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.
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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 :
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:
$\frac{125\pi}{12}$
$\frac{125\pi}{4}$
$\frac{125\pi}{24}$
$\frac{125\pi}{6}$
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 :
$\frac{4}{3}$
1
$\frac{2}{3}$
$\frac{8}{3}$
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
JBBOH
OBBJH
OBBHJ
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:
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:
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:
6
-$\frac{6}{5}$
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 :
16
20
14
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:
724
742
639
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
neither one-one nor onto.
one-one but not onto.
onto but not one-one.
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:
2
1
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:
1680
1640
1520
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:
$m^2 + n^2 - mn = 39$
$m^2 + n^2 - m - n = 46$
$m^2 + n^2 + m + n = 64$
$m^2 + n^2 + mn = 68$
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:
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:
r = 0
2r^2 - 4r + 1 = 0
2r^2 - 8r + 7 = 0
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
1021
2120
2012
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:
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:
8
4
10
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 _______
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
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
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 ________
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:
Statement I is false but Statement II is true
Statement I is true but Statement II is false
Both Statement I and Statement II are true
Both Statement I and Statement II are false
Answer: (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 $(V_0)$?
It is $1/e$ times the maximum kinetic energy of electrons emitted.
It increases with increase in intensity of the incident light.
It depends on the nature of emitter material.
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)
$r$
$\sqrt{r}$
$\frac{1}{\sqrt{r}}$
$\frac{1}{r}$
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 :
2
800
20
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:
To protect tyres from catching dirt from ground
To alert other vehicles
It is a custom
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:
$\sqrt{2n\pi d^2}$
$\frac{1}{\sqrt{2n\pi d^2}}$
$\frac{1}{\sqrt{2n\pi d^2}}$
$\frac{1}{\sqrt{2n^2\pi^2 d^2}}$
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:
$\frac{\pi}{4}$
$\frac{\pi}{6}$
$\frac{\pi}{2}$
$\frac{\pi}{3}$
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:
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{_____} $\mathrm{m/s^2}$.
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 _____.
$68\,\mathrm{A}$
$63\,\mathrm{A}$
$10\,\mathrm{A}$
$10\,\mathrm{mA}$
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.
$[M^{-1}L^5T^{3}]$
$[M^{5}L^7T^{4}]$
$[ML^2T^{-2}]$
$[M^0L^3T^{-2}]$
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 :
$A = 1$, $B = 0$
$A = 0$, $B = 1$
$A = 0$, $B = 0$
$A = 1$, $B = 1$
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:
$t$
$t^{3/2}$
$t^2$
$t^{2/3}$
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:
$(A)-(III), (B)-(II), (C)-(IV), (D)-(I)$
$(A)-(II), (B)-(I), (C)-(IV), (D)-(III)$
$(A)-(IV), (B)-(III), (C)-(II), (D)-(I)$
$(A)-(I), (B)-(III), (C)-(II), (D)-(IV)$
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 :
$\frac{5}{3}$
$\frac{9}{7}$
$\frac{3}{2}$
$\frac{7}{5}$
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 :
$(A)$–(IV), $(B)$–(II), $(C)$–(III), $(D)$–(I)
$(A)$–(III), $(B)$–(IV), $(C)$–(I), $(D)$–(II)
$(A)$–(II), $(B)$–(IV), $(C)$–(I), $(D)$–(III)
$(A)$–(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{_____} $\mathrm{mm}$.
0.5
2
5
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 :
1.47 N
147 N
14.7 N
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 )
The ratio of heat dissipated per second through the resistance $5\,\Omega$ and $10\,\Omega$ in the circuit given below is :
1 : 2
2 : 1
4 : 1
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{_____} $\mathrm{\mathring{A}}$.
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.
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.
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 $x$:
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 :
(A)-(IV), (B)-(I), $(C)$-(II), (D)-(III)
(A)-(I), (B)-(IV), $(C)$-(III), (D)-(II)
(A) -(IV), (B)-(III), $(C)$-(II), (D)-(I)
(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:
prevent the hydrolysis of ferrous sulphate
prevent the hydrolysis of ammonium sulphate
make the medium strongly acidic
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 :
2-formyl-4-hydroxyhept-7-enoic acid
2-formyl-4-hydroxyhept-6-enoic acid
2-carboxy-4-hydroxyhept-7-enal
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 :
Both (A) and $(R)$ are true and $(R)$ is the correct explanation of (A)
is false but $(R)$ is true
Both (A) and $(R)$ are true but $(R)$ is NOT the correct explanation of (A)
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 :
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: (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:
2
0
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 :
(A)-(II), (B)-(I), $(C)$-(IV), (D)-(III)
(A)-(III), (B)-(I), $(C)$-(II), (D)-(IV)
(A)-(I), (B)-(III), $(C)$-(IV), (D)-(II)
(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:
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 :
Both Statement I and Statement II are false
Statement I is incorrect but Statement II is true
Both Statement I and Statement II are true
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 $P$.
Methoxymethane
Methanoic acid
Methanal
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:
picric acid
acetic acid
adipic acid
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} ?
$\mathrm{M} + \mathrm{X} \rightarrow \mathrm{M}^{2+} + \mathrm{X}^{2-}$ is a spontaneous reaction
$\mathrm{E}_{cell} = 0.80 \, \mathrm{V}$
$\mathrm{E}_{cell} = -0.80 \, \mathrm{V}$
$\mathrm{M}^{2+} + \mathrm{X}^{2-} \rightarrow \mathrm{M} + \mathrm{X}$ 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)
0.35
0.5
0.75
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 $\mathrm{TiCl}_4$, $[\mathrm{MnO}_4]^-$, $[\mathrm{FeO}_4]^{2-}$, $[\mathrm{FeCl}_4]^-$, $[\mathrm{CoCl}_4]^{2-}$
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 $\mathrm{V}^{2+}$
2
3
1
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:
(C) and (E) only
(A), (C) and (E) only
(A), (B), (C) and (E) only
(A) and (C) only
Answer: (a)
Solution
Question 80
Chemistry · Biomolecules · Single correct
Coagulation of egg, on heating is because of:
The secondary structure of protein remains unchanged
Denaturation of protein occurs
Biological property of protein remains unchanged
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 $-286\,\text{kJ mol}^{-1}$.
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 \[ m_s=-\frac{1}{2}. \]
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.