JEE Main 29 July 2022 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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Maths · Relations and Functions · Single correct
Let $R$ be a relation from the set $\{1, 2, 3, \ldots, 60\}$ to itself such that $R = \{(a, b) : b = pq, \text{ where } p, q \geq 3 \text{ are prime numbers}\}$. Then, the number of elements in $R$ is:
Answer: (b)
Number of possible values of $a = 60$, for $b = pq$, If $p = 3$, $q = 3, 5, 7, 11, 13, 17, 19$. If $p = 5$, $q = 5, 7, 11$. If $p = 7$, $q = 7$. Total cases $= 60 \times 11 = 660$
Maths · Complex Numbers and Quadratic Equations · Single correct
If $z = 2 + 3i$, then $z^5 + \left( \overline{z} \right)^5$ is equal to:
Answer: (a)
Given $z^5 + (\overline{z})^5 = (2 + 3i)^5 + (2 - 3i)^5$. This equals $2 \left( \binom{5}{0} 2^5 + \binom{5}{2} 2^3 (3i)^2 + \binom{5}{4} 2^1 (3i)^4 \right)$. Simplifying, we have $2 (32 + 10 \times 8 \times (-9) + 5 \times 2 \times 81) = 244$.
Maths · Matrices · Single correct
Let A and B be two 3 $\times$ 3 non-zero real matrices such that AB is a zero matrix. Then
Answer: (b)
Given $AB = 0 \Rightarrow |AB| = 0$. $|A| \ |B| = 0$ $|A| = 0 |B| = 0$ If $|A| \neq 0$, $B = 0$ (not possible). If $|B| \neq 0$, $A = 0$ (not possible). Hence $|A| = |B| = 0$. $\Rightarrow AX = 0$ has infinitely many solutions.
Maths · Sequences and Series · Single correct
If $\frac{1}{(20-a)(40-a)} + \frac{1}{(40-a)(60-a)} + \ldots + \frac{1}{(180-a)(200-a)} = \frac{1}{256}$, then the maximum value of $a$ is:
Answer: (c)
By splitting $$\frac{1}{20} \left[ \left( \frac{1}{20-a} - \frac{1}{40-a} \right) + \left( \frac{1}{40-a} - \frac{1}{60-a} \right) + \ldots + \left( \frac{1}{180-a} - \frac{1}{200-a} \right) \right]$$ $$\Rightarrow \frac{1}{20} \left( \frac{1}{20-a} - \frac{1}{200-a} \right) = \frac{1}{256}$$ $$(20-a)(200-a) = 256 \times 9$$ $$a^2 - 220a + 1696 = 0$$ $$a = 8, 212$$ Hence maximum value of $a$ is 212.
Maths · Limits and Derivatives · Single correct
If $\lim_{x \to 0} \frac{\alpha e^x + \beta e^{-x} + \gamma \sin x}{x \sin^2 x} = \frac{2}{3}$, where $\alpha, \beta, \gamma \in \mathbb{R}$, then which of the following is NOT correct?
Given $\lim_{x\to0}$ $\frac{ \alpha\left( 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots \right) }{ x^3 }$ $+ \frac{ \beta\left( 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots \right) }{ x^3 }$ $+ \frac{ \gamma\left( x-\frac{x^3}{3!}+\cdots \right) }{ x^3 }$ constant terms should be zero $\Rightarrow\ \alpha+\beta=0$ coefficient of $x$ should be zero $\Rightarrow\ \alpha-\beta+\gamma=0$ coefficient of $x^2$ should be zero $\lim_{x\to0} \frac{ x^3\left( \frac{\alpha}{3!} -\frac{\beta}{3!} -\frac{\gamma}{3!} \right) }{ x^3 }$ $+ \frac{ x^4\left( \frac{\alpha}{3!} -\frac{\beta}{3!} -\frac{\gamma}{3!} \right) }{ x^3 } =\frac23$ $\Rightarrow\ \frac{\alpha}{2} + \frac{\beta}{2} =0$ $\frac{\alpha}{6} -\frac{\beta}{6} -\frac{\gamma}{6} =\frac{2}{3}$ $\Rightarrow\ \alpha=1,\ \beta=-1,\ \gamma=-2$
Maths · Integrals · Single correct
The integral $$\int_{0}^{\frac{\pi}{2}} \frac{1}{3 + 2 \sin x + \cos x} \, dx$$ is equal to:
Answer: (b)
Given $$I = \int_0^{\frac{\pi}{2}} \frac{dx}{3 + 2 \sin x + \cos x} = \int_0^{\frac{\pi}{2}} \frac{\sec^2 \frac{x}{2} \cdot dx}{2 \tan^2 \frac{x}{2} + 4 \tan \frac{x}{2} + 4}$$ Put $\tan \frac{x}{2} = t$, so $$I = \int_0^1 \frac{dt}{(t+1)^2 + 1} = \tan^{-1}(x+1) \bigg|_0^1 = \tan^{-1} 2 - \frac{\pi}{4}$$
Maths · Differential Equations · Single correct
Let the solution curve $y = y(x)$ of the differential equation $\left(1 + e^{2x}\right)\left(\frac{dy}{dx} + y\right) = 1$ pass through the point $\left(0, \frac{\pi}{2}\right)$. Then, $\lim_{x \to \infty} e^x y(x)$ is equal to:
Answer: (b)
Given $\($ $\frac{dy}{dx}$ + y = $\frac{1}{1 + e^{2x}}$ $\)$. So integrating factor is $\($ e^{$\int$ 1 $\,$ dx} = e^x $\)$. So solution is $\($ y $\cdot$ e^x = $\tan$^{-1}(e^x) + c $\)$. Now as curve is passing through $\($ $\left$( 0, $\frac{\pi}{2}$ $\right$) $\)$ so $\($ $\Rightarrow$ c = $\frac{\pi}{4}$ $\)$. $\($ $\Rightarrow$ $\lim$_{x $\to$ $\infty$} $\left$( y $\cdot$ e^x $\right$) = $\lim$_{x $\to$ $\infty$} $\left$( $\tan$^{-1}(e^x) + $\frac{\pi}{4}$ $\right$) = $\frac{3\pi}{4}$ $\)$
Maths · Conic Sections · Single correct
Let a line L pass through the point of intersection of the lines $bx + 10y - 8 = 0$ and $2x - 3y = 0$, $b \in \mathbb{R} - \left\{ \frac{4}{3} \right\}$. If the line L also passes through the point $(1, 1)$ and touches the circle $17 \left( x^2 + y^2 \right) = 16$, then the eccentricity of the ellipse $\frac{x^2}{5} + \frac{y^2}{b^2} = 1$ is:
Answer: (b)
Line is passing through intersection of $bx + 10y - 8 = 0$ and $2x - 3y = 0$ is $\left(bx + 10y - 8\right) + \lambda \left(2x - 3y\right) = 0$. As line is passing through $(1,1)$ so $\lambda = b + 2$. Now line $\left(3b + 4\right)x - \left(3b - 4\right)y - 8 = 0$ is tangent to circle $17\left(x^2 + y^2\right) = 16$. So $$\frac{8}{\sqrt{\left(3b + 4\right)^2 + \left(3b - 4\right)^2}} = \frac{4}{\sqrt{17}}$$ $$\Rightarrow b^2 = 2 \Rightarrow e = \sqrt{\frac{3}{5}}$$
Maths · Three Dimensional Geometry · Single correct
If the foot of the perpendicular from the point $A(-1, 4, 3)$ on the plane $P: 2x + my + nz = 4$, is $\left(-2, \frac{7}{2}, \frac{3}{2}\right)$, then the distance of the point $A$ from the plane $P$, measured parallel to a line with direction ratios $3, -1, -4$, is equal to:
Answer: (b)
Let B be the foot of the perpendicular. Coordinates of B are $\left(-2, \frac{7}{2}, \frac{3}{2}\right)$. Direction ratio of line AB is $\langle 2, 1, 3 \rangle$ so $m = 1$, $n = 3$. So the equation of AC is $$\frac{x+1}{3} = \frac{y-4}{-1} = \frac{z-3}{-4} = \lambda$$ So point C is $(3\lambda - 1, -\lambda + 4, -4\lambda + 3)$. But C lies on the plane, so $$6\lambda - 2 - \lambda + 4 - 12\lambda + 9 = 4$$ $$\Rightarrow \lambda = 1 \Rightarrow C(2, 3, -1)$$ $$\Rightarrow AC = \sqrt{26}$$
Maths · Vector Algebra · Single correct
Let $\vec{a} = 3\hat{i} + \hat{j}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$. Let $\vec{c}$ be a vector satisfying $\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} + \lambda \vec{c}$. If $\vec{b}$ and $\vec{c}$ are non-parallel, then the value of $\lambda$ is:
Answer: (a)
Given $\vec{a} = 3\hat{i} + \hat{j}$, $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$. As $\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} + \lambda \vec{c}$, we have: $$\Rightarrow \vec{a} \cdot \vec{c} (\vec{b}) - (\vec{a} \cdot \vec{b}) \vec{c} = \vec{b} + \lambda \vec{c}$$ $$\Rightarrow \vec{a} \cdot \vec{c} = 1, \vec{a} \cdot \vec{b} = -\lambda$$ $$\Rightarrow (3\hat{i} + \hat{j}) \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -\lambda$$ $$\Rightarrow \lambda = -5$$
Maths · Heights and Distances · Single correct
The angle of elevation of the top of a tower from a point A due north of it is $\alpha$ and from a point B at a distance of 9 units due west of A is $\cos^{-1}\left(\frac{3}{\sqrt{13}}\right)$. If the distance of the point B from the tower is 15 units, then $\cot \alpha$ is equal to:
Answer: (a)
Given $OB = 15$. $$\cos \beta = \frac{3}{\sqrt{13}}$$ For the triangle with sides $\sqrt{13}$, $3$, and $2$, we have: $$\tan \beta = \frac{2}{3}$$ In the triangle $BPO$, we have: $$\tan \beta = \frac{h}{15}$$ Equating the two expressions for $\tan \beta$: $$\frac{2}{3} = \frac{h}{15}$$ Solving for $h$ gives: $$10 = h$$ In the triangle $OAB$, we have: $$OA^2 + AB^2 = 225$$ Substituting $AB = 9$: $$OA^2 + 81 = 225$$ In the triangle $AO$, we have: $$\tan \alpha = \frac{10}{12}$$ Therefore, the cotangent is: $$\cot \alpha = \frac{12}{10} = \frac{6}{5}$$
Maths · Mathematical Reasoning · Single correct
The statement $(p \land q) \Rightarrow (p \land r)$ is equivalent to :
Answer: (d)
Given $(p \land q) \Rightarrow (p \land r)$. $$\sim (p \land q) \lor (p \land r)$$ $$(\sim p \lor \sim q) \lor (p \land r)$$ $$(\sim p \lor (p \land r)) \lor \sim q$$ $$(\sim p \lor p) \land (\sim p \lor r) \lor \sim q$$ $$(\sim p \lor r) \lor \sim q$$ $$(\sim p \lor \sim q) \lor r$$ $$\sim (p \land q) \lor r$$ $$(p \land q) \Rightarrow r$$
Maths · Properties of Triangles · Single correct
Let the circumcentre of a triangle with vertices $A(a, 3)$, $B(b, 5)$ and $C(a, b)$, $ab > 0$ be $P(1, 1)$. If the line $AP$ intersects the line $BC$ at the point $Q(k_1, k_2)$, then $k_1 + k_2$ is equal to:
Answer: (b)
Given $m_{AC} \to \infty$, $m_{PD} = 0$. The coordinates of $D$ are $\left( \frac{a+a}{2}, \frac{b+3}{2} \right)$. Therefore, $D \left( a, \frac{b+3}{2} \right)$. Since $m_{PD} = 0$, we have $$\frac{b+3}{2} - 1 = 0$$ $$b + 3 - 2 = 0$$ $$b = -1$$ The coordinates of $E$ are $\left( \frac{b+a}{2}, \frac{5+b}{2} \right) = \left( \frac{af}{2}, 2 \right)$. The condition $m_{CB} \cdot m_{EP} = -1$ gives $$\frac{5-b}{b-a} = \frac{2-1}{a-1} = -1$$ $$\frac{6}{-1-a} = \frac{2}{a-3} = -1$$ Solving, $$12 = (1+a)(a-3)$$ $$12 = a^2 - 3a + a - 3$$ $$a^2 - 2a - 15 = 0$$ $$(a-5)(a+3) = 0$$ $$a = 5 or a = -3$$ Given $ab > 0$, accept $a = 5$. The equation of line $AP$ is $y - 1 = \left( \frac{3-1}{-3-1} \right)(x-1)$, which simplifies to $$-2y + 2 = x - 1$$ $$x + 2y = 3$$ Applying equation (1), the line $BC$ with $B(-1, 5)$ and $C(-3, -1)$ is $$\left( y - 5 \right) = \frac{6}{2}(x + 1)$$ $$y = 3x + 8$$ Solving equations (1) and (2), $$x + 2(3x + 8) = 3$$ $$7x + 16 = 3$$ $$7x = -13$$ $$x = \frac{-13}{7}$$ $$y = 3\left( \frac{-13}{7} \right) + 8$$ $$= \frac{-39 + 56}{7}$$ $$y = \frac{17}{7}$$ Therefore, $$x + y = \frac{-13 + 17}{7} = \frac{4}{7}$$
Maths · Vector Algebra · Single correct
Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that the angle between them is $\frac{\pi}{4}$. If $\theta$ is the angle between the vectors $\left( \hat{a} + \hat{b} \right)$ and $\left( \hat{a} + 2\hat{b} + 2\left( \hat{a} \times \hat{b} \right) \right)$, then the value of $164 \cos^2 \theta$ is equal to:
Answer: (a)
Given $\hat{a} \cdot \hat{b} = \frac{\pi}{4} = \phi$. $\hat{a} \cdot \hat{b} = |\hat{a}| |\hat{b}| \cos \phi$. $\hat{a} \cdot \hat{b} = \cos \phi = \frac{1}{\sqrt{2}}$. $$\cos \theta = \frac{(\hat{a} + \hat{b}) \cdot (\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b}))}{|\hat{a} + \hat{b}| |\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})|}$$ $|\hat{a} + \hat{b}|^2 = (\hat{a} + \hat{b}) \cdot (\hat{a} + \hat{b})$. $|\hat{a} + \hat{b}|^2 = 2 + 2 \hat{a} \cdot \hat{b}$. $= 2 + \sqrt{2}$. $\hat{a} \times \hat{b} = |\hat{a}| |\hat{b}| \sin \phi \hat{n}$. $\hat{a} \times \hat{b} = \hat{n}$ when $\hat{n}$ is vector $\perp \hat{a}$ and $\hat{b}$. We know, $\vec{c} \cdot \vec{a} = 0$. $\vec{c} \cdot \vec{b} = 0$. $$|\hat{a} + 2\hat{b} + 2\vec{c}|^2$$ $= 1 + 4 + \frac{(4)}{2} + 4 \hat{a} \cdot \hat{b} + 8 \hat{b} \cdot \vec{c} + 4 \vec{c} \cdot \hat{a}$. $= 7 + \frac{4}{\sqrt{2}} = 7 + 2\sqrt{2}$. Now $(\hat{a} + \hat{b}) \cdot (\hat{a} + 2\hat{b} + 2\vec{c})$. $= |\hat{a}|^2 + 2 \hat{a} \cdot \hat{b} + 0 + \hat{b} \cdot \hat{a} + 2 |\hat{b}|^2 + 0$. $= 1 + \frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} + 2$. $= 3 + \frac{3}{\sqrt{2}}$. $$\cos \theta = \frac{3 + \frac{3}{\sqrt{2}}}{\sqrt{2} + \sqrt{2} \sqrt{7 + 2\sqrt{2}}}$$ $$\cos \theta = \frac{3 + \frac{3}{\sqrt{2}}}{\sqrt{2} + \sqrt{2} \sqrt{7 + 2\sqrt{2}}}$$ $$\cos^2 \theta = \frac{9(\sqrt{2} + 1)^2}{2(2 + \sqrt{2})(7 + 2\sqrt{2})}$$ $$\cos^2 \theta = \left(\frac{9}{2\sqrt{2}}\right) \left(\frac{(\sqrt{2} + 1)}{(7 + 2\sqrt{2})}\right)$$ $$164 \cos^2 \theta = \frac{(82)(9)}{\sqrt{2}} \left(\frac{(\sqrt{2} + 1)}{(7 + 2\sqrt{2})}\right) \left(\frac{(7 - 2\sqrt{2})}{(7 - 2\sqrt{2})}\right)$$ $= 90 + 27\sqrt{2}$
Maths · Integrals · Single correct
If f($\alpha$) = $\int_{1}^{\alpha}$ $\frac{\log_{10} t}{1+t}$ dt, $\alpha$ > 0, then f(e^3) + f(e^{-3}) is equal to:
Answer: (d)
Given $$f(e^3) = \int_1^{e^3} \frac{\ln t}{\ln 10 (1+t)} \, dt ......(1)$$ $$f(\alpha) = \int_1^{\alpha} \frac{\ln t}{(\ln 10)(1+t)} \, dt$$ Let $t = \frac{1}{x} \implies x = \frac{1}{t}$. Then $dt = -\frac{1}{x^2} \, dx$. Thus, $$= \int_1^{\frac{1}{\alpha}} \frac{-\ln x}{(\ln 10) \left(1 + \frac{1}{x}\right)} \left(-\frac{1}{x^2}\right) \, dx$$ $$f(\alpha) = \frac{1}{\ln 10} \int_1^{\frac{1}{\alpha}} \frac{\ln x}{x(x+1)} \, dx ......(2)$$ Now consider $f(e^3) + f(e^{-3})$. $$= \left(\frac{1}{\ln 10}\right) \int_1^{e^3} \frac{\ln t}{(1+t)} \left[1 + \frac{1}{t}\right] \, dt$$ $$= \left(\frac{1}{\ln 10}\right) \int_1^3 \frac{\ln t}{t} \, dt$$ Let $\ln t = r$. Then $\frac{dt}{t} = dr$. Thus, $$= \frac{1}{\ln 10} \int_0^3 r \, dr$$ $$= \left(\frac{1}{\ln 10}\right) \left[\frac{r^2}{2}\right]_0^3$$ $$= \left(\frac{1}{\log 10}\right) \left(\frac{9}{2}\right)$$ $$= \frac{9}{2 \log_e 10}$$
Maths · Applications of Integrals · Single correct
The area of the region \[ \left\{ (x,y): |x-1|\le y\le\sqrt{5-x^2} \right\} \] is equal to:
Answer: (d)
Given $|x - 1| < y < \sqrt{5 - x^2}$. When $|x - 1| = \sqrt{5 - x^2}$, $\[$ (x - 1)^2 = 5 - x^2 $\]$ $\[$ x^2 - x - 2 = 0 $\]$ $\[$ x = 2, -1 $\]$ Required Area = Area of $\triangle ABC$ + Area of region $BCD$. $\[$ = $\frac{1}{2}$ $\begin{vmatrix}$ 1 & 0 & 1 $\\$ 2 & 1 & 1 $\\$ -1 & 2 & 1 $\end{vmatrix}$ + $\frac{\pi}{4}$($\sqrt{5}$)^2 - $\frac{1}{2}$($\sqrt{5}$)^2 $\]$ $\[$ = $\frac{5\pi}{4}$ - $\frac{1}{2}$ $\]$
Maths · Conic Sections · Single correct
Let the focal chord of the parabola $P : y^2 = 4x$ along the line $L : y = mx + c, \ m > 0$ meet the parabola at the points $M$ and $N$. Let the line $L$ be a tangent to the hyperbola $H : x^2 - y^2 = 4$. If $O$ is the vertex of $P$ and $F$ is the focus of $H$ on the positive x-axis, then the area of the quadrilateral $OMFN$ is:
Focus $(ae, 0)$ $F\left(2\sqrt{2}, 0\right)$ Line $L : y = mx + c$ passes through $(1,0)$ $$0 = m + C ........(1)$$ Line $L$ is tangent to the hyperbola. $$\frac{x^2}{4} - \frac{y^2}{4} = 1$$ $$C = \pm \sqrt{a^2m^2 - \ell^2}$$ $$C = \pm \sqrt{4m^2 - 4}$$ From (1) $$-m = \pm \sqrt{4m^2 - 4}$$ Squaring $$m^2 = 4m^2 - 4$$ $$4 = 3m^2$$ $$\frac{2}{\sqrt{3}} = m (as $m > 0$)$$ $$C = -m$$ $$C = \frac{-2}{\sqrt{3}}$$ $$y = \frac{2x}{\sqrt{3}} - \frac{2}{\sqrt{3}}$$ $$y^2 = 4x$$ $$\Rightarrow \left(\frac{2x - 2}{\sqrt{3}}\right)^2 = 4x$$ $$\Rightarrow x^2 + 1 - 2x = 3x$$ $$\Rightarrow x^2 - 5x + 1 = 0$$ $$\Rightarrow y^2 - 2\sqrt{3}y - 4 = 0$$ Area $$\frac{1}{2}\begin{vmatrix} 0 & x_1 & 2\sqrt{2} & x_2 \\ 0 & y_1 & 0 & y_2 \end{vmatrix}$$ $$= \frac{1}{2}\left[-2\sqrt{2}y_1 + 2\sqrt{2}y_2\right]$$ $$= \sqrt{2}|y_2 - y_1| = \frac{(\sqrt{2})\sqrt{12 + 16}}{111}$$ $$= \sqrt{56}$$ $$= 2\sqrt{14}$$
Maths · Continuity and Differentiability · Single correct
The number of points, where the function $$f : \mathbb{R} \rightarrow \mathbb{R}, \ f(x) = |x - 1| \cos |x - 2| \sin |x - 1| + (x - 3) |x^2 - 5x + 4|,$$ is NOT differentiable, is :
Answer: (b)
Given $f(x) = |x - 1| \cos |x - 2| \sin |x - 1| + (x - 3)|x^2 - 5x + 4|$. This simplifies to $|x - 1| \cos |x - 2| \sin |x - 1| + (x - 3)|x - 1||x - 4|$. Further simplification gives $|x - 1| [\cos |x - 2| \sin |x - 1| + (x - 3) |x - 4|]$. The function is non-differentiable at $x = 1$ and $x = 4$.
Maths · Probability · Single correct
Let $S=\{1,2,3,\ldots,2022\}$. Then the probability that a randomly chosen number $n$ from the set $S$ is such that $\operatorname{HCF}(n,2022)=1$ is:
Answer: (d)
Total number of elements $=2022$ $2022=2\times3\times337$ $\operatorname{HCF}(n,2022)=1$ is feasible when $n$ and $2022$ have no common factor. $A=$ Numbers which are divisible by $2$ from $\{1,2,3,\ldots,2022\}$ $n(A)=1011$ $B=$ Numbers which are divisible by $3$ from $\{1,2,3,\ldots,2022\}$ $n(B)=674$ $A\cap B=$ Numbers divisible by $6$ from $\{1,2,3,\ldots,2022\}$ $337=n(A\cap B)$ $n(A\cup B)$ $=n(A)+n(B)-n(A\cap B)$ $=1011+674-337$ $=1348$ $C=$ Numbers divisible by $337$ from $\{1,\ldots,2022\}$ $C=\{337,674,1011,1348,1685,2022\}$ These are already counted in $A\cup B$ Total elements divisible by $2$ or $3$ or $337$ $=1348+2$ $=1350$ Favourable cases $=2022-1350$ $=672$ Required probability $=\frac{672}{2022}$ $=\frac{112}{337}$
Maths · Applications of Derivatives · Single correct
Let $f(x) = 3^{(x^2 - 2)^3 + 4}$, $x \in \mathbb{R}$. Then which of the following statements are true? P: $x = 0$ is a point of local minima of $f$ Q: $x = \sqrt{2}$ is a point of inflection of $f$ R: $f'$ is increasing for $x > \sqrt{2}$
Answer: (d)
Given $f(x) = 81.3^{(x^2 - 2)^3}$. Differentiating, $$f'(x) = 81.3^{(x^2 - 2)^3} \ln 3.3(x^2 - 2)^2 \cdot 2x.$$ Simplifying, $$(81 \times 6)3^{(x^2 - 2)^3} x(x^2 - 2)^2 \ln 3.$$ The sign chart is as follows: $$\begin{array}{c|c|c|c} & - & 0 & + \\ -\sqrt{2} & & & \\ 0 & & & \\ \sqrt{2} & & & \\ \end{array}$$ $x = 6$ is a point of local minimum. Now, $$f'(x) = \frac{(486 \ln 3)3^{(x^2 - 2)^3} x(x^2 - 2)^2}{k}$$ where $g(x)$ is $$g'(x) = 3^{(x^2 - 2)^3} (x^2 - 2) + x.3^{(x^2 - 2)^3} .4x.(x^2 - 2)$$ $$+ x.(x^2 - 2)^2 .3^{(x^2 - 2)^3} \ln 3.3(x^2 - 2)^2 .2x$$ Simplifying, $$= 3^{(x^2 - 2)^3} (x^2 - 2) \left[ x^2 - 2 + 4x^2 + 6x^2 \ln 3(x^2 - 2)^3 \right]$$ $$g'(x) = 3^{(x^2 - 2)} (x^2 - 2) \left[ 5x^2 - 2 + 6x^2 \ln 3(x^2 - 2)^3 \right]$$ Thus, $$f''(x) = k.g'(x)$$ At $x = \sqrt{2}$, $$f''(\sqrt{2}) = 0, f''(\sqrt{2}^+) > 0, f''(\sqrt{2}^-) \sqrt{2}$, $f'(x) > 0$ so $f(x)$ is increasing.
Maths · Trigonometric Functions · Numerical
Let $S = \{ \theta \in (0, 2\pi) : 7 \cos^2\theta - 3 \sin^2\theta - 2 \cos^2 2\theta = 2 \}$. Then, the sum of roots of all the equations $x^2 - 2 (\tan^2\theta + \cot^2\theta) x + 6 \sin^2\theta = 0$ where $\theta \in S$, is
Answer: 16
Given the equation $7 \cos^2 \theta - 3 \sin^2 \theta - 2 \cos^2 2\theta = 2$. Simplifying, we have $4 \cos^2 \theta + 3 \cos 2\theta - 2 \cos^2 2\theta = 2$. This can be rewritten as $2 (1 + \cos 2\theta) + 3 \cos 2\theta - 2 \cos^2 2\theta = 2$. Simplifying further, $2 \cos^2 2\theta - 5 \cos 2\theta = 0$. Factoring gives $\cos 2\theta (2 \cos 2\theta - 5) = 0$. Thus, $\cos 2\theta = 0$. Solving for $\theta$, $2\theta = (2n + 1) \frac{\pi}{2}$, which gives $\theta = (2n + 1) \frac{\pi}{4}$. Therefore, the set $S = \left\{ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\}$. For all four values of $\theta$, the equation $x^2 - 2 (\tan^2 \theta + \cot^2 \theta) x + 6 \sin^2 \theta = 0$ simplifies to $x^2 - 4x + 3 = 0$. The sum of roots of all four equations is $4 \times 4 = 16$.
Maths · Statistics · Fill in the blank
Let the mean and the variance of 20 observations $x_1, x_2, \ldots, x_{20}$ be 15 and 9, respectively. For $\alpha \in \mathbb{R}$, if the mean of $(x_1 + \alpha)^2, (x_2 + \alpha)^2, \ldots, (x_{20} + \alpha)^2$ is 178, then the square of the maximum value of $\alpha$ is equal to .
Answer: 4
Given $$\sum x_1 = 15 \times 20 = 300$$ ... (i) $$\frac{\sum x_1^2}{20} - (15)^2 = 9$$ ... (ii) $$\sum x_1^2 = 234 \times 20 = 4680$$ $$\frac{\sum (x_1 + \alpha)^2}{20} = 178 \Rightarrow \sum (x_1 + \alpha)^2 = 3560$$ $$\Rightarrow \sum x_1^2 + 2\alpha \sum x_1 + \sum \alpha^2 = 3560$$ $$4680 + 600\alpha + 20\alpha^2 = 3560$$ $$\Rightarrow \alpha^2 + 30\alpha + 56 = 0$$ $$\Rightarrow (\alpha + 28)(\alpha + 2) = 0$$ $$\alpha = -2, -28$$ Square of maximum value of $$\alpha$$ is 4
Maths · Three Dimensional Geometry · Fill in the blank
Let a line with direction ratios $a$, $-4a$, $-7$ be perpendicular to the lines with direction ratios $3$, $-1$, $2b$ and $b$, $a$, $-2$. If the point of intersection of the line $\frac{x + 1}{a^2 + b^2} = \frac{y - 2}{a^2 - b^2} = \frac{z}{1}$ and the plane $x - y + z = 0$ is $(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma$ is equal to _________.
Answer: 10
(a, -4a, -7) is perpendicular to (3, -1, 2b). $$a = 2b ...(i)$$ (a, -4a, -7) is perpendicular to (b, a, -2). $$3a + 4a - 14b = 0$$ $$ab - 4a^2 + 14 = 0 ...(ii)$$ From Equations (i) and (ii) $$2b^2 - 16b^2 + 14 = 0$$ $$b^2 = 1$$ $$a^2 = 4b^2 = 4$$ $$\frac{x + 1}{5} = \frac{y - 2}{3} = \frac{z}{1} = k$$ $$\alpha = 5k - 1, \beta = 3k + 2, \gamma = k$$ As $(\alpha, \beta, \gamma)$ satisfies $x - y + z = 0$ $$5k - 1 - (3k + 2) + k = 0$$ $$k = 1$$ $$\therefore \alpha + \beta + \gamma = 9k + 1 = 10$$
Maths · Sequences and Series · Fill in the blank
Let $a_1, a_2, a_3, \ldots$ be an A.P. If $$\sum_{r=1}^{\infty} \frac{a_r}{2^r} = 4$$, then $4a_2$ is equal to .
Answer: 16
Given $$S = \frac{a_1}{2} + \frac{a_2}{2^2} + \frac{a_3}{2^3} + \ldots$$ We have $$\frac{S}{2} = \frac{a_1}{2^2} + \frac{a_2}{2^3} + \ldots$$ Subtracting, $$\frac{S}{2} = \frac{a_1}{2} + d \left( \frac{1}{2^2} + \frac{1}{2^3} + \ldots \right)$$ This simplifies to $$\frac{S}{2} = \frac{a_1}{2} + d \left( \frac{1}{4}{\frac{1}{1 - \frac{1}{2}}} \right)$$ Therefore, $$S = a_1 + d = a_2 = 4$$ Or $$4a_2 = 16$$
Maths · Binomial Theorem · Numerical
Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of $$\left( \sqrt[4]{2} + \frac{1}{\sqrt[4]{3}} \right)^n$$, in the increasing powers of $$\frac{1}{\sqrt[4]{3}}$$ be $$\sqrt[4]{6} : 1$$. If the sixth term from the beginning is $$\frac{\alpha}{\sqrt[4]{3}}$$, then $$\alpha$$ is equal to .
Answer: 84
Given $\($ $\frac{T_5}{T_{n-3}}$ = $\frac{\binom{n}{4} (2^{1/4})^{n-4} (3^{-1/4})^4}{\binom{n}{n-4} (2^{1/4})^4 (3^{-1/4})^{n-4}}$ = $\frac{\sqrt[4]{6}}{1}$ $\)$. $\($ $\Rightarrow$ 2^{$\frac{n-8}{4}$} 3^{$\frac{n-8}{4}$} = 6^{1/4} $\)$ $\($ $\Rightarrow$ 6^{n-8} = 6 $\)$ $\($ $\Rightarrow$ n-8 = 1 $\Rightarrow$ n = 9 $\)$ $\($ T_6 = $\binom{9}{5}$ (2^{1/4})^4 (3^{-1/4})^5 = $\frac{84}{\sqrt[4]{3}}$ $\)$ Therefore, $\($ $\alpha$ = 84 $\)$
Maths · Matrices · Numerical
The number of matrices of order $3 \times 3$, whose entries are either 0 or 1 and the sum of all the entries is a prime number, is_____.
Answer: 282
Given the matrix $$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$ where $a_{ij} \in \{0, 1\}$. The sum $$\sum a_{ij} = 2, 3, 5, 7$$. The total number of matrices is given by $$\binom{9}{2} + \binom{9}{3} + \binom{9}{5} + \binom{9}{7}$$ which equals 282.
Maths · Determinants · Numerical
Let p and p + 2 be prime numbers and let $$\Delta = \begin{vmatrix} p! & (p+1)! & (p+2)! \\ (p+1)! & (p+2)! & (p+3)! \\ (p+2)! & (p+3)! & (p+4)! \end{vmatrix}$$ Then the sum of the maximum values of $\alpha$ and $\beta$, such that $p^\alpha$ and $(p + 2)^\beta$ divide $\Delta$, is ________.
Answer: 4
Given $$\Delta = \begin{vmatrix} P! & (P+1)! & (P+2)! \\ (P+1)! & (P+2)! & (P+3)! \\ (P+2)! & (P+3)! & (P+4)! \end{vmatrix}$$ We have $$\Delta = P!(P+1)!(P+2)! \begin{vmatrix} 1 & \frac{1}{P+1} & \frac{1}{P+2} \\ \frac{1}{(P+2)(P+1)} & \frac{1}{(P+3)(P+2)} & \frac{1}{(P+4)(P+3)} \end{vmatrix}$$ This simplifies to $$\Delta = 2P!(P+1)!(P+2)!$$ Which is divisible by $P^\alpha$ and $(P+2)^\beta$. Therefore, $\alpha = 3$, $\beta = 1$.
Maths · Sequences and Series · Numerical
If $\frac{1}{2 \times 3 \times 4} + \frac{1}{3 \times 4 \times 5} + \frac{1}{4 \times 5 \times 6} + \ldots + \frac{1}{100 \times 101 \times 102} = \frac{k}{101}$, then $34 \times k$ is equal to
Answer: 286
The given series is: $$\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \ldots + \frac{1}{100 \cdot 101 \cdot 102} = \frac{k}{101}$$ Rewriting each term, we have: $$\frac{4 - 2}{2 \cdot 3 \cdot 4} + \frac{5 - 3}{3 \cdot 4 \cdot 5} + \ldots + \frac{102 - 100}{100 \cdot 101 \cdot 102} = \frac{2k}{101}$$ This simplifies to: $$\frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4} + \frac{1}{3 \cdot 4} - \frac{1}{4 \cdot 5} + \ldots + \frac{1}{100 \cdot 101} - \frac{1}{101 \cdot 102} = \frac{2k}{101}$$ The series telescopes to: $$\frac{1}{2 \cdot 3} - \frac{1}{101 \cdot 102} = \frac{2k}{101}$$ Thus, $$2k = \frac{101}{6} - \frac{1}{102}$$ Therefore, $$34k = 286$$
Maths · Sets · Numerical
Let S = \{$4, 6, 9$\} and T = {9, 10, 11, $\ldots$, 1000}. If A = \{$a_1 + a_2$ + $\ldots$ + $a_k$ : k $\in$ $\mathbb{N}$, $a_1, a_2, a_3$, $\ldots$,$a_k$ $\in$ S\}, then the sum of all the elements in the set T - A is equal to .
Answer: 11
Given sets $S = \{4, 6, 9\}$ and $T = \{9, 10, 11, \ldots, 1000\}$. Define $A = \{a_1 + a_2 + \ldots + a_k : k \in \mathbb{N}\}$ where $a_i \in S$. Here, by the definition of set $A$, we have $A = \{a : a = 4x + 6y + 9z\}$. Except for the element 11, every element of set $T$ is of the form $4x + 6y + 9z$ for some $x, y, z \in \mathbb{W}$. Therefore, $T - A = \{11\}$.
Maths · Conic Sections · Numerical
Let the mirror image of a circle $c_1 : x^2 + y^2 - 2x - 6y + \alpha = 0$ in line $y = x + 1$ be $c_2 : 5x^2 + 5y^2 + 10gx + 10fy + 38 = 0$. If $r$ is the radius of circle $c_2$, then $\alpha + 6r^2$ is equal to
Answer: 12
Image of centre $c_1 \equiv (1,3)$ in $x - y + 1 = 0$ is given by $$\frac{x_1 - 1}{1} = \frac{y_1 - 3}{-1} = \frac{-2(1 - 3 + 1)}{1^2 + 1^2}$$ $$\Rightarrow x_1 = 2, y_1 = 2$$ Therefore, centre of circle $c_2 \equiv (2,2)$. Therefore, equation of $c_2$ be $x^2 + y^2 - 4x - 4y + \frac{38}{5} = 0$. Now radius of $c_2$ is $$\sqrt{4 + 4 - \frac{38}{5}} = \sqrt{\frac{2}{5}} = r$$ (radius of $c_1)^2 = (radius of c_2)^2$ $$\Rightarrow 10 - \alpha = \frac{2}{5} \Rightarrow \alpha = \frac{48}{5}$$ Therefore, $\alpha + 6r^2 = \frac{48}{5} + \frac{12}{5} = 12$
Physics · Physical World, Units and Measurements · Single correct
Given below are two statements : One is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is p and radius of the drop is r, then $T = k \sqrt{\frac{p r^3}{S^{3/2}}}$ is dimensionally correct, where K is dimensionless. Reason (R) : Using dimensional analysis we get R.H.S. having different dimension than that of time period. In the light of above statements, choose the correct answer from the options given below.
Answer: (d)
Given $T = k \sqrt{\frac{\rho r^3}{s^{3/2}}}$. Dimensions of RHS = $\left$[ M^{1/2} L^{-3/2} T^{3/2} $\right$] $\left$[ L^{3/2} $\right$] $\left$[ M T^{-2} $\right$]^{-3/4} = M^{1/8} L^0 T^{3/2}. Dimensions of L.H.S $\neq$ Dimensions of R.H.S Therefore, option (D)
Physics · Motion in a Straight Line · Single correct
A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height $h$. Find the ratio of the times in which it is at height $\frac{h}{3}$ while going up and coming down respectively.
Answer: (b)
Max. Height $= h = \frac{u^2}{2g}$ $\($ $\Rightarrow$ u = $\sqrt{2gh}$ $\)$ $S = ut + \frac{1}{2} at^2$ $$\frac{h}{3} = \sqrt{2gh}t + \frac{1}{2}(-g)t^2$$ $$\frac{gt^2}{2} - \sqrt{2gh}t + \frac{h}{3} = 0$$ (Roots are $t_1$ and $t_2$) $$\frac{t_2}{t_1} = \frac{\sqrt{2gh} + \sqrt{2gh - 4 \times \frac{g}{2} \times \frac{h}{3}}}{\sqrt{2gh} - \sqrt{2gh - 4 \times \frac{g}{2} \times \frac{h}{3}}} = \frac{\sqrt{2gh} + \frac{4gh}{3}}{\sqrt{2gh} - \frac{4gh}{3}} = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$$
Physics · Motion in a Straight Line · Single correct
If $t=\sqrt{x}+4$, then $\left(\frac{dx}{dt}\right)_{t=4}$ is:
Answer: (b)
Given $t = \sqrt{x} + 4$. Therefore, $x = (t - 4)^2 = t^2 - 8t + 16$. Differentiating with respect to $t$, we have $$\frac{dx}{dt} = 2t - 8.$$ Evaluating at $t = 4$, $$\frac{dx}{dt}\bigg|_{t=4} = 2 \times 4 - 8 = 0.$$
Physics · Laws of Motion · Single correct
A smooth circular groove has a smooth vertical wall as shown in figure. A block of mass $m$ moves against the wall with a speed $v$. Which of the following curve represents the correct relation between the normal reaction on the block by the wall $(N)$ and speed of the block $(v)$ ?
Answer: (a)
Given $$N = \frac{mv^2}{r}$$. The curve is a parabola. $$Y = kx^2$$.
Physics · Motion in a Plane · Single correct
A ball is projected with kinetic energy $E$, at an angle of $60^0$ to the horizontal. The kinetic energy of this ball at the highest point of its flight will become:
Answer: (c)
The initial energy is given by $$E = \frac{1}{2} m u^2$$. At the highest point, the velocity $V$ is $u \cos 60^\circ = \frac{u}{2}$. Therefore, the kinetic energy at the topmost point is $$\frac{1}{2} m \left( \frac{u}{2} \right)^2 = \frac{E}{4}.$$
Physics · System of Particles and Rotational Motion · Single correct
Two bodies of mass 1 kg and 3 kg have position vectors $\hat{i} + 2\hat{j} + \hat{k}$ and $-3\hat{i} - 2\hat{j} + \hat{k}$ respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector:
Answer: (a)
The center of mass position is given by $$\vec{r}_{com} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} = \frac{1(\hat{i} + 2\hat{j} + \hat{k}) + 3(-3\hat{i} - 2\hat{j} + \hat{k})}{1 + 3}$$ This simplifies to $$= -2\hat{i} - \hat{j} + \hat{k}$$ The magnitude is $$|2\hat{i} - \hat{j} + \hat{k}| = \sqrt{(2)^2 + (1)^2 + (1)^2} = \sqrt{6}$$
Physics · Mechanical Properties of Fluids · Single correct
Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Clothes containing oil or grease stains cannot be cleaned by water wash. Reason (R) : Because the angle of contact between the oil/ grease and water is obtuse. In the light of the above statements, choose the correct answer from the option given below.
Answer: (a)
For water oil interface, $\theta_c > 90^\circ$.
Physics · Mechanical Properties of Solids · Single correct
If the length of a wire is made double and radius is halved of its respective values. Then, the Young’s modules of the material of the wire will :
Answer: (a)
Y depends on material of wire
Physics · Oscillations · Single correct
The time period of oscillation of a simple pendulum of length $L$ suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination $\alpha$, is given by:
Answer: (a)
The effective acceleration $g_{eff}$ is given by $g \cos \alpha$. The diagram shows the forces acting on the object. The acceleration $a$ is equal to $g \sin \alpha$. The period $T$ is given by the formula: $$T = 2\pi \sqrt{\frac{L}{g \cos \alpha}}.$$
Physics · Electric Charges and Fields · Single correct
A spherically symmetric charge distribution is considered with charge density varying as $$\rho(r) = \begin{cases} \rho_0 \left( \frac{3}{4} - \frac{r}{R} \right) & for r \leq R \\ Zero & for r > R \end{cases}$$ Where, $r(r < R)$ is the distance from the centre $O$ (as shown in figure). The electric field at point $P$ will be :
Answer: (c)
By Gauss law $$\oint \vec{E} \cdot d\vec{s} = \frac{Q_{in}}{\varepsilon_0}$$ $$E \cdot 4 \pi r^2 = \frac{\rho_0 \int_0^r \left( \frac{3}{4} - \frac{r}{R} \right) 4 \pi r^2 \, dr}{\varepsilon_0}$$ $$E 4 \pi r^2 = \frac{\rho_0 4 \pi}{\varepsilon_0} \left( \frac{3}{4} \frac{r^3}{3} - \frac{r^4}{4R} \right)$$ $$E r^2 = \frac{\rho_0 r^3}{4 \varepsilon_0} \left\{ 1 - \frac{r}{R} \right\}$$ $$E = \frac{\rho_0 r}{4 \varepsilon_0} \left\{ 1 - \frac{r}{R} \right\}$$
Physics · Electric Charges and Fields · Single correct
Given below are two statements. Statement I : Electric potential is constant within and at the surface of each conductor. Statement II : Electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point. In the light of the above statements, choose the most appropriate answer from the options given below.
Answer: (a)
Statement – I, true as body of conductor acts as equipotential surface. Statement – 2 True, as conductor is equipotential. Tangential component of electric field should be zero. Therefore electric field should be perpendicular to surface.
Physics · Current Electricity · Single correct
Two metallic wires of identical dimensions are connected is series. If $\sigma_1$ and $\sigma_2$ are the conductivities of the these wires respectively, the effective conductivity of the combination is:
Answer: (b)
Let length of wire be $'\ell'$. Area of wire as $'A'$. For equivalent wire length $= 2 \ell$ and area will be $A$. Thermal resistance $$R_{eq} = R_1 + R_2$$ $$\frac{2\ell}{\sigma_{eq} \, A} = \frac{\ell}{\sigma_1 A} + \frac{\ell}{\sigma_2 A}$$ $$\frac{2\ell}{\sigma_{eq}} = \frac{\ell}{\sigma_1} + \frac{\ell}{\sigma_2} \implies \sigma_{eq} = \frac{2 \sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$$
Physics · Alternating Current · Single correct
An alternating emf $E = 440 \sin 100 \pi t$ is applied to a circuit containing an inductance of $\frac{\sqrt{2}}{\pi}$ H. If an a.c. ammeter is connected in the circuit, its reading will be:
Answer: (c)
Given $E = 440 \sin 100 \pi t$, $L = \frac{\sqrt{2}}{\pi} \, \mathrm{H}$. $X_L = \omega L = 100 \pi \frac{\sqrt{2}}{\pi} = 100 \sqrt{2} \, \Omega$. Peak current $I_0 = \frac{E_0}{X_L} = \frac{440}{100 \sqrt{2}} = 2.2 \sqrt{2} \, \mathrm{A}$. AC ammeter reads RMS value therefore reading will be $I_{\mathrm{rms}}$. $I_{\mathrm{rms}} = \frac{I_0}{\sqrt{2}} = 2.2 \, \mathrm{A}$.
Physics · Alternating Current · Single correct
A coil of inductance 1 H and resistance 100 $\Omega$ is connected to a battery of 6 V. Determine approximately: (a) The time elapsed before the current acquires half of its steady – state value (b) The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circuit is switched on. (Given ln2 = 0.693, $e^{-3/2} = 0.25$)
Answer: (c)
Given circuit is R-L growth circuit. $$i = \frac{E}{R} \left(1 - e^{-t/\tau}\right)$$ $$i = \frac{E}{2R} = \frac{E}{R} \left(1 - e^{-t/\tau}\right)$$ Solving $t = \tau \ln 2$ $$t = \frac{1}{R} \ln 2 = \frac{1}{100} \times 0.693 = 0.00693$$ $$= 7 \, ms$$ $$i(15 \, ms) = \frac{E}{R} \left(1 - e^{-\frac{15}{10}}\right)$$ $$i = \frac{6}{100} \left(1 - \frac{1}{4}\right) = \frac{3}{4} \times \frac{6}{100}$$ $$U = \frac{1}{2} LI^2,$$ by solving we get $U = 1 \, mJ$.
Physics · Electromagnetic Waves · Single correct
Match List - I with List - II Choose the correct answer from the options given below:
Answer: (b)
$(A)$ UV rays – used for water purification $(B)$ X-rays – used for diagnosing fractures $(C)$ Microwaves are used for mobile and radar communication $(D)$ Infrared waves show less scattering; therefore, they are used on foggy days. $(A)$–(ii), $(B)$–(i), $(C)$–(iii), $(D)$–(iv)
Physics · Dual Nature of Radiation and Matter · Single correct
The kinetic energy of emitted electron is E when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy, the incident light must have wavelength:
Answer: (b)
Given the equations: $$E = \frac{hc}{\lambda} - \phi (i)$$ $$2E = \frac{hc}{\lambda'} - \phi (ii)$$ Subtracting equation (i) from equation (ii): $$E = hc \left( \frac{1}{\lambda'} - \frac{1}{\lambda} \right)$$ Solving for $\lambda'$: $$\Rightarrow \lambda' = \frac{hc \lambda}{E \lambda + hc}$$
Physics · Atoms · Single correct
Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.
Answer: (a)
Given $E_n = \frac{-13.6}{n^2} \, eV$. Therefore, $$\frac{E_2 - E_1}{E_\infty - E_1} = \frac{13.6(1 - \frac{1}{4})}{13.6} = \frac{3}{4}$$
Physics · Communication Systems · Single correct
Find the modulation index of an AM wave having 8 $\,$ $\mathrm{V}$ variation where maximum amplitude of the AM wave is 9 $\,$ $\mathrm{V}$.
Answer: (a)
Modulation index: $m = \frac{A_m}{A_c}$ Given $2A_m = 8$ $A_m + A_c = 9 \Rightarrow A_c = 5$ Therefore, $m = \frac{4}{5} = 0.8$
Physics · Experimental Physics · Single correct
A travelling microscope has 20 divisions per cm on the main scale while its Vernier scale has total 50 divisions and 25 Vernier scale divisions are equal to 24 main scale divisions, what is the least count of the travelling microscope?
Answer: (c)
1 MSD = $\frac{1}{20}$ cm 1 VSD = $\frac{24}{25}$ MSD = $\frac{24}{25}$ $\times$ $\frac{1}{20}$ cm $\therefore$ Least count = $\frac{1}{20}$ $\left$( 1 - $\frac{24}{25}$ $\right$) cm = $\frac{1}{20}$ $\times$ $\frac{1}{25}$ = $\frac{1}{500}$ cm = 0.002 cm
Physics · Experimental Physics · Single correct
In an experiment to find out the diameter of wire using screw gauge, the following observation were noted: (a) Screw moves 0.5 mm on main scale in one complete rotation (b) Total divisions on circular scale = 50 (c) Main scale reading is 2.5 mm (d) 45^{th} division of circular scale is in the pitch line (e) Instrument has 0.03 mm negative error Then the diameter of wire is:
Answer: (c)
MSR $=2.5\,\mathrm{mm}$ CSR $=45\times\frac{0.5}{50}\,\mathrm{mm}$ $=0.45\,\mathrm{mm}$ Diameter reading $=\mathrm{MSR}+\mathrm{CSR}-\text{zero error}$ $=2.5+0.45-(-0.03)$ $=2.98\,\mathrm{mm}$
Physics · Motion in a Plane · Numerical
An object is projected in the air with initial velocity $u$ at an angle $\theta$. The projectile motion is such that the horizontal range $R$, is maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be ______ degree. (Mark the smallest angle possible)
Answer: 15
Given $\($ R_{max} = $\frac{u^2 \sin 2(45^\circ)}{g}$ = $\frac{u^2}{g}$ $\)$ \[\frac{R}{2} = \frac{u^2}{2g} = \frac{u^2 \sin 2\theta}{g} \] $\($ $\sin$ 2$\theta$ = $\frac{1}{2}$ $\)$ $\($ 2$\theta$ = $30^\circ$, $150^\circ$ $\)$ $\($ $\theta$ = $15^\circ$, $75^\circ$ $\)$ Ans. 15, 75
Physics · Gravitation · Numerical
If the acceleration due to gravity experienced by a point mass at a height $h$ above the surface of earth is same as that of the acceleration due to gravity at a depth $\alpha h$ ($h \ll R_e$) from the earth surface. The value of $\alpha$ will be ________. (use $R_e = 6400 \, \mathrm{km}$)
Answer: 2
Given the equation $$g \left( 1 - \frac{2h}{R} \right) = g \left( 1 - \frac{d}{R} \right)$$ we equate the terms: $$\frac{2h}{R} = \frac{d}{R}$$ Solving for $d$, we have $$\alpha h = d$$ Given $\alpha = 2$
Physics · Kinetic Theory · Numerical
The pressure $P_1$ and density $d_1$ of diatomic gas $\gamma = \frac{7}{5}$ changes suddenly to $P_2(>P_1)$ and $d_2$ respectively during an adiabatic process. The temperature of the gas increases and becomes times of its initial temperature. (given $\frac{d_2}{d_1} = 32$)
Answer: 4
Given $PV^\gamma = const$ and $d = \frac{m}{v}$. Therefore, $p \left( \frac{m}{d} \right)^\gamma = const$. This implies $\frac{p}{d^\gamma} = const$. Given $\frac{d_2}{d_1} = 32$, we have $$\frac{p_1}{p_2} = \left( \frac{d_1}{d_2} \right)^\gamma = \left( \frac{1}{32} \right)^{7/5} = \frac{1}{128}.$$ Therefore, $$\frac{T_1}{T_2} = \frac{P_1 V_1}{P_2 V_2} = \frac{1}{128} \times 32 = \frac{1}{4}.$$
Physics · Kinetic Theory · Numerical
One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is $\frac{\alpha^2}{4} R$ J/mol K; then the value of $\alpha$ will be ______. (Assume that the given diatomic gas has no vibrational mode.)
Answer: 3
The expression for the molar heat capacity at constant volume for a mixture is given by: $$C_V / mix = \frac{n_1 C_{V_1} + n_2 C_{V_2}}{n_1 + n_2}$$ Substituting the given values: $$= \frac{1 \cdot \frac{3R}{2} + 3 \cdot \frac{5R}{2}}{1 + 3}$$ Simplifying the expression: $$= \frac{9R}{4} = \frac{\alpha^2}{4} R$$ Solving for $\alpha$ gives: $$\alpha = 3$$
Physics · Current Electricity · Numerical
The current I flowing through the given circuit will be ________ A.
Answer: 2
Equivalent circuit $$I = \frac{6}{3} = 2 \, \mathrm{A}$$
Physics · Moving Charges and Magnetism · Numerical
A closely wounded circular coil of radius 5 cm produces a magnetic field of $37.68 \times 10^{-4} \, \mathrm{T}$ at its center. The current through the coil is _______ A. [Given, number of turns in the coil is 100 and $\pi = 3.14$]
Answer: 3
The magnetic field at the center is given by the formula: $$B_{centre} = \frac{N \mu_0 I}{2R}$$ Substituting the given values: $$37.68 \times 10^{-4} = \frac{100 \times 4 \pi \times 10^{-7} \times I}{2 \times 5 \times 10^{-2}}$$ Solving for $I$, we find: $$I = 3 \, A$$
Physics · Wave Optics · Numerical
Two light beams of intensities $4I$ and $9I$ interfere on a screen. The phase difference between these beams on the screen at point $A$ is zero and at point $B$ is $\pi$. The difference of resultant intensities, at the point $A$ and $B$, will be
Answer: 24
Given $I_{net} = I_1 + I_2 + 2 \sqrt{I_1} \sqrt{I_2} \cos \phi$. $I_{max}$ for $\phi = 0$ and $I_{min}$ for $\phi = \pi$. $I_{max} = \left( \sqrt{I_1} + \sqrt{I_2} \right)^2 = \left( \sqrt{9I} + \sqrt{4I} \right)^2 = 25I$. $I_{min} = \left( \sqrt{I_1} - \sqrt{I_2} \right)^2 = \left( \sqrt{9I} - \sqrt{4I} \right)^2 = I$. $I_{max} - I_{min} = 25I - I = 24I$.
Physics · Magnetism and Matter · Numerical
A wire of length 314 cm carrying current of 14 A is bent to form a circle. The magnetic moment of the coil is _______ A-m^2. [Given $\pi = 3.14$]
Answer: 11
Given $\frac{314}{100} = 2\pi R$ and $R = 0.5 \, \mathrm{m}$. The magnetic moment is given by $IA$. Magnetic Moment $= 14 \times \pi R^2$ $= 14 \times (3.14) \times \frac{1}{4}$ $= 10.99 \approx 11.00$
Physics · Ray Optics and Optical Instruments · Numerical
The X-Y plane be taken as the boundary between two transparent media $M_1$ and $M_2$. $M_1$ in $Z \geq 0$ has a refractive index of $\sqrt{2}$ and $M_2$ with $Z < 0$ has a refractive index of $\sqrt{3}$. A ray of light travelling in $M_1$ along the direction given by the vector $\vec{A} = 4\sqrt{3} \hat{i} - 3\sqrt{3} \hat{j} - 5 \hat{k}$, is incident on the plane of separation. The value of difference between the angle of incident in $M_1$ and the angle of refraction in $M_2$ will be ______ degree.
Answer: 15
Given $\vec{A} = 4\sqrt{3} \hat{i} - 3\sqrt{3} \hat{j} - 5 \hat{k}$. As incident vector $A$ makes $i$ angle with normal $z$-axis and refracted vector $R$ makes $r$ angle with normal $z$-axis with help of direction cosine, $$i = \cos^{-1} \left( \frac{A_z}{A} \right) = \cos^{-1} \left( \frac{5}{\sqrt{(4\sqrt{3})^2 + (3\sqrt{3})^2 + 5^2}} \right)$$ $$= \cos^{-1} \left( \frac{5}{10} \right) \Rightarrow i = 60^\circ$$ $$\sqrt{2} \sin 60 = \sqrt{3} \times \sin r$$ $$r = 45^\circ$$ Difference between $i$ and $r = 60 - 45 = 15$
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Numerical
If the potential barrier across a p-n junction is $0.6\,\mathrm{V}$. Then the electric field intensity, in the depletion region having the width of $6 \times 10^{-6}\,\mathrm{m}$, will be $x \times 10^5\,\mathrm{N/C}$.
Answer: 1
The electric field $E$ is given by the potential difference $V$ across the junction divided by the width of the depletion layer $d$. $$E = \frac{V}{d} = \frac{Potential barrier Across Junction}{width of Depletion layer}$$ Substituting the given values: $$E = \frac{0.6 \, \mathrm{V}}{6 \times 10^{-6} \, \mathrm{m}} = 1 \times 10^5 \, \mathrm{V/m}$$ Therefore, $$E = 1 \times 10^5 \, \mathrm{N/C}$$
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Which of the following pair of molecules contain odd electron molecule and an expanded octet molecule?
Answer: (b)
(A) $\mathrm{BCl_3}$ $\rightarrow$ Even Electron molecule $\mathrm{SF_6}$ $\rightarrow$ Expanded octet molecule (B) NO $\rightarrow$ Odd Electron molecule $\mathrm{H_2SO_4}$ $\rightarrow$ Expanded octet. (C) $\mathrm{SF_6}$ $\rightarrow$ Even Electron molecule $\mathrm{H_2SO_4}$ $\rightarrow$ Expanded octet. (D) $\mathrm{BCl_3}$ $\rightarrow$ Even Electron molecule NO $\rightarrow$ Odd Electron molecule S $\rightarrow$ $\mathrm{12e^-}$ in outer orbit
Chemistry · Some Basic Concepts of Chemistry · Single correct
Consider the above reaction, the limiting reagent of the reaction and number of moles of $NH_3$ formed respectively are:
Answer: (c)
The reaction is given by $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$. The weights are $W_2 = 20 \, \mathrm{g}$ and $5 \, \mathrm{g}$. The moles are calculated as $$n = \frac{20}{28} \frac{5}{2}$$. The stoichiometric amount is calculated as follows: $$\mathrm{N_2 \rightarrow \frac{20/28}{1} = \frac{20}{28}} \mathrm{H_2 \rightarrow \frac{5/2}{3} = \frac{5}{6}}$$ Therefore, $\mathrm{N_2}$ is the limiting reagent. Thus, $$n(\mathrm{NH_3}) = 2 \times n(\mathrm{N_2}) = 2 \times \frac{20}{28}$$ which equals $1.42$.
Chemistry · Surface Chemistry · Single correct
100 mL of 5% (w/v) solution of NaCl in water was prepared in 250 mL beaker. Albumin from the egg was poured into NaCl solution and stirred well. This resulted in a/ an:
Answer: (a)
Standard method for the preparation of lyophilic sol. (Discussed in lab Manual)
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The first ionization enthalpy of Na, Mg and Si, respectively, are: 496, 737 and 786 $\mathrm{kJ \, mol^{-1}}$. The first ionization enthalpy ($\mathrm{kJ \, mol^{-1}}$) of Al is:
Answer: (c)
I. E : Na < Al < Mg < Si Therefore, $496 < \mathrm{IE} (\mathrm{Al}) < 737$ Option $(C)$, matches the condition. i.e $\mathrm{IE} (\mathrm{Al}) = 577 \, \mathrm{kJmol^{-1}}$
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
In metallurgy the term "gangue" is used for:
Answer: (a)
Earthy and undesired materials present in the ore, other than the desired metal, is known as gangue.
Chemistry · The s-Block Elements · Single correct
The reaction of zinc with excess of aqueous alkali, evolves hydrogen gas and gives:
Answer: (c)
Zinc dissolves in excess of aqueous alkali $$\mathrm{Zn} + 2\mathrm{OH}^- + 2\mathrm{H_2O} \rightarrow [\mathrm{Zn(OH)}_4]^{2-} + \mathrm{H_2} \uparrow$$ Tetrahydroxozincate(II) ion However, this reaction in NCERT is given as $$\mathrm{Zn} + 2\mathrm{NaOH} \rightarrow \mathrm{Na_2ZnO_2} + \mathrm{H_2} \uparrow$$ $$\mathrm{ZnO_2^{2-}}$$ is anhydrous form of $$[\mathrm{Zn(OH)}_4]^{2-}$$ So in aqueous medium best answer of this question is $$[\mathrm{Zn(OH)}_4]^{2-}$$
Chemistry · The s-Block Elements · Single correct
Lithium nitrate and sodium nitrate, when heated separately, respectively, give :
Answer: (c)
$\mathrm{Li_2O,\ NaNO_2}$ As per NCERT Lithium nitrate when heated gives lithium oxide, $\mathrm{Li_2O}$, whereas other alkali metal nitrates decompose to give the corresponding nitrite. $4\mathrm{LiNO_3} \longrightarrow 2\mathrm{Li_2O} + 4\mathrm{NO_2} + \mathrm{O_2}$ $2\mathrm{NaNO_3} \longrightarrow 2\mathrm{NaNO_2} + \mathrm{O_2}$ However, the decomposition product of $\mathrm{NaNO_3}$ are temperature dependent process as shown in the below reaction. As temperature is not mentioned, we can go by
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Number of lone pairs of electrons in the central atom of $\mathrm{SCl_2}$, $\mathrm{O_3}$, $\mathrm{ClF_3}$ and $\mathrm{SF_6}$, respectively, are :
Answer: (b)
The molecule with the least number of lone pairs is $\mathrm{SF_6}$, which has 0 lone pairs.
Chemistry · The d-and f-Block Elements · Single correct
In following pairs, the one in which both transition metal ions are colourless is :
Answer: (a)
(A) $\mathrm{Sc^{3+}}$, $\mathrm{Zn^{2+}}$ $$3d^0 3d^{10}$$ (B) $\mathrm{Ti^{4+}}$, $\mathrm{Cu^{2+}}$ $$3d^0 3d^9$$ (C) $\mathrm{V^{2+}}$, $\mathrm{Ti^{3+}}$ $$3d^3 3d^1$$ (D) $\mathrm{Zn^{2+}}$, $\mathrm{Mn^{2+}}$ $$3d^{10} 3d^5$$ No d-d transitions in ions with $d^0$ and $d^{10}$ configuration. Therefore they are colourless.
Chemistry · The d-and f-Block Elements · Single correct
In neutral or faintly alkaline medium, $\mathrm{KMnO_4}$ being a powerful oxidant can oxidize, thiosulphate almost quantitatively, to sulphate. In this reaction overall change in oxidation state of manganese will be :
Answer: (d)
The reaction is: $$8\mathrm{MnO_4^-} + 3\mathrm{S_2O_3^{2-}} + \mathrm{H_2O} \rightarrow 8\mathrm{MnO_2} + 6\mathrm{SO_4^{2-}} + 2\mathrm{OH^-}$$ Change in oxidation state of Mn is from $+7$ to $+4$ which is $3$.
Chemistry · Environmental Chemistry · Single correct
Which among the following pairs has only herbicides?
Answer: (d)
Both sodium chlorate and sodium arsenite behave as herbicide.
Chemistry · Amines · Single correct
Which among the following is the strongest Bronsted base?
Answer: (d)
It is most basic because there is no amine inversion.
Chemistry · Hydrocarbons · Single correct
Which among the following pairs of the structures will give different products on ozonolysis? (Consider the double bonds in the structures are rigid and not delocalized.)
Answer: (c)
The ozonolysis of the given compound results in the formation of the following products. For the first compound: $$CH_3C = C - CH_3 \xrightarrow{Ozonolysis} O = C - C = O + 2 CHO$$ For the second compound: $$2 CH_3C = CHO + CHO$$
Chemistry · Haloalkanes and Haloarenes · Single correct
Considering the above reactions, the compound 'A' and compound 'B' respectively are:
Answer: (c)
In NaCN, carbon is more nucleophilic atom. Whereas in AgCN, Ag-C has covalent bond.
Chemistry · Alcohols, Phenols and Ethers · Single correct
Consider the above reaction sequence, the Product 'C' is :
Answer: (d)
The reaction sequence involves the bromination of the phenol compound to form compound (A). Then, compound (A) reacts with $\mathrm{NH_2OH}$ to form compound (B). Finally, compound (B) undergoes dehydration with $\mathrm{P_2O_5}$ to form compound $(C)$.
Chemistry · Amines · Single correct
Consider the above reaction, the compound 'A' is :
Answer: (c)
Chemistry · Amines · Single correct
Which among the following represent reagent 'A'?
Answer: (a)
The reaction involves the coupling of naphthol with a diazonium salt in a basic medium to form an azo compound.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Consider the following reaction sequence: The product 'B' is:
Answer: (b)
The reaction involves the conversion of the cyano group to an aldehyde group using diisobutylaluminium hydride (i-Bu)$_2$AlH followed by hydrolysis with water. The resulting compound undergoes a cross aldol condensation with acetaldehyde (CH$_3$CHO) in the presence of dilute NaOH and heat to form the final product.
Chemistry · Environmental Chemistry · Single correct
Which of the following compounds is an example of hypnotic drug?
Answer: (b)
Amytal is a hypnotic drug used to treat sleeping disorder.
Chemistry · Alcohols, Phenols and Ethers · Single correct
A compound 'X' is acidic and it is soluble in NaOH solution, but insoluble in $NaHCO_3$ solution. Compound 'X' also gives violet colour with neutral $FeCl_3$ solution. The compound 'X' is :
Answer: (b)
Phenol reacts with $\mathrm{FeCl_3}$ to form $[\mathrm{Fe(C_6H_5O)_6}]^{3-}$, which gives a violet colour.
Chemistry · Electrochemistry · Numerical
Resistance of a conductivity cell (cell constant $129 \, \mathrm{m}^{-1}$) filled with $74.5 \, \mathrm{ppm}$ solution of $\mathrm{KCl}$ is $100 \, \Omega$ (labelled as solution 1). When the same cell is filled with KCl solution of $149 \, \mathrm{ppm}$, the resistance is $50 \, \Omega$ (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. $\frac{\Lambda_1}{\Lambda_2} = x \times 10^{-3}$. The value of $x$ is _____. (Nearest integer) Given, molar mass of KCl is $74.5 \, \mathrm{g} \, \mathrm{mol}^{-1}$
Answer: 1000
Given $\frac{\ell}{A} = 129 \, \mathrm{m}^{-1}$. KCl solution 1: $74.5 \, \mathrm{ppm}$, $R_1 = 100 \, \Omega$ KCl solution 2: $149 \, \mathrm{ppm}$, $R_2 = 50 \, \Omega$ Here, $$\frac{\mathrm{ppm}_1}{\mathrm{ppm}_2} = \frac{M_1}{M_2} \left( \frac{\frac{W_1}{M_0}}{V} \times \frac{V}{\frac{W_2}{M_0}} \right)$$ $$\frac{\wedge_1}{\wedge_2} = \frac{\kappa_1 \times \frac{1000}{M_1}}{\kappa_2 \times \frac{1000}{M_2}}$$ $$= \frac{\kappa_1}{\kappa_2} \times \frac{M_2}{M_1}$$ $$= \frac{50}{100} \times 2$$ $$= \frac{\wedge_1}{\wedge_2} = 1,000 \times 10^{-3}$$
Chemistry · The Solid State · Numerical
Ionic radii of cation $\mathrm{A}^+$ and anion $\mathrm{B}^-$ are $102$ and $181 \, \mathrm{pm}$ respectively. These ions are allowed to crystallize into an ionic solid. This crystal has cubic close packing for $\mathrm{B}^-$. $\mathrm{A}^+$ is present in all octahedral voids. The edge length of the unit cell of the crystal $\mathrm{AB}$ is _____ $\mathrm{pm}$. (Nearest Integer)
Answer: 566
Given the equation for a: $$a = 2(r_+ + r_-)$$ Substitute the values: $$a = 2(102 + 181)$$ Calculate the sum: $$a = 2(283)$$ Finally, multiply to find: $$a = 566 \, \mathrm{pm}$$
Chemistry · Structure of Atom · Numerical
The minimum uncertainty in the speed of an electron in an one dimensional region of length $2a_0$ (Where $a_0 = Bohr radius \ 52.9 \, \mathrm{pm}$) is _____ $\mathrm{km \, s^{-1}}$. (Given : Mass of electron $= 9.1 \times 10^{-31} \, \mathrm{kg}$, Planck's constant $h = 6.63 \times 10^{-34} \, \mathrm{Js}$)
Answer: 548
According to Heisenberg's uncertainty principle, $$\Delta x \times \Delta p_x \geq \frac{h}{4\pi}$$ This implies $$2a_0 \times m \Delta v_x = \frac{h}{4\pi} (minimum)$$ Therefore, $$\Delta v_x = \frac{h}{4\pi} \times \frac{1}{2a_0} \times \frac{1}{m}$$ Substituting the values, $$= \frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 2 \times 52.9 \times 10^{-12} \times 9.1 \times 10^{-31}}$$ This results in $$= 548273 \, \mathrm{ms^{-1}}$$ Converting to kilometers per second, $$= 548.273 \, \mathrm{km \, s^{-1}}$$ Finally, $$= \boxed{548} \, \mathrm{km \, s^{-1}}$$
Chemistry · Thermodynamics · Numerical
When 600 mL of 0.2 M HNO$_3$ is mixed with 400 mL of 0.1M NaOH solution in a flask, the rise in temperature of the flask is _______ $\times 10^{-2}$ $^\circ$C. (Enthalpy of neutralisation = 57 kJ mol$^{-1}$ and Specific heat of water = 4.2 JK$^{-1}$g$^{-1}$) (Neglect heat capacity of flask)
Answer: 54
HNO$_3$ 600 mL $\times$ 0.2 M $= 120$ m mol NaOH 400 mL $\times$ 0.1 M $= 40$ m mol HNO$_3$ + NaOH $\rightarrow$ NaNO$_3$ + H$_2$O Bef. 120 40 Aft. 80 0 40 m mol $\Delta_r H = 40$ m mol $\times (57 \times 10^3) \frac{\mathrm{J}}{\mathrm{mol}}$ $= 40 \times 10^{-3} \mathrm{mol} \times 57 \times 10^3 \frac{\mathrm{J}}{\mathrm{mol}}$ $= 2280$ J $m \; S \Delta T = 2280$ $\Rightarrow 1000$ mL $\times \frac{1 \mathrm{gm}}{\mathrm{mL}} \times 4.2 \times \Delta T = 2280$ $$\Delta T = \frac{2280}{4.2 \times 10^{-3}}$$ $$= \frac{22800}{42 \times 10^{-3}}$$ $$= 542.86 \times 10^{-3}$$ $\Delta T = 54.286 \times 10^{-2}$ K $\Delta T = 54.286 \times 10^{-2}$ C Ans. 54.286 Answer mentioned as 54 (Closest integer)
Chemistry · Solutions · Numerical
If $\mathrm{O}_2$ gas is bubbled through water at $303 \, \mathrm{K}$, the number of millimoles of $\mathrm{O}_2$ gas that dissolve in $1 \, \mathrm{litre}$ of water is _______. (Nearest Integer) (Given : Henry's Law constant for $\mathrm{O}_2$ at $303 \, \mathrm{K}$ is $46.82 \, \mathrm{k \, bar}$ and partial pressure of $\mathrm{O}_2 = 0.920 \, \mathrm{bar}$) (Assume solubility of $\mathrm{O}_2$ in water is too small, nearly negligible)
Answer: 1
Given $p = K_H \times x$. $$0.920 \, \mathrm{bar} = 46.82 \times 10^3 \, \mathrm{bar} \times \frac{mol of O_2}{mol of H_2O}$$ $$0.920 = 46.82 \times 10^3 \times \frac{mol of O_2}{\frac{1000}{18}}$$ $$0.920 = 46.82 \times n_{o_2}$$ $$p = \frac{0.920}{46.82 \times 18} = n_{o_2}$$ $$\Rightarrow 1.09 \times 10^{-3} = n_{o_2}$$ $$\Rightarrow m mol of O_2 = 1$$
Chemistry · Equilibrium · Numerical
If the solubility product of PbS is $8 \times 10^{-28}$, then the solubility of PbS in pure water at $298 \, \mathrm{K}$ is $x \times 10^{-16} \, \mathrm{mol} \, \mathrm{L}^{-1}$. The value of $x$ is _______. (Nearest Integer) [Given $\sqrt{2} = 1.41$]
Answer: 282
Given $K_{sp} = S^2$. $S = \sqrt{K_{sp}} = \sqrt{8 \times 10^{-28}} = 2\sqrt{2} \times 10^{-14}$. $= 2.82 \times 10^{-14}$. $= 282 \times 10^{-16}$. Ans. = 282
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
The reaction between X and Y is first order with respect to X and zero order with respect to Y. \begin{tabular}{|c|c|c|c|} \hline Experiment & $[\mathrm{X}]\ \mathrm{mol\ L^{-1}}$ & $[\mathrm{Y}]\ \mathrm{mol\ L^{-1}}$ & Initial rate $\mathrm{mol\ L^{-1}\ min^{-1}}$ \\ \hline I. & 0.1 & 0.1 & $2 \times 10^{-3}$ \\ \hline II. & $L$ & 0.2 & $4 \times 10^{-3}$ \\ \hline III. & 0.4 & 0.4 & $M \times 10^{-3}$ \\ \hline IV. & 0.1 & 0.2 & $2 \times 10^{-3}$ \\ \hline \end{tabular} Examine the data of table and calculate ratio of numerical values of $M$ and $L$. (Nearest Integer)
Answer: 40
Given $r = k [x] [y]^0 = k [x]$. Using I & II $$\frac{4 \times 10^{-3}}{2 \times 10^{-3}} = \left( \frac{L}{0.1} \right) \Rightarrow L = 0.2$$ Using I & III $$\frac{M \times 10^{-3}}{2 \times 10^{-3}} = \frac{0.4}{0.1} \Rightarrow M = 8$$ $$\frac{M}{L} = \frac{8}{0.2} = 40$$ Ans. 40
Chemistry · Biomolecules · Numerical
In a linear tetrapeptide (Constituted with different amino acids), (number of amino acids) - (number of peptide bonds) is ______.
Answer: 1
In Tetrapeptide, No. of Amino Acids = 4 No. of Peptide bonds = 3 Hence Ans. = 1
Chemistry · Haloalkanes and Haloarenes · Numerical
In bromination of Propyne, with Bromine 1, 1, 2, 2-tetrabromopropane is obtained in 27$\%$ yield. The amount of 1, 1, 2, 2 tetrabromopropane obtained from 1 $\mathrm{g}$ of Bromine in this reaction is ______ $\times 10^{-1}$ $\mathrm{g}$. (Nearest integer) (Molar Mass : Bromine = 80 g/mol)
Answer: 3
$= \dfrac{1}{160} \times \dfrac{1}{2} \times 360 \times 0.27$ $= 0.30375$ $= 3.0375 \times 10^{-1}$ $\textbf{Ans.} = 3$
Chemistry · Co-ordination Compounds · Numerical
$\mathrm{[Fe(CN)_6]^{3-}}$ should be an inner orbital complex. Ignoring the pairing energy, the value of crystal field stabilization energy for this complex is $(-)\Delta_o$. (Nearest integer)
Answer: 2
$\mathrm{[Fe(CN)_6]^{3-}}$ $\mathrm{CN^-}$ is a strong field ligand. $\mathrm{Fe^{3+}}$ $3d^5$ $(t_{2g}^5\,e_g^0)$ CFSE $= 5 \times (-0.4\Delta_o)$ $= -2.0\Delta_o$