JEE Main 27 January 2024 Shift 2 question paper with solutions
JEE Main 27 January 2024 Shift 2: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
Register free to attempt this shift as a timed 180-minute test with instant scoring and chapter-wise analysis.
Maths
Question 1
Maths · Inverse Trigonometric Functions · Single correct
Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $\tan^{-1}(x) + \tan^{-1}(2x) = \frac{\pi}{4}$ is:
More than 2
1
2
0
Answer: (b)
Solution
Given $\tan^{-1} x + \tan^{-1} 2x = \frac{\pi}{4}$; $x > 0$. Therefore, $\tan^{-1} 2x = \frac{\pi}{4} - \tan^{-1} x$. Taking tan on both sides, we get $$2x = \frac{1-x}{1+x}$$ which simplifies to $$2x^2 + 3x - 1 = 0$$ Solving the quadratic equation, $$x = \frac{-3 \pm \sqrt{9 + 8}}{8} = \frac{-3 \pm \sqrt{17}}{8}$$ The only possible $x = \frac{-3 + \sqrt{17}}{8}$.
Question 2
Maths · Continuity and Differentiability · Single correct
Consider the function $f : (0, 2) \to \mathbb{R}$ defined by $f(x) = \frac{x}{2} + \frac{2}{x}$ and the function $g(x)$ defined by $$g(x) = \begin{cases} \min\{f(t)\}, & 0 < t \leq x and 0 < x \leq 1 \\ \frac{3}{2} + x, & 1 < x < 2 \end{cases}.$$ Then
$g$ is continuous but not differentiable at $x = 1$
$g$ is not continuous for all $x \in (0, 2)$
$g$ is neither continuous nor differentiable at $x = 1$
$g$ is continuous and differentiable for all $x \in (0, 2)$
Answer: (a)
Solution
Given $f : (0, 2) \to \mathbb{R}; \ f(x) = \frac{x}{2} + \frac{2}{x}$. The derivative is $f'(x) = \frac{1}{2} - \frac{2}{x^2}$. Therefore, $f(x)$ is decreasing in the domain.
Question 3
Maths · Three Dimensional Geometry · Single correct
Let the image of the point $(1, 0, 7)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$ be the point $(\alpha, \beta, \gamma)$. Then which one of the following points lies on the line passing through $(\alpha, \beta, \gamma)$ and making angles $\frac{2\pi}{3}$ and $\frac{3\pi}{4}$ with $y$-axis and $z$-axis respectively and an acute angle with $x$-axis ?
$(1, -2, 1 + \sqrt{2})$
$(1, 2, 1 - \sqrt{2})$
$(3, 4, 3 - 2\sqrt{2})$
$(3, -4, 3 + 2\sqrt{2})$
Answer: (c)
Solution
Given $L_1=\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda$. Let $P(1,0,7)$. $M(\lambda,1+2\lambda,2+3\lambda)$. $\overrightarrow{PM}=(\lambda-1)\hat{i}+(1+2\lambda)\hat{j}+(3\lambda-5)\hat{k}$. $\overrightarrow{PM}$ is perpendicular to line $L_1$. $\overrightarrow{PM}\cdot\overrightarrow{b}=0 \quad \left(\overrightarrow{b}=\hat{i}+2\hat{j}+3\hat{k}\right)$ $\Rightarrow \lambda-1+4\lambda+2+9\lambda-15=0$ $14\lambda=14 \Rightarrow \lambda=1$ Therefore, $M=(1,3,5)$. $\overrightarrow{Q}=2\overrightarrow{M}-\overrightarrow{P}\;[M\text{ is midpoint of }\overrightarrow{P}\ \&\ \overrightarrow{Q}]$ $\overrightarrow{Q}=2\hat{i}+6\hat{j}+10\hat{k}-\hat{i}-7\hat{k}$ $\overrightarrow{Q}=\hat{i}+6\hat{j}+3\hat{k}$ Therefore, $(\alpha,\beta,\gamma)=(1,6,3)$. Required line having direction cosine $(l,m,n)$. $l^2+m^2+n^2=1$ $\Rightarrow l^2+\left(-\frac{1}{2}\right)^2+\left(-\frac{1}{\sqrt{2}}\right)^2=1$ $l^2=\frac{1}{4}$ $\therefore l=\frac{1}{2}\;[\text{Line makes acute angle with x-axis}]$ Equation of line passing through $(1,6,3)$ will be $\overrightarrow{r}=(\hat{i}+6\hat{j}+3\hat{k})+\mu\left(\frac{1}{2}\hat{i}-\frac{1}{2}\hat{j}-\frac{1}{\sqrt{2}}\hat{k}\right)$ Option 3 satisfying for $\mu=4$
Question 4
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let $\mathbb{R}$ be the interior region between the lines 3x - y + 1 = 0 and x + 2y - 5 = 0 containing the origin. The set of all values of a, for which the points ($a^2$, a + 1) lie in $\mathbb{R}$, is:
(-3, -1) $\\cup( -\\frac{1}{3}, 1 )$
(-3, 0) $\\cup(\\frac{1}{3}, 1)$
(-3, 0) $\\cup( \\frac{2}{3}, 1 )$
(-3, -1) $\\cup(\\frac{1}{3}, 1)$
Answer: (b)
Solution
Given $P \left(a^2, a+1\right)$ and $L_1 = 3x - y + 1 = 0$. Origin and $P$ lies on the same side with respect to $L_1$. Therefore, $L_1(0) \cdot L_1(P) > 0$. Thus, $3 \left(a^2\right) - (a+1) + 1 > 0$. $$\Rightarrow 3a^2 - a > 0$$ $$a \in (-\infty, 0) \cup \left(\frac{1}{3}, \infty\right) \cdots (1)$$ Let $L_2 : x + 2y - 5 = 0$. Origin and $P$ lies on the same side with respect to $L_2$. Therefore, $L_2(0) \cdot L_2(P) > 0$. Thus, $a^2 + 2(a+1) - 5 < 0$. $$a^2 + 2a - 3 < 0$$ $$\Rightarrow (a+3)(a-1) < 0$$ Therefore, $a \in (-3, 1) \cdots (2)$ Intersection of $(1)$ and $(2)$: $$a \in (-3, 0) \cup \left(\frac{1}{3}, 1\right)$$
Question 5
Maths · Sequences and Series · Single correct
The $20^{\text{th}}$ term from the end of the progression $20,\ 19\frac{1}{4},\ 18\frac{1}{2},\ 17\frac{3}{4},\ \ldots,\ -129\frac{1}{4}$ is:
-118
-110
-115
-100
Answer: (c)
Solution
The sequence is 20, 19 $\frac{1}{4}$, 18 $\frac{1}{2}$, 17 $\frac{3}{4}$, $\ldots$, -129 $\frac{1}{4}$. This is an A.P. with common difference $$d_1 = -1 + \frac{1}{4} = -\frac{3}{4}$$ The sequence is -129 $\frac{1}{4}$, $\ldots$, 19 $\frac{1}{4}$, 20. This is also an A.P. with $a = -129 \frac{1}{4}$ and $d = \frac{3}{4}$. The required term is $$-129 \frac{1}{4} + (20 - 1) \left( \frac{3}{4} \right)$$ $$= -129 \frac{1}{4} + 15 - \frac{3}{4} = -115$$
Question 6
Maths · Relations and Functions · Single correct
Let f : $\mathbb{R} - \left\{ -\frac{1}{2} \right\} \to \mathbb{R}$ and $g : \mathbb{R} - \left\{ -\frac{5}{2} \right\} \to \mathbb{R}$ be defined as $f(x) = \frac{2x+3}{2x+1}$ and $g(x) = \frac{|x|+1}{2x+5}$. Then the domain of the function fog is :
Given $f(x) = \frac{2x + 3}{2x + 1}$, $x \neq -\frac{1}{2}$ and $g(x) = \frac{|x| + 1}{2x + 5}$, $x \neq -\frac{5}{2}$. Domain of $f(g(x))$: $$f(g(x)) = \frac{2g(x) + 3}{2g(x) + 1}$$ $x \neq -\frac{5}{2}$ and $\frac{|x| + 1}{2x + 5} \neq -\frac{1}{2}$. $x \in \mathbb{R} - \left\{ -\frac{5}{2} \right\}$ and $x \in \mathbb{R}$. Therefore, the domain will be $\mathbb{R} - \left\{ -\frac{5}{2} \right\}$.
Question 7
Maths · Integrals · Single correct
For $0 < a < 1$, the value of the integral $\int_{0}^{\pi} \frac{dx}{1 - 2a \cos x + a^2}$ is:
$\frac{\pi^2}{\pi + a^2}$
$\frac{\pi^2}{\pi - a^2}$
$\frac{\pi}{1 - a^2}$
$\frac{\pi}{1 + a^2}$
Answer: (c)
Solution
Given $$I = \int_0^\pi \frac{dx}{1 - 2a \cos x + a^2}; \; 0 < a < 1$$ We have $$I = \int_0^\pi \frac{dx}{1 + 2a \cos x + a^2}$$ Thus, $$2I = 2 \int_0^{\pi/2} \frac{2 \left( 1 + a^2 \right)}{\left( 1 + a^2 \right)^2 - 4a^2 \cos^2 x} \, dx$$ This implies $$\Rightarrow I = \int_0^{\pi/2} \frac{2 \left( 1 + a^2 \right) \cdot \sec^2 x}{\left( 1 + a^2 \right)^2 \cdot \sec^2 x - 4a^2} \, dx$$ Further simplifying, $$\Rightarrow I = \int_0^{\pi/2} \frac{2 \cdot \left( 1 + a^2 \right) \cdot \sec^2 x}{\left( 1 + a^2 \right)^2 \cdot \tan^2 x + \left( 1 - a^2 \right)^2} \, dx$$ This leads to $$\Rightarrow I = \int_0^{\pi/2} \frac{2 \cdot \sec^2 x}{\tan^2 x + \left( \frac{1-a^2}{1+a^2} \right)^2} \, dx$$ Finally, $$\Rightarrow I = \frac{2}{\left( 1 - a^2 \right)} \left[ \frac{\pi}{2} - 0 \right]$$ Thus, $$I = \frac{\pi}{1 - a^2}$$
Question 8
Maths · Applications of Derivatives · Single correct
Let $g(x) = 3f\left(\frac{x}{3}\right) + f(3-x)$ and $f''(x) > 0$ for all $x \in (0, 3)$. If $g$ is decreasing in $(0, \alpha)$ and increasing in $(\alpha, 3)$, then $8\alpha$ is
24
0
18
20
Answer: (c)
Solution
Given $g(x) = 3f\left(\frac{x}{3}\right) + f(3-x)$ and $f''(x) > 0 \forall x \in (0, 3)$ implies $f'(x)$ is an increasing function. $g'(x) = 3 \times \frac{1}{3} \cdot f'\left(\frac{x}{3}\right) - f'(3-x)$ $$= f'\left(\frac{x}{3}\right) - f'(3-x)$$ If $g$ is decreasing in $(0, \alpha)$ $g'(x) < 0$ $$f'\left(\frac{x}{3}\right) - f'(3-x) < 0$$ $$f'\left(\frac{x}{3}\right) < f'(3-x)$$ $$\Rightarrow \frac{x}{3} < 3-x$$ $$\Rightarrow x < \frac{9}{4}$$ Therefore $\alpha = \frac{9}{4}$. Then $8\alpha = 8 \times \frac{9}{4} = 18$
Question 9
Maths · Limits and Derivatives · Single correct
If $\lim_{x \to 0} \frac{3 + \alpha \sin x + \beta \cos x + \log_e (1-x)}{3 \tan^2 x} = \frac{1}{3}$, then $2\alpha - \beta$ is equal to :
2
7
5
1
Answer: (c)
Solution
Given the limit $$\lim_{x \to 0} \frac{3 + \alpha \sin x + \beta \cos x + \log_e(1-x)}{3 \tan^2 x} = \frac{1}{3}$$ we expand the terms as follows: $$3 + \alpha \left[x - \frac{x^3}{3!} + \ldots \right] + \beta \left[1 - \frac{x^2}{2!} + \frac{x^4}{4!} \ldots \right] + \left(-x - \frac{x^2}{2} - \frac{x^3}{3} \ldots \right)$$ This simplifies to: $$\Rightarrow \lim_{x \to 0} \frac{(3 + \beta) + (\alpha - 1)x + \left(-\frac{1}{2} - \frac{\beta}{2}\right)x^2 + \ldots}{3x^2} \times \frac{x^2}{\tan^2 x} = \frac{1}{3}$$ Solving the equations: $$\Rightarrow \beta + 3 = 0, \alpha - 1 = 0 and -\frac{1}{2} - \frac{\beta}{2} = \frac{1}{3}$$ We find: $$\Rightarrow \beta = -3, \alpha = 1$$ Therefore, $$2\alpha - \beta = 2 + 3 = 5$$
Question 10
Maths · Complex Numbers and Quadratic Equations · Single correct
If $\alpha$, $\beta$ are the roots of the equation, $x^2 - x - 1 = 0$ and $S_n = 2023 \alpha^n + 2024 \beta^n$, then
Let A and B be two finite sets with m and n elements respectively. The total number of subsets of the set A is 56 more than the total number of subsets of B. Then the distance of the point P(m, n) from the point Q(-2, -3) is
10
6
4
8
Answer: (a)
Solution
Given $2^m - 2^n = 56$. $2^n \left(2^{m-n} - 1\right) = 2^3 \times 7$ $2^n = 2^3$ and $2^{m-n} - 1 = 7$ $\Rightarrow n = 3$ and $2^{m-n} = 8$ $\Rightarrow n = 3$ and $m - n = 3$ $\Rightarrow n = 3$ and $m = 6$ $P(6, 3)$ and $Q(-2, -3)$ $PQ = \sqrt{8^2 + 6^2} = \sqrt{100} = 10$ Hence option (1) is correct
Question 12
Maths · Determinants · Single correct
The values of $\alpha$, for which $$\begin{vmatrix} 1 & \frac{3}{2} & \alpha + \frac{3}{2} \\ 1 & \frac{1}{3} & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0,$$ lie in the interval
An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability, that the first draw gives all white balls and the second draw gives all black balls, is :
$\frac{5}{256}$
$\frac{5}{715}$
$\frac{3}{715}$
$\frac{3}{256}$
Answer: (c)
Solution
Given $$\frac{{^6C_4}}{{^{15}C_4}} \times \frac{{^9C_4}}{{^{11}C_4}} = \frac{3}{715}$$ Hence option (3) is correct.
Question 14
Maths · Integrals · Single correct
The integral $\int \frac{(x^8 - x^2) \, dx}{(x^{12} + 3x^6 + 1) \tan^{-1} \left( x^3 + \frac{1}{x^3} \right)}$ equal to:
If $$2\tan^2\theta-5\sec\theta=1$$ has exactly $7$ solutions in the interval $$\left[0,\frac{n\pi}{2}\right],$$ for the least value of $n\in\mathbb{N}$, then $$\sum_{k=1}^{n}\frac{k}{2^k}$$ is equal to:
$$\frac{1}{2^{15}}\left(2^{14}-14\right)$$
$$\frac{1}{2^{14}}\left(2^{15}-15\right)$$
$$1 - \frac{15}{2^{15}}$$
$$\frac{1}{2^{15}}\left(2^{14}-15\right)$$
Answer: (d)
Solution
Given the equation $2 \tan^2 \theta - 5 \sec \theta - 1 = 0$. This implies $2 \sec^2 \theta - 5 \sec \theta - 3 = 0$. Further simplifying, we have $(2 \sec \theta + 1)(\sec \theta - 3) = 0$. Therefore, $\sec \theta = -\frac{1}{2}, 3$. This implies $\cos \theta = -2, \frac{1}{3}$. Thus, $\cos \theta = \frac{1}{3}$. For 7 solutions, $n = 13$. So, $$\sum_{k=1}^{13} \frac{k}{2^k} = S (say)$$ We have $$S = \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \ldots + \frac{13}{2^{13}}$$ $$\frac{1}{2} S = \frac{1}{2^2} + \frac{1}{2^3} + \ldots + \frac{12}{2^{13}} + \frac{13}{2^{14}}$$ Therefore, $$\Rightarrow \frac{S}{2} = \frac{1}{2} \cdot \frac{1 - \frac{1}{2^{13}}}{1 - \frac{1}{2}} - \frac{13}{2^{14}} \Rightarrow S = 2 \cdot \left( \frac{2^{13} - 1}{2^{13}} \right) - \frac{13}{2^{13}}$$
Question 16
Maths · Vector Algebra · Single correct
The position vectors of the vertices A, B and C of a triangle are $2\hat{i} - 3\hat{j} + 3\hat{k}$, $2\hat{i} + 2\hat{j} + 3\hat{k}$ and $-\hat{i} + \hat{j} + 3\hat{k}$ respectively. Let $l$ denotes the length of the angle bisector $AD$ of $\angle BAC$ where $D$ is on the line segment $BC$, then $2l^2$ equals :
49
42
50
45
Answer: (d)
Solution
Since D is the midpoint of BC, we have D $\left$( $\frac{1}{2}$, $\frac{3}{2}$, 3 $\right$). Therefore, $\ell$ = $\sqrt{\left( 2 - \frac{1}{2} \right)^2 + \left( -3 - \frac{3}{2} \right)^2 + (3 - 3)^2}$. Thus, $\ell$ = $\sqrt{\frac{45}{2}}$. Therefore, 2$\ell$^2 = 45.
Question 17
Maths · Differential Equations · Single correct
If $y = y(x)$ is the solution curve of the differential equation $(x^2 - 4) \, dy - (y^2 - 3y) \, dx = 0$, $x > 2, y(4) = \frac{3}{2}$ and the slope of the curve is never zero, then the value of $y(10)$ equals:
Let $e_1$ be the eccentricity of the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ and $e_2$ be the eccentricity of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $a > b$, which passes through the foci of the hyperbola. If $e_1 e_2 = 1$, then the length of the chord of the ellipse parallel to the x-axis and passing through $(0, 2)$ is:
$4\sqrt{5}$
$\frac{8\sqrt{5}}{3}$
$\frac{10\sqrt{5}}{3}$
$3\sqrt{5}$
Answer: (c)
Solution
Given $H : \frac{x^2}{16} - \frac{y^2}{9} = 1$ with $e_1 = \frac{5}{4}$. Therefore, $e_1 e_2 = 1 \Rightarrow e_2 = \frac{4}{5}$. Also, the ellipse is passing through $(\pm 5, 0)$. Thus, $a = 5$ and $b = 3$. The equation of the ellipse is $E : \frac{x^2}{25} + \frac{y^2}{9} = 1$. The endpoints of the chord are $\left( \pm \frac{5\sqrt{5}}{3}, 2 \right)$. Therefore, $LPQ = \frac{10\sqrt{5}}{3}$.
Question 19
Maths · Permutations and Combinations · Single correct
Let $\alpha = \frac{(4!)!}{(4!)^3!}$ and $\beta = \frac{(5!)!}{(5!)^4!}$. Then:
$\alpha \in \mathbb{N}$ and $\beta \notin \mathbb{N}$
$\alpha \notin \mathbb{N}$ and $\beta \in \mathbb{N}$
$\alpha \in \mathbb{N}$ and $\beta \in \mathbb{N}$
$\alpha \notin \mathbb{N}$ and $\beta \notin \mathbb{N}$
Answer: (c)
Solution
Given $\($ $\alpha$ = $\frac{(4!)!}{(4!)^3!}$, $\beta$ = $\frac{(5!)!}{(5!)^4!}$ $\)$ $\($ $\alpha$ = $\frac{(24)!}{(4!)^6}$, $\beta$ = $\frac{(120)!}{(5!)^{24}}$ $\)$ Let 24 distinct objects are divided into 6 groups of 4 objects in each group. No. of ways of formation of group = $\($ $\frac{24!}{(4!)^6 \cdot 6!}$ $\in$ $\mathbb{N}$ $\)$ Similarly, Let 120 distinct objects are divided into 24 groups of 5 objects in each group. No. of ways of formation of groups = $\($ $\frac{(120)!}{(5!)^{24} \cdot 24!}$ $\in$ $\mathbb{N}$ $\)$
Question 20
Maths · Three Dimensional Geometry · Single correct
Let the position vectors of the vertices $A, B$ and $C$ of a triangle be $2\hat{i} + 2\hat{j} + \hat{k}$, $\hat{i} + 2\hat{j} + 2\hat{k}$ and $2\hat{i} + \hat{j} + 2\hat{k}$ respectively. Let $l_1, l_2$ and $l_3$ be the lengths of perpendiculars drawn from the ortho center of the triangle on the sides $AB, BC$ and $CA$ respectively, then $l_1^2 + l_2^2 + l_3^2$ equals:
$\frac{1}{5}$
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{3}$
Answer: (b)
Solution
Triangle $\triangle ABC$ is equilateral. Orthocentre and centroid will be the same. $G \left( \frac{5}{3}, \frac{5}{3}, \frac{5}{3} \right)$. Mid-point of $AB$ is $D \left( \frac{3}{2}, 2, \frac{3}{2} \right)$. Therefore, $$ \ell_1 = \sqrt{\frac{1}{36} + \frac{1}{9} + \frac{1}{36}} $$ $$ \ell_1 = \sqrt{\frac{1}{6}} = \ell_2 = \ell_3 $$ Thus, $$ \ell_1^2 + \ell_2^2 + \ell_3^2 = \frac{1}{2} $$
Question 21
Maths · Statistics · Numerical
The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12. If $\mu$ and $\sigma^2$ denote the mean and variance of the correct observations respectively, then $15 \left( \mu + \mu^2 + \sigma^2 \right)$ is equal to
Answer: 2521
Solution
Let the incorrect mean be $\mu'$ and standard deviation be $\sigma'$. We have $$\mu' = \frac{\Sigma x_i}{15} = 12 \Rightarrow \Sigma x_i = 180$$ As per given information correct $\Sigma x_i = 180 - 10 + 12$ $$\Rightarrow \mu (correct mean) = \frac{182}{15}$$ Also $$\sigma' = \sqrt{\frac{\Sigma x_i^2}{15} - 144} = 3 \Rightarrow \Sigma x_i^2 = 2295$$ Correct $\Sigma x_i^2 = 2295 - 100 + 144 = 2339$ $$\sigma^2 (correct variance) = \frac{2339}{15} - \frac{182 \times 182}{15 \times 15}$$ Required value $$= 15 \left( \mu + \mu^2 + \sigma^2 \right)$$ $$= 15 \left( \frac{182}{15} + \frac{182 \times 182}{15 \times 15} + \frac{2339}{15} - \frac{182 \times 182}{15 \times 15} \right)$$ $$= 15 \left( \frac{182}{15} + \frac{2339}{15} \right)$$ $$= 2521$$
Question 22
Maths · Applications of Integrals · Numerical
If the area of the region $\{(x, y) : 0 \leq y \leq \min \{2x, 6x - x^2\}\}$ is $A$, then $12A$ is equal to.
Let A be a $2 \times 2$ real matrix and $I$ be the identity matrix of order 2. If the roots of the equation $|A - xI| = 0$ be $-1$ and $3$, then the sum of the diagonal elements of the matrix $A^2$ is.
Answer: 10
Solution
Given $|A - xI| = 0$. Roots are $-1$ and $3$. Sum of roots $= \mathrm{tr}(A) = 2$. Product of roots $= |A| = -3$. Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. We have $a + d = 2$. $ad - bc = -3$. $$A^2 = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \times \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{bmatrix}$$ We need $a^2 + bc + bc + d^2$. $$= a^2 + 2bc + d^2$$ $$= (a + d)^2 - 2ad + 2bc$$ $$= 4 - 2(ad - bc)$$ $$= 4 - 2(-3)$$ $$= 4 + 6$$ $$= 10$$
Question 24
Maths · Straight Lines and Pair of Straight Lines · Numerical
If the sum of squares of all real values of $\alpha$, for which the lines $2x - y + 3 = 0$, $6x + 3y + 1 = 0$ and $\alpha x + 2y - 2 = 0$ do not form a triangle is $p$, then the greatest integer less than or equal to $p$ is
Answer: 32
Solution
Given the equations: $$2x - y + 3 = 0$$ $$6x + 3y + 1 = 0$$ $$\alpha x + 2y - 2 = 0$$ The line $\alpha x + 2y - 2 = 0$ will not form a triangle if it is concurrent with $2x - y + 3 = 0$ and $6x + 3y + 1 = 0$ or parallel to either of them. Case-1: Concurrent lines $$\begin{vmatrix} 2 & -1 & 3 \\ 6 & 3 & 1 \\ \alpha & 2 & -2 \end{vmatrix} = 0 \implies \alpha = \frac{4}{5}$$ Case-2: Parallel lines $$-\frac{\alpha}{2} = -\frac{6}{3} or -\frac{\alpha}{2} = 2$$ $$\implies \alpha = 4 or \alpha = -4$$ $$P = 16 + 16 + \frac{16}{25}$$ $$[P] = \left[ 32 + \frac{16}{25} \right] = 32$$
Question 25
Maths · Binomial Theorem · Numerical
The coefficient of $x^{2012}$ in the expansion of $(1-x)^{2008}(1+x+x^2)^{2007}$ is equal to
Answer: 0
Solution
$(1-x)(1-x)^{2007}(1+x+x^2)^{2007}$ $(1-x)(1-x^3)^{2007}$ $(1-x)\left({}^{2007}C_0 - {}^{2007}C_1(x^3) + \ldots\right)$ General term $(1-x)\left((-1)^r {}^{2007}C_r x^{3r}\right)$ $(-1)^r {}^{2007}C_r x^{3r} - (-1)^r {}^{2007}C_r x^{3r+1}$ $3r = 2012$ $r \neq \dfrac{2012}{3}$ $3r + 1 = 2012$ $3r = 2011$ $r \neq \dfrac{2011}{3}$ Hence there is no term containing $x^{2012}$. So coefficient of $x^{2012} = 0$
Question 26
Maths · Differential Equations · Numerical
If the solution curve, of the differential equation $\frac{dy}{dx} = \frac{x+y-2}{x-y}$ passing through the point $(2, 1)$ is $$\tan^{-1}\left(\frac{y-1}{x-1}\right) - \frac{1}{\beta} \log_e\left(\alpha + \left(\frac{y-1}{x-1}\right)^2\right) = \log_e |x-1|,$$ then $5\beta + \alpha$ is equal to
Answer: 11
Solution
Given $\($ $\frac{dy}{dx}$ = $\frac{x+y-2}{x-y}$ $\)$. Let $\($ x = X + h, $\ $y = Y + k $\)$. Then, $\($ $\frac{dY}{dX}$ = $\frac{X+Y}{X-Y}$ $\)$. Solving $\($ h + k - 2 = 0 $\)$ and $\($ h - k = 0 $\)$, we find $\($ h = k = 1 $\)$. Thus, $\($ Y = vX $\)$. Substituting, we have $\($ v + $\frac{dv}{dX}$ = $\frac{1+v}{1-v}$ $\Rightarrow$ X - $\frac{dv}{dX}$ = $\frac{1+v^2}{1-v}$ $\)$. Rearranging gives $\($ $\frac{1-v}{1+v^2}$ dv = $\frac{dX}{X}$ $\)$. Integrating, we get $\($ $\tan$^{-1} v - $\frac{1}{2}$ $\ln$(1+v^2) = $\ln$ |X| + C $\)$. As the curve passes through $\($(2, 1)$\)$, $\($ $\tan$^{-1} $\left$( $\frac{y-1}{x-1}$ $\right$) - $\frac{1}{2}$ $\ln$ $\left$( 1 + $\left$( $\frac{y-1}{x-1}$ $\right$)^2 $\right$) = $\ln$ |x-1| $\)$. Therefore, $\($ $\alpha$ = 1 $\)$ and $\($ $\beta$ = 2 $\)$. Thus, $\($ 5$\beta$ + $\alpha$ = 11 $\)$.
Question 27
Maths · Integrals · Numerical
Let $f(x) = \int_0^x g(t) \log_e \left( \frac{1-t}{1+t} \right) dt$, where $g$ is a continuous odd function. If $\int_{-\pi/2}^{\pi/2} \left( f(x) + \frac{x^2 \cos x}{1+e^x} \right) dx = \left( \frac{\pi}{\alpha} \right)^2 - \alpha$, then $\alpha$ is equal to
Consider a circle $(x - \alpha)^2 + (y - \beta)^2 = 50$, where $\alpha, \beta > 0$. If the circle touches the line $y + x = 0$ at the point $P$, whose distance from the origin is $4\sqrt{2}$, then $(\alpha + \beta)^2$ is equal to
Answer: 100
Solution
The equation of the circle is given by $S : (x - \alpha)^2 + (y - \beta)^2 = 50$. The distance $CP = r$. The line $x + y = 0$ gives the condition $$\frac{|\alpha + \beta|}{\sqrt{2}} = 5\sqrt{2}$$ which implies $$(\alpha + \beta)^2 = 100.$$
Question 29
Maths · Three Dimensional Geometry · Numerical
The lines $\frac{x-2}{2} = \frac{y}{-2} = \frac{z-7}{16}$ and $\frac{x+3}{4} = \frac{y+2}{3} = \frac{z+2}{1}$ intersect at the point $P$. If the distance of $P$ from the line $\frac{x+1}{2} = \frac{y-1}{3} = \frac{z-1}{1}$ is $l$, then $14l^2$ is equal to
Answer: 108
Solution
Given $\frac{x-2}{1}=\frac{y}{-1}=\frac{z-7}{8}=\lambda$. $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}=k$ $\Rightarrow \lambda+2=4k-3$ $\lambda=3k-2$ $\Rightarrow k=1,\ \lambda=-1$ $8\lambda+7=k-2$ $\therefore P=(1,1,-1)$ Projection of $(2\hat{i}-2\hat{k})$ on $(2\hat{i}+3\hat{j}+\hat{k})$ is $\frac{4-2}{\sqrt{4+9+1}}=\frac{2}{\sqrt{14}}$ $\therefore l^2=8-\frac{4}{14}=\frac{108}{14}$ $\Rightarrow 14l^2=108$
Question 30
Maths · Complex Numbers and Quadratic Equations · Numerical
Let the complex numbers $\alpha$ and $\frac{1}{\alpha}$ lie on the circles $|z - z_0|^2 = 4$ and $|z - z_0|^2 = 16$ respectively, where $z_0 = 1 + i$. Then, the value of $100|\alpha|^2$ is.
Answer: 20
Solution
Given $|z - z_0|^2 = 4$. This implies $(\alpha - z_0)(\bar{\alpha} - \bar{z}_0) = 4$. Expanding, we get $\alpha \bar{\alpha} - \alpha \bar{z}_0 - z_0 \bar{\alpha} + |z_0|^2 = 4$. Thus, $|\alpha|^2 - \alpha \bar{z}_0 - z_0 \bar{\alpha} = 2 \ldots (1)$. Given $|z - z_0|^2 = 16$. This implies $\left(\frac{1}{\bar{\alpha}} - z_0\right)\left(\frac{1}{\alpha} - \bar{z}_0\right) = 16$. Expanding, we get $(1 - \bar{\alpha} z_0)(1 - \alpha \bar{z}_0) = 16 |\alpha|^2$. Thus, $1 - \bar{\alpha} z_0 - \alpha \bar{z}_0 + |\alpha|^2 |z_0|^2 = 16$. Therefore, $1 - \bar{\alpha} z_0 - \alpha \bar{z}_0 = 14 |\alpha|^2 \ldots (2)$. From (1) and (2), We have $5 |\alpha|^2 = 1$. Thus, $100 |\alpha|^2 = 20$.
Physics
Question 31
Physics · Physical World, Units and Measurements · Single correct
The equation of state of a real gas is given by $$\left( P + \frac{a}{V^2} \right)(V-b) = RT$$, where $P$, $V$ and $T$ are pressure, volume and temperature respectively and $R$ is the universal gas constant. The dimensions of $\frac{a}{b^2}$ is similar to that of:
$PV$
$P$
$RT$
$R$
Answer: (b)
Solution
Question 32
Physics · Current Electricity · Single correct
Wheatstone bridge principle is used to measure the specific resistance ($S_1$) of given wire, having length $L$, radius $r$. If $X$ is the resistance of wire, then specific resistance is : $S_1 = X \left( \frac{\pi r^2}{L} \right)$. If the length of the wire gets doubled then the value of specific resistance will be :
$\frac{S_1}{4}$
$2 S_1$
$\frac{S_1}{2}$
$S_1$
Answer: (d)
Solution
As specific resistance does not depend on the dimension of the wire, it will not change.
Question 33
Physics · Gravitation · Single correct
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The angular speed of the moon in its orbit about the earth is more than the angular speed of the earth in its orbit about the sun. Reason (R): The moon takes less time to move around the earth than the time taken by the earth to move around the sun. In the light of the above statements, choose the most appropriate answer from the options given below :
is correct but (R) is not correct
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)
is not correct but (R) is correct
Answer: (b)
Solution
Given $\omega = \frac{2\pi}{T}$, it follows that $\omega \propto \frac{1}{T}$. The period of the moon is $T_{moon} = 27 \, days$ and the period of the earth is $T_{earth} = 365 \, days \, 4 \, hour$. Therefore, $\omega_{moon} > \omega_{earth}$.
Question 34
Physics · Laws of Motion · Single correct
Given below are two statements: Statement (I) : The limiting force of static friction depends on the area of contact and independent of materials. Statement (II) : The limiting force of kinetic friction is independent of the area of contact and depends on materials. In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I is correct but Statement II is incorrect
Statement I is incorrect but Statement II is correct
Both Statement I and Statement II are incorrect
Both Statement I and Statement II are correct
Answer: (b)
Solution
Coefficient of friction depends on surface in contact. So, it depends on the material of the object.
Question 35
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
The expression for the circuit is given by $$Y = A \cdot \overline{B} + \overline{A} \cdot B$$. This is an XOR gate.
Question 36
Physics · Current Electricity · Single correct
A current of 200 $\mu A$ deflects the coil of a moving coil galvanometer through $60^\circ$. The current to cause deflection through $\frac{\pi}{10}$ radian is :
The atomic mass of $_6^{}C^{12}$ is $12.000000 \, \mathrm{u}$ and that of $_6^{}C^{13}$ is $13.003354 \, \mathrm{u}$. The required energy to remove a neutron from $_6^{}C^{13}$, if mass of neutron is $1.008665 \, \mathrm{u}$, will be:
$62.5 \, \mathrm{MeV}$
$6.25 \, \mathrm{MeV}$
$4.95 \, \mathrm{MeV}$
$49.5 \, \mathrm{MeV}$
Answer: (c)
Solution
Given the reaction: $$^6C^{13} + Energy \rightarrow ^6C^{12} + ^0n^1$$ The change in mass is given by $$\Delta m = (12.000000 + 1.008665) - 13.003354$$ $$= -0.00531 \, u$$ Therefore, the energy required is $$= 0.00531 \times 931.5 \, MeV$$ $$= 4.95 \, MeV$$
Question 38
Physics · Oscillations · Single correct
A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle ($\theta$) of thread deflection in the extreme position will be:
$\tan^{-1}(\sqrt{2})$
$2 \tan^{-1}\left(\frac{1}{2}\right)$
$\tan^{-1}\left(\frac{1}{2}\right)$
$2 \tan^{-1}\left(\frac{1}{\sqrt{5}}\right)$
Answer: (b)
Solution
Loss in kinetic energy = Gain in potential energy $$\Rightarrow \frac{1}{2} mv^2 = mg\ell (1 - \cos \theta)$$ $$\Rightarrow \frac{v^2}{\ell} = 2 \, g (1 - \cos \theta)$$ Acceleration at lowest point = $\($ $\frac{v^2}{\ell}$ $\)$ Acceleration at extreme point = $\($ g $\sin$ $\theta$ $\)$ Hence, $\($ $\frac{v^2}{\ell}$ = g $\sin$ $\theta$ $\)$ $\($ $\therefore$ $\sin$ $\theta$ = 2 (1 - $\cos$ $\theta$) $\)$ $$\Rightarrow \tan \frac{\theta}{2} = \frac{1}{2} \Rightarrow \theta = 2 \tan^{-1} \left( \frac{1}{2} \right)$$
Question 39
Physics · Current Electricity · Single correct
Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across $A$ and $B$, their readings are $V_1$, $V_2$ and $V_3$. Choose the correct option.
The total kinetic energy of 1 mole of oxygen at 27°C is : [Use universal gas constant (R) = 8.31 J/mole K ]
6845.5 J
5942.0 J
6232.5 J
5670.5 J
Answer: (c)
Solution
Kinetic energy is given by the formula $$\frac{f}{2} nRT$$. Substituting the values, we have $$= \frac{5}{2} \times 1 \times 8.31 \times 300 \, \mathrm{J}$$ $$= 6232.5 \, \mathrm{J}$$
Question 41
Physics · Physical World, Units and Measurements · Single correct
Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$. Assertion $(A)$: In Vernier calliper if positive zero error exists, then while taking measurements, the reading taken will be more than the actual reading. Reason $(R)$: The zero error in Vernier Calliper might have happened due to manufacturing defect or due to rough handling. 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 true but $(R)$ is false
$(A)$ is false but $(R)$ is true
Answer: (b)
Solution
Assertion & Reason both are correct Theory
Question 42
Physics · Alternating Current · Single correct
Primary side of a transformer is connected to $230 \, \mathrm{V}$, $50 \, \mathrm{Hz}$ supply. Turns ratio of primary to secondary winding is $10 : 1$. Load resistance connected to secondary side is $46 \, \Omega$. The power consumed in it is:
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio of $\frac{C_p}{C_v}$ for the gas is :
$\frac{5}{3}$
$\frac{3}{2}$
$\frac{7}{5}$
$\frac{9}{7}$
Answer: (b)
Solution
Given $P \propto T^3$, we have $PT^{-3} = const$. From $PV^\gamma = const$, we can write $P \left( \frac{nRT}{P} \right)^\gamma = const$. This implies $P^{1-\gamma} T^\gamma = const$. Therefore, $PT^{\frac{\gamma}{1-\gamma}} = const$. Solving $\frac{\gamma}{1-\gamma} = -3$, we get $\gamma = -3 + 3\gamma$. Simplifying gives $3 = 2\gamma$, so $\gamma = \frac{3}{2}$.
Question 44
Physics · Dual Nature of Radiation and Matter · Single correct
The threshold frequency of a metal with work function $6.63 \, \mathrm{eV}$ is:
Physics · Mechanical Properties of Solids · Single correct
Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$. Assertion $(A)$: The property of a body, by virtue of which it tends to regain its original shape when the external force is removed, is elasticity. Reason $(R)$: The restoring force depends upon the bonded interatomic and intermolecular forces of a solid. In the light of the above statements, choose the correct answer from the options given below:
Assertion $(A)$ is false but Reason $(R)$ is true.
Assertion $(A)$ is true but Reason $(R)$ is false.
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 $(A)$.
Answer: (c)
Solution
Theory
Question 46
Physics · Wave Optics · Single correct
When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of:
$60^\circ$
$30^\circ$
$90^\circ$
$45^\circ$
Answer: (d)
Solution
Let $I_0$ be intensity of unpolarised light incident on first polaroid. $I_1 =$ Intensity of light transmitted from 1st polaroid $$I_1 = \frac{I_0}{2}$$ $\theta$ be the angle between 1st and 2nd polaroid $\phi$ be the angle between 2nd and 3rd polaroid $$\theta + \phi = 90^\circ (as 1st and 3rd polaroid are crossed)$$ $$\phi = 90^\circ - \theta$$ $I_2 =$ Intensity from 2nd polaroid $$I_2 = I_1 \cos^2 \theta = \frac{I_0}{2} \cos^2 \theta$$ $I_3 =$ Intensity from 3rd polaroid $$I_3 = I_2 \cos^2 \phi$$ $$I_3 = I_1 \cos^2 \theta \cos^2 \phi$$ $$I_3 = \frac{I_0}{2} \cos^2 \theta \cos^2 \phi$$ $$\phi = 90^\circ - \theta$$ $$I_3 = \frac{I_0}{2} \cos^2 \theta \sin^2 \theta$$ $$I_3 = \frac{I_0}{2} \left[ \frac{2 \sin \theta \cos \theta}{2} \right]^2$$ $$I_3 = \frac{I_0}{8} \sin^2 2\theta$$ $I_3$ will be maximum when $\sin 2\theta = 1$ $$2\theta = 90^\circ$$ $$\theta = 45^\circ$$
Question 47
Physics · Electromagnetic Waves · Single correct
An object is placed in a medium of refractive index 3. An electromagnetic wave of intensity $6 \times 10^8 \, \mathrm{W/m^2}$ falls normally on the object and it is absorbed completely. The radiation pressure on the object would be (speed of light in free space $= 3 \times 10^8 \, \mathrm{m/s}$):
$36 \, \mathrm{Nm^{-2}}$
$18 \, \mathrm{Nm^{-2}}$
$6 \, \mathrm{Nm^{-2}}$
$2 \, \mathrm{Nm^{-2}}$
Answer: (c)
Solution
Radiation pressure is given by the formula $$\frac{I}{v}$$. This can be expressed as $$\frac{I \cdot \mu}{c}$$. Substituting the values, we have $$\frac{6 \times 10^8 \times 3}{3 \times 10^8}$$. This simplifies to $$6 \, \mathrm{N/m^2}$$.
Question 48
Physics · Electric Charges and Fields · Single correct
Given below are two statements : one is labelled a Assertion (A) and the other is labelled as Reason (R) Assertion (A) : Work done by electric field on moving a positive charge on an equipotential surface is always zero. Reason (R) : Electric lines of forces are always perpendicular to equipotential surfaces. In the light of the above statements, choose the most appropriate answer from the options given below :
Both (A) and (R) are correct but (R) is not the correct explanation of (A)
is correct but (R) is not correct
is not correct but (R) is correct
Both (A) and (R) are correct and (R) is the correct explanation of (A)
Answer: (d)
Solution
Electric lines of force are always perpendicular to equipotential surfaces, so the angle between force and displacement will always be $90^\circ$. So work done is equal to $0$.
Question 49
Physics · Laws of Motion · Single correct
A heavy iron bar of weight 12 kg is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle $60^\circ$ with the horizontal, the weight experienced by the man is :
Physics · Motion in a Straight Line · Single correct
A bullet is fired into a fixed target looses one third of its velocity after travelling 4 cm. It penetrates further $D \times 10^{-3} \, \mathrm{m}$ before coming to rest. The value of D is :
Physics · Moving Charges and Magnetism · Numerical
The magnetic field at the centre of a wire loop formed by two semicircular wires of radii $R_1 = 2\pi \mathrm{m}$ and $R_2 = 4\pi \mathrm{m}$ carrying current $I = 4 \, \mathrm{A}$ as per figure given below is $\alpha \times 10^{-7} \, \mathrm{T}$. The value of $\alpha$ is _____ (Centre $O$ is common for all segments)
Answer: 3
Solution
The expression for the magnetic field is given by: $$ \frac{\mu_0 i}{2R_2} \left( \frac{\pi}{2\pi} \right) \otimes + \frac{\mu_0 i}{2R_1} \left( \frac{\pi}{2\pi} \right) \otimes $$ This simplifies to: $$ \left( \frac{\mu_0 i}{4R_2} + \frac{\mu_0 i}{4R_1} \right) \otimes $$ Calculating further: $$ \frac{4\pi \times 10^{-7} \times 4}{4 \times 4\pi} + \frac{4\pi \times 10^{-7} \times 4}{4 \times 2\pi} $$ This results in: $$ = 3 \times 10^{-7} = \alpha \times 10^{-7} $$ Therefore, $\($ $\alpha$ = 3 $\)$.
Question 52
Physics · Electric Charges and Fields · Numerical
Two charges of $-4 \mu \mathrm{C}$ and $+4 \mu \mathrm{C}$ are placed at the points $A(1, 0, 4) \, \mathrm{m}$ and $B(2, -1, 5) \, \mathrm{m}$ located in an electric field $\vec{E} = 0.20 \hat{i} \, \mathrm{V/cm}$. The magnitude of the torque acting on the dipole is $8 \sqrt{\alpha} \times 10^{-5} \, \mathrm{Nm}$, Where $\alpha =$
Answer: 2
Solution
The torque $\vec{\tau}$ is given by $\vec{p} \times \vec{E}$. The dipole moment $\vec{p}$ is $q \vec{\ell}$. The electric field $\vec{E}$ is $0.2 \, \mathrm{V/cm} = 20 \, \mathrm{V/m}$. The dipole moment $\vec{p}$ is $4 \times (\hat{i} - \hat{j} + \hat{k})$. This simplifies to $(4 \hat{i} - 4 \hat{j} + 4 \hat{k}) \, \mu \mathrm{C} \cdot \mathrm{m}$. The torque $\vec{\tau}$ is $(4 \hat{i} - 4 \hat{j} + 4 \hat{k}) \times (20 \hat{i}) \times 10^{-6} \, \mathrm{Nm}$. This results in $(8 \hat{k} + 8 \hat{j}) \times 10^{-5} = 8 \sqrt{2} \times 10^{-5}$. Therefore, $\alpha = 2$.
Question 53
Physics · Waves · Numerical
A closed organ pipe 150 $\mathrm{\ cm}$ long gives 7 beats per second with an open organ pipe of length 350 $\mathrm{\ cm}$, both vibrating in fundamental mode. The velocity of sound is $\mathrm{m/s}$.
Answer: 294
Solution
Given $f_c = \frac{v}{4\ell_1}$ and $f_o = \frac{v}{2\ell_2}$. The condition is $|f_c - f_o| = 7$. Substituting the values, we have: $$\frac{v}{4 \times 150} - \frac{v}{2 \times 350} = 7$$ Simplifying, we get: $$\frac{v}{600 \, \mathrm{cm}} - \frac{v}{700 \, \mathrm{cm}} = 7$$ Converting to meters: $$\frac{v}{6 \, \mathrm{m}} - \frac{v}{7 \, \mathrm{m}} = 7$$ This simplifies to: $$v \left( \frac{1}{42} \right) = 7$$ Solving for $v$ gives: $$v = 42 \times 7$$ Thus, $v = 294 \, \mathrm{m/s}$.
Question 54
Physics · Motion in a Straight Line · Numerical
A body falling under gravity covers two points $A$ and $B$ separated by $80 \, \mathrm{m}$ in $2 \, \mathrm{s}$. The distance of upper point $A$ from the starting point is _____ $\mathrm{m}$ (use $g = 10 \, \mathrm{ms^{-2}}$)
Answer: 45
Solution
From A to B $$-80 = -v_1 t - \frac{1}{2} \times 10 t^2$$ $$-80 = -2v_1 - \frac{1}{2} \times 10 \times 2^2$$ $$-80 = -2v_1 - 20$$ $$-60 = -2v_1$$ $$v_1 = 30 \, \mathrm{m/s}$$ From O to A $$v^2 = u^2 + 2gS$$ $$30^2 = 0 + 2 \times (-10)(-S)$$ $$900 = 20S$$ $$S = 45 \, \mathrm{m}$$
Question 55
Physics · Mechanical Properties of Fluids · Numerical
The reading of pressure metre attached with a closed pipe is $4.5 \times 10^4 \, \mathrm{N/m^2}$. On opening the valve, water starts flowing and the reading of pressure metre falls to $2.0 \times 10^4 \, \mathrm{N/m^2}$. The velocity of water is found to be $\sqrt{V} \, \mathrm{m/s}$. The value of $V$ is _____
Physics · System of Particles and Rotational Motion · Numerical
A ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is $\frac{7}{x}$ where $x$ is _____
Answer: 7
Solution
In pure rolling work done by friction is zero. Hence potential energy is converted into kinetic energy. Since initially the ring and the sphere have same potential energy, finally they will have same kinetic energy too. Therefore, the ratio of kinetic energies is $1$. $$\frac{7}{x} = 1 \Rightarrow x = 7$$
Question 57
Physics · Wave Optics · Numerical
A parallel beam of monochromatic light of wavelength $5000 \cdot \mathrm{Å}$ is incident normally on a single narrow slit of width $0.001 \, \mathrm{mm}$. The light is focused by convex lens on screen, placed on its focal plane. The first minima will be formed for the angle of diffraction of _____ (degree).
Answer: 30
Solution
For first minima $$a \sin \theta = \lambda$$ $$\Rightarrow \sin \theta = \frac{\lambda}{a} = \frac{5000 \times 10^{-10}}{1 \times 10^{-6}} = \frac{1}{2}$$ $$\Rightarrow \theta = 30^\circ$$
Question 58
Physics · Electric Charges and Fields · Numerical
The electric potential at the surface of an atomic nucleus $(z = 50)$ of radius $9 \times 10^{-13} \, \mathrm{cm}$ is _____ $\times 10^6 \, \mathrm{V}$
If Rydberg's constant is $R$, the longest wavelength of radiation in Paschen series will be $\frac{\alpha}{7R}$, where $\alpha$ =.
Answer: 144
Solution
Longest wavelength corresponds to transition between $n = 3$ and $n = 4$ $$\frac{1}{\lambda} = RZ^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = RZ^2 \left( \frac{1}{9} - \frac{1}{16} \right)$$ $$= \frac{7RZ^2}{9 \times 16}$$ $$\Rightarrow \lambda = \frac{144}{7R} for Z = 1 \therefore \alpha = 144$$
Question 60
Physics · Alternating Current · Numerical
A series LCR circuit with $L = \frac{100}{\pi} \, \mathrm{mH}$, $C = \frac{10^{-3}}{\pi} \, \mathrm{F}$ and $R = 10 \Omega$, is connected across an ac source of 220 V, 50 Hz supply. The power factor of the circuit would be
Answer: 1
Solution
Given $X_c = \frac{1}{\omega C} = \frac{\pi}{2\pi \times 50 \times 10^{-3}} = 10 \Omega$. $X_L = \omega L = 2\pi \times 50 \times \frac{100}{\pi} \times 10^{-3} = 10 \Omega$. Since $X_C = X_L$, the circuit is in resonance. Therefore, the power factor $= \frac{R}{Z} = \frac{R}{R} = 1$.
Chemistry
Question 61
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The order of relative stability of the contributing structure is Choose the correct answer from the options given below:
I > II > III
II > I > III
I = II = III
III > II > I
Answer: (a)
Solution
I > II > III, since neutral resonating structures are more stable than charged resonating structure. II > III, since stability of structure with negative charge on more electronegative atom is higher.
Question 62
Chemistry · Haloalkanes and Haloarenes · Single correct
Which among the following halide/s will not show $S_N1$ reaction:
$\mathrm{H_2C = CH - CH_2Cl}$
$\mathrm{CH_3 - CH = CH - Cl}$
Answer: (d)
Solution
Since $\mathrm{CH_3-CH=CH^{+}}$ is very unstable, $\mathrm{CH_3-CH=CH-Cl}$ cannot give $\mathrm{S_N1}$ reaction.
Question 63
Chemistry · Electrochemistry · Single correct
Which of the following statements is not correct about rusting of iron?
Coating of iron surface by tin prevents rusting, even if the tin coating is peeling off.
When pH lies above 9 or 10, rusting of iron does not take place.
Dissolved acidic oxides $\mathrm{SO_2}$, $\mathrm{NO_2}$ in water act as catalyst in the process of rusting.
Rusting of iron is envisaged as setting up of electrochemical cell on the surface of iron object.
Answer: (a)
Solution
As tin coating is peeled off, then iron is exposed to atmosphere.
Question 64
Chemistry · The d-and f-Block Elements · Single correct
Given below are two statements: Statement (I) : In the Lanthanoids, the formation of Ce$^{+4}$ is favoured by its noble gas configuration. Statement (II) : Ce$^{+4}$ is a strong oxidant reverting to the common +3 state. In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I is false but Statement II is true
Both Statement I and Statement II are true
Statement I is true but Statement II is false
Both Statement I and Statement II are false
Answer: (b)
Solution
Statement (1) is true, $\mathrm{Ce}^{+4}$ has noble gas electronic configuration. Statement (2) is also true due to high reduction potential for $\mathrm{Ce}^{4+}/\mathrm{Ce}^{3+} (+1.74 \, \mathrm{V})$, and stability of $\mathrm{Ce}^{3+}$, $\mathrm{Ce}^{4+}$ acts as strong oxidizing agent.
Question 65
Chemistry · The d-and f-Block Elements · Single correct
Choose the correct option having all the elements with $d^{10}$ electronic configuration from the following:
The electronic configurations are given as follows: $$[\mathrm{Cr}] = [\mathrm{Ar}] 4s^1 3d^5$$ $$[\mathrm{Cd}] = [\mathrm{Kr}] 5s^2 4d^{10}$$ $$[\mathrm{Cu}] = [\mathrm{Ar}] 4s^1 3d^{10}$$ $$[\mathrm{Ag}] = [\mathrm{Kr}] 5s^1 4d^{10}$$ $$[\mathrm{Zn}] = [\mathrm{Ar}] 4s^2 3d^{10}$$
Question 66
Chemistry · Alcohols, Phenols and Ethers · Single correct
Phenolic group can be identified by a positive:
Phthalein dye test
Lucas test
Tollen's test
Carbylamine test
Answer: (a)
Solution
Carbylamine Test - Identification of primary amines. Lucas Test - Differentiation between $1^\circ$, $2^\circ$ and $3^\circ$ alcohols. Tollen's Test - Identification of Aldehydes. Phthalein Dye Test - Identification of phenols.
Question 67
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The molecular formula of second homologue in the homologous series of mono carboxylic acids is
$\mathrm{C_3H_6O_2}$
$\mathrm{C_2H_4O_2}$
$\mathrm{CH_2O}$
$\mathrm{C_2H_2O_2}$
Answer: (b)
Solution
The second homologue of $\mathrm{HCOOH}$ is $\mathrm{CH_3COOH}$.
Question 68
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The technique used for purification of steam volatile water immiscible substance is:
Fractional distillation
Fractional distillation under reduced pressure
Distillation
Steam distillation
Answer: (d)
Solution
Steam distillation is used for those liquids which are insoluble in water, containing non-volatile impurities and are steam volatile.
Question 69
Chemistry · Hydrocarbons · Single correct
The final product A, formed in the following reaction sequence is:
Answer: (d)
Solution
The reaction starts with the hydroboration-oxidation of $\mathrm{PhCH=CH_2}$ using $\mathrm{B_2H_6/H_2O_2, OH^-}$ to form $\mathrm{PhCH_2CH_2OH}$. Next, $\mathrm{PhCH_2CH_2OH}$ reacts with $\mathrm{HBr}$ to form $\mathrm{PhCH_2CH_2Br}$ and $\mathrm{H_2O}$ via $\mathrm{S_N^{GP}}$ mechanism. Then, $\mathrm{PhCH_2CH_2Br}$ is treated with $\mathrm{Mg/dry\ ether}$ to form $\mathrm{PhCH_2CH_2MgBr}$. Finally, $\mathrm{PhCH_2CH_2MgBr}$ reacts with (I) $\mathrm{H-C=O}$ and (II) $\mathrm{H_3O^+}$ to yield $\mathrm{PhCH_2CH_2CH_2OH}$.
Question 70
Chemistry · Co-ordination Compounds · Single correct
Match List-I with List-II Choose the correct answer from the options given below:
(A)-(IV), (B)-(I), (C)-(III), (D)-(II)
(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
(A)-(II), (B)-(I), (C)-(III), (D)-(IV)
(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Answer: (d)
Solution
The reaction shown is: $$\mathrm{PhO^- + CH_3Cl \rightarrow PhOCH_3 + Cl^-}$$ This is an example of the Williamson ether synthesis, which is not listed among the given options.
Question 71
Chemistry · Haloalkanes and Haloarenes · Single correct
Major product formed in the following reaction is a mixture of:
Answer: (d)
Solution
The reaction begins with the addition of $\mathrm{H-I}$ to the ether. The iodine ion $\mathrm{I^-}$ attacks the ether oxygen, leading to the formation of an oxonium ion. This is followed by the departure of the iodide ion, resulting in the formation of an alcohol and another iodide ion.
Question 72
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Bond line formula of HOCH(CN)_2 is:
Answer: (d)
Solution
The compound CH(OH)(CN)_2 can be represented as either of the following structures: Structure 1: $$\begin{array}{c} OH \\ | \\ CH \\ / \, \\ CN CN \end{array}$$ Structure 2: $$\begin{array}{c} OH \\ | \\ CH \\ / \, \\ NC CN \end{array}$$
Question 73
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Given below are two statements: Statement (I) : Oxygen being the first member of group 16 exhibits only -2 oxidation state. Statement (II) : Down the group 16 stability of +4 oxidation state decreases and +6 oxidation state increases. In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I is correct but Statement II is incorrect
Both Statement I and Statement II are correct
Both Statement I and Statement II are incorrect
Statement I is incorrect but Statement II is correct
Answer: (c)
Solution
Statement-I: Oxygen can have oxidation state from $-2$ to $+2$, so statement I is incorrect. Statement-II: On moving down the group stability of $+4$ oxidation state increases whereas stability of $+6$ oxidation state decreases down the group, according to inert pair effect. So both statements are wrong.
Question 74
Chemistry · Co-ordination Compounds · Single correct
Identify from the following species in which $d^2sp^3$ hybridization is shown by central atom:
$[Co(NH_3)_6]^{3+}$
$BrF_5$
$[Pt(Cl)_4]^{2-}$
$SF_6$
Answer: (a)
Solution
Q6 $[\mathrm{Co(NH_3)_6}]^{+3}$ has $d^2sp^3$ hybridization. $\mathrm{BrF_5}$ has $sp^3d^2$ hybridization. $[\mathrm{PtCl_4}]^{-2}$ has $dsp^2$ hybridization. $\mathrm{SF_6}$ has $sp^3d^2$ hybridization.
Question 75
Chemistry · Haloalkanes and Haloarenes · Single correct
Identify B formed in the reaction.
\quad \mathrm{H_2N-(CH_2)_4-NH_2}
\quad \mathrm{Cl^-\,H_3N^+-(CH_2)_4-NH_3^+\,Cl^-}
Answer: (b)
Solution
The reaction starts with $\mathrm{Cl} - (\mathrm{CH}_2)_4 - \mathrm{Cl}$ reacting with excess $\mathrm{NH}_3$ to form $\mathrm{Cl}^- \; \overset{\oplus}{\mathrm{NH}_3} - (\mathrm{CH}_2)_4 \overset{\oplus}{\mathrm{NH}_3} \mathrm{Cl}^-$. This intermediate (A) is then treated with $\mathrm{NaOH}$ to yield $2\mathrm{NaCl} + 2\mathrm{H}_2\mathrm{O} + \mathrm{NH}_2 - (\mathrm{CH}_2)_4 - \mathrm{NH}_2$.
Question 76
Chemistry · Solutions · Single correct
The quantity which changes with temperature is:
Molarity
Mass percentage
Molality
Mole fraction
Answer: (a)
Solution
Molarity is given by the formula: $$Molarity = \frac{Moles of solute}{Volume of solution}.$$ Since volume depends on temperature, molarity will change upon change in temperature.
Question 77
Chemistry · Biomolecules · Single correct
Which structure of protein remains intact after coagulation of egg white on boiling?
Primary
Tertiary
Secondary
Quaternary
Answer: (a)
Solution
Boiling an egg causes denaturation of its protein resulting in loss of its quaternary, tertiary and secondary structures.
Question 78
Chemistry · Redox Reactions · Single correct
Which of the following cannot function as an oxidising agent?
$\mathrm{N}^{3-}$
$\mathrm{SO}_4^{2-}$
$\mathrm{BrO}_3^{-}$
$\mathrm{MnO}_4^{-}$
Answer: (a)
Solution
In $\mathrm{N^{3-}}$ ion 'N' is present in its lowest possible oxidation state, hence it cannot be reduced further because of which it cannot act as an oxidizing agent.
Question 79
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The incorrect statement regarding conformations of ethane is:
Ethane has infinite number of conformations
The dihedral angle in staggered conformation is $60^\circ$
Eclipsed conformation is the most stable conformation.
The conformations of ethane are interconvertible to one-another.
Answer: (c)
Solution
Eclipsed conformation is the least stable conformation of ethane.
Question 80
Chemistry · Co-ordination Compounds · Single correct
Identity the incorrect pair from the following:
Photography - AgBr
Polythene preparation - TiCl_4, Al(CH_3)_3
Haber process - Iron
Wacker process – PtCl_2
Answer: (d)
Solution
The catalyst used in Wacker's process is $\mathrm{PdCl_2}$.
Question 81
Chemistry · Classification of Elements and Periodicity in Properties · Numerical
Total number of ions from the following with noble gas configuration is $Sr^{2+}$ (Z = 38), $Cs^{+}$ (Z = 55), $La^{2+}$ (Z = 57), $Pb^{2+}$ (Z = 82), $Yb^{2+}$ (Z = 70) and $Fe^{2+}$ (Z = 26)
Chemistry · Chemical Bonding and Molecular Structure · Fill in the blank
The number of non-polar molecules from the following is \_\_\_\_ $\mathrm{HF},\ \mathrm{H_2O},\ \mathrm{SO_2},\ \mathrm{H_2},\ \mathrm{CO_2},\ \mathrm{CH_4},\ \mathrm{NH_3},\ \mathrm{HCl},\ \mathrm{CHCl_3},\ \mathrm{BF_3}$
Answer: 4
Solution
The non-polar molecules are $\mathrm{CO_2}$, $\mathrm{H_2}$, $\mathrm{CH_4}$ and $\mathrm{BF_3}$.
Question 83
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
Time required for completion of 99.9$\%$ of a First order reaction is $\_$$\_$$\_$$\_$$\_$ times of half life $(t_{1/2})$ of the reaction.
The Spin only magnetic moment value of square planar complex $[\mathrm{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl} (\mathrm{NH}_2\mathrm{CH}_3)] \mathrm{Cl}$ is ______ B.M. (Nearest integer) (Given atomic number for $\mathrm{Pt} = 78$)
Answer: 0
Solution
For $\mathrm{Pt^{2+}}$ with $(d^8)$ configuration, the electrons are arranged in the 5d orbitals. The 6s and 6p orbitals are empty. The hybridization is $dsp^2$ and there are no unpaired electrons. Therefore, the magnetic moment is $0$.
Question 85
Chemistry · Thermodynamics · Numerical
For a certain thermochemical reaction M $\rightarrow$ N at $T$ = 400 \, $\mathrm{K}$, $\Delta H^\circ$ = 77.2 \, $\mathrm{kJ \, mol^{-1}}$, $\Delta S = 122 \, \mathrm{J \, K^{-1}}$, log equilibrium constant (log K) is ______ $\times 10^{-1}$.
Answer: 37
Solution
Given the equation for Gibbs free energy change: $$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$$ Substitute the given values: $$= 77.2 \times 10^3 - 400 \times 122 = 28400 \, \mathrm{J}$$ Now, using the relation: $$\Delta G^\circ = -2.303RT \log K$$ Substitute the values: $$\Rightarrow 28400 = -2.303 \times 8.314 \times 400 \log K$$ Solve for $\log K$: $$\Rightarrow \log K = -3.708 = -37.08 \times 10^{-1}$$
Question 86
Chemistry · Some Basic Concepts of Chemistry · Numerical
Volume of 3M $\mathrm{NaOH}$ (formula weight $40 \, \mathrm{g \, mol^{-1}}$) which can be prepared from $84 \, \mathrm{g}$ of $\mathrm{NaOH}$ is ______ $\times 10^{-1} \, \mathrm{dm^3}$.
Answer: 7
Solution
Given the molarity equation: $$M = \frac{n_{\mathrm{NaOH}}}{V_{\mathrm{sol}} (in L)}$$ which implies $$3 = \frac{(84/40)}{V}$$ leading to $$V = 0.7 \, \mathrm{L} = 7 \times 10^{-1} \, \mathrm{L}$$
Question 87
Chemistry · Redox Reactions · Numerical
1 mole of PbS is oxidised by " X " moles of $O_3$ to get " Y " moles of $O_2$. X + Y =
Answer: 8
Solution
The balanced chemical equation is: $$\mathrm{PbS} + 4\mathrm{O}_3 \rightarrow \mathrm{PbSO}_4 + 4\mathrm{O}_2$$ The values are $x = 4$, $y = 4$.
Question 88
Chemistry · Electrochemistry · Fill in the blank
The hydrogen electrode is dipped in a solution of pH $=3$ at $25^\circ\mathrm{C}$. The potential of the electrode will be $\_\_\_\_\times10^{-2}\,\mathrm{V}$. $\left(\frac{2.303RT}{F}=0.059\,\mathrm{V}\right)$
Answer: 18
Solution
The reaction is given by: $$2\mathrm{H}^+_{(\mathrm{aq.})} + 2e^- \rightarrow \mathrm{H}_2\,(\mathrm{g})$$ The cell potential is calculated using the Nernst equation: $$E_{cell} = E^0_{cell} - \frac{0.059}{2} \log \frac{P_{\mathrm{H}_2}}{[\mathrm{H}^+]^2}$$ Substituting the values, we have: $$= 0 - 0.059 \times 3 = -0.177 volts. = -17.7 \times 10^{-2} V.$$
Question 89
Chemistry · Amines · Numerical
9.3 $\mathrm{g}$ of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is 100$\%$ completed is $\times$ $10^{-1}$ $\mathrm{g}$. (Given molar mass in $\mathrm{g/mol^{-1}}$ N : 14, O : 16, C : 12, H : 1)