JEE Main 18 March 2021 Shift 1 question paper with solutions
JEE Main 18 March 2021 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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Maths
Question 1
Maths · Differential Equations · Single correct
The differential equation satisfied by the system of parabolas $y^2 = 4a(x + a)$ is:
Given $y^2 = 4ax + 4a^2$. Differentiate with respect to $x$: $$2y \frac{dy}{dx} = 4a$$ $$\Rightarrow a = \left( \frac{y}{2} \frac{dy}{dx} \right)$$ So, the required differential equation is $$y^2 = \left( 4 \times \frac{y}{2} \frac{dy}{dx} \right) x + 4 \left( \frac{y}{2} \frac{dy}{dx} \right)^2$$ $$\Rightarrow y^2 \left( \frac{dy}{dx} \right)^2 + 2xy \left( \frac{dy}{dx} \right) - y^2 = 0$$ $$\Rightarrow y \left( \frac{dy}{dx} \right)^2 + 2x \left( \frac{dy}{dx} \right) - y = 0$$
Question 2
Maths · Straight Lines and Pair of Straight Lines · Single correct
The number of integral values of $m$ so that the abscissa of point of intersection of lines $3x + 4y = 9$ and $y = mx + 1$ is also an integer, is:
1
2
3
0
Answer: (b)
Solution
Given the equations $3x + 4y = 9$ and $y = mx + 1$. Substituting $y$ in the first equation gives: $$3x + 4mx + 4 = 9$$ Simplifying, we have: $$(3 + 4m)x = 5$$ $x$ will be an integer when $3 + 4m = 5, -5, 1, -1$. Solving for $m$, we get: $$m = \frac{1}{2}, -2, -\frac{1}{2}, -1$$ Thus, the number of integral values of $m$ is 2.
Question 3
Maths · Binomial Theorem · Single correct
$$ (1+x+2x^2)^{20} = a_0 + a_1x + a_2x^2 + \ldots + a_{40}x^{40} $$ Then $$ a_1 + a_3 + a_5 + \ldots + a_{37} $$ is equal to
The solutions of the equation $$\left| \begin{array}{ccc} 1 + \sin^2 x & \sin^2 x & \sin^2 x \\ \cos^2 x & 1 + \cos^2 x & \cos^2 x \\ 4 \sin 2x & 4 \sin 2x & 1 + 4 \sin 2x \end{array} \right|=0,(0 < x < \pi) are$$
$\frac{\pi}{12}, \frac{\pi}{6}$
$\frac{\pi}{6}, \frac{5\pi}{6}$
$\frac{5\pi}{12}, \frac{7\pi}{12}$
$\frac{7\pi}{12}, \frac{11\pi}{12}$
Answer: (d)
Solution
Given the matrix equation: $$\begin{vmatrix} 1 + \sin^2 x & \sin^2 x & \sin^2 x \\ \cos^2 x & 1 + \cos^2 x & \cos^2 x \\ 4 \sin 2x & 4 \sin 2x & 1 + 4 \sin 2x \end{vmatrix} = 0$$ Use $R_1 \rightarrow R_1 + R_2 + R_3$. This gives: $$\Rightarrow (2 + 4 \sin 2x) \begin{vmatrix} 1 & 1 & 1 \\ \cos^2 x & 1 + \cos^2 x & \cos^2 x \\ 4 \sin 2x & 4 \sin 2x & 1 + 4 \sin 2x \end{vmatrix} = 0$$ Therefore, $$x = \frac{\pi}{2} + \frac{\pi}{12}, \pi - \frac{\pi}{12}$$
Question 5
Maths · Conic Sections · Single correct
Choose the correct statement about two circles whose equations are given below: $$x^2 + y^2 - 10x - 10y + 41 = 0$$ $$x^2 + y^2 - 22x - 10y + 137 = 0$$
circles have same centre
circles have no meeting point
circles have only one meeting point
circles have two meeting points
Answer: (c)
Solution
Given the equations of the circles: $$x^2 + y^2 - 10x - 10y + 41 = 0$$ Center $A(5, 5)$, $R_1 = 3$ $$x^2 + y^2 - 22x - 10y + 137 = 0$$ Center $B(11, 5)$, $R_2 = 3$ The distance $AB = 6 = R_1 + R_2$ The circles touch each other externally. Therefore, the circles have only one meeting point.
Question 6
Maths · Determinants · Single correct
Let $\alpha, \beta, \gamma$ be the real roots of the equation, $x^3 + ax^2 + bx + c = 0$, $(a, b, c \in \mathbb{R} and a, b \neq 0)$ If the system of equations (in, u, v, w ) given by $\alpha u + \beta v + \gamma w = 0$, $\beta u + \gamma v + \alpha w = 0$, $\gamma u + \alpha v + \beta w = 0$ has non-trivial solution, then the value of $\frac{a^2}{b}$ is
5
3
1
0
Answer: (b)
Solution
The determinant of the matrix is given by: $$\begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix} = 0$$ This implies: $$-(\alpha + \beta + \gamma) \left(a^2 + b^2 + \gamma^2 - \sum \alpha \beta \right) = 0$$ Simplifying further: $$-(-a) \left(a^2 - 2b - b \right) = 0$$ This leads to: $$a \left(a^2 - 3b \right) = 0$$ Thus: $$a^2 = 3b \implies \frac{a^2}{b} = 3$$
Question 7
Maths · Integrals · Single correct
The integral $$\int \frac{(2x-1) \cos \sqrt{(2x-1)^2+5}}{\sqrt{4x^2-4x+6}} \, dx$$ is equal to (where $c$ is a constant of integration)
$\frac{1}{2} \sin \sqrt{(2x-1)^2+5} + c$
$\frac{1}{2} \cos \sqrt{(2x+1)^2+5} + c$
$\frac{1}{2} \cos \sqrt{(2x-1)^2+5} + c$
$\frac{1}{2} \sin \sqrt{(2x+1)^2+5} + c$
Answer: (a)
Solution
Given the integral $$\int \frac{(2x-1) \cos \sqrt{(2x-1)^2+5}}{\sqrt{(2x-1)^2+5}} \, dx$$ let $$(2x-1)^2 + 5 = t^2.$$ Then $$2(2x-1) \, 2 \, dx = 2t \, dt.$$ Simplifying, we have $$2\sqrt{t^2 - 5} \, dx = t \, dt.$$ So, $$\int \frac{\sqrt{t^2 - 5} \cos t}{2\sqrt{t^2 - 5}} \, dt = \frac{1}{2} \sin t + c.$$ Therefore, $$= \frac{1}{2} \sin \sqrt{(2x-1)^2 + 5} + c.$$
Question 8
Maths · Straight Lines and Pair of Straight Lines · Single correct
The equation of one of the straight lines which passes through the point (1,3) and makes an angles $\tan^{-1}(\sqrt{2})$ with the straight line, $y + 1 = 3\sqrt{2}x$ is
$4\sqrt{2}x + 5y - (15 + 4\sqrt{2}) = 0$
$5\sqrt{2}x + 4y - (15 + 4\sqrt{2}) = 0$
$4\sqrt{2}x + 5y - 4\sqrt{2} = 0$
$4\sqrt{2}x - 5y - (5 + 4\sqrt{2}) = 0$
Answer: (a)
Solution
Given $y = mx + c$ and $3 = m + c$. $$\sqrt{2} = \left| \frac{m - 3\sqrt{2}}{1 + 3\sqrt{2}m} \right|$$ $$= 6m + \sqrt{2} = m - 3\sqrt{2}$$ $$= \sin = -4\sqrt{2} \rightarrow m = \frac{-4\sqrt{2}}{5}$$ $$= 6m - \sqrt{2} = m - 3\sqrt{2}$$ $$= 7m - 2\sqrt{2} \rightarrow m = \frac{2\sqrt{2}}{7}$$ According to options take $m = \frac{-4\sqrt{2}}{5}$. So $y = \frac{-4\sqrt{2}x}{5} + \frac{3 + 4\sqrt{2}}{5}$. $$4\sqrt{2}x + 5y - (15 + 4\sqrt{2}) = 0$$
Question 9
Maths · Limits and Derivatives · Single correct
If $\lim_{x \to 0} \frac{\sin^{-1} x - \tan^{-1} x}{3x^3}$ is equal to $L$, then the value of $(6L + 1)$ is
A vector $\mathbf{a}$ has components $3p$ and $1$ with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to new system, $\mathbf{a}$ has components $p + 1$ and $\sqrt{10}$, then a value of $p$ is equal to:
Maths · Complex Numbers and Quadratic Equations · Single correct
If the equation $a|z|^2 + \overline{\overline{\alpha z} + \alpha \overline{z}} + d = 0$ represents a circle where $a, d$ are real constants then which of the following condition is correct?
$|\alpha|^2 - ad \neq 0$
$|\alpha|^2 - ad > 0$ and $a \in \mathbb{R} - \{0\}$
$|\alpha|^2 - ad \geq 0$ and $a \in \mathbb{R}$
$\alpha = 0, a, d \in \mathbb{R}^+$
Answer: (b)
Solution
Given $az \bar{z} + \alpha \bar{z} + \bar{\alpha} z + d = 0 \rightarrow Circle$. Centre $= -\frac{\alpha}{a}$, $2 = \sqrt{\frac{\alpha \bar{\alpha}}{a^2} - \frac{d}{a}} = \sqrt{\frac{\alpha \bar{\alpha} - ad}{a^2}}$. So $|\alpha|^2 - ad > 0$ and $a \in \mathbb{R} - \{0\}$.
Question 12
Maths · Conic Sections · Single correct
For the four circles M, N, O and P, following four equations are given: Circle $M$: $x^2 + y^2 = 1$ Circle $N$: $x^2 + y^2 - 2x = 0$ Circle $O$: $x^2 + y^2 - 2x - 2y + 1 = 0$ Circle $P$: $x^2 + y^2 - 2y = 0$ If the centre of circle M is joined with centre of the circle N, further centre of circle N is joined with centre of the circle O, centre of circle O is joined with the centre of circle P and lastly, centre of circle P is joined with centre of circle M, then these lines form the sides of a:
Rhombus
Square
Rectangle
Parallelogram
Answer: (b)
Solution
Given the equations of the circles and lines: For $M$: $x^2 + y^2 = 1$ at $(0,0)$. For $N$: $x^2 + y^2 - 2x = 0$ at $(1,0)$. For $O$: $x^2 + y^2 - 2x - 2y + 1 = 0$ at $(1,1)$. For $P$: $x^2 + y^2 - 2y = 0$ at $(0,1)$. The diagram shows a rectangle with vertices $M(0,0)$, $N(1,0)$, $O(1,1)$, and $P(0,1)$. Each side of the rectangle is of length 1.
Question 13
Maths · Sequences and Series · Single correct
If $\alpha, \beta$ are natural numbers such that $100^\alpha - 199 \beta = (100)(100) + (99)(101) + (98)(102) + \ldots + (1)(199)$, then the slope of the line passing through $(\alpha, \beta)$ and origin is:
540
550
530
510
Answer: (b)
Solution
Given $$S = (100)(100) + (99)(101) + (98)(102) \ldots (2)(198) + (1)(199)$$ We have $$S = \sum_{x=0}^{99} (100-x)(100+x) = \sum 100^2 - x^2$$ This simplifies to $$= 100^3 - \frac{99 \times 100 \times 199}{6}$$ Given $$\alpha = 3 \beta = 1650$$ The slope is $$slope = \frac{1650}{3} = 550$$
Question 14
Maths · Relations and Functions · Single correct
The real valued function $f(x) = \frac{\csc^{-1} x}{\sqrt{x - \lfloor x \rfloor}}$, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$, is defined for all $x$ belonging to:
all reals except integers
all non-integers except the interval $[-1,1]$
all integers except $0,-1,1$
all reals except the Interval $[-1,1]$
Answer: (b)
Solution
Given $$f(x) = \frac{\csc^{-1} x}{\sqrt{\{x\}}}$$ Domain is $$(-\infty, -1] \cup [1, \infty)$$ $\($ $\{$x$\}$ $\neq$ 0 $\)$ so $\($ x $\neq$ $\)$ integers.
Question 15
Maths · Sequences and Series · Single correct
$\frac{1}{3^2-1}$ + $\frac{1}{5^2-1}$ + $\frac{1}{7^2-1}$ + $\cdots$ + $\frac{1}{(201)^2-1}$ is equal to
$\frac{101}{404}$
$\frac{25}{101}$
$\frac{101}{408}$
$\frac{99}{400}$
Answer: (b)
Solution
Given $$T_n = \frac{1}{(2n+1)^2 - 1} = \frac{1}{(2n+2)2n} = \frac{1}{4(n)(n+1)}$$ This simplifies to $$= \frac{(n+1) - n}{4n(n+1)} = \frac{1}{4} \left( \frac{1}{n} - \frac{1}{n+1} \right)$$ The sum is $$S = \frac{1}{4} \left( 1 - \frac{1}{101} \right) = \frac{1}{4} \left( \frac{100}{101} \right) = \frac{25}{101}$$
Question 16
Maths · Relations and Functions · Single correct
If the functions are defined as $f(x) = \sqrt{x}$ and $g(x) = \sqrt{1-x}$, then what is the common domain of the following functions: $f + g, f - g, f/g, g/f, g - f$ where $(f \pm g)(x) = f(x) \pm g(x), (f/g)(x) = \frac{f(x)}{g(x)}$
Maths · Continuity and Differentiability · Single correct
If $f(x) = \begin{cases} \frac{1}{|x|} & ; |x| \geq 1 \\ ax^2 + b & ; |x| < 1 \end{cases}$ is differentiable at every point of the domain, then the values of $a$ and $b$ are respectively:
$\frac{1}{2}, \frac{1}{2}$
$\frac{1}{2}, -\frac{3}{2}$
$\frac{5}{2}, -\frac{3}{2}$
$-\frac{1}{2}, \frac{3}{2}$
Answer: (d)
Solution
Given $$f(x) = \begin{cases} \frac{1}{|x|}, & |x| \geq 1 \\ ax^2 + b, & |x| < 1 \end{cases}$$ At $x = 1$, the function must be continuous. So, $1 = a + b$ ...(i) Differentiability at $x = 1$ $$\left(-\frac{1}{x^2}\right)_{x=1} = (2ax)_{x=1}$$ $$\Rightarrow -1 = 2a \Rightarrow a = -\frac{1}{2}$$ From (1) $\Rightarrow b = 1 + \frac{1}{2} = \frac{3}{2}$
Question 18
Maths · Matrices · Single correct
Let $A + 2B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix}$ and $2A - B = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix}$. If $\mathrm{Tr}(A)$ denotes the sum of all diagonal elements of the matrix $A$, then $\mathrm{Tr}(A) - \mathrm{Tr}(B)$ has value equal to
Maths · Permutations and Combinations · Single correct
The sum of all the 4-digit distinct numbers that can be formed with the digits 1, 2, 2 and 3 is:
26664
122664
122234
22264
Answer: (a)
Solution
Digits are 1, 2, 2, 3. Total distinct numbers $\frac{4!}{2!} = 12$. Total numbers when 1 at unit place is 3. 2 at unit place is 6, 3 at unit place is 3. So, sum $= (3 + 12 + 9) \left(10^3 + 10^2 + 10 + 1\right)$ $$= (1111) \times 24$$ $$= 26664$$
Question 20
Maths · Sequences and Series · Single correct
The value of $3+\cfrac{1}{4+\cfrac{1}{3+\cfrac{1}{4+\cfrac{1}{3+\cdots\infty}}}}$ is equal to
$1.5 + \sqrt{3}$
$2 + \sqrt{3}$
$3 + 2\sqrt{3}$
$4 + \sqrt{3}$
Answer: (a)
Solution
Let $x = 3 + \cfrac{1}{4 + \cfrac{1}{3 + \cfrac{1}{4 + \cfrac{1}{3 + \cdots}}}}$. So, $x = 3 + \cfrac{1}{4 + \cfrac{1}{x}} = 3 + \cfrac{1}{\cfrac{4x + 1}{x}}$. Therefore, $(x - 3) = \cfrac{x}{4x + 1}$. This implies $(4x + 1)(x - 3) = x$. Expanding gives $4x^2 - 12x + x - 3 = x$. Simplifying, $4x^2 - 12x - 3 = 0$. Solving for $x$, we have $$x = \cfrac{12 \pm \sqrt{(12)^2 + 12 \times 4}}{2 \times 4} = \cfrac{12 \pm \sqrt{12 \times 16}}{8}$$ $$= \cfrac{12 \pm 4 \times 2 \sqrt{3}}{8} = \cfrac{3 \pm 2 \sqrt{3}}{2}$$ $$x = \cfrac{3}{2} \pm \sqrt{3} = 1.5 \pm \sqrt{3}$$ But only the positive value is accepted. So, $x = 1.5 + \sqrt{3}$.
Question 21
Maths · Permutations and Combinations · Numerical
The number of times the digit 3 will be written when listing the integers from 1 to 1000 is
Let the plane $ax + by + cz + d = 0$ bisect the line joining the points $(4, -3, 1)$ and $(2, 3, -5)$ at the right angles. If $a, b, c, d$ are integers, then the minimum value of $(a^2 + b^2 + c^2 + d^2)$ is
Let $f(x)$ and $g(x)$ be two functions satisfying $f(x^2) + g(4-x) = 4x^3$ and $g(4-x) + g(x) = 0$, then the value of $$\int_{-4}^{4} f(x)^2 \, dx$$ is
Answer: 512
Solution
Given that the function is even, we have: $$I = 2 \int_0^4 f(x^2) \, dx \{ Even function \}$$ This simplifies to: $$= 2 \int_0^4 (4x^3 - g(4-x)) \, dx$$ Evaluating the integral, we get: $$= 2 \left( \frac{4x^4}{4} \bigg|_0^4 - \int_0^4 g(4-x) \, dx \right)$$ Simplifying further: $$= 2(256 - 0) = 512$$
Question 24
Maths · Permutations and Combinations · Numerical
The missing value in the following figure is Use the logic which gives answer in single digit
The equation of the planes parallel to the plane $x - 2y + 2z - 3 = 0$ which are at unit distance from the point $(1,2,3)$ is $ax + by + cz + d = 0$. If $(b - d) = K(c - a)$, then the positive value of $K$ is
Answer: 4
Solution
Let plane is $x - 2y + 2z + \lambda = 0$. Distance from $(1, 2, 3) = 1$. $$\frac{|\lambda + 3|}{5} = 1 \Rightarrow \lambda = 0, -6$$ $$\Rightarrow a = 1, b = -2, c = 2, d = -6 or 0$$ $$b - d = 4 or -2, c - a = 1$$ $$\Rightarrow k = 4 or -2$$
Question 27
Maths · Statistics · Numerical
The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If the mean age of the teachers in this school now is 39 years, then the age (in years) of the newly appointed teacher is ___.
Answer: 35
Solution
Given $\($ $\frac{\sum x_i}{25}$ = 40 $\)$ and $\($ $\frac{\sum x_i - 60 + N}{25}$ = 39 $\)$. Let age of newly appointed teacher is $\($ N $\)$. \[ 1000 - 60 + N = 975 \] \[ N = 35 years \]
Question 28
Maths · Integrals · Numerical
If $f(x) = \int \frac{5x^8 + 7x^6}{(x^2 + 1 + 2x^7)^2} \, dx$, $(x \geq 0)$, $f(0) = 0$ and $f(1) = \frac{1}{K}$, then the value of $K$ is
A square ABCD has all its vertices on the curve $x^2 y^2 = 1$. The midpoints of its sides also lie on the same curve. Then, the square of area of ABCD is
The number of solutions of the equation $|\cot x| = \cot x + \frac{1}{\sin x}$ in the interval $[0, 2\pi]$ is
Answer: 1
Solution
If $\cot x > 0$ then $\frac{1}{\sin x} = 0$ (Not possible). If $\cot x < 0$ then $2 \cot x + \frac{1}{\sin x} = 0$. $$\Rightarrow 2 \cos x = -1$$ $$\Rightarrow x = \frac{2\pi}{3} or \frac{4\pi}{3} (reject)$$
Physics
Question 31
Physics · Electric Charges and Fields · Single correct
An oil drop of radius 2 mm with a density 3 g $cm^{-3}$ is held stationary under a constant electric field 3.55 $\times$ $10^{5}$ $\mathrm{V}$ $\mathrm{m}^{-1}$ in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (consider g = 9.81 $\mathrm{m/s}^{2}$)
48.8 $\times$ 10^{11}
1.73 $\times$ 10^{10}
17.3 $\times$ 10^{10}
1.73 $\times$ 10^{12}
Answer: (b)
Solution
Given $qE = Mg$. We have $n e E = \rho \left( \frac{4}{3} \pi r^3 \right) \times g$. Substituting the values: $$n \times 1.6 \times 10^{-19} \times 3.55 \times 10^5$$ $$= 3 \times 10^3 \times \frac{4}{3} \times \pi \times \left( 2 \times 10^{-3} \right)^3 \times 9.81$$ Simplifying: $$n = 173 \times 10^{(3-9-5+19)}$$ Finally, we get: $$n = 1.73 \times 10^{10}$$
Question 32
Physics · Electromagnetic Waves · Single correct
Match List-I with List-II. List-I $\text{(a)}$ 10 km height over earth's surface $\text{(b)}$ 70 km height over earth's surface $\text{(c)}$ 180 km height over earth's surface $\text{(d)}$ 270 km height over earth's surface List-II (i) Thermosphere (ii) Mesosphere (iii) Stratosphere (iv) Troposphere
Order of atmosphere stratification from bottom Troposphere, Stratosphere, Mesosphere, Thermosphere $\text{(a)}$ $\rightarrow$ (iv) $\text{(b)}$ $\rightarrow$ (iii) $\text{(c)}$ $\rightarrow$ (ii) $\text{(d)}$ $\rightarrow$ (i)
Question 33
Physics · Atoms · Single correct
Imagine that the electron in a hydrogen atom is replaced by a muon ( $\mu$ ). The mass of muon particle is 207 times that of an electron and charge is equal to the charge of an electron. The ionization potential of this hydrogen atom will be :-
13.6$\mathrm{eV}$
2815.2$\mathrm{eV}$
331.2$\mathrm{eV}$
27.2$\mathrm{eV}$
Answer: (b)
Solution
Given $E \propto \frac{1}{r}$ and $r \propto \frac{1}{m}$. Therefore, $E \propto m$. The ionization potential is given by $$Ionization potential = 13.6 \times \left(\frac{Mass_{\mu}}{Mass_{e}}\right) eV$$ $$= 13.6 \times 207 eV = 2815.2 eV$$
Question 34
Physics · Electric Charges and Fields · Single correct
A plane electromagnetic wave of frequency 100 MHz is travelling in vacuum along the $x-$ direction. At a particular point in space and time, $\vec{B} = 2.0 \times 10^{-8} \hat{k} \, \mathrm{T}$. (where $\hat{k}$ is unit vector along $z$-direction) What is $\vec{E}$ at this point?
0.6 $\hat{j}$ V/m
6.0 $\hat{k}$ V/m
6.0 $\hat{j}$ V/m
0.6 $\hat{k}$ V/m
Answer: (c)
Solution
Given $E = BC = 6$. The direction of the wave is parallel to $\left( \vec{E} \times \vec{B} \right)$. We have $\hat{i} = \hat{j} \times \hat{k}$. Therefore, $\vec{E} = 6 \hat{j} \, \mathrm{V/m}$.
Question 35
Physics · System of Particles and Rotational Motion · Single correct
A thin circular ring of mass M and radius r is rotating about its axis with an angular speed $\omega$. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become:
$\omega \frac{M}{M+m}$
$\omega \frac{M+2m}{M}$
$\omega \frac{M}{M+2m}$
$\omega \frac{M-2m}{M+2m}$
Answer: (c)
Solution
Using conservation of angular momentum $$\left( M r^2 \right) \omega = \left( M r^2 + 2 m r^2 \right) \omega'$$ $$\omega' = \frac{M \omega}{M + 2m}$$
Question 36
Physics · Moving Charges and Magnetism · Single correct
Four identical long solenoids A, B, C and D are connected to each other as shown in the figure. If the magnetic field at the center of A is $3 \, \mathrm{T}$ the field at the center of C would be: (Assume that the magnetic field is confined with in the volume of respective solenoid).
12 T
6 T
9 T
1 T
Answer: (d)
Solution
The magnetic flux $\phi$ is proportional to the current $i$. Therefore, the magnetic field $\mathbf{B}$ is also proportional to $i$. Thus, the field at the center of C is $\frac{3}{3} = 1 \, \mathrm{T}$.
Question 37
Physics · Mathematics in Physics · Single correct
The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$. The measured value of the length of pendulum is 10 cm known to a 1 mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1 s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to the nearest integer is:-
2$\%$
3$\%$
5$\%$
4$\%$
Answer: (b)
Solution
Given $$g = \frac{4\pi^2 \ell}{T^2}$$ The relative error is given by $$\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta T}{T} = \frac{0.1}{10} + 2 \left( \frac{1}{200} \times 0.5 \right)$$ Simplifying, we have $$\frac{\Delta g}{g} = \frac{1}{100} + \frac{1}{50}$$ Thus, $$\frac{\Delta g}{g} \times 100 = 3\%$$
Question 38
Physics · Work, Energy and Power · Single correct
A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time ' $t$ ' is proportional to :-
$t^{2/3}$
$t^{3/2}$
$t$
$t^{1/2}$
Answer: (b)
Solution
Given $P = C$. Then $FV = C$. Therefore, $M \frac{dV}{dt} V = C$. Since $\frac{v^2}{2} \propto t$, we have $V \propto t^{1/2}$. Thus, $\frac{dx}{dt} \propto t^{1/2}$, leading to $x \propto t^{3/2}$.
Question 39
Physics · Kinetic Theory · Single correct
What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature $T$ ($k_B$ is Boltzmann constant)
$\frac{1}{2} k_B T$
$\frac{2}{3} k_B T$
$\frac{3}{2} k_B T$
$k_B T$
Answer: (a)
Solution
Energy associated with each degree of freedom per molecule is $\frac{1}{2} k_B T$.
Question 40
Physics · Nuclei · Single correct
A radioactive sample disintegrates via two independent decay processes having half lives $T_{1/2}^{(1)}$ and $T_{1/2}^{(2)}$ respectively. The effective half-life $T_{1/2}$ of the nuclei is:
The P-V diagram of a diatomic ideal gas system going under cyclic process as shown in figure. The work done during an adiabatic process CD is (use $\gamma = 1.4$):
$-500 \, \mathrm{J}$
$-400 \, \mathrm{J}$
$400 \, \mathrm{J}$
$200 \, \mathrm{J}$
Answer: (a)
Solution
Question 42
Physics · Wave Optics · Single correct
In Young's double slit arrangement, slits are separated by a gap of 0.5 mm, and the screen is placed at a distance of 0.5 m from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of 5890 $\AA$ is :-
$1178 \times 10^{-9} \, \mathrm{m}$
$1178 \times 10^{-6} \, \mathrm{m}$
$1178 \times 10^{-12} \, \mathrm{m}$
$5890 \times 10^{-7} \, \mathrm{m}$
Answer: (b)
Solution
Given $\($ $\beta$ = $\frac{\lambda D}{d}$ = $\frac{5890 \times 10^{-10} \times 0.5}{0.5 \times 10^{-3}}$ $\)$ $\[$ = 589 $\times$ 10^{-6} $\,$ m $\]$ Distance between first and third bright fringe is $\($ 2$\beta$ = 2 $\times$ 589 $\times$ 10^{-6} $\,$ m $\)$ $\[$ = 1178 $\times$ 10^{-6} $\,$ m $\]$ Ans.(b)
Question 43
Physics · Dual Nature of Radiation and Matter · Single correct
A particle is travelling 4 times as fast as an electron. Assuming the ratio of de-Broglie wavelength of a particle to that of electron is 2 : 1, the mass of the particle is :-
Physics · Motion in a Straight Line · Single correct
The position, velocity and acceleration of a particle moving with a constant acceleration can be represented by:
Answer: (b)
Solution
Option (b) represents the correct graph for a particle moving with constant acceleration, as for constant acceleration velocity time graph is a straight line with positive slope and $x - t$ graph should be an opening upward parabola.
Question 45
Physics · Current Electricity · Single correct
In the experiment of Ohm's law, a potential difference of 5.0 $\mathrm{V}$ is applied across the end of a conductor of length 10.0 $\mathrm{cm}$ and diameter of 5.00 $\mathrm{mm}$. The measured current in the conductor is 2.00 $\mathrm{A}$. The maximum permissible percentage error in the resistivity of the conductor is :-
3.9
8.4
7.5
3
Answer: (a)
Solution
Given $R = \frac{\rho \ell}{A} = \frac{V}{I}$. The resistivity $\rho$ is given by $\rho = \frac{AV}{I \ell} = \frac{\pi d^2 V}{4 I \ell}$ where $A = \frac{\pi d^2}{4}$. Therefore, $$\frac{\Delta \rho}{\rho} = \frac{2 \Delta d}{d} + \frac{\Delta V}{V} + \frac{\Delta I}{I} + \frac{\Delta \ell}{\ell}$$ Substituting the given values, $$\frac{\Delta \rho}{\rho} = 2 \left( \frac{0.01}{5.00} \right) + \frac{0.1}{5.0} + \frac{0.01}{2.00} + \frac{0.1}{10.0}$$ $$\frac{\Delta \rho}{\rho} = 0.004 + 0.02 + 0.005 + 0.01$$ $$\frac{\Delta \rho}{\rho} = 0.039$$ The percentage error is given by $$\% error = \frac{\Delta \rho}{\rho} \times 100 = 0.039 \times 100 = 3.90\%$$
Question 46
Physics · Alternating Current · Single correct
In a series LCR resonance circuit, if we change the resistance only, from a lower to higher value:
The bandwidth of resonance circuit will increase.
The resonance frequency will increase.
The quality factor will increase.
The quality factor and the resonance frequency will remain constant.
Answer: (a)
Solution
Bandwidth $= \frac{R}{L}$ Bandwidth $\propto R$ So bandwidth will increase
Question 47
Physics · Alternating Current · Single correct
An AC source rated 220 V, 50 Hz is connected to a resistor. The time taken by the current to change from its maximum to the rms value is:
2.5 ms
25 ms
2.5 s
0.25 ms
Answer: (a)
Solution
Given $i = i_0 \cos(\omega t)$. $i = i_0$ at $t = 0$. $i = \frac{i_0}{\sqrt{2}}$ at $\omega t = \frac{\pi}{4}$. $t = \frac{\pi}{4\omega} = \frac{\pi}{4(2\pi f)} = \frac{1}{8f}$. $t = \frac{1}{400} = 2.5 \, \mathrm{ms}$.
Question 48
Physics · Ray Optics and Optical Instruments · Single correct
Your friend is having eye sight problem. She is not able to see clearly a distant uniform window mesh and it appears to her as nonuniform and distorted. The doctor diagnosed the problem as:
Astigmatism
Myopia with Astigmatism
Presbyopia with Astigmatism
Myopia and hypermetropia
Answer: (b)
Solution
If distant objects are blurry then problem is Myopia. If objects are distorted then problem is Astigmatism.
Question 49
Physics · Moving Charges and Magnetism · Single correct
A loop of flexible wire of irregular shape carrying current is placed in an external magnetic field. Identify the effect of the field on the wire.
Loop assumes circular shape with its plane normal to the field.
Loop assumes circular shape with its plane parallel to the field.
Wire gets stretched to become straight.
Shape of the loop remains unchanged.
Answer: (a)
Solution
Every part $(d \ell)$ of the wire is pulled by force $i(d\ell)B$ acting perpendicular to current and magnetic field giving it a shape of circle.
Question 50
Physics · Gravitation · Single correct
The time period of a satellite in a circular orbit of radius $R$ is $T$. The period of another satellite in a circular orbit of radius $9R$ is:
A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac{1}{a}$ s. The value of 'a' to the nearest integer is ___
Answer: 6
Solution
Given \[ T = 2\,\text{s}. \] The time is \[ t = \frac{T}{12}. \] Therefore, \[ t = \frac{2}{12} = \frac{1}{6}\,\text{s}. \] Thus, the correct answer is \[ 6.00. \]
Question 52
Physics · Alternating Current · Numerical
The circuit shown in the figure consists of a charged capacitor of capacity $3 \, \mu \mathrm{F}$ and a charge of $30 \, \mu \mathrm{C}$. At time $t = 0$, when the key is closed, the value of current flowing through the $5 \, \mathrm{M}\Omega$ resistor is '$x' \, \mu$ - A. The value of 'x to the nearest integer is ___
The voltage across the $10\,\Omega$ resistor in the given circuit is $x$ volt.
Answer: 70
Solution
The equivalent resistance $R_{eq1}$ is calculated as follows: $$R_{eq1} = \frac{50 \times 20}{70} = \frac{100}{7}$$ The circuit is shown with resistances $10$ and $\frac{100}{7}$ in series with a voltage source of $170$. The equivalent resistance $R_{eq}$ is: $$R_{eq} = \frac{170}{7}$$ The voltage $v_1$ is calculated as: $$v_1 = \left[ \frac{170}{170/7} \right] \times 10 = 70\, V$$ Therefore, the answer is $70.00$.
Question 54
Physics · Mechanical Properties of Solids · Numerical
Two separate wires A and B are stretched by 2 $\mathrm{\ mm}$ and 4 $\mathrm{\ mm}$ respectively, when they are subjected to a force of 2 $\mathrm{\ N}$. Assume that both the wires are made up of same material and the radius of wire B is 4 times that of the radius of wire A. The length of the wires A and B are in the ratio of a : b. Then a/b can be expressed as 1/x where x is ___
A person is swimming with a speed of $10 \, \mathrm{m/s}$ at an angle of $120^\circ$ with the flow and reaches to a point directly opposite on the other side of the river. The speed of the flow is $x \, \mathrm{m/s}$. The value of 'x' to the nearest integer is ___
Answer: 5
Solution
Given the velocity vector with a magnitude of $10 \, \mathrm{m/s}$ at an angle of $30^\circ$, we need to find the horizontal component $x$. Using the sine function, we have: $$10 \sin 30^\circ = x$$ Calculating the value, we find: $$x = 5 \, \mathrm{m/s}$$
Question 56
Physics · Electrostatic Potential and Capacitance · Numerical
A parallel plate capacitor has plate area 100 m$^2$ and plate separation of 10 m. The space between the plates is filled up to a thickness 5 m with a material of dielectric constant of 10. The resultant capacitance of the system is 'x' pF. The value of $\varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{F} \cdot \mathrm{m}^{-1}$ The value of 'x' to the nearest integer is ___
Answer: 161
Solution
Given $A = 100 \, \mathrm{m}^2$. Using $C = \frac{k \varepsilon_0 A}{d}$. $C_1 = \frac{10 \varepsilon_0 (100)}{5}$ $= 200 \varepsilon_0$ $C_2 = \frac{\varepsilon_0 (100)}{5} = 20 \varepsilon_0$ $C_1$ and $C_2$ are in series so $C_{eqv.} = \frac{C_1 C_2}{C_1 + C_2}$. $= \frac{4000 \varepsilon_0}{220}$ $= 160.9 \times 10^{-12} \simeq 161 \, \mathrm{pF}$
Question 57
Physics · Work, Energy and Power · Numerical
A ball of mass 10 kg moving with a velocity 10$\sqrt{3}$ $\mathrm{m/s}$ along the x-axis, hits another ball of mass 20 $\mathrm{kg}$ which is at rest. After the collision, first ball comes to rest while the second ball disintegrates into two equal pieces. One piece starts moving along y-axis with a speed of 10 $\mathrm{m/s}$. The second piece starts moving at an angle of 30^$\circ$ with respect to the x-axis. The velocity of the ball moving at 30^$\circ$ with x-axis is $x \, \mathrm{m/s}$. The configuration of pieces after collision is shown in the figure below. The value of $x$ to the nearest integer is ___
Answer: 20
Solution
Let velocity of 2nd fragment is $\vec{v}$ then by conservation of linear momentum $$10(10\sqrt{3}) \hat{i} = (10)(10 \hat{j}) + 10 \vec{v}$$ $$\Rightarrow \vec{v} = 10\sqrt{3} \hat{i} - 10 \hat{j}$$ $$|\vec{v}| = \sqrt{300 + 100} = \sqrt{400} = 20 \, \mathrm{m/s}$$
Question 58
Physics · Work, Energy and Power · Numerical
As shown in the figure, a particle of mass 10 kg is placed at a point A. When the particle is slightly displaced to its right, it starts moving and reaches the point B. The speed of the particle at B is $x \, \mathrm{m/s}$. (Take $g = 10 \, \mathrm{m/s^2}$) The value of 'x' to the nearest integer is ___
Answer: 10
Solution
Using work energy theorem, $$W_g = \Delta K.E$$ $$(10)(g)(5) = \frac{1}{2}(10)v^2 - 0$$ $v = 10 \, \mathrm{m/s}$
An npn transistor operates as a common emitter amplifier with a power gain of $10^6$. The input circuit resistance is $100\,\Omega$ and the output load resistance is $10\,\mathrm{K}\Omega$. The common emitter current gain '$\beta$' will be ___ (Round off to the Nearest Integer)
Answer: 100
Solution
Given $10^6 = \beta^2 \times \frac{R_0}{R_i}$. Substituting $R_0 = 10^4$ and $R_i = 10^2$, we have: $$10^6 = \beta^2 \times \frac{10^4}{10^2}$$ Simplifying gives: $$\beta^2 = 10^4 \Rightarrow \beta = 100$$
Question 60
Physics · Laws of Motion · Numerical
A bullet of mass $0.1 \, \mathrm{kg}$ is fired on a wooden block to pierce through it, but it stops after moving a distance of $50 \, \mathrm{cm}$ into it. If the velocity of bullet before hitting the wood is $10 \, \mathrm{m/s}$ and it slows down with uniform deceleration, then the magnitude of effective retarding force on the bullet is 'x' N. The value of 'x' to the nearest integer is ___
Answer: 10
Solution
Given $v^2 = u^2 + 2as$. $$0 = (10)^2 + 2(-a) \left( \frac{1}{2} \right)$$ Solving for $a$, we get $a = 100 \, \mathrm{m/s^2}$. Then, $F = ma = (0.1)(100) = 10 \, \mathrm{N}$.
Chemistry
Question 61
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Considering the above reaction, $X$ and $Y$ respectively are
Answer: (b)
Solution
Question 62
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The ionic radius of $\mathrm{Na^+}$ ions is $1.02\,\mathrm{\AA}$. The ionic radii (in $\mathrm{\AA}$) of $\mathrm{Mg^{2+}}$ and $\mathrm{Al^{3+}}$, respectively, are
1.05 and 0.99
0.72 and 0.54
0.85 and 0.99
0.68 and 0.72
Answer: (b)
Solution
The ionic radii order is $\mathrm{Na^+} > \mathrm{Mg^{2+}} > \mathrm{Al^{3+}}$
Question 63
Chemistry · Amines · Single correct
Reaction of Grignard reagent, $C_2H_5MgBr$ with $C_8H_8O$ followed by hydrolysis gives compound "A" which reacts instantly with Lucas reagent to give compound B, $C_{10}H_{13}Cl$. The Compound B is :
Answer: (c)
Solution
(3)
Question 64
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Reagent, 1-naphthylamine and sulphanilic acid in acetic acid is used for the detection of
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Consider the above chemical reaction and identify product "A"
Answer: (c)
Solution
The reaction starts with a nitrile group $\mathrm{C \equiv N}$ attached to a cyclohexane ring. Upon partial hydrolysis with $\mathrm{H_2O^+}$, the major product 'A' is formed, which is an amide $\mathrm{C - NH_2}$ attached to the cyclohexane ring. Further complete hydrolysis with $\mathrm{H_2O^+}$ and heat $\Delta$ leads to the formation of a carboxylic acid $\mathrm{C - OH}$ attached to the cyclohexane ring.
Question 68
Chemistry · Biomolecules · Single correct
$$ \begin{array}{ll} \text{List-I} & \text{List-II} \\ (a) \text{ Chlorophyll} & (i) \text{ Ruthenium} \\ (b) \text{ Vitamin-B}_{12} & (ii) \text{ Platinum} \\ (c) \text{ Anticancer drug} & (iii) \text{ Cobalt} \\ (d) \text{ Grubbs catalyst} & (iv) \text{ Magnesium} \end{array} $$ Choose the most appropriate answer from the options given below:
a - iii, b - ii, c - iv, d - i
a - iv, b - iii, c - ii, d - i
a - iv, b - iii, c - i, d - ii
a - iv, b - ii, c - iii, d - i
Answer: (b)
Solution
Chlorophyll is a coordination compound of magnesium. Vitamin B-12, cyanocobalamine is a coordination compound of cobalt. Cisplatin is used as an anti-cancer drug and is a coordination compound of platinum. Grubbs catalyst is a compound of Ruthenium.
Question 69
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
(a) Alcoholic potassium hydroxide $\rightarrow$ used for $\beta$-elimination (b) $\mathrm{Pd/BaSO_4} \rightarrow$ Lindlar's catalyst (c) BHC (Benzene hexachloride) $\rightarrow$ Obtained by addition reactions (d) Polyacetylene $\rightarrow$ Electrodes in batteries
Question 70
Chemistry · Environmental Chemistry · Single correct
The statements that are TRUE: (A) Methane leads to both global warming and photochemical smog (B) Methane is generated from paddy fields (C) Methane is a stronger global warming gas than $\mathrm{CO}_2$ (D) Methane is a part of reducing smog Choose the most appropriate answer from the options given below:
(A), (B), (C) only
(A) and (B) only
(B), (C), (D) only
(A), (B), (D) only
Answer: (a)
Solution
Methane leads to both global warming and photochemical smog. Methane is generated in large amounts from paddy fields. $\mathrm{CO_2}$ can be absorbed by photosynthesis, or by formation of acid rain etc., while no such activities are there for methane. Hence methane is stronger global warming gas than $\mathrm{CH_4}$. Methane is not a part of reducing smog.
Question 71
Chemistry · Chemistry in Everyday Life · Single correct
Match List-I with List-II \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{\textbf{List-I}} & \multicolumn{2}{c|}{\textbf{List-II}} \\ \hline (a) & $\mathrm{Ca(OCl)_2}$ & (i) & Antacid \\ \hline (b) & $\mathrm{CaSO_4 \cdot \dfrac{1}{2}H_2O}$ & (ii) & Cement \\ \hline (c) & $\mathrm{CaO}$ & (iii) & Bleach \\ \hline (d) & $\mathrm{CaCO_3}$ & (iv) & Plaster of paris \\ \hline \end{tabular} Choose the most appropriate answer from the options given below:
a − i, b − iv, c − iii, d − ii
a-iii, b-ii, c − iv, d − i
a-iii, b-iv, c − ii, d − i
a-iii, b − ii, c − i, d − iv
Answer: (c)
Solution
$\mathrm{Ca(OCl)_2}$ is Bleach. $\mathrm{CaSO_4 \cdot \dfrac{1}{2}H_2O}$ is plaster of paris. $\mathrm{CaCO_3}$ is used as an antacid. $\mathrm{CaO}$ is major component of cement.
Question 72
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Compound with molecular formula $\mathrm{C_3H_6O}$ can show:
Positional isomerism
Both positional isomerism and metamerism
Metamerism
Functional group isomerism
Answer: (d)
Solution
Given $\mathrm{C_3H_6O} \Rightarrow \mathrm{CH_3 - CH_2 - CH = O}$ and $\mathrm{CH_3 - C - CH_3}$ with $\mathrm{O}$. They are functional group isomerism.
Question 73
Chemistry · Co-ordination Compounds · Single correct
The correct structures of trans-$[\mathrm{NiBr_2(PPh_3)_2}]$ and meridional-$[\mathrm{Co(NH_3)_3(NO_2)_3}]$, respectively, are
Answer: (d)
Solution
The compound trans-$[\mathrm{NiBr_2(PPh_3)_2}]$ is shown with the structure in which the bromine atoms and triphenylphosphine ligands are opposite to each other. The compound meridional-$[\mathrm{Co(NH_3)_3(NO_2)_3}]$ is shown with the structure in which the nitrito and ammine ligands are arranged in a meridional fashion around the cobalt centre.
Question 74
Chemistry · Structure of Atom · Single correct
A certain orbital has no angular nodes and two radial nodes. The orbital is:
Considering the above chemical reaction, identify the product ``X'':
Answer: (c)
Solution
The reaction involves the oxidation of the methyl group attached to the benzene ring using alkaline $\mathrm{KMnO_4}$ in the presence of $\mathrm{H^+}$. The methyl group is converted to a carboxylic acid group, resulting in the formation of $\mathrm{CO_2H}$ at the position where $\mathrm{CH_3}$ was originally attached.
Question 76
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Match List-I with List-II \begin{tabular}{|c|p{5.8cm}|c|l|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{(process)} & \multicolumn{2}{c|}{(catalyst)} \\ \hline (a) & Dacron's process & (i) & ZSM-5 \\ \hline (b) & Contact process & (ii) & CuCl$_2$ \\ \hline (c) & Cracking of hydrocarbons & (iii) & Particles 'Ni' \\ \hline (d) & Hydrogenation of vegetable oils & (iv) & V$_2$O$_5$ \\ \hline \end{tabular} Choose the most appropriate answer from the options given below -
a - ii, b - iv, c - i, d - iii
a - i, b - iii, c - ii, d - iv
a - iii, b - i, c - iv, d - ii
a - iv, b - ii, c - i, d - iii
Answer: (d)
Question 77
Chemistry · The s-Block Elements · Single correct
Given below are two statements: One is labelled as Assertion A and the other labelled as reason R Assertion A : During the boiling of water having temporary hardness, $\mathrm{Mg(HCO_3)_2}$ is converted to $\mathrm{MgCO_3}$ Reason R : The solubility product of $\mathrm{Mg(OH)_2}$ is greater than that of $\mathrm{MgCO_3}$. In the light of the above statements, choose the most appropriate answer from the options given below:
Both A and R are true but R is not the correct explanation of A
A is true but R is false
Both A and R are true and R is the correct explanation of A
A is false and R is also false
Answer: (d)
Solution
For temporary hardness, $\mathrm{Mg(HCO_3)_2} \xrightarrow{heating} \mathrm{Mg(OH)_2} \downarrow + 2\mathrm{CO_2} \uparrow$. Assertion is false. $\mathrm{MgCO_3}$ has high solubility product than $\mathrm{Mg(OH)_2}$. According to data of NCERT table 7.9 (Equilibrium chapter), the solubility product of magnesium carbonate is $3.5 \times 10^{-8}$ and solubility product of $\mathrm{Mg(OH)_2}$ is $1.8 \times 10^{-11}$. Hence Reason is incorrect.
Question 78
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
The number of ionisable hydrogens present in the product obtained from a reaction of phosphorus trichloride and phosphonic acid is:
3
0
2
1
Answer: (c)
Question 79
Chemistry · The Solid State · Single correct
In a binary compound, atoms of element A form a hcp structure and those of element M occupy 2/3 of the tetrahedral voids of the hcp structure. The formula of the binary compound is:
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
The chemical that is added to reduce the melting point of the reaction mixture during the extraction of aluminium is:
Cryolite
Bauxite
Calamine
Kaolite Official
Answer: (a)
Solution
To reduce the melting point of reaction mixture, cryolite is added.
Question 81
Chemistry · Chemical Bonding and Molecular Structure · Numerical
AX is a covalent diatomic molecule where A and X are second row elements of periodic table. Based on Molecular orbital theory, the bond order of AX is 2.5. The total number of electrons in AX is____( Round off to the Nearest Integer).
Answer: 15
Solution
AX is a covalent diatomic molecule. The molecule is NO. Total number of electrons is 15.
Question 82
Chemistry · Equilibrium · Numerical
In order to prepare a buffer solution of pH 5.74 sodium acetate is added to acetic acid. If the concentration of acetic acid in the buffer is 1.0 M, the concentration of sodium acetate in the buffer is ___ M.(Round off to the Nearest Integer). [Given: $pK_a$ (acetic acid) = 4.74]
Answer: 10
Solution
Given the equation for pH: $$\mathrm{pH} = \mathrm{pK_a} + \log \frac{[\mathrm{CB}]}{[\mathrm{WA}]}$$ Substitute the given values: $$5.74 = 4.74 + \log \frac{[\mathrm{CB}]}{1}$$ Solving for $[\mathrm{CB}]$: $$\Rightarrow [\mathrm{CB}] = 10\, \mathrm{M}$$
Question 83
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
2$\mathrm{NO}$(g) + $\mathrm{Cl}$_2(g) $\rightleftharpoons$ 2$\mathrm{NOCl}$(s) This reaction was studied at $-10^\circ \mathrm{C}$ and the following data was obtained \begin{tabular}{|l|l|l|l|} \hline run & $NO_0$ & $[\mathrm{Cl}_2]_0$ & $r_0$ \\ \hline 1 & 0.10 & 0.10 & 0.18 \\ \hline 2 & 0.10 & 0.20 & 0.35 \\ \hline 3 & 0.20 & 0.20 & 1.40 \\ \hline \end{tabular} $[\mathrm{NO}]_0$ and $[\mathrm{Cl}_2]_0$ are the initial concentrations and $r_0$ is the initial reaction rate. The overall order of the reaction is____. (Round off to the Nearest Integer).
Answer: 3
Solution
Given the rate equation $r = k[\mathrm{NO}]^m[\mathrm{Cl}_2]^n$. Substituting the given concentrations, we have: $$r = k(0.1)^m(0.1)^n$$ $$= k(0.1)^m(0.2)^n \cdots (1)$$ $$= k(0.2)^m(0.2)^n \cdots (3)$$ From the equations, we find $n = 1$ and $m = 2$. Therefore, $m + n = 3$.
Question 84
Chemistry · Equilibrium · Numerical
For the reaction $$\mathrm{C_2H_6 \rightarrow C_2H_4 + H_2}$$ the reaction enthalpy $\Delta_r H = \, \mathrm{kJ \, mol^{-1}}$ (Round off to the Nearest Integer). [Given : Bond enthalpies in $\mathrm{kJ \, mol^{-1}}$ : $\mathrm{C-C}$ : 347, $\mathrm{C=C}$ : 611; $\mathrm{C-H}$ : 414, $\mathrm{H-H}$ : 436]
Chemistry · Some Basic Concepts of Chemistry · Numerical
____ grams of 3-Hydroxy propanal (MW = 74) must be dehydrated to produce 7.8 g of acrolein (MW = 56) $(C_3H_4O)$ if the percentage yield is 64. (Round off to the Nearest Integer). [Given : Atomic masses: C: 12.0 u, H: 1.0 u, O : 16.0 u]
Answer: 16
Solution
The reaction is given as follows: $$\mathrm{(HO)H_2C-CH_2-CHO \xrightarrow{\Delta \, 64\%} C_3H_4O + H_2O}$$ The number of moles is calculated as: $$\frac{x}{74} \, mol$$ Using the given yield: $$\frac{x}{74} \times 0.64 = \frac{7.8}{56}$$ Solving for $x$ gives: $$x = 16.10$$ Rounding to two decimal places: $$\approx 16.00$$
Question 86
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical
A reaction of $0.1$ mole of Benzylamine with bromomethane gave $23\,\mathrm{g}$ of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are $n \times 10^{-1}$, when $n = \ldots$ (Round off to the Nearest Integer). (Given: Atomic masses: C: $12.0\,\mathrm{u}$, H: $1.0\,\mathrm{u}$, N: $14.0\,\mathrm{u}$, Br: $80.0\,\mathrm{u}$)
Answer: 3
Solution
Question 87
Chemistry · Co-ordination Compounds · Numerical
The total number of unpaired electrons present in the complex $\mathrm{K}_3[\mathrm{Cr}(oxalate)_3]$ is ____.
Answer: 3
Solution
Chromium is in +3 oxidation state. Number of unpaired electrons in $\mathrm{Cr^{+3}}$ will be 3.
Question 88
Chemistry · Solutions · Numerical
2 molal solution of a weak acid HA has a freezing point of $3.885^\circ \mathrm{C}$. The degree of dissociation of this acid is ___ $\times 10^{-3}$. (Round off to the Nearest Integer). [Given : Molal depression constant of water = $1.85 \, \mathrm{K \, kg \, mol^{-1}}$ Freezing point of pure water = $0^\circ \mathrm{C}$]
For the reaction $2\mathrm{Fe}^{3+}\mathrm{(aq)}+2\mathrm{I}^{-}\mathrm{(aq)}\rightarrow2\mathrm{Fe}^{2+}\mathrm{(aq)}+\mathrm{I}_2\mathrm{(s)}$, the magnitude of the standard molar free energy change, $\Delta_rG^\circ_m=-\underline{\hspace{1cm}}\,\mathrm{kJ}$ (Round off to the Nearest Integer).
Answer: 45
Solution
The reaction sequence is given as: $$\mathrm{Fe^{3+} \xrightarrow{E_1^0} Fe^{2+} \xrightarrow{E_2^0} Fe}$$ The overall reaction is: $$E_1^0 + 2E_2^0 = 3E_3^0$$ Calculating the values: $$E_1^0 = 3E_3^0 - 2E_2^0$$ $$= 3(-0.036) - 2(-0.44)$$ $$= + 0.772 \, \mathrm{V}$$ The cell potential is: $$E_cell^0 = E_{\mathrm{Fe^{3+}/Fe^{2+}}}^0 + E_{\Gamma_{1/2}}^0 = 0.233$$ The change in Gibbs free energy is: $$\Delta_r G^0 = -2 \times 96.5 \times 0.233 = -45 \, \mathrm{kJ}$$
Question 90
Chemistry · Some Basic Concepts of Chemistry · Numerical
Complete combustion of 3 g of ethane gives $x \times 10^{22}$ molecules of water. The value of $x$ is ____ (Round off to the Nearest Integer). [Use : $N_A = 6.023 \times 10^{23}$; Atomic masses in $u : C : 12.0; O : 16.0; H : 1.0$]
Answer: 18
Solution
The reaction is given as: $$\mathrm{C_2H_6} \rightarrow 3\mathrm{H_2O}$$ For $0.1$ mol of $\mathrm{C_2H_6}$, the amount of $\mathrm{H_2O}$ produced is $0.3$ mol. This is calculated as: $$0.3 = 0.3 \times 6 \times 10^{23} = 18 \times 10^{22}$$ The number of molecules is calculated as: $$No. of molecules = 0.3 \times 6.023 \times 10^{23}$$ $$= 18.069 \times 10^{22}$$