JEE Main 16 March 2021 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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Maths · Sets · Single correct
The number of elements in the set $$\{ x \in \mathbb{R} : (|x| - 3)|x + 4| = 6 \}$$ is equal to
Answer: (b)
Given $x \neq -4$, the equation is $(|x| - 3)(|x + 4|) = 6$. This implies $|x| - 3 = \frac{6}{|x+4|}$. The graph shows the functions $y = \frac{6}{|x+4|}$ and $y = |x| - 3$. The number of solutions is $2$.
Maths · Vector Algebra · Single correct
Let a vector $\alpha \hat{i} + \beta \hat{j}$ be obtained by rotating the vector $\sqrt{3} \hat{i} + \hat{j}$ by an angle $45^\circ$ about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices $(\alpha, \beta), (0, \beta)$ and $(0,0)$ is equal to
Answer: (a)
Area of $\triangle (OA'B) = \frac{1}{2} OA' \cos 15^\circ \times OA' \sin 15^\circ$ $$= \frac{1}{2} (OA')^2 \frac{\sin 30^\circ}{2}$$ $$= (3 + 1) \times \frac{1}{8} = \frac{1}{2}$$
Maths · Three Dimensional Geometry · Single correct
If for $a > 0$, the feet of perpendiculars from the points $A(a, -2a, 3)$ and $B(0, 4, 5)$ on the plane $lx + my + nz = 0$ are points $C(0, -a, -1)$ and $D$ respectively, then the length of line segment $CD$ is equal to:
Answer: (d)
$C$ lies on plane $\Rightarrow -ma-n=0$ $\Rightarrow \dfrac{m}{n}=-\dfrac{1}{a} \qquad \ldots (1)$ $\overrightarrow{CA}\parallel \hat{i}+m\hat{j}+n\hat{k}$ $\dfrac{a-0}{1}=\dfrac{-a}{m}=\dfrac{4}{n}$ $\Rightarrow \dfrac{m}{n}=-\dfrac{a}{4} \qquad \ldots (2)$ From (1) \& (2) $-\dfrac{1}{a}=-\dfrac{a}{4}$ $\Rightarrow a^2=4$ $\Rightarrow a=2 \qquad (\text{since } a>0)$ From (2) $\dfrac{m}{n}=-\dfrac{1}{2}$ Let $m=-t \Rightarrow n=2t$ $\dfrac{2}{1}=\dfrac{-2}{-t}$ $\Rightarrow t=1$ So plane : $t(x-y+2z)=0$ $BD=\dfrac{6}{\sqrt{6}}=\sqrt{6}$ $C\equiv(0,-2,-1)$ $CD=\sqrt{BC^2-BD^2}$ $=\sqrt{(0^2+6^2+6^2)-(\sqrt{6})^2}$ $=\sqrt{66}$
Maths · Statistics · Single correct
Consider three observations $a$, $b$ and $c$ such that $b = a + c$. If the standard deviation of $a + 2$, $b + 2$, $c + 2$ is $d$, then which of the following is true?
Answer: (d)
For $a, b, c$ mean $= \frac{a+b+c}{3} \left( = \bar{x} \right)$ $b = a + c$ $$\Rightarrow \bar{x} = \frac{2b}{3}$$ S.D. $(a+2, b+2, c+2) = S.D. \cdot (a, b, c) = d$ $$\Rightarrow d^2 = \frac{a^2 + b^2 + c^2}{3} - \left( \bar{x} \right)^2$$ $$\Rightarrow d^2 = \frac{a^2 + b^2 + c^2}{3} - \frac{4b^2}{9}$$ $$\Rightarrow 9 \, d^2 = 3 \left( a^2 + b^2 + c^2 \right) - 4b^2$$ $$\Rightarrow b^2 = 3 \left( a^2 + c^2 \right) - 9d^2$$
Maths · Trigonometric Functions · Single correct
If for $x \in \left(0, \frac{\pi}{2}\right)$, $\log_{10} \sin x + \log_{10} \cos x = -1$ and $\log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - 1)$, $n > 0$ then the value of $n$ is equal to:
Answer: (b)
Given $x \in \left(0, \frac{\pi}{2}\right)$. $$\log_{10} \sin x + \log_{10} \cos x = -1$$ This implies: $$\log_{10} \sin x \cdot \cos x = -1$$ Therefore: $$\sin x \cdot \cos x = \frac{1}{10}$$ Now, consider: $$\log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - 1)$$ This implies: $$\sin x + \cos x = 10^{\left(\log_{10} \sqrt{n - \frac{1}{2}}\right)} = \sqrt{\frac{n}{10}}$$ By squaring: $$1 + 2 \sin x \cdot \cos x = \frac{n}{10}$$ Thus: $$1 + \frac{1}{5} = \frac{n}{10} \implies n = 12$$
Maths · Determinants · Single correct
Let $A = \begin{bmatrix} i & -i \\ -i & i \end{bmatrix}$, $i = \sqrt{-1}$. Then, the system of linear equations $A^8 \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 64 \end{bmatrix}$ has:
Answer: (c)
Given $$A = \begin{bmatrix} i & -i \\ -i & i \end{bmatrix}$$ We calculate $$A^2 = \begin{bmatrix} -2 & 2 \\ 2 & -2 \end{bmatrix} = 2 \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}$$ Then $$A^4 = 2^2 \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = 8 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$ Next $$A^8 = 64 \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = 128 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$ We have $$A^8 \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 64 \end{bmatrix}$$ This implies $$128 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 64 \end{bmatrix}$$ Therefore $$128 \begin{bmatrix} x - y \\ -x + y \end{bmatrix} = \begin{bmatrix} 8 \\ 64 \end{bmatrix}$$ Thus $$x - y = \frac{1}{16}$$ and $$-x + y = \frac{1}{2}$$
Maths · Conic Sections · Single correct
If the three normals drawn to the parabola, $y^2 = 2x$ pass through the point $(a, 0), a \neq 0$, then 'a' must be greater than :
Answer: (d)
For a standard parabola, for more than 3 normals (on axis), $$x > \frac{L}{2}$$ where $L$ is the length of L.R. For $y^2 = 2x$, L.R. = 2. For $(a, 0)$, $$a > \frac{L.R.}{2} \implies a > 1$$
Maths · Three Dimensional Geometry · Single correct
Let the position vectors of two points P and Q be $3\hat{i} - \hat{j} + 2\hat{k}$ and $\hat{i} + 2\hat{j} - 4\hat{k}$, respectively. Let R and S be two points such that the direction ratios of lines PR and QS are $(4, -1, 2)$ and $(-2, 1, -2)$, respectively. Let lines PR and QS intersect at T. If the vector $\overrightarrow{TA}$ is perpendicular to both $\overrightarrow{PR}$ and $\overrightarrow{QS}$ and the length of vector $\overrightarrow{TA}$ is $\sqrt{5}$ units, then the modulus of a position vector of A is:
Answer: (b)
$P(3,-1,2)$ $Q(1,2,-4)$ $\overrightarrow{PR}=4\hat{i}-\hat{j}+2\hat{k}$ $\overrightarrow{QS}=-2\hat{i}+\hat{j}-2\hat{k}$ Direction ratios of normal to the plane containing $P,\ T$ and $Q$ will be proportional to $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 4 & -1 & 2\\ -2 & 1 & -2 \end{vmatrix}$ $\frac{\ell}{0}=\frac{m}{4}=\frac{n}{2}$ PT $\frac{x-3}{4}=\frac{y+1}{-1}=\frac{z-2}{2}=\lambda$ $\Rightarrow T=(4\lambda+3,\,-\lambda-1,\;2\lambda+2)$ QT $\frac{x-1}{-2}=\frac{y-1}{1}=\frac{z+4}{-2}=\mu$ $\Rightarrow T=(1-2\mu,\;1+\mu,\;-2\mu-4)$ $\therefore$ $4\lambda+3=1-2\mu$ $-\lambda-1=1+\mu$ $2\lambda+2=-2\mu-4$ $4\lambda+2\mu=-2$ $\lambda+\mu=-2$ $\lambda+\mu=-3$ $\Rightarrow \lambda=2,\ \mu=-5$ So point $T:(11,-3,6)$ $\overrightarrow{OA} =(11\hat{i}-3\hat{j}+6\hat{k}) +\frac{2\hat{j}+\hat{k}}{\sqrt{5}}\sqrt{5}$ $\overrightarrow{OA} =(11\hat{i}-3\hat{j}+6\hat{k}) +(2\hat{j}+\hat{k})$ $\overrightarrow{OA} =11\hat{i}-\hat{j}+7\hat{k}$ or $\overrightarrow{OA} =11\hat{i}-5\hat{j}+5\hat{k}$ $|\overrightarrow{OA}| =\sqrt{121+1+49} =\sqrt{171}$ or $\sqrt{81+25+25} =\sqrt{131}$
Maths · Continuity and Differentiability · Single correct
Let the functions $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ be defined as : $$f(x) = \begin{cases} x + 2, & x < 0 \\ x^2, & x \geq 0 \end{cases} and g(x) = \begin{cases} x^3, & x < 1 \\ 3x - 2, & x \geq 1 \end{cases}$$ Then, the number of points in $\mathbb{R}$ where $(f \circ g)(x)$ is NOT differentiable is
Answer: (b)
Given $$f(g(x)) = \begin{cases} g(x) + 2, & g(x) < 0 \\ (g(x))^2, & g(x) \geq 0 \end{cases}$$ This simplifies to $$= \begin{cases} x^3 + 2, & x < 0 \\ x^6, & x \in [0, 1) \\ (3x - 2)^2, & x \in [1, \infty) \end{cases}$$ The derivative is $$(f \circ g(x))' = \begin{cases} 3x^2, & x < 0 \\ 6x^5, & x \in (0, 1) \\ 2(3x - 2) \times 3, & x \in (1, \infty) \end{cases}$$ At 'O' L.H.L. $\neq$ R.H.L. (Discontinuous) At '1' L.H.D. $= 6 = $ R.H.D. $\Rightarrow \ f \circ g(x)$ is differentiable for $x \in \mathbb{R} - \{0\}$
Maths · Mathematical Reasoning · Single correct
Which of the following Boolean expression is a tautology ?
Answer: (d)
The truth table is given as follows: The expression $(p \land q) \to (p \to q)$ is a tautology.
Maths · Complex Numbers and Quadratic Equations · Single correct
Let a complex number $z$, $|z| \neq 1$, satisfy $\log_{\frac{1}{\sqrt{2}}} \left( \frac{|z| + 11}{(|z| - 1)^2} \right) \leq 2$. Then, the largest value of $|z|$ is equal to
Answer: (b)
Given the inequality: $$\log_{\frac{1}{\sqrt{2}}} \left( \frac{|z| + 11}{(|z| - 1)^2} \right) \leq 2$$ This implies: $$\frac{|z| + 11}{(|z| - 1)^2} \geq \frac{1}{2}$$ Simplifying, we get: $$2|z| + 22 \geq (|z| - 1)^2$$ Expanding the right side: $$2|z| + 22 \geq |z|^2 + 1 - 2|z|$$ Rearranging terms gives: $$|z|^2 - 4|z| - 21 \leq 0$$ Solving the quadratic inequality, we find: $$\Rightarrow |z| \leq 7$$ Therefore, the largest value of $|z|$ is 7.
Maths · Binomial Theorem · Single correct
If n is the number of irrational terms in the expansion of $\left(3^{1/4} + 5^{1/8}\right)^{60}$, then $(n - 1)$ is divisible by:
Answer: (a)
Consider the expression $$\left(3^{1/4} + 5^{1/8}\right)^{60}$$ and expand it using the binomial theorem: $$\binom{60}{r} (3^{1/4})^{60-r} \cdot (5^{1/8})^r.$$ For rational terms, we have $$\frac{r}{8} = k; 0 \leq r \leq 60$$ which implies $$0 \leq 8k \leq 60$$ and $$0 \leq k \leq \frac{60}{8}$$ leading to $$0 \leq k \leq 7.5.$$ Therefore, $$k = 0, 1, 2, 3, 4, 5, 6, 7.$$ The expression $$\frac{60 - 8k}{4}$$ is always divisible by 4 for all values of $k$. Total rational terms = 8. Total terms = 61. Irrational terms = 53. Thus, $$n - 1 = 53 - 1 = 52.$$ 52 is divisible by 26.
Maths · Three Dimensional Geometry · Single correct
Let P be a plane $lx + my + nz = 0$ containing the line, $\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$. If plane P divides the line segment AB joining points $A(-3, -6, 1)$ and $B(2, 4, -3)$ in ratio $k : 1$ then the value of $k$ is equal to:
Answer: (c)
Point $C$ is $\left(\dfrac{2k-3}{k+1},\ \dfrac{4k-6}{k+1},\ \dfrac{-3k+1}{k+1}\right)$ $\dfrac{x-1}{-1}=\dfrac{y+4}{2}=\dfrac{z+2}{3}$ Plane $lx+my+nz=0$ $l(-1)+m(2)+n(3)=0$ $-l+2m+3n=0 \qquad \ldots (1)$ It also satisfies point $(1,-4,-2)$ $l-4m-2n=0 \qquad \ldots (2)$ Solving (1) and (2) $2m+3n=4m+2n$ $\Rightarrow n=2m$ $l-4m-4m=0$ $\Rightarrow l=8m$ $\therefore \frac{l}{8}=\frac{m}{1}=\frac{n}{2}$ $\therefore l:m:n=8:1:2$ Plane is $8x+y+2z=0$ It will satisfy point $C$ $8\left(\dfrac{2k-3}{k+1}\right)+\left(\dfrac{4k-6}{k+1}\right)+2\left(\dfrac{-3k+1}{k+1}\right)=0$ $\Rightarrow 16k-24+4k-6-6k+2=0$ $\Rightarrow 14k-28=0$ $\therefore k=2$
Maths · Applications of Derivatives · Single correct
The range of $a \in \mathbb{R}$ for which the function $$f(x) = (4a - 3) \left(x + \log_e 5\right) + 2(a - 7) \cot\left(\frac{x}{2}\right) \sin^2\left(\frac{x}{2}\right)$$ $x + 2n\pi$, $n \in \mathbb{N}$, has critical points, is
Answer: (b)
Given $f(x) = (4a - 3) \left( x + \log_e 5 \right) + (a - 7) \sin x$. $f(x) = (4a - 3)(1) + (a - 7) \cos x = 0$ This implies $\cos x = \frac{3 - 4a}{a - 7}$. $$\frac{3a + 4}{a - 7} \leq 0$$ $$\frac{3 - 4a}{a - 7} + 1 \geq 0$$ $$\frac{3 - 4a + a - 7}{a - 7} \geq 0$$ $$\frac{-3a - 4}{a - 7} \geq 0$$ $$\frac{3 - 4a}{a - 7} 0$$ $$\frac{5(a - 2)}{a - 7} > 0$$ $\alpha \in \left[ -\frac{4}{3}, 2 \right)$ Check end point $\left[ -\frac{4}{3}, 2 \right)$
Maths · Probability · Single correct
A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :
Answer: (c)
Event $E_1$: Event denotes spade is missing. $\mathrm{P}(E_1) = \frac{1}{4}; \mathrm{P}(\overline{E_1}) = \frac{3}{4}$ Event $A$: Event drawn two cards are spade. $$\mathrm{P}(A) = \frac{\frac{1}{4} \times \binom{12}{2} + \frac{3}{4} \times \binom{13}{2} + \frac{3}{4} \times \binom{13}{2}}{\frac{1}{4} \times \binom{12}{2} + \frac{3}{4} \times \binom{13}{2}}$$ $$= \frac{39}{50}$$
Maths · Binomial Theorem · Single correct
Let [x] denote greatest integer less than or equal to x. If for $n \in \mathbb{N}$, $\left(1 - x + x^3\right)^n = \sum_{j=0}^{3n} a_j x^j$, then $\sum_{j=0}^{\left\lfloor \frac{3n}{2} \right\rfloor} a_{2j} + 4 \sum_{j=0}^{\left\lfloor \frac{3n-1}{2} \right\rfloor} a_{2j} + 1$ is equal to:
Answer: (c)
(1 - x + x^3)^n = $\sum$_{j=0}^{3n} a_j x^j (1 - x + x^3)^n = a_0 + a_1 x + a_2 x^2 + $\ldots$ + a_{3n} x^{3n} $\sum$_{j=0}^{$\left$$\lfloor$ $\frac{3n}{2}$ $\right$$\rfloor$} a_{2j} = Sum of a_0 + a_2 + a_4 + $\ldots$ $\sum$_{j=0}^{$\left$$\lfloor$ $\frac{3n-1}{2}$ $\right$$\rfloor$} a_{2j+1} = Sum of a_1 + a_3 + a_5 + $\ldots$ put x = 1 1 = a_0 + a_1 + a_2 + a_3 + $\ldots$ + a_{3n} $\ldots$ (A) Put x = -1 1 = a_0 - a_1 + a_2 - a_3 + $\ldots$ + (-1)^{3n} a_{3n} $\ldots$ (B) Solving (A) and (B) a_0 + a_2 + a_4 + $\ldots$ = 1 a_1 + a_3 + a_5 + $\ldots$ = 0 $\sum$_{j=0}^{$\left$$\lfloor$ $\frac{3n}{2}$ $\right$$\rfloor$} a_{2j} + 4 $\sum$_{j=0}^{$\left$$\lfloor$ $\frac{3n-1}{2}$ $\right$$\rfloor$} a_{2j+1} = 1
Maths · Differential Equations · Single correct
If $y = y(x)$ is the solution of the differential equation, $\frac{dy}{dx} + 2y \tan x = \sin x$, $y\left(\frac{\pi}{3}\right) = 0$, then the maximum value of the function $y(x)$ over $\mathbb{R}$ is equal to:
Answer: (d)
Given $\($ $\frac{dy}{dx}$ + 2y $\tan$ x = $\sin$ x $\)$. I.F. = $\($ e^{$\int$ 2 $\tan$ x $\,$ dx} = e^{2 $\ln$ $\sec$ x} $\)$. I.F. = $\($ $\sec$^2 x $\)$. $\($ y $\cdot$ ($\sec$^2 x) = $\int$ $\sin$ x $\cdot$ $\sec$^2 x $\,$ dx $\)$. $\($ y $\cdot$ ($\sec$^2 x) = $\int$ $\sec$ x $\tan$ x $\,$ dx $\)$. $\($ y $\cdot$ ($\sec$^2 x) = $\sec$ x + C $\)$. $\($ x = $\frac{\pi}{3}$ ; y = 0 $\)$. $\($ $\Rightarrow$ C = -2 $\)$. $\($ $\Rightarrow$ y = $\frac{\sec x - 2}{\sec^2 x}$ = $\cos$ x - 2 $\cos$^2 x $\)$. $\($ y = t - 2t^2 $\Rightarrow$ $\frac{dy}{dt}$ = 1 - 4t = 0 $\Rightarrow$ t = $\frac{1}{4}$ $\)$. $\($ $\therefore$ max = $\frac{1}{4}$ - $\frac{1}{8}$ = $\frac{2 - 1}{8}$ = $\frac{1}{8}$ $\)$.
Maths · Conic Sections · Single correct
The locus of the midpoints of the chord of the circle, $x^2 + y^2 = 25$ which is tangent to the hyperbola, $$\frac{x^2}{9} - \frac{y^2}{16} = 1$$ is:
Answer: (d)
Equation of chord $$y - k = -\frac{h}{k}(x - h)$$ $$ky - k^2 = -hx + h^2$$ $$hx + ky = h^2 + k^2$$ $$y = -\frac{hx}{k} + \frac{h^2 + k^2}{k}$$ tangent to $$\frac{x^2}{9} - \frac{y^2}{16} = 1$$ $$c^2 = a^2m^2 - b^2$$ $$\left( \frac{h^2 + k^2}{k} \right)^2 = 9 \left( -\frac{h}{k} \right)^2 - 16$$ $$(x^2 + y^2)^2 = 9x^2 - 16y^2$$
Maths · Relations and Functions · Single correct
The number of roots of the equation, $$(81)^{\sin^2 x} + (81)^{\cos^2 x} = 30$$ in the interval $[0, \pi]$ is equal to:
Answer: (b)
Given $81 \sin^2 x + 81 \cos^2 x = 30$. This simplifies to $81 \sin^2 x + \frac{81}{(18)^{\sin^2 x}} = 30$. Let $81 \sin^2 x = t$. Then $t + \frac{81}{t} = 30$. This can be factored as $(t - 3)(t - 27) = 0$. So, $81 \sin^2 x = 3^1$ or $81 \sin^2 x = 3^3$. This gives $3^{4 \sin^2 x} = 3^1$ or $3^{4 \sin^2 x} = 3^3$. Thus, $\sin^2 x = \frac{1}{4}$ or $\sin^2 x = \frac{3}{4}$. The total number of solutions is 4.
Maths · Inverse Trigonometric Functions · Single correct
Let $S_k = \sum_{r=1}^{k} \tan^{-1} \left( \frac{6^r}{2^{2r+1} + 3^{2r+1}} \right)$. Then $\lim_{k \to \infty} S_k$ equal to:
Answer: (c)
Given $$S_k = \sum_{r=1}^{k} \tan^{-1} \left( \frac{6^r}{2^{2r+1} + 3^{2r+1}} \right)$$ Divide by $3^{2x}$ $$\sum_{r=1}^{k} \tan^{-1} \left( \frac{\left( \frac{2}{3} \right)^r}{\left( \frac{2}{3} \right)^{2r} \cdot 2 + 3} \right)$$ $$\sum_{r=1}^{k} \tan^{-1} \left( \frac{\left( \frac{2}{3} \right)^r}{3 \left( \left( \frac{2}{3} \right)^{2r+1} + 1 \right)} \right)$$ Let $\left( \frac{2}{3} \right)^r = t$ $$\sum_{r=1}^{k} \tan^{-1} \left( \frac{\frac{t}{3}}{1 + \frac{3}{2} t^2} \right)$$ $$\sum_{r=1}^{k-1} \tan^{-1} \left( \frac{t - \frac{3t}{3}}{1 + \frac{2t}{3}} \right)$$ $$\sum_{r=1}^{k} \left( \tan^{-1} (t) - \tan^{-1} \left( 2t \left( \frac{2}{3} \right) \right) \right)$$ $$\sum_{r=1}^{k} \left( \tan^{-1} \left( \left( \frac{2}{3} \right)^r \right) - \tan^{-1} \left( \left( \frac{2}{3} \right)^{r+1} \right) \right)$$ $$S_k = \tan^{-1} \left( \frac{2}{3} \right) - \tan^{-1} \left( \left( \frac{2}{3} \right)^{k+1} \right)$$ $$S_\infty = \lim_{k \to \infty} \left( \tan^{-1} \left( \frac{2}{3} \right) - \tan^{-1} \left( \left( \frac{2}{3} \right)^{k+1} \right) \right)$$ $$= \tan^{-1} \left( \frac{2}{3} \right) - \tan^{-1} (0)$$ Therefore, $$S_\infty = \tan^{-1} \left( \frac{2}{3} \right) = \cot^{-1} \left( \frac{3}{2} \right)$$
Maths · Sequences and Series · Numerical
Consider an arithmetic series and a geometric series having four initial terms from the set {11, 8, 21, 16, 26, 32, 4}. If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these two series is equal to
Answer: 3
GP: 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192 AP: 11, 16, 21, 26, 31, 36 Common terms: 16, 256, 4096 only
Maths · Integrals · Numerical
Let $f : (0, 2) \to \mathbb{R}$ be defined as $f(x) = \log_2 \left( 1 + \tan \left( \frac{\pi x}{4} \right) \right)$ Then, $\lim_{n \to \infty} \frac{2}{n} \left( f \left( \frac{1}{n} \right) + f \left( \frac{2}{n} \right) + \ldots + f(1) \right)$ is equal to
Answer: 1
Given $$E = 2 \lim_{n \to \infty} \sum_{r=1}^{n-1} \frac{1}{n} f\left(\frac{r}{n}\right)$$ $$E = \frac{2}{\ln 2} \int_0^1 \ln\left(1 + \tan \frac{\pi x}{4}\right) \, dx$$ Replacing $x \to 1 - x$ $$E = \frac{2}{\ln 2} \int_0^1 \ln\left(1 + \tan \frac{\pi}{4} (1-x)\right) \, dx$$ $$E = \frac{2}{\ln 2} \int_0^1 \ln\left(1 + \tan\left(\frac{\pi}{4} - \frac{\pi}{4} x\right)\right) \, dx$$ $$E = \frac{2}{\ln 2} \int_0^1 \ln\left(1 + \frac{1 + \tan \frac{\pi}{4} x}{1 + \tan \frac{\pi}{4} x}\right) \, dx$$ $$E = \frac{2}{\ln 2} \int_0^1 \ln\left(\frac{2}{1 + \tan \frac{\pi x}{4}}\right) \, dx$$ $$E = \frac{2}{\ln 2} \int_0^1 \left(\ln 2 - \ln\left(1 + \tan \frac{\pi x}{4}\right)\right) \, dx$$ Equation (i) + (ii) $$E = 1$$
Maths · Conic Sections · Numerical
Let ABCD be a square of side of unit length. Let a circle $C_1$ centered at $A$ with unit radius is drawn. Another circle $C_2$ which touches $C_1$ and the lines $AD$ and $AB$ are tangent to it, is also drawn. Let a tangent line from the point $C$ to the circle $C_2$ meet the side $AB$ at $E$. If the length of $EB$ is $\alpha + \sqrt{3}\beta$, where $\alpha, \beta$ are integers, then $\alpha + \beta$ is equal to
Answer: 1
Here $AO + OD = 1$ or $(\sqrt{2} + 1)r = 1$ implies $r = \sqrt{2} - 1$. The equation of the circle is $(x - r)^2 + (y - r)^2 = r^2$. The equation of $CE$ is $y - 1 = m(x - 1)$. Rearranging gives $mx - y + 1 - M = 0$. It is tangent to the circle, therefore: $$\left| \frac{mr - r + 1 - m}{\sqrt{m^2 + 1}} \right| = r$$ $$\left| \frac{(m - 1)r + 1 - m}{\sqrt{m^2 + 1}} \right| = r$$ $$\frac{\sqrt{m^2 + 1}}{(m - 1)^2(r - 1)^2} = r^2$$ Put $r = \sqrt{2} - 1$. On solving, $m = 2 - \sqrt{3}, 2 + \sqrt{3}$. Taking the greater slope of $CE$ as $2 + \sqrt{3}$, we have $y - 1 = (2 + \sqrt{3})(x - 1)$. Put $y = 0$: $$-1 = (2 + \sqrt{3})(x - 1)$$ $$\frac{-1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = x - 1$$ $$x - 1 = \sqrt{3} - 1$$ Then $EB = 1 - x = 1 - (\sqrt{3} - 1)$, so $EB = 2 - \sqrt{3}$.
Maths · Limits and Derivatives · Numerical
If $\lim_{x \to 0} \frac{ae^x - b \cos x + ce^{-x}}{x \sin x} = 2$, then $a + b + c$ is equal to
Answer: 4
Given $\($ $\lim$_{x $\to$ 0} $\frac{ae^x - b \cos x + ce^{-x}}{x \sin x}$ = 2 $\)$. This implies: $$ \lim_{x \to 0} \frac{a \left( 1 + x + \frac{x^2}{2!} \cdots \right) - b \left( 1 - \frac{x^2}{2!} + \cdots \right) + c \left( 1 - x + \frac{x^2}{2!} \right)}{\left( \frac{x \sin x}{x} \right) x} = 2 $$ From this, we derive the equations: $\($ a - b + c = 0 $\)$ $\($ a - c = 0 $\)$ And $\($ $\frac{a + b + c}{2}$ = 2 $\)$ Thus, $\($ a + b + c = 4 $\)$
Maths · Matrices · Numerical
The total number of $3 \times 3$ matrices $A$ having entries from the set $(0,1,2,3)$ such that the sum of all the diagonal entries of $AA^T$ is $9$, is equal to
Answer: 766
Let $A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$. The diagonal elements of $AA^T$ are $a^2 + b^2 + c^2$, $d^2 + e^2 + f^2$, $g^2 + h^2 + i^2$. The sum is $a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2 + h^2 + i^2 = 9$. The elements $a, b, c, d, e, f, g, h, i \in \{0, 1, 2, 3\}$. The equation of the circle is $x^2 + y^2 = 25$. The total number of ways is $1 + 9 + 8 \times 63 + 63 \times 4 = 766$.
Maths · Matrices · Numerical
Let $$\mathbf{P} = \begin{bmatrix} -30 & 20 & 56 \\ 90 & 140 & 112 \\ 120 & 60 & 14 \end{bmatrix} and \mathbf{A} = \begin{bmatrix} 2 & 7 & \omega^2 \\ -1 & -\omega & 1 \\ 0 & -\omega & -\omega + 1 \end{bmatrix}$$ where $\omega = \frac{-1 + i \sqrt{3}}{2}$, and $\mathbf{I}_3$ be the identity matrix of order 3. If the determinant of the matrix $\left( \mathbf{P}^{-1} \mathbf{A} \mathbf{P} - \mathbf{I}_3 \right)^2$ is $\alpha \omega^2$, then the value of $\alpha$ is equal to
Answer: 36
Let $M = \left( P^{-1}AP - I \right)^2$. $$= \left( P^{-1}AP \right)^2 - 2P^{-1}AP + I$$ $$= P^{-1}A^2P - 2P^{-1}AP + I$$ $$PM = A^2P - 2AP + P$$ $$= \left( A^2 - 2\, A \cdot I + I^2 \right) P$$ Therefore, $\mathrm{Det}(PM) = \mathrm{Det}((A - I)^2 \times P)$. Thus, $\mathrm{Det}P \cdot \mathrm{Det}M = \mathrm{Det}(A - I)^2 \times \mathrm{Det}(P)$. Hence, $\mathrm{Det}M = (\mathrm{Det}(A - I))^2$. Now $A - I = \begin{bmatrix} 1 & 7 & w^2 \\ -1 & -w - 1 & 1 \\ 0 & -w & -w \end{bmatrix}$. $\mathrm{Det}(A - I) = (w^2 + w + w) + 7(-w) + w^3 = -6w$. $\mathrm{Det}((A - I))^2 = 36w^2$. Therefore, $\alpha = 36$.
Maths · Applications of Derivatives · Numerical
If the normal to the curve $y(x) = \int_0^x (2t^2 - 15t + 10) \, dt$ at a point $(a, b)$ is parallel to the line $x + 3y = -5, a > 1$, then the value of $|a + 6b|$ is equal to
Answer: 406
Given $$y(x) = \int_0^x (2t^2 - 15t + 10) \, dt$$ $$y'(x) \bigg|_{x=a} = \left[ 2x^2 - 15x + 10 \right]_a = 2a^2 - 15a + 10$$ Slope of normal = $-\frac{1}{3}$ $$\Rightarrow 2a^2 - 15a + 10 = 3 \Rightarrow a = 7$$ and $a = \frac{1}{2}$ (rejected) $$b = y(7) = \int_0^7 (2t^2 - 15t + 10) \, dt$$ $$= \left[ \frac{2t^3}{3} - \frac{15t^2}{2} + 10t \right]_0^7$$ $$\Rightarrow 6b = 4 \times 7^3 - 45 \times 49 + 60 \times 7$$ $$|a + 6b| = 406$$
Maths · Differential Equations · Numerical
Let the curve $y = y(x)$ be the solution of the differential equation, $\frac{dy}{dx} = 2(x + 1)$. If the numerical value of area bounded by the curve $y = y(x)$ and $x$-axis is $\frac{4\sqrt{8}}{3}$, then the value of $y(1)$ is equal to___
Answer: 2
Given $\($ $\frac{dy}{dx}$ = 2(x+1) $\)$ $\($ $\Rightarrow$ $\int$ dy = $\int$ 2(x+1) dx $\)$ $\($ $\Rightarrow$ y(x) = x^2 + 2x + C $\)$ Area = $\($ $\frac{4\sqrt{8}}{3}$ $\)$ $\($-1 + $\sqrt{1-C}$ $\)$ $\($ $\Rightarrow$ 2 $\int$_{-1}^{-1+$\sqrt{1-C}$} $\left$( -(x+1)^2 - C + 1 $\right$) dx = $\frac{4\sqrt{8}}{3}$ $\)$ $\($ $\Rightarrow$ 2 $\left$[ $\frac{-(x+1)^3}{3}$ - Cx + x $\right$]_{-1}^{-1+$\sqrt{1-C}$} = $\frac{4\sqrt{8}}{3}$ $\)$ $\($ $\Rightarrow$ -($\sqrt{1-C}$)^3 + 3c - 3C$\sqrt{1-C}$ $\)$ $\($-3 + 3$\sqrt{1-C}$ - 3C + 3 = 2$\sqrt{8}$ $\)$ $\($ $\Rightarrow$ C = -1 $\)$ $\($ $\Rightarrow$ f(x) = x^2 + 2x - 1, f(1) = 2 $\)$
Maths · Integrals · Numerical
Let $f : \mathbb{R} \to \mathbb{R}$ be a continuous function such that $f(x) + f(x + 1) = 2$, for all $x \in \mathbb{R}$. If $I_1 = \int_0^8 f(x) \, dx$ and $I_2 = \int_{-1}^3 f(x) \, dx$, then the value of $I_1 + 2I_2$ is equal to
Answer: 16
Given $f(x) + f(x + 1) = 2$. Therefore, $f(x)$ is periodic with period $= 2$. $I_1 = \int_0^8 f(x) \, dx = 4 \int_0^2 f(x) \, dx$ $= 4 \int_0^1 (f(x) + f(1 + x)) \, dx = 8$ Similarly $I_2 = 2 \times 2 = 4$ $I_1 + 2I_2 = 16$
Maths · Complex Numbers and Quadratic Equations · Numerical
Let $z$ and $w$ be two complex numbers such that $$w = z\overline{z} - 2z + 2,$$ $$\left| \frac{z+i}{z-3i} \right| = 1$$ and $\mathrm{Re}(w)$ has minimum value. Then, the minimum value of $n \in \mathbb{N}$ for which $w^n$ is real, is equal to .
Answer: 4
Given $\omega = z \bar{z} - 2z + 2$ and $\($ $\left$| $\frac{z+i}{z-3i}$ $\right$| = 1 $\)$. This implies $|z+i| = |z-3i|$. Therefore, $z = x + i$, where $x \in \mathbb{R}$. Then, $\omega = (x+i)(x-i) - 2(x+i) + 2 = x^2 + 1 - 2x - 2i + 2$. The real part of $\omega$ is $Re(\omega) = x^2 - 2x + 3$. For minimum $Re(\omega)$, let $x = 1$. Thus, $\omega = 2 - 2i = 2(1-i) = 2\sqrt{2} e^{-i \frac{\pi}{4}}$. Therefore, $\omega^n = (2\sqrt{2})^n e^{-i \frac{n \pi}{4}}$. For real and minimum value of $n$, $n = 4$.
Physics · Experimental Physics · Single correct
One main scale division of a vernier callipers is 'a' cm and $n^{th}$ division of the vernier scale coincide with $(n-1)^{th}$ division of the main scale. The least count of the callipers in mm is:
Answer: (d)
Given $$(n-1)a = n(a')$$ we have $$a' = \frac{(n-1)a}{n}$$. Therefore, the least count (L.C) is given by $$\mathrm{L.C} = 1\mathrm{MSD} - 1\mathrm{VSD}$$ which equals $$(a - a') \mathrm{cm}$$. Substituting the value of $a'$, we get: $$= a - \frac{(n-1)a}{n}$$ Simplifying further: $$= \frac{na - na + a}{n} = \frac{a}{n} \mathrm{cm}$$ Finally, converting to millimeters: $$= \left( \frac{10a}{n} \right) \mathrm{mm}$$
Physics · Electrostatic Potential and Capacitance · Single correct
For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} \, d$, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance ($C'$) in terms of original capacitance ($C_0$) is given by the following relation:
Answer: (c)
Given $C_0 = \frac{\varepsilon_0 A}{d}$. $C'$ is $C_1$ and $C_2$ in series. i.e. $\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2}$ $$\frac{1}{C'} = \frac{3d/4}{\varepsilon_0 KA} + \frac{d/4}{\varepsilon_0 A}$$ $$\frac{1}{C'} = \frac{d}{4\varepsilon_0 A} \left( \frac{3+K}{K} \right)$$ $$C' = \frac{4KC_0}{3+K}$$
Physics · Laws of Motion · Single correct
A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle $\theta$ as shown in figure. The coefficient of kinetic friction is $\mu_K$. Then, the block's acceleration 'a' is given by : (g is acceleration due to gravity)
Answer: (b)
The normal force is given by $N = mg - F \sin \theta$. The horizontal force equation is $F \cos \theta - \mu_k N = ma$. Substituting for $N$, we have $F \cos \theta - \mu_k (mg - F \sin \theta) = ma$. Solving for $a$, we get $a = \frac{F}{m} \cos \theta - \mu_k \left( g - \frac{F}{m} \sin \theta \right)$.
Physics · Mechanical Properties of Fluids · Single correct
The pressure acting on a submarine is $3 \times 10^5 \, \mathrm{Pa}$ at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be: (Assume that atmospheric pressure is $1 \times 10^5 \, \mathrm{Pa}$, density of water is $10^3 \, \mathrm{kg} \, \mathrm{m}^{-3}$, $g = 10 \, \mathrm{ms}^{-2}$)
Answer: (a)
Given $P_1 = \rho g d + P_0 = 3 \times 10^5 \, \mathrm{Pa}$. Therefore, $\rho g d = 2 \times 10^5 \, \mathrm{Pa}$. Now, $P_2 = 2 \rho g d + P_0$. This gives $P_2 = 4 \times 10^5 + 10^5 = 5 \times 10^5 \, \mathrm{Pa}$. The percentage increase is given by $$\% increase = \frac{P_2 - P_1}{P_1} \times 100$$ $$= \frac{5 \times 10^5 - 3 \times 10^5}{3 \times 10^5} \times 100 = \frac{200}{3}\%$$
Physics · Ray Optics and Optical Instruments · Single correct
The angle of deviation through a prism is minimum when
Answer: (a)
Deviation is minimum in a prism when: $i = e$, $r_1 = r_2$ and ray (2) is parallel to base of prism.
Physics · Electromagnetic Waves · Single correct
A plane electromagnetic wave of frequency 500 $\mathrm{MHz}$ is travelling in vacuum along y-direction. At a particular point in space and time, $\vec{B} = 8.0 \times 10^{-8} \hat{z} \, \mathrm{T}$. The value of electric field at this point is : (speed of light = $3 \times 10^8 \, \mathrm{ms}^{-1}$) $\hat{x}, \hat{y}, \hat{z}$ are unit vectors along x, y and Z direction.
Answer: (a)
Given $f = 5 \times 10^8 \, \mathrm{Hz}$. EM wave is travelling towards $+\hat{j}$. $$\vec{B} = 8.0 \times 10^{-8} \hat{z} \, \mathrm{T}$$ $$\vec{E} = \vec{B} \times \vec{C} = \left(8 \times 10^{-8} \hat{z}\right) \times \left(3 \times 10^8 \hat{y}\right)$$ $$= -24 \hat{x} \, \mathrm{V/m}$$
Physics · Gravitation · Single correct
The maximum and minimum distances of a comet from the Sun are $1.6 \times 10^{12} \, \mathrm{m}$ and $8.0 \times 10^{10} \, \mathrm{m}$ respectively. If the speed of the comet at the nearest point is $6 \times 10^{4} \, \mathrm{ms^{-1}}$, the speed at the farthest point is:
Answer: (c)
By angular momentum conservation: $$mv_1 r_1 = mv_2 r_2$$ $$v_1 = \frac{48 \times 10^{14}}{1.6 \times 10^{12}} = 3000 \, \mathrm{m/sec}$$ $$= 3 \times 10^3 \, \mathrm{m/sec}$$
Physics · Magnetism and Matter · Single correct
A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the center of the magnet. If $B_H = 0.4 \, \mathrm{G}$, the magnetic moment of the magnet is $(1 \, \mathrm{G} = 10^{-4} \, \mathrm{T})$
Answer: (c)
i.e. $\($ $\frac{2 \mu_0}{4 \pi}$ $\frac{m}{r^2}$ $\times$ $\frac{7}{r}$ = 0.4 $\times$ 10^{-4} $\)$ $\($ $\Rightarrow$ 2 $\times$ 10^{-7} $\times$ $\frac{m \times 7}{(7^2 + 18^2)^{3/2}}$ $\times$ 10^4 $\)$ $\($ = 0.4 $\times$ 10^{-4} $\)$ $\($ m = $\frac{4 \times 10^{-2} \times (373)^{3/2}}{14}$ $\)$ $\($ M = m $\times$ 14 $\,$ cm = m $\times$ $\frac{14}{100}$ $\)$ $\($ = $\frac{0.04 \times (373)^{3/2}}{14}$ $\times$ $\frac{14}{100}$ $\)$ $\($ = 4 $\times$ 10^{-4} $\times$ 7203.82 = 2.88 $\,$ J/T $\)$
Physics · Kinetic Theory · Single correct
The volume $V$ of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of nitrogen and 44 g of carbon dioxide at absolute temperature $T$. Consider $R$ as universal gas constant. The pressure of the mixture of gases is :
Answer: (c)
Given the equation $PV = (n_1 + n_2 + n_3)RT$. We have: $$P \times V = \left[ \frac{16}{32} + \frac{28}{28} + \frac{44}{44} \right] RT$$ Simplifying, we get: $$PV = \left[ \frac{1}{2} + 1 + 1 \right] RT$$
Physics · Thermodynamics · Single correct
In thermodynamics, heat and work are :
Answer: (a)
Physics · System of Particles and Rotational Motion · Single correct
Four equal masses, $m$ each are placed at the corners of a square of length $(l)$ as shown in the figure. The moment of inertia of the system about an axis passing through $A$ and parallel to $DB$ would be:
Answer: (c)
Moment of inertia of point mass is equal to mass times the square of the perpendicular distance from the axis. Moment of Inertia $$= m(0)^2 + m(l\sqrt{2})^2 + m\left(\frac{l}{\sqrt{2}}\right)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2$$ $$= 3 \, m l^2$$
Physics · Current Electricity · Single correct
A conducting wire of length $l$, area of crosssection $A$ and electric resistivity $\rho$ is connected between the terminals of a battery. A potential difference $V$ is developed between its ends, causing an electric current. If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be:
Answer: (a)
As per the question $$Resistance = \frac{\rho (2l)}{(A/2)} = \frac{4\rho l}{A}$$ Therefore, the current is given by $$Current = \frac{V}{R} = \frac{VA}{4\rho l}$$
Physics · Oscillations · Single correct
Time period of a simple pendulum is $T$ inside a lift when the lift is stationary. If the lift moves upwards with an acceleration $g/2$, the time period of pendulum will be:
Answer: (d)
When lift is stationary $$T = 2\pi \sqrt{\frac{L}{g}}$$ When lift is moving upwards $\Rightarrow$ Pseudo force acts downwards $\Rightarrow$ g_{eff} = g + $\frac{g}{2}$ = $\frac{3g}{2}$ $\Rightarrow$ New time period $$T' = 2\pi \sqrt{\frac{L}{g_{eff}}} = 2\pi \sqrt{\frac{2L}{3g}}$$ $$T' = \sqrt{\frac{2}{3}} \, T$$
Physics · Motion in a Straight Line · Single correct
The velocity-displacement graph describing the motion of a bicycle is shown in the figure. The acceleration-displacement graph of the bicycle's motion is best described by:
Answer: (a)
For $0 \leq x \leq 200$ $v = mx + C$ $v = \frac{1}{5}x + 10$ $$a = \frac{vdv}{dx} = \left( \frac{x}{5} + 10 \right) \left( \frac{1}{5} \right)$$ $$a = \frac{x}{25} + 2 \Rightarrow Straight line till x = 200$$ For $x > 200$ $v = constant$ $\Rightarrow a = 0$
Physics · Communication Systems · Single correct
A 25 $\mathrm{m}$ long antenna is mounted on an antenna tower. The height of the antenna tower is 75 $\mathrm{m}$. The wavelength (in meter) of the signal transmitted by this antenna would be:
Answer: (d)
Length of Antenna = $25 \, \mathrm{m} = \frac{\lambda}{4}$ Therefore, $\lambda = 100 \, \mathrm{m}$
Physics · Electromagnetic Waves · Single correct
For an electromagnetic wave travelling in free space, the relation between average energy densities due to electric ($U_e$) and magnetic ($U_m$) fields is :
Answer: (a)
In EMW, average energy density due to electric ($U_e$) and magnetic ($U_m$) fields is the same.
Physics · Alternating Current · Single correct
An RC circuit as shown in the figure is driven by a AC source generating a square wave. The output wave pattern monitored by CRO would look close to :
Answer: (c)
For $t_1 - t_2$ Charging graph $t_2 - t_3$ Discharging graph
Physics · Dual Nature of Radiation and Matter · Single correct
The stopping potential in the context of photoelectric effect depends on the following property of incident electromagnetic radiation:
Answer: (d)
Stopping potential changes linearly with frequency of incident radiation.
Physics · Laws of Motion · Single correct
A block of 200 g mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius 20 cm. If the block takes 40 s to complete one round, the normal force by the side walls of the groove is :
Answer: (d)
Given $N = m \omega^2 R$. $$N = m \left[ \frac{4 \pi^2}{T^2} \right] R$$ Given $m = 0.2 \, \mathrm{kg}$, $T = 40 \, \mathrm{S}$, $R = 0.2 \, \mathrm{m}$. Put values in equation (1) $$N = 9.859 \times 10^{-4} \, \mathrm{N}$$
Physics · Electromagnetic Induction · Single correct
A conducting bar of length $L$ is free to slide on two parallel conducting rails as shown in the figure Two resistors $R_1$ and $R_2$ are connected across the ends of the rails. There is a uniform magnetic field $\vec{B}$ pointing into the page. An external agent pulls the bar to the left at a constant speed $v$ The correct statement about the directions of induced currents $I_1$ and $I_2$ flowing through $R_1$ and $R_2$ respectively is:
Answer: (c)
Consider the circuit with resistors $R_1$ and $R_2$, and a battery with emf $\varepsilon$. The currents $I_1$ and $I_2$ flow through $R_1$ and $R_2$ respectively. By applying Kirchhoff's loop rule, we have: $$\varepsilon = I_1 R_1 + I_2 R_2.$$ Assuming the junction rule, $I_1 = I_2$, we can simplify the equation to: $$\varepsilon = I_1 (R_1 + R_2).$$ Solving for $I_1$, we get: $$I_1 = \frac{\varepsilon}{R_1 + R_2}.$$
Physics · Current Electricity · Numerical
In the figure given, the electric current flowing through the 5k$\Omega$ resistor is 'x' mA. The value of $x$ to the nearest integer is
Answer: 3
The circuit is simplified by combining the parallel resistors. The equivalent resistance of the parallel resistors is calculated as follows: $$R_{eq} = \left( \frac{1}{3} + \frac{1}{3} \right)^{-1} = 1 \, k\Omega.$$ The total resistance in the circuit is then $5 \, k\Omega + 1 \, k\Omega + 1 \, k\Omega = 7 \, k\Omega$. The current $I$ is calculated using Ohm's law: $$I = \frac{21}{5 + 1 + 1} = 3 \, mA.$$
Physics · Wave Optics · Numerical
A fringe width of $6 \, \mathrm{mm}$ was produced for two slits separated by $1 \, \mathrm{mm}$ apart. The screen is placed $10 \, \mathrm{m}$ away. The wavelength of light used is '$x$' $\mathrm{nm}$. The value of '$x$' to the nearest integer is ____
Answer: 600
Given $\beta = \frac{\lambda D}{d}$. $$\lambda = \frac{\beta d}{D}$$ Substituting the values: $$\lambda = \frac{6 \times 10^{-3} \times 10^{-3}}{10}$$ Calculating: $$\lambda = 6 \times 10^{-7} \, \mathrm{m} = 600 \times 10^{-9} \, \mathrm{m}$$ Therefore, $\lambda = 600 \, \mathrm{nm}$.
Physics · System of Particles and Rotational Motion · Numerical
Consider a 20 kg uniform circular disk of radius 0.2 m. It is pin supported at its center and is at rest initially. The disk is acted upon by a constant force $F = 20 \, \mathrm{N}$ through a massless string wrapped around its periphery as shown in the figure. Suppose the disk makes $n$ number of revolutions to attain an angular speed of $50 \, \mathrm{rad/s}$. The value of $n$, to the nearest integer, is ____ [Given : In one complete revolution, the disk rotates by $6.28 \, \mathrm{rad}$]
Answer: 20
Given \(Q2\) (20), $$\alpha = \frac{\tau}{I} = \frac{F \cdot R}{mR^2/2} = \frac{2F}{mR}.$$ $$\alpha = \frac{2 \times 200}{20 \times (0.2)} = 10\ \mathrm{rad/s^2}.$$ $$\omega^2 = \omega_0^2 + 2\alpha\Delta\theta.$$ $$50^2 = 0^2 + 2(10)\Delta\theta \Rightarrow \Delta\theta = \frac{2500}{20}.$$ $$\Delta\theta = 125\ \mathrm{rad}.$$ Number of revolutions $$= \frac{125}{2\pi} \approx 20\ \text{revolutions}.$$
Physics · Atoms · Numerical
The first three spectral lines of H -atom in the Balmer series are given $\lambda_1, \lambda_2, \lambda_3$ considering the Bohr atomic model, the wave lengths of first and third spectral lines $\left( \frac{\lambda_1}{\lambda_3} \right)$ are related by a factor of approximately 'x' $\times 10^{-1}$. The value of x, to the nearest integer, is ____
Answer: 15
For 1st line $$\frac{1}{\lambda_1} = \mathrm{Rz}^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$$ $$\frac{1}{\lambda_1} = \mathrm{Rz}^2 \frac{5}{36}$$ For 3rd line $$\frac{1}{\lambda_3} = \mathrm{Rz}^2 \left( \frac{1}{2^2} - \frac{1}{5^2} \right)$$ $$\frac{1}{\lambda_3} = \mathrm{Rz}^2 \frac{21}{100}$$ (ii) + (i) $$\frac{\lambda_1}{\lambda_3} = \frac{21}{100} \times \frac{36}{5} = 1.512 = 15.12 \times 10^{-1}$$ $$x \approx 15$$
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Numerical
The value of power dissipated across the zener diode ($V_z = 15 \, \mathrm{V}$) connected in the circuit as shown in the figure is $x \times 10^{-1}$ watt. The value of $x$, to the nearest integer, is ____
Answer: 5
Voltage across $R_S = 22 - 15 = 7 \, \mathrm{V}$ Current through $R_S = I = \frac{7}{35} = \frac{1}{5} \, \mathrm{A}$ Current through $90\Omega = I_2 = \frac{15}{90} = \frac{1}{6} \, \mathrm{A}$ Current through zener $= \frac{1}{5} - \frac{1}{6} = \frac{1}{30} \, \mathrm{A}$ Power through zener diode $P = VI$ $P = 15 \times \frac{1}{30} = 0.5 \, \mathrm{watt}$ $P = 5 \times 10^{-1} \, \mathrm{watt}$
Physics · Alternating Current · Numerical
A sinusoidal voltage of peak value 250 V is applied to a series LCR circuit, in which $R = 8\,\Omega$, $L = 24\,\mathrm{mH}$ and $C = 60\,\mu\mathrm{F}$. The value of power dissipated at resonant condition is 'x' kW. The value of x to the nearest integer is ____
Answer: 4
At resonance power ($P$) $$P = \frac{(V_{rms})^2}{R}$$ $$P = \frac{(250/\sqrt{2})^2}{8} = 3906.25 \, W$$ $$\approx 4 \, kW$$
Physics · Communication Systems · Numerical
In the logic circuit shown in the figure, if input $A$ and $B$ are 0 to 1 respectively, the output at $Y$ would be 'x'. The value of $x$ is ____
Answer: 0
Physics · Mathematics in Physics · Numerical
The resistance $R = \frac{V}{I}$, where $V = (50 \pm 2) \, \mathrm{V}$ and $I = (20 \pm 0.2) \, \mathrm{A}$. The percentage error in $R$ is '$x$'$\%$. The value of '$x$' to the nearest integer is ____
Answer: 5
The percentage error in $R$ is calculated as follows: $$\frac{\Delta R}{R} \times 100 = \frac{\Delta V}{V} \times 100 + \frac{\Delta I}{I} \times 100$$ Substituting the given values: $$\% error in R = \frac{2}{50} \times 100 + \frac{0.2}{20} \times 100$$ Calculating each term: $$\% error in R = 4 + 1$$ Therefore, the percentage error in $R$ is: $$\% error in R = 5\%$$
Physics · Motion in a Plane · Numerical
Consider a frame that is made up of two thin massless rods AB and AC as shown in the figure. A vertical force $\vec{P}$ of magnitude 100 N is applied at point A of the frame. Suppose the force is $\vec{P}$ resolved parallel to the arms AB and AC of the frame. The magnitude of the resolved component along the arm AC is xN. The value of x, to the nearest integer, is ____ [Given : $\sin$($35^\circ$) = 0.573, $\cos$($35^\circ$) = 0.819 $\sin$($110^\circ$) = 0.939, $\cos$($110^\circ$) = -0.342 ]
Answer: 82
Component along AC $$= 100 \cos 35^\circ \mathrm{N}$$ $$= 100 \times 0.819 \, \mathrm{N}$$ $$= 81.9 \, \mathrm{N}$$ $$\approx 82 \, \mathrm{N}$$
Physics · Work, Energy and Power · Numerical
A ball of mass 10 kg moving with a velocity 10$\sqrt{3}$ $\mathrm{ms^{-1}}$ along X-axis, hits another ball of mass 20 kg which is at rest. After collision, the first ball comes to rest and the second one disintegrates into two equal pieces. One of the pieces starts moving along Y-axis at a speed of 10 $\mathrm{m/s}$. The second piece starts moving at a speed of 20 $\mathrm{m/s}$ at an angle $\theta$ (degree) with respect to the $X$-axis. The configuration of pieces after collision is shown in the figure. The value of $\theta$ to the nearest integer is ____
Answer: 30
From conservation of momentum along x axis, $\vec{P}_i = \vec{P}_f$. $$10 \times 10\sqrt{3} = 200 \cos \theta$$ $$\cos \theta = \frac{\sqrt{3}}{2}$$ $$\theta = 30^\circ$$
Chemistry · The d-and f-Block Elements · Single correct
Given below are two statement: one is labelled as Assertion A and the other is labelled as Reason R: Assertion A: Size of $\mathrm{Bk}^{3+}$ ion is less than $\mathrm{Np}^{3+}$ ion. Reason R: The above is a consequence of the lanthanoid contraction. In the light of the above statements, choose the correct answer from the options given below:
Answer: (d)
Size of $^{97}\mathrm{Bk}^{3+}$ ion is less than that of $^{93}\mathrm{Np}^{3+}$ due to actinoid contraction. As we know that in a period from left to right ionic radius decreases and in actinide series it is due to actinoid contraction.
Chemistry · Biomolecules · Single correct
Which among the following pairs of Vitamins is stored in our body relatively for longer duration?
Answer: (b)
Vitamin-A & Vitamin-D
Chemistry · The s-Block Elements · Single correct
Given below are two statements: \textbf{Statement I:} Both $\mathrm{CaCl_2\cdot 6H_2O}$ and $\mathrm{MgCl_2\cdot 8H_2O}$ undergo dehydration on heating. \textbf{Statement II:} $\mathrm{BeO}$ is amphoteric whereas the oxides of other elements in the same group are acidic. In the light of the above statements, choose the correct answer from the options given below:
Answer: (b)
(a) $\mathrm{CaCl_2 \cdot 6H_2O} \xrightarrow{\Delta} \mathrm{CaCl_2} (Anmurovs) + 6\mathrm{H_2O}$ (b) $\mathrm{MgCl_2 \cdot 8H_2O} \xrightarrow{\Delta} \mathrm{MgO} + 2\mathrm{HCl} + 6\mathrm{H_2O}$ The dehydration of hydrated chloride of calcium can be achieved. The corresponding hydrated chloride of magnesium on heating suffer hydrolysis. (c) $\mathrm{BeO} \rightarrow Amphoteric$ $$ \begin{array}{l} \mathrm{MgO} \\ \mathrm{CaO} \\ \mathrm{SrO} \\ \mathrm{BaO} \end{array} \Rightarrow All are basic oxide $$
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The product "P" in the above reaction is:
Answer: (b)
DIBAL can not reduce double bond. It can reduce cyclic ester.
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Match List-I with List-II : \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{Industrial process} & \multicolumn{2}{c|}{Application} \\ \hline (a) & Haber's process & (i) & HNO$_3$ synthesis \\ \hline (b) & Ostwald's process & (ii) & Aluminium extraction \\ \hline (c) & Contact process & (iii) & NH$_3$ synthesis \\ \hline (d) & Hall-Heroult process & (iv) & H$_2$SO$_4$ synthesis \\ \hline \end{tabular} Choose the correct answer from the options given below :
Answer: (c)
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Among the following, the aromatic compounds are:
Answer: (b)
Q1 (2) (A) Non-Aromatic (B) Aromatic (C) Aromatic (D) Anti-Aromatic
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
In the above chemical reaction, intermediate "X" and reagent/condition "A" are :
Answer: (c)
The given reaction is a diazotisation reaction followed by hydrolysis. Aniline ($\mathrm{NH_2}$) reacts with $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ at $273 - 278 \, \mathrm{K}$ to form benzene diazonium chloride ($\mathrm{N_2^+Cl^-}$). This intermediate, when treated with water, undergoes hydrolysis to form phenol ($\mathrm{OH}$).
Chemistry · The d-and f-Block Elements · Single correct
Given below are two statements: Statement I : The $E^\circ$ value of $\mathrm{Ce}^{4+}/\mathrm{Ce}^{3+}$ is $+1.74 \, \mathrm{V}$ Statement II : Ce is more stable in $\mathrm{Ce}^{4+}$ state than $\mathrm{Ce}^{3+}$ state. In the light of the above statements, choose the most appropriate answer from the options given below:
Answer: (d)
The $E^\circ$ value for $\mathrm{Ce^{4+}/Ce^{3+}}$ is $+1.74 \, \mathrm{V}$ because the most stable oxidation state of lanthanide series elements is $+3$. It means $\mathrm{Ce^{3+}}$ is more stable than $\mathrm{Ce^{4+}}$.
Chemistry · Chemistry in Everyday Life · Single correct
The functions of antihistamine are:
Answer: (b)
Chemistry · Hydrocarbons · Single correct
Which of the following is Lindlar catalyst?
Answer: (d)
Partially deactivated palladised charcoal $(\mathrm{H_2}/\mathrm{Pd}/\mathrm{CaCO_3})$ is lindlar catalyst.
Chemistry · Haloalkanes and Haloarenes · Single correct
The product "A" and "B" formed in above reactions are
Answer: (c)
The reaction of the alcohol with 20$\%$ $\mathrm{H_3PO_4}$ under heat ($\Delta$) leads to an elimination reaction (E_1) producing the Saytzeff product. The reaction of the chloride with $\mathrm{Me_3COK}$ (a bulky base) leads to an elimination reaction (E_2) producing the Hoffmann product.
Chemistry · Hydrogen · Single correct
Given below are two statements: Statement I : $\mathrm{H_2O_2}$ can act as both oxidising and reducing agent in basic medium. Statement II : In the hydrogen economy, the energy is transmitted in the form of dihydrogen. In the light of the above statements, choose the correct answer from the options given below:
Answer: (b)
(a) $\mathrm{H_2O_2}$ can act as both oxidising and reducing agent in basic medium. (i) $2\mathrm{Fe^{2+}} + \mathrm{H_2O_2} \rightarrow 2\mathrm{Fe^{3+}} + 2\mathrm{OH^-}$ In this reaction, $\mathrm{H_2O_2}$ acts as oxidising agent. (ii) $2\mathrm{MnO_4^-} + 3\mathrm{H_2O_2} \rightarrow 2\mathrm{MnO_2} + 3\mathrm{O_2} + 2\mathrm{H_2O} + 2\mathrm{OH^-}$ In this reaction, $\mathrm{H_2O_2}$ acts as reducing agent. (b) The basic principle of hydrogen economy is the transportation and storage of energy in the form of liquids or gaseous dihydrogen. Advantage of hydrogen economy is that energy is transmitted in the form of dihydrogen and not as electric power.
Chemistry · Environmental Chemistry · Single correct
The type of pollution that gets increased during the day time and in the presence of $\mathrm{O}_3$ is:
Answer: (b)
In presence of ozone ($\mathrm{O_3}$), oxidising smog gets increased during the day time because automobiles and factories produce main components of the photochemical smog (oxidising smog) results from the action of sunlight on unsaturated hydrocarbon and nitrogen oxide. Ozone is strong oxidising agent and can react with the unburnt hydrocarbons in the polluted air to produce chemicals.
Chemistry · Alcohols, Phenols and Ethers · Single correct
Assertion A: Enol form of acetone $[\mathrm{CH_3COCH_3}]$ exists in $< 0.1\%$ quantity. However, the enol form of acetyl acetone $[\mathrm{CH_3COCH_2OCCH_3}]$ exists in approximately $15\%$ quantity. Reason R: enol form of acetyl acetone is stabilized by intramolecular hydrogen bonding, which is not possible in enol form of acetone. Choose the correct statement:
Answer: (b)
The keto form of acetone is represented as $\mathrm{CH_3C(O)CH_3}$, and the enol form is $\mathrm{CH_2=C(OH)CH_3}$. The enol form of acetone is very less, less than $0.1\%$. In the case of the compound $\mathrm{CH_3C(O)CH_2C(O)CH_3}$, the enol form is stabilized by intramolecular hydrogen bonding, resulting in more than $50\%$ enol content.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Which of the following reaction DOES NOT involve Hoffmann Bromamide degradation?
Answer: (c)
This reaction does not involve haffmann bromamide degradation. Rest all options involve haffmann bromamide degradation during the reaction of $\mathrm{Br_2} + \mathrm{NaOH}$ with amide.
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
The process that involves the removal of sulphur from the ores is:
Answer: (b)
In the roasting process, metal sulphide (MS) ore is converted into metal oxide and sulphur is removed in the form of $\mathrm{SO_2}$ gas. $$2\mathrm{MS} + 3\mathrm{O_2} \xrightarrow{\Delta} 2\mathrm{MO} + 2\mathrm{SO_2} \uparrow$$
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Match List-I with List-II : \begin{tabular}{|c|p{5cm}|c|l|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{Name of oxo acid} & \multicolumn{2}{c|}{Oxidation state of 'P'} \\ \hline (a) & Hypophosphorous acid & (i) & +5 \\ \hline (b) & Orthophosphoric acid & (ii) & +4 \\ \hline (c) & Hypophosphoric acid & (iii) & +3 \\ \hline (d) & Orthophosphorous acid & (iv) & +2 \\ \hline & & (v) & +1 \\ \hline \end{tabular} Choose the correct answer from the options given below:
Answer: (c)
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R : Assertion $\textbf{A}$ : The H – O – H bond angle in water molecule is $104.5^\circ$ Reason $\textbf{R}$ : The lone pair – lone pair repulsion of electrons is higher than the bond pair - bond pair repulsion.
Answer: (d)
The hybridisation of oxygen in the water molecule is $\mathrm{sp}^3$. So the electron geometry of the water molecule is tetrahedral and the bond angle should be $109^\circ 28''$, but as we know that lone pair-lone pair repulsion of electrons is higher than the bond pair-bond pair repulsion because the lone pair occupies more space around the central atom than that of the bond pair.
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
In chromatography technique, the purification of compound is independent of:
Answer: (d)
In chromatography technique, the purification of a compound is independent of the physical state of the pure compound.
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
A group 15 element, which is a metal and forms a hydride with strongest reducing power among group 15 hydrides. The element is :
Answer: (d)
Chemistry · Equilibrium · Numerical
For the reaction $\mathrm{A(g) \rightleftharpoons B(g)}$ at $495\,\mathrm{K}$, $\Delta_r G^\circ = -9.478\,\mathrm{kJ\,mol^{-1}}$ If the reaction is started in a closed container at $495\,\mathrm{K}$ with $22$ millimoles of $\mathrm{A}$, the amount of $\mathrm{B}$ in the equilibrium mixture is __ millimoles. (Round off to the nearest integer.) Given: $R = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ $\ln 10 = 2.303$
Answer: 20
Solution. $\Delta G^\circ = -RT \ln K_{eq}$ Given $\Delta G^\circ = -9.478 KJ/mole$ $T = 495 \, K$ $R = 8.314 \, J mol^{-1}$ So $-9.478 \times 10^3 = -495 \times 8.314 \times \ln K_{eq}$ $\ln K_{eq} = 2.303$ $= \ln 10$ So $K_{eq} = 10$ Now $A(g) \rightleftharpoons B(g)$ $t = 0$ $22$ $0$ $t = t$ $22 - x$ $x$ $K_{eq} = \frac{[B]}{[C]} = \frac{x}{22-x} = 10$ or $x = 20$ So millimoles of B $= 20$
Chemistry · Some Basic Concepts of Chemistry · Numerical
Complete combustion of 750 g of an organic compound provides 420 g of $CO_2$ and 210 g of $H_2O$. The percentage composition of carbon and hydrogen in organic compound is 15.3 and....respectively. (Round off to the Nearest Integer)
Answer: 3
44 $\mathrm{gmCO_2}$, have 12 $\mathrm{gm}$ carbon So, 420 $\mathrm{gmCO_2}$ $\Rightarrow$ $\frac{12}{44}$ $\times$ 420 $\Rightarrow$ $\frac{1260}{11}$ gm carbon $\Rightarrow$ 114.545 gram carbon So, $\%$ of carbon = $\frac{114.545}{750}$ $\times$ 100 $\simeq$ 15.3$\%$ 18 $\mathrm{gmH_2O}$ $\Rightarrow$ 2 $\mathrm{gmH_2}$ 210 $\mathrm{gm}$ $\Rightarrow$ $\frac{2}{18}$ $\times$ 210 = 23.33 $\mathrm{gmH_2}$ So, $\%$ $\mathrm{H_2}$ $\Rightarrow$ $\frac{23.33}{750}$ $\times$ 100 = 3.11$\%$ $\approx$ 3$\%$
Chemistry · Redox Reactions · Numerical
$2\mathrm{MnO_4^-} + b\mathrm{C_2O_4^{2-}} + c\mathrm{H^+} \rightarrow x\mathrm{Mn^{2+}} + y\mathrm{CO_2} + z\mathrm{H_2O}$ If the above equation is balanced with integer coefficients, the value of $c$ is \_\_\_\_ (Round off to the Nearest Integer).
Answer: 16
Writing the half reaction oxidation half reaction $$\mathrm{MnO_4^- \rightarrow Mn^{2+}}$$ Balancing oxygen $$\mathrm{MnO_4^- \rightarrow Mn^{2+} + 4H_2O}$$ Balancing hydrogen $$\mathrm{8H^+ + MnO_4^- \rightarrow Mn^{2+} + 4H_2O}$$ Balancing charge $$\mathrm{5e^- + 8H^+ + MnO_4^- \rightarrow Mn^{2+} + 4H_2O}$$ Reduction half $$\mathrm{C_2O_4^{2-} \rightarrow CO_2}$$ Balancing carbon $$\mathrm{C_2O_4^{2-} \rightarrow 2CO_2}$$ Balancing charge $$\mathrm{C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-}$$ Net equation $$\mathrm{16H^+ + 2MnO_4^- + 5C_2O_4^{2-} \rightarrow 10CO_2 + 2Mn^{2+} + 8H_2O}$$ So $c = 16$
Chemistry · Solutions · Numerical
$AB_2$ is $10\%$ dissociated in water to $A^{2+}$ and $B^-$. The boiling point of a $10.0$ molal aqueous solution of $AB_2$ is ____ °C. (Round off to the nearest integer.) [Given: Molal elevation constant of water $K_b = 0.5\ \mathrm{K\,kg\,mol^{-1}}$; boiling point of pure water = $100$ °C.]
Answer: 106
The reaction is given by $$\mathrm{AB_2} \rightarrow \mathrm{A^{2+}} + 2 \mathrm{B^-}$$ At time $t = 0$, the concentrations are $a$, $0$, $0$. At time $t = t$, the concentrations are $a - a\alpha$, $a\alpha$, $2a\alpha$. The total concentration $n_T$ is $$n_T = a - a\alpha + a\alpha + 2a\alpha$$ $$= a(1 + 2\alpha)$$ So $i = 1 + 2\alpha$. Now, the boiling point elevation $\Delta T_b$ is given by $$\Delta T_b = i \times m \times K_b$$ Substituting the values, $$\Delta T_b = (1 + 2\alpha) \times m \times K_b$$ Given $\alpha = 0.1$, $m = 10$, $K_b = 0.5$, $$\Delta T_b = 1.2 \times 10 \times 0.5$$ $$= 6$$ So the boiling point is $106$.
Chemistry · Co-ordination Compounds · Numerical
The equivalents of ethylene diamine required to replace the neutral ligands from the coordination sphere of the trans-complex of $\mathrm{CoCl_3 \cdot 4NH_3}$ is____. (Round off to the Nearest Integer).
Answer: 2
As we know that ethylene diamine is a bidentate ligand and ammonia is a mono dentate ligand. It means overall two ethylene diamine is required to replace all the neutral ligands (four ammonia) from the coordination sphere of this complex.
Chemistry · Some Basic Concepts of Chemistry · Numerical
A 6.50 molal solution of KOH (aq.) has a density of $1.89 \, \mathrm{g} \, \mathrm{cm}^{-3}$. The molarity of the solution is _____ mol dm$^{-3}$. (Round off to the Nearest Integer). [Atomic masses: K : 39.0u; O : 16.0u; H : 1.0u]
Answer: 9
6.5 molal KOH = 1000 $\mathrm{gm}$ solvent has 6.5 moles KOH so wt of solute = 6.5 $\times$ 56 = 364 $\mathrm{gm}$ wt of solution = 1000 + 364 = 1364 Volume of solution = $\frac{1364}{1.89}$ $\,$ $\mathrm{ml}$ Molarity = $\frac{mole of solute}{V_{solution in Litre}}$ = $\frac{6.5 \times 1.89 \times 1000}{1364}$ = 9.00
Chemistry · Structure of Atom · Numerical
When light of wavelength $248\,\mathrm{nm}$ falls on a metal of threshold energy $3.0\,\mathrm{eV}$, the de-Broglie wavelength of emitted electrons is ____ $\mathrm{\AA}$. (Round off to the Nearest Integer) [Use: $\sqrt{3} = 1.73$, $h = 6.63 \times 10^{-34}\,\mathrm{Js}$, $m_e = 9.1 \times 10^{-31}\,\mathrm{kg}$, $c = 3.0 \times 10^8\,\mathrm{ms}^{-1}$, $1\,\mathrm{eV} = 1.6 \times 10^{-19}\,\mathrm{J}$]
Answer: 9
Energy $= \frac{hc}{\lambda}$ $$= \frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{248 \times 10^{-9} \times 1.6 \times 10^{-19}} \, \mathrm{eV}$$ $$= \frac{6.63 \times 100}{248 \times 1.6}$$ $$= 0.05 \, \mathrm{eV} \times 100 = 5 \, \mathrm{eV}$$ Now using $E = \phi + \mathrm{K.E}$ $$5 = 3 + \mathrm{K.E}$$ $$\mathrm{K.E} = 2 \, \mathrm{eV} = 3.2 \times 10^{-19} \, \mathrm{J}$$ For de Broglie wavelength $\lambda = \frac{h}{mv}$ $$\mathrm{K.E} = \frac{1}{2} mv^2$$ So $v = \sqrt{\frac{2 \mathrm{KE}}{m}}$ Hence $\lambda = \frac{h}{\sqrt{2 \mathrm{KE} \times m}}$ $$= \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 3.2 \times 10^{-19} \times 9.1 \times 10^{-31}}}$$ $$= \frac{6.63}{7.6} \times 10^{-34 + 25} = \frac{66.3 \times 10^{-10}}{7.6}$$ $$= 8.72 \times 10^{-10} \, \mathrm{m}$$ $$\approx 9 \times 10^{-10} \, \mathrm{m}$$ $$= 9 \AA$$
Chemistry · Equilibrium · Numerical
Two salts $A_2X$ and $MX$ have the same value of solubility product of $4.0 \times 10^{-12}$. The ratio of their molar solubilities i.e. $\frac{S(A_2X)}{S(MX)}$ = . (Round off to the Nearest Integer).
Answer: 50
For $A_2X$: $$A_2X \rightarrow 2 \, A^+ + X^{2-}$$ $$2 \, S_1 S_1$$ $$K_{sp} = 4 \, S_1^3 = 4 \times 10^{-12}$$ $$S_1 = 10^{-4}$$ For $MX$: $$MX \rightarrow M^+ + X^-$$ $$S_2 S_2$$ $$K_{sp} = S_2^2 = 4 \times 10^{-12}$$ $$S_2 = 2 \times 10^{-6}$$ So, $$\frac{S_{A_2X}}{S_{MX}} = \frac{10^{-4}}{2 \times 10^{-6}} = 50$$
Chemistry · The Solid State · Numerical
A certain element crystallises in a bcc lattice of unit cell edge length $27 \, \mathrm{\AA}$. If the same element under the same conditions crystallises in the fcc lattice, the edge length of the unit cell in $\mathrm{\AA}$ will be ____. (Round off to the Nearest Integer). [Assume each lattice point has a single atom] [ Assume $\sqrt{3} = 1.73$, $\sqrt{2} = 1.41$]
Answer: 33
For BCC $\sqrt{3}a = 4r$ so $r = \frac{\sqrt{3}}{4} \times 27$ for FCC $a = 2\sqrt{2}r$ $$= 2 \times \sqrt{2} \times \frac{\sqrt{3}}{4} \times 27$$ $$= \frac{\sqrt{3}}{\sqrt{2}} \times 27$$ $$= 33$$
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 $\mathrm{K}$ is $1.0 \times 10^{-3} \, \mathrm{s}^{-1}$ and the activation energy $E_a = 11.488 \, \mathrm{kJ} \, \mathrm{mol}^{-1}$, the rate constant at 200 $\mathrm{K}$ is _____ $\times 10^{-5} \, \mathrm{s}^{-1}$. (Round of to the Nearest Integer). ( Given : $R = 8.314 \, \mathrm{J} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}$ )
Answer: 10
Given $K_{300} = 10^{-4}$ and $K_{200} = ?$. The activation energy $E_a = 11.488 \, \mathrm{KJ/mole}$ and the gas constant $R = 8.314 \, \mathrm{J/mole \cdot K}$. So, $$\ln \left( \frac{K_{300}}{K_{200}} \right) = \frac{E_a}{R} \left( \frac{1}{200} - \frac{1}{300} \right)$$ Therefore, $$\frac{K_{300}}{K_{200}} = 10$$ Now, $$\ln \left( \frac{K_{300}}{K_{200}} \right) = \frac{11.488 \times 1000 \times 100}{8.314 \times 200 \times 300}$$ This simplifies to $$= 2.303$$ Which equals $$\ln 10$$ Thus, $$K_{200} = \frac{1}{10} \times K_{300} = 10^{-4}$$ Finally, $$= 10 \times 10^{-5} \, \mathrm{sec^{-1}}$$