JEE Main 27 July 2022 Shift 2 question paper with solutions
JEE Main 27 July 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 · Inverse Trigonometric Functions · Single correct
The domain of the function \[ f(x)=\sin^{-1}\!\left([2x^2-3]\right)+\log_{2}\!\left(\log_{\frac12}\!\left(x^2-5x+5\right)\right), \] where $[t]$ is the greatest integer function, is:
Maths · Complex Numbers and Quadratic Equations · Single correct
Let S be the set of all $(\alpha, \beta)$, $\pi < \alpha$, $\beta < 2\pi$, for which the complex number $\frac{1 - i \sin \alpha}{1 + 2 i \sin \alpha}$ is purely imaginary and $\frac{1 + i \cos \beta}{1 - 2 i \cos \beta}$ is purely real. Let $Z_{\alpha \beta} = \sin 2\alpha + i \cos 2\beta$, $(\alpha, \beta) \in S$. Then $$\sum_{(\alpha, \beta) \in S} \left( i Z_{\alpha \beta} + \frac{1}{i\overline{Z_{\alpha\beta}}} \right)$$ is equal to:
Let the sum of an infinite G.P., whose first term is $a$ and the common ratio is $r$, be $5$. Let the sum of its first five terms be $\frac{98}{25}$. Then the sum of the first $21$ terms of an AP, whose first term is $10ar$, $n^{th}$ term is $a_n$ and the common difference is $10ar^2$, is equal to:
21 a_{11}
22 a_{11}
15 a_{16}
14 a_{16}
Answer: (a)
Solution
Given $$S_{21} = \frac{21}{2} [20ar + 20 \cdot 10 ar^2]$$ Simplifying, we have $$= 21 [10 ar + 100 ar^2]$$ Therefore, $$= 21 \cdot a_{11}$$
Question 8
Maths · Applications of Integrals · Single correct
The area of the region enclosed by $y \leq 4x^2$, $x^2 \leq 9y$ and $y \leq 4$, is equal to:
$\frac{40}{3}$
$\frac{56}{3}$
$\frac{112}{3}$
$\frac{80}{3}$
Answer: (d)
Solution
The area $\Delta$ is calculated as follows: $$\Delta = 2 \cdot \int_0^4 \left( 3 \sqrt{y} - \frac{\sqrt{y}}{2} \right) \, dy$$ Simplifying the integrand: $$= 2 \cdot \int_0^4 \frac{5}{2} \sqrt{y} \, dy = \frac{80}{3}$$
Question 9
Maths · Integrals · Single correct
\[ \int_{0}^{2} \left( \left|2x^2-3x\right| +\left[x-\frac{1}{2}\right] \right)\,dx, \] where \([t]\) is the greatest integer function, is equal to:
$\frac{7}{6}$
$\frac{19}{12}$
$\frac{31}{12}$
$\frac{3}{2}$
Answer: (b)
Solution
Evaluate the integral from 0 to 2 of the absolute value of $2x^2 - 3x$ with respect to $x$. $$\int_0^2 |2x^2 - 3x| \, dx$$ This can be split into two integrals: $$= \int_0^{\frac{3}{2}} (3x - 2x^2) \, dx + \int_{\frac{3}{2}}^2 (2x^2 - 3x) \, dx = \frac{19}{12}.$$ Now consider the integral from 0 to 2 of $\left[x - \frac{1}{2}\right]$ with respect to $x$: $$\int_0^2 \left[x - \frac{1}{2}\right] \, dx = \int_{-\frac{1}{2}}^{\frac{3}{2}} [t] \, dt$$ This can be evaluated as: $$= \int_{-\frac{1}{2}}^0 (-1) \, dt + \int_0^1 0 \cdot dt + \int_1^{\frac{3}{2}} 1 \cdot dt = 0.$$
Question 10
Maths · Applications of Derivatives · Single correct
Consider a curve $y = y(x)$ in the first quadrant as shown in the figure. Let the area $A_1$ is twice the area $A_2$. Then the normal to the curve perpendicular to the line $2x - 12y = 15$ does NOT pass through the point.
(6, 21)
(8, 9)
(10, -4)
(12, -15)
Answer: (c)
Solution
Given that $A_1 = 2A_2$ from the graph $A_1 + A_2 = xy - 8$ $$\Rightarrow \frac{3}{2} A_1 = xy - 8$$ $$\Rightarrow A_1 = \frac{2}{3} xy - \frac{16}{3}$$ $$\Rightarrow \int_4^x f(x) \, dx = \frac{2}{3} xy - \frac{16}{3}$$ $$\Rightarrow f(x) = \frac{2}{3} \left( x \frac{dy}{dx} + y \right)$$ $$\Rightarrow \frac{2}{3} x \frac{dy}{dx} = \frac{y}{3}$$ $$\Rightarrow 2 \int \frac{dy}{y} = \int \frac{dx}{x}$$ $$\Rightarrow 2 \ln y = \ln x + \ln c$$ $$\Rightarrow y^2 = cx$$ As $f(4) = 2 \Rightarrow c = 1$ so $y^2 = x$ slope of normal $= -6$ $$y = -6(x) - \frac{1}{2}(-6) - \frac{1}{4}(-6)^3$$ $$\Rightarrow y = -6x + 3 + 54$$ $$\Rightarrow y + 6x = 57$$ Now check options and (C) will not satisfy.
Question 11
Maths · Determinants · Single correct
The equations of the sides AB, BC and CA of a triangle ABC are $2x + y = 0$, $x + py = 39$ and $x - y = 3$ respectively and $P(2, 3)$ is its circumcentre. Then which of the following is NOT true :
$(AC)^2 = 9p$
$(AC)^2 + p^2 = 136$
$32 < area (\Delta ABC) < 36$
$34 < area (\Delta ABC) < 38$
Answer: (d)
Solution
Perpendicular bisector of AB is given by $x + y = 5$. Take the image of A. $$\frac{x - 1}{1} = \frac{y + 2}{1} = \frac{-2(-6)}{2} = 6$$ The coordinates are $(7, 4)$. Solving $7 + 4p = 39$, we find $p = 8$. Solving $x + 8y = 39$ and $y = -2x$, we get $$x = \frac{-39}{15}, y = \frac{78}{15}$$ For $AC^2 = 72 = 9p$, we have $$AC^2 + p^2 = 72 + 64 = 136$$ The area of $\Delta ABC$ is given by $$\Delta ABC = \frac{1}{2} \begin{vmatrix} 1 & -2 & 1 \\ 7 & 4 & 1 \\ -\frac{39}{15} & \frac{78}{15} & 1 \end{vmatrix}$$ $$= \frac{1}{2} \left[ 18 + 18 \times \frac{13}{5} \right]$$ $$= 9 \left[ \frac{18}{5} \right] = \frac{162}{5} = 32.4$$
Question 12
Maths · Conic Sections · Single correct
A circle $C_1$ passes through the origin $O$ and has diameter 4 on the positive $x$-axis. The line $y = 2x$ gives a chord $OA$ of a circle $C_1$. Let $C_2$ be the circle with $OA$ as a diameter. If the tangent to $C_2$ at the point $A$ meets the $x$-axis at $P$ and $y$-axis at $Q$, then $QA : AP$ is equal to:
1 : 4
1 : 5
2 : 5
1 : 3
Answer: (a)
Solution
Given $C_1: x^2 + y^2 - 4x = 0$ and $\tan \theta = 2$. The line $L: 2x - y = 0$ is tangent to the circle $C_1$ at point $A$. The circle $C_2$ is a circle with $OA$ as diameter. So, the tangent at $A$ on $C_2$ is perpendicular to $OR$. Let $OA = \ell$. Therefore, $$\frac{QA}{AP} = \frac{\ell \cot \theta}{\ell \tan \theta} = \frac{1}{\tan^2 \theta} = \frac{1}{4}.$$
Question 13
Maths · Conic Sections · Single correct
If the length of the latus rectum of a parabola, whose focus is $(a, a)$ and the tangent at its vertex is $x + y = a$, is 16, then $|a|$ is equal to:
$2\sqrt{2}$
$2\sqrt{3}$
$4\sqrt{2}$
4
Answer: (c)
Solution
Given the equation $x + y = a$. The absolute value of $|P|$ is given by $$|P| = \left| \frac{a}{\sqrt{2}} \right| = \frac{16}{4} = 4.$$ Therefore, $$|a| = 4\sqrt{2}.$$
Question 14
Maths · Three Dimensional Geometry · Single correct
If the length of the perpendicular drawn from the point $P(a, 4, 2)$, $a > 0$ on the line $$\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1}$$ is $2\sqrt{6}$ units and $Q(\alpha_1, \alpha_2, \alpha_3)$ is the image of the point $P$ in this line, then $a + \sum_{i=1}^{3} \alpha_i$ is equal to:
Maths · Three Dimensional Geometry · Single correct
If the line of intersection of the planes $ax + by = 3$ and $ax + by + cz = 0$, $a > 0$ makes an angle $30^\circ$ with the plane $y - z + 2 = 0$, then the direction cosines of the line are:
$\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0$
$\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0$
$\frac{1}{\sqrt{5}}, -\frac{2}{\sqrt{5}}, 0$
$\frac{1}{2}, -\frac{\sqrt{3}}{2}, 0$
Answer: (b)
Solution
Given $\mathbf{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a & b & 0 \\ a & b & c \end{vmatrix}$ $= bc\hat{i} - ac\hat{j}$ Direction ratios of line are $(b,\ -a,\ 0)$ Direction ratios of normal of the plane are $(0,\ 1,\ -1)$ $$\cos 60^\circ = \left|\frac{-a}{\sqrt{2}\sqrt{b^2 + a^2}}\right| = \frac{1}{2}$$ $$\Rightarrow \left|\frac{a}{\sqrt{a^2 + b^2}}\right| = \frac{1}{\sqrt{2}}$$ $\Rightarrow b = \pm\, a$ So, D.R.'s can be $(\pm\, a,\ -a,\ 0)$ $\therefore$ D.C.'s can be $\pm\left(\dfrac{\pm 1}{\sqrt{2}},\ -\dfrac{1}{\sqrt{2}},\ 0\right)$
Question 16
Maths · Probability · Single correct
Let X have a binomial distribution B(n, p) such that the sum and the product of the mean and variance of X are 24 and 128 respectively. If P(X > n - 3) = $\frac{k}{2^n}$, then k is equal to
A six faced die is biased such that $3 \times \mathrm{P}$(a prime number) $= 6 \times \mathrm{P}$(a composite number) = $2 \times \mathrm{P}(1)$. Let $X$ be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of $X$ is:
$\frac{3}{11}$
$\frac{5}{11}$
$\frac{7}{11}$
$\frac{8}{11}$
Answer: (d)
Solution
Let $\dfrac{P(\text{a prime number})}{2}=k$ $\dfrac{P(\text{a composite number})}{1}=k$ $\dfrac{P(1)}{3}=k$. So, $\($ P(a prime number) = 2k $\)$, $\($ P(a composite number) = k $\)$, and $\($ P(1) = 3k $\)$. And $\($ 3 $\times$ 2k + 2 $\times$ k + 3k = 1 $\)$. $\($ $\Rightarrow$ k = $\frac{1}{11}$ $\)$. $\($ P(success) = P(1 or 4) = 3k + k = $\frac{4}{11}$ $\)$. Number of trials, $\($ n = 2 $\)$. Therefore, mean $\($ = np = 2 $\times$ $\frac{4}{11}$ = $\frac{8}{11}$ $\)$.
Question 18
Maths · Heights and Distances · Single correct
The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizontal ground is $45^\circ$. Let R be a point on AQ and from a point B, vertically above R, the angle of elevation of P is $60^\circ$. If $\angle BAQ = 30^\circ$, AB = d and the area of the trapezium PQRB is $\alpha$, then the ordered pair (d, $\alpha$) is:
If the truth value of the statement $\left( P \land (\sim R) \right) \rightarrow \left( (\sim R) \land Q \right)$ is F, then the truth value of which of the following is F?
$P \lor Q \rightarrow R$
$R \lor Q \rightarrow \sim P$
$\sim (P \lor Q) \rightarrow R$
$\sim (R \lor Q) \rightarrow \sim P$
Answer: (d)
Solution
Given $X \Rightarrow Y$ is false when $X$ is true and $Y$ is false. So, $P \rightarrow T$, $Q \rightarrow F$, $R \rightarrow F$. (A) $P \lor Q \rightarrow \sim R$ is T (B) $R \lor Q \rightarrow \sim P$ is T (C) $\sim (P \lor Q) \rightarrow \sim R$ is T (D) $\sim (R \lor Q) \rightarrow \sim P$ is F
Question 21
Maths · Matrices · Numerical
Consider a matrix $A= \begin{bmatrix} \alpha & \beta & \gamma\\ \alpha^2 & \beta^2 & \gamma^2\\ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{bmatrix}$ where $\alpha,\beta,\gamma$ are three distinct natural numbers. If $\frac{\det(\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(A))))} {(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}} =2^{32}\times3^{16}$, then the number of such $3$-tuples $(\alpha,\beta,\gamma)$ is ______.
Answer: 42
Solution
Given the matrix $A = \begin{bmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ \beta + \gamma & \gamma + \alpha & \alpha + \beta \end{bmatrix}$. Perform the row operation $R_3 \rightarrow R_3 + R_1$. This gives: $$|A| = |\alpha + \beta + \gamma| \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ 1 & 1 & 1 \end{vmatrix}$$ Therefore, $$|A| = (\alpha + \beta + \gamma)(\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)$$ The adjugate of $A$ is given by $|\mathrm{adj} \, A| = |A|^{n-1}$. Thus, $$|\mathrm{adj} \, (\mathrm{adj} \, A)| = |A|^{(n-1)^2}$$ $$|\mathrm{adj} \, (\mathrm{adj} \, (\mathrm{adj} \, A)))| = |A|^{(n-1)^4} = |A|^{24} = |A|^{16}$$ Therefore, $$(\alpha + \beta + \gamma)^{16} = 2^{32} \cdot 3^{16}$$ $$\Rightarrow (\alpha + \beta + \gamma)^{16} = (2^2 \cdot 3)^{16} = (12)^{16}$$ $$\Rightarrow \alpha + \beta + \gamma = 12$$ Given $\alpha, \beta, \gamma \in \mathbb{N}$, we have $$ (\alpha - 1)+(\beta - 1)+(\gamma - 1)=9 $$ Number of all tuples $(\alpha,\beta,\gamma)$ $=\binom{11}{2}$ $=55$ There is $1$ case for $\alpha=\beta=\gamma$ and $12$ cases when any two of $\alpha,\beta,\gamma$ are equal. Therefore, $\text{No. of distinct tuples }(\alpha,\beta,\gamma)$ $=55-13$ $=42$
Question 22
Maths · Relations and Functions · Fill in the blank
The number of functions $f$, from the set $$A = \{ x \in \mathbb{N} : x^2 - 10x + 9 \leq 0 \}$$ to the set $$B = \{ n^2 : n \in \mathbb{N} \}$$ such that $f(x) \leq (x-3)^2 + 1$, for every $x \in A$, is .
Let for the $9^{\text{th}}$ term in the binomial expansion of $(3 + 6x)^n$, in the increasing powers of $6x$, to be the greatest for $x = \frac{3}{2}$, the least value of n is $n_0$. If k is the ratio of the coefficient of $x^6$ to the coefficient of $x^3$, then $k + n_0$ is equal to:
Answer: 24
Solution
Given $(3 + 6x)^n = \binom{n}{0} 3^n + \binom{n}{1} 3^{n-1} (6x)^1 + \ldots$ $T_{r+1} = \binom{n}{r} 3^{n-r} (6x)^r = \binom{n}{r} 3^{n-r} 6^r \cdot x^r$ $$= \binom{n}{r} 3^n \cdot 3^{-r} \cdot 2^r \cdot \left(\frac{3}{2}\right)^r = \binom{n}{r} 3^n \cdot 3^r$$ [for $x = \frac{3}{2}$] $T_9$ is greatest if $x = \frac{3}{2}$ So, $T_9 > T_{10}$ and $T_9 > T_8$ (concept of numerically greatest term) Here, $\frac{T_9}{T_{10}} > 1$ and $\frac{T_9}{T_8} > 1$ $$\Rightarrow \frac{\binom{n}{8} 3^n 3^8}{\binom{n}{9} 3^n 3^9} > 1 and \frac{\binom{n}{8} 3^n 3^8}{\binom{n}{7} 3^n 3^7} > 1$$ and $\frac{\binom{n}{8}}{\binom{n}{7}} > \frac{1}{3}$ $$\Rightarrow \frac{n-7}{8} > \frac{1}{3}$$ $$\Rightarrow \frac{29}{3} < n < 11 \Rightarrow n = 10 = n_0$$ So, in $(3 + 6x)^n$ for $n = n_0 = 10$ i.e., in $(3 + 6x)^{10}$, here $T_{r+1} = \binom{10}{r} 3^{10-r} 6^r x^r$ $T_7 = \binom{10}{6} 3^4 \cdot 6^6 x^6 = 210 \cdot 3^{10} \cdot 2^6 x^6$ $T_4 = \binom{10}{3} 3^7 \cdot 6^3 x^3 = 120 \cdot 3^{10} \cdot 2^3 x^3$ Ratio of coefficient of $x^6$ and coefficient of $x^3 = k$ $$\therefore k = \frac{210 \cdot 3^{10} \cdot 2^6}{120 \cdot 3^{10} \cdot 2^3} = \frac{7}{4} \times 2^3 = 14$$ So, $k + n_0 = 14 + 10 = 24$
Given $$T_n = \frac{2 \sum_{r=1}^{n} (2r)^3 - \left( \sum_{r=1}^{2n} r^3 \right)}{n(4n+3)}$$ This implies $$T_n = n$$ So, $$\sum_{n=1}^{15} T_n = 120$$
Question 25
Maths · Applications of Derivatives · Numerical
A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semi-vertical angle is $\tan^{-1} \frac{3}{4}$. Water is poured in it at a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, when the depth of water in the tank is 4 meters, is ________.
Maths · Continuity and Differentiability · Numerical
For the curve C : $(x^2 + y^2 - 3) + (x^2 - y^2 - 1)^5 = 0$, the value of $3y' - y^3y''$, at the point $(\alpha, \alpha)$, $\alpha > 0$, on C, is equal to .
Let $f(x)=\min\{[x-1],[x-2],\ldots,[x-10]\}$, where $[t]$ denotes the greatest integer $\le t$. Then $$ \int_{0}^{10} f(x)\,dx + \int_{0}^{10} (f(x))^{2}\,dx + \int_{0}^{10} |f(x)|\,dx $$ is equal to ______.
Answer: 385
Solution
Given $f(x) = [x] - 10$. Evaluate the integral from 0 to 10 of $f(x) \, dx$. $$\int_0^{10} f(x) \, dx = -10 - 9 - 8 - \ldots - 1$$ $$= -\frac{10 \cdot 11}{2} = -55$$ Evaluate the integral from 0 to 10 of $(f(x))^2 \, dx$. $$\int_0^{10} (f(x))^2 \, dx = 10^2 + 9^2 + 8^2 + \ldots + 1^2$$ $$= \frac{10 \cdot 11 \cdot 21}{6} = 385$$ Evaluate the integral from 0 to 10 of $|f(x)| \, dx$. $$\int_0^{10} |f(x)| \, dx = 10 + 9 + 8 + \ldots + 1$$ $$= \frac{10 \cdot 11}{2} = 55$$ Combine the results: $$-55 + 385 + 55 = 385$$
Question 28
Maths · Integrals · Numerical
Let f be a differentiable function satisfying $$f(x) = \frac{2}{\sqrt{3}} \int_{0}^{\sqrt{3}} f\left(\frac{\lambda^2 x}{3}\right) d\lambda, x > 0 and f(1) = \sqrt{3}.$$ If $y = f(x)$ passes through the point $(\alpha, 6)$, then $\alpha$ is equal to _____.
A common tangent T to the curves $C_1: \frac{x^2}{4} + \frac{y^2}{9} = 1$ and $C_2: \frac{x^2}{42} - \frac{y^2}{143} = 1$ does not pass through the fourth quadrant. If T touches $C_1$ at $(x_1, y_1)$ and $C_2$ at $(x_2, y_2)$, then $|2x_1 + x_2|$ is equal to ______.
Answer: 20
Solution
Let common tangents are $T_1: y = mx \pm \sqrt{4m^2 + 9}$ and $T_2: y = mx \pm \sqrt{42m^2 - 13}$ So, $4m^2 + 9 = 42m^2 - 143$ $$\Rightarrow 38m^2 = 152$$ $$\Rightarrow m = \pm 2$$ and $c = \pm 5$ For given tangent not pass through $4^{th}$ quadrant $T: y = 2x + 5$ Now, comparing with $\frac{xx_1}{4} + \frac{yy_1}{9} = 1$ We get, $\frac{x_1}{8} = -\frac{1}{5} \Rightarrow x_1 = -\frac{8}{5}$ $$\frac{xx_2}{42} - \frac{yy_2}{143} = 1$$ $2x - y = -5$ we have $$x_2 = -\frac{84}{5}$$
Question 30
Maths · Vector Algebra · Numerical
Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be three non-coplanar vectors such that $\vec{a}$ $\times$ $\vec{b}$ = 4 $\vec{c}$, $\vec{b}$ $\times$ $\vec{c}$ = 9 $\vec{a}$ and $\vec{c}$ $\times$ $\vec{a}$ = $\alpha$ $\vec{b}$, $\alpha$ > 0. If | $\vec{a}$ | + | $\vec{b}$ | + | $\vec{c}$ | = $\frac{1}{36}$, then $\alpha$ is equal to .
Answer: 36
Solution
Given $\vec{a} \times \vec{b} = 4 \vec{c}$, $\Rightarrow \vec{a} \cdot \vec{c} = 0 = \vec{b} \cdot \vec{c}$. $\vec{b} \times \vec{c} = 9 \vec{a}$, $\Rightarrow \vec{a} \cdot \vec{b} = 0 = \vec{a} \cdot \vec{c}$. Therefore, $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular set of vectors. $\Rightarrow |\vec{a}||\vec{b}| = 4|\vec{c}|$, $|\vec{b}||\vec{c}| = 9|\vec{a}|$ and $|\vec{c}||\vec{a}| = \alpha |\vec{b}|$. $\Rightarrow \frac{|\vec{a}|}{|\vec{c}|} = \frac{4}{9} \frac{|\vec{c}|}{|\vec{a}|}$. $\Rightarrow \frac{|\vec{c}|}{|\vec{a}|} = \frac{3}{2}$. If $|\vec{a}| = \lambda$, $|\vec{c}| = \frac{3 \lambda}{2}$ and $|\vec{b}| = 6$. Now $|\vec{a}| + |\vec{b}| + |\vec{c}| = \frac{1}{36}$. $\Rightarrow \frac{5}{2} \lambda + 6 = \frac{1}{36}$, $\lambda = \frac{-43}{18} = |\vec{a}|$. which gives negative value of $\lambda$ or $|\vec{a}|$ which is not possible and hence data seems to be wrong. But if $|\vec{a}| + |\vec{b}| + |\vec{c}| = 36$ $$\frac{5}{2} \lambda + 6 = 36$$ $$\lambda = 12$$ $$\alpha = \frac{|\vec{c}||\vec{a}|}{|\vec{b}|} = \frac{3 \times 12}{2} \times \frac{12}{6}$$ $$\alpha = 36$$
Physics
Question 31
Physics · Physical World, Units and Measurements · Single correct
An expression of energy density is given by $$u = \frac{\alpha}{\beta} \sin \left( \frac{\alpha x}{kt} \right),$$ where $\alpha$, $\beta$ are constants, $x$ is displacement, $k$ is Boltzmann constant and $t$ is the temperature. The dimensions of $\beta$ will be:
A body of mass 10 kg is projected at an angle of 45° with the horizontal. The trajectory of the body is observed to pass through a point (20, 10). If T is the time of flight, then its momentum vector, at time $t = \frac{T}{\sqrt{2}}$, is ________ [Take $g = 10 \, \mathrm{m/s^2}$]
Given $$y = x - \frac{10x^2}{2u^2 \left( \frac{1}{2} \right)} \Rightarrow 10 = 20 - \frac{(10)(100)}{u^2}$$ Let $u = 20$. Then $$T = \frac{(2)(20)}{\sqrt{2}(10)} = 2\sqrt{2}$$ Velocity $$\vec{v} = 10\sqrt{2} \hat{i} + (10\sqrt{2} - 10(2)) \hat{j}$$ Momentum $$\vec{p} = M \vec{v} = 100\sqrt{2} \hat{i} + (100\sqrt{2} - 200) \hat{j}$$
Question 33
Physics · Laws of Motion · Single correct
A block of mass M slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is $\theta$. The magnitude of the contact force will be:
$Mg$
$Mg \cos \theta$
$\sqrt{Mg \sin \theta + Mg \cos \theta}$
$Mg \sin \theta \sqrt{1 + \mu}$
Answer: (a)
Solution
Given the forces on the inclined plane, we have: The normal force is given by $N = Mg \cos \theta$. The frictional force is $f = Mg \sin \theta$. The resultant force $R$ is calculated as: $$R = \sqrt{N^2 + f^2}$$ Since $R = Mg$, the forces are balanced.
Question 34
Physics · Laws of Motion · Single correct
A block 'A' takes 2 s to slide down a frictionless incline of $30^\circ$ and length '$l$', kept inside a lift going up with uniform velocity '$v$'. If the incline is changed to $45^\circ$, the time taken by the block, to slide down the incline, will be approximately:
2.66 $\,$ $\mathrm{s}$
0.83 $\,$ $\mathrm{s}$
1.68 $\,$ $\mathrm{s}$
0.70 $\,$ $\mathrm{s}$
Answer: (c)
Solution
Given $a = g \sin \theta$. $$\ell = \frac{1}{2} g \sin 30^\circ (2)^2 \ldots (1)$$ $$\ell = \frac{1}{2} g \sin 45^\circ \ t^2 \ldots (2)$$ $$\left( \frac{1}{2} \right) (4) = \frac{1}{\sqrt{2}} t^2 \Rightarrow t = \sqrt{2} \sqrt{2} \approx 1.68$$
Question 35
Physics · System of Particles and Rotational Motion · Single correct
The velocity of the bullet becomes one third after it penetrates 4 cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at (4 + x) cm inside the block. The value of x is:
A body of mass $m$ is projected with velocity $\lambda v_e$ in vertically upward direction from the surface of the earth into space. It is given that $v_e$ is escape velocity and $\lambda < 1$. If air resistance is considered to be negligible, then the maximum height from the centre of earth, to which the body can go, will be (R : radius of earth)
$\frac{R}{1+\lambda^2}$
$\frac{R}{1-\lambda^2}$
$\frac{R}{1-\lambda}$
$\frac{\lambda^2 R}{1-\lambda^2}$
Answer: (b)
Solution
The equation for gravitational potential energy and kinetic energy is given by: $$ -\frac{GMm}{R} + \frac{1}{2} m \lambda^2 V_e^2 = -\frac{GMm}{h} $$ Substituting the expression for escape velocity $V_e^2 = \frac{2GM}{R}$, we have: $$ -\frac{GMm}{R} + \frac{1}{2} \lambda^2 \frac{2GMm}{R} = -\frac{GMm}{h} $$ Simplifying, we get: $$ \frac{\lambda^2}{R} - \frac{1}{R} = -\frac{1}{h} $$ Rearranging gives: $$ \frac{1}{h} = \frac{1 - \lambda^2}{R} $$ Thus, the height $h$ is: $$ h = \frac{R}{1 - \lambda^2} $$
Question 37
Physics · Mechanical Properties of Solids · Single correct
A steel wire of length 3.2 m $(Y_s = 2.0 \times 10^{11} \, \mathrm{Nm}^{-2})$ and a copper wire of length 4.4 M $(Y_c = 1.1 \times 10^{11} \, \mathrm{Nm}^{-2})$, both of radius 1.4 mm are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm. The load applied, in Newton, will be: (Given $\pi = \frac{22}{7}$)
360
180
1080
154
Answer: (d)
Solution
Given $y_{steel} = 2 \times 10^{11}$ and $y_{cm} = 1.1 \times 10^{11}$. The total change in length is given by $$\Delta \ell_1 + \Delta \ell_2 = \Delta \ell$$ $$\frac{F \ell_1}{A_1 y_1} + \frac{F \ell_2}{A_2 y_2} = \Delta \ell$$ Solving for $F$, we have $$F = \frac{\Delta \ell}{\frac{\ell_1}{A_1 y_1} + \frac{\ell_2}{A_2 y_2}} = 1.54 \times 10^2 = 154$$
Question 38
Physics · Thermodynamics · Single correct
In $1^{st}$ case, Carnot engine operates between temperatures $300 \, \mathrm{K}$ and $100 \, \mathrm{K}$. In $2^{nd}$ case, as shown in the figure, a combination of two engines is used. The efficiency of this combination (in $2^{nd}$ case) will be:
same as the $1^{st}$ case
always greater than the $1^{st}$ case
always less than the $1^{st}$ case
may increase or decrease with respect to the $1^{st}$ case
Which statements are correct about degrees of freedom? A. A molecule with $n$ degrees of freedom has $n^2$ different ways of storing energy. B. Each degree of freedom is associated with $\frac{1}{2}RT$ average energy per mole. C. A monoatomic gas molecule has 1 rotational degree of freedom where as diatomic molecule has 2 rotational degrees of freedom D. $\mathrm{CH}_4$ has a total to 6 degrees of freedom Choose the correct answer from the option given below:
B and C only
B and D only
A and B only
C and D only
Answer: (b)
Solution
Methane molecule is tetrahedron. Degree of freedom due to rotation = 3. Degree of freedom due to translation = 3.
Question 40
Physics · Electric Charges and Fields · Single correct
A charge of $4 \, \mu \mathrm{C}$ is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be:
$1\,µ\mathrm{C}$ and $3\,µ\mathrm{C}$
$2\,µ\mathrm{C}$ and $2\,µ\mathrm{C}$
$0\,µ\mathrm{C}$ and $4\,µ\mathrm{C}$
$1.5\,µ\mathrm{C}$ and $2.5\,µ\mathrm{C}$
Answer: (b)
Solution
The force $F$ is given by the equation $$F = \frac{Kq(4-q)}{d^2}.$$ Differentiating $F$ with respect to $q$, we have $$\frac{dF}{dq} = \frac{K}{d^2}[4 - 2q] = 0.$$ Solving for $q$, we find $$q = 2.$$
Question 41
Physics · Current Electricity · Single correct
A. The drift velocity of electrons decreases with the increase in the temperature of conductor. B. The drift velocity is inversely proportional to the area of cross-section of given conductor. C. The drift velocity does not depend on the applied potential difference to the conductor. D. The drift velocity of electron is inversely proportional to the length of the conductor. E. The drift velocity increases with the increase in the temperature of conductor. Choose the correct answer from the options given below:
A and B only
A and D only
B and E only
B and C only
Answer: (b)
Solution
Drift velocity = $\left$( $\frac{e \tau}{m}$ $\right$) E $$v_d = \left( \frac{e \tau}{m} \right) \left( \frac{\Delta V}{\ell} \right)$$ $\Delta V$ = Potential difference applied across the wire. As temperature increases, relaxation time decreases, hence $v_d$ decreases. As per formula, $v_d \propto \frac{1}{\ell}$ $$v_d = \frac{I}{neA}$$ As it is not mentioned that current is at steady state neither it is mentioned that $n$ is constant for given conductor. So it can't be said that $v_d$ is inversely proportional to $A$. $I = neAv_d = \frac{V}{R} = \frac{V}{\rho \ell} A$ $$v_d = \frac{V}{\rho ne} \left( E = \frac{V}{\ell} \right)$$ $$v_d = \frac{eE \tau}{m}$$ $\tau$ decreases with temperature increase. First and fourth statements are correct.
Question 42
Physics · Magnetism and Matter · Single correct
A compass needle of oscillation magnetometer oscillates 20 times per minute at a place $P$ of dip $30^\circ$. The number of oscillations per minute become 10 at another place $Q$ of $60^\circ$ dip. The ratio of the total magnetic field at the two places $(B_Q : B_P)$ is:
Physics · Moving Charges and Magnetism · Single correct
A cyclotron is used to accelerate protons. If the operating magnetic field is 1.0 T and the radius of the cyclotron 'dees' is 60 cm, the kinetic energy of the accelerated protons in MeV will be: [use $m_p = 1.6 \times 10^{-27} \, \mathrm{kg}$, $e = 1.6 \times 10^{-19} \, \mathrm{C}$]
12
18
16
32
Answer: (b)
Solution
Kinetic energy of electron in cyclotron $$= \left[ \frac{q^2 B^2 r_0^2}{2m} \right]$$ $$= 18 \, \mathrm{MeV}$$
Question 44
Physics · Alternating Current · Single correct
A series LCR circuit has $L = 0.01 \, \mathrm{H}$, $R = 10 \, \Omega$ and $C = 1 \, \mu \mathrm{F}$ and it is connected to ac voltage of amplitude $(V_m) \, 50 \, \mathrm{V}$. At frequency 60$\%$ lower than resonant frequency, the amplitude of current will be approximately:
Identify the correct statements from the following descriptions of various properties of electromagnetic waves. A. In a plane electromagnetic wave electric field and magnetic field must be perpendicular to each other and direction of propagation of wave should be along electric field or magnetic field. B. The energy in electromagnetic wave is divided equally between electric and magnetic fields. C. Both electric field and magnetic field are parallel to each other and perpendicular to the direction of propagation of wave. D. The electric field, magnetic field and direction of propagation of wave must be perpendicular to each other. E. The ratio of amplitude of magnetic field to the amplitude of electric field is equal to speed of light. Choose the most appropriate answer from the options given below:
D only
B and D only
B, C and E only
A, B and E only
Answer: (b)
Solution
Second and fourth statements are correct.
Question 46
Physics · Wave Optics · Single correct
Two coherent sources of light interfere. The intensity ratio of two sources is 1 : 4. For this interference pattern if the value of $\frac{I_{max} + I_{min}}{I_{max} - I_{min}}$ is equal to $\frac{2\alpha + 1}{\beta + 3}$, then $\frac{\alpha}{\beta}$ will be:
1.5
2
0.5
1
Answer: (b)
Solution
Given $\frac{I_1}{I_2} = \frac{1}{4}$. Therefore, $I_2 = 4I_1$. The maximum current $I_{max} = I_1 + 4I_1 + 2\sqrt{I_1 \cdot 4I_1} = 9I_1$. The minimum current $I_{min} = I_1 + 4I_1 - 2\sqrt{I_1 \cdot 4I_1} = I_1$. Thus, $\frac{9I_1 + I_1}{9I_1 - I_1} = \frac{10}{8} = \frac{5}{4} = \frac{2\alpha + 1}{\beta + 1}$. Solving gives $\alpha = 2$ and $\beta = 1$. Therefore, $\frac{\alpha}{\beta} = \frac{2}{1} = 2$.
Question 47
Physics · Dual Nature of Radiation and Matter · Single correct
With reference to the observations in photo-electric effect, identify the correct statements from below: A. The square of maximum velocity of photoelectrons varies linearly with frequency of incident light. B. The value of saturation current increases on moving the source of light away from the metal surface. C. The maximum kinetic energy of photo-electrons decreases on decreasing the power of LED (light emitting diode) source of light. D. The immediate emission of photo-electrons out of metal surface can not be explained by particle nature of light/electromagnetic waves. E. Existence of threshold wavelength can not be explained by wave nature of light/electromagnetic waves. Choose the correct answer from the options given below:
A and B only
A and E only
C and E only
D and E only
Answer: (b)
Solution
The equation for the photoelectric effect is given by $$\frac{1}{2} m V_{max}^2 = h f - \phi$$ where $m$ is the mass, $V_{max}$ is the maximum velocity, $h$ is Planck's constant, $f$ is the frequency, and $\phi$ is the work function. The photoelectric effect can be explained by the particle nature of light. The threshold $\lambda$ is the maximum wavelength at which emission takes place.
Question 48
Physics · Nuclei · Single correct
The activity of a radioactive material is $6.4 \times 10^{-4}$ curie. Its half life is 5 days. The activity will become $5 \times 10^{-6}$ curie after:
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
For a constant collector-emitter voltage of $8\,\mathrm{V}$, the collector current of a transistor reached to the value of $6\,\mathrm{mA}$ from $4\,\mathrm{mA}$, whereas base current changed from $20\,\mathrm{µA}$ to $25\,\mathrm{µA}$ value. If transistor is in active state, small signal current gain (current amplification factor) will be:
240
400
0.0025
200
Answer: (b)
Solution
Given $V_{\mathrm{CE}} = 8 \, \mathrm{V}$, $I_{\mathrm{C}} = 6 \, \mathrm{mA}$ from $4 \, \mathrm{mA}$, $I_{\mathrm{B}} = 20 \, \mu\mathrm{A}$ to $25 \, \mu\mathrm{A}$. Current gain $\beta_{\mathrm{av}}$ is calculated as follows: $$\beta_{\mathrm{av}} = \frac{I_{\mathrm{C}}' - I_{\mathrm{C}}}{I_{\mathrm{B}}' - I_{\mathrm{B}}} = \frac{2 \, \mathrm{mA}}{5 \, \mu\mathrm{A}}$$ $$\beta_{\mathrm{av}} = \frac{2}{5} \times 10^3 = \frac{2000}{5} = 400$$
Question 50
Physics · Communication Systems · Single correct
A square wave of the modulating signal is shown in the figure. The carrier wave is given by $C(t) = 5 \sin (8 \pi t)$ Volt. The modulation index is:
0.2
0.1
0.3
0.4
Answer: (a)
Solution
Modulation Index $\mu = \frac{A_m}{A_C} = \frac{1}{5} = 0.2$ $A_m =$ amp. of modulating signal $A_C =$ amp. of carrier wave
Question 51
Physics · Mechanical Properties of Solids · Numerical
In an experiment to determine the Young's modulus, steel wires of five different lengths (1, 2, 3, 4 and 5 m) but of same cross section ($2\,\mathrm{mm}^2$) were taken and curves between extension and load were obtained. The slope (extension/load) of the curves were plotted with the wire length and the following graph is obtained. If the Young's modulus of given steel wires is $x \times 10^{11} \, \mathrm{Nm}^{-2}$, then the value of $x$ is ________ .
Answer: 2
Solution
Slope = $\frac{\Delta l / w}{L}$ = $\frac{\Delta l / L}{w}$ = $\frac{1}{YA}$ $\Rightarrow$ Y = $\frac{1}{(slope)A}$ Y = $\frac{1}{2 \times 10^{-6} (0.25 \times 10^{-5})}$ Y = 2 $\times$ $10^{11}$ $\mathrm{N/m^2}$
Question 52
Physics · Current Electricity · Numerical
In the given figure of meter bridge experiment, the balancing length AC corresponding to null deflection of the galvanometer is 40 cm. The balancing length, if the radius of the wire AB is doubled, will be……….cm.
Answer: 40
Solution
Independent of area in case of uniform wire.
Question 53
Physics · Ray Optics and Optical Instruments · Numerical
A thin prism of angle $6^\circ$ and refractive index for yellow light $(n_Y)1.5$ is combined with another prism of angle $5^\circ$ and $n_Y = 1.55$. The combination produces no dispersion. The net average deviation $(\delta)$ produced by the combination is $\left( \frac{1}{x} \right)^\circ$. The value of $x$ is.......
A conducting circular loop is placed in X - Y plane in presence of magnetic field $\vec{B} = (3t^3 \hat{j} + 3t^2 \hat{k})$ in SI unit. If the radius of the loop is $1 \, \mathrm{m}$, the induced emf in the loop, at time, $t = 2 \, \mathrm{s}$ is $n \pi \, \mathrm{V}$. The value of $n$ is........
Answer: 12
Solution
The magnetic flux $\phi$ is given by $\phi = \mathbf{B} \cdot \mathbf{A}$. This is equal to $\phi = (3t^3 \hat{\mathbf{j}} + 3t^2 \hat{\mathbf{k}}) \cdot (\pi (1)^2 \hat{\mathbf{k}})$. Simplifying, we get $\phi = 3t^2 \pi$. The induced emf $\varepsilon_{IND}$ is given by $\varepsilon_{IND} = \left| \frac{d\phi}{dt} \right| = 6t\pi$. At $t = 2$, $\varepsilon_{IND} = 12$.
Question 55
Physics · Electrostatic Potential and Capacitance · Numerical
As shown in the figure, in steady state, the charge stored in the capacitor is....... $\times 10^{-6} \mathrm{C}$.
Answer: 10
Solution
Given the equation for charge, we have: $$q = C V_{100\Omega}$$ Substituting the values, we get: $$= (1.1 \times 10^{-6}) \left( \frac{10}{R + r} R \right)$$ Simplifying further: $$= 1.1 \times 10^{-6} \left( \frac{10}{110} \times 100 \right)$$ Finally, we find: $$= 10 \, \mu C$$
Question 56
Physics · Electrostatic Potential and Capacitance · Fill in the blank
A parallel plate capacitor with width 4 cm, length 8 cm and separation between the plates of 4 $\mathrm{mm}$ is connected to a battery of 20 $\mathrm{V}$. A dielectric slab of dielectric constant 5 having length 1 $\mathrm{cm}$, width 4 $\mathrm{cm}$ and thickness 4 $\mathrm{mm}$ is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be ........ $\varepsilon_0$ $\mathrm{J}$. (Where $\varepsilon_0$ is the permittivity of free space)
Answer: 240
Solution
The effective capacitance is given by $$C_{eff} = \left[ \frac{\varepsilon_0 (7 \times 4)}{4/10} + \frac{5 \varepsilon_0 (1 \times 4)}{4/10} \right] \times 10^{-2}$$ Simplifying, we find $$C_{eff} = 1.2 \varepsilon_0$$ The energy is calculated as $$Energy = \frac{1}{2} C_{eff} V^2$$ Substituting the values, we get $$= \frac{1}{2} (1.2) \varepsilon_0 (20)(20) = 240 \varepsilon_0$$
Question 57
Physics · Waves · Numerical
A wire of length 30 cm, stretched between rigid supports, has it's $n^{th}$ and $(n + 1)^{th}$ harmonics at 400 Hz and 450 Hz, respectively. If tension in the string is 2700 N, it's linear mass density is........kg/m.
Physics · Mechanical Properties of Fluids · Numerical
A spherical soap bubble of radius 3 cm is formed inside another spherical soap bubble of radius 6 cm. If the internal pressure of the smaller bubble of radius 3 cm in the above system is equal to the internal pressure of the another single soap bubble of radius $r$ cm. The value of $r$ is.......
Answer: 2
Solution
Given $P_2 - P_0 = \frac{4T}{6}$ and $P_1 - P_2 = \frac{4T}{3}$. Therefore, $$P_1 - P_0 = \frac{4T}{2} = 2$$
Question 59
Physics · Work, Energy and Power · Numerical
A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should by unbinding the strings to achieve a speed of $4 \, \mathrm{ms^{-1}}$, is.......cm. (take $g = 10 \, \mathrm{ms^{-2}}$)
Answer: 120
Solution
From energy conservation $$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$ $$mgh = \frac{1}{2}mv^2 + \frac{1}{2} \frac{mR^2}{2} \omega^2$$ $$10h = \frac{16}{2} + \frac{16}{4} \Rightarrow h = 1.2\, \mathrm{m} = 120\, \mathrm{cm}$$
Question 60
Physics · Motion in a Plane · Numerical
Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top Point B at a height 10 $\,$ $\mathrm{m}$. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point C from point A is $t(\sqrt{2} + 1) \, \mathrm{s}$. The value of $t$ is........(use $g = 10 \, \mathrm{m/s^2}$)
The correct decreasing order of energy for the orbitals having the following set of quantum numbers: (A) $n=3,\ l=0,\ m=0$ (B) $n=4,\ l=0,\ m=0$ (C) $n=3,\ l=1,\ m=0$ (D) $n=3,\ l=2,\ m=1$
$(D) > (B) > (C) > (A)$
$(B) > (D) > (C) > (A)$
$(C) > (B) > (D) > (A)$
$(B) > (C) > (D) > (A)$
Answer: (a)
Solution
\text{(A)}\quad $n+\ell$=3+0=3 \text{(B)}\quad $n+\ell$=4+0=4 \text{(C)}\quad $n+\ell$=3+1=4 \text{(D)}\quad $n+\ell$=3+2=5 Higher $n+\ell$ value, higher the energy \& if same $n+\ell$ value, then higher $n$ value, higher the energy. Thus : $D>B>C>A$.
Question 62
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Match List-I with List-II List-I \begin{tabular}{|c|c|c|c|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \hline (A) & $\Psi_{MO}=\Psi_A-\Psi_B$ & (I) & Dipole moment \\ \hline (B) & $\mu=Q\times r$ & (II) & Bonding molecular orbital \\ \hline (C) & $\dfrac{N_b-N_a}{2}$ & (III) & Anti-bonding molecular orbital \\ \hline (D) & $\Psi_{MO}=\Psi_A+\Psi_B$ & (IV) & Bond order \\ \hline \end{tabular}
(A) -(II), (B)-(I), $(C)$-(IV), (D)-(III)
(A) -(III), (B)-(IV), $(C)$-(I), (D)-(II)
(A) -(III), (B)-(I), $(C)$-(IV), (D)-(II)
(A) -(III), (B)-(IV), $(C)$-(II), (D)-(I)
Answer: (c)
Solution
(A) $\psi_{MO} = \psi_A - \psi_B$ (B) $\mu = Q \times r$ $(C)$ $\frac{N_b - N_a}{2}$ (D) $\psi_{MO} = \psi_A + \psi_B$ (III) ABMO (I) Dipole moment (IV) Bond order (II) BMO
Question 63
Chemistry · Equilibrium · Single correct
The Plot of pH-metric titration of weak base $\mathrm{NH_4OH}$ vs strong acid $\mathrm{HCl}$ looks like:
Answer: (a)
Solution
Titration curve of $\mathrm{NH_4OH}$ vs $\mathrm{HCl}$ (WB + SA).
Question 64
Chemistry · Electrochemistry · Single correct
Given below are two statements: Statement I: For KI, molar conductivity increases steeply with dilution. Statement II: For carbonic acid, molar conductivity increases slowly with dilution. 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: (b)
Solution
Statement I: KI is a strong electrolyte thus almost constant on dilution. Statement II: In weak electrolyte it increases sharply.
Question 65
Chemistry · Surface Chemistry · Single correct
Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$ Assertion $(A)$ : Dissolved substances can be removed from a colloidal solution by diffusion through a parchment paper. Reason $(R)$ : Particles in a true solution cannot pass through parchment paper but the colloidal particles can pass through the parchment paper. In the light of the above statements, choose the correct answer from the options given below:
Both (A) and $(R)$ are correct and $(R)$ is the correct explanation of (A)
Both (A) and $(R)$ are correct but $(R)$ is not the correct explanation of (A)
$(A)$ is correct but $(R)$ is not correct
$(A)$ is not correct but $(R)$ is correct
Answer: (c)
Solution
Assertion (A): Correct. Reason (R): Incorrect. Particles of true solution pass through parchment paper thus answer is (C).
Question 66
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The outermost electronic configurations of four elements A, B, C, D are given below: (A) $3s^2$ (B) $3s^2 3p^1$ (C) $3s^2 3p^3$ (D) $3s^2 3p^4$ The correct order of first ionization enthalpy for them is:
< (B) < $(C)$ < (D)
< (A) < (D) < $(C)$
< (D) < (A) < $(C)$
< (A) < $(C)$ < (D)
Answer: (b)
Solution
Given the options: (A) $3s^2 \rightarrow Mg$ (B) $3s^2 3p^1 \rightarrow Al$ (C) $3s^2 3p^3 \rightarrow P$ (D) $3s^2 3p^4 \rightarrow S$ We have: Therefore, the order is $C > D > A > B$.
Question 67
Chemistry · The s-Block Elements · Single correct
An element A of group 1 shows similarity to an element B belonging to group 2. If A has maximum hydration enthalpy in group 1 then B is:
Mg
Be
Ca
Sr
Answer: (a)
Solution
Diagonal relationship $\mathrm{Li^+}$ has maximum hydration enthalpy in group 1 due to small size. So 'B' is $\mathrm{Mg}$.
Question 68
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$ Assertion $(A)$ : Boron is unable to form $\mathrm{BF}_6^{3-}$ Reason $(R)$ : Size of B is very small. In the light of the above statements, choose the correct answer from the options given below:
Both (A) and $(R)$ are true and $(R)$ is the correct explanation of (A)
Both (A) and $(R)$ are true but $(R)$ is not the correct explanation of (A)
is true but $(R)$ is false
is false but $(R)$ is true
Answer: (b)
Solution
Assertion (A): True Reason $(R)$: True but not correct explanation. Correct explanation: Expansion of octet not possible for 'B'.
Question 69
Chemistry · The d-and f-Block Elements · Single correct
In neutral or alkaline solution, $\mathrm{MnO}_4^-$ oxidises thiosulphate to:
$\mathrm{S}_2\mathrm{O}_7^{2-}$
$\mathrm{S}_2\mathrm{O}_8^{2-}$
$\mathrm{SO}_3^{2-}$
$\mathrm{SO}_4^{2-}$
Answer: (d)
Solution
The reaction is given by: $$8\mathrm{MnO_4^-} + 3\mathrm{S_2O_3^{2-}} + \mathrm{H_2O} \xrightarrow{neutral or alk. solution} 8\mathrm{MnO_2} + 6\mathrm{SO_4^{2-}} + 2\mathrm{OH^-}$$
Question 70
Chemistry · Co-ordination Compounds · Single correct
Low oxidation state of metals in their complexes are common when ligands:
have good $\pi$-accepting character
have good $\sigma$-donor character
are having good $\pi$-donating ability
are having poor $\sigma$-donating ability
Answer: (a)
Solution
When metal is in low oxidation state then it forms complexes when ligands have good $\pi$-accepting character.
Question 71
Chemistry · Environmental Chemistry · Single correct
Given below are two statements: Statement I : The non bio-degradable fly ash and slag from steel industry can be used by cement industry. Statement II : The fuel obtained from plastic waste is lead free. In the light of the above statements, choose the most appropriate answer from the options given below:
Both Statement I and Statement II are correct
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
(I) Fly ash and slag from steel industry are utilised by cement industry. (II) Fuel obtained from plastic waste has high octane rating. It contains no lead and it is known as green fuel. Both statement (I) & (II) are correct.
Question 72
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The structure of A in the given reaction is:
Answer: (c)
Solution
The reaction involves the deprotonation of the ketone by $\mathrm{NaOH}$ to form an enolate ion. This enolate ion then undergoes an alkylation reaction with ethyl bromide to form the final product.
Question 73
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Major product 'B' of the following reaction sequence is:
Answer: (b)
Solution
The reaction starts with the compound $\mathrm{CH_3-C=CH-CH_3}$ with a $\mathrm{CH_3}$ group attached to the second carbon. It reacts with $\mathrm{Br_2}$ in $\mathrm{CH_3OH}$ to form $\mathrm{CH_3-CH(OCH_3)-CHBr-CH_3}$. This intermediate then reacts with $\mathrm{HI}$ to form the final product $\mathrm{CH_3-CH(I)-CHBr-CH_3}$.
Question 74
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Match List-I with List-II.\ List-II\ (I) Gatterman Koch reaction\ (II) Etard reaction\ (III) Stephen reaction\ (IV) Rosenmund reaction\ Choose the correct answer from the options given below:
(A) -(IV), (B)-(III), $(C)$-(II), (D)-(I)
(A)-(I), (B)-(II), $(C)$-(III), (D)-(IV)
(A)-(II), (B)-(III), $(C)$-(IV), (D)-(I)
(A)-(III), (B)-(II), $(C)$-(I), (D)-(IV)
Answer: (a)
Solution
Option (A) involves the conversion of benzoyl chloride to benzaldehyde using hydrogen gas and palladium on barium sulfate as a catalyst. This is known as the Rosenmund reaction. Option (B) involves the conversion of acetonitrile to acetaldehyde using tin(II) chloride and hydrochloric acid, followed by hydrolysis. This is known as the Stephen reaction. Option (C) involves the oxidation of toluene to benzaldehyde using chromyl chloride and acid. This is known as the Etard reaction. Option (D) involves the formylation of benzene to benzaldehyde using carbon monoxide and hydrochloric acid in the presence of anhydrous aluminum chloride. This is known as the Gattermann Koch reaction.
Question 75
Chemistry · Polymers · Single correct
Match List-I with List-II. \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{(Polymer)} & \multicolumn{2}{c|}{(Monomer)} \\ \hline (A) & Neoprene & (I) & Acrylonitrile \\ \hline (B) & Teflon & (II) & Chloroprene \\ \hline (C) & Acrilan & (III) & Tetrafluoroethene \\ \hline (D) & Natural rubber & (IV) & Isoprene \\ \hline \end{tabular} Choose the correct answer from the option given below:
(A) -(II), (B)-(III), $(C)$-(I), (D-(IV)
(A) -(II), (B)-(I), $(C)$-(III), (D-(IV)
(A) -(II), (B)-(I), $(C)$-(IV), (D-(III)
(A) -(I), (B)-(II), $(C)$-(III), (D-(IV)
Answer: (a)
Solution
Question 76
Chemistry · Amines · Single correct
An organic compound ‘A’ contains nitrogen and chlorine. It dissolves readily in water to give a solution that turns litmus red. Titration of compound ‘A’ with standard base indicates that the molecular weight of ‘A’ is $131 \pm 2$. When a sample of ‘A’ is treated with aq. NaOH, a liquid separates which contains N but not Cl. Treatment of the obtained liquid with nitrous acid followed by phenol gives orange precipitate. The compound ‘A’ is :
Answer: (d)
Solution
Question 77
Chemistry · Biomolecules · Single correct
Match List-I with List-II.\ List-I\ (A) Glucose + HI\ (B) Glucose + $Br_2$ water\ (C) Glucose + acetic anhydride\ (D) Glucose + $HNO_3$\ List-II\ (I) Gluconic acid\ (II) Glucose pentacetate\ (III) Saccharic acid\ (IV) Hexane\ Choose the correct answer from the options given below:
(A) -(IV), (B)-(I), $(C)$-(II), (D)-(III)
(A)-(IV), (B)-(III), $(C)$-(II), (D)-(I)
(A)-(III), (B)-(I), $(C)$-(IV), (D)-(II)
(A)-(I), (B)-(III), $(C)$-(IV), (D)-(II)
Answer: (a)
Solution
Option (A) shows the reaction of glucose with HI to form n-hexane. Option (B) shows the reaction of glucose with $\mathrm{Br_2}$ and $\mathrm{H_2O}$ to form gluconic acid. Option $(C)$ shows the reaction of glucose with 5 acetic anhydride to form glucose pentacetate. Option (D) shows the reaction of glucose with $\mathrm{HNO_3}$ to form saccharic acid.
Question 78
Chemistry · Chemistry in Everyday Life · Single correct
Which of the following enhances the lathering property of soap?
Sodium stearate
Sodium carbonate
Sodium rosinate
Trisodium phosphate
Answer: (c)
Solution
Rosin is added to soaps which forms sodium rosinate which lathers well.
Question 79
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Chemistry · Co-ordination Compounds · Single correct
$Fe^{3+}$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:
[$\mathrm{Fe(H_2O)_6}]_2$ $[\mathrm{Fe(CN)_6}$]
$\mathrm{Fe_2[Fe(CN)_6]_2}$
$\mathrm{Fe_3[Fe(OH)_2(CN)_4]_2}$
$\mathrm{Fe_4[Fe(CN)_6]_3}$
Answer: (d)
Solution
The reaction is given by: $$4 \mathrm{Fe}^{3+} + 3[\mathrm{Fe(CN)}_6]^{-4} \rightarrow \mathrm{Fe}_4[\mathrm{Fe(CN)}_6]_3$$ This forms Prussian Blue.
Question 81
Chemistry · Some Basic Concepts of Chemistry · Numerical
The normality of $\mathrm{H_2SO_4}$ in the solution obtained on mixing $100 \, \mathrm{mL}$ of $0.1 \, \mathrm{M} \, \mathrm{H_2SO_4}$ with $50 \, \mathrm{mL}$ of $0.1 \, \mathrm{M} \, \mathrm{NaOH}$ is _______ $\times 10^{-1} \, \mathrm{N}$. (Nearest Integer)
Answer: 1
Solution
No. of equivalents of $\mathrm{H_2SO_4} = 100 \times 0.1 \times 2 = 20$ No. of equivalents of $\mathrm{NaOH} = 50 \times 0.1 = 5$ No. of equivalents of $\mathrm{H_2SO_4}$ left $= 20 - 5 = 15$ $$150 \times x = 15$$ $$x = \frac{1}{10} = 0.1\, \mathrm{N} = 1 \times 10^{-1}\, \mathrm{N}$$
Question 82
Chemistry · States of Matter · Numerical
for a real gas at $25^\circ \mathrm{C}$ temperature and high pressure ($99 \mathrm{\ bar}$) the value of compressibility factor is 2, so the value of Vander Waal’s constant ‘b’ should be _____ $\times 10^{-2} \mathrm{\ L \ mol^{-1}}$ (Nearest integer) (Given $R = 0.083 \mathrm{L \ bar \ K}^{-1} \mathrm{mol}^{-1}$)
Answer: 25
Solution
For real gas under high pressure, $$Z = 1 + \frac{Pb}{RT}$$ implies $$b = \frac{RT}{P}$$ $$= \frac{0.083 \times 298}{99}$$ $$= 0.25 \times 10^{-2} \, \mathrm{L \, mol^{-1}}$$
Question 83
Chemistry · Thermodynamics · Numerical
A gas (Molar mass = 280 $\mathrm{g \, mol^{-1}}$) was burnt in excess $\mathrm{O_2}$ in a constant volume calorimeter and during combustion the temperature of calorimeter increased from 298.0 $\mathrm{K}$ to 298.45 $\mathrm{K}$. If the heat capacity of calorimeter is 2.5 $\mathrm{kJ \, K^{-1}}$ and enthalpy of combustion of gas is 9 $\mathrm{kJ \, mol^{-1}}$ then amount of gas burnt is $\mathrm{g}$. (Nearest Integer)
Answer: 35
Solution
Let $x \, \mathrm{g}$ is burnt. moles = $\frac{x}{280}$ heat released by $\frac{x}{280}$ mole = 2.5 $\times$ 0.45 $\,$ $\mathrm{kJ}$ heat released by 1 mole = $\frac{2.5 \times 0.45 \times 280}{x}$ $\,$ $\mathrm{kJ}$ $\Delta$ H = $\Delta$ U + $\Delta$ n g R T $\Delta$ H $\approx$ $\Delta$ U 9 = $\frac{2.5 \times 280 \times 0.45}{x}$ x = 35 $\,$ $\mathrm{g}$
Question 84
Chemistry · Solutions · Numerical
When a certain amount of solid A is dissolved in $100\,\mathrm{g}$ of water at $25°\mathrm{C}$ to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is $23.76\,\mathrm{mmHg}$. The number of moles of solute A added is \_\_\_\_. (Nearest Integer) Assume moles of A to be less than moles of B.
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
$$ \begin{array}{ccc} [A] & \rightarrow & [B] \\ \text{Reactant} & & \text{Product} \end{array} $$ If formation of compound [B] follows the first order of kinetics and after 70 minutes the concentration of [A] was found to be half of its initial concentration. Then the rate constant of the reaction is $x\times10^{-6}\,\mathrm{s}^{-1}$. The value of $x$ is ______. (Nearest Integer)
Answer: 165
Solution
K is calculated as follows: $$K = \frac{0.693}{t_{1/2}} = \frac{0.693}{70 \times 60}$$ This simplifies to: $$= \frac{6930}{7 \times 6} \times 10^{-6}$$ Finally, we have: $$= 165 \times 10^{-6} \, \mathrm{s^{-1}}$$
Question 86
Chemistry · General Principles and Processes of Isolation of Elements · Numerical
Among the following ores Bauxite, Siderite, Cuprite, Calamine, Haematite, Kaolinite, Malachite, Magnetite, Sphalerite, Limonite, Cryolite, the number of principal ores if (of) iron is ________.
Chemistry · The d-and f-Block Elements · Fill in the blank
The oxidation state of manganese in the product obtained in a reaction of potassium permanganate and hydrogen peroxide in basic medium is _______.
Answer: 4
Solution
In a basic medium, the reaction is given by: $$2 \mathrm{KMnO_4} + 3 \mathrm{H_2O_2} \xrightarrow{basic medium} 2 \mathrm{MnO_2} + 3 \mathrm{O_2} + 2 \mathrm{H_2O} + 2 \mathrm{KOH}.$$
Question 88
Chemistry · Chemical Bonding and Molecular Structure · Numerical
The number of molecule(s) or ion(s) from the following having non-planar structure is ______. $NO_3^-, H_2O_2, BF_3, PCl_3, XeF_4, SF_4, XeO_3, PH_4^+, SO_3, [Al(OH)_4]^-$
Answer: 6
Solution
$SO_3$ has $sp^2$ hybridization and is planar. $BF_3$ has $sp^2$ hybridization and is planar. $NO_3^-$ has $sp^2$ hybridization and is planar. $SF_4$ has $sp^3d$ hybridization and is non-planar. $H_2O_2$ has $sp^3$ hybridization and is non-planar. $PCl_3$ has $sp^3$ hybridization and is non-planar. $[Al(OH)_4]^-$ has $sp^3$ hybridization and is non-planar. $XeF_4$ has $sp^3d^2$ hybridization and is planar. $XeO_3$ has $sp^3$ hybridization and is non-planar. $PH_4^+$ has $sp^3$ hybridization and is non-planar.
Question 89
Chemistry · Co-ordination Compounds · Numerical
The spin only magnetic moment of the complex present in Fehling's reagent is $\dots\dots$ B.M. (Nearest integer).
Answer: 2
Solution
Fehling solution is a complex of $\mathrm{Cu^{++}}$. $$\mathrm{Cu^{++} = 3d^9}$$ Number of unpaired $e^- = 1$. $$\mathrm{M.M} = \sqrt{1(1+2)} = \sqrt{3} = 1.73 \, \mathrm{BM}$$
Question 90
Chemistry · Some Basic Concepts of Chemistry · Numerical
In the above reaction, 5 g of toluene is converted into benzaldehyde with 92% yield. The amount of benzaldehyde produced is _____ $\times$ $10^{-2}$ g. (Nearest integer)
Answer: 530
Solution
Moles $=\frac{5}{92}$ Moles of $\mathrm{CHO}=\frac{5}{92}\times\frac{92}{100}=5\times10^{-2}$ Mass of $\mathrm{CHO}=106\times5\times10^{-2}=5.3\,\mathrm{g}=530\times10^{-2}\,\mathrm{g}$