JEE Main 28 July 2022 Shift 1 question paper with solutions
JEE Main 28 July 2022 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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Maths
Question 1
Maths · Differential Equations · Single correct
Let the solution curve of the differential equation $$x \, dy = \left( \sqrt{x^2 + y^2} + y \right) \, dx, \; x > 0,$$ intersect the line $$x = 1$$ at $$y = 0$$ and the line $$x = 2$$ at $$y = \alpha$$. Then the value of $$\alpha$$ is:
$\frac{1}{2}$
$\frac{3}{2}$
$-\frac{3}{2}$
$\frac{5}{2}$
Answer: (b)
Solution
Given the equation $x \, dy = \left( \sqrt{x^2 + y^2 + y} \right) dx$. Rewriting, we have $$\frac{x \, dy}{x} = y \, dx - \frac{\sqrt{x^2 + y^2}}{x} dx.$$ This simplifies to $$\frac{d \left( \frac{y}{x} \right)}{\sqrt{1 + \left( \frac{y}{x} \right)^2}} = \frac{dx}{x}.$$ Integrating both sides, we get $$\ln \left( \frac{y}{x} + \sqrt{\left( \frac{y}{x} \right)^2 + 1} \right) = \ln x + R.$$ Thus, $$\frac{y + \sqrt{y^2 + x^2}}{x} = cx.$$ Therefore, $$y + \sqrt{y^2 + x^2} = cx^2.$$ Setting $x = 1, y = 0$, we find $0 + 1 = C \Rightarrow C = 1$. The curve is $y + \sqrt{x^2 + y^2} = x^2$. For $x = 2, y = \alpha$, we have $$2 + \sqrt{4 + \alpha^2} = 4.$$ Solving, $$4 + \alpha^2 = 16 + \alpha^2 = 8 \alpha.$$ Therefore, $\alpha = \frac{3}{2}$.
Question 2
Maths · Inverse Trigonometric Functions · Single correct
Considering only the principal values of the inverse trigonometric functions, the domain of the function $f(x) = \cos^{-1}\left(\frac{x^2 - 4x + 2}{x^2 + 3}\right)$ is:
Let the vectors $\vec{a} = (1+t)\hat{i} + (1-t)\hat{j} + \hat{k}$, $\vec{b} = (1-t)\hat{i} + (1+t)\hat{j} + 2\hat{k}$ and $\vec{c} = t\hat{i} - t\hat{j} + \hat{k}$, $t \in \mathbb{R}$ be such that for $\alpha, \beta, \gamma \in \mathbb{R}$, $\alpha \vec{a} + \beta \vec{b} + \gamma \vec{c} = \vec{0}$ $\Rightarrow \alpha = \beta = \gamma = 0$. Then, the set of all values of $t$ is:
Maths · Inverse Trigonometric Functions · Single correct
Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $\cos^{-1}(x) - 2\sin^{-1}(x) = \cos^{-1}(2x)$ is equal to:
0
1
$\frac{1}{2}$
$-\frac{1}{2}$
Solution
Given $\cos^{-1} x = 2 \sin^{-1} x = \cos^{-1} 2x$. $$\cos^{-1} x - 2 \left( \frac{\pi}{2} - \cos^{-1} x \right) = \cos^{-1} 2x$$ $$\cos^{-1} x - \pi + 2 \cos^{-1} x = \cos^{-1} 2x$$ $$3 \cos^2 x = \pi + \cos^{-1} 2x$$ $$\cos \left( 3 \cos^{-1} x \right) = \cos \left( \pi + \cos^{-1} 2x \right)$$ $$4x^3 - 3x = -2x$$ $$4x^3 = x \implies x = 0, \pm \frac{1}{2}$$ All satisfy the original equation. Sum $= -\frac{1}{2}$ to $+\frac{1}{2} = 0$
Question 5
Maths · Mathematical Reasoning · Single correct
Let the operations $*$, $\odot \in \{\land, \lor\}$. If $(p * q) \odot (p \odot \sim q)$ is a tautology, then the ordered pair $(*, \odot)$ is:
$(\lor, \land)$
$(\lor, \lor)$
$(\land, \land)$
$(\land, \lor)$
Solution
Well check each option. For A $\pi = \lor$ of $0 = \Lambda$. $$(p \lor q) \land (p \lor \sim q)$$ $$\equiv p \lor (q \land \sim q)$$ $$\equiv p \lor (c) \equiv p$$ For B: $* = \lor$, $O = \lor$. $$(p \lor q) \lor (p \lor \sim q) \equiv t$$ using Venn Diagrams.
Question 6
Maths · Vector Algebra · Single correct
Let a vector $\vec{a}$ has a magnitude 9. Let a vector $\vec{b}$ be such that for every $(x,y) \in \mathbb{R} \times \mathbb{R} - \{(0,0)\}$, the vector $(x\vec{a} + y\vec{b})$ is perpendicular to the vector $(6y\vec{a} - 18x\vec{b})$. Then the value of $|\vec{a} \times \vec{b}|$ is equal to:
For $t \in (0, 2\pi)$, if $ABC$ is an equilateral triangle with vertices $A(\sin t, -\cos t)$, $B(\cos t, \sin t)$ and $C(a, b)$ such that its orthocentre lies on a circle with centre $\left(1, \frac{1}{3}\right)$, then $(a^2 - b^2)$ is equal to:
$\frac{8}{3}$
8
$\frac{77}{9}$
$\frac{80}{9}$
Answer: (b)
Solution
Given $s \equiv \sin t$, $c \equiv \cos t$. Let orthocentre be $(h, k)$. Since it is an equilateral triangle, hence orthocentre coincides with centroid. Therefore, $a + s + c = 3h$, $b + s - c = 3k$. Thus, $$(3h - a)^2 + (3k - b)^2 = (s + c)^2 + (s - c)^2 = 2(s^2 + c^2) = 2$$ Therefore, $$\left( h - \frac{a}{3} \right)^2 + \left( k - \frac{b}{3} \right)^2 = \frac{2}{9},$$ circle centre at $\left( \frac{a}{3}, \frac{b}{3} \right)$. Gives, $\frac{a}{3} = 1$, $\frac{b}{3} = \frac{1}{3} \implies a = 3, b = 1$ Thus, $a^2 - b^2 = 8$
Question 8
Maths · Relations and Functions · Single correct
For $\alpha \in \mathbb{N}$, consider a relation $R$ on $\mathbb{N}$ given by $R = \{(x,y) : 3x + \alpha y$ is a multiple of $7\}$. The relation $R$ is an equivalence relation if and only if:
$\alpha = 14$
$\alpha$ is a multiple of $4$
$4$ is the remainder when $\alpha$ is divided by $10$
$4$ is the remainder when $\alpha$ is divided by $7$
Answer: (d)
Solution
For R to be reflexive $\Rightarrow \ x \, R \, x$ $$\Rightarrow \ 3x + \alpha \, x = 7x \Rightarrow (3 + \alpha) \, x = 7K$$ $$\Rightarrow \ 3 + \alpha = 7\lambda \Rightarrow \alpha = 7\lambda - 3 = 7N + 4, \, K, \, \lambda, \, N \in \mathbb{I}$$ Therefore, when $\alpha$ divided by 7, remainder is 4. R to be symmetric $xRy \Rightarrow yRx$ $$3x + \alpha y = 7N_1, \, 3y + \alpha x = 7N_2$$ $$\Rightarrow (3 + \alpha)(x + y) = 7(N_1 + N_2) = 7N_3$$ Which holds when $3 + \alpha$ is multiple of 7 Therefore, $\alpha = 7N + 4$ (as did earlier) R to be transitive $xRy \& \, yRz \Rightarrow xRz.$ $$3x + \alpha y = 7N_1 \& 3y + \alpha z = 7N_2 and$$ $$3x + \alpha z = 7N_3$$ Therefore, $3x + 7N_2 - 3y = 7N_3$ Therefore, $7N_1 - \alpha y + 7N_2 - 3y = 7N_3$ Therefore, $7(N_1 + N_2) - (3 + \alpha) y = 7N_3$ Therefore, $(3 + \alpha) y = 7N$ Which is true again when $3 + \alpha$ divisible by 7, i.e. when $\alpha$ divided by 7, remainder is 4.
Question 9
Maths · Probability · Single correct
Out of 60$\%$ female and 40$\%$ male candidates appearing in an exam, 60$\%$ candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability, that the chosen candidate is a female, is :
$\frac{2}{3}$
$\frac{11}{16}$
$\frac{23}{32}$
$\frac{13}{16}$
Answer: (a)
Solution
Probability that chosen candidate is female = $\frac{40}{60}$ = $\frac{2}{3}$
Question 10
Maths · Differential Equations · Single correct
If $$ y = y(x), \; x \in \left(0, \frac{\pi}{2}\right) \,$$ be the solution curve of the differential equation $$\left(\sin^2 2x\right) \frac{dy}{dx} + \left(8 \sin^2 2x + 2 \sin 4x\right) y = 2e^{-4x} \left(2 \sin 2x + \cos 2x\right), \; with \; y\left(\frac{\pi}{4}\right) = e^{-\pi},$$ then $$\, y\left(\frac{\pi}{6}\right) \,$$ is equal to:
$\frac{2}{\sqrt{3}} e^{-2\pi/3}$
$\frac{2}{\sqrt{3}} e^{2\pi/3}$
$\frac{1}{\sqrt{3}} e^{-2\pi/3}$
$\frac{1}{\sqrt{3}} e^{2\pi/3}$
Answer: (a)
Solution
Given differential equation can be re-written as $$\frac{dy}{dx} + \left(8 + 4 \cot 2x \right) y = \frac{2e^{-4x}}{\sin^2 2x} \left(2 \sin x + \cos 2x \right)$$ which is a linear diff. equation. I.f. = $$e^{\int (8 + 4 \cot 2x) \, dx} = e^{8x + 2 \ln(\sin 2x)}$$ = $$e^{8x} \cdot \sin^2 2x$$ Therefore, solution is $$y \left(e^{8x} \cdot \sin^2 2x \right) = \int 2e^{4x} \left(2 \sin 2x + \cos 2x \right) dx + C$$ = $$e^{4x} \cdot \sin 2x + C$$ Given $$y \left(\frac{\pi}{4} \right) = e^{-\pi} \implies C = 0$$ Therefore, $$y = \frac{e^{-4x}}{\sin 2x}$$ Thus, $$y \left(\frac{\pi}{6} \right) = \frac{e^{-\frac{4 \cdot \pi}{6}}}{\sin \left(2 \cdot \frac{\pi}{6} \right)} = \frac{2}{\sqrt{3}} e^{\frac{2\pi}{3}}$$
Question 11
Maths · Conic Sections · Single correct
If the tangents drawn at the points P and Q on the parabola $y^2 = 2x - 3$ intersect at the point R(0, 1), then the orthocentre of the triangle PQR is:
(0, 1)
(2, -1)
(6, 3)
(2, 1)
Answer: (b)
Solution
Given $y^2 = 2x - 3$ (i). Equation of chord of contact is $y = x - 3$ (2). From (1) and (2), $$(x \cdot 3)^2 = 2x - 3$$ $$x^2 - 8x + 12 = 0$$ $$(x - 2)(x - 6) = 0$$ $$x = 2 or 6$$ $$y = -1 or 3$$ MQR is $$\frac{2}{6} = \frac{1}{3}$$ MPR is $$\frac{2}{6} = \frac{1}{3}$$ MPR is $$\frac{2}{-2} = -1$$ Therefore, $MPQ \times MPR = -1 \implies PQ \perp PR$. Orthocentre is $P(2, -1)$.
Question 12
Maths · Conic Sections · Single correct
Let C be the centre of the circle $x^2 + y^2 - x + 2y = \frac{11}{4}$ and P be a point on the circle. A line passes through the point C, makes an angle of $\frac{\pi}{4}$ with the line CP and intersects the circle at the points Q and R. Then the area of the triangle PQR (in unit$^2$) is :
2
2$\sqrt{2}$
8\sin\left(\frac{\pi}{8}\right)
8\cos\left(\frac{\pi}{8}\right)
Answer: (b)
Solution
Given the equation $x^2 + y^2 - x + 2y = \frac{11}{4}$. Rewriting it, we have: $$\left(x - \frac{1}{2}\right)^2 + (y + 1)^2 = (2)^2$$ Or consider $\triangle PQR$. From the diagram, we have: $$= 4 \cdot 6 \sin \frac{\pi}{8}$$ $$PQ = QR \cos 22 \frac{1}{2}$$ $$= 4 \cos \frac{\pi}{8}$$ As $\triangle PQR = \frac{1}{2} PR \times PQ$, we have: $$= \frac{1}{2} \left(4^2 \sin \frac{\pi}{6}\right) \left(4 \cos \frac{\pi}{8}\right)$$ $$= 4 \sin \frac{\pi}{4} = \frac{4}{\sqrt{2}} = 2 \sqrt{2}$$
Question 13
Maths · Binomial Theorem · Single correct
The remainder when $7^{2022} + 3^{2022}$ is divided by 5 is:
Let the matrix $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$ and the matrix $B_0 = A^{49} + 2A^{98}$. If $B_n = Adj(B_{n-1})$ for all $n \geq 1$, then $\det(B_4)$ is equal to:
Maths · Three Dimensional Geometry · Single correct
The foot of the perpendicular from a point on the circle $x^2 + y^2 = 1$, $z = 0$ to the plane $2x + 3y + z = 6$ lies on which one of the following curves?
Maths · Applications of Derivatives · Single correct
If the minimum value of $f(x) = \frac{5x^2}{2} + \frac{\alpha}{x^5}$, $x > 0$, is 14, then the value of $\alpha$ is equal to:
32
64
128
256
Answer: (c)
Solution
Given the expression: $$\frac{x^2}{2} + \frac{x^2}{2} + \frac{x^2}{2} + \frac{x^2}{2} + \frac{\alpha}{2x^5} + \frac{\alpha}{2x^5}$$ is greater than or equal to $$7 \left( \frac{\alpha^2}{2^7} \right)^{\frac{1}{7}}$$. Solving, we have: $$\frac{7 \cdot (\alpha)^{2/7}}{2} = 14$$ $$\left(\alpha^2\right)^{1/7} = 2^2$$ $$\alpha = \left(2^2\right)^{7/2} = 2^7$$ Therefore, $$\alpha = 128$$
Question 18
Maths · Relations and Functions (Advanced) · Single correct
Let $\alpha$, $\beta$ and $\gamma$ be three positive real numbers. Let $f(x) = \alpha x^5 + \beta x^3 + \gamma x$, $x \in \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ be such that $g(f(x)) = x$ for all $x \in \mathbb{R}$. If $a_1, a_2, a_3, \ldots, a_n$ be in arithmetic progression with mean zero, then the value of $f\left(g\left(\dfrac{1}{n}\sum_{i=1}^{n} f(a_i)\right)\right)$ is equal to:
0
3
9
27
Answer: (a)
Solution
Consider a case when $\alpha = \beta = 0$ then $f(x) = yx$ $g(x) = \frac{x}{y}$ $$\frac{1}{n} \sum_{i=1}^{n} f(a_i) \Rightarrow \frac{y}{n} (a_1 + a_2 + \ldots + a_n)$$ $$= 0$$ $$\Rightarrow f(g(0)) \Rightarrow f(0)$$ $$\Rightarrow 0$$
Question 19
Maths · Sequences and Series · Single correct
Consider the sequence $a_1, a_2, a_3, \ldots$ such that $$a_1 = 1, \ a_2 = 2 \ and \ a_{n+2} = \frac{2}{a_{n+1}} + a_n \ for \ n = 1, 2, 3, \ldots$$ If $$\left( \frac{a_1 + \frac{1}{a_2}}{a_3} \right) \cdot \left( \frac{a_2 + \frac{1}{a_3}}{a_4} \right) \cdot \left( \frac{a_3 + \frac{1}{a_4}}{a_5} \right) \cdot \ldots \cdot \left( \frac{a_{30} + \frac{1}{a_{31}}}{a_{32}} \right) = 2^\alpha \left( ^{61}C_{31} \right),$$ then $\alpha$ is equal to:
The minimum value of the twice differentiable function $f(x) = \int_{0}^{x} e^{x-t} f'(t) \, dt - (x^2 - x + 1) e^x$, $x \in \mathbb{R}$, is:
-$\frac{2}{\sqrt{e}}$
-2$\sqrt{e}$
-$\sqrt{e}$
$\frac{2}{\sqrt{e}}$
Answer: (a)
Solution
Given $$f(x) = e^x \cdot \int_0^x \frac{f'(t)}{e^t} \, dt$$ We have $$f'(x) = e^x \cdot \int_0^x \frac{f'(t)}{e^t} \, dt + e^x \cdot \frac{f'(x)}{e^x}$$ This simplifies to $$-\left[ (2x-1) \cdot e^x + (x^2 - x + 1) \cdot e^x \right]$$ The integral $$\int_0^x \frac{f'(t)}{e^t} \, dt = x^2 + x$$ Thus, $$\frac{f'(x)}{e^x} = 2x + 1$$ So, $$f'(x) = (2x + 1) \cdot e^x$$ Setting $$f'(x) = 0 \Rightarrow x = -\frac{1}{2}$$ Then, $$f(x) = (2x + 1) \cdot e^x - 2e^x + C$$ Given $$f(0) = -1$$ We have $$-1 = 1 - 2 + C$$ Thus, $$C = 0$$ Therefore, $$f(x) = e^x (2x - 1)$$ Finally, $$f\left(-\frac{1}{2}\right) = \frac{-2}{\sqrt{e}}$$
Question 21
Maths · Permutations and Combinations · Fill in the blank
Let S be the set of all passwords which are six to eight characters long, where each character is either an alphabet from $\{$A, B, C, D, E$\}$ or a number from {1, 2, 3, 4, 5} with the repetition of characters allowed. If the number of passwords in S whose at least one character is a number from {1, 2, 3, 4, 5} is $\alpha \times 5^6$, then $\alpha$ is equal to _____.
Maths · Three Dimensional Geometry · Fill in the blank
Let P(-2, -1, 1) and Q $(\frac{56}{17}, \frac{43}{17}, \frac{111}{17})$ be the vertices of the rhombus PRQS. If the direction ratios of the diagonal RS are $\alpha$, -1, $\beta$, where both $\alpha$ and $\beta$ are integers of minimum absolute values, then $\alpha^2$ + $\beta^2$ is equal to ______.
Let $f : [0, 1] \to \mathbb{R}$ be a twice differentiable function in $(0, 1)$ such that $f(0) = 3$ and $f(1) = 5$. If the line $y = 2x + 3$ intersects the graph of $f$ at only two distinct points in $(0, 1)$, then the least number of points $x \in (0, 1)$, at which $f''(x) = 0$, is ______.
Answer: 2
Solution
Given $f'(a) = f'(b) = f'(c) = 2$. This implies $f''(x)$ is zero for at least $x_1 \in (a, b)$ and $x_2 \in (b, c)$.
Question 24
Maths · Integrals · Numerical
If $$\int_{0}^{\sqrt{3}} \frac{15x^3}{\sqrt{1+x^2 + \sqrt{\left(1+x^2\right)^3}}} \, dx = \alpha \sqrt{2} + \beta \sqrt{3}$$, where $\alpha$, $\beta$ are integers, then $\alpha + \beta$ is equal to
Let $A = \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}$, $\alpha, \beta \in \mathbb{R}$. Let $\alpha_1$ be the value of $\alpha$ which satisfies $(A+B)^2 = A^2 + \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$ and $\alpha_2$ be the value of $\alpha$ which satisfies $(A+B)^2 = B^2$. Then $|\alpha_1 - \alpha_2|$ is equal to _______.
For $p, q \in \mathbb{R}$, consider the real valued function $f(x) = (x - p)^2 - q$, $x \in \mathbb{R}$ and $q > 0$. Let $a_1, a_2, a_3$ and $a_4$ be in an arithmetic progression with mean $p$ and positive common difference. If $\left| f(a_i) \right| = 500$ for all $i = 1, 2, 3, 4$, then the absolute difference between the roots of $f(x) = 0$ is
For the hyperbola H : $x^2-y^2=1$ and the ellipse E : $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $a>b>0$, let the (1) eccentricity of E be reciprocal of the eccentricity of H, and (2) the line $y=\sqrt{\frac{5}{2}}x+K$ be a common tangent of E and H. Then $4(a^2+b^2)$ is equal to ______.
Answer: 3
Solution
Given $e_E = \sqrt{1 - \frac{b^2}{a^2}}$, $e_H = \sqrt{2}$. If $e_E = \frac{1}{e_H}$, then $$\frac{a^2 - b^2}{a^2} = \frac{1}{2}$$ $$2a^2 - 2b^2 = a^2$$ $$a^2 = 2b^2$$ and $y = \sqrt{\frac{5}{2}} x + k$ is tangent to the ellipse, then $$K^2 = a^2 \times \frac{5}{2} + b^2 = \frac{3}{2}$$ $$6b^2 = \frac{3}{2} \Rightarrow b^2 = \frac{1}{4} and a^2 = \frac{1}{2}$$ Therefore, $4(a^2 + b^2) = 3$.
Question 28
Maths · Sequences and Series · Numerical
Let $x_1, x_2, x_3, \ldots, x_{20}$ be in geometric progression with $x_1 = 3$ and the common ratio $\frac{1}{2}$. A new data is constructed replacing each $x_i$ by $(x_i - i)^2$. If $\overline{x}$ is the mean of new data, then the greatest integer less than or equal to $\overline{x}$ is
Physics · Physical World, Units and Measurements · Single correct
The dimensions of $\frac{B^2}{\mu_0}$ will be: (if $\mu_0$: permeability of free space and B: magnetic field)
$[M\,L^{2}\,T^{-2}]$
$[M\,L\,T^{-2}]$
$[M\,L^{-1}\,T^{-2}]$
$[M\,L^{2}\,T^{-2}\,A^{-1}]$
Answer: (c)
Solution
Given $$u = \frac{B^2}{2\mu_0}$$ $u$ represents energy per unit volume. The dimensional formula is $$\left[ \frac{B^2}{\mu_0} \right] = [u] = \left[ \frac{\mathrm{ML^2T^{-2}}}{\mathrm{L^3}} \right] = [\mathrm{ML^{-1}T^{-2}}]$$
Question 32
Physics · Motion in a Plane · Single correct
A NCC parade is going at a uniform speed of 9 km/h under a mango tree on which a monkey is sitting at a height of 19.6 m. At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is: (Given g = 9.8 m/s²)
5 m
10 m
19.8 m
24.5 m
Answer: (a)
Solution
Monkey Time taken by mango = $\sqrt{\frac{2n}{g}}$ = $\sqrt{\frac{2 \times 19.6}{9.8}}$ = 2 second Distance = vt = 9 $\times$ $\frac{5}{18}$ $\times$ 2 = 5 m
Question 33
Physics · Laws of Motion · Single correct
In two different experiments, an object of mass $5 \, \mathrm{kg}$ moving with a speed of $25 \, \mathrm{ms}^{-1}$ hits two different walls and comes to rest within (i) $3 \, \mathrm{seconds}$, (ii) $5 \, \mathrm{seconds}$, respectively. Choose the correct option out of the following:
Impulse and average force acting on the object will be same for both the cases.
Impulse will be same for both the cases but the average force will be different.
Average force will be same for both the cases but the impulse will be different.
Average force and impulse will be different for both the cases.
Answer: (b)
Solution
Impulse equals change in momentum. $I = \Delta P$ $F_{avg} = \frac{\Delta P}{\Delta t}$ $\Delta t_1 = 3$ $\Delta t_2 = 5$ $\Delta P_1 = \Delta P_2$ $I_1 = I_2$ $F_{avg}$ in case (i) is more than (ii).
Question 34
Physics · Laws of Motion · Single correct
A balloon has mass of 10 g in air. The air escapes from the balloon at a uniform rate with velocity $4.5 \, \mathrm{cm/s}$. If the balloon shrinks in $5 \, \mathrm{s}$ completely. Then, the average force acting on that balloon will be (in dyne).
3
9
12
18
Answer: (b)
Solution
The force is given by the equation $$ F = \frac{dm}{dt} v $$. Substituting the given values, we have $$ = \frac{10 \, \mathrm{g}}{5 \, \mathrm{s}} \left( \frac{4.5 \, \mathrm{cm}}{\mathrm{s}} \right) = 9 \, \frac{\mathrm{g} \cdot \mathrm{cm}}{\mathrm{s}^2} = 9 \, \mathrm{dyne} $$.
Question 35
Physics · Gravitation · Single correct
If the radius of earth shrinks by 2$\%$ while its mass remains same. The acceleration due to gravity on the earth’s surface will approximately :
decrease by 2$\%$
decrease by 4$\%$
increase by 2$\%$
increase by 4$\%$
Answer: (d)
Solution
Given $g = \frac{GM}{R^2}$. Since $M = constant$, $g \propto \frac{1}{R^2}$. Therefore, $$100 \frac{\Delta g}{g} = -2 \frac{\Delta R}{R} 100$$ The percentage change is $-2 (-2)$. Thus, the percentage change in $g$ is $4\%$. This results in an increase by $4\%$.
Question 36
Physics · Mechanical Properties of Solids · Single correct
The force required to stretch a wire of cross-section $1 \, \mathrm{cm}^2$ to double its length will be: (Given Yong's modulus of the wire $= 2 \times 10^{11} \, \mathrm{N/m}^2$)
$1 \times 10^7 \, \mathrm{N}$
$1.5 \times 10^7 \, \mathrm{N}$
$2 \times 10^7 \, \mathrm{N}$
$2.5 \times 10^7 \, \mathrm{N}$
Answer: (c)
Solution
The force is given by the formula: $$F = \gamma A \frac{\Delta \ell}{\ell}$$ Substituting the values: $$= 2 \times 10^{11} \times 10^{-4} \left( \frac{2\ell - \ell}{\ell} \right)$$ Simplifying gives: $$= 2 \times 10^7 \, \mathrm{N}$$
Question 37
Physics · Thermodynamics · Single correct
A Carnot engine has efficiency of 50$\%$. If the temperature of sink is reduced by $40^\circ$$\mathrm{C}$, its efficiency increases by 30$\%$. The temperature of the source will be:
$166.7\,\mathrm{K}$
$255.1\,\mathrm{K}$
$266.7\,\mathrm{K}$
$367.7\,\mathrm{K}$
Answer: (c)
Solution
Question 38
Physics · Kinetic Theory · Single correct
Given below are two statements : Statement I : The average momentum of a molecule in a sample of an ideal gas depends on temperature. Statement II : The rms speed of oxygen molecules in a gas is $v$. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will become $2v$. In the light of the above statements, choose the correct answer from the options given below :
Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Statement I is false but Statement II is true
Answer: (d)
Solution
Given $[P_{avg} = 0]$ (due to random motion) $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ $$T_{new} = 2T$$ $$M_{new} = \frac{M}{2}$$ $$\frac{v_{new}}{v} = \frac{\sqrt{\frac{2T}{M/2}}}{\sqrt{\frac{T}{M}}}$$ $$v_{new} = 2v$$
Question 39
Physics · Waves · Single correct
In the wave equation $$y = 0.5 \sin \frac{2\pi}{\lambda} (400\, t - x)\, \mathrm{m}$$ the velocity of the wave will be:
200 $\mathrm{m/s}$
200$\sqrt{2}$ $\mathrm{m/s}$
400 $\mathrm{m/s}$
400$\sqrt{2}$ $\mathrm{m/s}$
Answer: (c)
Solution
Given the equation of the wave: $$y = 0.5 \sin \left( \frac{2\pi}{\lambda} 400t - \frac{2\pi}{\lambda} x \right)$$ The angular frequency is given by: $$\omega = \frac{2\pi}{\lambda} 400$$ The wave number is: $$K = \frac{2\pi}{\lambda}$$ The velocity of the wave is: $$v = \frac{\omega}{k}$$ Thus, $$v = 400 \, \mathrm{m/s}$$
Question 40
Physics · Electrostatic Potential and Capacitance · Single correct
Two capacitors, each having capacitance $40 \, \mu\mathrm{F}$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24 \, \mu\mathrm{F}$. The value of $K$ will be:
1.5
2.5
1.2
3
Answer: (a)
Solution
The equivalent capacitance is given by $$C_{eq} = \frac{C(KC)}{C + KC} = \frac{KC}{K + 1}.$$ Solving for the given condition, $$24 = \frac{K \cdot 40}{K + 1}.$$ From this, we find $$K = 1.5.$$
Question 41
Physics · Current Electricity · Single correct
A wire of resistance $R_1$ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:
Physics · Moving Charges and Magnetism · Multiple correct
The current sensitivity of a galvanometer can be increased by: (A) decreasing the number of turns (B) increasing the magnetic field (C) decreasing the area of the coil (D) decreasing the torsional constant of the spring Choose the most appropriate answer from the options given below:
Physics · Moving Charges and Magnetism · Single correct
As shown in the figure, a metallic rod of linear density $0.45 \, \mathrm{kg \, m^{-1}}$ is lying horizontally on a smooth incline plane which makes an angle of $45^\circ$ with the horizontal. The minimum current flowing in the rod required to keep it stationary, when $0.15 \, \mathrm{T}$ magnetic field is acting on it in the vertical upward direction, will be: $\{$Use $g = 10 \, \mathrm{m/s^2}$$\}$
The equation of current in a purely inductive circuit is $5\sin(49\pi t - 30^\circ)$. If the inductance is $30\,\mathrm{mH}$ then the equation for the voltage across the inductor, will be: $$\left\{\text{Let } \pi = \frac{22}{7}\right\}$$
$1.47 \sin(49 \pi t - 30^\circ)$
$1.47 \sin(49 \pi t + 60^\circ)$
$23.1 \sin(49 \pi t - 30^\circ)$
$23.1 \sin(49 \pi t + 60^\circ)$
Answer: (d)
Solution
Given $v_0 = i_0 x_L$. $$= i_0 (wL)$$ $$= (5)(49\pi)(30 \times 10^{-3})$$ $$= 23.1$$ Voltage will lead current by $90^\circ$. Therefore, $V = 23.1 \sin(49 \pi t + 60^\circ)$.
Question 45
Physics · Ray Optics and Optical Instruments · Single correct
As shown in the figure, after passing through the medium 1. The speed of light $v_2$ in medium 2 will be: (Given $c = 3 \times 10^8 \, \mathrm{ms}^{-1}$)
Physics · Ray Optics and Optical Instruments · Single correct
In normal adjustment, for a refracting telescope, the distance between objective and eye piece is 30 $\mathrm{cm}$. The focal length of the objective, when the angular magnification of the telescope is 2, will be:
Physics · Dual Nature of Radiation and Matter · Single correct
The equation $\lambda = \frac{1.227}{x} \, \mathrm{nm}$ can be used to find the de-Brogli wavelength of an electron. In this equation $x$ stands for: Where, m = mass of electron P = momentum of electron K = Kinetic energy of electron V = Accelerating potential in volts for electron
$\sqrt{mK}$
$\sqrt{P}$
$\sqrt{K}$
$\sqrt{V}$
Answer: (d)
Solution
Given the de-Broglie's wavelength, $$\lambda = \frac{h}{m \nu}$$. We have $$\lambda = \frac{h}{\sqrt{2m(K \cdot E)}}$$. This can be rewritten as $$h = \frac{h}{\sqrt{2mqV}}$$. Putting the values of $m$ and $q$, we get $$\lambda = \frac{1 \cdot 22}{\sqrt{V}} \, \mathrm{nm}$$.
Question 48
Physics · Nuclei · Single correct
The half life period of a radioactive substance is 60 days. The time taken for $\frac{7}{8}$ of its original mass to disintegrate will be:
120 days
130 days
180 days
20 days
Answer: (c)
Solution
7/8 disintegrates means 1/8 remains. Or $$\left( \frac{1}{2} \right)^3$$. Therefore, 3 half lives = 180 days.
Question 49
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
Identify the solar cell characteristics from the following options:
Answer: (b)
Solution
Conceptual / theory
Question 50
Physics · Communication Systems · Single correct
In the case of amplitude modulation to avoid distortion the modulation index ($\mu$) should be:
$\mu \leq 1$
$\mu \geq 1$
$\mu = 2$
$\mu = 0$
Answer: (a)
Solution
Given $$\mu = \frac{A_m}{A_c}$$ $$\mu \leq 1$$ to avoid distortion because $$\mu > 1$$ will result in interference between carrier frequency and message frequency.
Question 51
Physics · Mathematics in Physics · Numerical
If the projection of $2\hat{i} + 4\hat{j} - 2\hat{k}$ on $\hat{i} + 2\hat{j} + \alpha \hat{k}$ is zero. Then, the value of $\alpha$ will be
A freshly prepared radioactive source of half life 2 hours 30 minutes emits radiation which is 64 times the permissible safe level. The minimum time, after which it would be possible to work safely with source, will be ______ hours.
In a Young's double slit experiment, a laser light of 560 $\mathrm{nm}$ produces an interference pattern with consecutive bright fringes' separation of 7.2 $\mathrm{mm}$. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of 8.1 $\mathrm{mm}$. The wavelength of second light is $\mathrm{nm}$.
The frequencies at which the current amplitude in an LCR series circuit becomes $\frac{1}{\sqrt{2}}$ times its maximum value, are $212 \, \mathrm{rad} \, \mathrm{s}^{-1}$ and $232 \, \mathrm{rad} \, \mathrm{s}^{-1}$. The value of resistance in the circuit is $R = 5 \, \Omega$. The self inductance in the circuit is ______ mH.
As shown in the figure, a potentiometer wire of resistance $20\,\Omega$ and length $300\,\mathrm{cm}$ is connected with resistance box (R.B.) and a standard cell of emf $4\,\mathrm{V}$. For a resistance ‘R’ of resistance box introduced into the circuit, the null point for a cell of $20\,\mathrm{mV}$ is found to be $60\,\mathrm{cm}$. The value of ‘R’ is _________ $\Omega$.
Answer: 780
Solution
Given the equation for E: $$E = \frac{AC}{AB} (V_A - V_B)$$ Therefore, $$20 \times 10^{-3} = \frac{60}{300} \times \frac{4 \times 20}{R + 20}$$ Therefore, $$R = \boxed{780} \, \Omega$$
Question 56
Physics · Electric Charges and Fields · Numerical
Two electric dipoles of dipole moments $1.2 \times 10^{-30} \, \mathrm{cm}$ and $2.4 \times 10^{-30} \, \mathrm{cm}$ are placed in two different uniform electric fields of strengths $5 \times 10^{4} \, \mathrm{NC}^{-1}$ and $15 \times 10^{4} \, \mathrm{NC}^{-1}$ respectively. The ratio of maximum torque experienced by the electric dipoles will be $\frac{1}{x}$. The value of $x$ is _______.
The frequency of echo will be $\underline{\hspace{1cm}}$ Hz if the train blowing a whistle of frequency $320$ Hz is moving with a velocity of $36$ km/h towards a hill from which an echo is heard by the train driver. Velocity of sound in air is $330$ m/s.
Answer: 340
Solution
The hill will be a secondary source. $f_1$ = frequency of the car with respect to the hill $$f_1 = \left( \frac{v}{v - v_s} \right) f = \left( \frac{330}{320} \right) \times 320 = 330 \, \mathrm{Hz}$$ $f_2$ = Frequency of the sound reflected by the hill with respect to the car (echo) $$f_2 = \left( \frac{v + v_0}{v} \right) f_1 = \left( \frac{330 + 10}{330} \right) \times 330 = 340 \, \mathrm{Hz}$$
Question 58
Physics · Mechanical Properties of Fluids · Numerical
The diameter of an air bubble which was initially 2 $\mathrm{mm}$, rises steadily through a solution of density $1750 \, \mathrm{kg \, m^{-3}}$ at the rate of $0.35 \, \mathrm{cm \, s^{-1}}$. The coefficient of viscosity of the solution is ______ poise (in nearest integer). (the density of air is negligible).
Answer: 11
Solution
As the bubble is rising steadily the net force acting on it will be zero. (Because of density of air the value of mg can be neglected) So $B = F \implies \frac{4\pi}{3} R^3 \rho g = 6\pi \eta R v$ Putting $R = 1 \, \mathrm{mm} = 10^{-3} \, \mathrm{m}$ $\rho = 1.75 \times 10^3 \, \mathrm{kg/m^3}$ $g = 10 \, \mathrm{m/s^2}$ $v = 0.35 \times 10^{-2} \, \mathrm{m/s}$ $\eta = \frac{10}{9} \cong 1.11 \, \mathrm{SI \, unit} = 11 \, \mathrm{poise \, (CGS)}$
Question 59
Physics · Work, Energy and Power · Numerical
A block of mass 'm' (as shown in figure) moving with kinetic energy $E$ compresses a spring through a distance $25 \, \mathrm{cm}$ when, its speed is halved. The value of spring constant of used spring will be $nE \, \mathrm{Nm}^{-1}$ for n = _____________.
Answer: 24
Solution
Using work-energy theorem $$W_{net} = (K_f - K_i)$$ $$\Rightarrow -\frac{1}{2} K x^2 = \frac{1}{2} m \left( \frac{v}{2} \right)^2 - \frac{1}{2} m v^2 = \frac{E}{4} - E$$ $$\Rightarrow \frac{1}{2} K x^2 = \frac{3E}{4} \Rightarrow K = \frac{3E}{2x^2}$$ $$\Rightarrow K = \frac{3E}{2 \times \left( \frac{1}{4} \right)^2} = 24E$$ $n = 24$
Question 60
Physics · System of Particles and Rotational Motion · Numerical
Four identical discs each of mass 'M' and diameter 'a' are arranged in a small plane as shown in figure. If the moment of inertia of the system about OO' is $\frac{x}{4}Ma^2$. Then, the value of $x$ will be _____.
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Match List-I with List-II. \begin{tabular}{|c|p{7cm}|c|l|} \hline \multicolumn{2}{|c|}{\textbf{List-I}} & \multicolumn{2}{c|}{\textbf{List-II}} \\ \multicolumn{2}{|c|}{\textbf{Reaction}} & \multicolumn{2}{c|}{\textbf{Catalyst}} \\ \hline (A) & $4NH_3(g)+5O_2(g)\rightarrow4NO(g)+6H_2O(g)$ & (I) & NO(g) \\ \hline (B) & $N_2(g)+3H_2(g)\rightarrow2NH_3(g)$ & (II) & H$_2$SO$_4(l)$ \\ \hline (C) & $C_{12}H_{22}O_{11}(aq)+H_2O(l)\rightarrow C_6H_{12}O_6$ (Glucose) $+$ C$_6$H$_{12}$O$_6$ (Fructose) & (III) & Pt(s) \\ \hline (D) & $2SO_2(g)+O_2(g)\rightarrow2SO_3(g)$ & (IV) & Fe(s) \\ \hline \end{tabular} Choose the correct answer from the options given below:
(A) - (II), (B) - (III), $(C)$ - (I), (D) - (IV)
(A) - (III), (B) - (II), $(C)$ - (I), (D) - (IV)
(A) - (III), (B) - (IV), $(C)$ - (II), (D) - (I)
(A) - (III), (B) - (II), $(C)$ - (IV), (D) - (I)
Answer: (c)
Solution
(a) $4\mathrm{NH_3}(g) + 5\mathrm{O_2}(g) \xrightarrow{\mathrm{Pt(s)}} 4\mathrm{NO}(g) + 6\mathrm{H_2O}(g)$ Ostwald process 500 K (b) $\mathrm{N_2} + 3\mathrm{H_2} \xrightarrow{\mathrm{Fe(s)}} 2\mathrm{NH_3}(g)$ Haber’s process $(c)$ $\mathrm{C_{12}H_{22}O_{11}}(aq.) + \mathrm{H_2O}(\ell) \xrightarrow{\mathrm{H^+}} \mathrm{C_6H_{12}O_6} + \mathrm{C_6H_{12}O_6}$ Inversion of sugar cane (d) $2\mathrm{SO_2}(g) + \mathrm{O_2}(g) \xrightarrow{\mathrm{NO(g)}} 2\mathrm{SO_3}(g)$
Question 65
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
In which of the following pairs, electron gain enthalpies of constituent elements are nearly the same or identical? (A) Rb and Cs (B) Na and K $(C)$ Ar and Kr (D) I and At Choose the correct answer from the options given below:
$(A)$ and (B) only
$(B)$ and $(C)$ only
$(A)$ and $(C)$ only
$(C)$ and (D) only
Answer: (c)
Solution
Rb and Cs have nearly same electron gain enthalpy. Electron gain enthalpy is $-46 \, \mathrm{kJ/mol}$. Ar and Kr have same $\Delta H_{eq}$. Value is $+96 \, \mathrm{kJ/mol}$.
Question 66
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Which of the reaction is suitable for concentrating ore by leaching process?
Chemistry · The d-and f-Block Elements · Single correct
Which of the following has least tendency to liberate $\mathrm{H}_2$ from mineral acids?
Cu
Mn
Ni
Zn
Answer: (a)
Solution
Copper is least electropositive among the given metals and it lies below H in reactivity series.
Question 71
Chemistry · Environmental Chemistry · Single correct
Given below are two statements : Statement I : In polluted water values of both dissolved oxygen and BOD are very low. Statement II : Eutrophication results in decrease in the amount of dissolved oxygen. In the light of the above statements, choose the most appropriate answer from the options given below :
Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Statement I is false but Statement II is true
Answer: (d)
Solution
Since eutrophication is the result of excessive growth of weed in water bodies, which consume dissolved oxygen of water bodies. Therefore, eutrophication decreases the amount of dissolved oxygen in water bodies. Polluted water has a low value of dissolved oxygen, but a high value of BOD (Biological oxygen demand), since chemical and organic matter requires dissolved oxygen to decompose.
Question 72
Chemistry · Co-ordination Compounds · Single correct
Match List-I with List-II. Choose the correct answer from the options given below:
(A) - (II), (B) - (I), ($C$) - (IV), (D) - (III)
(A) - (IV), (B) - (III), ($C$) - (I), (D) - (II)
(A) - (III), (B) - (IV), ($C$) - (I), (D) - (II)
(A) - (IV), (B) - (III), ($C$) - (II), (D) - (I)
Answer: (c)
Solution
Question 73
Chemistry · Hydrocarbons · Single correct
Choose the correct option for the following reactions.
'A' and 'B' are both Markovnikov addition products.
'A' is Markovnikov product and 'B' is anti-Markovnikov product.
'A' and 'B' are both anti-Markovnikov products.
'B' is Markovnikov and 'A' is anti-Markovnikov product.
Answer: (b)
Solution
The reaction of the given alkene with $\mathrm{Hg(OAc)_2}$ and $\mathrm{H_2O}$ followed by $\mathrm{NaBH_4}$ results in the Markovnikov product (A). The reaction with $\mathrm{B_2H_6}$ and $\mathrm{H_2O_2/ OH^-}$ results in the anti-Markovnikov product (B).
Question 74
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Among the following marked proton of which compound shows lowest $pK_a$ value?
Answer: (c)
Solution
Question 75
Chemistry · Alcohols, Phenols and Ethers · Single correct
Identify the major product A and B for the below given reaction sequence.
Answer: (b)
Solution
The reaction sequence involves the following steps: 1. Friedel-Crafts alkylation of benzene with isopropyl chloride in the presence of AlCl3 to form isopropylbenzene. 2. Oxidation of isopropylbenzene with oxygen in acidic medium to form phenol (P). 3. Bromination of phenol in carbon disulfide to form bromophenol (B). 4. Oxidation of phenol with Na2Cr2O7 and H2SO4 to form benzoquinone (A).
Question 76
Chemistry · Amines · Single correct
Identify the correct statement for the below given transformation.
Ethane-1, 2-diol and Benzene-1, 3 dicarboxylic acid
Propane-1, 2-diol and Benzene-1, 4 dicarboxylic acid
Ethane-1, 2-diol and Benzene-1, 4 dicarboxylic acid
Ethane-1, 2-diol and Benzene-1, 2 dicarboxylic acid
Answer: (c)
Solution
Question 78
Chemistry · Biomolecules · Single correct
For the below given cyclic hemiacetal (X), the correct pyranose structure is:
Answer: (d)
Solution
The correct pyranose structure is shown. The corresponding aldopyranose is depicted in the cyclic form as a hemiacetal.
Question 79
Chemistry · Chemistry in Everyday Life · Single correct
Statements about Enzyme Inhibitor Drugs are given below : (A) There are Competitive and Non-competitive inhibitor drugs. (B) These can bind at the active sites and allosteric sites. (C) Competitive Drugs are allosteric site blocking drugs. (D) Non-competitive Drugs are active site blocking drugs. Choose the correct answer from the options given below :
$(A), (D)$ only
$(A), (C)$ only
$(A), (B)$ only
$(A), (B), (C)$ only
Answer: (c)
Solution
Enzyme inhibitors can be competitive inhibitors (inhibit the attachment of substrate on active site of enzyme) and non-competitive inhibitor (changes the active site of enzyme after binding at allosteric site).
Question 80
Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct
For kinetic study of the reaction of iodide ion with $H_2O_2$ at room temperature: (A) Always use freshly prepared starch solution. (B) Always keep the concentration of sodium thiosulphate solution less than that of KI solution. (C) Record the time immediately after the appearance of blue colour. (D) Record the time immediately before the appearance of blue colour. (E) Always keep the concentration of sodium thiosulphate solution more than that of KI solution. Choose the correct answer from the options given below:
, (B), $(C)$ only
, (D), (E) only
, (E) only
, (B), (E) only
Answer: (a)
Solution
The is recorded immediately after the blue colour appears. $\mathrm{Na_2S_2O_3}$ is kept in limited amount.
Question 81
Chemistry · Some Basic Concepts of Chemistry · Numerical
In the given reaction, $$\mathrm{X + Y + 3Z \rightleftharpoons XYZ_3}$$ if one mole of each of X and Y with 0.05 mol of Z gives compound XYZ$_3$. (Given : Atomic masses of X, Y and Z are 10, 20 and 30 amu, respectively). The yield of XYZ$_3$ is _________ g. (Nearest integer)
Answer: 2
Solution
Given the reaction: $$\mathrm{X + Y + 3Z \rightleftharpoons XYZ_3}$$ with initial moles: $1 mol$ of X, $1 mol$ of Y, and $0.05 mol$ of Z. Z is the limiting reagent (L.R.). The calculation for moles of $\mathrm{XYZ_3}$ is: $$\frac{0.05}{3} = 1 mole of \mathrm{XYZ_3}$$ The mass of $\mathrm{XYZ_3}$ is calculated as: $$Mass of \mathrm{XYZ_3} = \frac{0.05}{3} \times (10 + 20 + 30 \times 3)$$ This simplifies to: $$= 2 g$$
Question 82
Chemistry · The Solid State · Numerical
An element M crystallises in a body centred cubic unit cell with a cell edge of 300 pm. The density of the element is $6.0 \, \mathrm{g \, cm^{-3}}$. The number of atoms present in 180 g of the element is _____ $\times \, 10^{23}$. (Nearest integer)
Answer: 22
Solution
M is body centred cubic, therefore $Z = 2$. Let mass of 1 atom of M be $A$. Edge length $= 300 \, \mathrm{pm}$. Density $= 6 \, \mathrm{g/cm^3}$. Therefore, $$6 \, \mathrm{g/cm^3} = \frac{Z \times A}{(300 \times 10^{-10})^3} = \frac{2 \times A}{27 \times 10^{-24}}$$ $A = 81 \times 10^{-24} \, \mathrm{g}$. Therefore, Atomic mass $= 48.6 \, \mathrm{g}$. Therefore, Mole in $180 \, \mathrm{g} = \frac{180}{48.6} = 3.7$ moles. Atoms of M $= 3.7 \times 6 \times 10^{23} = 22.22 \times 10^{23}$ atoms.
Question 83
Chemistry · Chemical Bonding and Molecular Structure · Numerical
The number of paramagnetic species among the following is __________. $\mathrm{B_2}$, $\mathrm{Li_2}$, $\mathrm{C_2}$, $\mathrm{C_2^-}$, $\mathrm{O_2^{2-}}$, $\mathrm{O_2}$ and $\mathrm{He_2^+}$
$150\,\mathrm{g}$ of acetic acid was contaminated with $10.2\,\mathrm{g}$ of ascorbic acid ($\mathrm{C_6H_8O_6}$) to lower down its freezing point by $(x \times 10^{-1})°\mathrm{C}$. The value of $x$ is \_\_\_\_. (Nearest integer) [Given $K_f = 3.9\,\mathrm{K\,kg\,mol^{-1}}$; Molar mass of ascorbic acid $= 176\,\mathrm{g\,mol^{-1}}$]
$K_a$ for butyric acid $(\mathrm{C_3H_7COOH})$ is $2 \times 10^{-5}$. The pH of $0.2 \, \mathrm{M}$ solution of butyric acid is $\_\_\_ \times 10^{-1}$. (Nearest integer) [Given $\log 2 = 0.30$]
Answer: 27
Solution
Given $K_a$ of Butyric acid $\Rightarrow 2 \times 10^{-5}$, $\mathrm{p}K_a = 4.7$. Find the pH of a $0.2 \, \mathrm{M}$ solution. $$\mathrm{pH} = \frac{1}{2} \mathrm{p}K_a - \frac{1}{2} \log C$$ Substitute the values: $$= \frac{1}{2} (4.7) - \frac{1}{2} \log (0.2)$$ Calculate: $$= 2.35 + 0.35 = 2.7$$ Thus, $$\mathrm{pH} = 27 \times 10^{-1}$$
Question 86
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
For the given first order reaction A $\rightarrow$ B the half life of the reaction is 0.3010 min. The ratio of the initial concentration of reactant to the concentration of reactant at time 2.0 min will be equal to . (Nearest integer)
Answer: 100
Solution
Given the reaction $\mathrm{A} \rightarrow \mathrm{B}$ with $t_{1/2} = 0.3010 \, \mathrm{min}$. We need to find $A_0/A_t$ at time $2 \, \mathrm{min}$. The rate constant $K$ is given by $$K = \frac{2.303}{t} \log \left[ \frac{A_0}{A_t} \right].$$ Thus, $$\frac{0.693}{t_{1/2}} = \frac{2 \cdot 2.303}{2} \log \left( \frac{A_0}{A_t} \right).$$ Or, $$\frac{2.303 \times 0.3010}{0.3010} = \frac{2.303}{2} \log \frac{A_0}{A_t}.$$ This simplifies to $$\log \frac{A_0}{A_t} = 2.$$ Therefore, $$\frac{A_0}{A_t} = 10^2 = 100.$$
Question 87
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Numerical
The number of interhalogens from the following having square pyramidal structure is : $ClF_3, IF_7, BrF_5, BrF_3, I_2Cl_6, IF_5, ClF, ClF_5$
Answer: 3
Solution
Square pyramidal structures are $\mathrm{BrF_5}$, $\mathrm{IF_5}$ and $\mathrm{ClF_5}$.
Question 88
Chemistry · The d-and f-Block Elements · Numerical
The disproportionation of $\mathrm{MnO}_4^{2-}$ in acidic medium resulted in the formation of two manganese compounds A and B. If the oxidation state of Mn in B is smaller than that of A, then the spin-only magnetic moment ($\mu$) value of B in BM is ___________. (Nearest integer)
Answer: 4
Solution
The reaction is given by: $$\mathrm{MnO_4^{2-} \xrightarrow{H^+} MnO_4^- + MnO_2}$$ The number of unpaired electrons is 3. Therefore, the magnetic moment $\mu$ is calculated as: $$\mu = \sqrt{15} = 3.877$$ The nearest integer is 4.
Question 89
Chemistry · Co-ordination Compounds · Numerical
Total number of relatively more stable isomer(s) possible for octahedral complex $[Cu(en)_2(SCN)_2]$ will be __________.
Answer: 3
Solution
The compound $[\mathrm{Cu(en)_2(SCN)_2}]$ can exhibit different geometrical isomers. The diagrams show possible configurations of the ligands around the copper center. The ethylenediamine (en) ligands are bidentate, and the thiocyanate (SCN) ligands can bind in different orientations, leading to isomerism.
Question 90
Chemistry · Some Basic Concepts of Chemistry · Numerical
On complete combustion of $0.492$ g of an organic compound containing C, H and O, $0.7938$ g of $\mathrm{CO_2}$ and $0.4428$ g of $\mathrm{H_2O}$ was produced. The % composition of oxygen in the compound is _____.
Answer: 46
Solution
$0.492\,\mathrm{g}$ of $\mathrm{C_xH_yO_z}$ gives $0.7938\,\mathrm{g}$ of $\mathrm{CO_2}=0.018$ moles. $0.4428\,\mathrm{g}$ of $\mathrm{H_2O}=0.0246$ moles. So moles of C $=0.018\Rightarrow0.216\,\mathrm{g}$. Moles of H $=0.049\Rightarrow0.049\,\mathrm{g}$. $\therefore$ wt. of Oxygen $=0.492-0.216-0.049=0.227\,\mathrm{g}$. $\%$ of Oxygen $=\frac{0.227}{0.492}\times100\approx46$.