reaction to occur.

Question 71

Chemistry · Classification of Elements and Periodicity in Properties · Single correct

Given below are two statements : Statement I : The metallic radius of Na is 1.86 $\mathrm{\AA}$ and the ionic radius of $\mathrm{Na^+}$ is lesser than 1.86 $\mathrm{\AA}$ . Statement II : Ions are always smaller in size than the corresponding elements. In the light of the above statements, choose the correct answer from the options given below :

  1. Both Statement I and Statement II are false
  2. Statement I is incorrect but Statement II is true
  3. Both Statement I and Statement II are true
  4. Statement I is correct but Statement II is false

Answer: (d)

Solution

Given $r_{\mathrm{Na}} > r_{\mathrm{Na}^+}$. So, Statement (I) is correct but size of anions are greater than size of neutral atoms. So statement (II) is incorrect.

Question 72

Chemistry · Alcohols, Phenols and Ethers · Single correct

Consider the above reaction sequence and identify the major product $P$.

  1. Methoxymethane
  2. Methanoic acid
  3. Methanal
  4. Methane

Answer: (d)

Solution

The reaction starts with $\mathrm{CH_3CH_2OH}$, which is oxidized using Joner reagent ($\mathrm{CrO_3 + H^+}$) or $\mathrm{KMnO_4}$ to form $\mathrm{CH_3COOH}$. This acetic acid undergoes a soda lime process with $\mathrm{NaOH}$ and $\mathrm{CaI}$ under heat ($\Delta$) to produce $\mathrm{CH_4 + Na_2CO_3}$.

Question 73

Chemistry · Hydrocarbons · Single correct

Consider the given chemical reaction: Product "A" is:

  1. picric acid
  2. acetic acid
  3. adipic acid
  4. oxalic acid

Answer: (c)

Solution

The reaction shown is the oxidation of cyclohexene using potassium permanganate ($\mathrm{KMnO_4}$) in the presence of sulfuric acid ($\mathrm{H_2SO_4}$). This reaction converts the alkene into a dicarboxylic acid. The product formed is hexanedioic acid, also known as adipic acid.

Question 74

Chemistry · Electrochemistry · Single correct

For the electro chemical cell $\mathrm{M}|\mathrm{M}^{2+}||\mathrm{X}|\mathrm{X}^{2-}$ If $\mathrm{E}^0_{(\mathrm{M}^{2+}/\mathrm{M})} = 0.46 \, \mathrm{V}$ and $\mathrm{E}^0_{(\mathrm{X}/\mathrm{X}^{2-})} = 0.34 \, \mathrm{V}$. Which of the following is \textbf{correct} ?

  1. $\mathrm{M} + \mathrm{X} \rightarrow \mathrm{M}^{2+} + \mathrm{X}^{2-}$ is a spontaneous reaction
  2. $\mathrm{E}_{cell} = 0.80 \, \mathrm{V}$
  3. $\mathrm{E}_{cell} = -0.80 \, \mathrm{V}$
  4. $\mathrm{M}^{2+} + \mathrm{X}^{2-} \rightarrow \mathrm{M} + \mathrm{X}$ is a spontaneous reaction

Answer: (d)

Solution

M | $\mathrm{M^{+2}}$ || $\mathrm{X/X^{2-}}$ $$E^\circ_{cell} = E^\circ_{\mathrm{M/M^{+2}}} + E^\circ_{\mathrm{X/X^{2-}}}$$ $$= -0.46 + 0.34 = -0.12 \, \mathrm{V}$$ As $E^\circ_{cell}$ is negative so anode becomes cathode and cathode becomes anode. Spontaneous reaction will be $$\mathrm{M^{+2} + X^{2-} \rightarrow M + X}$$

Question 75

Chemistry · Some Basic Concepts of Chemistry · Single correct

The number of moles of methane required to produce 11 $\mathrm{gCO}_2(g)$ after complete combustion is: (Given molar mass of methane in $\mathrm{g \ mol}^{-1}$: 16)

  1. 0.35
  2. 0.5
  3. 0.75
  4. 0.25

Answer: (d)

Solution

For the reaction: $$\mathrm{C_nH_{2n+2} + \frac{3n+1}{2}O_2 \rightarrow nCO_2 + (n+1)H_2O}$$ Consider the specific case: $$\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$$ Given: 4 gm of $\mathrm{CH_4}$ corresponds to 0.25 mole. 11 gm of $\mathrm{CO_2}$ corresponds to 0.25 mole. Therefore, 0.25 mol $\mathrm{CH_4}$ gives 0.25 mole (or 11 gm) $\mathrm{CO_2}$.

Question 76

Chemistry · Co-ordination Compounds · Single correct

The number of complexes from the following with no electrons in the $t_2$ orbital is $\mathrm{TiCl}_4$, $[\mathrm{MnO}_4]^-$, $[\mathrm{FeO}_4]^{2-}$, $[\mathrm{FeCl}_4]^-$, $[\mathrm{CoCl}_4]^{2-}$

  1. 1
  2. 4
  3. 3
  4. 2

Answer: (c)

Solution

Q7 $\mathrm{TiCl_4} \Rightarrow \mathrm{Ti}^{+4} e^{0}t_{2}^{0}$ $\mathrm{MnO_4}^{-} \Rightarrow \mathrm{Mn}^{+7} e^{0}t_{2}^{0}$ $\mathrm{FeO_4}^{2-} \Rightarrow \mathrm{Fe}^{+6} e^{2}t_{2}^{0}$ $\mathrm{FeCl_4}^{2-} \Rightarrow \mathrm{Fe}^{+2} e^{3}t_{2}^{3}$ $\mathrm{CoCl_4}^{2-} \Rightarrow \mathrm{Co}^{+2} e^{4}t_{2}^{3}$

Question 77

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

The number of ions from the following that have the ability to liberate hydrogen from a dilute acid is _____. $\mathrm{Ti}^{2+}$, $\mathrm{Cr}^{2+}$ and $\mathrm{V}^{2+}$

  1. 2
  2. 3
  3. 1
  4. 0

Answer: (b)

Solution

The ions $\mathrm{Ti^{+2}}$, $\mathrm{V^{+2}Cr^{+2}}$ are strong reducing agents and will liberate hydrogen from a dilute acid, e.g. $$2\mathrm{Cr^{+2}_{(aq.)}} + 2\mathrm{H^{+}_{(aq.)}} \rightarrow 2\mathrm{Cr^{+3}_{(aq.)}} + \mathrm{H_2\,(g)}$$

Question 78

Chemistry · Hydrocarbons · Single correct

Identify $A$ and $B$ in the given chemical reaction sequence:

Answer: (c)

Solution

The reaction begins with the Friedel-Crafts acylation of benzene using an acyl chloride in the presence of $\mathrm{AlCl_3}$, leading to compound (A). Compound (A) undergoes Clemmensen reduction using $\mathrm{Zn, Hg}$ and $\mathrm{HCl}$ to form compound (B). Finally, compound (B) undergoes electrophilic substitution reaction (ESR) in the presence of $\mathrm{H^+}$ to form the final product.

Question 79

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

The correct statements from the following are: (A) The decreasing order of atomic radii of group 13 elements is Tl > In > Ga > Al > B. (B) Down the group 13 electronegativity decreases from top to bottom. (C) Al dissolves in dil. HCl and liberate $\mathrm{H}_2$, but conc. $\mathrm{HNO}_3$ renders Al passive by forming a protective oxide layer on the surface. (D) All elements of group 13 exhibits highly stable +1 oxidation state. (E) Hybridisation of Al in $[\mathrm{Al}(\mathrm{H}_2\mathrm{O})_6]^{3+}$ ion is $\mathrm{sp}^3\mathrm{d}^2$. Choose the correct answer from the options given below:

  1. (C) and (E) only
  2. (A), (C) and (E) only
  3. (A), (B), (C) and (E) only
  4. (A) and (C) only

Answer: (a)

Solution

Question 80

Chemistry · Biomolecules · Single correct

Coagulation of egg, on heating is because of:

  1. The secondary structure of protein remains unchanged
  2. Denaturation of protein occurs
  3. Biological property of protein remains unchanged
  4. Breaking of the peptide linkage in the primary structure of protein occurs

Answer: (b)

Solution

Coagulation of egg gives primary structure of protein, which is known as denaturation of protein.

Question 81

Chemistry · Thermodynamics · Numerical

Combustion of 1 mole of benzene is expressed at $C_6H_6(l) + \frac{15}{2} O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)$ The standard enthalpy of combustion of 2 mol of benzene is $-x\,\text{kJ}$. $x = \underline{\hspace{1.5cm}}$ \textbf{Given:} 1. Standard enthalpy of formation of 1 mol of $\mathrm{C_6H_6(l)}$, for the reaction $\mathrm{6C(graphite) + 3H_2(g) \rightarrow C_6H_6(l)}$ is $48.5\,\text{kJ mol}^{-1}$. 2. Standard enthalpy of formation of 1 mol of $\mathrm{CO_2(g)}$, for the reaction $\mathrm{C(graphite) + O_2(g) \rightarrow CO_2(g)}$ is $-393.5\,\text{kJ mol}^{-1}$. 3. Standard enthalpy of formation of 1 mol of $\mathrm{H_2O(l)}$, for the reaction $\mathrm{H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)}$ is $-286\,\text{kJ mol}^{-1}$.

Answer: 6534

Solution

6C(graphite) + 3$\mathrm{H_2}$($\mathrm{g}$) $\rightarrow$ $\mathrm{C_6H_6}$($\ell$); $\Delta$ H = 48.5 $\mathrm{kJ/mol}$ C(graphite) + $\mathrm{O_2}$($\mathrm{g}$) $\rightarrow$ $\mathrm{CO_2}$($\mathrm{g}$); $\Delta$ H = -393.5 $\mathrm{kJ/mol}$ $\mathrm{H_2^{(g)}}$ + $\frac{1}{2}$ $\mathrm{(g)}$ $\rightarrow$ $\mathrm{H_2O}$($\ell$); $\Delta$ H = -286 $\mathrm{kJ/mol}$ Equation - (1) $\times$ 1 + (2) $\times$ 6 + (3) $\times$ 3 -48.5 - 6 $\times$ 393.5 - 3 $\times$ 286 = -3267.5 $\mathrm{kJ}$ for 1 mol = -6535 $\mathrm{kJ}$ for 2 mol Ans. 6535 kJ

Question 82

Chemistry · The d-and f-Block Elements · Numerical

The fusion of chromite ore with sodium carbonate in the presence of air leads to the formation of products $A$ and $B$ along with the evolution of $\mathrm{CO}_2$. The sum of spin-only magnetic moment values of $A$ and $B$ is \text{_____} B.M. (Nearest integer) [Given atomic number : $\mathrm{C} : 6, \mathrm{Na} : 11, \mathrm{O} : 8, \mathrm{Fe} : 26, \mathrm{Cr} : 24$]

Answer: 6

Solution

Spin only magnetic moment. For $\mathrm{Na_2CrO_4}$, $\mu_B = 0$. For $\mathrm{Fe_2O_3}$, $\mu_B = 5.9$ sum $= 5.9$.

Question 83

Chemistry · Amines · Numerical

X g of ethanamine was subjected to reaction with NaNO$_2$/HCl followed by hydrolysis to liberate N$_2$ and HCl. The HCl generated was completely neutralised by 0.2 moles of NaOH. X is _____ g.

Answer: 9

Solution

The reaction involves ethylamine reacting with $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ to form a diazonium salt, which then reacts with water to form ethanol, nitrogen gas, and hydrochloric acid. Given $0.2$ mole of ethylamine with a molecular weight of $45$, the mass is calculated as $45 \times 0.2 = 9$ gm. The reaction produces $0.2$ mole of ethanol.

Question 84

Chemistry · Structure of Atom · Numerical

In an atom, total number of electrons having quantum numbers $n = 4$, $|m_l| = 1$ and $m_s = -\frac{1}{2}$ is

Answer: 6

Solution

Given \[ n=4 \] \[ \begin{array}{c|c} \ell & m_\ell \\ \hline 0 & 0 \\ 1 & -1,\;0,\;+1 \\ 2 & -2,\;-1,\;0,\;+1,\;+2 \\ 3 & -3,\;-2,\;-1,\;0,\;+1,\;+2,\;+3 \end{array} \] So, the number of orbitals associated with \[ n=4,\quad |m_\ell|=1 \] are \(6\). Now each orbital contains one \(e^{-}\) with \[ m_s=-\frac{1}{2}. \]

Question 85

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Numerical

Using the given figure, the ratio of $R_f$ values of sample A and sample C is $x \times 10^{-2}$. Value of $x$ is _____

Answer: 50

Solution

The $R_f$ of A is given by $$R_f of A = \frac{5}{12.5}$$ The $R_f$ of C is given by $$R_f of C = \frac{10}{12.5}$$ The ratio is calculated as $$Ratio = \frac{R_{f(A)}}{R_{f(C)}} = \frac{1}{2} = 0.5 or 50 \times 10^{-2}$$

Question 86

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical

In the Claisen-Schmidt reaction to prepare 351 g of dibenzalacetone using 87 g of acetone, the amount of benzaldehyde required is ______ g. (Nearest integer)

Answer: 318

Solution

Claisen Schmidt reaction mw of benzaldehyde = 106 $$106 \times 3 = 318 \, \mathrm{gm}.$$ Benzaldehyde is required to give 1.5 mole (or 351 gm) product.

Question 87

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

Consider the following single step reaction in gas phase at constant temperature. $$2 \mathrm{A}_{(g)} + \mathrm{B}_{(g)} \rightarrow \mathrm{C}_{(g)}$$ The initial rate of the reaction is recorded as $r_1$ when the reaction starts with $1.5 \, \mathrm{atm}$ pressure of A and $0.7 \, \mathrm{atm}$ pressure of B. After some time, the rate $r_2$ is recorded when the pressure of C becomes $0.5 \, \mathrm{atm}$. The ratio $r_1 : r_2$ is ______ $\times 10^{-1}$. (Nearest integer)

Answer: 315

Solution

Given the reaction: $$2 \mathrm{A} (\mathrm{g}) + \mathrm{B} (\mathrm{g}) \rightarrow \mathrm{C} (\mathrm{g})$$ For $r_1$: $1.5 \, \mathrm{atm}$ and $0.7 \, \mathrm{atm}$ For $r_2$: $0.5 \, \mathrm{atm}$, $0.2 \, \mathrm{atm}$, $0.5 \, \mathrm{atm}$ Therefore, $r = K[P_A]^2[P_B]$ For $r_1$: $$r_1 = K[1.5]^2[0.7]$$ For $r_2$: $$r_2 = K[0.5]^2[0.2]$$ The ratio is given by: $$\frac{r_1}{r_2} = 9 \times \frac{7}{2} = 31.5 = 315 \times 10^{-1}$$ The answer is 315.

Question 88

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical

The product $(C)$ in the following sequence of reactions has _____ $\pi$ bonds.

Answer: 4

Solution

All structures A, B, and C have 4 $\($ $\pi$ $\)$ bonds.

Question 89

Chemistry · Solutions · Numerical

Considering acetic acid dissociates in water, its dissociation constant is $6.25 \times 10^{-5}$. If $5 \, \mathrm{mL}$ of acetic acid is dissolved in $1 \, \mathrm{litre}$ water, the solution will freeze at $-x \times 10^{-2} \degree \mathrm{C}$, provided pure water freezes at $0 \degree \mathrm{C}$. $x =$ \text{_____}. (Nearest integer) Given : $\left(K_f\right)_{\mathrm{water}} = 1.86 \, \mathrm{K \, kg \, mol}^{-1}$. density of acetic acid is $1.2 \, \mathrm{g \, mL}^{-1}$. molar mass of water $= 18 \, \mathrm{g \, mol}^{-1}$. molar mass of acetic acid $=60 \, \mathrm{g \, mol}^{-1}$. density of water $= 1 \, \mathrm{g \, cm}^{-3}$. Acetic acid dissociates as $\mathrm{CH_3COOH \rightleftharpoons CH_3COO^- + H^+}$

Answer: 19

Solution

Mass of $\mathrm{CH_3COOH} = V \times d$ $= 5 \, \mathrm{ml} \times 1.2 \, \mathrm{g/ml}$ $= 6 \, \mathrm{gm}$

Question 90

Chemistry · Chemical Bonding and Molecular Structure · Numerical

Number of compounds from the following with zero dipole moment is $HF, H_2, H_2 S, CO_2, NH_3, BF_3, CH_4, CHCl_3, SiF_4, H_2O, BeF_2$

Answer: 6

Solution

$H_2, CO_2, BF_3, CH_4, SiF_4, BeF_2$ are symmetric molecules so the dipole moment is zero.