JEE Main 31 August 2021 Shift 1 question paper with solutions
JEE Main 31 August 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
Question 1
Maths · Mathematical Reasoning · Single correct
Let $\star$, $\square$ $\in$ $\{$$\land$, $\lor$$\}$ be such that the Boolean expression $(p \star \sim q) \Rightarrow (p \square q)$ is a tautology. Then:
$\star$ = $\lor$, $\square$ = $\lor$
$\star$ = $\land$, $\square$ = $\land$
$\star$ = $\land$, $\square$ = $\lor$
$\star$ = $\lor$, $\square$ = $\land$
Answer: (c)
Solution
To determine if $(p \land \sim q) \rightarrow (p \lor q)$ is a tautology, we construct a truth table. The columns represent the truth values of $p$, $q$, $\sim q$, $p \land \sim q$, $p \lor q$, and $(p \land \sim q) \rightarrow (p \lor q)$. For each row: 1. If $p = T$ and $q = T$, then $\sim q = F$, $p \land \sim q = F$, $p \lor q = T$, and $(p \land \sim q) \rightarrow (p \lor q) = T$. 2. If $p = T$ and $q = F$, then $\sim q = T$, $p \land \sim q = T$, $p \lor q = T$, and $(p \land \sim q) \rightarrow (p \lor q) = T$. 3. If $p = F$ and $q = T$, then $\sim q = F$, $p \land \sim q = F$, $p \lor q = T$, and $(p \land \sim q) \rightarrow (p \lor q) = T$. 4. If $p = F$ and $q = F$, then $\sim q = T$, $p \land \sim q = F$, $p \lor q = F$, and $(p \land \sim q) \rightarrow (p \lor q) = T$. Since $(p \land \sim q) \rightarrow (p \lor q)$ is true for all possible truth values of $p$ and $q$, it is a tautology.
Question 2
Maths · Continuity and Differentiability · Single correct
The number of real roots of the equation $e^{4x} + 2e^{3x} - e^x - 6 = 0$ is:
2
4
1
0
Answer: (c)
Solution
Let $e^x = t > 0$. $f(t) = t^4 + 2t^3 - t - 6 = 0$. $f'(t) = 4t^3 + 6t^2 - 1$. $f''(t) = 12t^2 + 12t > 0$. $f(0) = -6$, $f(1) = -4$, $f(2) = 24$. Therefore, the number of real roots is 1.
Question 3
Maths · Sequences and Series · Single correct
The sum of 10 terms of the series $\frac{3}{1^2 \times 2^2} + \frac{5}{2^2 \times 3^2} + \frac{7}{3^2 \times 4^2} + \ldots$ is:
1
$\frac{120}{121}$
$\frac{99}{100}$
$\frac{143}{144}$
Answer: (b)
Solution
Given $$S = \frac{2^2 - 1^2}{1^2 \times 2^2} + \frac{3^2 - 2^2}{2^2 \times 3^2} + \frac{4^2 - 3^2}{3^2 \times 4^2} + \cdots$$ This can be rewritten as $$= \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] + \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] + \left[ \frac{1}{3^2} - \frac{1}{4^2} \right] + \cdots + \left[ \frac{1}{10^2} - \frac{1}{11^2} \right]$$ Simplifying, we get $$= 1 - \frac{1}{121}$$ Finally, $$= \frac{120}{121}$$
Question 4
Maths · Three Dimensional Geometry · Single correct
Let the equation of the plane, that passes through the point $(1, 4, -3)$ and contains the line of intersection of the planes $3x - 2y + 4z - 7 = 0$ and $x + 5y - 2z + 9 = 0$, be $\alpha x + \beta y + \gamma z + 3 = 0$, then $\alpha + \beta + \gamma$ is equal to:
Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If $\int_0^x \sqrt{1 - (f'(t))^2} \, dt = \int_0^x f(t) \, dt$, $0 \leq x \leq 1$ and $f(0) = 0$, then $\lim_{x \to 0} \frac{1}{x^2} \int_0^x f(t) \, dt$ :
equals 0
equals 1
does not exist
equals $\frac{1}{2}$
Answer: (d)
Solution
Given $$\int_0^x \sqrt{1 - (f'(t))^2} \, dt = \int_0^x f(t) \, dt$$ for $$0 \leq x \leq 1$$. Differentiating both sides, we have $$\sqrt{1 - (f'(x))^2} = f(x)$$. This implies $$1 - (f'(x))^2 = f^2(x)$$. Therefore, $$\frac{f'(x)}{\sqrt{1 - f^2(x)}} = 1$$. Integrating, we get $$\sin^{-1} f(x) = x + C$$. Since $$f(0) = 0$$, it follows that $$C = 0$$, and thus $$f(x) = \sin x$$. Now, $$\lim_{x \to 0} \frac{\int_0^x \sin t \, dt}{x^2} = \frac{0}{0} = \frac{1}{2}$$.
Question 6
Maths · Vector Algebra · Single correct
Let $\vec{a}$ and $\vec{b}$ be two vectors such that $$|2\vec{a} + 3\vec{b}| = |3\vec{a} + \vec{b}|$$ and the angle between $\vec{a}$ and $\vec{b}$ is $60^\circ$. If $\frac{1}{8} \vec{a}$ is a unit vector, then $|\vec{b}|$ is equal to:
Three numbers are in an increasing geometric progression with common ratio $r$. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference $d$. If the fourth term of GP is $3r^2$, then $r^2 - d$ is equal to:
$7 - 7\sqrt{3}$
$7 + \sqrt{3}$
$7 - \sqrt{3}$
$7 + 3\sqrt{3}$
Answer: (b)
Solution
Let numbers be $\frac{a}{r}$, $a$, $ar \to G.P$ $\frac{a}{r}$, $2a$, $ar \to A \cdot P \Rightarrow 4a = \frac{a}{r} + ar \Rightarrow r + \frac{1}{r} = 4$ $r = 2 \pm \sqrt{3}$ $4^{th}$ form of $G.P = 3r^2 \Rightarrow ar^2 = 3r^2 \Rightarrow a = 3$ $r = 2 + \sqrt{3}$, $a = 3$, $d = 2a - \frac{a}{r} = 3\sqrt{3}$ $r^2 - d = (2 + \sqrt{3})^2 - 3\sqrt{3}$ $= 7 + 4\sqrt{3} - 3\sqrt{3}$ $= 7 + \sqrt{3}$
Question 9
Maths · Sets · Single correct
Which of the following is not correct for relation R on the set of real numbers?
$(x, y) \in \mathbb{R} \iff 0 < |x| - |y| \leq 1$ is neither transitive nor symmetric.
$(x, y) \in \mathbb{R} \iff 0 < |x - y| \leq 1$ is symmetric and transitive.
$(x, y) \in \mathbb{R} \iff |x| - |y| \leq 1$ is reflexive but not symmetric.
$(x, y) \in \mathbb{R} \iff |x - y| \leq 1$ is reflexive and symmetric.
Answer: (b)
Solution
Note that $(1, 2)$ and $(2, 3)$ satisfy $0 < |x - y| \leq 1$ but $(1, 3)$ does not satisfy it so $0 \leq |x - y| \leq 1$ is symmetric but not transitive. So, $(2)$ is correct.
Question 10
Maths · Integrals · Single correct
The integral $\int \frac{1}{4\sqrt{(x-1)^3(x+2)^5}} \, dx$ is equal to: (where $C$ is a constant of integration)
Given $$ \int \frac{\mathrm{dx}}{(x-1)^{3/4}(x+2)^{5/4}} $$ This is equal to $$ \int \frac{\mathrm{dx}}{\left(\frac{x+2}{x-1}\right)^{5/4} \cdot (x-1)^2} $$ Put $$ \frac{x+2}{x-1} = t $$ This becomes $$ = -\frac{1}{3} \int \frac{\mathrm{dt}}{t^{5/4}} $$ Which simplifies to $$ = \frac{4}{3} \cdot \frac{1}{t^{1/4}} + C $$ Finally, $$ = \frac{4}{3} \left( \frac{x-1}{x+2} \right)^{1/4} + C $$
Question 11
Maths · Straight Lines and Pair of Straight Lines · Single correct
If $p$ and $q$ are the lengths of the perpendiculars from the origin on the lines, $x \csc \alpha - y \sec \alpha = k \cot 2\alpha$ and $x \sin \alpha + y \cos \alpha = k \sin 2\alpha$ respectively, then $k^2$ is equal to:
$4p^2 + q^2$
$2p^2 + q^2$
$p^2 + 2q^2$
$p^2 + 4q^2$
Answer: (a)
Solution
First line is $\frac{x}{\sin \alpha} - \frac{y}{\cos \alpha} = \frac{k \cos 2\alpha}{\sin 2\alpha}$ $$\Rightarrow x \cos \alpha - y \sin \alpha = \frac{k}{2} \cos 2\alpha$$ $$\Rightarrow p = \left| \frac{k}{2} \cos \alpha \right| \Rightarrow 2p = \left| k \cos 2\alpha \right| ...(i)$$ Second line is $x \sin \alpha + y \cos \alpha = k \sin 2\alpha$ $$\Rightarrow q = \left| k \sin 2\alpha \right| ...(ii)$$ Hence $4p^2 + q^2 = k^2$ (From (i) $\&$ (ii))
Question 12
Maths · Complex Numbers and Quadratic Equations · Single correct
cosec $18^\circ$ is a root of the equation:
$x^2 + 2x - 4 = 0$
$4x^2 + 2x - 1 = 0$
$x^2 - 2x + 4 = 0$
$x^2 - 2x - 4 = 0$
Answer: (d)
Solution
cosec $18^\circ = \frac{1}{\sin 18^\circ} = \frac{4}{\sqrt{5} - 1} = \sqrt{5} + 1$ Let cosec $18^\circ = x = \sqrt{5} + 1$ $$\Rightarrow x - 1 = \sqrt{5}$$ Squaring both sides, we get $$x^2 - 2x + 1 = 5$$ $$\Rightarrow x^2 - 2x - 4 = 0$$
Question 13
Maths · Determinants · Single correct
If the following system of linear equations $$2x + y + z = 5$$ $$x - y + z = 3$$ $$x + y + az = b$$ has no solution, then :
The length of the latus rectum of a parabola, whose vertex and focus are on the positive $x$-axis at a distance $R$ and $S(> R)$ respectively from the origin, is:
$4(S + R)$
$2(S - R)$
$4(S - R)$
$2(S + R)$
Answer: (c)
Solution
V $\rightarrow$ Vertex F $\rightarrow$ focus VF = S - R So latus rectum = 4(S - R)
Question 15
Maths · Continuity and Differentiability · Single correct
If the function $f(x) = \begin{cases} \frac{1}{x} \log_e \left( \frac{1 + \frac{x}{a}}{1 - \frac{x}{b}} \right), & x 0 \end{cases}$ is continuous at $x = 0$, then $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ is equal to:
If $\frac{dy}{dx} = \frac{2^{x + y} - 2^x}{2^y}$, $y(0) = 1$, then $y(1)$ is equal to:
$\log$_2(2 + e)
$\log$_2(1 + e)
$\log$_2(2e)
$\log$_2(1 + e^2)
Answer: (b)
Solution
Given $\($ $\frac{dy}{dx}$ = $\frac{2^x 2^y - 2^x}{2^y}$ $\)$ $\($ 2^y $\frac{dy}{dx}$ = 2^x (2^y - 1) $\)$ $\($ $\int$ $\frac{2^y}{2^y - 1}$ $\,$ dy = $\int$ 2^x $\,$ dx $\)$ $\($ $\frac{\ln(2^y - 1)}{\ln 2}$ = $\frac{2^x}{\ln 2}$ + C $\)$ $\($ $\Rightarrow$ $\log$_2(2^y - 1) = 2^x $\log$_2 e + C $\)$ $\($ $\therefore$ y(0) = 1 $\Rightarrow$ 0 = $\log$_2 e + C $\)$ $\($ C = -$\log$_2 e $\)$ $\($ $\Rightarrow$ $\log$_2(2^y - 1) = (2^x - 1) $\log$_2 e $\)$ Put $\($ x = 1 $\)$, $\($ $\log$_2(2^y - 1) = $\log$_2 e $\)$ $\($ 2^y = e + 1 $\)$ $\($ y = $\log$_2(e + 1) $\)$ Ans.
Question 17
Maths · Limits and Derivatives · Single correct
\[ \lim_{x\to 0} \frac{\sin^2\!\left(\pi\cos^4 x\right)} {x^4} \] is equal to :
$\pi^2$
2$\pi^2$
4$\pi^2$
4$\pi$
Answer: (c)
Solution
Given the limit expression: $$\lim_{x \to 0} \frac{\sin^2(\pi \cos^4 x)}{x^4}$$ We can rewrite it as: $$\lim_{x \to 0} \frac{1 - \cos(2\pi \cos^4 x)}{2x^4}$$ Further simplifying, we have: $$\lim_{x \to 0} \frac{1 - \cos(2\pi - 2\pi \cos^4 x)}{[2\pi (1 - \cos^4 x)]^2} \cdot 4\pi^2 \cdot \frac{\sin^4 \frac{4}{2x^4}}{2x^4} \left(1 + \cos^2 x\right)^2$$ This evaluates to: $$= \frac{1}{2} \cdot 4\pi^2 \cdot \frac{1}{2}(2)^2 = 4\pi^2$$
Question 18
Maths · Heights and Distances · Single correct
A vertical pole fixed to the horizontal ground is divided in the ratio 3 : 7 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground 18 m away from the base of the pole, then the height of the pole (in meters) is :
12$\sqrt{15}$
12$\sqrt{10}$
8$\sqrt{10}$
6$\sqrt{10}$
Answer: (b)
Solution
Let height of pole = 10$\ell$. $$\tan \alpha = \frac{3\ell}{18} = \frac{\ell}{6}$$ $$\tan 2\alpha = \frac{10\ell}{18}$$ $$\frac{2 \tan \alpha}{1 - \tan^2 \alpha} = \frac{10\ell}{18}$$ Use $\tan \alpha = \frac{\ell}{6}$, which implies $\ell = \sqrt{\frac{72}{5}}$. Height of pole = $10\ell = 12\sqrt{10}$.
Question 19
Maths · Determinants · Single correct
If $a_r = \cos \frac{2\pi r}{9} + i \sin \frac{2\pi r}{9}$, $r = 1, 2, 3, \ldots$, $i = \sqrt{-1}$, then the determinant $$\begin{vmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{vmatrix}$$ is equal to :
The line $12x \cos \theta + 5y \sin \theta = 60$ is tangent to which of the following curves?
$x^2 + y^2 = 169$
$144x^2 + 25y^2 = 3600$
$25x^2 + 12y^2 = 3600$
$x^2 + y^2 = 60$
Answer: (b)
Solution
Given the equation $12x \cos \theta + 5y \sin \theta = 60$. This can be rewritten as $$\frac{x \cos \theta}{5} + \frac{y \sin \theta}{12} = 1$$ which is tangent to the ellipse $$\frac{x^2}{25} + \frac{y^2}{144} = 1.$$ Therefore, the equation of the ellipse is $$144x^2 + 25y^2 = 3600.$$
Question 21
Maths · Integrals · Numerical
Let $[t]$ denote the greatest integer $\leq t$. Then the value of $8 \cdot \int_{-\frac{1}{2}}^{1} ([2x] + |x|) \, dx$ is .
Answer: 5
Solution
Given $$I = \int_{-1/2}^{1} ([2x] + |x|) \, dx$$ We can split the integral as follows: $$= \int_{-1/2}^{1} [2x] \, dx + \int_{-1/2}^{1} |x| \, dx$$ This becomes: $$= 0 + \int_{-1/2}^{0} (-x) \, dx + \int_{0}^{1} x \, dx$$ Evaluating the integrals: $$= \left( -\frac{x^2}{2} \right)_{-1/2}^{0} + \left( \frac{x^2}{2} \right)_{0}^{1}$$ Simplifying: $$= \left( 0 + \frac{1}{8} \right) + \frac{1}{2}$$ Thus: $$= \frac{5}{8}$$ Finally, multiplying by 8: $$8I = 5$$
Question 22
Maths · Complex Numbers and Quadratic Equations · Numerical
A point $z$ moves in the complex plane such that $\arg \left( \frac{z-2}{z+2} \right) = \frac{\pi}{4}$, then the minimum value of $|z - 9\sqrt{2} - 2i|^2$ is equal to .
Answer: 98
Solution
Let $z = x + iy$. $$\arg \left( \frac{x-2+iy}{x+2+iy} \right) = \frac{\pi}{4}$$ $$\arg(x-2+iy) - \arg(x+2+iy) = \frac{\pi}{4}$$ $$\tan^{-1} \left( \frac{y}{x-2} \right) - \tan^{-1} \left( \frac{y}{x+2} \right) = \frac{\pi}{4}$$ $$\frac{\frac{y}{x-2} - \frac{y}{x+2}}{1 + \left( \frac{y}{x-2} \right) \cdot \left( \frac{y}{x+2} \right)} = \tan \frac{\pi}{4} = 1$$ $$\frac{xy + 2y - xy + 2y}{x^2 - 4 + y^2} = 1$$ $$4y = x^2 - 4 + y^2$$ $$x^2 + y^2 - 4y - 4 = 0$$ The locus is a circle with center $(0, 2)$ and radius $= 2\sqrt{2}$. Minimum value $= (AP)^2 = (OP - OA)^2$ $$= (9\sqrt{2} - 2\sqrt{2})^2$$ $$= (7\sqrt{2})^2 = 98$$
Question 23
Maths · Three Dimensional Geometry · Numerical
The square of the distance of the point of intersection of the line $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+1}{6}$ and the plane $2x - y + z = 6$ from the point $(-1, -1, 2)$ is .
Answer: 61
Solution
Given $\($ $\frac{x-1}{2}$ = $\frac{y-2}{3}$ = $\frac{z+1}{6}$ = $\lambda$ $\)$. We have $\($ x = 2$\lambda$ + 1, y = 3$\lambda$ + 2, z = 6$\lambda$ - 1 $\)$. For the point of intersection of the line and plane, $\($ 2(2$\lambda$ + 1) - (3$\lambda$ + 2) + (6$\lambda$ - 1) = 6 $\)$. Simplifying gives $\($ 7$\lambda$ = 7 $\Rightarrow$ $\lambda$ = 1 $\)$. The point is $\($ (3, 5, 5) $\)$. The distance squared is $\($ (distance)^2 = (3 + 1)^2 + (5 + 1)^2 + (5 - 2)^2 $\)$. This simplifies to $\($ 16 + 36 + 9 = 61 $\)$.
Question 24
Maths · Applications of Derivatives · Numerical
If $'R'$ is the least value of $'a'$ such that the function $f(x) = x^2 + ax + 1$ is increasing on $[1, 2]$ and $'S'$ is the greatest value of $'a'$ such that the function $f(x) = x^2 + ax + 1$ is decreasing on $[1, 2]$, then the value of $|R - S|$ is .
Answer: 2
Solution
Given $f(x) = x^2 + ax + 1$. The derivative is $f'(x) = 2x + a$. When $f(x)$ is increasing on $[1, 2]$, we have: $$2x + a \geq 0 \forall x \in [1, 2]$$ This implies $a \geq -2x \forall x \in [1, 2]$. Therefore, $R = -4$. When $f(x)$ is decreasing on $[1, 2]$, we have: $$2x + a \leq 0 \forall x \in [1, 2]$$ This implies $a \leq -2 \forall x \in [1, 2]$. Therefore, $S = -2$. The absolute difference is $|R - S| = |-4 + 2| = 2$.
Question 25
Maths · Sequences and Series · Numerical
The mean of 10 numbers $$7 \times 8, 10 \times 10, 13 \times 12, 16 \times 14, \ldots$$ is.
Answer: 398
Solution
The sequence is given by: $$7 \times 8, 10 \times 10, 13 \times 12, 16 \times 14 \ldots$$ The general term $T_n$ is: $$T_n = (3n + 4)(2n + 6) = 2(3n + 4)(n + 3)$$ Simplifying, we have: $$= 2(3n^2 + 13n + 12) = 6n^2 + 26n + 24$$ The sum $S_{10}$ is: $$S_{10} = \sum_{n=1}^{10} T_n = 6 \sum_{n=1}^{10} n^2 + 26 \sum_{n=1}^{10} n + 24 \sum_{n=1}^{10} 1$$ Calculating each part: $$= \frac{6(10 \times 11 \times 21)}{6} + 26 \times \frac{10 \times 11}{2} + 24 \times 10$$ Simplifying further: $$= 10 \times 11(21 + 13) + 240$$ Finally, we get: $$= 3980$$ The mean is: $$Mean = \frac{S_{10}}{10} = \frac{3980}{10} = 398.$$
Question 26
Maths · Conic Sections · Numerical
If the variable line $3x + 4y = \alpha$ lies between the two circles $(x - 1)^2 + (y - 1)^2 = 1$ and $(x - 9)^2 + (y - 1)^2 = 4$ without intercepting a chord on either circle, then the sum of all the integral values of $\alpha$ is .
Answer: 165
Solution
Both centres should lie on either side of the line as well as line can be tangent to circle. $$(3 + 4 - \alpha) \cdot (27 + 4 - \alpha) < 0$$ $$(7 - \alpha) \cdot (31 - \alpha) < 0 \implies \alpha \in (7, 31) \ldots (1)$$ $d_1 =$ distance of $(1, 1)$ from line $d_2 =$ distance of $(9, 1)$ from line $$d_1 \geq r_1 \implies \frac{|7 - \alpha|}{5} \geq 1 \implies \alpha \in (-\infty, 2] \cup [12, \infty) \ldots (2)$$ $$d_2 \geq r_2 \implies \frac{|31 - \alpha|}{5} \geq 2 \implies \alpha \in (-\infty, 21] \cup [41, \infty) \ldots (3)$$ $$(1) \cap (2) \cap (3) \implies \alpha \in [12, 21]$$ Sum of integers $= 165$
Question 27
Maths · Permutations and Combinations · Numerical
The number of six letter words (with or without meaning), formed using all the letters of the word 'VOWELS', so that all the consonants never come together, is .
Answer: 576
Solution
All consonants should not be together. Total minus all consonants together, $$= 6! - 3!4! = 576$$
Question 28
Maths · Differential Equations · Numerical
If $x \phi(x) = \int_{5}^{x} \left(3t^2 - 2\phi'(t)\right) \, dt$, $x > -2$, and $\phi(0) = 4$ then $\phi(2)$ is.
If $\left(\frac{3^{6}}{4^{4}}\right)\cdot k$ is the term independent of $x,$ in the binomial expansion of $\left(\frac{x}{4}-\frac{12}{x^{2}}\right)^{12},$ then $k$ is equal to :
An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is 0.9 and that of the second unit is 0.8. The instrument is switched on and it fails to operate. If the probability that only the first unit failed and second unit is functioning is $p$, then $98p$ is equal to .
Answer: 28
Solution
Given $I_1 =$ first unit is functioning. $I_2 =$ second unit is functioning. $\mathrm{P}(I_1) = 0.9$, $\mathrm{P}(I_2) = 0.8$. $\mathrm{P}(\overline{I_1}) = 0.1$, $\mathrm{P}(\overline{I_2}) = 0.2$. $$\mathrm{P} = \frac{0.8 \times 0.1}{0.1 \times 0.2 + 0.9 \times 0.2 + 0.1 \times 0.8} = \frac{8}{28}$$ $$98 \mathrm{P} = \frac{8}{28} \times 98 = 28$$
Physics
Question 31
Physics · Motion in a Plane · Single correct
A helicopter is flying horizontally with a speed 'v' at an altitude 'h' has to drop a food packet for man on the ground. What is the distance o helicopter from the man when the food packet is dropped?
$\sqrt{\frac{2ghv^2+1}{h^2}}$
$\sqrt{2ghv^2 + h^2}$
$\sqrt{\frac{2v^2h}{g} + h^2}$
$\sqrt{\frac{2gh}{v^2} + h^2}$
Answer: (c)
Solution
Given the diagram, we have the following equations. The range $R$ is given by $$R = \sqrt{\frac{2h}{g}} \cdot v.$$ The distance $D$ is given by $$D = \sqrt{R^2 + h^2}.$$ Substituting for $R$, we have $$D = \sqrt{\left(\sqrt{\frac{2h}{g}} \cdot v\right)^2 + h^2}.$$ Simplifying, we get $$D = \sqrt{\frac{2hv^2}{g} + h^2}.$$
Question 32
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
In the following logic circuit the sequence of the inputs A, B are (0, 0), (0, 1), (1, 0) and (1, 1). The output Y for this sequence will be :
1, 0, 1, 0
0, 1, 0, 1
1, 1, 1, 0
0, 0, 1, 1
Answer: (c)
Solution
The expression for Y is given by $$Y = \overline{(A \cdot B) \cdot (A + B)}$$. Evaluating the expression for different inputs: $Y_{(0,0)} = 1$ $Y_{(0,1)} = 1$ $Y_{(1,0)} = 1$ $Y_{(1,1)} = 0$ Option (3) is correct.
Question 33
Physics · Electric Charges and Fields · Single correct
Two particles A and B having charges $20\mu\mathrm{C}$ and $-5\mu\mathrm{C}$ respectively are held fixed with a separation of $5\,\mathrm{cm}$. At what position a third charged particle should be placed so that it does not experience a net electric force?
At $5\,\mathrm{cm}$ from $20\mu\mathrm{C}$ on the left side of system
At $5\,\mathrm{cm}$ from $-5\mu\mathrm{C}$ on the right side
At $1.25\,\mathrm{cm}$ from $-5\mu\mathrm{C}$ between two charges
At midpoint between two charges
Answer: (b)
Solution
Null point is possible only on the right side of $-5\mu\mathrm{C}$. $$E_N = \frac{k(-5\mu\mathrm{C})}{x^2} + \frac{k(20\mu\mathrm{C})}{(5+x)^2} = 0$$ $x = 5 \, \mathrm{cm}$ Therefore, option (2) is correct.
Question 34
Physics · Thermodynamics · Single correct
A reversible engine has an efficiency of $\frac{1}{4}$. If the temperature of the sink is reduced by $58^\circ \mathrm{C}$, its efficiency becomes double. Calculate the temperature of the sink:
$174^\circ \mathrm{C}$
$280^\circ \mathrm{C}$
$180.4^\circ \mathrm{C}$
$382^\circ \mathrm{C}$
Answer: (a)
Solution
Question 35
Physics · Ray Optics and Optical Instruments · Single correct
An object is placed at the focus of concave lens having focal length $f$. What is the magnification and distance of the image from the optical centre of the lens?
1, $\infty$
Very high, $\infty$
$\frac{1}{2}$, $\frac{f}{2}$
$\frac{1}{4}$, $\frac{f}{4}$
Answer: (c)
Solution
Given $U = -f$. Using the lens formula: $$\frac{1}{V} - \frac{1}{U} = \frac{1}{-f} \implies \frac{1}{V} = -\frac{2}{f}$$ Therefore, $$V = -\frac{f}{2}$$ The magnification $m$ is given by $$m = \frac{V}{U} = \frac{1}{2}$$ The distance is $$\frac{f}{2}$$
Question 36
Physics · Nuclei · Single correct
A sample of a radioactive nucleus $A$ disintegrates to another radioactive nucleus $B$, which in turn disintegrates to some other stable nucleus $C$. Plot of a graph showing the variation of number of atoms of nucleus $B$ versus time is: (Assume that at $t = 0$, there are no $B$ atoms in the sample)
Answer: (b)
Solution
Initially no. of atoms of B = 0 after $t = 0$, no. atoms of B will start increasing and reaches maximum value when rate of decay of B = rate of formation of B. After that maximum value, no. of atoms will start decreasing as growth and decay both are exponential functions, so best possible graph is (2) Option (2)
Question 37
Physics · Moving Charges and Magnetism · Single correct
A coil having N turns is wound tightly in the form of a spiral with inner and outer radii 'a' and 'b' respectively. Find the magnetic field at centre, when a current I passes through coil:
The number of turns in $dx$ width is given by $$\frac{N}{b-a} \, dx.$$ The integral of $dB$ is $$\int dB = \int_a^b \left( \frac{N}{b-a} \right) dx \frac{\mu_0 i}{2x}.$$ Therefore, $$B = \frac{N \mu_0 i}{2(b-a)} \ln \left( \frac{b}{a} \right).$$
Question 38
Physics · System of Particles and Rotational Motion · Single correct
A body of mass $M$ moving at speed $V_0$ collides elastically with a mass '$m$' at rest. After the collision, the two masses move at angles $\theta_1$ and $\theta_2$ with respect to the initial direction of motion of the body of mass $M$. The largest possible value of the ratio $M/m$, for which the angles $\theta_1$ and $\theta_2$ will be equal, is:
4
1
3
2
Answer: (c)
Solution
Given $\theta_1 = \theta_2 = \theta$. From momentum conservation: In $x$-direction $MV_0 = MV_1 \cos \theta + mV_2 \cos \theta$. In $y$-direction $0 = MV_1 \sin \theta - mV_2 \sin \theta$. Solving above equations: $$V_2 = \frac{MV_1}{m}, V_0 = 2V_1 \cos \theta$$ From energy conservation: $$\frac{1}{2}MV_0^2 = \frac{1}{2}MV_1^2 + \frac{1}{2}mV_2^2$$ Substituting value of $V_2$ and $V_0$, we will get: $$\frac{M}{m} + 1 = 4 \cos^2 \theta \leq 4$$ $$\frac{M}{m} \leq 3$$ Option (3)
Question 39
Physics · Gravitation · Single correct
The masses and radii of the earth and moon are $(M_1, R_1)$ and $(M_2, R_2)$ respectively. Their centres are at a distance 'r' apart. Find the minimum escape velocity for a particle of mass 'm' to be projected from the middle of these two masses:
$V = \frac{1}{2} \sqrt{\frac{4G(M_1+M_2)}{r}}$
$V = \sqrt{\frac{4G(M_1+M_2)}{r}}$
$V = \frac{1}{2} \sqrt{\frac{2G(M_1+M_2)}{r}}$
$V = \sqrt{\frac{2G(M_1+M_2)}{r}}$
Answer: (b)
Solution
The equation for the system is given by $$\frac{1}{2} m V^2 - \frac{G M_1 m}{r/2} - \frac{G M_2 m}{r/2} = 0.$$ Simplifying, we have $$\frac{1}{2} m V^2 = \frac{2 G m}{r} (M_1 + M_2).$$ Solving for $V$, we get $$V = \sqrt{\frac{4 G (M_1 + M_2)}{r}}.$$
Question 40
Physics · Electromagnetic Induction · Single correct
A small square loop of side 'a' and one turn is placed inside a larger square loop of side b and one turn (b >> a). The two loops are coplanar with their centres coinciding. If a current I is passed in the square loop of side 'b', then the coefficient of mutual inductance between the two loops is :
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
Choose the correct waveform that can represent the voltage across R of the following circuit, assuming the diode is ideal one:
Answer: (c)
Solution
When $V_i > 3 volt$, $V_R > 0$. Because diode will be in forward biased state. When $V_i \leq 3 volt$; $V_R = 0$. Because diode will be in reverse biased state.
Question 42
Physics · Mechanical Properties of Solids · Single correct
A uniform heavy rod of weight $10\,\mathrm{kg\,m\,s^{-2}}$, cross-sectional area $100\,\mathrm{cm^2}$, and length $20\,\mathrm{cm}$ is hanging from a fixed support. The Young's modulus of the material of the rod is $2 \times 10^{11}\,\mathrm{N\,m^{-2}}$. Neglecting lateral contraction, find the elongation of the rod due to its own weight.
Physics · Ray Optics and Optical Instruments · Single correct
Two plane mirrors $M_1$ and $M_2$ are at right angle to each other shown. A point source 'P' is placed at 'a' and '2a' meter away from $M_1$ and $M_2$ respectively The shortest distance between the images thus formed is : (Take $\sqrt{5} = 2.3$)
3a
4.6a
2.3a
2$\sqrt{10}$a
Answer: (b)
Solution
Shortest distance is $2a$ between $I_1$ and $I_3$. But answer given is for $I_1$ and $I_2$. $$\sqrt{(4a)^2 + (2a)^2}$$ $$a \sqrt{20}$$ $$4.47a$$ Option (2)
Question 44
Physics · Physical World, Units and Measurements · Single correct
Match List-I with List-II. $$ \begin{array}{ll} \text{List-I} & \text{List-II} \\ \hline (a) \text{ Torque} & (i) \, MLT^{-1} \\ (b) \text{ Impulse} & (ii) \, MT^{-2} \\ (c) \text{ Tension} & (iii) \, ML^{2} T^{-2} \\ (d) \text{ Surface Tension} & (iv) \, MLT^{-2} \end{array} $$ Choose the most appropriate answer from the option given below:
For an ideal gas the instantaneous change in pressure 'p' with volume 'v' is given by the equation $\frac{dp}{dv} = -ap$. If $p = p_0$ at $v = 0$ is the given boundary condition, then the maximum temperature one mole of gas can attain is: (Here R is the gas constant)
$\frac{p_0}{aeR}$
$\frac{ap_0}{eR}$
infinity
$0^\circ \mathrm{C}$
Answer: (a)
Solution
Given $\($ $\int$_{p_0}^{p} $\frac{dp}{P}$ = -a $\int$_{0}^{v} dv $\)$. $\($ $\ln$ $\left$( $\frac{p}{p_0}$ $\right$) = -av $\)$. $\($ p = p_0 e^{-av} $\)$. For temperature maximum, $\($ p - v $\)$ product should be maximum. $\($ T = $\frac{pv}{nR}$ = $\frac{p_0 ve^{-av}}{R}$ $\)$. $\($ $\frac{dT}{dv}$ = 0 $\Rightarrow$ $\frac{p_0}{R}$ $\{$ e^{-av} + ve^{-av}(-a) $\}$ $\)$. $\($ $\frac{p_0 e^{-av}}{R}$ $\{$ 1 - av $\}$ = 0 $\)$. $\($ v = $\frac{1}{a}$, $\infty$ $\)$. $\($ T = $\frac{p_0}{Rae}$ = $\frac{p_0}{Rae}$ $\)$. At $\($ v = $\infty$ $\)$, $\($ T = 0 $\)$.
Question 46
Physics · Physical World, Units and Measurements · Single correct
Which of the following equations is dimensionally incorrect? Where t = time, h = height, s = surface tension, $\theta$ = angle, $\rho$ = density, a, r = radius, g = acceleration due to gravity, v = volume, p = pressure, W = work done, $\Gamma$ = torque, $\epsilon$ = permittivity, $E$ = electric field, J = current density, L = length.
Physics · Dual Nature of Radiation and Matter · Single correct
A moving proton and electron have the same deBroglie wavelength. If K and P denote the K.E. and momentum respectively. Then choose the correct option :
$K_p < K_e$ and $P_p = P_e$
$K_p = K_e$ and $P_p = P_e$
$K_p < K_e$ and $P_p < P_e$
$K_p > K_e$ and $P_p = P_e$
Answer: (a)
Solution
Given $\lambda_P = \frac{h}{P_p}$ and $\lambda_e = \frac{h}{P_e}$. Therefore, $\lambda_P = \lambda_c$. This implies $P_p = P_e$. The kinetic energy for the proton is $$(K)_P = \frac{P_p^2}{2m_p}$$ and for the electron is $$(K)_e = \frac{P_e^2}{2m_e}.$$ Since $m_p > m_e$, it follows that $K_P < K_e$.
Question 50
Physics · Current Electricity · Single correct
Consider a galvanometer shunted with $5\,\Omega$ resistance and $2\%$ of current passes through it. What is the resistance of the given galvanometer?
$300\,\Omega$
$344\,\Omega$
$245\,\Omega$
$226\,\Omega$
Answer: (c)
Solution
Given the circuit, we have the equation: $$0.02i R_g = 0.98i \times 5$$ Solving for $R_g$, we get: $$R_g = 245 \Omega$$
Question 51
Physics · Mechanical Properties of Solids · Numerical
When a rubber ball is taken to a depth of ..... m in deep sea, its volume decreases by 0.5%. (The bulk modulus of rubber = $9.8 \times 10^8 \, \mathrm{N/m^{-2}}$ Density of sea water = $10^3 \, \mathrm{kg/m^{-3}}$ $g = 9.8 \, \mathrm{m/s^2}$)
A particle of mass 1 kg is hanging from a spring of force constant 100 $\mathrm{Nm}$^{-1}. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is $\frac{T}{x}$. The value of x is ...
Answer: 8
Solution
Given that the kinetic energy (KE) is equal to the potential energy (PE), we have: $$y = \frac{A}{\sqrt{2}} = A \sin \omega t$$ From the diagram, the angle is $45^\circ$, and the displacement is $\frac{A}{\sqrt{2}}$. The time $t$ is given by: $$t = \frac{T}{8} = \frac{T}{x}$$ Solving for $x$, we find: $$x = 8$$
Question 53
Physics · Communication Systems · Numerical
If the sum of the heights of transmitting and receiving antennas in the line of sight of communication is fixed at 160 m, then the maximum range of LOS communication is .....km. (Take radius of Earth = 6400 km )
Answer: 64
Solution
Given $h_T = h_R = 160 \ldots (i)$. The distance $d$ is given by: $$d = \sqrt{2R h_T} + \sqrt{2R h_R}$$ Substituting $h_T$ and $h_R$: $$d = \sqrt{2R} \left[ \sqrt{h_T} + \sqrt{h_R} \right]$$ Replacing $h_T$ and $h_R$ with $x$ and $160 - x$: $$d = \sqrt{2R} \left[ \sqrt{x} + \sqrt{160 - x} \right]$$ To find the maximum distance, set the derivative to zero: $$\frac{d(d)}{dx} = 0$$ This gives: $$\frac{1}{2\sqrt{x}} + \frac{1(-1)}{2\sqrt{160-x}} = 0$$ Simplifying: $$\frac{1}{\sqrt{x}} = \frac{1}{\sqrt{160-x}}$$ Solving for $x$ gives: $$x = 80 \, \mathrm{m}$$ The maximum distance $d_{max}$ is: $$d_{max} = \sqrt{2 \times 6400} \left[ \sqrt{\frac{80}{1000}} + \sqrt{\frac{20}{1000}} \right]$$ Simplifying further: $$= \frac{80 \sqrt{2} \times 2 \sqrt{80}}{10 \sqrt{10}}$$ Finally: $$= 8 \times 2 \times \sqrt{2} \times 2 \sqrt{2} = 64 \, \mathrm{km}$$
Question 54
Physics · Current Electricity · Numerical
A square shaped wire with resistance of each side $3\,\Omega$ is bent to form a complete circle. The resistance between two diametrically opposite points of the circle in unit of $\Omega$ will be
Answer: 3
Solution
The equivalent resistance of the square circuit is calculated by considering the resistors in series and parallel. Each side of the square has a resistance of $3\,\Omega$. The equivalent resistance $R_{eq}$ is given as $3\,\Omega$.
Question 55
Physics · Waves · Numerical
A wire having a linear mass density $9.0 \times 10^{-4} \, \mathrm{kg/m}$ is stretched between two rigid supports with a tension of $900 \, \mathrm{N}$. The wire resonates at a frequency of $500 \, \mathrm{Hz}$. The next higher frequency at which the same wire resonates is $550 \, \mathrm{Hz}$. The length of the wire is ..... m.
The voltage drop across 15$\Omega$ resistance in the given figure will be V.
Answer: 6
Solution
The effective circuit diagram will be as shown. The current $i = 1 \, \mathrm{A}$ flows through the $4 \, \Omega$ and $6 \, \Omega$ resistors. The current $i_0 = 2 \, \mathrm{A}$ flows through the $12 \, \mathrm{V}$ source and $1 \, \Omega$ resistor. Point drop across $6 \, \Omega = 1 \times 6 = 6 = V_{AB}$. Hence, point drop across $15 \, \Omega = 6 \, volt = V_{AB}$.
Question 57
Physics · Work, Energy and Power · Numerical
A block moving horizontally on a smooth surface with a speed of $40 \, \mathrm{ms^{-1}}$ splits into two equal parts. If one of the parts moves at $60 \, \mathrm{ms^{-1}}$ in the same direction, then the fractional change in the kinetic energy will be $x : 4$ where $x = \ldots$
Answer: 1
Solution
Initial momentum $P_i$ is equal to final momentum $P_f$. $$m \times 40 = \frac{m}{2} \times v + \frac{m}{2} \times 60$$ Solving for $v$: $$40 = \frac{v}{2} + 30$$ $$\Rightarrow v = 20$$ The initial kinetic energy $(\mathrm{K.E.})_1$ is: $$(\mathrm{K.E.})_1 = \frac{1}{2} m \times (40)^2 = 800 \, m$$ The final kinetic energy $(\mathrm{K.E.})_f$ is: $$(\mathrm{K.E.})_f = \frac{1}{2} m \cdot (20)^2 + \frac{1}{2} \cdot \frac{m}{2} (60)^2 = 1000 \, m$$ The change in kinetic energy $|\Delta \mathrm{K.E.}|$ is: $$|\Delta \mathrm{K.E.}| = |1000 \, m - 800 \, m| = 200 \, m$$ The ratio of the change in kinetic energy to the initial kinetic energy is: $$\frac{\Delta \mathrm{K.E}}{(\mathrm{K.E.})_i} = \frac{200 \, m}{800 \, m} = \frac{1}{4} = \frac{x}{4}$$ Solving for $x$ gives: $$x = 1$$
Question 58
Physics · Electromagnetic Waves · Numerical
The electric field in an electromagnetic wave is given by $\mathbf{E} = \left( 50 \, \mathrm{NC}^{-1} \right) \sin \omega (t - x/c)$ The energy contained in a cylinder of volume $V$ is $5.5 \times 10^{-12} \, \mathrm{J}$. The value of $V$ is $\mathrm{cm}^3$. ( given $\epsilon_0 = 8.8 \times 10^{-12} \mathrm{C}^2 \, \mathrm{N}^{-1} \, \mathrm{m}^{-2}$ )
Answer: 500
Solution
Given $\mathbf{E} = 50 \sin \left( \omega t - \frac{\omega}{c} \cdot \mathbf{x} \right)$. Energy density is $\frac{1}{2} \varepsilon_0 E_0^2$. Energy for volume $V = \frac{1}{2} \varepsilon_0 E_0^2 \cdot V = 5.5 \times 10^{-12}$. $$\frac{1}{2} 8.8 \times 10^{-12} \times 2500 \cdot V = 5.5 \times 10^{-12}$$ $$V = \frac{5.5 \times 2}{2500 \times 8.8} = 0.0005 \, \mathrm{m^3}$$ $$= 0.0005 \times 10^6 \, (\mathrm{c.m})^3$$ $$= 500 \, (\mathrm{c.m})^3$$
Question 59
Physics · Electrostatic Potential and Capacitance · Numerical
A capacitor of $50 \, \mu \mathrm{F}$ is connected in a circuit as shown in figure. The charge on the upper plate of the capacitor is ..... $\mu \mathrm{C}$.
Answer: 100
Solution
Potential difference across each resistor is $2 \, \mathrm{V}$. $q = CV$ $= 50 \times 10^{-6} \times 2 = 100 \times 10^{-6} = 100 \, \mu \mathrm{C}$
Question 60
Physics · Laws of Motion · Numerical
A car is moving on a plane inclined at $30^\circ$ to the horizontal with an acceleration of 10 $\mathrm{ms^{-2}}$ parallel to the plane upward. A bob is suspended by a string from the roof of the car. The angle in degrees which the string makes with the vertical is (Take g = 10 $\mathrm{ms^{-2}}$)
Chemistry · Haloalkanes and Haloarenes · Single correct
The correct order of reactivity of the given chlorides with acetate in acetic acid is :
Answer: (a)
Solution
As it is example of $\mathrm{S_N^1}$. So carbocation stability $\uparrow$, reaction rate $\uparrow$.
Question 62
Chemistry · Surface Chemistry · Single correct
Select the graph that correctly describes the adsorption isotherms at two temperatures $T_1$ and $T_2$ ($T_1 > T_2$) for a gas : ($x$ – mass of the gas adsorbed ; $m$ – mass of adsorbent ; $P$ - pressure)
Answer: (d)
Solution
Given $\frac{x}{m} \propto P^{1/n}$ with $\left(0 < \frac{1}{n} < 1\right)$. On increasing temperature, $\frac{x}{m}$ decreases. Therefore, adsorption is generally exothermic.
Question 63
Chemistry · The s-Block Elements · Single correct
The major component/ingredient of Portland Cement is :
tricalcium aluminate
tricalcium silicate
dicalcium aluminate
dicalcium silicate
Answer: (b)
Solution
Major component of portland cement is "Tricalcium silicate" (51$\%$, 3CaO $\cdot$ SiO_2).
Question 64
Chemistry · The d-and f-Block Elements · Single correct
In the structure of the dichromate ion, there is a :
linear symmetrical $\mathrm{Cr} - \mathrm{O} - \mathrm{Cr}$ bond.
Dichromate ion contains a non-linear symmetrical Cr-O-Cr bond.
Question 65
Chemistry · Biomolecules · Single correct
Which one of the following compounds contains $\beta - \mathrm{C}_1 - \mathrm{C}_4$ glycosidic linkage?
Lactose
Sucrose
Maltose
Amylose
Answer: (a)
Solution
In Lactose it is $\beta C_1 - C_4$ glycosidic linkage. In Maltose, Amylose $\alpha C_1 - C_4$ glycosidic linkage is present.
Question 66
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The major products A and B in the following set of reactions are :
Answer: (c)
Solution
The reaction sequence starts with the conversion of the alcohol to a nitrile using $\mathrm{H_3O^+}$. The nitrile is then reduced to an amine using $\mathrm{LiAlH_4}$. The amine is converted to a carboxylic acid using $\mathrm{H_2SO_4}$.
Question 67
Chemistry · The d-and f-Block Elements · Single correct
Which one of the following lanthanides exhibits $+2$ oxidation state with diamagnetic nature ? (Given $Z$ for Nd $= 60$, Yb $= 70$, La $= 57$, Ce $= 58$ )
Nd
Yb
La
Ce
Answer: (b)
Solution
Ytterbium shows $+2$ oxidation state with diamagnetic nature. So ans is 2.
Question 68
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Aluminium is extracted from bauxite by the electrolysis of molten mixture of $\mathrm{Al_2O_3}$ with cryolite. Reason (R): The oxidation state of Al in cryolite is $+3$. In the light of the above statements, choose the most appropriate answer from the options given below:
is true but (R) is false.
is false but (R) is true.
Both (A) and (R) are correct and (R) is the correct explanation of (A).
Both (A) and (R) are correct but (R) is not the correct explanation of (A).
Answer: (d)
Solution
(A) Aluminium is reactive metal so Aluminium is extracted by electrolysis of Alumina with molten mixture of Cryolite. (B) Cryolite, $\mathrm{Na_3AlF_6}$ Here Al is in $+3 \ O. \ S.$ So Answer is 4.
Question 69
Chemistry · Alcohols, Phenols and Ethers · Single correct
The major product formed in the following reaction is :
Answer: (b)
Solution
The reaction begins with the protonation of the alcohol group, leading to the formation of a carbocation. The structure is: $$CH_3C^+(CH_3)CH_2CH_3$$. A 1,2-shift of a methyl group occurs, resulting in a more stable carbocation: $$H_3CC^+(CH_3)CH_2CH_3$$. Finally, deprotonation leads to the formation of the alkene: $$H_3CC = C(CH_3)_2$$.
Question 70
Chemistry · Polymers · Single correct
Monomer of Novolac is :
3-Hydroxybutanoic acid
phenol and melamine
o-Hydroxymethylphenol
1,3-Butadiene and styrene
Answer: (c)
Solution
Monomer of Novolac is O-hydroxy methyl phenol.
Question 71
Chemistry · Hydrogen · Single correct
Given below are two statements: Statement - I: The process of producing syn-gas is called gasification of coal. Statement - II: The composition of syn-gas is $CO + CO_2 + H_2 (1 : 1 : 1)$ In the light of the above statements, choose the most appropriate answer from the options given below:
Statement - I is false but Statement - II is true
Statement - I is true but Statement - II is false
Both Statement - I and Statement - II are false
Both Statement - I and Statement - II are true
Answer: (b)
Solution
The process of producing syn-gas from coal is called gasification of coal. Syn-gas having composition of $\mathrm{CO}$ and $\mathrm{H_2}$ in $1 : 1$.
Question 72
Chemistry · Alcohols, Phenols and Ethers · Single correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Treatment of bromine water with propene yields 1-bromopropan-2-ol. Reason (R): Attack of water on bromonium ion follows Markovnikov rule and results in 1-bromopropan-2-ol. In the light of the above statements, choose the most appropriate answer from the options given below:
Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(A) is false but (R) is true.
Both (A) and (R) are true and (R) is the correct explanation of (A).
(A) is true but (R) is false.
Answer: (c)
Solution
The reaction sequence is as follows: Its IUPAC name is 1-bromopropan-2-ol. A and R are true and (R) is the correct explanation of (A).
Question 73
Chemistry · Co-ordination Compounds · Single correct
The denticity of an organic ligand, biuret is:
2
4
3
6
Answer: (a)
Solution
Biuret: Bidentate ligand. The denticity of organic ligand is 2.
Question 74
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason ($R$). Assertion (A) : Metallic character decreases and non-metallic character increases on moving from left to right in a period. Reason ($R$) : It is due to increase in ionisation enthalpy and decrease in electron gain enthalpy, when one moves from left to right in a period. In the light of the above statements, choose the most appropriate answer from the options given below:
(A) is false but ($R$) is true.
(A) is true but ($R$) is false.
Both (A) and ($R$) are correct and ($R$) is the correct explanation of (A).
Both (A) and ($R$) are correct but ($R$) is not the correct explanation of (A).
Answer: (b)
Solution
From left to right in the periodic table: Metallic character decreases Non-metallic character increases It is due to an increase in ionization enthalpy and an increase in electron gain enthalpy.
Question 75
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Which one of the following is the correct PV vs P plot at constant temperature for an ideal gas ? (P and V stand for pressure and volume of the gas respectively)
Answer: (a)
Solution
Given the equation $PV = nRT$ with $n$ and $T$ constant, we have $PV = constant$. The graph shows a horizontal line indicating that $PV$ is constant as $P$ changes.
Question 77
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R): Assertion (A): A simple distillation can be used to separate a mixture of propanol and propanone. Reason (R): Two liquids with a difference of more than 20$^\circ$C in their boiling points can be separated by simple distillations. In the light of the above statements, choose the most appropriate answer from the options given below:
is false but (R) is true.
Both (A) and (R) are correct but (R) is not the correct explanation of (A).
is true but (R) is false.
Both (A) and (R) are correct and (R) is the correct explanation of (A).
Answer: (d)
Solution
Both assertion and reason are correct and (R) is the correct explanation of (A).
Question 78
Chemistry · Solutions · Single correct
Which one of the following 0.10M aqueous solutions will exhibit the largest freezing point depression?
hydrazine
glucose
glycine
KHSO_4
Answer: (d)
Solution
Van't Hoff factor is highest for $\mathrm{KHSO_4}$. Therefore, colligative property $\left( \Delta T_f \right)$ will be highest for $\mathrm{KHSO_4}$.
Question 79
Chemistry · Environmental Chemistry · Single correct
BOD values (in ppm ) for clean water (A) and polluted water (B) are expected respectively :
A > 50, B < 27
A > 25, B < 17
A 17
A > 15, B > 47
Answer: (c)
Solution
BOD values of clean water (A) is less than 5 ppm. So $A 17$. So Ans. is 3.
Question 80
Chemistry · Haloalkanes and Haloarenes · Single correct
The structure of product C, formed by the following sequence of reactions is : $CH_3COOH+SOCl_2 \longrightarrow A \xrightarrow[\mathrm{AlCl_3}]{\mathrm{Benzene}} B \xrightarrow{\mathrm{KCN}} C$
Answer: (a)
Solution
Question 81
Chemistry · Electrochemistry · Numerical
Consider the following cell reaction: \[ \mathrm{Cd}(s)+\mathrm{Hg_2SO_4}(s)+\frac{9}{5}\,\mathrm{H_2O}(l) \rightleftharpoons \mathrm{CdSO_4}\cdot\frac{9}{5}\,\mathrm{H_2O}(s) +2\,\mathrm{Hg}(l) \] The value of $E^\circ_{\mathrm{cell}}$ is $4.315\,\mathrm{V}$ at $25^\circ\mathrm{C}$. If \[ \Delta H^\circ=-825.2\,\mathrm{kJ\,mol^{-1}}, \] the standard entropy change $\Delta S^\circ$ in $\mathrm{J\,K^{-1}\,}$ is $\underline{\qquad}$. (Nearest integer) Given: Faraday constant $=96487\,\mathrm{C\,mol^{-1}}$.
Chemistry · Some Basic Concepts of Chemistry · Numerical
The molarity of the solution prepared by dissolving 6.3 g of oxalic acid ($H_2C_2O_4$ $\cdot$ $2H_2O$) in 250 $\mathrm{mL}$ of water in $\mathrm{mol}$ $\mathrm{L}^{-1}$ is x $\times 10^{-2}$. The value of x is ____ . (Nearest integer) [ Atomic mass : H : 1.0, C : 12.0, O : 16.0 ]
Answer: 20
Solution
The concentration of $[\mathrm{H_2C_2O_4 \cdot 2H_2O}]$ is given by the formula: $$[\mathrm{H_2C_2O_4 \cdot 2H_2O}] = \frac{weight / M_w}{V(\mathrm{L})}$$ This implies: $$x \times 10^{-2} = \frac{6.3 / 126}{250 / 1000}$$ Solving for $x$, we find: $$x = 20$$
Question 83
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Numerical
Consider the sulphides $HgS, PbS, CuS, Sb_2S_3, As_2S_3$ and $CdS$. Number of these sulphides soluble in 50$\%$ $HNO_3$ is ____.
Answer: 4
Solution
$PbS, CuS, As_2S_3, Cds$ are soluble in 50$\%$ $\mathrm{HNO_3}$. $HgS, Sb_2S_3$ are insoluble in 50$\%$ $\mathrm{HNO_3}$. So Answer is 4.
Question 84
Chemistry · Amines · Numerical
The total number of reagents from those given below, that can convert nitrobenzene into aniline is ____. (Integer answer) I. Sn – HCl II. Sn – $NH_4OH$ III. Fe - HCl IV. Zn – HCl V. $H_2$ – Pd VI. $H_2$ – Raney Nickel
Answer: 5
Solution
The conversion of a nitro group ($\mathrm{NO_2}$) to an amino group ($\mathrm{NH_2}$) can be achieved using various reagents. The reagents used can be: (i) $\mathrm{Sn} + \mathrm{HCl}$ (ii) $\mathrm{Fe} + \mathrm{HCl}$ (iii) $\mathrm{Zn} + \mathrm{HCl}$ (iv) $\mathrm{H_2} - \mathrm{Pd}$ (v) $\mathrm{H_2}$ (Raney Ni)
Question 85
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Numerical
The number of halogen (s) forming halic (V) acid is ____.
Answer: 3
Solution
The number of halogen forming halic (V) acid $\mathrm{HClO_3}$ $\mathrm{HBrO_3}$ $\mathrm{HIO_3}$ So Answer is 3.
Question 86
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
For a first order reaction, the ratio of the time for 75$\%$ completion of a reaction to the time for 50$\%$ completion is ____. (Integer answer)
Answer: 2
Solution
Given $$k = \frac{2.303}{t} \log \frac{a}{a-x}$$ Substituting the values, we have: $$\frac{2.303}{t_{50\%}} \log \frac{100}{100-50} = \frac{2.303}{t_{75\%}} \log \frac{100}{100-75}$$ This implies: $$t_{75\%} = 2t_{50\%}$$
Question 87
Chemistry · Hydrogen · Numerical
The number of hydrogen bonded water molecule(s) associated with stoichiometry $\mathrm{CuSO_4 \cdot 5H_2O}$ is ____.
Answer: 1
Solution
One hydrogen bonded $\mathrm{H_2O}$ molecule
Question 88
Chemistry · Thermodynamics · Numerical
According to the following figure, the magnitude of the enthalpy change of the reaction $$A + B \rightarrow M + N$$ in kJ mol$^{-1}$ is equal to ____. (Integer answer)
Answer: 45
Solution
Given the reaction coordinate diagram, we have the following values: $x = 20 \, \mathrm{kJ \, mol^{-1}}$, $y = 45 \, \mathrm{kJ \, mol^{-1}}$, $z = 15 \, \mathrm{kJ \, mol^{-1}}$. The change in enthalpy $\Delta H$ is calculated as follows: $$\Delta H = E_{at} - E_{ab}$$ $$= 20 - 65$$ $$= -45 \, \mathrm{kJ/mol}$$ The absolute value of the change in enthalpy is: $$|\Delta H| = 45 \, \mathrm{kJ/mol}$$
Question 89
Chemistry · Structure of Atom · Numerical
Ge(Z = 32) in its ground state electronic configuration has x completely filled orbitals with $m_l = 0$. The value of x is ____.
Answer: 7
Solution
Completely filled orbital with $m_\ell = 0$ are $$= 1 + 1 + 1 + 1 + 1 + 1 + 1$$ $$= 7$$ So Answer is 7
Question 90
Chemistry · Equilibrium · Fill in the blank
$\mathrm{A_3B_2}$ is a sparingly soluble salt of molar mass M $\left(\mathrm{g \, mol^{-1}}\right)$ and solubility $x \, \mathrm{g \, L^{-1}}$. The solubility product satisfies $K_{sp} = a\left(\frac{x}{M}\right)^5$. The value of $a$ is ______. (Integer answer)