JEE Main 23 January 2025 Shift 2 question paper with solutions
JEE Main 23 January 2025 Shift 2: all 75 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 · Binomial Theorem · Single correct
If in the expansion of $(1+x)^p(1-x)^q$, the coefficients of $x$ and $x^2$ are 1 and -2, respectively, then $p^2 + q^2$ is equal to:
Let \[ A=\{(x,y)\in\mathbb{R}\times\mathbb{R}:|x+y|\geq 3\} \] and \[ B=\{(x,y)\in\mathbb{R}\times\mathbb{R}:|x|+|y|\leq 3\}. \] If \[ C=\{(x,y)\in A\cap B:x=0\text{ or }y=0\}, \] then \[ \sum_{(x,y)\in C}|x+y| \] is:
15
24
18
12
Answer: (d)
Solution
Given $$A = \{(x, y) \in \mathbb{R} \times \mathbb{R} : |x + y| \geq 3\}$$ and $$B = \{(x, y) \in \mathbb{R} \times \mathbb{R} : |x| + |y| \leq 3\}$$ $$C = \{(x, y) \in A \cap B : x = 0 or y = 0\}$$ $A \cap B$ will have only common points lying on the line $PQ$ and $RS$. Now, $$C = \{(-3, 0), (3, 0), (0, 3), (0, -3)\}$$ $$\sum_{(x,y) \in C} |x + y| = 3 + 3 + 3 + 3 = 12$$
Question 3
Maths · Determinants · Single correct
The system of equations $$x + y + z = 6$$ $$x + 2y + 5z = 9,$$ $$x + 5y + \lambda z = \mu,$$ has no solution if
$\lambda = 15, \mu \neq 17$
$\lambda \neq 17, \mu \neq 18$
$\lambda = 17, \mu \neq 18$
$\lambda = 17, \mu = 18$
Answer: (c)
Solution
Given the determinant equation: $$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 1 & 5 & \lambda \end{vmatrix} = 0$$ Solving for $\lambda$, we find: $$\lambda = 17$$ For the determinant $D_z$: $$D_z = \begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 9 \\ 1 & 5 & \mu \end{vmatrix} \neq 0$$ This implies: $$\mu \neq 18$$
Question 4
Maths · Integrals · Single correct
Let $\int x^3 \sin x\, dx = g(x) + C$, where $C$ is the constant of integration. If $8\left(g\left(\dfrac{\pi}{2}\right) + g'\left(\dfrac{\pi}{2}\right)\right) = \alpha\pi^3 + \beta\pi^2 + \gamma$, $\alpha, \beta, \gamma \in \mathbb{Z}$, then $\alpha + \beta - \gamma$ equals :
48
55
62
47
Answer: (b)
Solution
Given $$\int x^3 \sin x \, dx = -x^3 \cos x + \int 3x^2 \cos x \, dx$$ $$= -x^3 \cos x + 3x^2 \sin x - \int 6x \sin x \, dx$$ $$= -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x + c$$ So $$g(x) = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x$$ $$g\left(\frac{\pi}{2}\right) = \frac{3\pi^2}{4} - 6$$ $$g'(x) = -3x^2 \cos x + x^3 \sin x + 6 \cos x - 6 \cos x$$ $$g'\left(\frac{\pi}{2}\right) = \frac{\pi^3}{8}$$ $$8 \left( g\left(\frac{\pi}{2}\right) + g'\left(\frac{\pi}{2}\right) \right) = \pi^3 + 6\pi^2 - 48$$ So $$\alpha + \beta - \gamma = 55$$
Question 5
Maths · Straight Lines and Pair of Straight Lines · Single correct
A rod of length eight units moves such that its ends $A$ and $B$ always lie on the lines $x - y + 2 = 0$ and $y + 2 = 0$, respectively. If the locus of the point $P$, that divides the rod $AB$ internally in the ratio $2 : 1$ is $9 \left( x^2 + \alpha y^2 + \beta xy + \gamma x + 28y \right) - 76 = 0$, then $\alpha - \beta - \gamma$ is equal to :
Maths · Three Dimensional Geometry · Single correct
The distance of the line $\($ $\frac{x-2}{2}$ = $\frac{y-6}{3}$ = $\frac{z-3}{4}$ $\)$ from the point $\($(1, 4, 0)$\)$ along the line $\($ $\frac{x}{1}$ = $\frac{y-2}{2}$ = $\frac{z+3}{3}$ $\)$ is:
$\($ $\sqrt{17}$ $\)$
$\($ $\sqrt{15}$ $\)$
$\($ $\sqrt{14}$ $\)$
$\($ $\sqrt{13}$ $\)$
Answer: (c)
Solution
Line passing through $(1,4,0)$ and parallel to $\dfrac{x}{1}=\dfrac{y-2}{2}=\dfrac{z+3}{3}$ is $L:\dfrac{x-1}{1}=\dfrac{y-4}{2}=\dfrac{z}{3}$ Any point on $L$ : $(\lambda+1,\;2\lambda+4,\;3\lambda)$ Any point on $\dfrac{x-2}{2}=\dfrac{y-6}{3}=\dfrac{z-3}{4}$ is $(2\mu+2,\;3\mu+6,\;4\mu+3)$ $\lambda+1=2\mu+2$ $2\lambda+4=3\mu+6$ $3\lambda=4\mu+3$ $\Rightarrow\lambda=1,\;\mu=0$ Point: $(2,6,3)$ Distance $=\sqrt{(2-1)^2+(6-4)^2+(3-0)^2}$ $=\sqrt{1+4+9}=\sqrt{14}$
Question 7
Maths · Three Dimensional Geometry · Single correct
Let the point A divide the line segment joining the points $P(-1, -1, 2)$ and $Q(5, 5, 10)$ internally in the ratio $r : 1 (r > 0)$. If $O$ is the origin and $\left( \overrightarrow{OQ} \cdot \overrightarrow{OA} \right) - \frac{1}{5} \left| \overrightarrow{OP} \times \overrightarrow{OA} \right|^2 = 10$, then the value of $r$ is:
Maths · Applications of Integrals · Single correct
If the area of the region $\{(x, y) : -1 \leq x \leq 1, 0 \leq y \leq a + e^{|x|} - e^{-x}, a > 0 \}$ is $\frac{e^{2} + 8e + 1}{e}$, then the value of $a$ is :
8
7
5
6
Answer: (c)
Solution
Given $y \in \left[0, a + e^{|x|} - e^{-x}\right]$. (i) If $x \geq 0 \Rightarrow y \in \left(0, a + e^x - \frac{1}{e^x}\right)$ If $x < 0 \Rightarrow y \in \left(0, a + e^{-x} - e^{-x}\right) \Rightarrow y \in (0, a)$. Area $= (a) + \int_0^1 \left(a + e^x - e^{-x}\right) \, dx = \frac{e^2 + 8e + 1}{e}$. $= a + \left(ax + e^x + e^{-x}\right)\bigg|_0^1 = e + 8 + \frac{1}{e}$. $= a + \left(a + e + \frac{1}{e} - 2\right) = e + \frac{1}{e} + 8$. $\Rightarrow 2a - 2 = 8 \Rightarrow a = 5$
Question 9
Maths · Applications of Derivatives · Single correct
A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is $1 \, \mathrm{cm}$, the ice-cream melts at the rate of $81 \, \mathrm{cm}^3/\mathrm{min}$ and the thickness of the ice-cream layer decreases at the rate of $\frac{1}{4\pi} \, \mathrm{cm/min}$. The surface area (in $\mathrm{cm}^2$) of the chocolate ball (without the ice-cream layer) is:
$196\pi$
$256\pi$
$225\pi$
$128\pi$
Answer: (b)
Solution
The volume is given by $v = \frac{4}{3} \pi r^3$. Differentiating with respect to time, we have $\frac{dv}{dt} = 4 \pi r^2 \frac{dr}{dt}$. Given $81 = 4 \pi r^2 \times \frac{1}{4 \pi}$, we find $r^2 = 81$, so $r = 9$. The surface area of the chocolate is $4 \pi (r - 1)^2 = 256 \pi$.
Question 10
Maths · Probability · Single correct
A board has 16 squares as shown in the figure: Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is:
7/10
4/5
23/30
3/5
Answer: (b)
Solution
Total = $\binom{16}{2}$ Required ways = Total - (adjacent square) = $\binom{16}{2}$ - [3 pair in vertical $\&$ horizontal for each row and column] = $\binom{16}{2}$ - [3 $\times$ 4 + 3 $\times$ 4] = 96 Probability = $\frac{96}{120}$ = $\frac{4}{5}$
Question 11
Maths · Differential Equations · Single correct
Let $x = x(y)$ be the solution of the differential equation $y = \left( x - y \frac{dx}{dy} \right) \sin \left( \frac{x}{y} \right), y > 0$ and $x(1) = \frac{\pi}{2}$. Then $\cos(x(2))$ is equal to:
$1 - 2(\log_e 2)^2$
$1 - 2 (\log_e 2)$
$2 (\log_e 2) - 1$
$2(\log_e 2)^2 - 1$
Answer: (d)
Solution
Given $y \, dy = (x \, dy - y \, dx) \sin \left( \frac{x}{y} \right)$. We have $$\frac{dy}{y} = \left( \frac{x \, dy - y \, dx}{y^2} \right) \sin \left( \frac{x}{y} \right)$$ which simplifies to $$\frac{dy}{y} = \sin \left( \frac{x}{y} \right) d \left( -\frac{x}{y} \right)$$ Integrating both sides, we get $$\ln y = \cos \frac{x}{y} + C$$ Given $x(1) = \frac{\pi}{2}$, we have $$0 = \cos \frac{\pi}{2} + C \Rightarrow C = 0$$ Thus, $$\ln y = \cos \frac{x}{y}$$ But $y = 2 \Rightarrow \cos \frac{x}{2} = \ln 2$. Therefore, $$\cos x = 2 \cos^2 \frac{x}{2} - 1$$ which simplifies to $$= 2(\ln 2)^2 - 1$$
Question 12
Maths · Relations and Functions · Single correct
Let the range of the function $f(x) = 6 + 16 \cos x \cdot \cos \left( \frac{\pi}{3} - x \right) \cdot \cos \left( \frac{\pi}{3} + x \right) \cdot \sin 3x \cdot \cos 6x, x \in \mathbb{R}$ be $[\alpha, \beta]$. Then the distance of the point $(\alpha, \beta)$ from the line $3x + 4y + 12 = 0$ is :
11
8
10
9
Answer: (a)
Solution
Given $$f(x) = 6 + 16 \cos x \cdot \cos \left( \frac{\pi}{3} - x \right)$$ Using the identity: $$\cos \left( \frac{\pi}{3} + x \right) \cdot \sin 3x \cdot \cos 6x$$ We have: $$f(x) = 6 + 4 \cos 3x \cdot \sin 3x \cdot \cos 6x$$ Therefore, $$f(x) = 6 + \sin 12x$$ Thus, the range of $f(x)$ is $[5, 7]$. Hence, $[\alpha, \beta] = [5, 7]$. The distance of the point from $3x + 4y + 12 = 0$ is given by: $$= \left| \frac{3 \cdot 5 + 4 \cdot 7 + 12}{\sqrt{3^2 + 4^2}} \right|$$ This simplifies to: $$= 11 units$$
Question 13
Maths · Conic Sections · Single correct
Let the shortest distance from $(a, 0), a > 0$, to the parabola $y^2 = 4x$ be $4$. Then the equation of the circle passing through the point $(a, 0)$ and the focus of the parabola, and having its centre on the axis of the parabola is:
$x^2 + y^2 - 10x + 9 = 0$
$x^2 + y^2 - 6x + 5 = 0$
$x^2 + y^2 - 4x + 3 = 0$
$x^2 + y^2 - 8x + 7 = 0$
Answer: (b)
Solution
Normal at $P$ $y + tx = 2t + t^3$ $(a, 0)$ at $= 2t + t^3$ $a = 2 + t^2$ $R (2 + t^2, 0)$ $PR = 4 \Rightarrow 4 + 4t^2 = 16$ $4t^2 = 12 \Rightarrow t^2 = 3$ $a = 5 \Rightarrow R(5, 0)$ Focus $(1, 0)$ $(1, 0) \& (5, 0)$ will be the end pts. of diameter $\Rightarrow$ Eqn of circle is $(x - 1)(x - 5) + y^2 = 0$ $x^2 + y^2 - 6x + 5 = 0$
Question 14
Maths · Sets · Single correct
Let $X = \mathbb{R} \times \mathbb{R}$. Define a relation $R$ on $X$ as : $$(a_1, b_1) \, R \, (a_2, b_2) \iff b_1 = b_2$$ Statement I : $R$ is an equivalence relation. Statement II : For some $(a, b) \in X$, the set $S = \{(x, y) \in X : (x, y)R(a, b)\}$ represents a line parallel to $y = x$. In the light of the above statements, choose the correct answer from the options given below :
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Both Statement I and Statement II are true
Statement I is false but Statement II is true
Answer: (b)
Solution
Reflexive: $(a_1, b) \, R \, (a_1, b_1) \Rightarrow b_1 = b_1$ True Symmetric: $(a_1, b_1) \, R \, (a_2, b_2) \Rightarrow b_1 = b_2$ $(a_2, b_2) \, R \, (a_1, b_1) \Rightarrow b_2 = b_1$ True Transitive: $(a_1, b_1) \, R \, (a_2, b_2) \Rightarrow b_1 = b_2$ $\&$ $(a_2, b_2) \, R \, (a_3, b_3) \Rightarrow b_2 = b_3$ $\Rightarrow (a_1, b_1) \, R \, (a_3, b_3) \Rightarrow True \} b_1 = b_3$ Hence Relation $R$ is an equivalence relation Statement-I is true. For statement-II $\Rightarrow y = b$ so False
Question 15
Maths · Conic Sections · Single correct
The length of the chord of the ellipse $\frac{x^2}{4} + \frac{y^2}{2} = 1$, whose mid-point is $\left(1, \frac{1}{2}\right)$, is:
$\frac{5}{3}\sqrt{15}$
$\frac{1}{3}\sqrt{15}$
$\frac{2}{3}\sqrt{15}$
$\sqrt{15}$
Answer: (c)
Solution
Given $T = S_1$. $$\frac{x \cdot 1}{4} + \frac{y}{4} = \frac{1}{4} + \frac{1}{8}$$ This implies $2x + 2y = 3$. $$\frac{x^2}{4} + \left(\frac{3 - 2x}{2}\right)^2 = 1$$ This implies $$x = \frac{12 \pm \sqrt{120}}{12}$$ and $$y = \frac{1}{2} \mp \frac{\sqrt{120}}{12}$$ So the length of the chord is $$\frac{2\sqrt{15}}{3}$$
Question 16
Maths · Matrices · Single correct
Let $A = [a_{ij}]$ be $3 \times 3$ matrix such that $A \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, A \begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$ and $A \begin{bmatrix} 2 \\ 3 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$, then $a_{23}$ equals:
-1
2
1
0
Answer: (a)
Solution
Let $$A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$$ $$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$ Therefore, $b = 0$, $e = 0$, $h = 1$. And $$\begin{bmatrix} a & 0 & c \\ d & 0 & f \\ g & 1 & i \end{bmatrix} \begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$$ $$\begin{aligned} 4a + 3c &= 0 \\ 4d + 3f &= 1 \\ 4g + 1 + 3i &= 0 \end{aligned} \cdots (1)$$ And $$\begin{bmatrix} a & 0 & c \\ d & 0 & f \\ g & 1 & i \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$ $$\begin{aligned} 2a + 2c &= 1 \\ 2d + 2f &= 0 \\ 2g + 1 + 2i &= 0 \end{aligned} \cdots (2)$$ From equation (1) and (2) we get $d = 1$, $f = -1$. Therefore, $a_{23} = -1$.
Question 17
Maths · Complex Numbers and Quadratic Equations · Single correct
The number of complex numbers $z$, satisfying $|z| = 1$ and $\left| \frac{z}{\bar{z}} + \frac{\bar{z}}{z} \right| = 1$, is :
4
8
10
6
Answer: (b)
Solution
Given $|z| = 1$ and $\($ $\left$| $\frac{z}{\bar{z}}$ + $\frac{\bar{z}}{z}$ $\right$| = 1 $\)$. This implies $\($ |z^2 + ($\bar{z}$)^2| = 1 $\)$. Let $z = x + iy$. Then $\($ |(x + iy)^2 + (x - iy)^2| = 1 $\)$. This simplifies to $\($ |2x^2 - 2y^2| = 1 $\)$, which gives $\($ |x^2 - y^2| = $\frac{1}{2}$ $\)$. Therefore, $\($ x^2 - y^2 = $\pm$ $\frac{1}{2}$ $\)$. Also, $x^2 + y^2 = 1$. Case I: $x^2 - y^2 = \frac{1}{2}$ Case II: $x^2 - y^2 = -\frac{1}{2}$ Each case gives 4 points, hence we get 8 complex numbers.
Question 18
Maths · Three Dimensional Geometry · Single correct
If the square of the shortest distance between the lines $\frac{x-2}{1} = \frac{y-1}{2} = \frac{z+3}{-3}$ and $\frac{x+1}{2} = \frac{y+3}{4} = \frac{z+5}{-5}$ is $\frac{m}{n}$, where $m, n$ are coprime numbers, then $m + n$ is equal to:
If $I=\int_{0}^{\frac{\pi}{2}}\frac{\sin^{\frac{3}{2}}x}{\sin^{\frac{3}{2}}x+\cos^{\frac{3}{2}}x}\,dx$, then $\int_{0}^{21}\frac{x\sin x\cos x}{\sin^4x+\cos^4x}\,dx$ equals:
$\frac{\pi^2}{12}$
$\frac{\pi^2}{4}$
$\frac{\pi^2}{16}$
$\frac{\pi^2}{8}$
Answer: (c)
Solution
Given $$I = \int_0^{\frac{\pi}{2}} (\sin x)^3 \, dx = \int_0^{\frac{\pi}{2}} \sin^3 \left( \frac{\pi}{2} - x \right) \, dx$$ $$= \int_0^{\frac{\pi}{2}} \sin^3 \left( \frac{\pi}{2} - x \right) + \cos^3 \left( \frac{\pi}{2} - x \right)$$ Adding $$2I = \int_0^{\frac{\pi}{2}} (\sin x)^3 + (\cos x)^3 \, dx = \frac{\pi}{2}$$ Let $$I_0 = \int_0^{\frac{\pi}{2}} x \sin x \cos x \, dx = \int_0^{\frac{\pi}{2}} \left( \frac{\pi}{2} - x \right) \sin x \cos x \, dx$$ $$= \int_0^{\frac{\pi}{2}} \frac{\sin^4 x + \cos^4 x}{(\sin x)^4 + (\cos x)^4} \, dx$$ Adding, $$2I_0 = \int_0^{\frac{\pi}{2}} \pi \sin x \cos x \, dx$$ $$= \int_0^{\frac{\pi}{4}} \tan x (\sec^2 x) \, dx$$ $$= \int_0^{\frac{\pi}{4}} \frac{\tan x (\sec^2 x)}{1 + \tan^4 x} \, dx$$ Put $\tan^2 x = t$ so $2 \tan x \sec^2 x \, dx = dt$. Thus, $$I_0 = \frac{\pi}{4} \int_0^{\infty} \frac{dt}{1 + t^2} = \frac{\pi}{4} \left( \tan^{-1} t \right) \bigg|_0^{\infty} = \frac{\pi}{8} \left( \frac{\pi}{2} - 0 \right)$$ Therefore, $$I_0 = \frac{\pi^2}{16}$$
Question 20
Maths · Limits and Derivatives · Single correct
$\displaystyle \lim_{x \to \infty} \frac{(2x^2-3x+5)(3x-1)^{\frac{x}{2}}} {(3x^2+5x+4)\sqrt{(3x+2)^x}}$ is equal to:
The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is
Answer: 17280
Solution
A: number of ways that all boys sit together $= 5! \times 5!$ B: number of ways if no 2 boys sit together $= 4! \times 5!$ $A \cap B = \emptyset$ Required no. of ways $= 5! \times 5! + 4! \times 5! = 17280$
Question 22
Maths · Complex Numbers and Quadratic Equations · Fill in the blank
Let $\alpha$, $\beta$ be the roots of the equation $x^2 - ax - b = 0$ with $\mathrm{Im}(\alpha) < \mathrm{Im}(\beta)$. Let $P_n = \alpha^n - \beta^n$. If $P_3 = -5\sqrt{7}i, P_4 = -3\sqrt{7}i, P_5 = 11\sqrt{7}i$ and $P_6 = 45\sqrt{7}i$, then $|\alpha^4 + \beta^4|$ is equal to .
The focus of the parabola $y^2 = 4x + 16$ is the centre of the circle $\mathcal{C}$ of radius $5$. If the values of $\lambda$, for which $\mathcal{C}$ passes through the point of intersection of the lines $3x - y = 0$ and $x + \lambda y = 4$, are $\lambda_1$ and $\lambda_2$, $\lambda_1 < \lambda_2$, then $12\lambda_1 + 29\lambda_2$ is equal to
Answer: 15
Solution
Given $y^2 = 4(x + 4)$. Equation of circle $(x + 3)^2 + y^2 = 25$. Passes through the point of intersection of two lines $3x - y = 0$ and $x + \lambda y = 4$. $$\left( \frac{4}{3\lambda + 1}, \frac{12}{3\lambda + 1} \right),$$ we get $$\lambda = -\frac{7}{6}, 1$$ $$12\lambda_1 + 29\lambda_2$$ $$-14 + 29 = 15$$
Question 24
Maths · Statistics · Numerical
The variance of the numbers 8, 21, 34, 47, $\ldots$, 320 is
Answer: 8788
Solution
Given the equation $8 + (n-1)13 = 320$. Solving for $n$, we have: $$13n = 325$$ $$n = 25$$ The number of terms is $25$. The mean is given by: $$mean = \frac{\sum x_i}{n} = \frac{8 + 21 + \ldots + 320}{25} = \frac{\frac{25}{2}(8 + 320)}{25}$$ The variance $\sigma^2$ is: $$variance \sigma^2 = \frac{\sum x_i^2}{n} - (mean)^2$$ Calculating: $$= \frac{8^2 + 21^2 + \ldots + 320^2}{13} - (164)^2$$ $$= 8788$$
Question 25
Maths · Sequences and Series · Fill in the blank
The roots of the quadratic equation $3x^2 - px + q = 0$ are $10^{th}$ and $11^{th}$ terms of an arithmetic progression with common difference $\frac{3}{2}$. If the sum of the first 11 terms of this arithmetic progression is 88, then $q - 2p$ is equal to
A ball having kinetic energy KE, is projected at an angle of 60^$\circ$ from the horizontal. What will be the kinetic energy of ball at the highest point of its flight?
$\frac{(KE)}{8}$
$\frac{(KE)}{2}$
$\frac{(KE)}{16}$
$\frac{(KE)}{4}$
Answer: (d)
Solution
Initial K.E, K.E. = $\frac{1}{2}$ mu^2 Speed at highest point V = u $\cos$ 60^$\circ$ = $\frac{u}{2}$ $\therefore$ KE_2 = $\frac{1}{2}$ m $\left$( $\frac{u}{2}$ $\right$)^2 = $\frac{1}{4}$ $\times$ $\frac{1}{2}$ mu^2 = $\frac{KE}{4}$
Question 27
Physics · Electric Charges and Fields · Single correct
Two charges $7\mu\mathrm{C}$ and $-4\mu\mathrm{C}$ are placed at $(-7\ \mathrm{cm}, 0, 0)$ and $(7\ \mathrm{cm}, 0, 0)$ respectively. Given, $\epsilon_0 = 8.85 \times 10^{-12}\mathrm{C}^2\ \mathrm{N}^{-1}\ \mathrm{m}^{-2}$, the electrostatic potential energy of the charge configuration is :
Physics · Ray Optics and Optical Instruments · Single correct
The refractive index of the material of a glass prism is $\sqrt{3}$. The angle of minimum deviation is equal to the angle of the prism. What is the angle of the prism?
The equation of a transverse wave travelling along a string is $y(x,t) = 4.0 \sin [20 \times 10^{-3} x + 600t]$ mm, where $x$ is in mm and $t$ is in second. The velocity of the wave is:
$-60 \, \mathrm{m/s}$
$-30 \, \mathrm{m/s}$
$+30 \, \mathrm{m/s}$
$+60 \, \mathrm{m/s}$
Answer: (b)
Solution
Given $k = 20 \times 10^{-3} \, \mathrm{mm}^{-1} = 20 \, \mathrm{m}^{-1}$. $w = 600 \, \mathrm{s}^{-1}$ $$v = \frac{W}{k} = \frac{600}{20} = 30 \, \mathrm{m/s}$$ and $x$ and $t$ carry the same sign. Therefore $v = -30 \, \mathrm{m/s}$
Question 30
Physics · Physical World, Units and Measurements · Single correct
The energy of a system is given as $E(t) = \alpha^3 e^{-\beta t}$, where $t$ is the time and $\beta = 0.3 \, \mathrm{s}^{-1}$. The errors in the measurement of $\alpha$ and $t$ are $1.2\%$ and $1.6\%$, respectively. At $t = 5 \, \mathrm{s}$, maximum percentage error in the energy is:
6$\%$
8.4$\%$
11.6$\%$
4$\%$
Answer: (a)
Solution
Given $E = \alpha^3 e^{-\beta t}$. Taking the natural logarithm, we have $\ln E = 3 \ln \alpha - \beta t$. The maximum fractional change is given by $$\left( \frac{\mathrm{d}E}{E} \right)_{\max} = \frac{3 \, \mathrm{d}\alpha}{\alpha} + \frac{\beta \, \mathrm{d}t}{t} \times t.$$ Substituting the given values, $$= 3 \times 1.2\% + (0.3 \times 1.6 \times 5)\%$$ $$= 6\%$$
Question 31
Physics · Dual Nature of Radiation and Matter · Single correct
In photoelectric effect an EM-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 $\mathrm{eV}$ and stopping potential is 2 $\mathrm{V}$, what is the wavelength of the EM-wave ? (Given $\mathrm{hc}$ = 1242 $\mathrm{eVnm}$ where h is the Planck's constant and c is the speed of light in vacuum.)
300 $\mathrm{nm}$
400 $\mathrm{nm}$
600 $\mathrm{nm}$
200 $\mathrm{nm}$
Answer: (a)
Solution
Given $\phi = 2.14$ and $V_S = 2 \, \mathrm{V}$. Using the photoelectric equation. $$\frac{hc}{\lambda} = 2.14 + 2 = 4.14 \, \mathrm{eV}$$ $$\lambda = \frac{1242}{4.14} = 300 \, \mathrm{nm}$$
Question 32
Physics · System of Particles and Rotational Motion · Single correct
A circular disk of radius $R$ meter and mass $M$ kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that $\theta(t) = 5t^2 - 8t$, where $\theta(t)$ is the angular position of the rotating disc as a function of time $t$. How much power is delivered by the applied torque, when $t = 2 \, \mathrm{s}$?
Physics · Mechanical Properties of Fluids · Single correct
Water flows in a horizontal pipe whose one end is closed with a valve. The reading of the pressure gauge attached to the pipe is $P_1$. The reading of the pressure gauge falls to $P_2$ when the valve is opened. The speed of water flowing in the pipe is proportional to
Physics · Physical World, Units and Measurements · Single correct
Match List - I with List - II. List - I (A) Permeability of free space (B) Magnetic field (C) Magnetic moment (D) Torsional constant List - II (I) $\left[ M L^2 T^{-2} \right]$ (II) $\left[ M T^{-2} A^{-1} \right]$ (III) $\left[ M L T^{-2} A^{-2} \right]$ (IV) $\left[ L^2 A \right]$ Choose the correct answer from the options given below :
If a satellite orbiting the Earth is 9 times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon = 27 days and gravitational attraction between the satellite and the moon is neglected.
Physics · Electric Charges and Fields · Single correct
Two point charges $-4\mu c$ and $4\mu c$, constituting an electric dipole, are placed at $(-9, 0, 0)\, \mathrm{cm}$ and $(9, 0, 0)\, \mathrm{cm}$ in a uniform electric field of strength $10^4\, \mathrm{NC}^{-1}$. The work done on the dipole in rotating it from the equilibrium through $180^\circ$ is:
A galvanometer having a coil of resistance $30\,\Omega$ need $20\,\mathrm{mA}$ of current for full-scale deflection. If a maximum current of $3\,\mathrm{A}$ is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be $\frac{30}{X}\,\Omega$, where $X$ is
596
149
298
447
Answer: (b)
Solution
Given $(I - I_g)R = I_g G$. $(3 - 0.02) \times R = 0.02 \times G \Rightarrow R = 30$. Therefore, $149 = Required X$.
Question 38
Physics · Wave Optics · Single correct
The width of one of the two slits in Young's double slit experiment is $d$ while that of the other slit is $x \, d$. If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is $9 : 4$ then what is the value of $x$? (Assume that the field strength varies according to the slit width)
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason ($R$). Assertion (A) : The binding energy per nucleon is found to be practically independent of the atomic number A, for nuclei with mass numbers between 30 and 170. Reason ($R$): Nuclear force is long range. In the light of the above statements, choose the correct answer from the options given below :
(A) is true but ($R$) is false
(A) is false but ($R$) is true
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: (a)
Solution
From the graph between B.E/N and A, we can see B.E/N is almost constant, which implies it is correct. Reason is incorrect as nuclear forces are short range forces.
Question 40
Physics · Thermal Properties of Matter · Single correct
Water of mass $m$ gram is slowly heated to increase the temperature from $T_1$ to $T_2$. The change in entropy of the water, given specific heat of water is $1 \, \mathrm{Jkg^{-1} \, K^{-1}}$, is:
$m \ln \left( \frac{T_2}{T_1} \right)$
zero
$m \ln \left( \frac{T_1}{T_2} \right)$
$m (T_2 - T_1)$
Answer: (a)
Solution
Given $dQ = msdT$. 1. $dS = \frac{dQ}{T} = \frac{msdT}{T}$. 2. $\Delta S = \int \frac{msdT}{T} = ms \ln \frac{T_f}{T_i}$. 3. $\Delta S = m \ln \frac{T_2}{T_1}$.
Question 41
Physics · Current Electricity · Single correct
What is the current through the battery in the circuit shown below
1.5 A
0.5 A
0.25 A
1.0 A
Answer: (b)
Solution
Both are forward biased, hence $R_{eq} = 10\, \Omega$. $$i = \frac{V}{R} = \frac{5}{10} = \frac{1}{2} \, A$$
Question 42
Physics · Electromagnetic Waves · Single correct
A plane electromagnetic wave of frequency 20 MHz travels in free space along the $+x$ direction. At a particular point in space and time, the electric field vector of the wave is $E_y = 9.3 \, \mathrm{V m^{-1}}$. Then, the magnetic field vector of the wave at that point is
$B_z = 6.2 \times 10^{-8} \, \mathrm{T}$
$B_z = 3.1 \times 10^{-8} \, \mathrm{T}$
$B_z = 1.55 \times 10^{-8} \, \mathrm{T}$
$B_z = 9.3 \times 10^{-8} \, \mathrm{T}$
Answer: (b)
Solution
Given $E = BC$. $$9.3 = B \times 3 \times 10^8$$ Solving for $B$: $$B = \frac{9.3}{3 \times 10^8} = 3.1 \times 10^{-8} \, \mathrm{T}$$
Question 43
Physics · Kinetic Theory · Single correct
Using the given P - V diagram, the work done by an ideal gas along the path ABCD is :
3P_0 V_0
-4P_0 V_0
-3P_0 V_0
4P_0 V_0
Answer: (c)
Solution
Area under graph will be magnitude of graph and being counterclockwise graph it would be negative. Area $= 2P_0 \times V_0 + P_0 V_0 = 3P_0 V_0$. $W = -3P_0 V_0$.
Question 44
Physics · Ray Optics and Optical Instruments · Single correct
A concave mirror of focal length $f$ in air is dipped in a liquid of refractive index $\mu$. Its focal length in the liquid will be:
$\mu f$
$f$
$\frac{f}{(\mu - 1)}$
$\frac{f}{\mu}$
Answer: (b)
Solution
Focal length of mirror will not change because focal length of mirror doesn't depend on medium.
Question 45
Physics · Mechanical Properties of Solids · Single correct
A massless spring gets elongated by amount $x_1$ under a tension of 5 N. Its elongation is $x_2$ under the tension of 7 N. For the elongation of $(5x_1 - 2x_2)$, the tension in the spring will be,
39 N
15 N
11 N
20 N
Answer: (c)
Solution
Given $kx_1 = 5 \, \mathrm{N}$ and $kx_2 = 7 \, \mathrm{N}$. We calculate $k(5x_1 - 2x_2) = 5kx_1 - 2kx_2$. Substituting the values, we get $$= 5 \times 5 - 2 \times 7 = 11 \, \mathrm{N}.$$
Question 46
Physics · Mechanical Properties of Fluids · Numerical
An air bubble of radius 1.0 $\mathrm{mm}$ is observed at a depth of 20 $\mathrm{cm}$ below the free surface of a liquid having surface tension 0.095 $\mathrm{J/m^2}$ and density 10^3 $\mathrm{kg/m^3}$. The difference between pressure inside the bubble and atmospheric pressure is ______ $\mathrm{N/m^2}$. (Take g = 10 $\,$ $\mathrm{m/s^2}$)
Answer: 2190
Solution
The pressure difference is given by $\Delta P = P_{in} - P_0$. This can be expressed as $\Delta P = \rho gh + \frac{2T}{R}$. Substituting the values, we have: $$\Delta P = 1000 \times 10 \times 20 + \frac{2 \times 0.095}{10^{-3}}$$ This simplifies to: $$= 2000 + 190$$ $$= 2190$$
Question 47
Physics · Gravitation · Fill in the blank
A satellite of mass $\frac{M}{2}$ is revolving around earth in a circular orbit at a height of $\frac{R}{3}$ from earth surface. The angular momentum of the satellite is $M \sqrt{\frac{GMR}{x}}$. The value of $x$ is _____, where $M$ and $R$ are the mass and radius of earth, respectively. ( $G$ is the gravitational constant)
Answer: 3
Solution
Orbital velocity $v_0 = \sqrt{\frac{GM}{4R/3}} = \sqrt{\frac{3GM}{4R}}$. Angular momentum of satellite $= \frac{M}{2} v_0 \frac{4R}{3}$. $$= \frac{M}{2} \cdot \sqrt{\frac{3GM}{4R}} \cdot \frac{4R}{3}$$ $$= M \sqrt{\frac{GMR}{3}}$$ $x = 3$
Question 48
Physics · Electrostatic Potential and Capacitance · Fill in the blank
At steady state the charge on the capacitor, as shown in the circuit below, is ________ $\mu \mathrm{C}$.
Answer: 16
Solution
Given the circuit, the current $i$ is calculated as $$i = \left( \frac{5}{25} \right).$$ The charge $Q$ is given by $Q = CV$. Substituting the values, we have $$Q = \left( 8 \times 10^{-6} \right) \left( \frac{5}{25 \times 10} \right).$$ Simplifying further, $$Q = \left( \frac{8 \times 5 \times 10^{-2}}{25} \right) = 16 \, \mu \mathrm{C}.$$
Question 49
Physics · Electrostatic Potential and Capacitance · Numerical
A time varying potential difference is applied between the plates of a parallel plate capacitor of capacitance $2.5 \, \mu \mathrm{F}$. The dielectric constant of the medium between the capacitor plates is $1$. It produces an instantaneous displacement current of $0.25 \, \mathrm{mA}$ in the intervening space between the capacitor plates, the magnitude of the rate of change of the potential difference will be _____ $\mathrm{Vs}^{-1}$.
Answer: 100
Solution
Given $C \frac{dV}{dt} = I_d$. Rearranging, we have: $$\frac{dV}{dt} = \frac{I_d}{C}$$ Substituting the values: $$\frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}}$$ This simplifies to: $$= 100$$
Question 50
Physics · Alternating Current · Numerical
In a series LCR circuit, a resistor of 300 $\Omega$, a capacitor of 25 $\mathrm{nF}$ and an inductor of 100 $\mathrm{mH}$ are used. For maximum current in the circuit, the angular frequency of the ac source is $ \times 10^4$ radians $\mathrm{s}^{-1}$.
The effect of temperature on spontaneity of reactions are represented as The incorrect combinations are
and (C) only
and (D) only
and (D) only
and (C) only
Answer: (b)
Solution
Therefore, $\Delta G = \Delta H - T \Delta S$. For spontaneity of reaction: $\Delta G = -ve$.
Question 52
Chemistry · Electrochemistry · Single correct
Standard electrode potentials for a few half cells are mentioned below : $E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = 0.34 \, \mathrm{V}, E^\circ_{\mathrm{Zn}^{2+}/\mathrm{Zn}} = -0.76 \, \mathrm{V}$ $E^\circ_{\mathrm{Ag}^+/\mathrm{Ag}} = 0.80 \, \mathrm{V}, E^\circ_{\mathrm{Mg}^{2+}/\mathrm{Mg}} = -2.37 \, \mathrm{V}$ Which one of the following cells gives the most negative value of $\Delta G^\circ$ ?
Zn $\vert$ Zn$^{2+}$ (1M) $\parallel$ Ag$^+$ (1M) $\vert$ Ag
The $\alpha$-Helix and $\beta$ - Pleated sheet structures of protein are associated with its:
tertiary structure
quaternary structure
secondary structure
primary structure
Answer: (c)
Solution
The $\alpha$-helix and $\beta$-pleated sheet belong to the secondary structure of proteins, which have hydrogen bonds.
Question 54
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Given below are two statements : Consider the following reaction
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
Both Statement I and Statement II are true
Answer: (d)
Solution
$k_{eq}=2280$ is for HCHO $k_{eq}=2000$ is for chloral Both data are given in the Clayden and Warren book. $k_{eq}>1$ because HCHO and chloral are more electrophilic.
Question 55
Chemistry · Equilibrium · Single correct
Consider the reaction $$\mathrm{X_2Y\,(g) = X_2\,(g) + \frac{1}{2}Y_2\,(g)}$$ The equation representing correct relationship between the degree of dissociation $(x)$ of $\mathrm{X_2Y\,(g)}$ with its equilibrium constant $K_p$ is _______. Assume $x$ to be very very small.
$x=\sqrt[3]{\dfrac{2Kp}{p}}$
$x=\sqrt[3]{\dfrac{2Kp^2}{p}}$
$x=\sqrt[3]{\dfrac{Kp}{2}}$
$x=\sqrt[3]{\dfrac{Kp}{2p}}$
Answer: (b)
Solution
Given the reaction: $\[$ $\mathrm{X_2Y(g) \rightarrow \frac{XX_2(g)}{x} + \frac{1}{2}Y_2(g)}$ $\]$ Let $\($ x $\)$ be the extent of reaction. Therefore, $\($ 2 $\)$ $\[$ P_{$\mathrm{X_2Y}$} = $\frac{1-x}{1+\frac{x}{2}}$ $\times$ p $\]$ $\[$ P_{$\mathrm{X_2}$} = $\frac{x}{1+\frac{x}{2}}$ $\times$ p $\]$ $\[$ P_{$\mathrm{Y_2}$} = $\frac{x/2}{1+\frac{x}{2}}$ $\times$ p $\]$ Therefore, $\[$ K_p = $\left$( $\frac{x}{1+\frac{x}{2}}$ $\times$ p $\right$) $\left$( $\frac{x}{2\left(1+\frac{x}{2}\right)}$ $\times$ p $\right$)^{1/2} $\]$ $\[$ K_p = $\left$( $\frac{1-x}{1+\frac{x}{2}}$ $\times$ p $\right$) $\]$ Therefore, $\[$ K_p = $\left$( $\frac{x}{1-x}$ $\right$) $\left$( $\frac{x}{2\left(1+\frac{x}{2}\right)}$ $\right$)^{1/2} $\times$ p^{1/2} $\]$ Let $\($ x $\)$ be very small. Therefore, $\[$ K_p = $\frac{x^{3/2}}{2^{(1/2)}}$ $\times$ p^{1/2} $\]$ $\[$ x^{3/2} = K_p $\times$ 2^{1/2} $\]$ $\[$ x^3 = $\frac{K_p^2 \times 2}{p}$ $\]$ $\[$ x = $\left$( $\frac{K_p^2 \times 2}{p}$ $\right$)^{1/3} $\]$
Question 56
Chemistry · Analytical Chemistry · Single correct
Identify $A$, $B$ and $C$ in the given below reaction sequence. $A \xrightarrow{\mathrm{HNO_3}} \mathrm{Pb(NO_3)_2} \xrightarrow{\mathrm{H_2SO_4}} B \xrightarrow[(2)\ \mathrm{Acetic\ acid}\\(3)\ \mathrm{K_2CrO_4}] {(1)\ \mathrm{Ammonium\ acetate}} C$
PbCl_2, PbSO_4, PbCrO_4
PbS, PbSO_4, Pb(CH_3COO)_2
PbCl_2, Pb(SO_4)_2, PbCrO_4
PbS, PbSO_4, PbCrO_4
Answer: (d)
Solution
Question 57
Chemistry · Alcohols, Phenols and Ethers · Single correct
Given below are two statements: Statement (I): The boiling points of alcohols and phenols increase with increase in the number of C-atoms. Statement (II): The boiling points of alcohols and phenols are higher in comparison to other class of compounds such as ethers, haloalkanes. In the light of the above statements, choose the correct answer from the options given below:
Both Statement I and Statement II are false
Both Statement I and Statement II are true
Statement I is false but Statement II is true
Statement I is true but Statement II is false
Answer: (b)
Solution
Statement I is correct as boiling point of alcohol phenols increase with increase in the number of C-atoms due to increase in van der Waals forces. Statement II is correct, since alcohols phenols have intermolecular H-bonding therefore their boiling points are higher in comparison to other class of compounds such as ethers, haloalkanes.
Question 58
Chemistry · Solutions · Single correct
When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg. The mole fraction of the solute in the solution is 0.2. What would be the mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg?
0.8
0.4
0.2
0.6
Answer: (d)
Solution
Given $P^\circ - P \propto X_{solute}$ and $10 \propto 0.2$, therefore $20 \propto 0.4$. Thus, $X_{solvent} = 1 - X_{solute} = 1 - 0.4 = 0.6$.
Question 59
Chemistry · Structure of Atom · Single correct
Given below are two statements: Statement (I): For a given shell, the total number of allowed orbitals is given by $n^2$. Statement (II): For any subshell, the spatial orientation of the orbitals is given by $-l$ to $+l$ values including zero. In the light of the above statements, choose the correct answer from the options given below:
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Both Statement I and Statement II are true
Statement I is false but Statement II is true
Answer: (d)
Solution
For a shell, the total number of orbitals is $n^2$. Magnetic quantum numbers have values from $-\ell$ to $+\ell$, including $0$.
Question 60
Chemistry · Haloalkanes and Haloarenes · Single correct
The ascending order of relative rate of solvolysis of following compounds is:
$(C)< (B) < (A) < (D)$
$(D) < (A) < (B) < (C)$
$(D)< (B) < (A) < (C)$
$(C)< (D) < (B) < (A)$
Answer: (b)
Solution
Solvolysis or SN$_1 \propto$ stability of carbocation
Question 61
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Match List - I with List - II. Choose the correct answer from the options given below:
(A)-(I), (B)-(IV), (C)-(III), (D)-(II)
(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
Answer: (c)
Solution
The ozonolysis of the given compound results in the cleavage of the double bond, leading to the formation of carbonyl compounds. The correct ozonolysis product corresponds to option (B).
Question 62
Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct
Which of the following graphs most appropriately represents a zero order reaction?
Answer: (a)
Solution
Given $[A]_t = [A]_0 - kt$. A straight line with negative slope is shown in the graph. This is the graph of reactant concentration versus time for a zero order reaction.
Question 63
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Match List - I with List - II. \begin{tabular}{|l|l|} \hline \textbf{List - I} & \textbf{List - II} \\ \hline (A) Bronze & (I) Cu, Ni \\ \hline (B) Brass & (II) Fe, Cr, Ni, C \\ \hline (C) UK silver coin & (III) Cu, Zn \\ \hline (D) Stainless steel & (IV) Cu, Sn \\ \hline \end{tabular} Choose the correct answer from the options given below:
(A) - (IV), (B)-(II), ($C$)-(III), (D)-(I)
(A) - (IV), (B)-(III), ($C$)-(I), (D)-(II)
(A) - (III), (B)-(IV), ($C$)-(II), (D)-(I)
(A) - (III), (B)-(I), ($C$)-(IV), (D)-(II)
Answer: (b)
Solution
Bronze is an alloy of copper and Tin. (A-IV) Brass is an alloy of copper and Zinc. (B-III) UK Silver coin is an alloy of copper and Nickel. (C-I) Stainless steel is an alloy of Fe, Cr, Ni, C. (D-II)
Question 64
Chemistry · Co-ordination Compounds · Single correct
Identify the coordination complexes in which the central metal ion has $d^4$ configuration.
, (C) and (D) only
and (E) only
and (D) only
, (B) and (E) only
Answer: (c)
Solution
For Fe^{+6}, the electronic configuration is $[\mathrm{Ar}] 3d^2$. For Mn^{+3}, the electronic configuration is $[\mathrm{Ar}] 3d^4$. For Fe^{+3}, the electronic configuration is $[\mathrm{Ar}] 3d^5$. For Cr^{+2}, the electronic configuration is $[\mathrm{Ar}] 3d^4$. For Ni^{+4}, the electronic configuration is $[\mathrm{Ar}] 3d^6$.
Question 65
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
Given below are the atomic numbers of some group 14 elements. The atomic number of the element with lowest melting point is :
6
82
14
50
Answer: (d)
Solution
Order of melting points of group 14: $\mathrm{C} > \mathrm{Si} > \mathrm{Ge} > \mathrm{Pb} > \mathrm{Sn}$. Element and melting points ($^\circ \mathrm{C}$): $Z = 6 = \mathrm{C}$: $3730$ $Z = 14 = \mathrm{Si}$: $1410$ $Z = 32 = \mathrm{Ge}$: $937$ $Z = 50 = \mathrm{Sn}$: $232$ $Z = 82 = \mathrm{Pb}$: $327$
Question 66
Chemistry · Equilibrium · Single correct
pH of water is 7 at $25^{\circ} \mathrm{C}$. If water is heated to $80^{\circ} \mathrm{C}.$, it's pH will :
At $25^\circ \mathrm{C}$, pure water has pH $= 7$. As temperature increased, water molecules dissociate more into hydrogen ions ($\mathrm{H^+}$) and hydroxide ions ($\mathrm{OH^-}$). This increased dissociation leads to slightly decrease in pH. At $80^\circ \mathrm{C}$, pH $\approx 6.93$.
Question 67
Chemistry · Haloalkanes and Haloarenes · Single correct
Identify the products $[A]$ and $[B]$, respectively in the following reaction:
Answer: (b)
Solution
The reaction involves the conversion of chlorobenzene to phenol using NaOH at 623 K and 300 atm, followed by acidification with $\mathrm{H^+}$. The phenol is then oxidized using $\mathrm{Na_2CrO_4}$ and $\mathrm{H_2SO_4}$ to form benzoquinone.
Question 68
Chemistry · Solutions · Single correct
Consider a binary solution of two volatile liquid components 1 and 2. $x_1$ and $y_1$ are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of $\frac{1}{x_1}$ vs $\frac{1}{y_1}$ are given respectively as:
Given below are two statements about X-ray spectra of elements: Statement (I): A plot of $\sqrt{\nu}$ ($\nu$ = frequency of X-rays emitted) vs atomic mass is a straight line. Statement (II): A plot of $\nu$ ($\nu$ = frequency of X-rays emitted) vs atomic number is a straight line. In the light of the above statements, choose the correct answer from the options given below:
Both Statement I and Statement II are true
Statement I is false but Statement II is true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Answer: (b)
Solution
The graph on the left shows a step-like increase of $\sqrt{\nu}$ with respect to atomic mass. The graph on the right shows a linear increase of $\sqrt{\nu}$ with respect to $Z$.
Question 70
Chemistry · The d-and f-Block Elements · Single correct
Consider the following reactions $\mathrm{K_2Cr_2O_7} \xrightarrow[\mathrm{-H_2O}]{\mathrm{KOH}} [A] \xrightarrow[\mathrm{-H_2O}]{\mathrm{H_2SO_4}} [B]+\mathrm{K_2SO_4}$ The products $[A]$ and $[B]$, respectively are :
Chemistry · Some Basic Concepts of Chemistry · Numerical
0.01 mole of an organic compound $(X)$ containing 10$\%$ hydrogen, on complete combustion produced 0.9 g $\mathrm{H_2O}$. Molar mass of $(X)$ is _____ g $\mathrm{mol^{-1}}$.
Answer: 100
Solution
Organic compound undergoes combustion to form $\mathrm{H_2O}$ with $0.9 \, \mathrm{gm}$. Therefore, mole of $\mathrm{H_2O} = \frac{0.9}{18} = 0.05$ mole. Mole of H in $\mathrm{H_2O} = 0.05 \times 2 = 0.1$ mole, which is equal to the mole of H in 0.01 mole of the organic compound. Therefore, the weight of H atom in 0.01 mole compound is $0.1 \times 1 = 0.1 \, \mathrm{gm}$. The weight of H atom in one mole compound is $\frac{0.1}{0.01} = 10 \, \mathrm{gm}$. Therefore, the weight percentage of H is given by $$\frac{wt. of H in one mole compound}{Molar mass of compound} \times 1 = \frac{10}{M} \times 100.$$ Solving, we get $$10 = \frac{10}{M} \times 100.$$ Therefore, $M = 100$.
Question 72
Chemistry · Amines · Numerical
Consider the following sequence of reactions. The total number of $sp^3$ hybridised carbon atoms in the major product $C$ formed is ________.
Answer: 4
Solution
The reaction starts with the conversion of the amine group to a diazonium salt using $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ at $0-5^\circ \mathrm{C}$. This forms the diazonium ion $\mathrm{N_2^+Cl^-}$. The diazonium ion then reacts with phenol in the presence of $\mathrm{NaOH}$ to form an azo compound. Upon dilution with $\mathrm{HCl}$, the azo compound is formed with an $\mathrm{OH}$ group. The final step involves the reaction with $\mathrm{NaOH}$ and $\mathrm{H_3CCH_2Br}$ to form the ether linkage. The correct answer is option (4).
Question 73
Chemistry · Some Basic Concepts of Chemistry · Numerical
When 81.0 g of aluminum is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is _______ - (Nearest integer) Given: Molar mass of Al is $27.0 \, \mathrm{g \, mol^{-1}}$ Molar mass of O is $16.0 \, \mathrm{g \, mol^{-1}}$
Answer: 153
Solution
The reaction is given by $$4\mathrm{Al} + 3\mathrm{O}_2 \rightarrow 2\mathrm{Al}_2\mathrm{O}_3$$ For aluminum: $$\frac{81}{27} = 3 mole$$ For oxygen: $$\frac{128}{32} = 4 mole$$ Limiting reagent is aluminum. Therefore, moles of $\mathrm{Al}_2\mathrm{O}_3$ formed is $$\frac{1}{2} \times 3 mole$$ Therefore, weight of $\mathrm{Al}_2\mathrm{O}_3$ formed is $$\frac{3}{2} \times 102$$ which equals 153 gm.
Question 74
Chemistry · Thermodynamics · Numerical
The bond dissociation enthalpy of $X_2\Delta H_{bond}$ calculated from the given data is ______ kJ mol$^{-1}$. (Nearest integer) $\mathrm{M^+X^- (s) \rightarrow M^+ (g) + X^- (g)} \Delta H^*_{lattice} = 800 \, \mathrm{kJ \, mol^{-1}}$ $\mathrm{M (s) \rightarrow M (g)} \Delta H^\circ_{sub} = 100 \, \mathrm{kJ \, mol^{-1}}$ $\mathrm{M (g) \rightarrow M^+ (g) + e^- (g)} \Delta H_i = 500 \, \mathrm{kJ \, mol^{-1}}$ $\mathrm{X (g) + e^- (g) \rightarrow X^- (g)} \Delta H^*_{eg} = -300 \, \mathrm{kJ \, mol^{-1}}$ $\mathrm{M (s) + \frac{1}{2} X_2 (g) \rightarrow M^+X^- (s)} \Delta H_f = -400 \, \mathrm{kJ \, mol^{-1}}$ [Given : $\mathrm{M^+X^-}$ is a pure ionic compound and $\mathrm{X}$ forms a diatomic molecule $\mathrm{X_2}$ in gaseous state]
A compound 'X' absorbs 2 moles of hydrogen and 'X' upon oxidation with $\mathrm{KMnO_4} \mid \mathrm{H^+}$ gives The total number of $\sigma$ bonds present in the compound 'X' is______