JEE Main 28 June 2022 Shift 2 question paper with solutions
JEE Main 28 June 2022 Shift 2: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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Maths
Question 1
Maths · Relations and Functions · Single correct
Let $R_1 = \{(a, b) \in \mathbb{N} \times \mathbb{N} : |a - b| \leq 13\}$ and $R_2 = \{(a, b) \in \mathbb{N} \times \mathbb{N} : |a - b| \neq 13\}$. Then on $\mathbb{N}$:
Both $R_1$ and $R_2$ are equivalence relations
Neither $R_1$ nor $R_2$ is an equivalence relation
$R_1$ is an equivalence relation but $R_2$ is not
$R_2$ is an equivalence relation but $R_1$ is not
Answer: (b)
Solution
For $R_1$: i) Reflexive relation: $(a, a) \in \mathbb{N} \times \mathbb{N} : |a - a| \leq 13$ ii) Symmetric relation: $(a, b) \in R_1,\ (b, a) \in R_1 : |b - a| \leq 13$ iii) Transitive relation: $(a, b) \in R_1,\ (b, c) \in R_1,\ (a, c) \in R_1$: $(1, 3) \in R_1,\ (3, 16) \in R_1,\ \text{but } (1, 16) \notin R_1$ For $R_2$: i) Reflexive relation: $(a, a) \in \mathbb{N} \times \mathbb{N} : |a - a| \neq 13$ ii) Symmetric relation: $(b, a) \in \mathbb{N} \times \mathbb{N} : |b - a| \neq 13$ iii) Transitive relation: $(a, b) \in R_2,\ (b, c) \in R_2,\ (a, c) \in R_2$: $(1, 3) \in R_2,\ (3, 14) \in R_2,\ \text{but } (1, 14) \notin R_2$
Question 2
Maths · Complex Numbers and Quadratic Equations · Single correct
Let f(x) be a quadratic polynomial such that f(-2) + f(3) = 0. If one of the roots of f(x) = 0 is -1, then the sum of the roots of f(x) = 0 is equal to:
$\frac{11}{3}$
$\frac{7}{3}$
$\frac{13}{3}$
$\frac{14}{3}$
Answer: (a)
Solution
Given $f(-2) + f(3) = 0$. $f(x) = (x + 1)(ax + b)$. $f(-2) + f(3) = -1(-2a + b) + 4(3a + b) = 0$. $2a - b + 12a + 4b = 0$. $14a + 3b = 0$. $$\frac{-b}{a} = \frac{14}{3}$$ Sum of roots = $$\left(-1 + \frac{-b}{a}\right) = -1 + \frac{14}{3} = \frac{11}{3}$$
Question 3
Maths · Permutations and Combinations · Single correct
The number of ways to distribute 30 identical candies among four children $C_1$, $C_2$, $C_3$ and $C_4$ so that $C_2$ receives atleast 4 and atmost 7 candies, $C_3$ receives atleast 2 and atmost 6 candies, is equal to
The term independent of $x$ in the expression of $$\left(1 - x^2 + 3x^3\right) \left(\frac{5}{2} x^3 - \frac{1}{5x^2}\right)^{11}, x \neq 0$$ is
$\frac{7}{40}$
$\frac{33}{200}$
$\frac{39}{200}$
$\frac{11}{50}$
Answer: (b)
Solution
The expression $\left(1 - x^2 + 3x^3\right) \left(\frac{5}{2} x^3 - \frac{1}{5x^2}\right)^{11}$ is given. The general term of $\left(\frac{5}{2} x^3 - \frac{1}{5x^2}\right)^{11}$ is $$^{11}C_r \left(\frac{5}{2} x^3\right)^{11-r} \left(-\frac{1}{5x^2}\right)^r.$$ Simplifying, the general term is $$^{11}C_r \left(\frac{5}{2}\right)^{11-r} \left(-\frac{1}{5}\right)^r x^{33-5r}.$$ Now, consider the term independent of $x$. The coefficient of $x^0$ in $\left(\frac{5}{2} x^3 - \frac{1}{5x^2}\right)^{11}$ is $$1 \times coefficient of x^0$$ The coefficient of $x^{-2}$ in $\left(\frac{5}{2} x^3 - \frac{1}{5x^2}\right)^{11}$ is $$-1 \times coefficient of x^{-2}$$ The coefficient of $x^{-3}$ in $\left(\frac{5}{2} x^3 - \frac{1}{5x^2}\right)^{11}$ is $$3 \times coefficient of x^{-3}$$ For the coefficient of $x^0$, $33 - 5r = 0$, which is not possible. For the coefficient of $x^{-2}$, $33 - 5r = -2$, which gives $r = 7$. For the coefficient of $x^{-3}$, $33 - 5r = -3$, which is not possible. Thus, the term independent of $x$ is $$(-1)^{11}C_7 \left(\frac{5}{2}\right)^4 \left(-\frac{1}{5}\right)^7 = \frac{33}{200}.$$
Question 5
Maths · Sequences and Series · Single correct
If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1 : 7 and a + n = 33, then the value of n is
21
22
23
24
Answer: (c)
Solution
Given $\($ d = $\frac{100-a}{n+1}$ $\)$ $\($ A_1 = a + d $\)$ $\($ A_n = 100 - d $\)$ $\($ $\Rightarrow$ $\frac{A_1}{A_n}$ = $\frac{1}{7}$ $\Rightarrow$ $\frac{a+d}{100-d}$ = $\frac{1}{7}$ $\)$ $\($ $\Rightarrow$ 7a + 8d = 100 $\)$ $\($ $\Rightarrow$ 7a + 8 $\left$( $\frac{100-a}{n+1}$ $\right$) = 100 $\)$ $\($ $\ldots$ (1) $\)$ $\($ $\therefore$ a + n = 33 $\)$ $\($ $\ldots$ (2) $\)$ Now, by Eq. (1) and (2) $\($ 7n^2 - 132n - 667 = 0 $\)$ $\($ n = 23 $\)$ and $\($ n = $\frac{-29}{7}$ $\)$ reject.
Question 6
Maths · Continuity and Differentiability · Single correct
Let $f, g : \mathbb{R} \to \mathbb{R}$ be functions defined by $$f(x) = \begin{cases} \lfloor x \rfloor, & x < 0 \\ |1 - x|, & x \geq 0 \end{cases}$$ and $$g(x) = \begin{cases} e^x - x, & x < 0 \\ (x - 1)^2 - 1, & x \geq 0 \end{cases}$$ where $[x]$ denote the greatest integer less than or equal to $x$. Then, the function $f \circ g$ is discontinuous at exactly:
one point
two points
three points
four points
Answer: (b)
Solution
Check continuity at $x = 0$ and also check continuity at those $x$ where $g(x) = 0$. $g(x) = 0$ at $x = 0, 2$. $$\mathrm{fog}(0^+) = -1$$ $$\mathrm{fog}(0) = 0$$ Hence, discontinuous at $x = 0$. $$\mathrm{fog}(2^+) = 1$$ $$\mathrm{fog}(2^-) = -1$$ Hence, discontinuous at $x = 2$.
Question 7
Maths · Applications of Integrals · Single correct
Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f \left( \frac{\pi}{4} \right) = \sqrt{2}$, $f \left( \frac{\pi}{2} \right) = 0$ and $f' \left( \frac{\pi}{2} \right) = 1$ and let $g(x) = \int_{x}^{\pi/4} \left( f'(t) \sec t + \tan t \sec t \ f(t) \right) dt$ for $x \in \left[ \frac{\pi}{4}, \frac{\pi}{2} \right)$. Then $\lim_{x \to \left( \frac{\pi}{2} \right)^{-}} g(x)$ is equal to
2
3
4
-3
Answer: (b)
Solution
Given $$g(x) = \int_{x}^{\pi/4} \left( f'(t) \sec t + \tan t \sec t f(t) \right) \, dt$$ We have $$g(x) = \int_{x}^{\pi/4} d(f(t) \cdot \sec t) = f(t) \sec t \bigg|_{x}^{\pi/4}$$ Thus, $$g(x) = f\left(\frac{\pi}{4}\right) \sec \frac{\pi}{4} - f(x) \cdot \sec x$$ Simplifying, $$g(x) = 2 - f(x) \sec x = 2 - \left( \frac{f(x)}{\cos x} \right)$$ Taking the limit as $x$ approaches $\frac{\pi}{2}$, $$\lim_{x \to \frac{\pi}{2}} g(x) = 2 - \lim_{x \to \frac{\pi}{2}} \left( \frac{f(x)}{\cos x} \right)$$ Using L'Hopital's Rule, $$= 2 - \lim_{x \to \frac{\pi}{2}} \frac{f'(x)}{-\sin x}$$ This simplifies to $$= 2 + \frac{f'\left(\frac{\pi}{2}\right)}{\sin \frac{\pi}{2}} = 2 + \frac{1}{1} = 3$$
Question 8
Maths · Integrals · Single correct
Let $f : \mathbb{R} \to \mathbb{R}$ be continuous function satisfying $f(x) + f(x + k) = n$, for all $x \in \mathbb{R}$ where $k > 0$ and $n$ is a positive integer. If $I_1 = \int_{0}^{4nk} f(x) \, dx$ and $I_2 = \int_{-k}^{3k} f(x) \, dx$, then
Maths · Applications of Integrals · Single correct
The area of the bounded region enclosed by the curve $y = 3 - \left| x - \frac{1}{2} \right| - |x + 1|$ and the x-axis is
$\frac{9}{4}$
$\frac{45}{16}$
$\frac{27}{8}$
$\frac{63}{16}$
Answer: (c)
Solution
Given $$y = \begin{cases} 3 + (x + 1) + \left(x - \frac{1}{2}\right), & x < -1 \\ 3 - (x + 1) + \left(x - \frac{1}{2}\right), & -1 \leq x < \frac{1}{2} \\ 3 - (x + 1) - \left(x - \frac{1}{2}\right), & \frac{1}{2} \leq x \end{cases}$$ Simplifying, we have $$y = \begin{cases} \frac{7}{2} + 2x, & x < -1 \\ \frac{3}{2}, & -1 \leq x < \frac{1}{2} \\ \frac{5}{2} - 2x, & \frac{1}{2} \leq x \end{cases}$$ The area bounded is ar ABF + ar BCEF + ar CDE. $$= \frac{1}{2} \left(\frac{3}{4}\right) \left(\frac{3}{2}\right) + \left(\frac{3}{2}\right) \left(\frac{3}{2}\right) + \frac{1}{2} \left(\frac{3}{4}\right) \left(\frac{3}{2}\right)$$ $$= \frac{27}{8}$$ sq. units.
Question 10
Maths · Differential Equations · Single correct
Let $x = x(y)$ be the solution of the differential equation $2y e^{x/y^2} \, dx + \left( y^2 - 4x e^{x/y^2} \right) \, dy = 0$ such that $x(1) = 0$. Then, $x(e)$ is equal to
$e \log_e (2)$
$-e \log_e (2)$
$e^2 \log_e (2)$
$-e^2 \log_e (2)$
Answer: (d)
Solution
$2ye^{x/y^2}\,dx + (y^2 - 4xe^{x/y^2})\,dy = 0$ $2e^{x/y^2}[y\,dx - 2x\,dy] + y^2\,dy = 0$ $2e^{x/y^2}\left[\dfrac{y^2\,dx - x\cdot(2y)\,dy}{y}\right] + y^2\,dy = 0$ Divide by $y^3$ $2e^{x/y^2}\left[\dfrac{y^2\,dx - x\cdot(2y)\,dy}{y^4}\right] + \dfrac{1}{y}\,dy = 0$ $2e^{x/y^2}\,d\left(\dfrac{x}{y^2}\right) + \dfrac{1}{y}\,dy = 0$ Integrating $\int 2e^{x/y^2}\,d\left(\dfrac{x}{y^2}\right) + \int\dfrac{1}{y}\,dy = 0$ $2e^{x/y^2} + \ln y + c = 0$ $(0, 1)$ lies on it. $2e^0 + \ln 1 + c = 0 \Rightarrow c = -2$ Required curve : $\boxed{2e^{x/y^2} + \ln y - 2 = 0}$ For $x(e)$ $2e^{x/e^2} + \ln e - 2 = 0 \Rightarrow x = -e^2\log_e 2$
Question 11
Maths · Applications of Derivatives · Single correct
Let the slope of the tangent to a curve $y=f(x)$ at $(x,y)$ be given by \[ 2\tan x\,(\cos x-y). \] If the curve passes through the point $\left(\frac{\pi}{4},0\right)$, then the value of \[ \int_{0}^{\pi/2} y\,dx \] is equal to
$(2-\sqrt{2})+\frac{\pi}{\sqrt{2}}$
$2-\frac{\pi}{\sqrt{2}}$
$(2+\sqrt{2})+\frac{\pi}{\sqrt{2}}$
$2+\frac{\pi}{\sqrt{2}}$
Answer: (b)
Solution
Given $\($ $\frac{dy}{dx}$ = 2 $\tan$ x $\cos$ x - 2 $\tan$ x $\cdot$ y $\)$ $\($ $\frac{dy}{dx}$ + (2 $\tan$ x) y = 2 $\sin$ x $\)$ Integrating factor $\($ = e^{$\int$ 2 $\tan$ x $\,$ dx} = $\frac{1}{\cos^2 x}$ $\)$ $\($ y $\left$( $\frac{1}{\cos^2 x}$ $\right$) = $\int$ $\frac{2 \sin x}{\cos^2 x}$ $\,$ dx $\)$ $\($ y $\sec$^2 x = $\frac{2}{\cos x}$ + C $\)$ $\($ y = 2 $\cos$ x + C $\cos$^2 x $\)$ Passes through $\($ $\left$( $\frac{\pi}{4}$, 0 $\right$) $\)$ $\($ 0 = $\sqrt{2}$ + $\frac{C}{2}$ $\Rightarrow$ C = -2 $\sqrt{2}$ $\)$ $\($ f(x) = 2 $\cos$ x - 2 $\sqrt{2}$ $\cos$^2 x $\)$ : Required curve $\($ $\int$_{0}^{$\pi$/2} y $\,$ dx = 2 $\int$_{0}^{$\pi$/2} $\cos$ x $\,$ dx - 2 $\sqrt{2}$ $\int$_{0}^{$\pi$/2} $\cos$^2 x $\,$ dx $\)$ $\($ = $\left$[ 2 $\sin$ x $\right$]_{0}^{$\pi$/2} - 2 $\sqrt{2}$ $\left$[ $\frac{x}{2}$ + $\frac{\sin 2x}{4}$ $\right$]_{0}^{$\pi$/2} $\)$ $\($ = 2 - $\frac{\pi}{\sqrt{2}}$ $\)$
Question 12
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let a triangle be bounded by the lines $L_1 : 2x + 5y = 10$; $L_2 : -4x + 3y = 12$ and the line $L_3$, which passes through the point $P(2, 3)$, intersect $L_2$ at $A$ and $L_1$ at $B$. If the point $P$ divides the line-segment $AB$, internally in the ratio $1 : 3$, then the area of the triangle is equal to
Let $a > 0$, $b > 0$. Let $e$ and $\ell$ respectively be the eccentricity and length of the latus rectum of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Let $e'$ and $\ell'$ respectively the eccentricity and length of the latus rectum of its conjugate hyperbola. If $e^2 = \frac{11}{14} \ell$ and $(e')^2 = \frac{11}{8} \ell'$, then the value of $77a + 44b$ is equal to
100
110
120
130
Answer: (d)
Solution
Given $e = \sqrt{1 + \frac{b^2}{a^2}}, \ell = \frac{2b^2}{a}$. Given $e^2 = \frac{11}{14} \ell$. $$1 + \frac{b^2}{a^2} = \frac{11}{14} \cdot \frac{2b^2}{a}$$ $$\frac{a^2 + b^2}{a^2} = \frac{11}{7} \cdot \frac{b^2}{a} ......(1)$$ Also $e' = \sqrt{1 + \frac{a^2}{b^2}}, \ell' = \frac{2a^2}{b}$. Given $(e')^2 = \frac{11}{8} \ell'$. $$1 + \frac{a^2}{b^2} = \frac{11}{8} \cdot \frac{2a^2}{b}$$ $$\frac{a^2 + b^2}{b^2} = \frac{11}{4} \cdot \frac{a^2}{b} ......(2)$$ New (1) $\div$ (2) $$\frac{b^2}{a^2} = \frac{4}{7} \cdot \frac{b^3}{a^3}$$ $$\therefore 7a = 4b ...... (3)$$ From (2) $$\frac{16b^2}{49} + b^2 = \frac{11}{4} \cdot \frac{16b^2}{49b}$$ $$\frac{65}{49} = \frac{11}{4} \cdot \frac{16}{49} \cdot b$$ $$\therefore b = \frac{4 \times 65}{11 \times 16} ..... (4)$$ We have to find value of $77a + 44b$
Question 14
Maths · Vector Algebra · Single correct
Let $\vec{a}=\alpha\hat{i}+2\hat{j}-\hat{k}$ and $\vec{b}=-2\hat{i}+\alpha\hat{j}+\hat{k},$ where $\alpha\in\mathbb{R}$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$ is $\sqrt{15(\alpha^2+4)}$, then the value of $2|\vec{a}|^2+(\vec{a}\cdot\vec{b})|\vec{b}|^2$ is equal to $\underline{\hspace{2cm}}$
If vertex of a parabola is $(2, -1)$ and the equation of its directrix is $4x - 3y = 21$, then the length of its latus rectum is
2
8
12
16
Answer: (b)
Solution
Given the equation $4x - 3y = 21$ and the point $(2, -1)$. Calculate $a$ using the formula: $$a = \frac{|18 + 3 - 21|}{5} = \frac{10}{5} = 2$$ Therefore, the latus rectum is $4a = 8$.
Question 16
Maths · Three Dimensional Geometry · Single correct
Let the plane ax + by + cz = d pass through (2, 3, -5) and is perpendicular to the planes 2x + y - 5z = 10 and 3x + 5y - 7z = 12. If a, b, c, d are integers d > 0 and gcd (|a|, |b|, |c|, d) = 1, then the value of a + 7b + c + 20d is equal to
18
20
24
22
Answer: (d)
Solution
DR'S normal of plane $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -5 \\ 3 & 5 & -7 \end{vmatrix} = 18\hat{i} - \hat{j} + 7\hat{k}$$ Therefore, equation of plane $$18x - y + 7z = d$$ It passes through $(2, 3, -5)$ $$36 - 3 - 35 = d \therefore d = -2$$ Therefore, equation of plane $$18x - y + 7z = -2$$ $$-18x + y - 7z = 2$$ Therefore, $a = -18$, $b = 1$, $c = -7$, $d = 2$ $$a + 7b + c + 20d = -18 + 7 - 7 + 40 = 22$$
Question 17
Maths · Probability · Single correct
The probability that a randomly chosen one-one function from the set {a, b, c, d} to the set {1, 2, 3, 4, 5} satisfies $f(a) + 2f(b) - f(c) = f(d)$ is :
$\frac{1}{24}$
$\frac{1}{40}$
$\frac{1}{30}$
$\frac{1}{20}$
Answer: (d)
Solution
The number of sample points is given by $n(s) = \binom{5}{4} \times 4! = 120$. The table shows the values for $f(a)$, $2f(b)$, $f(c)$, and $f(d)$. The calculations are as follows: For $f(a) = 5$, $2f(b) = 2 \times 1$, $f(c) = 3$, $f(d) = 4$. For $f(a) = 4$, $2f(b) = 2 \times 2$, $f(c) = 3$, $f(d) = 5$. For $f(a) = 1$, $2f(b) = 2 \times 3$, $f(c) = 2$, $f(d) = 5$. The number of favorable outcomes is $n(A) = 2 \times 3 = 6$. Therefore, the probability $P(A)$ is given by $$P(A) = \frac{n(A)}{n(s)} = \frac{6}{120} = \frac{1}{20}.$$
Question 18
Maths · Limits and Derivatives · Single correct
The value of $\lim_{n \to \infty} 6 \tan \left\{ \sum_{r=1}^{n} \tan^{-1} \left( \frac{1}{r^2 + 3r + 3} \right) \right\}$ is equal to
1
2
3
6
Solution
Given $T_r = \tan^{-1} \left[ \frac{(r+2) - (r+1)}{1 + (r+2)(r+1)} \right]$. This simplifies to $\tan^{-1}(r+2) - \tan^{-1}(r+1)$. For $T_1$, we have $T_1 = \tan^{-1} 3 - \tan^{-1} 2$. For $T_2$, we have $T_2 = \tan^{-1} 4 - \tan^{-1} 3$. For $T_n$, we have $T_n = \tan^{-1} (n+2) - \tan^{-1} (n+1)$. The sum $S_n = \tan^{-1} (n+2) - \tan^{-1} 2$ $= \tan^{-1} \left( \frac{n+2-2}{1+2(n+2)} \right)$. This simplifies to $\tan^{-1} \left( \frac{n}{2n+5} \right)$. We then evaluate $\lim_{n \to \infty} 6 \tan \left( \tan^{-1} \left( \frac{n}{2n+5} \right) \right)$. This limit is $\lim_{n \to \infty} \frac{6n}{2n+5} = \frac{6}{2} = 3$.
Question 19
Maths · Vector Algebra · Single correct
Let $\vec{a}$ be a vector which is perpendicular to the vector $3\hat{i} + \frac{1}{2}\hat{j} + 2\hat{k}$. If $\vec{a} \times (2\hat{i} + \hat{k}) = 2\hat{i} - 13\hat{j} - 4\hat{k}$, then the projection of the vector $\vec{a}$ on the vector $2\hat{i} + 2\hat{j} + \hat{k}$ is
If $\cot \alpha = 1$ and $\sec \beta = -\frac{5}{3}$, where $\pi < \alpha < \frac{3\pi}{2}$ and $\frac{\pi}{2} < \beta < \pi$, then the value of $\tan(\alpha + \beta)$ and the quadrant in which $\alpha + \beta$ lies, respectively are
Let the image of the point P(1, 2, 3) in the line $$L: \frac{x-6}{3} = \frac{y-1}{2} = \frac{z-2}{3}$$ be Q. let $$R(\alpha, \beta, \gamma)$$ be a point that divides internally the line segment PQ in the ratio 1 : 3. Then the value of $$22(\alpha + \beta + \gamma)$$ is equal to
Answer: 125
Solution
Let M be the mid-point of PQ. Therefore, $\mathrm{M} = (3\lambda + 6, \, 2\lambda + 1, \, 3\lambda + 2)$. Now, $\overrightarrow{\mathrm{PM}} = (3\lambda + 5)\hat{i} + (2\lambda - 1)\hat{j} + (3\lambda - 1)\hat{k}$. Therefore, $\overrightarrow{\mathrm{PM}} \perp (3\hat{i} + 2\hat{j} + 3\hat{k})$. Thus, $3(3\lambda + 5) + 2(2\lambda - 1) + 3(3\lambda - 1) = 0$. $\lambda = \frac{-5}{11}$. Therefore, $\mathrm{M} \left( \frac{51}{11}, \frac{1}{11}, \frac{7}{11} \right)$. Since R is mid-point of PM, $22(\alpha + \beta + \gamma) = 125$.
Question 22
Maths · Mathematical Reasoning · Numerical
Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62, and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is
Answer: 0
Solution
Given $$20 = \frac{\sum_{i=1}^{7} |x_i - 62|^2}{7}$$ which implies $$|x_1 - 62|^2 + |x_2 - 62|^2 + \ldots + |x_7 - 62|^2 = 140$$ If $$x_1 = 49$$ then $$|49 - 62|^2 = 169$$ Therefore, $$|x_2 - 62|^2 + \ldots + |x_7 - 62|^2 = $$ Negative Number which is not possible, therefore, no student can fail.
Question 23
Maths · Conic Sections · Numerical
If one of the diameters of the circle \[ x^2 + y^2 - 2\sqrt{2}\,x - 6\sqrt{2}\,y + 14 = 0 \] is a chord of the circle \[ (x - 2\sqrt{2})^2 + (y - 2\sqrt{2})^2 = r^2, \] then the value of \(r^2\) is equal to
Answer: 10
Solution
PQ is the diameter of the circle. The equation of the circle is given by $S: x^2 + y^2 - 2\sqrt{2}x - 6\sqrt{2}y + 14 = 0$. The coordinates of $C$ and $O$ are $C(\sqrt{2}, 3\sqrt{2})$ and $O(2\sqrt{2}, 2\sqrt{2})$. The radius $r_1$ is $\sqrt{6}$. The equation of the circle $S_1$ is $(x - 2\sqrt{2})^2 + (y - 2\sqrt{2})^2 = r^2$. Now, in $\triangle OCQ$, we have $|OC|^2 + |CQ|^2 = |OQ|^2$. Therefore, $4 + 6 = r^2$, which gives $r^2 = 10$.
Question 24
Maths · Limits and Derivatives · Numerical
If $\lim_{{x \to 1}} \frac{{\sin(3x^2 - 4x + 1) - x^2 + 1}}{{2x^3 - 7x^2 + ax + b}} = -2$, then the value of $(a - b)$ is equal to
Answer: 11
Solution
Given $\($ $\lim$_{{x $\to$ 1}} $\frac{\sin(3x^2 - 4x + 1) - x^2 + 1}{2x^3 - 7x^2 + ax + b}$ = -2 $\)$. For finite limit $\($ a + b - 5 = 0 $\)$ $\ldots$(1) Apply L'Hôpital's rule $\($ $\lim$_{{x $\to$ 1}} $\frac{\cos(3x^2 - 4x + 1)(6x - 4) - 2x}{(6x^2 - 14x + a)}$ = -2 $\)$ For finite limit $\($ 6 - 14 + a = 0 $\)$ $\($ a = 8 $\)$ From (1) $\($ b = -3 $\)$ Now $\($ (a - b) = 11 $\)$
Question 25
Maths · Sequences and Series · Numerical
Let for $n = 1, 2, \ldots, 50$, $S_n$ be the sum of the infinite geometric progression whose first term is $n^2$ and whose common ratio is $\frac{1}{(n+1)^2}$. Then the value of $$\frac{1}{26} + \sum_{n=1}^{50} \left( S_n + \frac{2}{n+1} - n - 1 \right)$$ is equal to
If the system of linear equations $$2x - 3y = \gamma + 5,$$ $$\alpha x + 5y = \beta + 1,$$ where $\alpha, \beta, \gamma \in \mathbb{R}$ has infinitely many solutions, then the value of $|9\alpha + 3\beta + 5\gamma|$ is equal to
Answer: 58
Solution
Given the equations: $$2x - 3y = \gamma + 5$$ $$\alpha x + 5y = \beta + 1$$ For infinite many solutions: $$\frac{\alpha}{2} = \frac{5}{-3} = \frac{\beta + 1}{\gamma + 5}$$ Solving for $\alpha$: $$\alpha = \frac{-10}{3}$$ And for $\gamma$: $$5\gamma + 25 = -3\beta - 3$$ Simplifying: $$9\alpha = -30, 3\beta + 5\gamma = -28$$ Now: $$9\alpha + 3\beta + 5\gamma = -58$$ Taking the absolute value: $$|9\alpha + 3\beta + 5\gamma| = 58$$
Question 27
Maths · Matrices · Numerical
Let $A = \begin{pmatrix} 1+i & 1 \\ -i & 0 \end{pmatrix}$ where $i = \sqrt{-1}$. Then, the number of elements in the set $$\{ n \in \{1,2,\ldots,100\} : A^n = A \}$$ is
Answer: 25
Solution
Given $$A = \begin{bmatrix} 1+i & 1 \\ -i & 0 \end{bmatrix}$$ Calculate $$A^2 = \begin{bmatrix} 1+i & 1 \\ -i & 0 \end{bmatrix} \begin{bmatrix} 1+i & 1 \\ -i & 0 \end{bmatrix}$$ This results in $$A^2 = \begin{bmatrix} i & 1+i \\ -i+1 & -i \end{bmatrix}$$ Next, calculate $$A^4 = \begin{bmatrix} i & 1+i \\ -i+1 & -i \end{bmatrix} \begin{bmatrix} i & 1+i \\ -i+1 & -i \end{bmatrix}$$ This results in $$A^4 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$ Therefore, $$A^{4n+1} = A$$ For $$n = 1, 5, 9, \ldots, 97$$ The total elements in the set is 25.
Question 28
Maths · Complex Numbers and Quadratic Equations · Numerical
Sum of squares of modulus of all the complex numbers $z$ satisfying $\overline{z} = i z^2 + z^2 - z$ is equal to
Answer: 2
Solution
Given $z + \overline{z} = i \overline{z}^2 + z^2$. Consider $z = x + iy$. Then $2x = (i + 1)(x^2 - y^2 + 2xyi)$. This implies $2x = x^2 - y^2 - 2xy$ and $x^2 - y^2 + 2xy = 0$. Therefore, $2x = -4xy$. This implies $x = 0$ or $y = -\frac{1}{2}$. Case 1: $x = 0$ implies $y = 0$, hence $z = 0$. Case 2: $y = -\frac{1}{2}$. This gives $4x^2 - 4x - 1 = 0$. $$(2x - 1)^2 = 2$$ $2x - 1 = \pm \sqrt{2}$. Thus, $x = \frac{1 \pm \sqrt{2}}{2}$. Here $z = \frac{1 + \sqrt{2}}{2} - \frac{i}{2}$ or $z = \frac{1 - \sqrt{2}}{2} - \frac{i}{2}$. Sum of squares of modulus of $z$ is $$0 + \frac{(1 + \sqrt{2})^2 + 1}{4} + \frac{(1 - \sqrt{2})^2 + 1}{4} = \frac{8}{4} = 2$$
Question 29
Maths · Relations and Functions (Advanced) · Numerical
Let S = {1, 2, 3, 4}. Then the number of elements in the set $\{$f : S $\times$ S $\to$ S : f is onto and f(a, b) = f(b, a) $\ge$ a $\forall$ (a, b) $\in$ S $\times$ S$\}$ is
Answer: 37
Solution
(1, 1), (1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3), (4, 4) – all have one choice for image. (2, 1), (1, 2), (2, 2) – all have three choices for image (3, 2), (2, 3), (3, 1), (1, 3), (3, 3) – all have two choices for image. So the total functions = 3 $\times$ 3 $\times$ 2 $\times$ 2 $\times$ 2 = 72. Case 1: None of the pre-images have 3 as image. Total functions = 2 $\times$ 2 $\times$ 1 $\times$ 1 $\times$ 1 = 4. Case 2: None of the pre-images have 2 as image. Total functions = 2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 2 = 32. Case 3: None of the pre-images have either 3 or 2 as image. Total functions = 1 $\times$ 1 $\times$ 1 $\times$ 1 $\times$ 1 = 1. Therefore, total onto functions = 72 - 4 - 32 + 1 = 37.
Question 30
Maths · Mathematical Reasoning · Numerical
The maximum number of compound propositions, out of $p \lor r \lor s$, $p \lor r \lor \sim s$, $p \lor \sim q \lor s$, $\sim p \lor \sim r \lor s$, $\sim p \lor \sim r \lor \sim s$, $\sim p \lor q \lor \sim s$, $q \lor r \lor \sim s$, $q \lor \sim r \lor \sim s$, $\sim p \lor \sim q \lor \sim s$ that can be made simultaneously true by an assignment of the truth values to $p$, $q$, $r$ and $s$, is equal to
Answer: 9
Solution
If we take The truth value of all the propositions will be true.
Physics
Question 31
Physics · Motion in a Straight Line · Single correct
Velocity (v) and acceleration (a) in two systems of units 1 and 2 are related as $v_2 = \frac{n}{m^2} v_1$ and $a_2 = \frac{a_1}{mn}$ respectively. Here $m$ and $n$ are constants. The relations for distance and time in two systems respectively are:
$\frac{n^3}{m^3} L_1 = L_2$ and $\frac{n^2}{m} T_1 = T_2$
$\frac{n^4}{m^2} L_2 = L_2$ and $T_1 = \frac{n^2}{m} T_2$
$L_1 = \frac{n^2}{m} L_2$ and $T_1 = \frac{n^4}{m^2} T_2$
$\frac{n^2}{m} L_1 = L_2$ and $\frac{n^4}{m^2} T_1 = T_2$
A ball is spun with angular acceleration $\alpha = 6t^2 - 2t$ where $t$ is in second and $\alpha$ is in $\mathrm{rads}^{-2}$. At $t = 0$, the ball has angular velocity of $10 \, \mathrm{rads}^{-1}$ and angular position of $4 \, \mathrm{rad}$. The most appropriate expression for the angular position of the ball is:
A block of mass 2 kg moving on a horizontal surface with speed of 4 $\mathrm{m\,s^{-1}}$ enters a rough surface ranging from $x = 0.5 \mathrm{\, m}$ to $x = 1.5 \mathrm{\, m}$. The retarding force in this range of rough surface is related to distance by $F = -kx$ where $k = 12 \mathrm{\, Nm^{-1}}$. The speed of the block as it just crosses the rough surface will be:
Zero
1.5 $\mathrm{m\,s^{-1}}$
2.0 $\mathrm{m\,s^{-1}}$
2.5 $\mathrm{m\,s^{-1}}$
Answer: (c)
Solution
Given $a = \frac{-kx}{2} = \frac{-12x}{2} = -6x$. We have $v \frac{dv}{dx} = -6x$. Integrating both sides: $$\int_{4}^{v} v \, dv = -\int_{\frac{1}{2}}^{\frac{3}{2}} 6x \, dx$$ This gives: $$\frac{v^2 - 4^2}{2} = -\frac{6}{2} \left[ \left( \frac{3}{2} \right)^2 - \left( \frac{1}{2} \right)^2 \right]$$ Simplifying, we get: $$v^2 - 16 = -6 \left( \frac{9}{4} - \frac{1}{4} \right)$$ Further simplification gives: $$v^2 = 16 - 6 \times 2 = 4$$ Thus, $V = 2 \, \mathrm{m/s}$.
Question 34
Physics · System of Particles and Rotational Motion · Single correct
A $\sqrt{34}$ $\,$ $\mathrm{m}$ long ladder weighing 10 \, $\mathrm{kg}$ leans on a frictionless wall. Its feet rest on the floor 3 \, $\mathrm{m}$ away from the wall as shown in the figure. If F_f and F_w are the reaction forces of the floor and the wall, then ratio of $\frac{F_w}{F_f}$will be: (Use g = 10 \, $\mathrm{m/s^2}$)
$\frac{6}{\sqrt{110}}$
$\frac{3}{\sqrt{113}}$
$\frac{3}{\sqrt{109}}$
$\frac{2}{\sqrt{109}}$
Answer: (c)
Solution
Given $f = N_2$ and $N_1 = mg$. The equation $N_2 \times \ell \sin \theta = mg \frac{\ell}{2} \cos \theta$ gives $$N_2 = \frac{mg}{2} \cot \theta$$ The ratio of forces is $$\frac{F_w}{F_f} = \frac{\frac{mg}{2} \cot \theta}{\sqrt{(mg)^2 + \left(\frac{mg}{2} \cot \theta\right)^2}}$$ Simplifying, we have $$= \frac{1}{\sqrt{1 + \frac{4}{\cot^2 \theta}}}$$ Finally, $$= \frac{3}{\sqrt{109}}$$
Question 35
Physics · Gravitation · Single correct
Water fall from a 40 m high dam at the rate of $9 \times 10^4 \, \mathrm{kg}$ per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydroelectric energy number of 100W lamps, that can be lit, is: (Take $g = 10 \, \mathrm{ms^{-2}}$)
25
50
100
18
Answer: (b)
Solution
Given $$\frac{9 \times 10^4 \times g \times 40}{3600} \times 0.5 = n \times 100$$. Simplifying, we have: $$\frac{10^4 \times 0.5}{100} = n$$ $$100 \times 0.5 = n$$ $$n = 50$$
Question 36
Physics · Gravitation · Single correct
Two objects of equal masses placed at certain distance from each other attracts each other with a force of $F$. If one-third mass of one object is transferred to the other object, then the new force will be:
$\frac{2}{9} F$
$\frac{16}{9} F$
$\frac{8}{9} F$
$F$
Answer: (c)
Solution
Given the formula for force: $$F = \frac{Gm^2}{r^2}$$ The modified force is: $$F' = \frac{G \left( \frac{4m}{3} \right) \times \left( \frac{2m}{3} \right)}{r^2}$$ Simplifying gives: $$F' = \frac{8}{9} F$$
Question 37
Physics · Mechanical Properties of Fluids · Single correct
A water drop of radius 1 $\mu \mathrm{m}$ falls in a situation where the effect of buoyant force is negligible. Coefficient of viscosity of air is $1.8 \times 10^{-5} \, \mathrm{Nsm^{-2}}$ and its density is negligible as compared to that of water $10^{6} \, \mathrm{gm^{-3}}$. Terminal velocity of the water drop is: (Take acceleration due to gravity $= 10 \, \mathrm{ms^{-2}}$)
145.4 \times 10^{-6}\,\mathrm{m\,s^{-1}}
118.0 \times 10^{-6}\,\mathrm{m\,s^{-1}}
132.6 \times 10^{-6}\,\mathrm{m\,s^{-1}}
123.4 \times 10^{-6}\,\mathrm{m\,s^{-1}}
Answer: (d)
Solution
The force due to viscosity is given by $F_v = 6 \pi \eta r v_t$. The gravitational force is $mg = \frac{4}{3} \pi r^3 \rho g$. Equating the forces, we have: $$6 \pi \eta r v_t = \frac{4}{3} \pi r^3 \rho g$$ Solving for $v_t$, we get: $$v_t = \frac{4}{3} \times \frac{\pi r^3 \rho g}{6 \pi \eta r}$$ Substituting the given values: $$v_t = \frac{4}{3} \times \frac{\pi r^3 \rho g}{6 \pi \eta r} = \frac{2 \times 10^{-12} \times 10^3 \times 10}{9 \times 1.8 \times 10^{-5}}$$ This simplifies to: $$= 123.4 \times 10^{-6} \, \mathrm{m/s}$$
Question 38
Physics · Thermodynamics · Single correct
A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 \, $\mathrm{J}$ of heat during the part AB, no heat during BC and rejects 60 \, $\mathrm{J}$ of heat during CA. A work 50 \, $\mathrm{J}$ is done on the gas during the part BC. The internal energy of the gas at A is 1560 \, $\mathrm{J}$. The work done by the gas during the part CA is:
20 \, $\mathrm{J}$
30 \, $\mathrm{J}$
-30 \, $\mathrm{J}$
-60 \, $\mathrm{J}$
Answer: (b)
Solution
Question 39
Physics · Kinetic Theory · Single correct
What will be the effect on the root mean square velocity of oxygen molecules if the temperature is doubled and oxygen molecule dissociates into atomic oxygen?
Physics · Electric Charges and Fields · Single correct
Two point charges A and B of magnitude $+8 \times 10^{-6} \, \mathrm{C}$ and $-8 \times 10^{-6} \, \mathrm{C}$ respectively are placed at a distance $d$ apart. The electric field at the middle point $O$ between the charges is $6.4 \times 10^{4} \, \mathrm{NC}^{-1}$. The distance ‘$d$’ between the point charges A and B is:
Resistance of the wire is measured as $2\, \Omega$ and $3\, \Omega$ at $10^\circ \mathrm{C}$ and $30^\circ \mathrm{C}$ respectively. Temperature coefficient of resistance of the material of the wire is:
$0.033^\circ \mathrm{C}^{-1}$
$-0.033^\circ \mathrm{C}^{-1}$
$0.011^\circ \mathrm{C}^{-1}$
$0.055^\circ \mathrm{C}^{-1}$
Answer: (a)
Solution
Given the equation for resistance, $R = R_0 (1 + \alpha \Delta T)$. For the first condition, $3 = R_0 \left(1 + \alpha (30 - 0)\right)$. For the second condition, $2 = R_0 \left(1 + \alpha (10 - 0)\right)$. Equating the two conditions, we have: $$\frac{3}{2} = \frac{1 + 30\alpha}{1 + 10\alpha}$$ Solving for $\alpha$, we find: $$\alpha = \frac{1}{30} = 0.033$$
Question 42
Physics · Magnetism and Matter · Single correct
The space inside a straight current carrying solenoid is filled with a magnetic material having magnetic susceptibility equal to $1.2 \times 10^{-5}$. What is fractional increase in the magnetic field inside solenoid with respect to air as medium inside the solenoid?
$1.2 \times 10^{-5}$
$1.2 \times 10^{-3}$
$1.8 \times 10^{-3}$
$2.4 \times 10^{-5}$
Answer: (a)
Solution
Given $\chi = 1.2 \times 10^{-5}$. $\mu_r = 1 + \chi = 1 + 1.2 \times 10^{-5}$. Fractional Change $$\frac{\Delta B}{B} = \frac{\mu_0 \mu_r n i - \mu_0 n i}{\mu_0 n i} = (\mu_r - 1)$$ $$= 1.2 \times 10^{-5}$$
Question 43
Physics · Moving Charges and Magnetism · Single correct
Two parallel, long wires are kept 0.20 m apart in vacuum, each carrying current of $x$ A in the same direction. If the force of attraction per meter of each wire is $2 \times 10^{-6} \, \mathrm{N}$, then the value of $x$ is approximately:
Physics · Moving Charges and Magnetism · Single correct
A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be: (Assume the coil to be short circuited.)
An EM wave propagating in x-direction has a wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60 $\,$ $\mathrm{Vm}$^{-1}. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum:
$E_y = 60 \sin \left[ \frac{\pi}{4} \times 10^3 \left( x - 3 \times 10^8 t \right) \right] \hat{\jmath} \, \mathrm{Vm}^{-1}$, $B_z = 2 \sin \left[ \frac{\pi}{4} \times 10^3 \left( x - 3 \times 10^8 t \right) \right] \hat{k} \, \mathrm{T}$
Given $B_0 = \frac{E_0}{c} = \frac{60}{3 \times 10^8} = 2 \times 10^{-7} \, \mathrm{T}$. $\hat{\mathbf{E}} \times \hat{\mathbf{B}}$ must be the direction of propagation. So, $\hat{\mathbf{B}} \rightarrow z-axis$. $$\mathbf{k} = \frac{2\pi}{\lambda} = \frac{\pi}{4} \times 10^3 \, \mathrm{m}^{-1}$$ $$E_y = 60 \sin \left[ \frac{\pi}{4} \times 10^3 \left( x - 3 \times 10^8 t \right) \right] \hat{\mathbf{j}} \, \mathrm{V/m}$$ $$B_z = 2 \times 10^{-7} \sin \left[ \frac{\pi}{4} \times 10^3 \left( x - 3 \times 10^8 t \right) \right] \hat{\mathbf{k}} \, \mathrm{T}$$
Question 46
Physics · Wave Optics · Single correct
In young's double slit experiment performed using a monochromatic light of wavelength $\lambda$, when a glass plate $(\mu = 1.5)$ of thickness $x\lambda$ is introduced in the path of the one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of $x$ will be:
3
2
1.5
0.5
Answer: (b)
Solution
Path difference at O = $(\mu - 1)t$. If the intensity at O remains (maximum) unchanged, path difference must be $n \lambda$. $$\Rightarrow (\mu - 1)t = n \lambda$$ $$(1.5 - 1)x \lambda = n \lambda$$ $$\Rightarrow x = 2n$$ For $n = 1$, $x = 2$
Question 47
Physics · Dual Nature of Radiation and Matter · Single correct
Let $K_1$ and $K_2$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $\lambda_1$ and $\lambda_2$, respectively are incident on a metallic surface. If $\lambda_1 = 3\lambda_2$ then:
$K_1 > \frac{K_2}{3}$
$K_1 < \frac{K_2}{3}$
$K_1 = \frac{K_2}{3}$
$K_2 = \frac{K_1}{3}$
Answer: (b)
Solution
Given the equations: $$\frac{hc}{\lambda_1} - \phi = K_1$$ $$\frac{hc}{\lambda_2} - \phi = K_2$$ We know that $\lambda_1 = 3\lambda_2$. Substituting, we have: $$3K_1 = \frac{hc}{\lambda_2} - 3\phi$$ $$3K_1 = K_2 - 2\phi$$ Thus, $$3K_1 < K_2$$ Therefore, $$K_1 < \frac{K_2}{3}$$
Question 48
Physics · Nuclei · Single correct
Following statements related to radioactivity are given below: (A) Radioactivity is a random and spontaneous process and is dependent on physical and chemical conditions. (B) The number of un-decayed nuclei in the radioactive sample decays exponentially with time. ($C$) Slope of the graph of $\log_e$(no. of undecayed nuclei) Vs. time represents the reciprocal of mean life time ($\tau$). (D) Product of decay constant ($\lambda$) and half-life time ($T_{1/2}$) is not constant. Choose the most appropriate answer from the options given below:
(A) and (B) only
(B) and (D) only
(B) and ($C$) only
($C$) and (D) only
Answer: (c)
Solution
c
Question 49
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
In the given circuit the input voltage $V_{in}$ is shown in figure. The cut-in voltage of p–n junction diode ($D_1$ or $D_2$) is $0.6 \, \mathrm{V}$. Which of the following output voltage ($V_0$) waveform across the diode is correct?
Answer: (d)
Solution
In +ve half cycle $D_1 \rightarrow F.B.; D_2 \rightarrow R.B.$ $0 - 0.6 \, V$ $V_{out}$ same as $V_{in}$ In -ve half cycle $D_2 \rightarrow F.B.; D_1 \rightarrow R.B.$
Question 50
Physics · Communication Systems · Single correct
Amplitude modulated wave is represented by $$V_{AM} = 10 \left[ 1 + 0.4 \cos \left( 2 \pi \times 10^4 t \right) \right] \cos \left( 2 \pi \times 10^7 t \right).$$ The total bandwidth of the amplitude modulated wave is :
A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22 mm, 1.23 mm, 1.19 mm and 1.20 mm. The percentage error is $\frac{x}{121}\%$. The value of $x$ is
Answer: 150
Solution
Given $$X = \frac{1.22 \, \mathrm{mm} + 1.23 \, \mathrm{mm} + 1.19 \, \mathrm{mm} + 1.20 \, \mathrm{mm}}{4}$$ We find $$X = 1.21 \, \mathrm{mm}$$ The uncertainty is $$\Delta x = \frac{0.01 + 0.02 + 0.02 + 0.01}{4} = \frac{0.06}{4} = 0.015$$ The percentage error is $$Percentage error = \frac{0.015}{1.21} \times 100$$ Finally, $$X = 150$$
Question 52
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Fill in the blank
A Zener of breakdown voltage $V_Z = 8 \, \mathrm{V}$ and maximum zener current, $I_{ZM} = 10 \, \mathrm{mA}$ is subjected to an input voltage $V_i = 10 \, \mathrm{V}$ with series resistance $R = 100 \, \Omega$. In the given circuit $R_L$ represents the variable load resistance. The ratio of maximum and minimum value of $R_L$ is ________
Answer: 2
Solution
Given $I = \frac{2}{100} = 20 \, \mathrm{mA}$. The voltage across the load is $V_L = I_L R_L$. Therefore, $8 = 10 \times 10^{-3} \times R_{L_{max}}$. Solving for $R_{L_{max}}$ gives: $$\frac{4}{5} \times 10^3 = R_{L_{max}}$$ $$800 = R_{L_{max}}$$ For the minimum load resistance, $I = I_Z + I_L$ where $I_L = 10 \, \mathrm{mA}$. If $I_Z = 0$, then $I_{L_{max}} = 20 \, \mathrm{mA}$. The voltage across the load is $V_L = I_{L_{max}} \times R_{L_{min}}$. Therefore, $$\frac{8}{20} \times 10^3 = R_{L_{min}}$$ $$400 = R_{L_{min}}$$ The ratio of maximum to minimum load resistance is: $$\frac{R_{L_{max}}}{R_{L_{min}}} = \frac{800}{400} = 2$$
Question 53
Physics · Wave Optics · Numerical
In a Young's double slit experiment, an angular width of the fringe is $0.35^\circ$ on a screen placed at $2 \, \mathrm{m}$ away for particular wavelength of $450 \, \mathrm{nm}$. The angular width of the fringe, when whole system is immersed in a medium of refractive index $7/5$, is $\frac{1}{\alpha}$. The value of $\alpha$ is
In the given circuit, the magnitude of $V_L$ and $V_C$ are twice that of $V_R$. Given that $f = 50 \, \mathrm{Hz}$, the inductance of the coil is $\frac{1}{K \pi} \, \mathrm{mH}$. The value of $K$ is ______.
Answer: 0
Solution
Given $V_L = V_C = 2V_R$ and $X_L = X_C = 2R$. We have $X_L = 10 \, \Omega$. Therefore, $\omega L = 10$ and $2 \pi f L = 10$. Solving for $L$, we get $$L = \frac{10}{2 \pi f} = \frac{1}{10 \pi} \, \mathrm{H} = \frac{1000}{10 \pi} \, \mathrm{mH}.$$ Further simplifying, $$L = \frac{1}{100 \pi}; K = \frac{1}{100} = 0.01 \approx 0.$$
Question 55
Physics · Current Electricity · Numerical
All resistances in figure are $1\,\Omega$ each. The value of current ‘I’ is $\frac{a}{5}$ A. The value of $a$ is
Answer: 8
Solution
The equivalent resistance is given by $$R_{eq} = \frac{15R}{8} = \frac{15}{8} \, \Omega$$ The current is calculated as $$I = \frac{3}{\frac{15}{8}} = \frac{8}{5} \, A$$ Therefore, $a = 8$.
Question 56
Physics · Electrostatic Potential and Capacitance · Numerical
A capacitor $C_1$ of capacitance $5\,\mu\mathrm{F}$ is charged to a potential of $30\,\mathrm{V}$ using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor $C_2$ of capacitance $10\,\mu\mathrm{F}$ as shown in figure. When the switch is closed charge flows between the capacitors. At equilibrium, the charge on the capacitor $C_2$ is $\mu\mathrm{C}$.
Answer: 100
Solution
Before closing the switch $Q = C_1 V_0 = 5 \times 30 = 150 \, \mu \mathrm{C}$ After closing the switch $$V = \frac{Q}{C_1 + C_2} = \frac{150}{10 + 5} = 10 \, \mathrm{V}$$ $Q_2 = C_2 V = 10 \times 10 = 100 \, \mu \mathrm{C}$
Question 57
Physics · Waves · Fill in the blank
A tuning fork of frequency 340 Hz resonates in the fundamental mode with an air column of length 125 cm in a cylindrical tube closed at one end. When water is slowly poured in it, the minimum height of water required for observing resonance once again is $\underline{\hspace{1cm}}$ cm. (Velocity of sound in air is 340 ms$^{-1}$)
Answer: 50
Solution
Assumption: Ignore word "fundamental mode" in question. $$\lambda = \frac{V}{f} = \frac{340}{340} = 1 \, \mathrm{m}$$ First resonating length = $$\frac{\lambda}{4} = 25 \, \mathrm{cm}$$ Second resonating length = $$\frac{3\lambda}{4} = 75 \, \mathrm{cm}$$ Third resonating length = $$\frac{5\lambda}{4} = 125 \, \mathrm{cm}$$ Height of water required = $$125 - 75 = 50 \, \mathrm{cm}$$
Question 58
Physics · Mechanical Properties of Fluids · Numerical
A liquid of density $750 \, \mathrm{kg/m^3}$ flows smoothly through a horizontal pipe that tapers in cross-sectional area from $A_1 = 1.2 \times 10^{-2} \, \mathrm{m^2}$ to $A_2 = \frac{A_1}{2}$. The pressure difference between the wide and narrow sections of the pipe is $4500 \, \mathrm{Pa}$. The rate of flow of liquid is ______ $\times 10^{-3} \, \mathrm{m^3/s}$.
Answer: 24
Solution
Given $A_2 = \frac{A_1}{2}$. The pressure difference is $P_1 - P_2 = 4500 \, \mathrm{Pa}$. Using Bernoulli's equation: $$P_1 + \frac{1}{2} \rho V_1^2 + \rho gh = P_2 + \frac{1}{2} \rho V_2^2 + \rho gh$$ $$P_1 - P_2 = \frac{1}{2} \rho (V_2^2 - V_1^2) ...(1)$$ And $A_1 V_1 = A_2 V_2$ $$\Rightarrow V_2 = 2V_1 ...(2)$$ Substituting into equation (1): $$4500 = \frac{1}{2} \times 750 \times 3V_1^2$$ Solving for $V_1$: $$V_1 = 2 \, \mathrm{m/s}$$ The volume flow rate is $A_1 V_1 = 24 \times 10^{-3} \, \mathrm{m^3 \, s^{-1}}$
Question 59
Physics · System of Particles and Rotational Motion · Fill in the blank
A uniform disc with mass $M = 4 \, \mathrm{kg}$ and radius $R = 10 \, \mathrm{cm}$ is mounted on a fixed horizontal axle as shown in figure. A block with mass $m = 2 \, \mathrm{kg}$ hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is _______N. (Take $g = 10 \, \mathrm{ms^{-2}}$)
Answer: 10
Solution
Given the equations: $$2g - T = 2a ...(1)$$ $$TR = \frac{MR^2}{2} \alpha ...(2)$$ $$\alpha = \frac{a}{R} ...(3)$$ From equation (1), we have: $$T = 2a$$ Substituting back into equation (1): $$2g - T = 2a$$ Solving for $T$, we get: $$T = g = 10 \, \mathrm{N}$$
Question 60
Physics · Motion in a Straight Line · Numerical
A car covers $AB$ distance with first one–third at velocity $v_1 \, \mathrm{ms^{-1}}$, second one–third at $v_2 \, \mathrm{ms^{-1}}$ and last one–third at $v_3 \, \mathrm{ms^{-1}}$. If $v_3 = 3v_1$, $v_2 = 2v_1$ and $v_1 = 11 \, \mathrm{ms^{-1}}$ then the average velocity of the car is ________ $\mathrm{ms^{-1}}$.
Answer: 18
Solution
The average velocity is given by the formula: $$ \vec{v} = \frac{Displacement}{time} $$ Let the displacement be $l$. The expression becomes: $$ = \frac{l}{\left( \frac{l}{V_3} + \frac{l}{V_2} + \frac{l}{V_1} \right) \frac{1}{3}} $$ Simplifying further: $$ = \frac{3}{\frac{1}{V_1} + \frac{1}{V_2} + \frac{1}{V_3}} = \frac{3}{\frac{1}{11} + \frac{1}{22} + \frac{1}{33}} $$ The result is: $$ = 18 \, m/s $$
Chemistry
Question 61
Chemistry · Some Basic Concepts of Chemistry · Single correct
Compound A contains 8.7$\%$ Hydrogen, 74$\%$ Carbon and 17.3$\%$ Nitrogen. The molecular formula of the compound is, Given : Atomic masses of C, H and N are 12, 1 and 14 amu respectively. The molar mass of the compound A is 162 $\mathrm{g}$ $\mathrm{mol^{-1}}$.
Consider the following statements : (A) The principal quantum number ‘n’ is a positive integer with values of ‘n’ = 1, 2, 3, …. (B) The azimuthal quantum number ‘l’ for a given ‘n’ (principal quantum number) can have values as ‘l’ = 0, 1, 2, …. n $(C)$ Magnetic orbital quantum number ‘m’ for a particular ‘l’ (azimuthal quantum number) has (2l + 1) values. (D) ±1/2 are the two possible orientations of electron spin. (E) For l = 5, there will be a total of 9 orbital. Which of the above statements are correct?
$(A), (B)$ and $(C)$
$(A), (C), (D)$ and $(E)$
$(A), (C)$ and $(D)$
$(A), (B), (C)$ and $(D)$
Answer: (c)
Solution
Number of values of $n = 1, 2, 3, \ldots, \infty$. Number of values of $\ell = 0$ to $(n-1)$. Number of values of $m = -\ell$ to $+\ell$. Total values $= 2\ell + 1$. Values of spin $= \pm \frac{1}{2}$. For $\ell = 5$, number of orbitals $= 2\ell + 1 = 11$.
Question 63
Chemistry · Chemical Bonding and Molecular Structure · Single correct
In the structure of $\mathrm{SF}_4$, the lone pair of electrons on S is in.
equatorial position and there are two lone pair-bond pair repulsions at $90^\circ$
equatorial position and there are three lone pair-bond pair repulsions at $90^\circ$
axial position and there are three lone pair-bond pair repulsion at $90^\circ$
axial position and there are two lone pair-bond pair repulsion at $90^\circ$
Answer: (a)
Solution
The molecular geometry is see-saw with hybridization $sp^3d$.
Question 64
Chemistry · Equilibrium · Single correct
A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of $\frac{[\mathrm{CH}_3\mathrm{CH}_2\mathrm{COO}^-]}{[\mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH}]}$ required to make buffer is .......... Given : $K_a(\mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH}) = 1.3 \times 10^{-5}$
Chemistry · Co-ordination Compounds · Single correct
Match List-I with List-II. Choose the correct answer from the options given below :
$(A)- (II), (B) - (III), (C) - (IV), (D) - (I)$
$(A)- (II), (B) - (I), (C) - (III), (D) - (IV)$
$(A)- (II), (B) - (III), (C) - (I), (D) - (IV)$
$(A)- (I), (B) - (III), (C) - (II), (D) - (IV)$
Answer: (c)
Solution
Negative charged sol = CdS (II) Macromolecular colloid = starch (III) Positively charged sol = $Fe_2O_3.xH_2O$ (I) Cheese = gel (IV)
Question 66
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Match List-I with List-II. \begin{tabular}{|c|c|} \hline List-I (Oxide) & List-II (Nature) \\ \hline (A) $\mathrm{Cl_2O_7}$ & (I) Amphoteric \\ \hline (B) $\mathrm{Na_2O}$ & (II) Basic \\ \hline (C) $\mathrm{Al_2O_3}$ & (III) Neutral \\ \hline (D) $\mathrm{N_2O}$ & (IV) Acidic \\ \hline \end{tabular} Choose the correct answer from the options given below:
(A) - (IV), (B) - (III), (C) - (I), (D) - (II)
(A) - (IV), (B) - (II), (C) - (I), (D) - (III)
(A) - (II), (B) - (IV), (C) - (III), (D) - (I)
(A) - (I), (B) - (II), (C) - (III), (D) - (IV)
Answer: (b)
Solution
The properties of the given oxides are as follows: $\mathrm{Cr_2O_7}$ is acidic, $\mathrm{Na_2O}$ is basic, $\mathrm{Al_2O_3}$ is amphoteric, and $\mathrm{N_2O}$ is neutral.
Question 67
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
In the metallurgical extraction of copper, following reaction is used: $$\mathrm{FeO} + \mathrm{SiO}_2 \rightarrow \mathrm{FeSiO}_3$$ FeO and FeSiO$_3$ respectively are.
gangue and flux
flux and slag
slag and flux
gangue and slag
Answer: (d)
Solution
FeO = Gangue , FeSiO_3 = Slag
Question 68
Chemistry · Hydrogen · Single correct
Hydrogen has three isotopes : protium ($^1\mathrm{H}$), deuterium ($^2\mathrm{H}$ or D) and tritium ($^3\mathrm{H}$ or T). They have nearly same chemical properties but different physical properties. They differ in
number of protons
atomic number
electronic configuration
atomic mass
Answer: (d)
Solution
They have different neutrons and mass number.
Question 69
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Among the given oxides of nitrogen; $\mathrm{N_2O}$, $\mathrm{N_2O_3}$, $\mathrm{N_2O_4}$ and $\mathrm{N_2O_5}$, the number of compound(s) having $\mathrm{N{-}N}$ bond is :
1
2
3
4
Answer: (c)
Solution
Question 71
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Which of the following oxoacids of sulphur contains "S" in two different oxidation states?
H_2S_2O_3
H_2S_2O_6
H_2S_2O_7
H_2S_2O_8
Answer: (a)
Solution
Question 72
Chemistry · Environmental Chemistry · Single correct
Correct statement about photo-chemical smog is:
It occurs in humid climate.
It is a mixture of smoke, fog and $\mathrm{SO}_2$
It is reducing smog.
It results from reaction of unsaturated hydrocarbons.
Answer: (d)
Solution
Photochemical smog results from the action of sunlight on unsaturated hydrocarbons and nitrogen oxide.
Question 73
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The correct IUPAC name of the following compound is :
4-methyl-2-nitro-5-oxohept-3-enal
4-methyl-5-oxo-2-nitrohept-3-enal
4-methyl-6-nitro-3-oxohept-4-enal
6-formyl-4-methyl-2-nitrohex-3-enal
Answer: (c)
Solution
4-Methyl-6-nitro-3-oxohept-4-enal
Question 74
Chemistry · Alcohols, Phenols and Ethers · Single correct
The major product (P) of the given reaction is (where, Me is $-CH_3$)
Answer: (c)
Solution
The reaction begins with the protonation of the alcohol group, forming a good leaving group. This leads to the formation of a carbocation after the departure of water. A 1,2-methyl shift occurs to stabilize the carbocation, resulting in a more stable carbocation. Finally, deprotonation leads to the formation of the final product, which is an alkene.
Question 75
Chemistry · Haloalkanes and Haloarenes · Single correct
A $\xrightarrow{(i) \mathrm{Cl_2, \Delta} (ii) \mathrm{CN^-} (iii) \mathrm{H_2O/H^+}}$ 4-Bromophenyl acetic acid. In the above reaction 'A' is
Answer: (c)
Solution
The reaction sequence starts with the bromination of toluene using $\mathrm{Cl_2/\Delta}$ to form benzyl chloride. This is followed by a nucleophilic substitution reaction with $\mathrm{CN^-}$ to form benzyl cyanide. Finally, hydrolysis with $\mathrm{H_3O^+}$ converts the cyanide group to a carboxylic acid, resulting in the formation of phenylacetic acid.
Question 76
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Isobutyraldehyde on reaction with formaldehyde and $K_2CO_3$ gives compound ‘A’. Compound ‘A’ reacts with KCN and yields compound ‘B’, which on hydrolysis gives a stable compound ‘C’. The compound ‘C’ is :
Answer: (c)
Solution
Question 77
Chemistry · Amines · Single correct
With respect to the following reaction, consider the given statements:
o-Nitroaniline and p-nitroaniline are the predominant products
p-Nitroaniline and m-nitroaniline are the predominant products
HNO$_3$ acts as an acid
H$_2$SO$_4$ acts as an acid
Answer: (c)
Solution
Question 78
Chemistry · Polymers · Single correct
Given below are two statements, one is Assertion (A) and other is Reason ( R). Assertion (A): Natural rubber is a linear polymer of isoprene called cis-polyisoprene with elastic properties. Reason ( R): The cis-polyisoprene molecules consist of various chains held together by strong polar interactions with coiled structure. In the light of the above statements, choose the correct one from the options given below:
Both (A) and (R ) are true and ( R) is the correct explanation of (A)
Both (A) and (R ) are true but (R ) is not the correct explanation of (A).
is true but (R ) is false.
is false but (R ) is true.
Answer: (c)
Solution
Natural rubber is a linear polymer of isoprene (2-methyl-1,3-butadiene) and is also called cis-1,4-polyisoprene. The cis-polyisoprene molecules consist of various chains held together by weak Van der Waal's interactions and has a coiled structure.
Question 79
Chemistry · Biomolecules · Single correct
When sugar 'X' is boiled with dilute $\mathrm{H_2SO_4}$ in alcoholic solution, two isomers 'A' and 'B' are formed. 'A' on oxidation with $\mathrm{HNO_3}$ yields saccharic acid where as 'B' is laevorotatory. The compound 'X' is :
Maltose
Sucrose
Lactose
Strach
Answer: (b)
Solution
Question 80
Chemistry · Chemistry in Everyday Life · Single correct
The drug tegamet is :
Answer: (c)
Solution
Tegamet is the brand name of Cimetidine.
Question 81
Chemistry · States of Matter · Numerical
100 $\mathrm{g}$ of an ideal gas is kept in a cylinder of 416 $\mathrm{L}$ volume at $27^\circ$ $\mathrm{C}$ under 1.5 $\mathrm{bar}$ pressure. The molar mass of the gas is _____ $\mathrm{g}$ \, $\mathrm{mol^{-1}}$. (Nearest integer) (Given : R = 0.083 $\mathrm{L}$ \, $\mathrm{bar}$ \, $\mathrm{K^{-1}}$ \, $mol^{-1}$)
Answer: 4
Solution
Given the equation: $$1.5 \times 416 = \frac{100}{M} \times 0.083 \times 300$$ Solving for $M$ gives: $$M = 3.99$$
Question 82
Chemistry · Thermodynamics · Numerical
For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, $\Delta_C$ $H^\Theta$ = -601.70 \, $\mathrm{kJ}$ \, $\mathrm{mol}^{-1}$, the magnitude of change in internal energy for the reaction is _____ kJ. (Nearest integer) (Given : R = 8.3 J $K^{-1}$ $mol^{-1}$)
Answer: 600
Solution
The reaction is given by: $$\mathrm{Mg(s) + \frac{1}{2}O_2(g) \rightarrow MgO(s)}$$ The relation is: $$\Delta H = \Delta U + \Delta n_g RT$$ Substituting the values: $$-601.70 \times 10^3 = \Delta U - \frac{1}{2} \times 8.3 \times 300$$ Simplifying: $$-601.70 \, \mathrm{kJ} = \Delta U - 1.245 \, \mathrm{kJ}$$ Therefore: $$\Delta U = -600.455 \, \mathrm{kJ}$$ The answer is 600.
Question 83
Chemistry · Solutions · Numerical
2.5 $\mathrm{g}$ of protein containing only glycine ($\mathrm{C_2H_5NO_2}$) is dissolved in water to make 500 $\mathrm{mL}$ of solution. The osmotic pressure of this solution at 300 $\mathrm{K}$ is found to be 5.03 $\times 10^{-3}$ $\mathrm{bar}$. The total number of glycine units present in the protein is (Given: R=0.083 L bar $k^{-1}$ $mol^{-1}$)
Answer: 330
Solution
Given $\pi = CRT$. $$5.03 \times 10^{-3} = C \times 0.083 \times 300$$ $$C = 0.202 \times 10^{-3} \, \mathrm{M}$$ Moles of protein $= 0.202 \times 10^{-3} \times 0.5$ $$= 10^{-4} \times 1.01$$ $$1.01 \times 10^{-4} = \frac{2.5}{M}$$ M (molar mass of protein) $= 24752$ Therefore, the number of glycine units $= \frac{24752}{75} = 330.03$
Question 84
Chemistry · Electrochemistry · Numerical
For the given reactions $$\mathrm{Sn^{2+} + 2e^- \rightarrow Sn}$$ $$\mathrm{Sn^{4+} + 4e^- \rightarrow Sn}$$ The electrode potentials are; $E^\circ_{\mathrm{Sn^{2+}/Sn}} = -0.140 \, \mathrm{V}$ and $E^\circ_{\mathrm{Sn^{4+}/Sn}} = 0.010 \, \mathrm{V}$. The magnitude of standard electrode potential for $\mathrm{Sn^{4+}/Sn^{2+}}$ i.e. $E^\circ_{\mathrm{Sn^{4+}/Sn^{2+}}}$ is ______ $\times \, 10^{-2} \, \mathrm{V}$. (Nearest integer)
Answer: 16
Solution
For the reaction $\mathrm{Sn^{2+} + 2e^- \rightarrow Sn}$, $\Delta G_1^0 = +2 \times 0.140 \times \mathrm{F}$. For the reaction $\mathrm{Sn^{4+} + 4e^- \rightarrow Sn}$, $\Delta G_2^0 = -4 \times 0.01 \times \mathrm{F}$. For the reaction $\mathrm{Sn^{4+} + 2e^- \rightarrow Sn^{2+}}$, $\Delta G_3^0 = -2 \times E^0_{\mathrm{Sn^{4+}/Sn^{2+}}} \times \mathrm{F}$. We have $\Delta G_3^0 = \Delta G_2^0 - \Delta G_1^0$. Therefore, $-2 \times E^0 \times \mathrm{F} = -(0.04 + 0.28) \times \mathrm{F}$. Solving for $E^0$, we get $E^0 = 0.16 volt = 16 \times 10^{-2} V$.
Question 85
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is _______. (Nearest integer) (Given : antilog 0.125 = 1.333, antilog 0.693 = 4.93)
Answer: 75
Solution
Given the equation: $$t = \frac{t_{1/2}}{0.3} \log \frac{[A]_0}{[A]_t}$$ Substitute the values: $$83 = \frac{200}{0.3} \log \frac{[A]_0}{[A]_t}$$ Solve for the logarithm: $$0.125 = \log \frac{[A]_0}{[A]_t}$$ Convert the logarithm to a fraction: $$\frac{[A]_0}{[A]_t} = 1.333 \approx \frac{4}{3}$$ Calculate the percentage: $$\therefore \frac{[A]_t}{[A]_0} \times 100 = \frac{3}{4} \times 100 = 75\%$$
Question 86
Chemistry · Co-ordination Compounds · Numerical
[$\mathrm{Fe(CN)}_6]^{4-}$ [$\mathrm{Fe(CN)}_6]^{3-}$ [$\mathrm{Ti(CN)}_6]^{3-}$ [$\mathrm{Ni(CN)}_4]^{2-}$ [$\mathrm{Co(CN)}_6]^{3-}$ Among the given complexes, number of paramagnetic complexes is .
Answer: 2
Solution
The complex $[\mathrm{Fe(CN)_6}]^{4-}$ is diamagnetic. The complex $[\mathrm{Fe(CN)_6}]^{3-}$ is paramagnetic with 1 unpaired electron. The complex $[\mathrm{Ti(CN)_6}]^{3-}$ is paramagnetic with 1 unpaired electron. The complex $[\mathrm{Ni(CN)_4}]^{2-}$ is diamagnetic. The complex $[\mathrm{Co(CN)_6}]^{3-}$ is diamagnetic.
Question 87
Chemistry · Co-ordination Compounds · Numerical
Number of complex(es) which will exist in cis-trans is/are
Chemistry · Some Basic Concepts of Chemistry · Numerical
The complete combustion of $0.492\ \mathrm{g}$ of an organic compound containing 'C', 'H' and 'O' gives $0.793\ \mathrm{g}$ of $\mathrm{CO_2}$ and $0.442\ \mathrm{g}$ of $\mathrm{H_2O}$. The percentage of oxygen composition in the organic compound is ________. (nearest integer)
Answer: 46
Solution
Mole of CO$_2$ = Moles of C = $\frac{0.793}{44}$ Weight of 'C' = $\frac{0.793}{44} \times 12 = 0.216$ gm Moles of 'H' = $\frac{0.442}{18} \times 2$ Weight of 'H' = $\frac{0.442}{18} \times 2 \times 1 = 0.049$ gm Therefore, Weight of 'O' = $0.492 - 0.216 - 0.049 = 0.227$ gm % of 'O' = $\frac{0.227}{0.492} \times 100 = 46.13\%$
Question 89
Chemistry · Alcohols, Phenols and Ethers · Numerical
The major product of the following reaction contains bromine atom(s).
Answer: 1
Solution
No. of Br atoms = 1
Question 90
Chemistry · Redox Reactions · Numerical
$0.01\,\mathrm{M}$ $\mathrm{KMnO_4}$ solution was added to $20.0\,\mathrm{mL}$ of $0.05\,\mathrm{M}$ Mohr's salt solution through a burette. The initial reading of $50\,\mathrm{mL}$ burette is zero. The volume of $\mathrm{KMnO_4}$ solution left in the burette after the end point is \_\_\_\_ $\mathrm{mL}$. (nearest integer)