JEE Main 29 June 2022 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 · Complex Numbers and Quadratic Equations · Single correct
Let $\alpha$ be a root of the equation $1 + x^2 + x^4 = 0$. Then the value of $\alpha^{1011} + \alpha^{2022} - \alpha^{3033}$ is equal to:
Answer: (a)
Given $x^4 + x^2 + 1 = 0$. This implies $(x^2 + x + 1)(x^2 - x + 1) = 0$. Thus, $x = \pm \omega, \pm \omega^2$ where $\omega = 1^{1/3}$ and imaginary. So $\alpha^{1011} + \alpha^{2022} - \alpha^{3033} = 1 + 1 - 1 = 1$.
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $\arg$ (z) represent the principal argument of the complex number z. The, $|z| = 3$ and $\arg (z - 1) - \arg (z + 1) = \frac{\pi}{4}$ intersect:
Answer: (c)
The given diagram shows a circle with radius $3$ centered at the origin. The point $(0, 1)$ is marked on the circle. The argument of the expression $\frac{z+1}{z-1}$ is given as $\frac{\pi}{4}$. The point $(-1, 0)$ is also marked on the circle. The arc from $(-1, 0)$ to $(0, 1 + \sqrt{2})$ is shown. The correct answer is option C.
Maths · Matrices · Single correct
Let $A = \begin{pmatrix} 2 & -1 \\ 0 & 2 \end{pmatrix}$. If $B = I - \binom{5}{1} \ (adjA) + \binom{5}{2} \ (adjA)^2 - \ldots \binom{5}{5} \ (adjA)^5$, then the sum of all elements of the matrix $B$ is:
Answer: (c)
Given $$B = (I - adjA)^5 = \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix}^5 = \begin{bmatrix} -1 & -5 \\ 0 & -1 \end{bmatrix}$$. Sum of its all elements = -7.
Maths · Sequences and Series · Single correct
The sum of the infinite series $$1 + \frac{5}{6} + \frac{12}{6^2} + \frac{22}{6^3} + \frac{35}{6^4} + \frac{51}{6^5} + \frac{70}{6^6} + \ldots$$ is equal to:
Answer: (c)
Given $$S = 1 + \frac{5}{6} + \frac{12}{6^2} + \frac{22}{6^3} + \frac{35}{6^4} + \ldots$$ We have $$\frac{S}{6} = \frac{1}{6} + \frac{5}{6^2} + \frac{12}{6^3} + \frac{22}{6^4} + \ldots$$ On subtraction $$\frac{5}{6} S = 1 + \frac{4}{6} + \frac{7}{6^2} + \frac{10}{6^3} + \frac{13}{6^4} + \ldots$$ Thus, $$\frac{5}{36} S = 1 + \frac{4}{6^2} + \frac{7}{6^3} + \frac{10}{6^4} + \frac{13}{6^5} + \ldots$$ On subtraction $$\frac{25}{36} S = 1 + \frac{3}{6} + \frac{3}{6^2} + \frac{3}{6^3} + \ldots = \frac{8}{5}$$ Therefore, $$S = \frac{288}{125}$$
Maths · Limits and Derivatives · Single correct
The value of $\lim_{x \to 1} \frac{(x^2 - 1) \sin^2(\pi x)}{x^4 - 2x^3 + 2x - 1}$ is equal to:
The limit is given by $$\lim_{x \to 1} \frac{(x^2 - 1) \sin^2 \pi x}{(x^2 - 1)(x - 1)^2}$$ $$= \lim_{x \to 1} \left( \frac{\sin((1-x)\pi)}{\pi(1-x)} \right)^2 \pi^2 = \pi^2.$$
Maths · Applications of Derivatives · Single correct
Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by $$f(x) = (x-3)^{n_1} (x-5)^{n_2}, \ n_1, \ n_2 \in \mathbb{N}.$$ The, which of the following is NOT true?
Answer: (c)
Given $$f'(x) = (x-3)^{n_1-1}(x-5)^{n_2-1}(n_1+n_2) \left( \frac{x - \frac{5n_1 + 3n_2}{n_1 + n_2}}{n_1 + n_2} \right)$$ Option (3) is incorrect since for $n_1 = 3$, $n_2 = 5$ $$f'(x) = 8(x-3)^2(x-5)^4 \left( x - \frac{30}{8} \right)$$ minima at $x = \frac{30}{8}$
Maths · Integrals · Single correct
Let f be a real valued continuous function on [0,1] and $f(x) = x + \int_{0}^{1} (x-t) f(t) \, dt$. Then which of the following points (x,y) lies on the curve $y = f(x)$?
Answer: (d)
Given $$f(x) = \left(1 + \int_0^1 f(t) \, dt \right)x - \int_0^1 tf(t) \, dt$$ Let $$f(x) = Ax - B ...(i)$$ Then $$A = 1 + \int_0^1 f(t) \, dt = 1 + \int_0^1 (At - B) \, dt$$ This implies $$A = 2(1 - B) ...(ii)$$ Also $$B = \int_0^1 tf(t) \, dt = \int_0^1 (At^2 - Bt) \, dt$$ Thus $$A = \frac{9}{2} B ...(iii)$$ From (ii) and (iii), $$A = \frac{18}{13}, B = \frac{4}{13}$$ Therefore, $$f(6) = 8$$
Maths · Integrals · Single correct
If $$\int_{0}^{2} \left( \sqrt{2x} - \sqrt{2x - x^2} \right) dx = \int_{0}^{1} \left( 1 - \sqrt{1-y^2} - \frac{y^2}{2} \right) dy + \int_{1}^{2} \left( 2 - \frac{y^2}{2} \right) dy + I$$
Answer: (c)
LHS = $$\int_0^2 \left( \sqrt{2x} - \sqrt{2x - x^2} \right) \, dx = \frac{8}{3} - \frac{\pi}{2}$$ RHS = $$\int_0^1 \left( 1 - \sqrt{1-y^2} - \frac{y^2}{2} \right) \, dy + I \int_1^2 \left( 2 - \frac{y^2}{2} \right) \, dy + I$$ $$I + \frac{5}{3} - \frac{\pi}{4}$$ So, $$I = 1 - \frac{\pi}{4} = \int_0^1 \left( 1 - \sqrt{1-y^2} \right) \, dy$$
Maths · Differential Equations · Single correct
If $y = y \ (x)$ is the solution of the differential equation $\left(1 + e^{2x}\right) \frac{dy}{dx} + 2\left(1 + y^2\right) e^x = 0$ and $y(0) = 0$, then $6 \left( y'(0) + \left( y \log_e \sqrt{3} \right)^2 \right)$ is equal to:
Answer: (c)
Given $\dfrac{dy}{1+y^2} + \dfrac{2e^x}{1+e^{2x}}\,dx = 0$ $\quad$ (i). On integration: $\tan^{-1} y + 2\tan^{-1} e^x = c$ Since $y(0) = 0$, so $C = \dfrac{\pi}{2} \Rightarrow \tan^{-1} y + 2\tan^{-1} e^x = \dfrac{\pi}{2}$ From eq. (i), $\left(\dfrac{dy}{dx}\right)_{x=0} = -1$ $y(\ln\sqrt{3}) = -\dfrac{1}{\sqrt{3}}$ $6\left[y'(0) + \left(y(\ln\sqrt{3})\right)^2\right] = 6\left[-1 + \frac{1}{3}\right] = -4$
Maths · Conic Sections · Single correct
Let $P: y^2 = 4ax, a > 0$ be a parabola with focus $S$. Let the tangents to the parabola $P$ make an angle of $\frac{\pi}{4}$ with the line $y = 3x + 5$ touch the parabola $P$ at $A$ and $B$. Then the value of $a$ for which $A,B$ and $S$ are collinear is:
Lines making angle $\frac{\pi}{4}$ with $y = 3x + 5$ have slope $-2$ and $1/2$. Which are perpendicular to each other so, $A$, $S$, $B$ are collinear for all $a > 0$.
Maths · Conic Sections · Single correct
Let a triangle $ABC$ be inscribed in the circle $x^2 - \sqrt{2}(x+y) + y^2 = 0$ such that $\angle BAC = \frac{\pi}{2}$. If the length of side $AB$ is $\sqrt{2}$, then the area of the $\Delta ABC$ is equal to:
Radius of given circle is 1. $BC = diameter = 2$, $AB = \sqrt{2}$ $$AC = \sqrt{BC^2 - AB^2} = \sqrt{2}$$ $$\Delta ABC = \frac{1}{2} AB \cdot AC = 1$$
Maths · Three Dimensional Geometry · Single correct
Let $\frac{x-2}{3} = \frac{y+1}{-2} = \frac{z+3}{-1}$ lie on the plane $px - qy + z = 5$, for some $p, q \in \mathbb{R}$. The shortest distance of the plane from the origin is:
Answer: (b)
(2, -1, -3) satisfy the given plane. So $2p + q = 8$ (i) Also given line is perpendicular to normal plane so $3p + 2q - 1 = 0$ (ii) $\Rightarrow p = 15, q = -22$ Eq. of plane $15x - 22y + z - 5 = 0$ its distance from origin $= \frac{6}{\sqrt{710}} = \sqrt{\frac{5}{142}}$
Maths · Straight Lines and Pair of Straight Lines · Single correct
The distance of the origin from the centroid of the triangle whose two sides have the equations $x - 2y + 1 = 0$ and $2x - y - 1 = 0$ and whose orthocenter is $\left( \frac{7}{3}, \frac{7}{3} \right)$ is:
Answer: (a)
AB is given by $x - 2y + 1 = 0$. AC is given by $2x - y - 1 = 0$. So $A(1, 1)$. Altitude from B is $BH = x + 2y - 7 = 0 \Rightarrow B(3, 2)$. Altitude from C is $CH = 2x + y - 7 = 0 \Rightarrow C(2, 3)$. Centroid of $\triangle ABC = E(2, 2)$, $OE = 2\sqrt{2}$.
Maths · Three Dimensional Geometry · Single correct
Let Q be the mirror image of the point P(1, 2, 1) with respect to the plane x + 2y + 2z = 16. Let T be a plane passing through the point Q and contains the line $\vec{r} = -\hat{k} + \lambda (\hat{i} + \hat{j} + 2\hat{k}), \lambda \in \mathbb{R}$. Then, which of the following points lies on T?
Answer: (d)
Image of $P(1, 2, 1)$ in $x + 2y + 2z - 16 = 0$ is given by $Q(4, 8, 7)$. Eq. of plane $T = \begin{vmatrix} x & y & z + 1 \\ 4 & 8 & 6 \\ 1 & 1 & 2 \end{vmatrix} = 0$. Therefore, $2x - z = 1$ so $B(1, 2, 1)$ lies on it.
Maths · Vector Algebra · Single correct
Let A, B, C be three points whose position vectors respectively are: $$\vec{a} = \hat{i} + 4\hat{j} + 3\hat{k}$$ $$\vec{b} = 2\hat{i} + \alpha \hat{j} + 4\hat{k}, \alpha \in \mathbb{R}$$ $$\vec{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}$$ If $\alpha$ is the smallest positive integer for which $\vec{a}, \vec{b}, \vec{c}$ are non-collinear, then the length of the median, in $\triangle ABC$, through A is:
Answer: (a)
Given $\overrightarrow{AB} \parallel \overrightarrow{AC}$ if $\frac{1}{2} = \frac{\alpha - 4}{-6} = \frac{1}{2} \Rightarrow \alpha = 1$. $\vec{a}, \vec{b}, \vec{c}$ are non-collinear for $\alpha = 2$ (smallest positive integer). Mid-point of $BC = M \left( \frac{5}{2}, 0, \frac{9}{2} \right)$. AM $= \sqrt{\frac{9}{4} + 16 + \frac{9}{4}} = \frac{\sqrt{82}}{2}$.
Maths · Relations and Functions · Single correct
The probability that a relation R from $\{$x,y$\}$ to $\{$x,y$\}$ is both symmetric and transitive, is equal to:
Answer: (a)
Total number of relations = $2^{2 \times 2} = 16$ Favored relation = $\emptyset$, $\{(x, x)\}$, $\{(y, y)\}$, $\{(x, x)(y, y)\}$ $\{(x, x), (y, y), (x, y)(y, x)\}$ Probability = $\($ $\frac{5}{16}$ $\)$
Maths · Statistics · Single correct
The number of values of $a \in \mathbb{N}$ such that the variance of $3, 7, 12, a, 43 - a$ is a natural number is:
Answer: (a)
Mean = 13 Variance = $\frac{9 + 49 + 144 + a^2 + (43 - a)^2}{5}$ - 13^2 $\in$ $\mathbb{N}$ $\Rightarrow$ $\frac{2a^2 - a + 1}{5}$ $\in$ $\mathbb{N}$ $\Rightarrow$ 2a^2 - a + 1 - 5n = 0 must have solution as natural numbers its D = 40n - 7 always has 3 at unit place $\Rightarrow$ D can't be perfect square So, a can't be integer.
Maths · Heights and Distances · Single correct
From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is $60^\circ$. The pole subtends an angle $30^\circ$ at the top of the tower. Then the height of the tower is:
Answer: (d)
Given $\mathrm{PT} = \frac{h}{\sqrt{3}} = \mathrm{AB}$. $$\frac{\mathrm{AB}}{h - 20} = \sqrt{3}$$ Solving for $h$, we have: $$h = 3(h - 20)$$ Simplifying gives: $$h = 30$$
Maths · Mathematical Reasoning · Single correct
Negation of the Boolean statement $(p\vee q)\Rightarrow((\sim r)\vee p)$ is equivalent to:
Given $P \lor q \Rightarrow (\sim r \lor p)$. This is equivalent to $\sim (P \lor q) \lor (\sim r \lor p)$. This is equivalent to $(\sim p \land \sim q) \lor (p \lor \sim r)$. This is equivalent to $[\sim p \lor p] \land (\sim q \lor p) \lor \sim r$. This is equivalent to $[\sim q \lor p] \lor \sim r$. Its negation is $\sim p \land q \land r$.
Maths · Binomial Theorem · Single correct
Let $n \geq 5$ be an integer. If $9^n - 8n - 1 = 64 \, \alpha$ and $6^n - 5n - 1 = 25 \, \beta$, then $\alpha - \beta$ is equal to:
Answer: (c)
Given $$\alpha = \frac{(1+8)^n - 8n - 1}{64} = \binom{n}{2} + \binom{n}{3} 8 + \binom{n}{4} 8^2 + \ldots$$ and $$\beta = \binom{n}{2} + \binom{n}{3} 5 + \binom{n}{4} 5^2 + \ldots$$ option (3) will be the answer.
Maths · Vector Algebra · Numerical
Let \[ \vec{a}=\hat{i}-2\hat{j}+3\hat{k},\qquad \vec{b}=\hat{i}+\hat{j}+\hat{k}, \] and $\vec{c}$ be a vector such that \[ \vec{a}+(\vec{b}\times\vec{c})=\vec{0} \] and \[ \vec{b}\cdot\vec{c}=5. \] Then, the value of \[ 3(\vec{c}\cdot\vec{a}) \] is equal to \[ \underline{\hspace{2cm}} \]
Answer: 10
Given $\vec{a} + \vec{b} \times \vec{c} = 0$ $\vec{a} \times \vec{b} + |\vec{b}|^2 \vec{c} - 5 \vec{b} = 0$ It gives $\vec{c} = \frac{1}{3} (10 \hat{i} + 3 \hat{j} + 2 \hat{k})$ so $3 \vec{a} \cdot \vec{c} = 10$ But it does not satisfy $\vec{a} + \vec{b} \times \vec{c} = 0$. This question has data error. Alternate (Explanation): According to given $\vec{a}$ and $\vec{b}$ $\vec{a} \cdot \vec{b} = 1 - 2 + 3 = 2 \ldots (i)$ but given equation $\vec{a} = - (\vec{b} \times \vec{c})$ $\Rightarrow \vec{a} \perp \vec{b} \Rightarrow \vec{a} \cdot \vec{b} = 0$ which contradicts.
Maths · Differential Equations · Numerical
Let $y = y(x)$, $x > 1$, be the solution of the differential equation $(x-1) \frac{dy}{dx} + 2xy = \frac{1}{x-1}$, with $y(2) = \frac{1 + e^4}{2e^4}$. If $y(3) = \frac{e^\alpha + 1}{\beta e^\alpha}$, then the value of $\alpha + \beta$ is equal to.
Answer: 14
Given $\dfrac{dy}{dx} + \dfrac{2x}{x-1}\cdot y = \dfrac{1}{(x-1)^2}$. $$y = \frac{1}{(x-1)^2}\left[\frac{e^{2x}+1}{2e^{2x}}\right]$$ $$y(3) = \frac{e^6+1}{8e^6}$$ $\alpha + \beta = 14$
Maths · Sequences and Series · Numerical
Let $3, 6, 9, 12, \ldots$ upto $78$ terms and $5, 9, 13, 17, \ldots$ upto $59$ terms be two series. Then, the sum of the terms common to both the series is equal to ___.
Answer: 351
For series of common terms a = 9, d = 12, n = 19 $$S_{19} = \frac{19}{2} [2(9) + 18(12)] = 2223$$
Maths · Trigonometric Functions · Numerical
The number of solutions of the equation $\sin x = \cos^2 x$ in the interval $(0,10)$ is __.
Answer: 4
Given the equation $\sin^2 x + \sin x - 1 = 0$. Solving for $\sin x$, we have: $$\sin x = \frac{-1 + \sqrt{5}}{2} = +ve$$ There are only 4 roots.
Maths · Applications of Integrals · Numerical
For real numbers $a,b$ $(a > b > 0)$, let $$Area\left\{(x,y) : x^2 + y^2 \leq a^2 and \frac{x^2}{a^2} + \frac{y^2}{b^2} \geq 1 \right\} = 30\pi$$ and $$Area\left\{(x,y) : x^2 + y^2 \geq b^2 and \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1 \right\} = 18\pi$$ Then the value of $(a-b)^2$ is equal to __.
Answer: 12
Given $\pi a^2 - \pi ab = 30\pi$ and $\pi ab - \pi b^2 = 18\pi$. On subtracting, we get $(a-b)^2 = a^2 - 2ab + b^2 = 12$.
Maths · Applications of Derivatives · Fill in the blank
Let $f$ and $g$ be twice differentiable even functions on $(-2,2)$ such that \[ f\left(\frac{1}{4}\right)=0,\quad f\left(\frac{1}{2}\right)=0,\quad f(1)=1 \] and \[ g\left(\frac{3}{4}\right)=0,\quad g(1)=2. \] Then, the minimum number of solutions of \[ f(x)g''(x)+f'(x)g'(x)=0 \] in $(-2,2)$ is equal to ______.
Answer: 4
Let $h(x) = f(x) \, g'(x) \rightarrow 5$ roots. Therefore, $f(x)$ is even implies $$f\left(\frac{1}{4}\right) = f\left(\frac{1}{2}\right) = f\left(-\frac{1}{2}\right) = f\left(\frac{1}{4}\right) = 0.$$ $g(x)$ is even implies $$g\left(\frac{3}{4}\right) = g\left(-\frac{3}{4}\right) = 0.$$ $g'(x) = 0$ has minimum one root. $h'(x)$ has at least 4 roots.
Maths · Binomial Theorem · Numerical
Let the coefficients of $x^{-1}$ and $x^{-3}$ in the expansion of $$\left(2x^{\frac{1}{5}} - \frac{1}{x^{\frac{1}{5}}}\right)^{15}$$, $x > 0$, be $m$ and $n$ respectively. If $r$ is a positive integer such that $mn^2 = \binom{15}{r} \cdot 2^r$, then the value of $r$ is equal to __.
Answer: 5
Given $$T_{r+1} = (-1)^r \cdot \binom{15}{r} \cdot 2^{15-r} \cdot \frac{15-2r}{5} \cdot x$$. Let $$m = \binom{15}{10} \cdot 2^5$$ and $$n = -1$$. So, $$mn^2 = \binom{15}{5} \cdot 2^5$$.
Maths · Permutations and Combinations · Numerical
The total number of four digit numbers such that each of the first three digits is divisible by the last digit, is equal to _____.
Answer: 1086
Let the number be abcd, where a, b, c are divisible by d. For $d = 1$, the number of such numbers is $9 \times 10 \times 10 = 900$. For $d = 2$, the number of such numbers is $4 \times 5 \times 5 = 100$. For $d = 3$, the number of such numbers is $3 \times 4 \times 4 = 48$. For $d = 4$, the number of such numbers is $2 \times 3 \times 3 = 18$. For $d = 5$, the number of such numbers is $1 \times 2 \times 2 = 4$. For $d = 6, 7, 8, 9$, the number of such numbers is $4 \times 4 = 16$. The total is 1086.
Maths · Matrices · Numerical
Let $M= \begin{bmatrix} 0 & -\alpha\\ \alpha & 0 \end{bmatrix}$, where $\alpha$ is a non-zero real number and $N=\sum_{k=1}^{49}M^{2k}$. If $(I-M^2)N=-2I$, then the positive integral value of $\alpha$ is ______.
Answer: 1
Given $$\mathbf{M} = \begin{bmatrix} 0 & -\alpha \\ \alpha & 0 \end{bmatrix}; \mathbf{M}^2 = \begin{bmatrix} -\alpha^2 & 0 \\ 0 & -\alpha^2 \end{bmatrix} = -\alpha^2 \mathbf{I}$$ $$\mathbf{N} = \mathbf{M}^2 + \mathbf{M}^4 + \ldots + \mathbf{M}^{98} = [-\alpha^2 + \alpha^4 - \alpha^6 + \ldots] \mathbf{I}$$ $$= -\alpha^2 \left( \frac{1 - (-\alpha^2)^{49}}{1 + \alpha^2} \right) \mathbf{I}$$ $$\mathbf{I} - \mathbf{M}^2 = (1 + \alpha^2) \mathbf{I}$$ $$(\mathbf{I} - \mathbf{M}^2) \mathbf{N} = -\alpha^2 (\alpha^{98} + 1) = -2$$ $$\alpha = 1$$
Maths · Relations and Functions · Fill in the blank
Let $f(x)$ and $g(x)$ be two real polynomials of degree 2 and 1 respectively. If $f(g(x)) = 8x^2 - 2x$, and $g(f(x)) = 4x^2 + 6x + 1$, then the value of $f(2) + g(2)$ is_____.
Answer: 18
Given $f(g(x)) = 8x^2 - 2x$ and $g(f(x)) = 4x^2 + 6x + 1$. So, $g(x) = 2x - 1$ and $f(x) = 2x^2 + 3x + 1$. We find $f(2) = 8 + 6 + 1 = 15$. The answer is 18.
Physics · Motion in a Straight Line · Single correct
A small toy starts moving from the position of rest under a constant acceleration. If it travels a distance of $10\,\mathrm{m}$ in $t\,\mathrm{s}$, the distance travelled by the toy in the next $t\,\mathrm{s}$ will be:
Answer: (c)
Given $u = 0$, say acceleration is $a$. For $t$ seconds: $$10 = \frac{1}{2} a t^2$$ For $2t$ seconds: $$10 + x = \frac{1}{2} a (2t)^2$$ $$\frac{10 + x}{10} = \frac{4}{1}$$ Thus, $x = 30 \, \mathrm{m}$.
Physics · Thermal Properties of Matter · Single correct
At what temperature a gold ring of diameter $6.230 \, \mathrm{cm}$ be heated so that it can be fitted on a wooden bangle of diameter $6.241 \, \mathrm{cm}$? Both the diameters have been measured at room temperature ($27^\circ \mathrm{C}$). (Given: coefficient of linear thermal expansion of gold $\alpha_L = 1.4 \times 10^{-5} \, \mathrm{K}^{-1}$)
Answer: (d)
The force $F$ is given by $$F = \frac{KQq}{\left( x^2 + \frac{d^2}{4} \right)}.$$ The net force on $g$ is $2F \cos \theta$. The net force $F_{net}$ is $$F_{net} = \frac{2KQqx}{\left( x^2 + \frac{d^2}{4} \right)^{3/2}}.$$ For maximum $F_{net}$, $$\frac{d F_{net}}{dx} = 0.$$ Solving, we get $$x = \frac{d}{2\sqrt{2}}.$$
Physics · Electric Charges and Fields · Single correct
Two point charges $Q$ each are placed at a distance $d$ apart. A third point charge $q$ is placed at a distance $x$ from the mid-point on the perpendicular bisector. The value of $x$ at which charge $q$ will experience the maximum Coulomb's force is:
F=\frac{KQq}{x^2+\frac{d^2}{4}} Net force on $q$ is $F_{\mathrm{net}}=2F\cos\theta$ F_{\mathrm{net}}=\frac{2KQqx}{\left(x^2+\frac{d^2}{4}\right)^{3/2}} For maximum $F_{\mathrm{net}}$, \frac{dF_{\mathrm{net}}}{dx}=0 we get x=\frac{d}{2\sqrt{2}}
Physics · Ray Optics and Optical Instruments · Single correct
The speed of light in media 'A' and 'B' are $2.0 \times 10^{10} \, \mathrm{cm/s}$ and $1.5 \times 10^{10} \, \mathrm{cm/s}$ respectively. A ray of light enters from the medium B to A at an incident angle '$\theta$'. If the ray suffers total internal reflection, then
Answer: (d)
Given $\sin i_c = \frac{n_r}{n_d} = \frac{C_d}{C_r} = \frac{1.5 \times 10^{10}}{2 \times 10^{10}}$. $\sin i_c = \frac{3}{4}$. $i_c = \sin^{-1} \left( \frac{3}{4} \right)$. For TIR, $\theta > i_c$. $\theta > \sin^{-1} \left( \frac{3}{4} \right)$.
Physics · Nuclei · Single correct
In the following nuclear reaction, $$D \xrightarrow{\alpha} D_1 \xrightarrow{\beta^-} D_2 \xrightarrow{\alpha} D_3 \xrightarrow{\gamma} D_4$$ Mass number of D is 182 and atomic number is 74. Mass number and atomic number of $D_4$ respectively will be___.
Answer: (a)
Say for $\mathrm{D_4}$ Atomic No = $Z$. Mass Number = $A$. $$A = 182 - 4 - 4 = 174$$ $$Z = 74 - 2 + 1 - 2 = 71$$
Physics · Dual Nature of Radiation and Matter · Single correct
The electric field at the point associated with a light wave is given by E = 200 [$\sin$(6 $\times$ $10^{15}$) t + $\sin$(9 $\times$ $10^{15}$) t] $\mathrm{Vm}^{-1}$ Given : h = 4.14 $\times$ $10^{-15}$ $\mathrm{eVs}$ If this light falls on a metal surface having a work function of 2.50 $\mathrm{eV}$, the maximum kinetic energy of the photoelectrons will be :
Answer: (d)
For maximum KE we will take higher frequency $$f = \frac{9 \times 10^{15}}{2\pi} \, \mathrm{Hz}$$ $$K_{max} = hf - \phi$$ $$= \frac{9 \times 10^{15} \times 4.14 \times 10^{-15}}{2\pi} - 2.50$$ 3.43 eV nearest is 3.42 eV
Physics · Electrostatic Potential and Capacitance · Single correct
A capacitor is discharging through a resistor R. Consider in time $t_1$, the energy stored in the capacitor reduces to half of its initial value and in time $t_2$, the charge stored reduces to one eighth of its initial value. The ratio $t_1/t_2$ will be:
Answer: (d)
In $t_1$ time energy becomes half so charge will become $\frac{1}{\sqrt{2}}$ time. $$q = Q_0 e^{-\frac{t_1}{RC}} = \frac{Q_0}{\sqrt{2}}$$ and $$q = Q_0 e^{-\frac{t_1}{RC}} = \frac{Q_0}{8} = \left(\frac{Q_0}{\sqrt{2}}\right)^6$$ $$t_2 = 6t_1$$ $$\frac{t_1}{t_2} = \frac{1}{6}$$
Physics · Thermodynamics · Single correct
Starting with the same initial conditions, an ideal gas expands from volume $V_1$ to $V_2$ in three different ways. The work done by the gas is $W_1$ if the process is purely isothermal, $W_2$ if the process is purely adiabatic and $W_3$ if the process is purely isobaric. Then, choose the correct option
Answer: (d)
Physics · Moving Charges and Magnetism · Single correct
Two long current carrying conductors are placed parallel to each other at a distance of 8 cm between them. The magnitude of magnetic field produced at mid-point between the two conductors due to current flowing in them is 300 $\mu T$. The equal current flowing in the two conductors is :
Answer: (b)
B at O = 2 $\frac{\mu_0 I}{2 \pi r}$ $$\frac{2 \times 4 \pi \times 10^{-7} I}{2 \pi \times 4 \times 10^{-2}} = 3 \times 10^{-4} \, \mathrm{T}$$ I = 30 $\,$ $\mathrm{A}$ in opp. direction
Physics · Gravitation · Single correct
The time period of a satellite revolving around earth in a given orbit is 7 hours. If the radius of orbit is increased to three times its previous value, then approximate new time period of the satellite will be:
Answer: (b)
The formula for the period is given by $$T = \frac{2\pi}{\sqrt{GM}} r^{3/2}$$ The ratio of periods is $$\frac{T_1}{T_2} = \left( \frac{r_1}{r_2} \right)^{3/2} = \left( \frac{1}{3} \right)^{3/2}$$ Thus, $$T_2 = T_1 \cdot 3\sqrt{3} = 21 \sqrt{3} hours$$ This is approximately 36 hours.
Physics · Communication Systems · Single correct
The TV transmission tower at a particular station has a height of 125 m. For doubling the coverage of its range, the height of the tower should be increased by:
Answer: (c)
Range $d = \sqrt{2Rh}$. $d_2 = 2d_1$ $$\sqrt{2Rh_2} = 2\sqrt{2Rh_1}$$ $h_2 = 4h_1 = 500 \, \mathrm{m}$ $\Delta h = 500 \, \mathrm{m} - 125 \, \mathrm{m} = 375 \, \mathrm{m}$
Physics · Oscillations · Single correct
The motion of a simple pendulum excuting S.H.M. is represented by following equation. Y = A $\sin$ ($\pi$ t + $\phi$), where time is measured in second. The length of pendulum is :
Answer: (c)
Given $\omega = \sqrt{\frac{g}{\ell}} = \pi$. Then, $\frac{g}{\ell} = \pi^2 \Rightarrow \ell = \frac{g}{\pi^2}$. Therefore, $\ell = \frac{980}{\pi^2} \approx 99.4 \, \mathrm{cm}$.
Physics · Kinetic Theory · Single correct
A vessel contains 16 g of hydrogen and 128 g of oxygen at standard temperature and pressure. The volume of the vessel in cm³ is:
Answer: (c)
No of moles of $\mathrm{H_2} = 8$ moles. No of moles of $\mathrm{O_2} = 4$ moles. Total moles $= 12$ moles. At STP 1 mole occupy $= 22.4 \ell = 22.4 \times 10^3 \, \mathrm{cm^3}$. 12 moles will occupy $= 12 \times 22.4 \times 10^3 \, \mathrm{cm^3} \approx 26.8 \times 10^4 \, \mathrm{cm^3}$.
Physics · Electromagnetic Induction · Single correct
Given below are two statements : Statement I: The electric force changes the speed of the charged particle and hence changes its kinetic energy; whereas the magnetic force does not change the kinetic energy of the charged particle. Statement II: The electric force accelerates the positively charged particle perpendicular to the direction of electric field. The magnetic force accelerates the moving charged particle along the direction of magnetic field. In the light of the above statements, choose the most appropriate answer from the options given below:
Answer: (c)
Electric field can change speed and kinetic energy but magnetic field can not change speed or kinetic energy. Because magnetic force is always perpendicular to velocity.
Physics · Laws of Motion · Single correct
A block of mass 40 $\mathrm{kg}$ slides over a surface, when a mass of 4 $\mathrm{kg}$ is suspended through an inextensible massless string passing over frictionless pulley as shown below. The coefficient of kinetic friction between the surface and block is 0.02. The acceleration of block is. (Given $g = 10 \, \mathrm{ms^{-2}}$.)
Answer: (d)
For 4 kg block $$4g - T = 4a$$ For 40 kg block $$T - 40g \times 0.02 = 40a$$ Adding both equations. $$40 - 8 = 44a$$ $$a = \frac{32}{44} = \frac{8}{11} \, \mathrm{m/s^2}$$
Physics · Work, Energy and Power · Single correct
In the given figure, the block of mass $m$ is dropped from the point 'A'. The expression for kinetic energy of block when it reaches point 'B' is:
Answer: (d)
Work done by gravity = $K_B - K_A$ $$mgy_0 = K_B - 0$$ $$K_B = mgy_0$$
Physics · Laws of Motion · Single correct
A block of mass $M$ placed inside a box descends vertically with acceleration 'a'. The block exerts a force equal to one-fourth of its weight on the floor of the box. The value of 'a' will be:
Answer: (c)
The normal force is given by $N = \frac{mg}{4}$. Using the equation of motion, we have $mg - N = ma$. Substituting for $N$, we get $a = g - \frac{g}{4}$. Simplifying, we find $a = \frac{3g}{4}$.
Physics · Electric Charges and Fields · Single correct
If the electric potential at any point $(x, y, z) \mathrm{m}$ in space is given by $V = 3x^2$ volt. The electric field at the point $(1, 0, 3) \mathrm{m}$ will be:
Answer: (d)
Given $E_x = -\frac{\partial V}{\partial x} = -6x$. At $(1, 0, 3)$, $\vec{E} = -6 \, \mathrm{V/m} \, \hat{i}$.
Physics · Current Electricity · Single correct
The combination of two identical cells, whether connected in series or parallel combination provides the same current through an external resistance of $2\Omega$. The value of internal resistance of each cell is :
Answer: (a)
Given the circuit, we have the current $I_1$ as $$I_1 = \frac{2E}{2r + 2}.$$ For the modified circuit, the current $I_2$ is $$I_2 = \frac{E}{\frac{r}{2} + 2} = \frac{2E}{r + 4}.$$ Setting $I_1 = I_2$, we get $$\frac{2E}{2r + 2} = \frac{2E}{r + 4}.$$ Solving for $r$, we have $$2r + 2 = r + 4.$$ Simplifying gives $$2r - r = 2\Omega \Rightarrow r = 2\Omega.$$
Physics · Motion in a Plane · Single correct
A person can throw a ball upto a maximum range of 100 m. How high above the ground he can throw the same ball?
Answer: (b)
Given $R = \frac{u^2 \sin 2\theta}{g}$. $R_{\max} = \frac{u^2}{g} = 100$. $H_{\max} = \frac{u^2}{2g} = \frac{100}{2} = 50 \, \mathrm{m}$.
Physics · Experimental Physics · Numerical
The vernier constant of Vernier callipers is $0.1 \, \mathrm{mm}$ and it has zero error of $(-0.05) \, \mathrm{cm}$. While measuring diameter of a sphere, the main scale reading is $1.7 \, \mathrm{cm}$ and coinciding vernier division is $5$. The corrected diameter will be _____ $\times 10^{-2} \, \mathrm{cm}$.
Answer: 180
Measured diameter = MSR + VSR $\times$ VC $$= 1.7 + 0.01 \times 5$$ $$= 1.75$$ Corrected = Measured - Error $$= 1.75 - (-0.05)$$ $$= 1.80 \, \mathrm{cm}$$ $$= 180 \times 10^{-2} \, \mathrm{cm}$$ 180
Physics · Mechanical Properties of Fluids · Numerical
A small spherical ball of radius 0.1 mm and density $10^4 \, \mathrm{kg\ m^{-3}}$ falls freely under gravity through a a distance $h$ before entering a tank of water. If after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value of $h$ wil be_____m. (Given $g = 10 \, \mathrm{ms^{-2}}$, viscosity of water $= 1.0 \times 10^{-5} \, \mathrm{N\ sm^{-2}}$).
Answer: 20
Speed after falling through height $h$ should be equal to terminal velocity. $$\sqrt{2gh} = \frac{2}{9} \frac{r^2 (d - \rho) g}{\eta}$$ $$\sqrt{2gh} = \frac{2}{9} \frac{10^{-8} (10000 - 1000) \times 10}{10^{-5}}$$ $$= \frac{2}{9} \times 10^{-8} \frac{9 \times 10^4}{10^{-5}} = 20$$ $$2 \times 10 \times h = 400$$ $$h = 20 \, \mathrm{m}$$
Physics · Waves · Numerical
In an experiment to determine the velocity of sound in air at room temperature using a resonance is observed when the air column has a length of 20.0 $\mathrm{cm}$ for a tuning fork of frequency 400 $\mathrm{Hz}$ is used. The velocity of the sound at room temperature is 336 $\mathrm{ms^{-1}}$. The third resonance is observed when the air column has a length of $\mathrm{cm}$.
Answer: 104
For first resonance $$\ell_1 + e = \frac{\lambda}{4}$$ $$\lambda = \frac{336}{400} \times 100 \, \mathrm{cm} = 84 \, \mathrm{cm} \implies \frac{\lambda}{4} = 21 \, \mathrm{cm}$$ $$e = 21 - 20 = 1 \, \mathrm{cm}$$ For third resonance $$\ell_3 + e = \frac{5\lambda}{4} = 105 \, \mathrm{cm} \implies \ell_3 = 104 \, \mathrm{cm}$$
Physics · Current Electricity · Numerical
Two resistors are connected in series across a battery as shown in figure. If a voltmeter of resistance $2000 \, \Omega$ is used to measure the potential difference across $500 \, \Omega$ resistor, the reading of the voltmeter will be ____ V.
Answer: 8
The current $I$ is calculated as follows: $$I = \frac{20}{1000} A$$ The voltage $V_1$ across the $400 \, \Omega$ resistor is given by: $$V_1 = I \times 400 = \frac{20}{1000} \times 400$$ This simplifies to: $$= 8 \, V$$
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Fill in the blank
A potential barrier of 0.4 V exists across a p-n junction. An electron enters the junction from the n-side with a speed of $6.0 \times 10^5 \, \mathrm{ms^{-1}}$. The speed with which electron enters the p side will be $\frac{x}{3} \times 10^5 \, \mathrm{ms^{-1}}$ the value of $x$ is _______. (Given mass of electron = $9 \times 10^{-31} \, \mathrm{kg}$, charge on electron = $1.6 \times 10^{-19} \, \mathrm{C}$.)
Answer: 14
Work done by Electric field = $K_f - K_i$ $$\frac{1}{2} mv^2 - \frac{1}{2} mu^2 = -1.6 - 10^{-19} \times 0.4$$ $$\frac{1}{2} \times 9 \times 10^{-31} (v^2 - u^2) = -0.64 \times 10^{-19}$$ $$u^2 - v^2 = \frac{2 \times 0.64 \times 10^{12}}{9}$$ $$v^2 = \left(36 - \frac{128}{9}\right) \times 10^{10}$$ $$v = \frac{14}{3} \times 10^5 \, \mathrm{m/s}$$ $$x = 14$$
Physics · Electromagnetic Waves · Numerical
The displacement current of $4.425\,\mu\mathrm{A}$ is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of $10^6\,\mathrm{Vs}^{-1}$. The area of each plate of the capacitor is $40\,\mathrm{cm}^2$. The distance between each plate of the capacitor is $x \times 10^{-3}\,\mathrm{m}$. The value of $x$ is, (Permittivity of free space, $\varepsilon_0 = 8.85 \times 10^{-12}\,\mathrm{C^2\,N^{-1}\,m^{-2}}$)
Answer: 8
Displacement Current = Conduction Current $$= \frac{dq}{dt}$$ $$I_d = \frac{\varepsilon_0 \, A}{d} \frac{dV}{dt}$$ $$d = \frac{8.85 \times 10^{-12} \times 4 \times 10^{-3} \times 10^6}{4.425 \times 10^{-6}}$$ $$= 8 \, \mathrm{mm}$$ $$X = 8$$
Physics · System of Particles and Rotational Motion · Numerical
The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is $I_1$. The same rod is bent into a ring and its moment of inertia about a diameter is $I_2$. If $\frac{I_1}{I_2}$ is $\frac{x \pi^2}{3}$, then the value of $x$ will be______.
Answer: 8
Given \[ I_1 = \frac{m\ell^2}{3}. \] Also, \[ \ell = 2\pi r \implies \frac{\ell}{r} = 2\pi. \] And, \[ I_2 = \frac{mr^2}{2}. \] Therefore, \[ \frac{I_1}{I_2} = \frac{2}{3}\left(\frac{\ell}{r}\right)^2 \] \[ = \frac{2}{3}\times (2\pi)^2 = \frac{2}{3}\times 4\pi^2 = \frac{8\pi^2}{3}. \] Hence, \[ x = 8. \]
Physics · Nuclei · Numerical
The half life of a radioactive substance is $5$ years. After $x$ years a given sample of the radioactive substance gets reduced to $6.25\%$ of its initial value of $x$ is _______.
Answer: 20
Given $T_{1/2} = 5$ year. $N = N_0 \left( \frac{1}{2} \right)^{No of half lives}$ $$\frac{N}{N_0} = \frac{1}{16} = \left( \frac{1}{2} \right)^4$$ Time $= 4$ half lives $= 20$ years.
Physics · Wave Optics · Numerical
In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by $5 \times 10^{-2} \, \mathrm{m}$ towards the slits, the change in fringe width is $3 \times 10^{-3} \, \mathrm{cm}$. If the distance between the slits is $1 \, \mathrm{mm}$, then the wavelength of the light will be ______ $\mathrm{nm}$.
Answer: 600
Given $\beta = \frac{\lambda D}{d}$. $\Delta \beta = \frac{\lambda}{d} \Delta D$. $\lambda = \frac{\Delta \beta \cdot d}{\Delta D}$. $$= \frac{3 \times 10^{-5} \times 1 \times 10^{-3}}{5 \times 10^{-2}}$$ $$= 60 \times 10^{-8} = 600 \times 10^{-9} \, \mathrm{m}$$ $$= 600 \, \mathrm{nm}$$
Physics · Alternating Current · Numerical
An inductor of $0.5\,\mathrm{mH}$, a capacitor of $200\,\mu\mathrm{F}$ and a resistor of $2\,\Omega$ are connected in series with a $220\,\mathrm{V}$ AC source. If the current is in phase with the emf, the frequency of AC source will be ___ $\times 10^2\,\mathrm{Hz}$.
Answer: 5
If current is in phase with emf, then the frequency of source is given by the resonant frequency: $$\frac{1}{2\pi \sqrt{LC}}$$. Substituting the values, we have: $$\frac{1}{2\pi \sqrt{\frac{1}{2} \times 10^{-3} \times 2 \times 10^{-4}}}$$. Simplifying, we get: $$= \frac{1}{2\pi} \times \sqrt{10} \times 1000 = 500 \, Hz$$.
Chemistry · Some Basic Concepts of Chemistry · Single correct
Using the rules for significant figures, the correct answer for the expression $\($ $\frac{0.0258 \times 0.112}{0.5702}$ $\)$ will be:
Answer: (c)
Reported answer should not be more precise than least precise term in calculations, so there should be three significant figures in reported answer.
Chemistry · Structure of Atom · Single correct
Which of the following is the correct plot for the probability density $\psi^2(r)$ as a function of distance 'r' of the electron form the nucleus for 2s orbital?
Answer: (b)
For 2s, number of radial nodes = $2 - 0 - 1 = 1$ and value of $\psi^2$ is always positive.
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Consider the species $CH_4, NH_4^+$ and $BH_4^-$. Choose the correct option with respect to the three species:
Answer: (b)
All are tetrahedral and each have 10 electrons.
Chemistry · Equilibrium · Single correct
4.0 moles of argon and 5.0 moles of $\mathrm{PCl}_5$ are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The $K_p$ for the reaction is [Given : $R = 0.082 \, \mathrm{L} \, \mathrm{atm} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1}$]
Answer: (a)
Given $\mathrm{PCl_5} = 5 mole$ and $\mathrm{Ar} = 4 mole$. The total pressure $P_{Total}$ is calculated as: $$P_{Total} = \frac{9 \times 0.82 \times 610}{100} = 4.5 \, atm$$ The partial pressures are: $$P_{\mathrm{PCl_5}} = \frac{5 \times 4.5}{9} = 2.5$$ $$P_{\mathrm{Ar}} = \frac{4 \times 4.5}{9} = 2$$ The reaction is: $$\mathrm{PCl_5} \rightleftharpoons \mathrm{PCl_3} + \mathrm{Cl_2}$$ The pressures are: $$2.5 - P P P$$ The total pressure is: $$P_{total} = 2.5 - P + P + P + P_{\mathrm{Ar}} = 6$$ Solving for $P$: $$P = 1.5$$ The equilibrium constant $K_p$ is: $$K_p = \frac{1.5 \times 1.5}{1} = 2.25$$
Chemistry · Surface Chemistry · Single correct
A 42.12$\%$ (w/v) solution of NaCl causes precipitation of a certain sol in 10 hours. The coagulating value of NaCl for the sol is [Given : Molar mass : Na = 23.0 g mol$^{-1}$; Cl = 35.5 g mol$^{-1}$]
Answer: (d)
Data insufficient.
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : The first ionization enthalpy for oxygen is lower than that of nitrogen. Reason R : The four electrons in 2p orbitals of oxygen experience more electron-electron repulsion. In the light of the above statements, choose the correct answer from the options given below.
Answer: (a)
Ionisation energy $= \mathrm{N} > \mathrm{O}$. In oxygen atom, 2 of the 4 $2p$ electrons must occupy the same $2p$ orbital resulting in an increased electron electron-repulsion.
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Match List I with List II. Choose the correct answer from the options given below:
Answer: (a)
Siderite is $\mathrm{FeCO_3}$. Malachite is $\mathrm{CuCO_3} \cdot \mathrm{Cu(OH)_2}$. Calamine is $\mathrm{ZnCO_3}$. Sphalerite is $\mathrm{ZnS}$.
Chemistry · Co-ordination Compounds · Single correct
Given below are two statements. Statement I: In $\mathrm{CuSO_4.5H_2O}$, Cu–O bonds are present. Statement II: In $\mathrm{CuSO_4.5H_2O}$, ligands coordinating with Cu(II) ion are O-and S-based ligands. In the light of the above statements, choose the correct answer from the options given below
Answer: (c)
The structure shown is a complex ion with copper at the center. It is surrounded by water molecules and sulfate ions. The copper ion is coordinated by four water molecules and forms hydrogen bonds with the sulfate ion.
Chemistry · The s-Block Elements · Single correct
Amongst baking soda, caustic soda and washing soda carbonate anion is present in:
Answer: (a)
Baking soda $\rightarrow \mathrm{NaHCO_3}$ Washing soda $\rightarrow \mathrm{Na_2CO_3 \cdot 10H_2O}$ Caustic soda $\rightarrow \mathrm{NaOH}$
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Number of lone pair (s) of electrons on central atom and the shape of $\mathrm{BrF_3}$ molecule respectively, are :
Answer: (c)
Steric no. = 5 (sp^3d), lone pair = 2. Bent T shape.
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
Aqueous solution of which of the following boron compounds will be strongly basic in nature?
Answer: (d)
$Na_2B_4O_7$ gives $H_3BO_3$ and NaOH (strong base) in water.
Chemistry · Environmental Chemistry · Single correct
Sulphur dioxide is one of the components of polluted air. $\mathrm{SO}_2$ is also a major contributor to acid rain. The correct and complete reaction to represent acid rain caused by $\mathrm{SO}_2$ is:
Answer: (d)
The reaction is given by: $$2\mathrm{SO_2} + \mathrm{O_2} + 2\mathrm{H_2O} \rightarrow 2\mathrm{H_2SO_4} (Acid rain)$$
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Which of the following carbocations is most stable?
Answer: (d)
Is most stable carbocation
Chemistry · Haloalkanes and Haloarenes · Single correct
Major Product The stable carbocation formed in the above reaction is:
Answer: (c)
The carbocation $\overset{\oplus}{\mathrm{CH_3CHCH_3}}$ is formed in the above reaction.
Chemistry · Haloalkanes and Haloarenes · Single correct
Two isomers $(\mathrm{A})$ and $(\mathrm{B})$ with molar mass $184\ \mathrm{g\,mol^{-1}}$ and elemental composition: $\mathrm{C}$, $52.2\%$; $\mathrm{H}$, $4.9\%$ and $\mathrm{Br}$, $42.9\%$ gave benzoic acid and $p$-bromobenzoic acid, respectively, on oxidation with $\mathrm{KMnO_4}$. Isomer $\mathrm{A}$ is optically active and gives a pale yellow precipitate when warmed with alcoholic $\mathrm{AgNO_3}$. Isomers $\mathrm{A}$ and $\mathrm{B}$ are, respectively,
Answer: (c)
Chemistry · Hydrocarbons · Single correct
In Friedel-Crafts alkylation of aniline, one gets:
Answer: (d)
Chemistry · Polymers · Single correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Dacron is an example of polyester polymer. Reason R: Dacron is made up of ethylene glycol and terephthalic acid monomers. In the light of the above statements, choose the most appropriate answer from the options given below.
Answer: (a)
Ethylene glycol reacts with terephthalic acid to form Dacron. The reaction is as follows:
Chemistry · Biomolecules · Single correct
The structure of protein that is unaffected by heating is :
Answer: (c)
Primary structure of protein is unaffected by physical or chemical changes.
Chemistry · Environmental Chemistry · Single correct
The mixture of chloroxylenol and terpineol is an example of:
Answer: (a)
Antiseptic Dettol is a mixture of chloroxylenol and terpineol.
Chemistry · Analytical Chemistry · Single correct
A white precipitate was formed when BaCl$_2$ was added to water extract of an inorganic salt. Further, a gas 'X' with characteristic odour was released when the formed white precipitate was dissolved in dilute HCl. The anion present in the inorganic salt is :
Answer: (b)
$BaCl_2 + SO_3^{2-} \rightarrow BaSO_3 \downarrow \xrightarrow{\text{dil. HCl}} SO_2 \uparrow$ White precipitate with burning sulphur-like smell.
Chemistry · Equilibrium · Numerical
A box contains $0.90\ \mathrm{g}$ of liquid water in equilibrium with water vapour at $27°\mathrm{C}$. The equilibrium vapour pressure of water at $27°\mathrm{C}$ is $32.0\ \mathrm{Torr}$. When the volume of the box is increased, some of the liquid water evaporates to maintain the equilibrium pressure. If all the liquid water evaporates, then the volume of the box must be \_\_\_\_ litre. [nearest integer] (Given: $R = 0.082\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$) (Ignore the volume of the liquid water and assume water vapours behave as an ideal gas.)
Answer: 29
Given the equation for volume, we have: $$V = \frac{nRT}{P} = \frac{0.90 \times 0.82 \times 300 \times 760}{18 \times 32} = 29.21$$
Chemistry · Thermodynamics · Numerical
2.2 $\mathrm{g}$ of nitrous oxide ($\mathrm{N}_2\mathrm{O}$) gas is cooled at a constant pressure of 1 $\mathrm{atm}$ from 310 $\mathrm{K}$ to 270 $\mathrm{K}$ causing the compression of the gas from 217.1 $\mathrm{mL}$ to 167.75 $\mathrm{mL}$. The change in internal energy of the process, $\Delta$ U is '-x' J. The value of 'x' is _____. [nearest integer] (Given: atomic mass of N = 14 $\mathrm{g}$ \, $\mathrm{mol^{-1}}$ and of O = 16 $\mathrm{g}$ \, $\mathrm{mol^{-1}}$. Molar heat capacity of $\mathrm{N}_2\mathrm{O}$ is 100 $\mathrm{J}$ \, $K^{-1}$ \, $mol^{-1}$)
Answer: 195
Given $\mathrm{N_2O}$ moles $= \frac{2.2}{44} = \frac{1}{20}$. $\Delta H = nC_p \Delta T = \frac{1}{20} \times 100(-40) = -200 \, \mathrm{J}$. $\Delta U = q_p + w$. $w = -P_{ext} \Delta V$. $W = -1 \left( \frac{167.75 - 217.1}{1000} \right) \times 101.3 \, \mathrm{J}$. $w = +5 \, \mathrm{J}$. $\Delta U = -200 + 5 = -195 \, \mathrm{J}$.
Chemistry · Solutions · Numerical
Elevation in boiling point for 1.5 molal solution of glucose in water is 4K. The depression in freezing point for 4.5 molal solution of glucose in water is 4K. The ratio of molal elevation constant to molal depression constant ($K_b/K_f$) is ___.
Answer: 3
Given the equations for boiling point elevation and freezing point depression: $$\Delta T_b = i K_b m$$ $$\Delta T_f = i K_f m$$ We have the ratio: $$\frac{4}{4} = \frac{K_b \cdot 1.5}{K_f \cdot 4.5}$$ Solving for the ratio of $K_b$ to $K_f$ gives: $$\frac{K_b}{K_f} = 3$$
Chemistry · Electrochemistry · Numerical
The cell potential for the given cell at $298\,\mathrm{K}$, $\mathrm{Pt}\,|\,\mathrm{H}_2(g,1\,\mathrm{bar})\,|\,\mathrm{H}^{+}(aq)\,||\,\mathrm{Cu}^{2+}(aq)\,|\,\mathrm{Cu}(s)$, is $0.31\,\mathrm{V}$. The pH of the acidic solution is found to be $3$, whereas the concentration of $\mathrm{Cu}^{2+}$ is $10^{-x}\,\mathrm{M}$. The value of $x$ is $\underline{\hspace{1cm}}$. Given: $E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}}=0.34\,\mathrm{V}$ and $\dfrac{2.303RT}{F}=0.06\,\mathrm{V}$.
Answer: 7
The reaction is given by: $$\mathrm{H_2(g) + Cu^{2+}(aq.) \rightarrow 2H^+(aq.) + Cu(s)}$$ The equation is: $$0.31 = 0.34 - \frac{0.06}{2} \log \frac{[\mathrm{H^+}]^2}{[\mathrm{Cu^{2+}}]}$$ Given: $$[\mathrm{Cu^{2+}}] = 10^{-7} \, \mathrm{M}$$ The value of $x$ is: $$x = 7$$
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
The equation $$k = \left( 6.5 \times 10^{12} \, \mathrm{s}^{-1} \right) e^{-26000K/T}$$ is followed for the decomposition of compound A. The activation energy for the reaction is _____ kJ mol$^{-1}$. [nearest integer] (Given: R = 8.314 J K$^{-1}$ mol$^{-1}$)
Answer: 216
Given $$K = Ae^{-E_a/RT} = (6.5 \times 10^{12} \, \mathrm{s^{-1}}) e^{-26000K/T}$$ $$\frac{E_a}{8.314} = 26000$$ $$E_a = 216.164 \, \mathrm{kJ/mol}.$$
Chemistry · Co-ordination Compounds · Numerical
Spin only magnetic moment of $[\mathrm{MnBr}_6]^{4-}$ is____ B.M. (round off to the closest integer)
The electronic configuration for $\mathrm{Mn^{2+}}$ is $t_{2g}^{11} e_g^{11}$. The spin-only magnetic moment $\mu_s$ is given by $\mu_s = \sqrt{35}$. This evaluates to $5.91$, which is approximately $6$.
Chemistry · Redox Reactions · Numerical
For the reaction given below: $$\mathrm{CoCl_3} \cdot x\mathrm{NH_3} + \mathrm{AgNO_3(aq)} \rightarrow$$ If two equivalents of $\mathrm{AgCl}$ precipitate out, then the value of $x$ will be _____.
Answer: 5
CoCl_3.x$\mathrm{NH_3}$ + $\mathrm{AgNO_3}$ $\rightarrow$ $\mathrm{AgCl}$ $\downarrow$ 2 mol [$\mathrm{Co(NH_3)_5Cl}$]$\mathrm{Cl_2}$ + $\mathrm{AgNO_3}$ $\rightarrow$ $\mathrm{AgCl}$ $\downarrow$ 2 mol x = 5
Chemistry · Alcohols, Phenols and Ethers · Fill in the blank
The number of chiral alcohol(s) with molecular formula $\mathrm{C_4H_{10}O}$ is ______.
Answer: 1
$CH_3$-$CH_2$-$CH_2$-$CH_2$-OH Out of which only two are chiral
Chemistry · Alcohols, Phenols and Ethers · Numerical
In the given reaction the number of $sp^2$ hybridised carbon (s) in compound 'X' is .
Answer: 8
The reaction sequence involves the following steps: 1. The cyclohexanol is oxidized to cyclohexanone using $K_2Cr_2O_7$. 2. The cyclohexanone undergoes a Grignard reaction with $C_6H_5MgBr$ in the presence of water to form a tertiary alcohol. 3. The tertiary alcohol is then dehydrated using $H^+$ and heat to form an alkene with a phenyl group.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical
In the given reaction, The number of $\pi$ electrons present in the product 'P' is _____.
Answer: 4