JEE Main 26 July 2022 Shift 2: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
Register free to attempt this shift as a timed 180-minute test with instant scoring and chapter-wise analysis.
Maths · Complex Numbers and Quadratic Equations · Single correct
The minimum value of the sum of the squares of the roots of $x^2 + (3-a)x + 1 = 2a$ is:
Answer: (c)
Given $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha \beta$. Let $f(a) = (3 - a)^2 - 2(1 - 2a)$. Then, $f(a) = a^2 - 2a + 7$. We can rewrite this as $f(a) = (a - 1)^2 + 6$. Therefore, the minimum value of $f(a)$ is $6$.
Maths · Complex Numbers and Quadratic Equations · Single correct
If $z = x + iy$ satisfies $|z| - 2 = 0$ and $|z - i| - |z + 5i| = 0$, then
Answer: (c)
Given $|z - i| - |z + 5i| = 0$. This implies $|x + (y - 1)i| = |x + (y + 5)i|$. Therefore, $x^2 + (y - 1)^2 = x^2 + (y + 5)^2$. Simplifying, $(y - 1)^2 - (y + 5)^2 = 0$. This gives $(2y + 4)(-6) = 0$. Solving for $y$, we get $y = -2$. Thus, $x^2 + (-2)^2 = 4$. Solving for $x$, we find $x = 0$. Therefore, $Z \equiv (0, -2)$, check options.
Maths · Matrices · Single correct
Let $A = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ and $B = \begin{bmatrix} 9^2 & -10^2 & 11^2 \\ 12^2 & 13^2 & -14^2 \\ -15^2 & 16^2 & 17^2 \end{bmatrix}$, then the value of $A'BA$ is:
Answer: (d)
Given $$A'BA = \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 9^2 & -10^2 & 11^2 \\ 12^2 & 13^2 & -14^2 \\ -15^2 & 16^2 & 17^2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$$ This simplifies to $$= \begin{bmatrix} 9^2 + 12^2 - 15^2 & -10^2 + 13^2 + 16^2 & 11^2 - 14^2 + 17^2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$$ Further simplifying, we get $= \begin{bmatrix} 9^2 + 12^2 - 15^2 - 10^2 + 13^2 + 16^2 + 11^2 - 14^2 + 17^2 \end{bmatrix}$ Finally, $$= \begin{bmatrix} 539 \end{bmatrix}$$
Maths · Binomial Theorem · Single correct
$\sum_{i,j=0}^{n} \binom{n}{i} \binom{n}{j}$ is equal to
Answer: (a)
The expression is given by $$\sum_{i,j=0, \, i \neq j}^{n} \binom{n}{i} \binom{n}{j}$$ This can be rewritten as $$= \sum_{i=0}^{n} \binom{n}{i} \cdot \sum_{j=0}^{n} \binom{n}{j} - \sum_{i=j=0}^{n} \left( \binom{n}{i} \right)^2$$ Simplifying further, $$= (2^n)(2^n) - 2^n \binom{n}{n}$$ Finally, $$= 2^{2n} - 2^n \binom{n}{n}$$
Maths · Conic Sections · Single correct
Let P and Q be any points on the curves $(x-1)^2+(y+1)^2=1$ and $y = x^2$, respectively. The distance between P and Q is minimum for some value of the abscissa of P in the interval
Answer: (c)
$Q=(t,t^2)$ $m_{CQ}$ $=m_{\text{normal}}$ $\frac{t^2+1}{t-1}$ $=-\frac1{2t}$ Let $f(t)=2t^3+3t-1$ $f\!\left(\frac14\right)\cdot f\!\left(\frac13\right)<0$ $\Rightarrow t\in\left(\frac14,\frac13\right)$ $P= (1+\cos(90+\theta), -1+\sin(90+\theta))$ $P= (1-\sin\theta, -1+\cos\theta)$ $m_{\text{normal}}$ $=m_{CP}$ $\Rightarrow -\frac1{2t}$ $=\frac{\cos\theta}{-\sin\theta}$ $\Rightarrow \tan\theta=2t$ $x=1-\sin\theta$ $=1-\frac{2t}{\sqrt{1+4t^2}}$ $=g(t)$ (let) $\Rightarrow g'(t)<0$ $\Rightarrow g(t)$ is a decreasing function $t\in\left(\frac14,\frac13\right)$
Maths · Applications of Derivatives · Single correct
If the maximum value of $a$, for which the function $f_a(x) = \tan^{-1} 2x - 3ax + 7$ is non-decreasing in $$\left( -\frac{\pi}{6}, \frac{\pi}{6} \right)$$, is $\bar{a}$, then $f_a\left( \frac{\pi}{8} \right)$ is equal to
Answer: (d)
Given $f_a(x) = \tan^{-1} 2x - 3ax + 7$. The derivative is $$f'_a(x) = \frac{2}{1 + 4x^2} - 3a \geq 0$$ which implies $$a \leq \left( \frac{2}{3(1 + 4x^2)} \right)_{\min.}$$ at $x = \pm \frac{\pi}{6}$. Therefore, $$a_{\max} = \bar{a} = \frac{6}{9 + \pi^2}$$ Evaluating at $x = \frac{\pi}{8}$, $$f_a\left(\frac{\pi}{8}\right) = \tan^{-1} \frac{\pi}{4} - 3 \frac{6}{9 + \pi^2} \frac{\pi}{8} + 7 = \tan^{-1} \frac{\pi}{4} - \frac{9\pi}{4(\pi^2 + 9)} + 7$$
Maths · Limits and Derivatives · Single correct
Let $\beta = \lim_{x \to 0} \frac{\alpha x - \left(e^{3x} - 1\right)}{\alpha x \left(e^{3x} - 1\right)}$ for some $\alpha \in \mathbb{R}$. Then the value of $\alpha + \beta$ is :
Answer: (c)
Given $$\beta = \lim_{x \to 0} \frac{\alpha x - (e^{3x} - 1)}{\alpha x (e^{3x} - 1)}$$ $$\beta = \lim_{x \to 0} \frac{1 + \alpha x - \left[1 + 3x + \frac{9x^2}{2!} + \ldots \right]}{(\alpha x) \frac{(e^{3x} - 1)}{3x} 3x}$$ $$\beta = \lim_{x \to 0} \frac{(\alpha x - 3x) - \frac{9x^2}{2!} - \ldots}{3 \alpha x^2}$$ For existence of limit $$\alpha - 3 = 0$$ $$\alpha = 3$$ Limit $$\beta = \frac{-3}{2\alpha}$$ $$\beta = -\frac{1}{2}$$ Now, $$\alpha + \beta = \frac{5}{2}$$
Maths · Continuity and Differentiability · Single correct
The value of $\log_e 2 \frac{d}{dx}(\log_{\cos x} \csc x)$ at $x = \frac{\pi}{4}$ is
Answer: (d)
Given $\log_e 2 \frac{d}{dx} \left( \log_{\cos x} \csc x \right)$. Let, $y = \log_{\cos x} \csc x$. Then, $y = -\frac{\ln(\sin x)}{\ln(\cos x)}$. The derivative is $$\frac{dy}{dx} = -\frac{[\cot x \cdot \ln(\cos x) + \tan x \cdot \ln(\sin x)]}{(\ln(\cos x))^2}$$ Evaluating at $x = \frac{\pi}{4}$, $$\left. \frac{dy}{dx} \right|_{x = \frac{\pi}{4}} = \frac{4}{\ln 2}$$ Now, $$\Rightarrow \log_e 2 \cdot \frac{4}{\ln 2} = 4$$
Maths · Integrals · Single correct
$$\int_{0}^{20\pi} (|\sin x| + |\cos x|)^2 \, dx$$ is equal to :-
Answer: (d)
Given $$I = \int_0^{20\pi} (|\sin x| + |\cos x|)^2 \, dx$$ Using the Jack property, we have $$I = 40 \int_0^{\pi/2} (\sin x + \cos x)^2 \, dx$$ Simplifying further, $$I = 40 \int_0^{\pi/2} (1 + \sin 2x) \, dx$$ Finally, $$I = 20[\pi + 2]$$
Maths · Differential Equations · Single correct
Let the solution curve $y = f(x)$ of the differential equation $\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^4 + 2x}{\sqrt{1-x^2}}, x \in (-1,1)$ pass through the origin. Then $\int_{\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) \, dx$ is equal to
Answer: (b)
Given $\($ $\frac{dy}{dx}$ + $\frac{xy}{x^2 - 1}$ = $\frac{x^4 + 2x}{\sqrt{1-x^2}}$ $\)$. The integrating factor (I.F) is given by $$ I.F = e^{\int \frac{x}{x^2 - 1} \, dx} $$ which simplifies to $$ I.F = \sqrt{1-x^2} $$ Solution of the differential equation: $\($ y $\cdot$ $\sqrt{1-x^2}$ = $\int$ $\frac{x^4 + 2x}{\sqrt{1-x^2}}$ $\,$ dx $\)$ This simplifies to $$ y \cdot \sqrt{1-x^2} = \int (x^4 + 2x) \, dx $$ Integrating, we get $$ y \cdot \sqrt{1-x^2} = \frac{x^5}{5} + x^2 + C $$ At $\($ x = 0, y = 0 $\)$, we find $\($ C = 0 $\)$. Thus, $$ y = \frac{x^5}{5\sqrt{1-x^2}} + \frac{x^2}{\sqrt{1-x^2}} $$ Now, $$ \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) \, dx = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^5}{5\sqrt{1-x^2}} \, dx + \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} \, dx $$ This simplifies to $$ \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) \, dx = 0 + 2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} \, dx $$ Finally, $$ \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) \, dx = \frac{\pi}{2} \cdot \frac{\sqrt{3}}{4} $$
Maths · Conic Sections · Single correct
The acute angle between the pair of tangents drawn to the ellipse $2x^2 + 3y^2 = 5$ from the point $(1,3)$ is
Answer: (b)
Equation of tangent to the ellipse $2x^2 + 3y^2 = 5$ is $$y = mx \pm \sqrt{\frac{5}{2}m^2 + \frac{5}{3}}$$ It passes through $(1, 3)$ $$3 = m \pm \sqrt{\frac{5}{2}m^2 + \frac{5}{3}}$$ $$3m^2 + 12m - \frac{44}{3} = 0$$ Let $\theta$ be the angle between the tangents $$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$ $$\tan \theta = \left| \frac{3 \sqrt{320}}{-35} \right|$$ $$\theta = \tan^{-1} \left( \frac{24}{7 \sqrt{5}} \right)$$
Maths · Conic Sections · Single correct
The equation of a common tangent to the parabolas $y = x^2$ and $y = -(x-2)^2$ is
Answer: (b)
Equation of tangent of $y = x^2$ be $$tx = y + at^2 \ldots (1)$$ $$y = tx - \frac{t^2}{4}$$ Solve with $y = -(x-2)^2$ $$tx - \frac{t^2}{4} = -(x-2)^2$$ $$x^2 + x(t-4) - \frac{t^2}{4} + 4 = 0$$ $D = 0$ $$(t-4)^2 - 4 \cdot \left(4 - \frac{t^2}{4}\right) = 0$$ $$t^2 - 4t = 0$$ $t = 0$ or $t = 4$ From eq. (1), required common tangent is $$y = 4(x-1)$$
Maths · Complex Numbers and Quadratic Equations · Single correct
Let the abscissae of the two points P and Q on a circle be the roots of $x^2 - 4x - 6 = 0$ and the ordinates of P and Q be the roots of $y^2 + 2y - 7 = 0$. If PQ is a diameter of the circle $x^2 + y^2 + 2ax + 2by + c = 0$, then the value of $(a+b-c)$ is
Answer: (a)
Equation of circle diameter form $$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$ (where $x_1, x_2$ are the roots of $x^2 - 4x - 6 = 0$ and $y_1, y_2$ are the roots of $y^2 + 2y - 7 = 0$) $$x^2 + y^2 - 4x + 2y - 13 = 0$$ Now, Compare it with the given equation, we get $a = -2, b = 1, c = -13$ Now $a + b - c = 12$
Maths · Conic Sections · Single correct
If the line $x - 1 = 0$, is a directrix of the hyperbola $kx^2 - y^2 = 6$, then the hyperbola passes through the point
Answer: (c)
Given $\($ $\frac{x^2}{6/k}$ - $\frac{y^2}{6}$ = 1 $\)$ $\($ $\ldots$ (1) $\)$ $\($ e^2 = 1 + $\frac{6}{6/k}$ $\)$ $\($ e = $\sqrt{1 + k}$ $\)$ $\($ a = $\sqrt{\frac{6}{k}}$ $\)$ Equation of directrix $\($ x = $\frac{a}{e}$ $\Rightarrow$ x = $\sqrt{\frac{6}{k(k+1)}}$ $\)$ $\($ $\frac{6}{k(k+1)}$ = 1 $\)$ $\($ k = 2 $\)$ From equation (1), we get $\($ 2x^2 - y^2 = 6 $\)$ Check options
Maths · Three Dimensional Geometry · Single correct
A vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}, \hat{i} + \hat{j}$ and the plane determined by the vectors $\hat{i} - \hat{j}, \hat{i} + \hat{k}$. The obtuse angle between $\vec{a}$ and the vector $\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}$ is
Answer: (a)
Given $\mathbf{n}_1 = \mathbf{i} \times (\mathbf{i} + \mathbf{j}) = \mathbf{k}$. $\mathbf{n}_2 = (\mathbf{i} + \mathbf{k}) \times (\mathbf{i} - \mathbf{j}) = \mathbf{i} + \mathbf{j} - \mathbf{k}$. Line of intersection along $\mathbf{n}_1 \times \mathbf{n}_2 = \mathbf{k} \times (\mathbf{i} + \mathbf{j} - \mathbf{k}) = -\mathbf{i} + \mathbf{j}$. D.R of $\mathbf{a} = -\mathbf{i} + \mathbf{j}$. D.R of $\mathbf{b} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$. $\mathbf{a} \cdot \mathbf{b} = -3$ and $\left( \mathbf{a} \wedge \mathbf{b} \right) = 0$. $\cos \theta = \frac{-3}{\sqrt{2 \times 3}}$. $\theta = \frac{3\pi}{4}$.
Maths · Inverse Trigonometric Functions · Single correct
If $0 < x < \frac{1}{\sqrt{2}}$ and $\frac{\sin^{-1} x}{\alpha} = \frac{\cos^{-1} x}{\beta}$, then a value of $\sin \left( \frac{2 \pi \alpha}{\alpha + \beta} \right)$ is
Answer: (b)
Given $\($ $\frac{\sin^{-1} x}{\alpha}$ = $\frac{\cos^{-1} x}{\beta}$ = k $\)$. $\($ $\sin$^{-1} x = k $\alpha$ $\)$ $\($ $\cos$^{-1} x = k $\beta$ $\)$ $\($ k = $\frac{\pi}{2(\alpha + \beta)}$ $\)$ ....(i) $\($ $\sin$ $\left$( $\frac{2 \pi \alpha}{\alpha + \beta}$ $\right$) = $\sin$ (4 $\sin$^{-1} x) $\)$ $\($ = 2 $\sin$(2 $\sin$^{-1} x) $\cos$(2 $\sin$^{-1} x) $\)$ $\($ = 4x $\sqrt{1-x^2}$ (1-2x^2) $\)$
Maths · Mathematical Reasoning · Single correct
Negation of the Boolean expression $p \iff (q \Rightarrow p)$ is
Given $\sim(p\leftrightarrow(q\rightarrow p))$ $\sim(p\leftrightarrow q) =(p\land\sim q)\lor(q\land\sim p)$ $\sim(p\leftrightarrow(q\rightarrow p))$ $=(p\land\sim(q\rightarrow p)) \lor((q\rightarrow p)\land\sim p)$ $(p\land\sim(q\rightarrow p)) =p\land(q\land\sim p)$ $=(p\land\sim p)\land q=c$ $(q\rightarrow p)\land\sim p$ $=(\sim q\lor p)\land\sim p$ $=\sim p\land(\sim q\lor p)$ $=(\sim p\land\sim q)$ $\lor(\sim p\land p)$ $=\sim p\land\sim q$ $\sim(p\leftrightarrow(q\rightarrow p))$ $=c\lor(\sim p\land\sim q)$ $=\sim p\land\sim q$
Maths · Probability · Single correct
Let X be a binomially distributed random variable with mean 4 and variance $\frac{4}{3}$. Then $54 \mathrm{P}(X \leq 2)$ is equal to
Answer: (b)
Given $np = 4$ and $npq = 4/3$. Let $n = 6$, $p = 2/3$, $q = 1/3$. Calculate $54(\mathrm{P}(X = 2) + \mathrm{P}(X = 1) + \mathrm{P}(X = 0))$. $$54 \left( \binom{6}{2} \left( \frac{2}{3} \right)^2 \left( \frac{1}{3} \right)^4 + \binom{6}{1} \left( \frac{2}{3} \right)^1 \left( \frac{1}{3} \right)^5 + \binom{6}{0} \left( \frac{2}{3} \right)^0 \left( \frac{1}{3} \right)^6 \right)$$ $$= \frac{146}{27}$$
Maths · Integrals · Single correct
The integral $$\int \frac{\left(1 - \frac{1}{\sqrt{3}}\right)(\cos x - \sin x)}{1 + \frac{2}{\sqrt{3}} \sin 2x} \, dx$$ is equal to
$I=\int \frac{\left(1-\frac1{\sqrt3}\right)(\cos x-\sin x)} {\left(1+\frac1{\sqrt3}\sin2x\right)}\,dx$ $=\frac{\sqrt3-1}{2} \int \frac{\cos x-\sin x} {\frac{\sqrt3}{2}+\sin2x}\,dx$ $=\left(\frac{\sqrt3}{2}-\frac12\right) \int \frac{\cos x-\sin x} {\sin60^\circ+\sin2x}\,dx$ $=\int \frac{\cos\left(x-\frac{\pi}{6}\right)-\sin\left(x+\frac{\pi}{6}\right)} {2\sin\left(x+\frac{\pi}{6}\right)\cos\left(x-\frac{\pi}{6}\right)} \,dx$ $=\frac12\int \frac{\cos\left(x-\frac{\pi}{6}\right)} {\sin\left(x+\frac{\pi}{6}\right)} \,dx -\frac12\int \frac{\sin\left(x+\frac{\pi}{6}\right)} {\cos\left(x-\frac{\pi}{6}\right)} \,dx$ $=\frac12\int \cot\left(x+\frac{\pi}{6}\right)\,dx -\frac12\int \tan\left(x-\frac{\pi}{6}\right)\,dx$ $=\frac12\ln\left| \tan\left(\frac{x}{2}+\frac{\pi}{12}\right) \right| +\frac12\ln\left| \cos\left(x-\frac{\pi}{6}\right) \right|+C$
Maths · Applications of Integrals · Single correct
The area bounded by the curves $y = |x^2 - 1|$ and $y = 1$ is
Answer: (d)
Given $y = |x^2 - 1|$. The area is $Area = ABCDEA$. $$= 2 \left( \int_{0}^{1} \left( 1 - (1 - x^2) \right) \, dx + \int_{1}^{\sqrt{2}} \left( 1 - (x^2 - 1) \right) \, dx \right)$$ $$= \frac{8}{3}(\sqrt{2} - 1)$$
Maths · Sets · Numerical
Let $A = \{1,2,3,4,5,6,7\}$ and $B = \{3,6,7,9\}$. Then the number of elements in the set $\{C \subseteq A : C \cap B \neq \emptyset\}$ is
Answer: 112
Given sets $A = \{1, 2, 3, 4, 5, 6, 7\}$ and $B = \{3, 6, 7, 9\}$. The total number of subsets of $A$ is $2^7 = 128$. The intersection $C \cap B = \emptyset$ when set $C$ contains the elements $1, 2, 4, 5$. Therefore, $S = \{C \subseteq A; C \cap B \neq \emptyset\}$. The total is calculated as Total $- (C \cap B = \emptyset) = 128 - 2^4 = 112$.
Maths · Three Dimensional Geometry · Numerical
The largest value of $a$, for which the perpendicular distance of the plane containing the lines $\vec{r} = (\hat{i} + \hat{j}) + \lambda (\hat{i} + a\hat{j} - \hat{k})$ and $\vec{r} = (\hat{i} + \hat{j}) + \mu (-\hat{i} + \hat{j} - a\hat{k})$ from the point $(2,1,4)$ is $\sqrt{3}$, is ________.
Answer: 2
Given $\vec{r} = (\hat{i} + \hat{j}) + \lambda (\hat{i} + a\hat{j} - \hat{k})$ and $\vec{r} = (\hat{i} + \hat{j}) + \mu (-\hat{i} + \hat{j} - a\hat{k})$. The direction ratios of the plane containing these lines is given by the determinant: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & a & -1 \\ -1 & 1 & -a \end{vmatrix} = \hat{i}(1 - a^2) - \hat{j}(-a - 1) + \hat{k}(1 + a)$$ Thus, $\vec{n} = (1-a)\hat{i} + \hat{j} + \hat{k}$. One point in the plane is $(1, 1, 0)$. Therefore, the equation of the plane is $$(1-a)(x-1) + (y-1) + (z-0) = 0$$ Simplifying, $$(1-a)x + y + z + a - 2 = 0$$ The distance $D$ is given by $$D = \frac{|(1-a)2 + 1 + 4 + a - 2|}{\sqrt{(1-a)^2 + 1 + 1}}$$ This simplifies to $$|5-a| = \sqrt{3} \cdot \sqrt{a^2 - 2a + 3}$$ Solving $a^2 + 2a - 8 = 0$, we find $$a = 2, -4$$ The largest value of $a$ is $2$.
Maths · Permutations and Combinations · Numerical
Numbers are to be formed between 1000 and 3000, which are divisible by 4, using the digits 1,2,3,4,5 and 6 without repetition of digits. Then the total number of such numbers is .
Answer: 30
Here 1st digit is 1 or 2 only. Case-I If first digit is 1 Then last two digits can be 24, 32, 36, 52, 56, 64 $$1 \times 3 \times 6 = 18 ways$$ Case-II If first digit is 2 then last two digit can be 16, 36, 56, 64 $$1 \times 3 \times 4 = 12 ways$$ Total ways = 12 + 18 = 30 ways
Maths · Sequences and Series · Numerical
If $$\sum_{k=1}^{10} \frac{k}{k^4 + k^2 + 1} = \frac{m}{n}$$, where m and n are co-prime, then m + n is equal to
Answer: 166
Given $\displaystyle\sum_{k=1}^{10} \frac{k}{k^4 + k^2 + 1} = \frac{m}{n}$. $$\Rightarrow \frac{1}{2} \sum_{k=1}^{10} \frac{(k^2+k+1)-(k^2-k+1)}{(k^2+k+1)(k^2-k+1)}$$ $$\Rightarrow \frac{1}{2}\left(\sum_{k=1}^{10}\left(\frac{1}{k^2-k+1} - \frac{1}{k^2+k+1}\right)\right)$$ $$\Rightarrow \frac{55}{111} = \frac{m}{n}$$ $m + n = 166$
Maths · Trigonometric Functions · Numerical
If the sum of solutions of the system of equations $2 \sin^2 \theta - \cos 2\theta = 0$ and $2 \cos^2 \theta + 3 \sin \theta = 0$ in the interval $[0, 2\pi]$ is $k\pi$, then $k$ is equal to ______.
Answer: 3
Given the equation $2 \sin^2 \theta - \cos 2\theta = 0$. Rewriting, we have $2 \sin^2 \theta - (1 - 2 \sin^2 \theta) = 0$. This implies $$\sin^2 \theta = \left(\frac{1}{2}\right)^2$$ Therefore, $$\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$$ Now consider $2 \cos^2 \theta + 3 \sin \theta = 0$. This leads to $$2 \sin^2 \theta - 3 \sin \theta - 2 = 0$$ Thus, $$\sin \theta = -\frac{1}{2}$$ Therefore, $$\theta = \frac{7\pi}{6}, \frac{11\pi}{6}$$ So, the common solution is $$\theta = \frac{7\pi}{6}, \frac{11\pi}{6}$$ Hence, $K = 3$
Maths · Statistics · Fill in the blank
The mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of these observations 12 and 10 were wrongly recorded. If $\sigma$ is the standard deviation of the data after omitting the two wrong observations from the data, then $38\sigma^2$ is equal to ________.
Answer: 238
Wrong mean = $\mu_1 = 30$ Wrong S.D = $\sigma_1 = 5$ $$\frac{\sum x_i}{40} = 30$$ $$\Rightarrow \sum x_i = 1200$$ $$\sigma_1^2 = 25$$ $$\Rightarrow \frac{\sum x_i^2}{40} - 30^2 = 25$$ $$\Rightarrow \sum x_i^2 = 925 \times 40 = 37000$$ New sum = $\sum x_i' = 1200 - 10 - 12 = 1178$ New mean = $\mu_1' = \frac{1178}{38} = 31$ New $\sum x_i'^2 = 37000 - (10)^2 - (12)^2 = 36756$ New S.D, $\sigma_1' = \sqrt{\frac{36756}{38} - (31)^2} = \sigma$ $$36756 - (31)^2 \times 38 = 38\sigma^2$$ $$\Rightarrow 38\sigma^2 = 238$$
Maths · Three Dimensional Geometry · Numerical
The plane passing through the line L: $\ell \ x-y+3(1-\ell) \ z = 1$, $x+2y - z = 2$ and perpendicular to the plane $3x+2y+z = 6$ is $3x-8y+7z=4$. If $\theta$ is the acute angle between the line L and the y-axis, then $415 \cos^2 \theta$ is equal to.
Answer: 125
Given $\mathbf{n}_1 = \ell \hat{\imath} - \hat{\jmath} + 3(1-\ell) \hat{k}$ and $\mathbf{n}_2 = \hat{\imath} + 2 \hat{\jmath} - \hat{k}$. The direction ratio of the line is given by the determinant: $$\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \ell & -1 & 3(1-\ell) \\ 1 & 2 & -1 \end{vmatrix}$$ This simplifies to: $$(6\ell - 5) \hat{\imath} + (3 - 2\ell) \hat{\jmath} + (2\ell + 1) \hat{k}$$ The line $3x - 8y + 7z = 4$ will contain the line $(6\ell - 5) \hat{\imath} + (3 - 2\ell) \hat{\jmath} + (2\ell + 1) \hat{k}$. The normal of $3x - 8y + 7z = 4$ will be perpendicular to the line: $$3(6\ell - 5) + (3 - 2\ell)(-8) + 7(2\ell + 1) = 0$$ Solving gives $\ell = \frac{2}{3}$. Therefore, the direction ratio of the line is: $$\left(-1, \frac{5}{3}, \frac{7}{3}\right)$$ The angle with the $y$-axis is given by: $$\cos \theta = \frac{5/3}{\sqrt{1 + \frac{25}{9} + \frac{49}{9}}}$$ This simplifies to: $$\cos \theta = \frac{5}{\sqrt{83}}$$ Thus, $415 \cos^2 \theta = \frac{25}{83} \times 415 = 125$
Maths · Differential Equations · Fill in the blank
Suppose $y = y(x)$ be the solution curve to the differential equation $\frac{dy}{dx} - y = 2 - e^{-x}$ such that $\lim_{x \to \infty} y(x)$ is finite. If $a$ and $b$ are respectively the x- and y-intercepts of the tangent to the curve at $x=0$, then the value of $a - 4b$ is equal to ________.
Answer: 3
Given $\($ $\frac{dy}{dx}$ = 2 - e^{-x} $\)$. I.F. = $\($ e^{$\int$ -dx} = e^{-x} $\)$ Therefore, the solution of D.E. $\($ y $\cdot$ e^{-x} = $\int$ (2e^{-x} - e^{-2x}) $\,$ dx $\)$ $\($ $\Rightarrow$ y = -2 + $\frac{e^{-x}}{2}$ + C $\cdot$ e^{x} $\)$ Therefore, $\($ $\lim$_{x $\to$ $\infty$} y $\)$ is finite. $\($ $\therefore$ $\lim$_{x $\to$ $\infty$} $\left$( -2 + $\frac{e^{-x}}{2}$ + C $\cdot$ e^{x} $\right$) $\to$ finite $\)$ This is possible only when $\($ C = 0 $\)$. Therefore, $\($ y = y(x) = -2 + $\frac{e^{-x}}{2}$ $\)$ $\($ $\frac{dy}{dx}$ = -$\frac{1}{2}$ e^{-x} $\)$ $\($ $\left$. $\frac{dy}{dx}$ $\right$|_{x=0} = -$\frac{1}{2}$ = m $\)$, $\($ y(0) = -2 + $\frac{1}{2}$ = -$\frac{3}{2}$ $\)$ Therefore, equation of tangent $\($ a = -3, $\ $b = -$\frac{3}{2}$ $\)$ $\($ a - 4b = -3 + 6 = 3 $\)$
Maths · Sequences and Series · Numerical
Different A.P.'s are constructed with the first term 100, the last term 199, And integral common differences. The sum of the common differences of all such, A.P's having at least 3 terms and at most 33 terms is.
Answer: 53
First term $= 100 = a$. Last term $= 199 = \ell$. If 3 terms are $a$, $a + d$, $a + 2d$. $a_n = \ell = a + (n - 1)d$. $n \rightarrow$ number of terms. For $n = 3$, $d_1 = \frac{199 - 100}{2} = \frac{99}{2} \notin \mathbb{I}$. For $n = 4$, $d_2 = \frac{99}{3} = 33 \in \mathbb{I}$. For $n = 10$, $d_3 = \frac{99}{9} = 11 \in \mathbb{I}$. For $n = 12$, $d_4 = \frac{99}{11} = 9 \in \mathbb{I}$. Therefore, $\sum d_i = 33 + 11 + 9 = 53$.
Maths · Matrices · Numerical
The number of matrices $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, where $a, b, c, d \in \{-1, 0, 1, 2, 3, \ldots, 10\}$, such that $A = A^{-1}$, is _______.
Answer: 50
Given $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Given $A = A^{-1}$. Therefore, $A^2 = A \cdot A^{-1} = I$. $$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ Thus, $a^2 + bc = 1 \ldots (1)$ $ab + bd = 0 \ldots (2)$ $ac + cd = 0 \ldots (3)$ $bc + d^2 = 1 \ldots (4)$ (1) - (4) gives $$a^2 - d^2 = 0$$ Therefore, $(a + d) = 0$ or $a - d = 0$. Case - I $a + d = 0 \Rightarrow (a, d) = (-1, 1), (0, 0), (1, -1)$ (a) $(a, d) = (-1, 1)$ Therefore, from equation (1) $1 + bc = 1 \Rightarrow bc = 0$ $b = 0 C = 12$ possibilities $c = 0 b = 12$ possibilities but $(0, 0)$ is repeated Therefore, $2 \times 12 = 24$ $24 - 1$ (repeated) $= 23$ pairs (b) $(a, d) = (1, -1) \Rightarrow bc = 0 \rightarrow 23$ pairs (c) $(a, d) = (0, 0) \Rightarrow bc = 1$ Therefore, $(b, c) = (1, 1)$ and $(-1, -1)$, 2 pairs Case - II $a = d$ from (2) and (3) $$ab + bd = 0 ac + cd = 0$$ $(a, d) = (1, 1), (-1, -1) \rightarrow 2$ pairs Therefore, Total $= 23 + 23 + 2 + 2 = 50$ pairs
Physics · Motion in a Straight Line · Single correct
Two projectiles are thrown with same initial velocity making an angle of 45^$\circ$ and 30^$\circ$ with the horizontal respectively. The ratio of their respective ranges will be
Answer: (c)
Let projection speed is $u$. $$R_1 = \frac{u^2 \sin(90^\circ)}{g}; R_2 = \frac{u^2 \sin(60^\circ)}{g}$$ $$\frac{R_1}{R_2} = \frac{2}{\sqrt{3}}$$
Physics · Experimental Physics · Single correct
In a Vernier Calipers. 10 divisions of Vernier scale is equal to the 9 divisions of main scale. When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and $4^{th}$ Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to 1 mm. While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and $6^{th}$ Vernier scale division exactly coincides with the main scale reading. The diameter of the spherical body will be:
Answer: (c)
1 M.S.D = 1 $\mathrm{mm}$ 9 M.S.D = 10 V.S.D 1 V.S.D = 0.9 M.S.D = 0.9 $\mathrm{mm}$ L.C of vernier caliper = 1 - 0.9 = 0.1 $\mathrm{mm}$ = 0.01 $\mathrm{cm}$ zero error = (10 - 4) $\times$ 0.1 $\mathrm{mm}$ = -0.6 $\mathrm{mm}$ Reading = M.S.R + V.S.R - Zero error = 3 $\mathrm{cm}$ + 6 $\times$ 0.01 - [-0.06] = 3 + 0.06 + 0.06 = 3.12 $\mathrm{cm}$ Nearest given answer in the options is 3.10
Physics · Laws of Motion · Single correct
A ball of mass 0.15 kg hits the wall with its initial speed of 12 $\mathrm{ms}^{-1}$ and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is 100 $\mathrm{N}$, calculate the time duration of the contact of ball with the wall.
Answer: (b)
The initial momentum $\vec{P}_i = 0.15 \times 12 \left( \hat{i} \right)$. The final momentum $\vec{P}_f = 0.15 \times 12 \left( -\hat{i} \right)$. The change in momentum $|\Delta \vec{P}| = 3.6 \, \mathrm{kg \cdot m/s}$. We have $3.6 = F \Delta t$. Solving for $\Delta t$, we get $3.6 = 100 \, \Delta t$. Therefore, $\Delta t = 0.036 \, \mathrm{sec}$.
Physics · Work, Energy and Power · Single correct
A body of mass 8 kg and another of mass 2 kg are moving with equal kinetic energy. The ratio of their respective momenta will be :
Answer: (b)
The kinetic energy is given by $$\mathrm{K.E} = \frac{P^2}{2m}$$. For the first case, $$K_1 = \frac{P_1^2}{2(8)}$$ and for the second case, $$K_2 = \frac{P_2^2}{2(2)}$$. Since $$K_1 = K_2$$, we have $$4P_2^2 = P_1^2$$. Therefore, $$\frac{P_1}{P_2} = 2$$.
Physics · Electric Charges and Fields · Single correct
Two uniformly charged spherical conductors A and B of radii 5 mm and 10 mm are separated by a distance of 2 cm. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitudes of the electric fields at the surface of the sphere A and B will be:
Answer: (b)
Given $V_A = V_B$. $$\frac{KQ_A}{R_A} = \frac{KQ_B}{R_B}$$ $$\frac{Q_A}{Q_B} = \frac{R_A}{R_B} = \frac{1}{2}$$ $$E_A = \frac{KQ_A}{R_A^2}; E_B = \frac{KQ_B}{R_B^2}$$ $$\frac{E_A}{E_B} = \frac{Q_A}{Q_B} \times \frac{R_B^2}{R_A^2} = \frac{R_B}{R_A} = \frac{2}{1}$$
Physics · Electromagnetic Waves · Single correct
The oscillating magnetic field in a plane electromagnetic wave is given by $B_y = 5 \times 10^{-6} \sin 1000 \pi \left(5x - 4 \times 10^8 \, t\right)\, \mathrm{T}$. The amplitude of electric field will be:
Answer: (d)
Given $B_0 = 5 \times 10^{-6}$. $v = Speed of wave = \frac{4 \times 10^8}{5} = 8 \times 10^7 \left[ \therefore v = \frac{\mathbf{w}}{\mathbf{k}} \right]$ $E_0 = v B_0 = 40 \times 10^1$ $= 4 \times 10^2 \, \mathrm{V/m}$
Physics · Ray Optics and Optical Instruments · Single correct
Light travels in two media $M_1$ and $M_2$ with speeds $1.5 \times 10^8 \, \mathrm{ms}^{-1}$ and $2.0 \times 10^8 \, \mathrm{ms}^{-1}$ respectively. The critical angle between them is:
Answer: (a)
Given $v = \frac{c}{n}$. $n_d \sin i_c = n_r \sin 90^\circ$. Therefore, $\sin i_c = \frac{n_r}{n_d} = \frac{v_d}{v_r}$. Calculating $\sin i_c$: $$\sin i_c = \frac{1.5 \times 10^8}{2 \times 10^8} = \frac{1.5}{2}$$ Thus, $\sin i_c = \frac{3}{4}$. Now, $\tan i_c = \frac{3}{\sqrt{4^2 - 3^2}} \implies \frac{3}{\sqrt{7}}$. Finally, $i_c = \tan^{-1}\left(\frac{3}{\sqrt{7}}\right)$.
Physics · Gravitation · Single correct
A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be: (Take radius of earth $= 6400 \, \mathrm{km}$ and $g = 10 \, \mathrm{ms^{-2}}$)
Answer: (a)
The escape velocity is given by $v_e = \sqrt{\frac{2Gm}{R}}$. The equation for energy conservation is: $$\frac{-GMm}{R} + \frac{1}{2}m \frac{v_e^2}{9} = \frac{-GMm}{R+h}$$ Simplifying, we have: $$\frac{GM}{R+h} = \frac{GM}{R} - \frac{v_e^2}{18}$$ Substituting $v_e^2 = \frac{2GM}{R}$, we get: $$\frac{GM}{R+h} = \frac{GM}{R} - \frac{GM}{9R}$$ This simplifies to: $$\frac{GM}{R+h} = \frac{8GM}{9R}$$ Therefore: $$\frac{1}{R+h} = \frac{8}{9R}$$ Solving for $h$, we have: $$9R = 8R + 8h$$ Thus: $$h = \frac{R}{8} \implies \frac{6400}{8} \implies 800 \, \mathrm{km}$$
Physics · Communication Systems · Single correct
The maximum and minimum voltage of an amplitude modulated signal are $60 \, \mathrm{V}$ and $20 \, \mathrm{V}$ respectively. The percentage modulation index will be:
Answer: (b)
Given $V_{max} = 60$ and $V_{min} = 20$. The percentage modulation is calculated as follows: $$\left( \frac{V_{max} - V_{min}}{V_{max} + V_{min}} \right) 100 \Rightarrow \left( \frac{60 - 20}{60 + 20} \right) 100 \Rightarrow \left( \frac{40}{80} \right) 100$$ This simplifies to $50\%$.
Physics · Atoms · Single correct
A nucleus of mass $M$ at rest splits into two parts having masses $\frac{M'}{3}$ and $\frac{2M'}{3}$ $(M' < M)$. The ratio of de Broglie wavelength of two parts will be:
Answer: (c)
Here $\vec{P}$ is momentum. So $\lambda = \frac{h}{P}$. Hence both will have same de Broglie wavelength.
Physics · Thermal Properties of Matter · Single correct
An ice cube of dimensions $60\,\mathrm{cm} \times 50\,\mathrm{cm} \times 20\,\mathrm{cm}$ is placed in an insulation box of wall thickness $1\,\mathrm{cm}$. The box keeping the ice cube at $0^\circ\mathrm{C}$ of temperature is brought to a room of temperature $40^\circ\mathrm{C}$. The rate of melting of ice is approximately: (Latent heat of fusion of ice is $3.4 \times 10^5\,\mathrm{J\,kg^{-1}}$ and thermal conductivity of insulation wall is $0.05\,\mathrm{W\,m^{-1}\,^\circ C^{-1}}$.)
Answer: (b)
Given $\frac{dQ}{dt} = \frac{KA \Delta T}{\ell}$. $A = 2\left(0.6 \times 0.5 + 0.5 \times 0.2 + 0.2 \times 0.6\right)$ $= 2(0.3 + 0.1 + 0.12)$ $= 2(0.4 + 0.12)$ $= 2(0.52)$ $= 1.04\,\mathrm{m^2}$ $R_{th} = \frac{\ell}{KA} = \frac{1 \times 10^{-2}}{0.05 \times 1.04} = \frac{10^{-2}}{0.052}$ $\frac{dQ}{dt} = \frac{\Delta T}{R_{th}} = \frac{40 \times 0.052}{10^{-2}} = 2.08 \times 10^2\,\mathrm{J/s}$ $2.08 \times 10^2 = m \times 3.4 \times 10^5$ $m = \frac{2.08}{3.4 \times 10^3} = 0.61 \times 10^{-3}\,\mathrm{kg/s}$ $= 61 \times 10^{-5}\,\mathrm{kg/s}$
Physics · Kinetic Theory · Single correct
A gas has $n$ degrees of freedom. The ratio of specific heat of gas at constant volume to the specific heat of gas at constant pressure will be:
Answer: (a)
Given the expressions for specific heats: $$C_v = \frac{nR}{2}$$ $$C_p = \frac{(n+2)R}{2}$$ The ratio of specific heats is given by: $$\frac{C_v}{C_p} = \frac{n}{n+2}$$
Physics · Waves · Single correct
A transverse wave is represented by $y = 2 \sin (\omega t - kx)$ cm. The value of wavelength (in cm) for which the wave velocity becomes equal to the maximum particle velocity, will be ;
Answer: (a)
Given $y = 2 \sin(\omega t - kx)$. Maximum particle velocity $= A \omega$. Wave velocity $= \frac{\omega}{k}$. $$\frac{\omega}{k} = A \omega$$ $$k = \frac{1}{A} = \frac{2\pi}{\lambda}$$ $$\lambda = 2\pi A$$ $$= 4 \pi \, \mathrm{cm}$$
Physics · Current Electricity · Single correct
A battery of 6 V is connected to the circuit as shown below. The current I drawn from the battery is:
Answer: (a)
Balanced wheatstone bridge in circuit so there is no current in $5 \, \Omega$ resistor so it can be removed from the circuit. $$R_{eq} = \frac{6 \times 12}{6 + 12} + 2$$ $$= \frac{6 \times 12}{18} + 2$$ $$R_{eq} = 6 \, \Omega$$ $$I = \frac{V}{R_{eq}} = \frac{6}{6} = 1 \, Amp.$$
Physics · Electrostatic Potential and Capacitance · Single correct
A source of potential difference $V$ is connected to the combination of two identical capacitors as shown in the figure. When key 'K' is closed, the total energy stored across the combination is $E_1$. Now key ‘K’ is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now $E_2$. The ratio $E_1/E_2$ will be:
Answer: (c)
(1) Switch is closed $C_{eq} = 2C$ Energy $E_1 = \frac{1}{2} C_{eq} V^2$ $$= \frac{1}{2} 2C \times V^2$$ $E_1 = CV^2$ (ii) When switch is opened charge on right capacitor remain $CV$ while potential on left capacitor remain same Dielectric $K = 5$ $C' = KC$ $C' = 5C$ $$E_2 = \frac{1}{2} (5C) V^2 + \frac{(CV)^2}{2(5C)}$$ $$E_2 = \frac{5CV^2}{2} + \frac{CV^2}{10}$$ $$E_2 = \frac{13CV^2}{5}$$ $$\frac{E_1}{E_2} = \frac{CV^2}{\frac{13CV^2}{5}} = \frac{5}{13}$$ $$\frac{E_1}{E_2} = \frac{5}{13}$$
Physics · Moving Charges and Magnetism · Single correct
Two concentric circular loops of radii $r_1 = 30 \, \mathrm{cm}$ and $r_2 = 50 \, \mathrm{cm}$ are placed in X-Y plane as shown in the figure. A current $I = 7 \, \mathrm{A}$ is flowing through them in the direction as shown in figure. The net magnetic moment of this system of two circular loops is approximately:
Answer: (b)
Magnetic moment $$\vec{M} = -i \pi (0.5)^2 \hat{k} + i \pi (0.3)^2 \hat{k}$$ $$\vec{M} = -7 \times \frac{22}{7} \left( \frac{25}{100} - \frac{9}{100} \right) \hat{k}$$ $$= -22 \left( \frac{16}{100} \right) \hat{k}$$ $$\vec{M} = -3.52 \hat{k} \, \mathrm{Am^2}$$ $$= -\frac{7}{2} \hat{k} \, \mathrm{Am^2}$$
Physics · Electromagnetic Induction · Single correct
A velocity selector consists of electric field $\vec{E} = E\hat{k}$ and magnetic field $\vec{B} = B\hat{j}$ with $B = 12 \, \mathrm{mT}$. The value $E$ required for an electron of energy $728 \, \mathrm{eV}$ moving along the positive x-axis to pass undeflected is: (Given, mass of electron $= 9.1 \times 10^{-31} \, \mathrm{kg}$)
Answer: (a)
Given $\vec{E} = E \hat{k}$ and $B = 12 \, \mathrm{mT}$. $\vec{B} = B \hat{j}$ and Energy $= 728 \, \mathrm{eV}$. Energy $= \frac{1}{2} mv^2$. $728 \, \mathrm{eV} = \frac{1}{2} \times 9.1 \times 10^{-31} \times v^2$. $728 \times 1.6 \times 10^{-19} = \frac{1}{2} \times 9.1 \times 10^{-31} \times v^2$. $v = 16 \times 10^6 \, \mathrm{m/s}$. $E = vB$. $E = 16 \times 10^6 \times 12 \times 10^{-3}$. $E = 192 \times 10^3 \, \mathrm{V/m}$.
Physics · Laws of Motion · Single correct
Two masses $M_1$ and $M_2$ are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass $M_2$ is twice that of $M_1$, the acceleration of the system is $a_1$. When the mass $M_2$ is thrice that of $M_1$, the acceleration of the system is $a_2$. The ratio $\frac{a_1}{a_2}$ will be:
Answer: (b)
Given $a = \frac{m_2 g - m_1 g}{m_1 + m_2}$. Case 1: $M_2 = 2m_1$ $$a_1 = \frac{2m_1 g - m_1 g}{3m_1}$$ $$a_1 = \frac{g}{3}$$ Case 2: $M_2 = 3m_1$ $$a_2 = \frac{3m_1 g - m_1 g}{4m_1}$$ $$a_2 = \frac{g}{2}$$ $$\frac{a_1}{a_2} = \frac{\frac{g}{3}}{\frac{g}{2}} = \frac{2}{3}$$
Physics · Atoms · Single correct
Mass numbers of two nuclei are in the ratio of 4:3. Their nuclear densities will be in the ratio of
Answer: (c)
Radius of nucleus $R = R_0 A^{\frac{1}{3}}$. Density of nucleus $= \frac{Mass of nucleus}{volume of nucleus}$. $$\rho = \frac{m \times A}{\frac{4}{3} \pi R^3}$$ Where $m$: mass of proton or neutron. $$\rho = \frac{m \times A}{\frac{4}{3} \pi R_0^3 A}$$ $$\rho \propto A^0$$ Hence density of nucleus is independent of mass number.
Physics · Mechanical Properties of Solids · Single correct
The area of cross section of the rope used to lift a load by a crane is $2.5 \times 10^{-4} \, \mathrm{m}^2$. The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross section of the rope should be: (take $g = 10 \, \mathrm{ms}^{-2}$)
Answer: (a)
Since breaking stress (Maximum lifting capacity) is the property of material so it will remain same. Breaking stress is given by the formula: breaking stress = \[ \frac{\text{Maximum lifting capacity}}{\text{Area of cross section of rope}} \] Substituting the given values: $$\frac{10}{2.5 \times 10^{-4}} = \frac{25}{A}$$ Solving for $A$: $$A = 625 \times 10^{-6}$$ Which simplifies to: $$= 6.25 \times 10^{-4} \, \mathrm{m^2}$$
Physics · Mathematics in Physics · Numerical
If $\vec{A} = (2\hat{i} + 3\hat{j} - \hat{k})$ m and $\vec{B} = (\hat{i} + 2\hat{j} + 2\hat{k})$ m. The magnitude of component of vector $\vec{A}$ along vector $\vec{B}$ will be _______ m.
Answer: 2
Given $\vec{A} = (2\hat{i} + 3\hat{j} - \hat{k}) \, \mathrm{m}$ and $\vec{B} = (\hat{i} + 2\hat{j} + 2\hat{k}) \, \mathrm{m}$. The magnitude of component of vector $\vec{A}$ along vector $\vec{B}$ will be ________ m.
Physics · System of Particles and Rotational Motion · Fill in the blank
The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through the center will be _______ m. Given, the length of the rod is $10\sqrt{3}$ m.
Answer: 5
The moment of inertia is given by $$I = \frac{m \ell^2}{12} = m k^2 \Rightarrow k^2 = \frac{\ell^2}{12} \Rightarrow k = \frac{\ell}{\sqrt{12}} = \frac{\ell}{2\sqrt{3}} = \frac{10\sqrt{3}}{2\sqrt{3}} = 5.$$
Physics · Ray Optics and Optical Instruments · Numerical
In the given figure, the face AC of the equilateral prism is immersed in a liquid of refractive index 'n'. For incident angle 60^$\circ$ at the side AC, the refracted light beam just grazes along face AC. The refractive index of the liquid n = $\frac{\sqrt{x}}{4}$. The value of x is_________. (Given refractive index of glass = 1.5)
Answer: 27
Using Snell's law at face AC $$1.5 \sin 60^\circ = n \times \sin 90^\circ$$ $$1.5 \times \frac{\sqrt{3}}{2} = n = \frac{\sqrt{x}}{4}$$ $$3\sqrt{3} = \sqrt{x}$$ $$x = 27$$
Physics · Nuclei · Numerical
Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below: $$^{2}_{1}X + ^{2}_{1}X = ^{4}_{2}Y$$ The binding energies per nucleon $^{2}_{1}X$ and $^{4}_{2}Y$ are $1.1 \, \mathrm{MeV}$ and $7.6 \, \mathrm{MeV}$ respectively. The energy released in this process is MeV.
Answer: 26
Energy released in the given process = Binding energy of product - Binding energy of reactants $$= 7.6 \times 4 - (1.1 \times 2) \times 2$$ $$= 30.4 - 4.4$$ $$= 26 \, MeV$$
Physics · Mechanical Properties of Solids · Numerical
A uniform heavy rod of mass 20 kg. Cross sectional area 0.4 $\mathrm{m}^2$ and length 20 $\mathrm{m}$ is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is $x \times 10^{-9}$ m. The value of $x$ is _______. (Given. Young’s modulus $Y=2 \times 10^{11} \mathrm{Nm}^{-2}$ and $g=10 \mathrm{ms}^{-2}$)
Answer: 25
Given $Y = \frac{T}{A} \frac{dx}{dy}$. $m = 20 \, \mathrm{kg}$ $A = 0.4 \, \mathrm{m^2}$ $l = 20 \, \mathrm{m}$ Let the extension be $dy$ in length $dx$. $Y = \frac{stress}{strain}$ $Y = \frac{A}{dy} = \frac{T}{A} \frac{dx}{dy}$ $dy = \frac{T dx}{AY}$ Tension at a distance $x$ from the lower end $= \frac{mg}{\ell} x$ So, $$\int_0^{\Delta \ell} dy = \int_0^{\ell} \frac{mg}{\ell} x \frac{dx}{AY}$$ $$\Delta \ell = \frac{mg}{\ell AY} \left[ \frac{x^2}{2} \right]_0^{\ell}$$ $$= \frac{mg \ell}{2AY}$$
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Numerical
The typical transfer characteristic of a transistor in CE configuration is shown in figure. A load resistor of 2 \, $\mathrm{k\Omega}$ is connected in the collector branch of the circuit used. The input resistance of the transistor is 0.50 \, $\mathrm{k\Omega}$. The voltage gain of the transistor is
Answer: 200
Current gain in C-E configuration. Therefore, $$\beta = \frac{\Delta I_C}{\Delta I_B}$$ Given $R_C = 2 \, \mathrm{k\Omega}$, $R_B = 0.50 \, \mathrm{k\Omega}$. Voltage gain is given by $$Voltage gain = \frac{\Delta I_C R_C}{\Delta I_B R_B} = \frac{5 \times 10^{-3}}{100 \times 10^{-6}} \times \frac{2}{0.5}$$ $$= \frac{10^{-2}}{5 \times 10^{-5}} = \frac{1000}{5} = 200$$
Physics · Electric Charges and Fields · Numerical
Three point charges of magnitude $5\,\mu\mathrm{C}$, $0.16\,\mu\mathrm{C}$ and $0.3\,\mu\mathrm{C}$ are located at the vertices $A$, $B$, $C$ of a right angled triangle whose sides are $AB = 3\,\mathrm{cm}$, $BC = 3\sqrt{2}\,\mathrm{cm}$ and $CA = 3\,\mathrm{cm}$ and point $A$ is the right angle corner. Charge at point $A$ experiences $\mathrm{N}$ of electrostatic force due to the other two charges.
Answer: 17
Given the charges and distances, we calculate the forces. For $F_1$: $$F_1 = \frac{k \times 5 \times 0.3 \times 10^{-12}}{9 \times 10^{-4}}$$ Substituting the values: $$= \frac{9 \times 10^9 \times 5 \times 0.3 \times 10^{-12}}{9 \times 10^{-4}}$$ This simplifies to: $$= 1.5 \times 10 = 15 \, \mathrm{N}$$ For $F_2$: $$F_2 = \frac{9 \times 10^9 \times 5 \times 0.16 \times 10^{-12}}{9 \times 10^{-4}} = 8 \, \mathrm{N}$$ The force experienced by the charge at $A$ is: $$A = \sqrt{F_1^2 + F_2^2}$$ Calculating: $$= \sqrt{15^2 + 8^2}$$ This results in: $$= \sqrt{289} = 17 \, \mathrm{N}$$
Physics · Magnetism and Matter · Numerical
In a coil of resistance $8\, \Omega$, the magnetic flux due to an external magnetic field varies with time as $\phi = \frac{2}{3}(9 - t^2)$. The value of total heat produced in the coil, till the flux becomes zero, will be
Answer: 2
Given $$\phi = \frac{2}{3}(9 - t^2) = 0$$ We find $$t = 3 \, sec$$ The expression for $$e$$ is given by $$e = -\frac{d\phi}{dt} = -\frac{2}{3}(0 - 2t) = \frac{4t}{3}$$ Heat produced in 3 sec is $$\int e \, dt = \int_0^3 \frac{16t^2}{9 \times 8} \, dt = 2 \, J$$
Physics · Current Electricity · Numerical
A potentiometer wire of length 300 $\mathrm{\ cm}$ is connected in series with a resistance 780 $\Omega$ and a standard cell of emf 4 $\mathrm{V}$. A constant current flows through potentiometer wire. The length of the null point for cell of emf 20 $\mathrm{mV}$ is found to be 60 $\mathrm{\ cm}$. The resistance of the potentiometer wire is____ $\Omega$.
Answer: 20
Let resistance of potentiometer's wire be $R$. $$i = \frac{4}{R + 780}$$ Potential difference across $AB$ $$= \frac{4R}{R + 780}$$ Potential difference across $AC$ $$= \frac{4R \times 60}{(R + 780) \times 300} = \frac{4R}{5(R + 780)}$$ This should be equal to $20 \, \mathrm{mV}$ $$\frac{4R}{5(R + 780)} = 20 \times 10^{-3} = 2 \times 10^{-2}$$ $$4R = 10^{-1}(R + 780)$$ $$4R = \frac{R}{10} + 78$$
Physics · Oscillations · Numerical
As per given figures, two springs of spring constants $K$ and $2K$ are connected to mass $m$. If the period of oscillation in figure (a) is $3 \, \mathrm{s}$, then the period of oscillation in figure (b) will be $\sqrt{x} \, \mathrm{s}$. The value of $x$ is_________.
Answer: 2
For figure (a): $$K_{eq} = \frac{K \times 2K}{K + 2K} = \frac{2K}{3}$$ $$T = 2\pi \sqrt{\frac{m}{K_{eq}}} = 2\pi \sqrt{\frac{m}{2K/3}} = 2\pi \sqrt{\frac{3m}{2K}}$$ For figure (b): $$K_{eq} = 3K, T' = 2\pi \sqrt{\frac{m}{3K}}$$ $$\frac{T'}{T} = \sqrt{\frac{m \times 2K}{3K \times 3m}} = \frac{\sqrt{2}}{3}$$ $$T' = \sqrt{2}$$ $$x = 2$$
Chemistry · Some Basic Concepts of Chemistry · Single correct
Hemoglobin contains 0.34$\%$ of iron by mass. The number of Fe atoms in 3.3 $\mathrm{g}$ of hemoglobin is : (Given : Atomic mass of Fe is 56 $\mathrm{u}$, $N_A$ in 6.022 $\times$ $10^{23}$ $\mathrm{mol^{-1}}$)
Answer: (c)
Number of Fe atoms is calculated as follows: $$\frac{0.34}{100} \times \frac{3.3}{56} \times 6.022 \times 10^{23}$$ This equals: $$1.206 \times 10^{20}$$
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Arrange the following in increasing order of their covalent character. $(A) CaF_2 (B) CaCl_2 (C) CaBr_2 (D) CaI_2$ Choose the correct answer from the options given below.
Answer: (b)
According to Fajan's rule, covalent character $\propto$ size of anion.
Chemistry · Equilibrium · Single correct
Class XII students were asked to prepare one litre of buffer solution of pH 8.26 by their chemistry teacher. The amount of ammonium chloride to be dissolved by the student in 0.2 $\mathrm{M}$ ammonia solution to make one litre of the buffer is (Given $\mathrm{p}K_b$ (NH$_3$) $= 4.74$; Molar mass of NH$_3 = 17 \, \mathrm{g \, mol}^{-1}$; Molar mass of NH$_4\mathrm{Cl} = 53.5 \, \mathrm{g \, mol}^{-1}$)
Answer: (c)
POH = 14 - 8.26 = $\mathrm{pK_b}$ + $\log$ $\frac{[\mathrm{NH_4^+}]}{[\mathrm{NH_3}]}$ = 5.74 = 4.74 + $\log$ $\frac{[\mathrm{NH_4^+}]}{0.2}$ $\implies$ [$\mathrm{NH_4^+}$] = 2 Hence $$\mathrm{NH_4Cl} = 2 \times 53.5 = 107 \, \mathrm{g}$$
Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct
At $30°\mathrm{C}$, the half life for the decomposition of $\mathrm{AB_2}$ is $200\,\mathrm{s}$ and is independent of the initial concentration of $\mathrm{AB_2}$. The time required for $80\%$ of the $\mathrm{AB_2}$ to decompose is (Given: $\log 2 = 0.30$; $\log 3 = 0.48$)
Answer: (c)
Given $T_{1/2} = 200 \, \mathrm{s}$ and first order reaction. $$K = \frac{2.303 \log 2}{200} = \frac{2.303}{t} \log \frac{A_0}{0.2 A_0}$$ $$\frac{\log 2}{200} = \frac{1}{t} \log 5$$ $$t = \frac{7}{3} \times 200 = 466.67 \, \mathrm{s} = 467 \, \mathrm{s}$$
Chemistry · Surface Chemistry · Single correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Finest gold is red in colour, as the size of the particles increases, it appears purple then blue and finally gold. Assertion R : The colour of the colloidal solution depends on the wavelength of light scattered by the dispersed particles. In the light of the above statements, choose the most appropriate answer from the options given below;
Answer: (a)
A
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
The metal that has very low melting point and its periodic position is closer to a metalloid is:
Answer: (b)
\begin{tabular}{l c} & \textbf{Melting point} \\[2pt] Al $\rightarrow$ & 933 K \\ Ga $\rightarrow$ & 303 K \\ In $\rightarrow$ & 430 K \\ Se $\rightarrow$ & 490 K \\ \end{tabular}
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
The metal that is not extracted from its sulphide ore is:
Answer: (a)
Al is extracted from $\mathrm{Al_2O_3 \cdot 2H_2O}$ i.e., Bauxite ore.
Chemistry · The d-and f-Block Elements · Single correct
The products obtained from a reaction of hydrogen peroxide and acidified potassium permanganate are
Answer: (d)
The balanced chemical equation is: $$6\mathrm{H}^+ + 2\mathrm{MnO}_4^- + 5\mathrm{H}_2\mathrm{O}_2 \rightarrow 2\mathrm{Mn}^{+2} + 8\mathrm{H}_2\mathrm{O} + 5\mathrm{O}_2$$
Chemistry · The s-Block Elements · Single correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : $\mathrm{LiF}$ is sparingly soluble in water. Reason R : The ionic radius of $\mathrm{Li^+}$ ion is smallest among its group members, hence has least hydration enthalpy. In the light of the above statements, choose the most appropriate answer from the options given below.
Answer: (c)
Due to high lattice energy $\mathrm{LiF}$ is sparingly soluble in water. $\mathrm{Li^+}$ has high hydration energy among its group members due to smallest size.
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Boric acid is a weak acid Reason R : Boric acid is not able to release $\mathrm{H}^+$ ion on its own. It receives $\mathrm{OH}^-$ ion from water and releases $\mathrm{H}^+$ ion. In the light of the above statements, choose the most appropriate answer from the options given below.
Answer: (a)
Chemistry · Co-ordination Compounds · Single correct
The metal complex that is diamagnetic is (Atomic number : Fe, 26; Cu, 29)
Answer: (a)
Given $\mathrm{K_3[Cu(CN)_4]}$. The oxidation number of copper is $\mathrm{Cu^{+1}}$. $$\mathrm{Cu^{+1} = [Ar]3d^{10}} \Rightarrow Diamagnetic$$
Chemistry · Co-ordination Compounds · Single correct
Match List I with List II Choose the correct answer from the options given below :
Answer: (a)
List I contains pollutants and List II contains their sources. The correct matches are: A. Microorganisms - Domestic sewage, B. Plant nutrients - Chemical fertilizer, C. Toxic heavy metals - Chemical factory, D. Sediment - Strip mining.
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The correct decreasing order of priority of functional groups in naming an organic compound as per IUPAC system of nomenclature is:
Answer: (b)
The order of groups is $SO_3H > COCl > CONH_2 > CN$.
Chemistry · Hydrocarbons · Single correct
Which of the following is not an example of benzenoid compound?
Answer: (b)
Answer : B
Chemistry · Amines · Single correct
Hydrolysis of which compound will give carbolic acid?
Answer: (b)
The reaction involves the conversion of a diazonium salt to a phenol using water. The diazonium group $\mathrm{N_2^+Cl^-}$ is replaced by an $\mathrm{OH}$ group, releasing $\mathrm{N_2}$ and $\mathrm{HCl}$ as byproducts.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Consider the above reaction and predict the major product.
Answer: (a)
Chemistry · Amines · Single correct
The correct sequential order of the reagents for the given reaction is :
Answer: (b)
Chemistry · Polymers · Single correct
Vulcanization of rubber is carried out by heating a mixture of:
Answer: (c)
Vulcanization of rubber is carried out by heating a mixture of isoprene and sulphur.
Chemistry · Biomolecules · Single correct
Animal starch is the other name of:
Answer: (c)
Glycogen
Chemistry · Redox Reactions · Single correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Phenolphthalein is a pH dependent indicator, remains colourless in acidic solution and gives pink colour in basic medium Reason R : Phenolphthalein is a weak acid. It doesn’t dissociate in basic medium. In the light of the above statements, choose the most appropriate answer from the options given below :
Answer: (c)
Phenolphthalein dissociates in basic medium. $$\mathrm{HPh(aq)} \rightleftharpoons \mathrm{H^+} + \mathrm{Ph^-}$$ (colourless) (Pink)
Chemistry · States of Matter · Numerical
A 10 $\mathrm{g}$ mixture of hydrogen and helium is contained in a vessel of capacity 0.0125 $\mathrm{m^3}$ at 6 $\mathrm{bar}$ and $27^\circ\mathrm{C}$. The mass of helium in the mixture is ______ $\mathrm{g}$. (nearest integer) Given : R = 8.3 $\mathrm{JK^{-1}}$ $\mathrm{mol^{-1}}$ (Atomic masses of H and He are 1u and 4u, respectively)
Answer: 8
Given $PV = n_{mix}RT$. $$n_{mix} = \frac{6 \times 12.5}{0.083 \times 300} \approx 3$$ Let mole of He = $x$. Mole of $\mathrm{H_2} = 3 - x$. $$4x + 2(3 - x) = 10$$ $$x = 2 \, \mathrm{mol}$$ Mass of He = $8 \, \mathrm{g}$
Chemistry · Structure of Atom · Numerical
Consider an imaginary ion $^{48}_{22}\mathrm{X}^{3-}$. The nucleus contains ‘a’% more neutrons than the number of electrons in the ion. The value of ‘a’ is ____. [nearest integer]
Answer: 4
No. of neutrons = 26 No. of electrons = 25 % of extra neutrons than electrons = $\($ $\frac{26 - 25}{25}$ $\times$ 100 = 4 $\)$
Chemistry · Thermodynamics · Numerical
For the reaction $$\mathrm{H_2F_2(g) \rightarrow H_2(g) + F_2(g)}$$ $$\Delta U = -59.6 \, \mathrm{kJ \, mol^{-1}}$$ at 27°C. The enthalpy change for the above reaction is (−) ___ kJ mol⁻¹ [nearest integer] Given : R = 8.314 JK⁻¹ mol⁻¹.
Answer: 57
Given $\Delta H=\Delta U+\Delta n_gRT$. $\Delta H=-59.6+1\times8.314\times300\times10^{-3}=-57.10$
Chemistry · Solutions · Numerical
The elevation in boiling point for 1 molal solution of non-volatile solute A is 3 K. The depression in freezing point for 2 molal solution of A in the same solvent is 6 K. The ratio of $K_b$ and $K_f$ i.e., $K_b/K_f$ is 1 : X. The value of X is [nearest integer]
Answer: 1
Given $\Delta T_b = i K_b m_1$ and $\Delta T_f = i K_f m_2$. The ratio is given by $$\frac{\Delta T_b}{\Delta T_f} = \frac{K_b \times 1}{K_f \times 2} \Rightarrow \frac{3}{6} = \frac{1}{2} = \frac{K_b}{K_f} \times \frac{1}{2}$$ Thus, $$\frac{K_b}{K_f} = \frac{1}{1} \Rightarrow x = 1$$
Chemistry · Redox Reactions · Numerical
20 mL of 0.02 M hypo solution is used for the titration of 10 mL of copper sulphate solution, in the presence of excess of KI using starch as an indicator. The molarity of $\mathrm{Cu}^{2+}$ is found to be _____ $\times 10^{-2}$ M [nearest integer] Given: $2\mathrm{Cu}^{2+} + 4\mathrm{I}^- \rightarrow \mathrm{Cu}_2\mathrm{I}_2 + \mathrm{I}_2$ $\mathrm{I}_2 + 2\mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow 2\mathrm{I}^- + \mathrm{S}_4\mathrm{O}_6^{2-}$
Answer: 4
The equivalent of $I_2$ is equal to the equivalent of $\mathrm{Na_2S_2O_3}$, which is $20 \times 0.002 \times 1$. Therefore, $2 \times n_{mol}$ of $I_2 = 0.4$. The moles of $I_2$ are $0.2$ mmol. The moles of $\mathrm{Cu^{+2}}$ are $0.2 \times 2 \times 10^{-3}$. The concentration of $\mathrm{Cu^{+2}}$ is given by: $$[\mathrm{Cu^{+2}}] = \frac{0.4 \times 10^{-3}}{10 \times 10^{-3}} = 0.04 = 4 \times 10^{-2}$$
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Numerical
The number of non-ionisable protons present in the product B obtained from the following reaction is _____. $C_2H_5OH + PCl_3$ $\rightarrow$ $C_2H_5Cl + A$ $A + PCl_3$ $\rightarrow$ $B$
Answer: 2
$C_2H_5OH+PCl_3\longrightarrow C_2H_5Cl+H_3PO_3$ $H_3PO_3+PCl_3\longrightarrow H_4P_2O_5+HCl$
Chemistry · The d-and f-Block Elements · Numerical
The spin-only magnetic moment value of the compound with strongest oxidizing ability among $\mathrm{MnF_4}$, $\mathrm{MnF_3}$ and $\mathrm{MnF_2}$ is _____ B.M. [nearest integer]
Answer: 5
The electronic configurations are as follows: MnF_4 has an oxidation state of +4 with E.C = [$\mathrm{Ar}$] 3d^3. MnF_3 has an oxidation state of +3 with [$\mathrm{Ar}$] 3d^4. MnF_2 has an oxidation state of +2 with [$\mathrm{Ar}$] 3d^5. Hence MnF_3 is the strongest oxidizing agent. The magnetic moment is calculated as follows: $$\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.89 \approx 5$$
Chemistry · Hydrocarbons · Numerical
Total number of isomers (including stereoisomers) obtain on monochlorination of methylcyclohexane is _______.
Answer: 12
The number of isomers for each compound is given below each structure. The first compound has 1 isomer, the second compound has 1 isomer, the third compound has 4 isomers, and the fourth compound has 2 isomers.
Chemistry · Alcohols, Phenols and Ethers · Numerical
A 100 $\mathrm{mL}$ solution of $\mathrm{CH_3CH_2MgBr}$ on treatment with methanol produces 2.24 $\mathrm{mL}$ of a gas at STP. The weight of gas produced is ______ mg. [nearest integer]
Answer: 3
The reaction is given by: $$\mathrm{CH_3{-}CH_2{-}MgBr + CH_3OH \rightarrow}$$ $$\mathrm{CH_3{-}CH_3 +}$$ $$\mathrm{Mg-OCH_3Br}$$ The number of moles $n$ is calculated as: $$n = \frac{2.24 \times 10^{-3}}{22.4} = 10^{-4}$$ The weight $W$ is given by: $$W = n \times M$$ $$= 10^{-4} \times 30 = 3 \, \mathrm{mg}$$
Chemistry · Chemistry in Everyday Life · Numerical
How many of the following drugs is/are example(s) of broad spectrum antibiotic? Ofloxacin, Penicillin G, Terpineol, Salvarsan
Answer: 1
Ofloxacin