JEE Main 24 June 2022 Shift 2 question paper with solutions
JEE Main 24 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
Question 1
Maths · Complex Numbers and Quadratic Equations · Single correct
The sum of all the real roots of the equation $\left(e^{2x} - 4\right) \left(6e^{2x} - 5e^x + 1\right) = 0$ is
$\log_e 3$
$-\log_e 3$
$\log_e 6$
$-\log_e 6$
Answer: (b)
Solution
Given $\left(e^{2x} - 4\right) \left(6e^{2x} - 3e^x - 2e^x + 1\right) = 0$. $\left(e^{2x} - 4\right) \left(3e^x - 1\right) \left(2e^x - 1\right) = 0$. $e^{2x} = 4$ or $e^x = \frac{1}{3}$ or $e^x = \frac{1}{2}$. Therefore, the sum of real roots is $= \frac{1}{2} \ln 4 + \ln \frac{1}{3} + \ln \frac{1}{2}$. $= -\ln 3$.
Question 2
Maths · Basics Of Mathematics · Single correct
Let $x, y > 0$. If $x^3 y^2 = 2^{15}$, then the least value of $3x + 2y$ is
30
32
36
40
Answer: (d)
Solution
Using AM ≥ GM $$\frac{x + x + x + y + y}{5} \geq \left(x^3 \cdot y^2\right)^{\frac{1}{5}}$$ $$\frac{3x + 2y}{5} \geq \left(2^{15}\right)^{\frac{1}{5}}$$ $$(3x + 2y)_{\min} = 40$$
Question 3
Maths · Trigonometric Functions · Single correct
The number of solutions of the equation $$ \cos \left( x + \frac{\pi}{3} \right) \cos \left( \frac{\pi}{3} - x \right) = \frac{1}{4} \cos^2 2x, \ x \in [-3\pi, 3\pi] $$ is:
Let the area of the triangle with vertices A(1, $\alpha$), B($\alpha$, 0) and C(0, $\alpha$) be 4 sq. units. If the point ($\alpha$, $-\alpha$), ($-\alpha$, $\alpha$) and ($\alpha^2$, $\beta$) are collinear, then $\beta$ is equal to
64
-8
-64
512
Answer: (c)
Solution
Given $\dfrac{1}{2} \begin{vmatrix} \alpha & 0 & 1 \\ 1 & \alpha & 1 \\ 0 & \alpha & 1 \end{vmatrix} = \pm\, 4$. $\alpha = \pm\, 8$ Now given points $(8, -8),\ (-8, 8),\ (64, \beta)$ or $(-8, 8),\ (8, -8),\ (64, \beta)$ are collinear $\Rightarrow$ Slope $= -1$ $\beta = -64$ Ans. (C)
Question 5
Maths · Conic Sections · Single correct
A particle is moving in the xy-plane along a curve C passing through the point (3, 3). The tangent to the curve C at the point P meets the x-axis at Q. If the y-axis bisects the segment PQ, then C is a parabola with
length of latus rectum 3
length of latus rectum 6
focus $\left( \frac{4}{3}, 0 \right)$
focus $\left( 0, \frac{3}{4} \right)$
Answer: (a)
Solution
Let Point P(x,y) Y - y = y'(X - x) Y = 0 $\Rightarrow$ X = x - $\frac{y}{y'}$ Q $\left$( x - $\frac{y}{y'}$, 0 $\right$) Mid Point of PQ lies on y axis x - $\frac{y}{y'}$ + x = 0 $$y' = \frac{y}{2x} \Rightarrow 2 \frac{dy}{y} = \frac{dx}{x}$$ $$2\ln y = \ln x + \ln k$$ $$y^2 = kx$$ It passes through (3, 3) $\Rightarrow$ k = 3 curve c $\Rightarrow$ y^2 = 3x Length of L.R. = 3 Focus = $\left$( $\frac{3}{4}$, 0 $\right$) Ans. (A)
Question 6
Maths · Conic Sections · Single correct
Let the maximum area of the triangle that can be inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$, $a > 2$, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be $6\sqrt{3}$. Then the eccentricity of the ellipse is:
$\frac{\sqrt{3}}{2}$
$\frac{1}{2}$
$\frac{1}{\sqrt{2}}$
$\frac{\sqrt{3}}{4}$
Answer: (a)
Solution
Given the ellipse with points $(a \cos \theta, 2 \sin \theta)$, $(a \cos \theta, -2 \sin \theta)$, and $(a, 0)$, where $b = 2$. The area $A$ is given by: $$A = \frac{1}{2} a (1 - \cos \theta) (4 \sin \theta)$$ Simplifying, we have: $$A = 2a(1 - \cos \theta) \sin \theta$$ Differentiating $A$ with respect to $\theta$: $$\frac{dA}{d\theta} = 2a (\sin^2 \theta + \cos \theta - \cos^2 \theta)$$ Setting $\frac{dA}{d\theta} = 0$, we get: $$1 + \cos \theta - 2 \cos^2 \theta = 0$$ Solving for $\cos \theta$, we find $\cos \theta = 1$ (Reject) or: $$\cos \theta = -\frac{1}{2} \implies \theta = \frac{2\pi}{3}$$ The second derivative is: $$\frac{d^2A}{d\theta^2} = 2a (2 \sin^2 \theta - \sin \theta)$$ Evaluating at $\theta = \frac{2\pi}{3}$, we find: $$\frac{d^2A}{d\theta^2} < 0$$ Now, the maximum area $A_{\max}$ is: $$A_{\max} = \frac{3\sqrt{3}}{2} a = 6 \sqrt{3}$$ Given $a = 4$, we find the eccentricity $e$: $$e = \sqrt{\frac{a^2 - b^2}{a^2}} = \frac{\sqrt{3}}{2}$$ Answer: (A)
Question 7
Maths · Mathematical Reasoning · Single correct
Consider the following statements : A : Rishi is a judge. B : Rishi is honest. C : Rishi is not arrogant. The negation of the statement "if Rishi is a judge and he is not arrogant, then he is honest" is
B $\rightarrow$ (A $\lor$ C)
($\sim$ B) $\land$ (A $\land$ C)
B $\rightarrow$ (($\sim$ A) $\lor$ ($\sim$ C))
B $\rightarrow$ (A $\land$ C)
Answer: (b)
Solution
Given $\sim ((A \land C) \rightarrow B)$. This is equivalent to $\sim (\sim (A \land C) \lor B)$. Using De-Morgan's law, we have $$(A \land C) \land (\sim B).$$ Option B is correct.
Question 8
Maths · Determinants · Single correct
Let the system of linear equations x + y + $\alpha$ z = 2 3x + y + z = 4 x + 2z = 1 have a unique solution (x^*, y^*, z^*). If ($\alpha$, x^*), (y^*, $\alpha$) and (x^*, -y^*) are collinear points, then the sum of absolute values of all possible values of $\alpha$ is :
Maths · Continuity and Differentiability · Single correct
\[ f(x)= \left\{ \begin{array}{ll} \dfrac{\sin(x-[x])}{x-[x]} & x\in(-2,-1), \\[6pt] \max\{2x,3[[x]]\} & |x|<1, \\[6pt] 1 & \text{otherwise}. \end{array} \right. \] where $[t]$ denotes greatest integer $\leq t$. If $m$ is the number of points where $f$ is not continuous and $n$ is the number of points where $f$ is not differentiable, then the ordered pair \(m,n\) is :
(3, 3)
(2, 4)
(2, 3)
(3, 4)
Answer: (c)
Solution
Given $$f(x) = \begin{cases} \frac{\sin(x+2)}{x+2}, & x \in (-2, -1) \\ \max \{2x, 0\}, & x \in (-1, 1) \\ 1, & otherwise \end{cases}$$ We have $$f(-2^+) = \lim_{h \to 0} f(-2 + h) = \lim_{h \to 0} \frac{\sinh}{h} = 1$$ Thus, $f$ is continuous at $x = -2$. Next, $$f(-1^-) = \lim_{h \to 0} \frac{\sin(-1 - h + 2)}{-1 - h + 2} = \sin 1$$ And $$f(-1) = f(-1^+) = 0$$ Since $f(1^+) = 1$ and $f(1^-) = 0$, $f$ is not continuous at $x = 1$. $f$ is continuous but not differentiable at $x = 0$. Therefore, $f$ is discontinuous at $x = -1$ and $1$, and $f$ is not differentiable at $x = -1$, $0$, and $1$. Thus, $$m = 2$$ $$n = 3$$
Question 11
Maths · Continuity and Differentiability · Single correct
If $y = \tan^{-1}(\sec x^3 - \tan x^3)$. $\frac{\pi}{2} < x^3 < \frac{3\pi}{2}$, then
$xy'' + 2y' = 0$
$x^2y'' - 6y + \frac{3\pi}{2} = 0$
$x^2y'' - 6y + 3\pi = 0$
$xy'' - 4y' = 0$
Answer: (b)
Solution
Given $y = \tan^{-1} \left( \sec x^3 - \tan x^3 \right)$. This can be rewritten as: $$y = \tan^{-1} \left( \frac{1 - \sin x^3}{\cos x^3} \right)$$ Further simplifying: $$= \tan^{-1} \left( \frac{1 - \cos \left( \frac{\pi}{2} - x^3 \right)}{\sin \left( \frac{\pi}{2} - x^3 \right)} \right)$$ This becomes: $$= \tan^{-1} \left( \tan \left( \frac{\pi}{4} - \frac{x^3}{2} \right) \right)$$ Since $\frac{\pi}{4} - \frac{x^3}{2} \in \left( -\frac{\pi}{2}, 0 \right)$, we have: $$y = \left( \frac{\pi}{4} - \frac{x^3}{2} \right)$$ Differentiating, we get: $$y' = \frac{-3x^2}{2}, y'' = -3x$$ Substituting into the equation: $$4y = \pi - 2x^3$$ Multiplying by 3: $$12y = 3\pi + 2x^3 y''$$ Finally, we have: $$x^2 y'' - 6y + \frac{3\pi}{2} = 0$$
Question 12
Maths · Applications of Derivatives · Single correct
The number of distinct real roots of the equation $x^7 - 7x - 2 = 0$ is
5
7
1
3
Answer: (d)
Solution
Given the equation $x^7 - 7x - 2 = 0$. Rewriting, we have $x^7 - 7x = 2$. Let $f(x) = x^7 - 7x$ (which is odd) and $y = 2$. Then, $f(x) = x \left(x^2 - 7^{1/3}\right) \left(x^4 + x^2 \cdot 7^{1/3} + 7^{2/3}\right)$. The derivative is $f'(x) = 7(x^6 - 1) = 7 \left(x^2 - 1\right) \left(x^4 + x^2 + 1\right)$. Setting $f'(x) = 0$ implies $x = \pm 1$. The function $f(x) = 2$ has 3 real distinct solutions.
Question 13
Maths · Applications of Derivatives · Single correct
Let $\lambda^*$ be the largest value of $\lambda$ for which the function $f_\lambda(x) = 4\lambda x^3 - 36x^2 + 36x + 48$ is increasing for all $x \in \mathbb{R}$. Then $f_\lambda^*(1) + f_\lambda^*(-1)$ is equal to :
36
48
64
72
Answer: (d)
Solution
Given $f_\lambda(x) = 4\lambda x^3 - 36\lambda x^2 + 36x + 48$. Differentiating, we have $f_\lambda'(x) = 12\lambda x^2 - 72\lambda x + 36$. Setting $f_\lambda'(x) = 12(\lambda x^2 - 6\lambda x + 3) \geq 0$. Thus, $\lambda > 0$ and $D \leq 0$. Solving $36\lambda^2 - 4 \times \lambda \times 3 \leq 0$, we get $9\lambda^2 - 3\lambda \leq 0$. This simplifies to $3\lambda (3\lambda - 1) \leq 0$. Therefore, $\lambda \in \left[0, \frac{1}{3}\right]$. Thus, $\lambda_{largest} = \frac{1}{3}$. Substituting, $f(x) = \frac{4}{3} x^3 - 12x^2 + 36x + 48$. Therefore, $f(1) + f(1) = 72$.
Question 14
Maths · Integrals · Single correct
The value of the integral $$\int_{-\pi/2}^{\pi/2} \frac{dx}{(1+e^x)(\sin^6 x + \cos^6 x)}$$ is equal to
$2\pi$
$0$
$\pi$
$\frac{\pi}{2}$
Answer: (c)
Solution
Let $$I = \int_{-\pi/2}^{0} \frac{dx}{(1+e^x)(\sin^6 x + \cos^6 x)} + \int_{0}^{\pi/2} \frac{dx}{(1+e^x)(\sin^6 x + \cos^6 x)}.$$ Put $x = -t$ $$= \int_{\pi/2}^{0} \frac{-dt}{(1+e^{-t})(\sin^6 t + \cos^6 t)} + \int_{0}^{\pi/2} \frac{dx}{(1+e^x)(\sin^6 x + \cos^6 x)}$$ $$= \int_{0}^{\pi/2} \frac{(e^x + 1) \, dx}{(1+e^x)(\sin^6 x + \cos^6 x)}$$ $$= \int_{0}^{\pi/2} \frac{dx}{(\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)}$$ $$= \int_{0}^{\pi/2} \frac{(1+\tan^2 x) \sec^2 x \, dx}{(\tan^4 x - \tan^2 x + 1)}.$$ Put $\tan x = t$ $$= \int_{0}^{\infty} \frac{(1+t^2) \, dt}{(t^4 - t^2 + 1)}$$ $$= \int_{0}^{\infty} \left[ \frac{1+\frac{1}{t^2}}{t^2 - 1 + \frac{1}{t^2}} \right] dt = \int_{0}^{\infty} \left[ \frac{1+\frac{1}{t^2}}{t - \frac{1}{t}}^2 + 1 \right] dt.$$ Put $t - \frac{1}{t} = z$ $$\left(1 + \frac{1}{t^2}\right) dt = dz$$ $$= \int_{-\infty}^{\infty} \frac{dz}{1+z^2} = (\tan^{-1} z) \bigg|_{-\infty}^{\infty}$$ $$= \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi.$$
The limit is given by $$ \lim_{n \to \infty} \left( \sum_{r=1}^{n} \frac{n^2}{n^2 + r^2} \frac{1}{n+r} \right) $$ which simplifies to $$ \lim_{n \to \infty} \left( \sum_{r=1}^{n} \frac{1}{n \left( 1 + \left( \frac{r}{n} \right)^2 \right) \left( 1 + \frac{r}{n} \right)} \right). $$ This is equal to $$ \int_{0}^{1} \frac{dx}{(1+x^2)(1+x)} = \frac{1}{2} \int_{0}^{1} \frac{1-x}{1+x^2} \, dx + \frac{1}{2} \int_{0}^{1} \frac{1}{1+x} \, dx. $$ Further simplifying, we have $$ = \frac{1}{2} \left( \int_{0}^{1} \frac{1}{1+x^2} - \frac{x}{1+x^2} \, dx \right) + \frac{1}{2} \left( \ln(1+x) \right) \bigg|_{0}^{1}. $$ This results in $$ = \frac{1}{2} \left[ \tan^{-1} x - \frac{1}{2} \ln(1+x^2) \right]_{0}^{1} + \frac{1}{2} \ln 2. $$ Evaluating the integrals, we get $$ = \frac{1}{2} \left[ \frac{\pi}{4} - \frac{1}{2} \ln 2 \right] + \frac{1}{2} \ln 2. $$ Finally, the result is $$ = \frac{\pi}{8} + \frac{1}{4} \ln 2. $$
Question 16
Maths · Differential Equations · Single correct
The slope of normal at any point $(x, y)$, $x > 0$, $y > 0$ on the curve $y = y(x)$ is given by $\frac{x^2}{xy - x^2y^2 - 1}$. If the curve passes through the point $(1, 1)$, then $e.y(e)$ is equal to
$\frac{1 - \tan(1)}{1 + \tan(1)}$
$\tan(1)$
1
$\frac{1 + \tan(1)}{1 - \tan(1)}$
Answer: (d)
Solution
Slope of normal = $-\frac{dx}{dy} = \frac{x^2}{xy - x^2y^2 - 1}$ $x^2y^2 dx + dx - xydx = x^2 dy$ $x^2y^2 dx + dx = x^2 dy + xydx$ $x^2y^2 dx + dx = x(xdy + ydx)$ $x^2y^2 dx + dx = x d(xy)$ $$\frac{dx}{x} = \frac{d(xy)}{1 + x^2y^2}$$ $\ln kx = \tan^{-1}(xy)$ (i) passes through $(1, 1)$ $\ln k = \frac{\pi}{4} \implies k = e^{\frac{\pi}{4}}$ Equation (i) becomes $$\frac{\pi}{4} + \ln x = \tan^{-1}(xy)$$ $$xy = \tan\left(\frac{\pi}{4} + \ln x\right)$$ $$xy = \left(\frac{1 + \tan(\ln x)}{1 - \tan(\ln x)}\right) (ii)$$ Put $x = e$ in (ii) $$\therefore \, ey(e) = \frac{1 + \tan 1}{1 - \tan 1}$$
Question 17
Maths · Vector Algebra · Single correct
Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that $\left| \left( \hat{a} + \hat{b} \right) + 2 \left( \hat{a} \times \hat{b} \right) \right| = 2$. If $\theta \in (0, \pi)$ is the angle between $\hat{a}$ and $\hat{b}$, then among the statements: (S1) : $2 \left| \hat{a} \times \hat{b} \right| = \left| \hat{a} - \hat{b} \right|$ (S2) : The projection of $\hat{a}$ on $\left( \hat{a} + \hat{b} \right)$ is $\frac{1}{2}$
Only (S1) is true
Only (S2) is true
Both (S1) and (S2) are true
Both (S1) and (S2) are false
Answer: (c)
Solution
Let the angle be $\theta$ between $\hat{a}$ and $\hat{b}$. $2 + 2\cos\theta + 4\sin^2\theta = 4$ $2 + 2\cos\theta - 4\cos^2\theta = 0$ Let $\cos\theta = t$ then $$2t^2 - t - 1 = 0$$ $$2t^2 - 2t + t - 1 = 0$$ $$2t(t - 1) + (t - 1) = 0$$ $$(2t + 1)(t - 1) = 0$$ $t = -\frac{1}{2}$ or $t = 1$ $\cos\theta = -\frac{1}{2}$ not possible as $\theta \in (0, \pi)$ $$\theta = \frac{2\pi}{3}$$ Now, $S_1$, $2|\hat{a} \times \hat{b}| = 2\sin\left(\frac{2\pi}{3}\right)$ $$|\hat{a} - \hat{b}| = \sqrt{1 + 1 - 2\cos\left(\frac{2\pi}{3}\right)}$$ $$= \sqrt{2 - 2 \times \left(-\frac{1}{2}\right)}$$ $$= \sqrt{3}$$ $S_1$ is correct. $S_2$, projection of $\hat{a}$ on $(\hat{a} + \hat{b})$. $$\frac{\hat{a} \cdot (\hat{a} + \hat{b})}{|\hat{a} + \hat{b}|} = \frac{1 + \cos\left(\frac{2\pi}{3}\right)}{\sqrt{2 + 2\cos\frac{2\pi}{3}}}$$ $$= \frac{1 - \frac{1}{2}}{\sqrt{1}}$$ $$= \frac{1}{2}$$ C Option is true.
Question 18
Maths · Three Dimensional Geometry · Single correct
If the shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{\lambda}$ and $\frac{x-2}{1} = \frac{y-4}{4} = \frac{z-5}{5}$ is $\frac{1}{\sqrt{3}}$, then the sum of all possible values of $\lambda$ is:
Maths · Three Dimensional Geometry · Single correct
Let the points on the plane P be equidistant from the points (-4, 2, 1) and (2, -2, 3). Then the acute angle between the plane P and the plane 2x + y + 3z = 1 is
A random variable X has the following probability distribution: \begin{tabular}{|l|l|l|l|l|l|} \hline X & 0 & 1 & 2 & 3 & 4\\ \hline P(X) & $k$ & $2k$ & $4k$ & $6k$ & $86$\\ \hline \end{tabular} The value of P(1 < X < 4 $\mid$ X $\leq$ 2) is equal to:
$\frac{4}{7}$
$\frac{2}{3}$
$\frac{3}{7}$
$\frac{4}{5}$
Answer: (a)
Solution
The probability is given by $$ \mathrm{P} \left( \frac{1 < x < 4}{x \leq 2} \right) = \frac{\mathrm{P}(1 < x < 4 \cap x \leq 2)}{\mathrm{P}(x \leq 2)} $$ This simplifies to $$ \frac{\mathrm{P}(1 < x \leq 2)}{\mathrm{P}(x \leq 2)} = \frac{\mathrm{P}(x = 2)}{\mathrm{P}(x \leq 2)} $$ Further simplifying gives $$ \frac{4k}{k + 2k + 4k} = \frac{4}{7} $$
Question 21
Maths · Complex Numbers and Quadratic Equations · Numerical
Let\[ S=\left\{z\in\mathbb{C}: |z-3|\le 1 \text{ and } z(4+3i)+\bar{z}(4-3i)\le 24 \right\}. \] If α + iβ is the point in \(S\) which is closest to \(4i\), then \[25(\alpha+\beta)\] is equal to ______.
Answer: 80
Solution
Given $|z - 3| \leq 1$, represent point $z$ inside of a circle of radius 1 centered at $(3, 0)$. $2(4 + 3y) + 4x - 3y \leq 24$ $(x + y + 4) + 3(y - (x - 6)) = 3(0) \leq 24$ $4x + 3x + 4y + 3y + 4x - 3x - 4y - 3y \leq 24$ $8x - 6y \leq 24$ $4x - 3y \leq 12$ Minimum of $(0, 4)$ from circle $= \sqrt{3^2 + 4^2} - 1 = 4$ will lie along line joining $(0, 4)$ and $(3, 0)$. Equation of line $$\frac{x}{3} + \frac{y}{4} = 1 \Rightarrow 4x + 3y = 12 \ldots (i)$$ Equation of circle $(x - 3)^2 + y^2 = 1 \ldots (ii)$ $$\left( \frac{12 - 3y}{4} \right)^2 + y^2 = 1$$ $$\left( \frac{3y}{4} \right)^2 + y^2 = 1$$ $$\frac{25y^2}{16} = 1 \Rightarrow y = \frac{4}{5}$$ For minimum distance $y = \frac{4}{5}$ $$\therefore x = \frac{12}{5}$$ $$\therefore 25(\alpha + \beta) = 25 \left( \frac{4}{5} + \frac{12}{5} \right)$$ $$= 16 \times 5 = 80$$
Question 22
Maths · Permutations and Combinations · Numerical
The number of 7-digit numbers which are multiples of 11 and are formed using all the digits 1, 2, 3, 4, 5, 7 and 9 is
Answer: 576
Solution
Digits are 1, 2, 3, 4, 5, 7, 9. Multiple of 11 implies the difference of the sum at even and odd places is divisible by 11. Let the number be of the form abcdefg. Therefore, $(a + c + e + g) - (b + d + f) = 11x$. $a + b + c + d + e + f = 31$. Therefore, either $a + c + e + g = 21$ or $10$. Therefore, $b + d + f = 10$ or $21$. Case-1: $a + c + e + g = 21$ $b + d + f = 10$ $(b, d, f) \in \{(1, 2, 7), (2, 3, 5), (1, 4, 5)\}$ $(a, c, e, g) \in \{(1, 4, 7, 9), (3, 4, 5, 9), (2, 3, 7, 9)\}$ Therefore, total number in case-1 $= (3! \times 3) (4!) = 432$ Case-2: $a + c + e + g = 10$ $b + d + f = 21$ $(a, b, e, g) \in \{1, 2, 3, 4\}$ $(b, d, f) \in \{(5, 7, 9)\}$ Therefore, total number in case-2 $= 3! \times 4! = 144$ Total numbers $= 144 + 432 = 576$
Question 23
Maths · Binomial Theorem · Numerical
The remainder on dividing $1 + 3 + 3^2 + 3^3 + \ldots + 3^{2021}$ by $50$ is .
Let a circle C : (x - h)^2 + (y - k)^2 = r^2, k > 0, touch the x-axis at (1, 0). If the line x + y = 0 intersects the circle C at P and Q such that the length of the chord PQ is 2, then the value of h + k + r is equal to _____.
Answer: 7
Solution
Given $k = r$ and $h = 1$. We have $OP = r$, $PR = 1$. The expression for $OR$ is $\left| \frac{r+1}{\sqrt{2}} \right|$. The equation for $r^2$ is: $$r^2 = 1 + \frac{(r+1)^2}{2}$$ Expanding and simplifying: $$2r^2 = 2 + r^2 + 1 + 2r$$ $$r^2 - 2r - 3 = 0$$ Factoring gives: $$(r - 3)(r + 1) = 0$$ Thus, $r = 3$ or $r = -1$. Since $r$ must be positive, $r = 3$. Now, $h + k + r = 1 + 3 + 3 = 7$.
Question 25
Maths · Conic Sections · Fill in the blank
Let $P_1$ be a parabola with vertex $(3, 2)$ and focus $(4, 4)$ and $P_2$ be its mirror image with respect to the line $x + 2y = 6$. Then the directrix of $P_2$ is $x + 2y =$ .
Answer: 10
Solution
For $P_1$: Directrix is given by $x + 2y = k$. The equation $x + 2y - k = 0$ is considered. We have: $$\frac{|3 + 4 - K|}{\sqrt{5}} = \sqrt{5}$$ This implies $|7 - k| = 5$. Solving gives $7 - K = 5$ or $7 - K = -5$. Thus, $k = 2$ or $k = 12$. $k = 2$ is accepted, while $k = 12$ is rejected. The directrix passes through the focus. For $D_1 = x + 2y = 2$, $\ell = x + 2y = 6$ implies $\Rightarrow d$. For $D_2 = x + 2y = C$, $\Rightarrow d$ implies $c = 10$.
Question 26
Maths · Conic Sections · Numerical
Let the hyperbola H : $\frac{x^2}{a^2} - y^2 = 1$ and the ellipse $E : 3x^2 + 4y^2 = 12$ be such that the length of latus rectum of H is equal to the length of latus rectum of E. If $e_H$ and $e_E$ are the eccentricities of H and E respectively, then the value of $12(e_H^2 + e_E^2)$ is equal to _____.
Answer: 42
Solution
Given the equations $\($ $\frac{x^2}{a^2}$ - $\frac{y^2}{1}$ = 1 $\)$ and $\($ $\frac{x^2}{4}$ + $\frac{y^2}{3}$ = 1 $\)$. The eccentricity for the hyperbola is $\($ e_H = $\sqrt{1 + \frac{1}{a^2}}$ $\)$. The eccentricity for the ellipse is $\($ e_E = $\sqrt{1 - \frac{3}{4}}$ = $\frac{1}{2}$ $\)$. The relation $\($ $\ell$ R = $\frac{2}{a}$ $\)$ and $\($ $\ell$ R = $\frac{2 \times 3}{2}$ = 3 $\)$. Thus, $\($ $\frac{2}{a}$ = 3 $\)$. Solving for $\($ a $\)$, we get $\($ a = $\frac{2}{3}$ $\)$. Now, $\($ e_H = $\sqrt{1 + \frac{9}{4}}$ = $\frac{\sqrt{13}}{2}$ $\)$. Calculating $\($ 12(e_H^2 + e_E^2) = 12 $\left$( $\frac{13}{4}$ + $\frac{1}{4}$ $\right$) $\)$. This simplifies to $\($ $\frac{12 \times 14}{4}$ = 42 $\)$.
Question 27
Maths · Permutations and Combinations · Numerical
The sum of all the elements of the set $\{ \alpha \in \{ 1, 2, \ldots, 100 \} : \mathrm{HCF} (\alpha, 24) = 1 \}$ is _____.
Answer: 1633
Solution
HCF $(\alpha,24)=1$ Now, $24=2^3\cdot3$ $\Rightarrow\ \alpha$ is not the multiple of $2$ or $3$ Sum of values of $\alpha$ $=S(U)-\{S(\text{multiple of }2)+S(\text{multiple of }3)\}$ $+S(\text{multiple of }6)$ $=(1+2+3+\cdots+100)$ $-(2+4+6+\cdots+100)$ $-(3+6+9+\cdots+99)$ $+(6+12+\cdots+96)$ $=\frac{100\times101}{2}$ $-50\times51$ $-\frac{33}{2}(3+99)$ $+\frac{16}{2}(6+96)$ $=5050-2550-1683+816$ $=1633$
Question 28
Maths · Matrices · Numerical
Let $S = \left\{ \begin{pmatrix} -1 & a \\ 0 & b \end{pmatrix} ; a, b \in \{1, 2, 3, \ldots, 100\} \right\}$ and let $T_n = \{ A \in S : A^{n(n+1)} = I \}$. Then the number of elements in $\bigcap_{n=1}^{100} T_n$ is .
Answer: 100
Solution
Given $$A = \begin{bmatrix} -1 & a \\ 0 & b \end{bmatrix}$$ Then $$A^2 = \begin{bmatrix} -1 & a \\ 0 & b \end{bmatrix} \begin{bmatrix} -1 & a \\ 0 & b \end{bmatrix}$$ This simplifies to $$= \begin{bmatrix} 1 & -a + ab \\ 0 & b^2 \end{bmatrix}$$ Therefore, $$T_n = \{ A \in S; \; A^{(n+1)} = I \}$$ Thus, $b$ must be equal to 1. In this case $A^2$ will become the identity matrix and $a$ can take any value from 1 to 100. The total number of common elements will be 100.
Question 29
Maths · Applications of Integrals · Numerical
The area (in sq. units) of the region enclosed between the parabola $y^2 = 2x$ and the line $x + y = 4$ is _____.
Answer: 18
Solution
Given the equations: $x = 4 - y$ $y^2 = 2(4 - y)$ $y^2 = 8 - 2y$ $y^2 + 2y - 8 = 0$ $y = -4, y = 2$ $x = 8, x = 2$ The integral to find the area is: $$\int_{-4}^{2} \left[ (4 - y) - \frac{y^2}{2} \right] \, dy$$ Evaluating the integral: $$= \left[ 4y - \frac{y^2}{2} - \frac{y^3}{6} \right]_{-4}^{2}$$ $$= 8 - 2 - \frac{8}{6} + 16 + \frac{16}{2} - \frac{64}{6}$$ $$= 22 + 8 - \frac{72}{6}$$ $$= 30 - 12 = 18$$
Question 30
Maths · Probability · Numerical
In an examination, there are 10 true-false type questions. Out of 10, a student can guess the answer of 4 questions correctly with probability $\frac{3}{4}$ and the remaining 6 questions correctly with probability $\frac{1}{4}$. If the probability that the student guesses the answers of exactly 8 questions correctly out of 10 is $\frac{27k}{4^{10}}$, then k is equal to _____.
Physics · Physical World, Units and Measurements · Single correct
Identify the pair of physical quantities that have same dimensions:
velocity gradient and decay constant
wien's constant and Stefan constant
angular frequency and angular momentum
wave number and Avogadro number
Answer: (a)
Solution
Velocity gradient is given by $$\frac{\mathrm{d}V}{\mathrm{d}x} = \frac{1}{S}$$ Therefore, $$\lambda = \frac{1}{S}$$
Question 32
Physics · Motion in a Straight Line · Single correct
An object of mass 5 kg is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of 10 N throughout the motion. The ratio of time of ascent to the time of descent will be equal to: [Use $g = 10 \, \mathrm{ms^{-2}}$]
1 : 1
$\sqrt{2}$ : $\sqrt{3}$
$\sqrt{3}$ : $\sqrt{2}$
2 : 3
Answer: (b)
Solution
The equation given is $6t_1^2 = 4t_2^2$.
Question 33
Physics · Laws of Motion · Single correct
A stone of mass $m$, tied to a string is being whirled in a vertical circle with a uniform speed. The tension in the string is :
the same throughout the motion
minimum at the highest position of the circular path
minimum at the lowest position of the circular path
minimum when the rope is in the horizontal position
Answer: (b)
Solution
Theory
Question 34
Physics · Work, Energy and Power · Single correct
Potential energy as a function of r is given by $$U = \frac{A}{r^{10}} - \frac{B}{r^5}$$, where r is the interatomic distance, A and B are positive constants. The equilibrium distance between the two atoms will be:
$\left(\frac{A}{B}\right)^{\frac{1}{5}}$
$\left(\frac{B}{A}\right)^{\frac{1}{5}}$
$\left(\frac{2A}{B}\right)^{\frac{1}{5}}$
$\left(\frac{B}{2A}\right)^{\frac{1}{5}}$
Answer: (c)
Solution
Given the equation $$\frac{-10A}{r^{11}} + \frac{5B}{r^6} = 0$$ we solve for $r$. Rearranging terms gives $$\frac{10A}{r^{11}} = \frac{5B}{r^6}.$$ Cross-multiplying yields $$r^5 = \frac{10A}{5B} = \frac{2A}{B}.$$
Question 35
Physics · System of Particles and Rotational Motion · Single correct
A fly wheel is accelerated uniformly from rest and rotates through 5 rad in the first second. The angle rotated by the fly wheel in the next second, will be :
7.5 rad
15 rad
20 rad
30 rad
Answer: (b)
Solution
Given the equation $5 = \frac{1}{2} \alpha (1)^2$. Then, $\theta = \frac{1}{2} \alpha (2)^2$. Solving for $\theta$, we have $\theta - 5 = 15$.
Question 36
Physics · Gravitation · Single correct
The distance between Sun and Earth is $R$. The duration of year if the distance between Sun and Earth becomes $3R$ will be:
$\sqrt{3}$ years
3 years
9 years
$3\sqrt{3}$ years
Answer: (d)
Solution
Given $$T' = T \left( \frac{3R}{R} \right)^{3/2} = 3 \sqrt{3} T$$
Question 37
Physics · Thermal Properties of Matter · Single correct
A $100\,\mathrm{g}$ iron nail is hit by a $1.5\,\mathrm{kg}$ hammer striking at a velocity of $60\,\mathrm{ms^{-1}}$. What will be the rise in the temperature of the nail if one fourth of the energy of the hammer goes into heating the nail? [Specific heat capacity of iron $= 0.42\,\mathrm{J\,g^{-1}\,^\circ C^{-1}}$]
A Carnot engine take 5000 kcal of heat from a reservoir at 727°C and gives heat to a sink at 127°C. The work done by the engine is :
$3 \times 10^6\,\mathrm{J}$
Zero
$12.6 \times 10^6\,\mathrm{J}$
$8.4 \times 10^6\,\mathrm{J}$
Answer: (c)
Solution
Question 39
Physics · Oscillations · Single correct
Two massless springs with spring constants $2 \, k$ and $2 \, k$, carry $50 \, \mathrm{g}$ and $100 \, \mathrm{g}$ masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then, the ratio of their respective amplitudes will be:
Physics · Electric Charges and Fields · Single correct
Two identical charged particles each having a mass $10 \, \mathrm{g}$ and charge $2.0 \times 10^{-7} \, \mathrm{C}$ are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is $0.25$, find the value of L. [Use $g = 10 \, \mathrm{ms^{-2}}$]
$12\,\mathrm{cm}$
$10\,\mathrm{cm}$
$8\,\mathrm{cm}$
$5\,\mathrm{cm}$
Answer: (a)
Solution
Given $\($ $\frac{kq^2}{L^2}$ = $\mu$ mg $\)$, it follows that $\($ L = $\sqrt{\frac{k}{\mu mg}}$ q $\)$.
Question 42
Physics · Electric Charges and Fields · Single correct
A long cylindrical volume contains a uniformly distributed charge of density $\rho$. The radius of cylindrical volume is $R$. A charge particle $(q)$ revolves around the cylinder in a circular path. The kinetic of the particle is :
$\frac{\rho q R^2}{4 \varepsilon_0}$
$\frac{\rho q R^2}{2 \varepsilon_0}$
$\frac{q \rho}{4 \varepsilon_0 R^2}$
$\frac{4 \varepsilon_0 R^2}{q \rho}$
Answer: (a)
Solution
Given the equation for electric field, $$E = 2 \pi r \ell = \frac{\rho \pi r^2 \ell}{\varepsilon_0}$$ we have the expression for charge times electric field, $$qE = \frac{q \rho R^2}{2 \varepsilon_0 r} = \frac{mv^2}{r}$$ Solving for $mv^2$, we get $$mv^2 = \frac{q \rho R^2}{2 \varepsilon_0}$$
Question 43
Physics · Electrostatic Potential and Capacitance · Single correct
If the charge on a capacitor is increased by 2 \, $\mathrm{C}$, the energy stored in it increases by 44$\%$. The original charge on the capacitor is (in $\mathrm{C}$):
10
20
30
40
Answer: (a)
Solution
Given $U \propto q^2$. Therefore, $q_f = 1.2 \, q$. We have $q_f - q = 2$. Substituting, $1.2 \, q - q = 2$. Solving, $q = 10$.
Question 44
Physics · Current Electricity · Single correct
What will be the most suitable combination of three resistors $A = 2\,\Omega$, $B = 4\,\Omega$, $C = 6\,\Omega$ so that $\left(\frac{22}{3}\right)\,\Omega$ is equivalent resistance of combination?
Parallel combination of A and C connected in series with B.
Parallel combination of A and B connected in series with C.
Series combination of A and C connected in parallel with B.
Series combination of B and C connected in parallel with A.
The soft-iron is a suitable material for making an electromagnet. This is because soft-iron has:
low coercively and high retentively
low coercively and low permeability
high permeability and low retentively
high permeability and high retentively
Answer: (c)
Solution
Theory
Question 46
Physics · Moving Charges and Magnetism · Single correct
A proton, a deuteron and an $\alpha$-particle with same kinetic energy enter into a uniform magnetic field at right angle to magnetic field. The ratio of the radii of their respective circular paths is :
1 : $\sqrt{2}$ : $\sqrt{2}$
1 : 1 : $\sqrt{2}$
$\sqrt{2}$ : 1 : 1
1 : $\sqrt{2}$ : 1
Answer: (d)
Solution
Given $$R = \frac{\sqrt{2km}}{qB} \propto \frac{\sqrt{m}}{q}$$. The ratio is $$\frac{\sqrt{m}}{e} : \frac{\sqrt{2m}}{e} : \frac{\sqrt{4m}}{2e}$$ which simplifies to $$1 : \sqrt{2} : 1$$.
Question 47
Physics · Alternating Current · Single correct
Given below are two statements: Statement-I: The reactance of an ac circuit is zero. It is possible that the circuit contains a capacitor and an inductor. Statement-II: In ac circuit, the average power delivered by the source never becomes zero. In the light of the above statements, choose the correct answer from the options given below:
Both Statement I and Statement II are true.
Both Statement I and Statement II are false.
Statement I is true but Statement II is false.
Statement I is false but Statement II is true.
Answer: (c)
Solution
If $R = 0$, $P = 0$.
Question 48
Physics · Electromagnetic Waves · Single correct
An electric bulb is rated as 200 W. What will be the peak magnetic field at 4 m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5$\%$ efficiency.
Physics · Dual Nature of Radiation and Matter · Single correct
The light of two different frequencies whose photons have energies $3.8 \, \mathrm{eV}$ and $1.4 \, \mathrm{eV}$ respectively, illuminate a metallic surface whose work function is $0.6 \, \mathrm{eV}$ successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be:
1 : 1
2 : 1
4 : 1
1 : 4
Answer: (b)
Solution
The expression is simplified as follows: $$\sqrt{\frac{3.8 - 0.6}{1.4 - 0.6}} = \sqrt{\frac{3.2}{0.8}} = 2$$
Question 50
Physics · Atoms · Single correct
In Bohr's atomic model of hydrogen, let K, P and E are the kinetic energy, potential energy and total energy of the electron respectively. Choose the correct option when the electron undergoes transitions to a higher level:
All K, P and E increase.
K decreases, P and E increase.
P decreases, K and E increase.
K increases, P and E decrease.
Answer: (b)
Solution
Based on theory
Question 51
Physics · Motion in a Plane · Numerical
A body is projected from the ground at an angle of $45^\circ$ with the horizontal. Its velocity after $2\, \mathrm{s}$ is $20\, \mathrm{ms}^{-1}$. The maximum height reached by the body during its motion is ______m. (use $g = 10\, \mathrm{ms}^{-2}$)
Physics · Thermal Properties of Matter · Numerical
In an experiment to verify Newton's law of cooling, a graph is plotted between, the temperature difference ($\Delta T$) of the water and surroundings and time as shown in figure. The initial temperature of water is taken as $80^\circ \mathrm{C}$. The value of $t_2$ as mentioned in the graph will be _______.
Answer: 16
Solution
Given the equation $T - T_0 = (T_i - T_0) e^{-\frac{Bt}{ms}}$. We have $6\lambda = \ln 1.5$. From $40 = 60e^{-\lambda(6)}$, it follows that $6\lambda = \ln 1.5$. From $20 = 60e^{-\lambda t_2}$, it follows that $t_2 \lambda = \ln 3$. Therefore, $$\frac{t_2}{6} = \frac{\ln 3}{\ln 1.5}$$ Thus, $t_2 = 16.25$ min. So $\approx 16$.
Question 53
Physics · Thermodynamics · Fill in the blank
A monoatomic gas performs a work of $\frac{Q}{4}$ where $Q$ is the heat supplied to it. The molar heat capacity of the gas will be ______R during this transformation. Where $R$ is the gas constant.
Answer: 2
Solution
Question 54
Physics · Waves · Numerical
Two travelling waves of equal amplitudes and equal frequencies move in opposite directions along a string. They interfere to produce a stationary wave whose equation is given by $$y = (10 \cos \pi x \sin \frac{2\pi t}{T}) cm$$ The amplitude of the particle at $x = \frac{4}{3}$ cm will be
Answer: 5
Solution
Calculate the expression: $$10 \cos \left( \frac{4\pi}{3} \right)$$
Question 55
Physics · Current Electricity · Numerical
A potentiometer wire of length 10 m and resistance 20 $\Omega$ is connected in series with a 25 V battery and an external resistance 30 $\Omega$ . A cell of emf E in secondary circuit is balanced by 250 cm long potentiometer wire. The value of E (in volt) is $\frac{x}{10}$. The value of x is
Answer: 25
Solution
Given the circuit, the current $I$ is calculated as follows: $$I = \frac{25}{50} = \frac{1}{2} \, \mathrm{A}$$ Therefore, the voltage difference $\Delta V$ is $10 \, \mathrm{V}$. For $10 \, \mathrm{m}$, the voltage is $10 \, \mathrm{V}$. For $2.5 \, \mathrm{m}$, the voltage is $2.5 \, \mathrm{V}$.
Question 56
Physics · Electromagnetic Induction · Numerical
A circular coil of 1000 turns each with area $1 \, \mathrm{m}^2$ is rotated about its vertical diameter at the rate of one revolution per second in a uniform horizontal magnetic field of $0.07 \, \mathrm{T}$. The maximum voltage generation will be _______ V.
Physics · Ray Optics and Optical Instruments · Fill in the blank
A ray of light is incident at an angle of incidence $60^\circ$ on the glass slab of refractive index $\sqrt{3}$. After refraction, the light ray emerges out from other parallel faces and lateral shift between incident ray and emergent ray is $4\sqrt{3} \, \mathrm{cm}$. The thickness of the glass slab is _______ cm.
Answer: 12
Solution
Given $$\ell = t \sin i \left[ 1 - \frac{\cos i}{\sqrt{\mu^2 - \sin^2 i}} \right]$$ It follows that $$4 \sqrt{3} = t \sin 60^\circ \left[ 1 - \frac{\cos 60^\circ}{\sqrt{3 - \frac{3}{4}}} \right]$$
Question 58
Physics · Nuclei · Numerical
A sample contains $10^{-2}$ kg each of two substances A and B with half lives 4 s and 8 s respectively. The ratio of then atomic weights is 1 : 2. The ratio of the amounts of A and B after 16 s is $\frac{x}{100}$. The value of $x$ is _______.
Answer: 25
Solution
Given $$N_t = N_0 \left(0.5\right)^{\frac{t}{t_{1/2}}}$$ This can be rewritten as $$= \frac{m}{M} \times N_A \left(0.5\right)^{\frac{t}{t_{1/2}}}$$ For the ratio $$\frac{N_1}{N_2} = \frac{M_2}{M_1} \left(0.5\right)^t \left[\frac{1}{T_A} - \frac{1}{T_B}\right]$$ Simplifying further, $$= 2 \left(0.5\right)^{16 \times \frac{1}{8}} = \frac{2}{4} = \frac{1}{2} = \frac{x}{100}$$
In the given circuit- the value of current $I_L$ will be _____ mA. (When $R_L = 1 \, \mathrm{k}\Omega$)
Answer: 5
Solution
Given $I_L = \frac{5}{1000} = 5 \, \mathrm{mA}$.
Question 60
Physics · Communication Systems · Numerical
An antenna is placed in a dielectric medium of dielectric constant 6.25. If the maximum size of that antenna is $5.0 \, \mathrm{mm}$, it can radiate a signal of minimum frequency of _____ GHz. (Given $\mu_r = 1$ for dielectric medium)
Answer: 6
Solution
C' = $\frac{C}{\sqrt{\mu_r \varepsilon_r}}$ = $\frac{3 \times 10^8}{\sqrt{6.25}}$ = $\frac{3 \times 10^8}{2.5}$ f $\lambda$ = 1.25 $\times$ 10^8 $\,$ $\mathrm{s}$ $\Rightarrow$ f $\left$(5 $\times$ 10^{-3} $\times$ 4$\right$) = 1.25 $\times$ 10^8 f = 6.25 $\,$ $\mathrm{GHz}$ So f $\approx$ 6
Chemistry
Question 61
Chemistry · Some Basic Concepts of Chemistry · Single correct
120 of an organic compound that contains only carbon and hydrogen gives 330g of $CO_2$ and 270g of water on complete combustion. The percentage of carbon and hydrogen, respectively are.
25 and 75
40 and 60
60 and 40
75 and 25
Answer: (d)
Solution
Given mass of organic compound = 120 mass of $\mathrm{CO_2(g)} = 330 \, \mathrm{g}$ mass of $\mathrm{H_2O (\ell)} = 270 \, \mathrm{g}$ mass of carbon = $n_{\mathrm{CO_2}} \times 12$ $$= \frac{330}{44} \times 12 = 90 \, \mathrm{g}$$ % of carbon = $\frac{90}{120} \times 100 = 75\%$ mass of hydrogen = $n_{\mathrm{H_2O}} \times 2$ $$= \frac{270}{18} \times 2 = 30 \, \mathrm{g}$$ % of hydrogen = $\frac{30}{120} \times 100 = 25\%$
Question 62
Chemistry · Structure of Atom · Single correct
The energy of one mole of photons of radiation of wavelength 300 nm is (Given : h = 6.63 $\times$ $10^{-34}$ $\mathrm{Js}$, $N_A = 6.02 \times 10^{23}$ $\mathrm{mol}^{-1}$, c = 3 $\times$ $10^8$ $\mathrm{ms}^{-1}$)
235 $\mathrm{kJ}$ $\mathrm{mol}^{-1}$
325 $\mathrm{kJ}$ $\mathrm{mol}^{-1}$
399 $\mathrm{kJ}$ $\mathrm{mol}^{-1}$
435 $\mathrm{kJ}$ $\mathrm{mol}^{-1}$
Answer: (c)
Solution
Energy of one mole of photons $=\frac{hc}{\lambda}\times N_A$ $=\frac{6.63\times10^{-34}\times3\times10^8}{300\times10^{-9}}\times6.02\times10^{23}$ $=399.13\times10^3\ \mathrm{Joule/mole}$ $=399\ \mathrm{kJ/mole}$
Question 63
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Metals generally melt at very high temperature. Amongst the following, the metal with the highest melting point will be
Hg
Ag
Ga
Cs
Answer: (b)
Solution
Hg, Ga, Cs are liquid near room temperature. But Ag (silver) is solid.
Question 64
Chemistry · Chemical Bonding and Molecular Structure · Single correct
The correct order of bond orders of $\mathrm{C}_2^{2-}$, $\mathrm{N}_2^{2-}$ and $\mathrm{O}_2^{2-}$ is, respectively.
At $25^\circ\mathrm{C}$ and $1\,\mathrm{atm}$ pressure, the enthalpies of combustion are as given below: $$ \begin{array}{c|ccc} \text{Substance} & \mathrm{H_2} & \mathrm{C(graphite)} & \mathrm{C_2H_6(g)}\\ \hline \Delta_c H^\circ\,(\mathrm{kJ\,mol^{-1}}) & -286.0 & -394.0 & -1560.0 \end{array} $$ The enthalpy of formation of ethane is
$+54.0\,\mathrm{kJ\,mol^{-1}}$
$-68.0\,\mathrm{kJ\,mol^{-1}}$
$-86.0\,\mathrm{kJ\,mol^{-1}}$
$+97.0\,\mathrm{kJ\,mol^{-1}}$
Answer: (c)
Question 66
Chemistry · The s-Block Elements · Single correct
Which one of the following compounds is used as a chemical in certain type of fire extinguishers?
Baking Soda
Soda ash
Washing Soda
Caustic Soda
Answer: (a)
Solution
Sodium hydrogencarbonate (Baking soda), $\mathrm{NaHCO_3}$, is used in the fire extinguishers.
Question 67
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Arrange the following carbocations in decreasing order of stability.
A > C > B
A > B > C
C > B > A
C > A > B
Answer: (a)
Solution
Carbocation is stabilised by resonance with lone pairs on oxygen atom and +H effect of 2 hydrogens. B > A > C
Question 68
Chemistry · Hydrocarbons · Single correct
Given below are two statements. Statement I : The presence of weaker $\pi$- bonds make alkenes less stable than alkanes. Statement II : The strength of the double bond is greater than that of carbon-carbon single bond. In the light of the above statements, choose the correct answer from the options given below.
Both Statement I and Statement II are correct.
Both Statement I and Statement II are incorrect.
Statement I is correct but Statement II is incorrect.
Statement I is incorrect but Statement II is correct.
Answer: (a)
Solution
answer is a
Question 69
Chemistry · Hydrocarbons · Single correct
Which of the following reagents/ reactions will convert 'A' to 'B'?
PCC oxidation
Ozonolysis
BH$_3$, H$_2$O$_2$/OH$^-$ followed by PCC oxidation
HBr, hydrolysis followed by oxidation by K$_2$Cr$_2$O$_7$.
Answer: (c)
Solution
Question 70
Chemistry · Environmental Chemistry · Single correct
Some gases are responsible for heating of atmosphere (green house effect). Identify from the following the gaseous species which does not cause it.
CH_4
O_3
H_2O
N_2
Answer: (d)
Solution
$\mathrm{CH_4}$, $\mathrm{O_3}$ and $\mathrm{H_2O}$ cause global warming at the tropospheric level, whereas $\mathrm{N_2}$ does not.
Question 71
Chemistry · Hydrogen · Single correct
In the industrial production of which of the following, molecular hydrogen is obtained as a byproduct?
NaOH
NaCl
Na metal
Na$_2$CO$_3$
Answer: (a)
Solution
Sodium hydroxide is generally prepared commercially by electrolysis of sodium chloride in castner Kellner cell. At cathode: $\mathrm{Na} + e^- \xrightarrow{\mathrm{Hg}} \mathrm{Na} - amalgam$ Anode: $\mathrm{Cl}^- \rightarrow \frac{1}{2} \mathrm{Cl}_2 + e^-$ The Na-amalgam is treated with water to give sodium hydroxide and hydrogen gas: $$2\mathrm{Na} (amalgam) + \mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{NaOH} + \mathrm{H}_2 + 2\mathrm{Hg}$$
Question 72
Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct
For a first order reaction, the time required for completion of $90\%$ reaction is $x$ times the half life of the reaction. The value of $x$ is (Given: $\ln 10=2.303$ and $\log 2=0.3010$)
$1.12$
$2.43$
$3.32$
$33.31$
Answer: (c)
Question 73
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Which of the following chemical reactions represents Hall-Heroult Process?
Hall Heroult process is the major industrial process for extraction of aluminium.
Question 74
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
$PCl_5$ is well known, but $NCl_5$ is not. Because.
nitrogen is less reactive than phosphorous.
nitrogen doesn't have d-orbitals in its valence shell.
catenation tendency is weaker in nitrogen than phosphorous.
size of phosphorous is larger than nitrogen.
Answer: (b)
Solution
$PCl_5$ forms five bonds by using the d–orbitals to "expand the octet". But $NCl_5$ does not exist because there are no d–orbitals in the valence shell ($2^{nd}$ shell). Therefore there is no way to expand the octet.
Question 75
Chemistry · Co-ordination Compounds · Single correct
Transition metal complex with highest value of crystal field splitting ($\Delta_0$) will be
CFSE of octahedral complexes with water is greater for 5d series metal centre ion as compared to 3d and 4d series metal centre.
Question 76
Chemistry · Alcohols, Phenols and Ethers · Single correct
Hex-4-ene-2-ol on treatment with PCC gives 'A'. 'A' on reaction with sodium hypoiodite gives 'B', which on further heating with soda lime gives 'C'. The compound 'C' is
2-pentene
propionaldehyde
2-butene
4-methylpent-2-ene
Answer: (c)
Solution
The reaction sequence is as follows: Starting with $\mathrm{CH_3{-}CH{=}CH{-}CH_2{-}CH(OH){-}CH_3}$, the first step involves PCC oxidation: $$\mathrm{CH_3{-}CH{=}CH{-}CH_2{-}CH(OH){-}CH_3} \xrightarrow{\mathrm{PCC}} \mathrm{CH_3{-}CH{=}CH{-}CH_2{-}C(=O){-}CH_3} (\mathrm{A})$$ Next, the compound reacts with $\mathrm{NaOI}$: $$\mathrm{CH_3{-}CH{=}CH{-}CH_2{-}C(=O){-}CH_3} \xrightarrow{\mathrm{NaOI}} \mathrm{CH_3{-}CH{=}CH{-}CH_2{-}COOH + CHI_3} (\mathrm{B})$$ Finally, decarboxylation occurs with $\mathrm{NaOH + CaO}$: $$\mathrm{CH_3{-}CH{=}CH{-}CH_2{-}COOH} \xrightarrow{\mathrm{NaOH + CaO}} \mathrm{CH_3{-}CH{=}CH{-}CH_3} (\mathrm{C})$$ The final product is But-2-ene.
Question 77
Chemistry · Amines · Single correct
The conversion of propan-1-ol to n-butylamine involves the sequential addition of reagents. The correct sequential order of reagents is.
SOCl$_2$ (ii) KCN (iii) H$_2$/Ni,Na(Hg)/C$_2$H$_5$OH
HCl (ii) H$_2$/Ni, Na(Hg)/C$_2$H$_5$OH
SOCl$_2$ (ii) KCN (iii) CH$_3$NH$_2$
HCl (ii) CH$_3$NH$_2$
Answer: (a)
Solution
Question 78
Chemistry · Polymers · Single correct
Which of the following is not an example of a condensation polymer?
Nylon 6,6
Decron
Buna-N
Silicone
Answer: (c)
Solution
Buna-N is an addition copolymer of 1,3-butadiene and acrylonitrile.
Question 79
Chemistry · Chemistry in Everyday Life · Single correct
The structure shown below is of which well-known drug molecule?
Ranitidine
Seldane
Cimetidine
Codeine
Answer: (c)
Solution
Cimetidine
Question 80
Chemistry · The s-Block Elements · Single correct
In the flame test of a mixture of salts, a green flame with blue centre was observed. Which one of the following cations may be present?
$\mathrm{Cu}^{2+}$
$\mathrm{Sr}^{2+}$
$\mathrm{Ba}^{2+}$
$\mathrm{Ca}^{2+}$
Answer: (a)
Solution
Ion and Colour of the flame: (A) $\mathrm{Cu^{+2}}$: green flame with blue centre (B) $\mathrm{Sr^{2+}}$: Crimson Red (C) $\mathrm{Ba^{2+}}$: Apple green
Question 81
Chemistry · States of Matter · Numerical
At 300 $\mathrm{K}$, a sample of 3.0 $\mathrm{g}$ of gas A occupies the same volume as 0.2 $\mathrm{g}$ of hydrogen at 200 $\mathrm{K}$ at the same pressure. The molar mass of gas A is ____ $\mathrm{g}$ $\mathrm{mol^{-1}}$ (nearest integer) Assume that the behaviour of gases as ideal. (Given: The molar mass of hydrogen $(H_2)$ gas is 2.0 $\mathrm{g}$ $\mathrm{mol^{-1}}$)
Answer: 45
Solution
Given: Ideal gas A and $\mathrm{H_2}$ gas at same pressure and volume. From ideal gas equation $pv = nRT$ $$n_1 T_1 = n_2 T_2$$ $$\frac{3}{GMM of A} \times 300 = \frac{0.2}{2} \times 200$$ GMM of A = 45 g/mole
Question 82
Chemistry · Equilibrium · Numerical
PCl$_5$ dissociates as PCl$_5$ (g) $\rightleftharpoons$ PCl$_3$ (g) + Cl$_2$ (g) 5 moles of PCl$_5$ are placed in a 200 litre vessel which contains 2 moles of N$_2$ and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant $K_p$ for the dissociation of PCl$_5$ is ______ $\times 10^{-3}$. (nearest integer) (Given: R = 0.082 L atm K$^{-1}$ mol$^{-1}$ : Assume ideal gas behaviour)
Answer: 1107
Solution
Given: 2 mole of $\mathrm{N_2}$ gas was present as inert gas. Equilibrium pressure = 2.46 atm. $\mathrm{PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2 (g)}$ At $t = 0$: 5, 0, 0 At $t = Eqm$: $5-x$, $x$, $x$ From ideal gas equation: $$PV = nRT$$ $$2.46 \times 200 = (5 - x + x + x + 2) \times 0.082 \times 600$$ $$x = 3$$ $$K_p = \frac{n_{\mathrm{PCl_3}} \times n_{\mathrm{Cl_2}}}{n_{\mathrm{PCl_5}}} \times \left[ \frac{P_{total}}{n_{total}} \right]$$ $$\frac{3 \times 3}{2} \times \frac{2.46}{10} = 1.107 = 1107 \times 10^{-3}$$
Question 83
Chemistry · Redox Reactions · Numerical
Manganese (VI) has ability to disproportionate in acidic solution. The difference in oxidation states of two ions it forms in acidic solution is
Answer: 3
Solution
$\mathrm{MnO_4^{2-}}$ disproportionates in a neutral or acidic solution to give $\mathrm{MnO_4^-}$ and $\mathrm{Mn^{4+}}$. $$3\mathrm{MnO_4^{2-}} + 3\mathrm{H^+} \longrightarrow 2\mathrm{MnO_4^-} + \mathrm{MnO_2} + 2\mathrm{H_2O}$$ O.S. of Mn in $\mathrm{MnO_4^-} = +7$ O.S. of Mn in $\mathrm{MnO_2} = +4$ difference $= 3$
0.2 $\mathrm{g}$ of an organic compound was subjected to estimation of nitrogen by Dumas method in which volume of N_2 evolved (at STP) was found to be 22.400 $\mathrm{mL}$. The percentage of nitrogen in the compound is . [nearest integer] (Given: Molar mass of $N_2$ is 28 $\mathrm{mol^{-1}}$. Molar volume of $N_2$ at STP : 22.4 $\mathrm{L}$)
Answer: 14
Solution
Weight of organic compound = 0.2 $\mathrm{g}$. Mass of $\mathrm{N_2(g)}$ evolved = $\frac{22.4 \times 10^{-3}}{22.4}$ $\times$ 28. = 28 $\times$ 10^{-3} $\mathrm{g}$. $\%$ of $\mathrm{N}$ = $\frac{28 \times 10^{-3}}{0.2}$ $\times$ 100 = 14.
Question 85
Chemistry · Solutions · Numerical
A company dissolves 'X' amount of $\mathrm{CO_2}$ at $298 \, \mathrm{K}$ in $1 \, \mathrm{litre}$ of water to prepare soda water $X =.... \times 10^{-3} \, \mathrm{g}$. (nearest integer) (Given: partial pressure of $\mathrm{CO_2}$ at $298 \, \mathrm{K} = 0.835 \, \mathrm{bar}$. Henry's law constant for $\mathrm{CO_2}$ at $298 \, \mathrm{K} = 1.67 \, \mathrm{kbar}$. Atomic mass of H,C and O is $1, 12$ and $6 \, \mathrm{g} \, \mathrm{mol}^{-1}$, respectively)
The resistance of conductivity cell containing 0.01 M KCl solution at 298 K is 1750 Ω. If the conductivity of 0.01 M KCl solution at 298 K is $0.152 \times 10^{-3} \, \mathrm{S} \, \mathrm{cm}^{-1}$, then the cell constant of the conductivity cell is _______ $\times 10^{-3} \, \mathrm{cm}^{-1}$.
Answer: 266
Solution
Given the equation for conductivity, $$K = \frac{1}{R} \times cell constant$$. Substituting the given values, $$0.152 \times 10^{-3} = \frac{1}{1750} \times cell constant$$ Solving for the cell constant, $$cell constant = 266 \times 10^{-3}$$
Question 87
Chemistry · Surface Chemistry · Numerical
When 200 $mL$ of 0.2 M acetic acid is shaken with 0.6 g of wood charcoal, the final concentration of acetic after adsorption is 0.1 M. The mass of acetic acid adsorbed per gram of carbon is_____ g.
Answer: 2
Solution
Weight of wood charcoal = 0.6 $\mathrm{g}$ Mass of acetic acid adsorbed = $\frac{M_1 V_1 - M_2 V_2}{1000}$ $\times$ 60 $$= \frac{0.2 \times 200 - 0.1 \times 200}{1000} \times 60$$ $$= 1.2 \, \mathrm{g}$$ Mass of acetic acid adsorbed per gram of carbon = $\frac{1.2}{0.6}$ = 2
Question 88
Chemistry · General Principles and Processes of Isolation of Elements · Numerical
Baryte, Galena, Zinc blende and Copper pyrites. How many of these minerals are sulphide based?
Baryte
Galena
Zinc blende
Copper pyrites
Answer: (c)
Solution
The question asks which of the following are sulphide ($\mathrm{S^{2-}}$) ores: (1) Baryte: $\mathrm{BaSO_4}$ (2) Galena: $\mathrm{PbS}$ (3) Zinc blende: $\mathrm{ZnS}$ (4) Copper pyrite: $\mathrm{CuFeS_2}$ Galena, Zinc blende, and Copper pyrite are sulphide ores.
Question 89
Chemistry · Haloalkanes and Haloarenes · Numerical
Consider the above reaction. The number of $\pi$ electrons present in the product 'P' is____.
Answer: 2
Solution
Number of $\pi$ electron $= 2$
Question 90
Chemistry · Biomolecules · Numerical
In alanylglycylleucylalanylvaline, the number of peptide linkages is .
Answer: 4
Solution
There are five amino acids and four peptide linkages.