JEE Main 26 February 2021 Shift 1 question paper with solutions

JEE Main 26 February 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 · Vector Algebra · Single correct

If $\vec{a} \& \vec{b}$ are perpendicular vectors, then $\vec{a} \times \left( \vec{a} \times \left( \vec{a} \times \left( \vec{a} \times \vec{b} \right) \right) \right)$ is equal to

  1. $\frac{1}{2} |\vec{a}|^4 \vec{b}$
  2. $\vec{a} \times \vec{b}$
  3. $|\vec{a}|^4 \vec{b}$
  4. $\vec{0}$

Answer: (c)

Solution

Given $$\overline{\mathbf{a}} \times \left( \overline{\mathbf{a}} \times \left( (\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}) \overline{\mathbf{a}} - |\overline{\mathbf{a}}|^2 \overline{\mathbf{b}} \right) \right)$$ $$\overline{\mathbf{a}} \times \left( -|\overline{\mathbf{a}}|^2 (\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \right) = -|\overline{\mathbf{a}}|^2 \left( (\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}) \overline{\mathbf{a}} - |\overline{\mathbf{a}}|^2 \overline{\mathbf{b}} \right)$$ $$= -\left( (\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}) \overline{\mathbf{a}} |\overline{\mathbf{a}}|^2 + |\overline{\mathbf{a}}|^4 \overline{\mathbf{b}} \right)$$ $$= |\overline{\mathbf{a}}|^4 \overline{\mathbf{b}} (\because \overline{\mathbf{a}} \cdot \overline{\mathbf{b}} = 0)$$

Question 2

Maths · Probability · Single correct

A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to probability of getting 9 heads, then the probability of getting 2 heads is :

  1. $\frac{15}{2^{12}}$
  2. $\frac{15}{2^{13}}$
  3. $\frac{15}{2^{14}}$
  4. $\frac{15}{2^{8}}$

Answer: (b)

Solution

Given $p(x = 9) = p(x = 7)$ $$\binom{n}{9} \left( \frac{1}{2} \right)^{n-9} \times \left( \frac{1}{2} \right)^9 = \binom{n}{7} \left( \frac{1}{2} \right)^{n-7} \times \left( \frac{1}{2} \right)^7$$ Simplifying, we have: $$\binom{n}{9} \times \left( \frac{1}{2} \right)^2 = \left( \frac{1}{2} \right)^2 \times \binom{n}{7}$$ Thus, $x + y = n \implies n = 16$. Now, $p(x = 2)$ is given by: $$p(x = 2) = \binom{16}{2} \times \left( \frac{1}{2} \right)^{14} \times \left( \frac{1}{2} \right)^2$$ Simplifying further: $$= \binom{16}{2} \times \left( \frac{1}{2} \right)^{16} = \frac{15}{2^{13}}$$

Question 3

Maths · Matrices · Single correct

Let A be a symmetric matrix of order 2 with integer entries. If the sum of the diagonal elements of $A^2$ is 1, then the possible number of such matrices is:

  1. 6
  2. 1
  3. 4
  4. 12

Answer: (c)

Solution

Let $A = \begin{bmatrix} a & b \\ b & c \end{bmatrix}$. $$A^2 = \begin{bmatrix} a & b \\ b & c \end{bmatrix} \begin{bmatrix} a & b \\ b & c \end{bmatrix} = \begin{bmatrix} a^2 + b^2 & ab + bc \\ ab + bc & c^2 + b^2 \end{bmatrix}$$ $$= a^2 + 2b^2 + c^2 = 1$$ $a = 1, b = 0, c = 0$ $a = 0, b = 0, c = 1$ $a = -1, b = 0, c = 0$ $c = -1, b = 0, a = 0$

Question 4

Maths · Sequences and Series · Single correct

In an increasing, geometric series, the sum of the second and the sixth term is $\frac{25}{2}$ and the product of the third and fifth term is $25$. Then, the sum of $4^{th}$, $6^{th}$ and $8^{th}$ terms is equal to:

  1. 35
  2. 30
  3. 26
  4. 32

Answer: (a)

Solution

Given $ar + ar^5 = \frac{25}{2}$ and $an^2 \times ar^4 = 25$. Then $a^2r^6 = 25$ and $ar^3 = 5$. Therefore, $a = \frac{5}{r^3}$. Now, $\frac{5r}{r^3} + \frac{5r^5}{r^3} = \frac{25}{2}$. This implies $\frac{1}{r^2} + r^2 = \frac{5}{2}$. Put $n^2 = t$. Then $\frac{t^2 + 1}{t} = \frac{5}{2}$. Solving $2t^2 - 5t + 2 = 0$, we get $2t^2 - 4t - t + 2 = 0$. Factoring gives $(2t - 1)(t - 2) = 0$. Thus, $t = \frac{1}{2}, 2 \Rightarrow n^2 = \frac{1}{2}, 2$. Therefore, $r = \sqrt{2}$. Now, $ar^3 + ar^5 + ar^7 = ar^3 (1 + r^2 + r^4)$. This simplifies to $5[1 + 2 + 4] = 35$.

Question 5

Maths · Integrals · Single correct

The value of $\sum_{n=1}^{100} \int_{n-1}^{n} e^{x - \lfloor x \rfloor} dx$, where $\lfloor x \rfloor$ is the greatest integer $\leq x$, is:

  1. 100(e - 1)
  2. 100e
  3. 100(1 - e)
  4. 100(1 + e)

Answer: (a)

Solution

The given expression is $$\sum_{n=1}^{100} \int_{n-1}^{n} e^{x - \lfloor x \rfloor} dx$$ which can be rewritten as $$\int_{0}^{1} e^{\{x\}} dx + \int_{1}^{2} e^{\{x\}} dx + \int_{2}^{3} e^{\{x\}} dx + \ldots + \int_{99}^{100} e^{\{x\}} dx$$ where $\{x\} = x - \lfloor x \rfloor$. This simplifies to $$e^x \bigg|_0^1 + e^{(x-1)} \bigg|_1^2 + e^{(x-2)} \bigg|_2^3 + \ldots + e^{(x-99)} \bigg|_{99}^{100}$$ which equals $$(e - 1) + (e - 1) + (e - 1) + \ldots + (e - 1)$$ resulting in $$100(e - 1).$$

Question 6

Maths · Conic Sections · Single correct

In the circle given below, let $OA = 1$ unit, $OB = 13$ unit and $PQ \perp OB$. Then, the area of the triangle $PQB$ (in square units) is:

  1. $26\sqrt{3}$
  2. $24\sqrt{2}$
  3. $24\sqrt{3}$
  4. $26\sqrt{2}$

Answer: (c)

Solution

Given $OC = \frac{13}{2} = 6.5$. $AC = CO - AO$ $$= 6.5 - 1$$ $$= 5.5$$ In $\triangle PAC$, $$PA = \sqrt{6.5^2 - 5.5^2}$$ $$PA = \sqrt{12}$$ $$\Rightarrow PQ = 2PA = 2\sqrt{12}$$ Now, area of $\triangle PQB = \frac{1}{2} \times PQ \times AB$ $$= \frac{1}{2} \times 2\sqrt{12} \times 12$$ $$= 12\sqrt{12}$$ $$= 24\sqrt{3}$$

Question 7

Maths · Sequences and Series · Single correct

The sum of the infinite series $1 + \frac{2}{3} + \frac{7}{3^2} + \frac{12}{3^3} + \frac{17}{3^4} + \frac{22}{3^5} + \ldots$ is equal to

  1. $\frac{9}{4}$
  2. $\frac{15}{4}$
  3. $\frac{13}{4}$
  4. $\frac{11}{4}$

Answer: (c)

Solution

Given $$s = 1 + \frac{2}{3} + \frac{7}{3^2} + \frac{12}{3^3} + \frac{17}{3^4} + \frac{22}{3^5} + \cdots$$ We have $$\frac{s}{3} = \frac{1}{3} + \frac{2}{3^2} + \frac{7}{3^3} + \cdots \infty$$ Then $$\frac{2s}{3} = 1 + \frac{1}{3} + \frac{5}{3^2} + \frac{5}{3^3} + \cdots \infty$$ Thus, $$\frac{2s}{3} = \frac{4}{3} + \frac{5}{3} \left\{ \frac{1/3}{1 - 1/3} \right\} = \frac{5}{6} + \frac{4}{3} = \frac{13}{6}$$ Therefore, $$s = \frac{13}{4}$$

Question 8

Maths · Limits and Derivatives · Single correct

The value of $\lim_{h \to 0} \left\{ \frac{\sqrt{3} \sin\left(\frac{\pi}{6} + h\right) - \cos\left(\frac{\pi}{6} + h\right)}{\sqrt{3}h(\sqrt{3} \cosh - \sinh)} \right\}$ is:

  1. $\frac{3}{4}$
  2. $\frac{2}{\sqrt{3}}$
  3. $\frac{4}{3}$
  4. $\frac{2}{3}$

Answer: (c)

Solution

Given the limit expression: $$ \lim_{h \to 0} 2 \times 2 \left\{ \frac{\sin\left(\frac{\pi}{6} + h - \frac{\pi}{6}\right)}{2\sqrt{3}h \left(\cos\left(h + \frac{\pi}{6}\right)\right)} \right\} $$ Simplifying, we have: $$ = \frac{2}{\sqrt{3}} \times \frac{2}{\sqrt{3}} = \frac{4}{3} $$

Question 9

Maths · Binomial Theorem · Single correct

The maximum value of the term independent of 't' in the expansion of $$\left( tx^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{10}$$ where $x \in (0, 1)$ is:

  1. $\($ $\frac{10!}{\sqrt{3}(5!)^2}$ $\)$
  2. $\($ $\frac{2 \cdot 10!}{3(5!)^2}$ $\)$
  3. $\($ $\frac{10!}{3(5!)^2}$ $\)$
  4. $\($ $\frac{2 \cdot 10!}{3 \sqrt{3}(5!)^2}$ $\)$

Answer: (d)

Solution

Given $$T_{r+1} = 10C_r \left( tx^{1/5} \right)^{10-r} \left[ \frac{(1-x)^{1/10}}{t} \right]^r$$ This simplifies to $$= 10C_r t^{(10-2r)} \times x^{\frac{10-r}{5}} \times (1-x)^{\frac{r}{10}}$$ Setting the equation $$10 - 2r = 0 \Rightarrow r = 5$$ Then $$T_6 = 10C_5 x \sqrt{1-x}$$ Differentiating with respect to $x$, $$\frac{dT_6}{dx} = 10C_5 \left[ \sqrt{1-x} - \frac{x}{2\sqrt{1-x}} \right] = 0$$ Solving gives $$1-x = x/2 \Rightarrow 3x = 2$$ Thus, $$x = 2/3$$ Finally, $$T_{6|\max} = \frac{10!}{5!5!} \times \frac{2}{3\sqrt{3}}$$

Question 10

Maths · Differential Equations · Single correct

The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time $t = 0$. The number of bacteria is increased by 20$\%$ in 2 hours. If the population of bacteria is 2000 after $\frac{k}{\log_e \left( \frac{6}{5} \right)}$ hours, then $\left( \frac{k}{\log_e 2} \right)^2$ is equal to

  1. 4
  2. 2
  3. 16
  4. 8

Answer: (a)

Solution

Given $\($ $\frac{dx}{dt}$ $\propto$ x $\)$ $\($ $\frac{dx}{dt}$ = $\lambda$ x $\)$ $\($ $\int$_{1000}^{x} $\frac{dx}{x}$ = $\int$_{0}^{t} $\lambda$ dt $\)$ $\($ $\ln$ x - $\ln$ 1000 = $\lambda$ t $\)$ $\($ $\ln$ $\left$( $\frac{x}{1000}$ $\right$) = $\lambda$ t $\)$ Put $\($ t = 2, x = 1200 $\)$ $\($ $\ln$ $\left$( $\frac{12}{10}$ $\right$) = 2$\lambda$ $\Rightarrow$ $\lambda$ = $\frac{1}{2}$ $\ln$ $\frac{6}{5}$ $\)$ Now $\($ $\ln$ $\left$( $\frac{x}{1000}$ $\right$) = $\frac{t}{2}$ $\ln$ $\left$( $\frac{6}{5}$ $\right$) $\)$ $\($ x = 1000 e^{$\frac{1}{2}$ t $\ln$ $\left$( $\frac{6}{5}$ $\right$)} $\)$ $\($ x = 2000 $\)$ at $\($ t = -$\frac{k}{\ln \left( \frac{6}{5} \right)}$ $\)$ $\($ $\Rightarrow$ 2000 = 1000 e^{$\frac{-k}{2 \ln (6/5)}$} $\times$ $\ln$ (6/5) $\)$ $\($ $\Rightarrow$ 2 = e^{k/2} $\)$ $\($ $\Rightarrow$ $\ln$ 2 = $\frac{k}{2}$ $\)$ $\($ $\Rightarrow$ $\frac{k}{\ln 2}$ = 2 $\)$ $\($ $\Rightarrow$ $\left$( $\frac{k}{\ln 2}$ $\right$)^2 = 4 $\)$

Question 11

Maths · Vector Algebra · Single correct

If $(1, 5, 35)$, $(7, 5, 5)$, $(1, \lambda, 7)$ and $(2\lambda, 1, 2)$ are coplanar, then the sum of all possible values of $\lambda$ is:

  1. -$\frac{44}{5}$
  2. $\frac{39}{5}$
  3. -$\frac{39}{5}$
  4. $\frac{44}{5}$

Answer: (d)

Solution

Let $P(1, 5, 35)$, $Q(7, 5, 5)$, $R(1, \lambda, 7)$, $S(2\lambda, 1, 2)$. $$\begin{vmatrix} \overrightarrow{PQPR} & \overrightarrow{PS} \end{vmatrix} = 0$$ $$\begin{vmatrix} 6 & 0 & -30 \\ 0 & \lambda - 5 & -28 \\ 2\lambda - 1 & -4 & -33 \end{vmatrix} = 0$$ $$\begin{vmatrix} 1 & 0 & -5 \\ 0 & \lambda - 5 & -28 \\ 2\lambda - 1 & -4 & -33 \end{vmatrix} = 0$$ $\{-33\lambda + 165 - 112\} + 5(\lambda - 5)(2\lambda - 1) = 0$ $53 - 33\lambda + 5\{2\lambda^2 - 11\lambda + 5\} = 0$ $$16\lambda^2 - 88\lambda + 78 = 0$$ $$5\lambda^2 - 44\lambda + 39 = 0 < \frac{\lambda_1}{\lambda_2}$$ $\Rightarrow \lambda_1 + \lambda_2 = \frac{44}{5}$

Question 12

Maths · Inverse Trigonometric Functions · Single correct

If $\frac{\sin^{-1} x}{a}$ = $\frac{\cos^{-1} x}{b}$ = $\frac{\tan^{-1} y}{c}$; $\;$ 0 < x < 1, then the value of $\cos(\frac{\pi c}{a+b})$ is:

  1. $\frac{1-y^2}{2y}$
  2. $\frac{1-y^2}{1+y^2}$
  3. $1 - y^2$
  4. $\frac{1-y^2}{y\sqrt{y}}$

Answer: (b)

Solution

Given $\($ $\frac{\sin^{-1} x}{a}$ = $\frac{\cos^{-1} x}{b}$ = $\frac{\tan^{-1} y}{c}$ $\)$. $\($ $\frac{\sin^{-1} x}{a}$ = $\frac{\cos^{-1} x}{b}$ = $\frac{\sin^{-1} x + \cos^{-1} x}{a+b}$ = $\frac{\pi}{2(a+b)}$ $\)$ Now, $\($ $\frac{\tan^{-1} y}{c}$ = $\frac{\pi}{2(a+b)}$ $\)$ $\($ 2 $\tan$^{-1} y = $\frac{\pi c}{a+b}$ $\)$ $\($ $\Rightarrow$ $\cos$ $\left$( $\frac{\pi c}{a+b}$ $\right$) = $\cos$ $\left$( 2 $\tan$^{-1} y $\right$) = $\frac{1-y^2}{1+y^2}$ $\)$

Question 13

Maths · Permutations and Combinations · Single correct

The number of seven digit integers with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only is

  1. 77
  2. 42
  3. 35
  4. 82

Answer: (a)

Solution

CASE-I: 1, 1, 1, 1, 1, 2, 3 WAYS = $\frac{7!}{5!}$ = 42 CASE-II: 1, 1, 1, 1, 2, 2, 2 WAYS = $\frac{7!}{4! \cdot 3!}$ = 35 TOTAL WAYS = 42 + 35 = 77

Question 14

Maths · Continuity and Differentiability · Single correct

Let $f$ be any function defined on $\mathbb{R}$ and let it satisfy the condition: $$|f(x) - f(y)| \leq |(x - y)^2|, \forall (x, y) \in \mathbb{R}$$ If $f(0) = 1$, then:

  1. $f(x) < 0, \forall x \in \mathbb{R}$
  2. $f(x)$ can take any value in $\mathbb{R}$
  3. $f(x) = 0, \forall x \in \mathbb{R}$
  4. $f(x) > 0, \forall x \in \mathbb{R}$

Answer: (d)

Solution

Given $|f(x) - f(y)| \leq |(x-y)^2|$, $\forall (x, y) \in \mathbb{R}$. We have $$\left| \frac{f(x) - f(y)}{x-y} \right| \leq |x-y|$$ $$\lim_{x \to y} \left| \frac{f(x) - f(y)}{x-y} \right| \leq 0$$ $$|f'(y)| \leq 0 \Rightarrow f'(y) = 0$$ Therefore, $f(y) = C$. Thus, $$c = 1$$ $$\Rightarrow f(x) = 1$$

Question 15

Maths · Applications of Derivatives · Single correct

The maximum slope of the curve $y = \frac{1}{2}x^4 - 5x^3 + 18x^2 - 19x$ occurs at the point:

  1. (2,9)
  2. (2,2)
  3. $\left(3, \frac{21}{2}\right)$
  4. (0,0)

Answer: (b)

Solution

Given $\($ $\frac{dy}{dx}$ = 2x^3 - 15x^2 + 36x - 19 $\)$. Let $\($ f(x) = 2x^3 - 15x^2 + 36x - 19 $\)$. Differentiating, we have $\($ f'(x) = 6x^2 - 30x + 36 = 0 $\)$. Solving $\($ x^2 - 5x + 6 = 0 $\)$, we find $\($ x = 2, 3 $\)$. The second derivative is $\($ f''(x) = 12x - 30 $\)$. Since $\($ f''(x) < 0 $\)$ for $\($ x = 2 $\)$, at $\($ x = 2 $\)$, $\($ y = 8 - 40 + 72 - 38 $\)$. Simplifying gives $\($ y = 72 - 70 = 2 $\)$. Therefore, the point is $\($ (2, 2) $\)$.

Question 16

Maths · Straight Lines and Pair of Straight Lines · Single correct

The intersection of three lines $x - y = 0$, $x + 2y = 3$ and $2x + y = 6$ is a:

  1. Equilateral triangle
  2. None of the above
  3. Isosceles triangle
  4. Right angled triangle

Answer: (c)

Solution

Question 17

Maths · Three Dimensional Geometry · Single correct

Consider the three planes $P_1 : 3x + 15y + 21z = 9$ $P_2 : x - 3y - z = 5$, and $P_3 : 2x + 10y + 14z = 5$ Then, which one of the following is true?

  1. $P_1$ and $P_3$ are parallel.
  2. $P_2$ and $P_3$ are parallel.
  3. $P_1$ and $P_2$ are parallel.
  4. $P_1, P_2$ and $P_3$ all are parallel.

Answer: (a)

Solution

Given the planes: $$P_1 = x + 5y + 7z = 3$$ $$P_2 = x - 3y - z = 5$$ $$P_3 = x + 5y + 7z = \frac{5}{2}$$ It follows that $P_1$ is parallel to $P_3$.

Question 18

Maths · Determinants · Single correct

The value of $$\begin{vmatrix} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{vmatrix}$$ is

  1. -2
  2. (a+1)(a+2)(a+3)
  3. 0
  4. (a+2)(a+3)(a+4)

Answer: (a)

Solution

Perform the column operations $C_1 \to c_1 - c_2$, $c_2 \to c_2 - c_3$ on the matrix: $$\begin{vmatrix} (a+2)a & a+1 & 1 \\ (a+3)(a+1) & a+2 & 1 \\ (a+4)(a+2) & a+3 & 1 \end{vmatrix}$$ This results in: $$\begin{vmatrix} a^2 + 2a & a+1 & 1 \\ 2a+3 & 1 & 0 \\ 4a+8 & 2 & 0 \end{vmatrix}$$ Now perform the row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$: $$\begin{vmatrix} a^2 + 2a & a+1 & 1 \\ 2a+3 & 1 & 0 \\ 4a+8 & 2 & 0 \end{vmatrix}$$ This results in: $$\begin{vmatrix} a^2 + 2a & a+1 & 1 \\ 2a+3 & 1 & 0 \\ 4a+8 & 2 & 0 \end{vmatrix}$$ Finally, calculate the determinant: $6 - 8 = -2$.

Question 19

Maths · Integrals · Single correct

The value of $$\int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1+3^x} \, dx$$ is:

  1. $2\pi$
  2. $4\pi$
  3. $\frac{\pi}{2}$
  4. $\frac{\pi}{4}$

Answer: (d)

Solution

Let $I = \int_{-\pi/2}^{5\pi/2} \frac{\cos^2 x}{1+3^x} \, dx$. $I = \int_{-\pi/2}^{\pi/2} \frac{3^x \cos^2 x}{1+3^x} \, dx$. $2I = \int_{-\pi/2}^{\pi/2} \cos^2 x \, dx$. $I = \int_{0}^{\pi/2} \cos^2 x \, dx = \frac{\pi}{4}$.

Question 20

Maths · Relations and Functions · Single correct

Let $R = \{ P, Q \} \mid P$ and $Q$ are at the same distance from the origin be a relation, then the equivalence class of $(1, -1)$ is the set:

  1. $S = \{ (x, y) \mid x^2 + y^2 = 1 \}$
  2. $S = \{ (x, y) \mid x^2 + y^2 = 4 \}$
  3. $S = \{ (x, y) \mid x^2 + y^2 = \sqrt{2} \}$
  4. $S = \{ (x, y) \mid x^2 + y^2 = 2 \}$

Answer: (d)

Solution

P(a, b), Q(c, d), PO = QO implies $$a^2 + b^2 = c^2 + d^2$$ R(x, y) s = (1, -1) implies RO = SO (therefore: equivalence class) $$x^2 + y^2 = 2$$

Question 21

Maths · Differential Equations · Numerical

The difference between degree and order of differential equation that represents the family of curves given by $$y^2 = a \left( x + \frac{\sqrt{a}}{2} \right), \ a > 0$$ is

Answer: 2

Solution

Given $y^2 = a \left( x + \frac{\sqrt{a}}{2} \right)$. Then $2yy' = a$. Substituting, $y^2 = 2yy' \left( x + \frac{\sqrt{2yy'}}{2} \right)$. This simplifies to $y = 2y' \left( x + \frac{\sqrt{yy'}}{\sqrt{2}} \right)$. Rearranging gives $y - 2xy' = \sqrt{2}y' \sqrt{yy'}$. Finally, $(y - 2x \frac{dy}{dx})^2 = 2y \left( \frac{dy}{dx} \right)^3$. Here, $D = 3$ and $O = 1$. Thus, $D - O = 3 - 1 = 2$.

Question 22

Maths · Trigonometric Functions · Numerical

The number of integral values of 'k' for which the equation $3 \sin x + 4 \cos x = k + 1$ has a solution, $k \in \mathbb{R}$ is

Answer: 11

Solution

Given $3 \sin x + 4 \cos x = k + 1$. The range is $-5 \leq k + 1 \leq 5$. This implies $-6 \leq k \leq 4$. The integral values are $-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4$. Therefore, there are 11 integral values.

Question 23

Maths · Basics Of Mathematics · Numerical

The number of solutions of the equation $\log_4(x - 1) = \log_2(x - 3)$ is

Answer: 1

Solution

$\dfrac{1}{2}\log_2(x-1)=\log_2(x-3)$ $x-1=(x-3)^2$ $x^2-6x+9=x-1$ $x^2-7x+10=0$ $x=2,5$ $X=2$ Not possible as $\log_2(x-3)$ is not defined $\Rightarrow$ No. of solution $=1$

Question 24

Maths · Complex Numbers and Quadratic Equations · Numerical

The sum of $162^{th}$ power of the roots of the equation $x^3 - 2x^2 + 2x - 1 = 0$ is

Answer: 3

Solution

Let roots of $x^3 - 2x^2 + 2x - 1 = 0$ are $\alpha, \beta, \gamma$. $$(x - 1) \left(x^2 - x + 1\right) = 0$$ Now $\alpha^{162} + \beta^{162} + \gamma^{162}$ $$= 1 + \omega^{162} + (\omega^2)^{162}$$ $$= 1 + (\omega^3)^{54} + (\omega^3)^{108}$$ $$= 3$$

Question 25

Maths · Binomial Theorem · Numerical

Let m, n $\in$ N and gcd (2, n) = 1. If 30 $\binom{30}{0}$ + 29 $\binom{30}{1}$ + $\ldots$ + 2 $\binom{30}{28}$ + 1 $\binom{30}{29}$ = n $\cdot$ $2^m$, then n + m is equal to

Answer: 45

Solution

Let $S = \sum_{r=0}^{30} (30 - r) \binom{30}{r}$. $$= 30 \sum_{r=0}^{30} \binom{30}{r} - \sum_{r=0}^{30} r \binom{30}{r}$$ $$= 20 \times 2^{30} - \sum_{r=1}^{30} r \cdot \frac{30}{4} \cdot 29 \binom{30}{r-1}$$ $$= 30 \times 2^{30} - 30 \cdot 2^{29}$$ $$= (30 \times 2 - 30) \cdot 2^{29} = 30 \cdot 2^{29} \Rightarrow 15 \cdot 2^{30}$$ $$= n = 15, m = 30$$ $$n + m = 45$$

Question 26

Maths · Differential Equations · Numerical

If $y = y(x)$ is the solution of the equation $e^{\sin y} \cos y \frac{dy}{dx} + e^{\sin y} \cos x = \cos x, y(0) = 0$; then $$1 + y\left(\frac{\pi}{6}\right) + \frac{\sqrt{3}}{2} y\left(\frac{\pi}{3}\right) + \frac{1}{\sqrt{2}} y\left(\frac{\pi}{4}\right)$$ is equal to

Answer: 1

Solution

Given $e^{\sin y} \cos y \frac{dy}{dx} + e^{\sin y} \cos x = \cos x$. Put $e^{\sin y} = t$. Then $e^{\sin y} \times \cos y \frac{dy}{dx} = \frac{dt}{dx}$. This implies $$\frac{dt}{dx} + t \cos x = \cos x.$$ The integrating factor is $I.F. = e^{\int \cos x \, dx} = e^{\sin x}$. Solution of the differential equation: $$t \cdot e^{\sin x} = \int e^{\sin x} \cdot \cos x \, dx$$ $$e^{\sin y} \cdot e^{\sin x} = e^{\sin x} + c$$ At $x = 0$, $y = 0$: $$1 = 1 + c \implies c = 0$$ Thus, $e^{\sin x + \sin y} = e^{\sin x}$. This implies $\sin x + \sin y = \sin x$. Therefore, $y = 0$. This gives $\Rightarrow y\left(\frac{\pi}{6}\right) = 0$, $y\left(\frac{\pi}{3}\right) = 0$, $y\left(\frac{\pi}{4}\right) = 0$. Thus, $1 + 0 + 0 + 0 = 1$.

Question 27

Maths · Three Dimensional Geometry · Numerical

Let $(\lambda, 2, 1)$ be a point on the plane which passes through the point $(4, -2, 2)$. If the plane is perpendicular to the line joining the point $(-2, -21, 29)$ and $(-1, -16, 23)$, then $$\left( \frac{\lambda}{11} \right)^2 - \frac{4\lambda}{11} - 4$$ is equal to

Answer: 8

Solution

Given $\overrightarrow{AB} \perp \overrightarrow{PQ}$. $$[(4 - \lambda)\hat{i} - 4\hat{j} + \hat{k}] \cdot [\hat{i} + 5\hat{j} - 6\hat{k}] = 0$$ $$4 - \lambda - 20 - 6 = 0$$ $$N = -22$$ Now, $\frac{\lambda}{11} = -2$ $$\Rightarrow \left(\frac{\lambda}{11}\right)^2 - \frac{4\lambda}{11} - 4$$ $$\Rightarrow 4 + 8 - 4 = 8$$

Question 28

Maths · Applications of Integrals · Subjective

The area bounded by the lines $y = ||x - 1| - 2|$ is

Answer: 8

Solution

Question 29

Maths · Integrals · Numerical

The value of the integral $\int_{0}^{\pi} |\sin 2x| \, dx$ is

Answer: 2

Solution

Given $$I = \int_0^{\pi} |\sin 2x| \, dx$$ We have $$I = 2 \int_0^{\pi/2} |\sin 2x| \, dx = 2 \int_0^{\pi/2} \sin 2x \, dx$$ Evaluating the integral, $$I = 2 \left[ \frac{-\cos(2x)}{2} \right]_0^{\pi/2} = 2$$

Question 30

Maths · Trigonometric Functions · Numerical

If $\sqrt{3} \left( \cos^2 x \right) = \left( \sqrt{3} - 1 \right) \cos x + 1$, the number of solutions of the given equation when $x \in \left[ 0, \frac{\pi}{2} \right]$ is

Answer: 1

Solution

Given $\sqrt{3}t^2 - (\sqrt{3} - 1)t - 1 = 0$ (where $t = \cos x$). Now, $t = \frac{(\sqrt{3} - 1) \pm \sqrt{4 + 2\sqrt{3}}}{2\sqrt{3}}$. $t = \cos x = 1$ or $-\frac{1}{\sqrt{3}} \to$ rejected as $x \in \left[0, \frac{\pi}{2}\right]$. Therefore, $\cos x = 1$. Hence, the number of solutions is 1.

Physics

Question 31

Physics · Gravitation · Single correct

Find the gravitational force of attraction between the ring and sphere as shown in the diagram, where the plane of the ring is perpendicular to the line joining the centres. If $\sqrt{8}R$ is the distance between the centres of a ring (of mass 'm') and a sphere (mass 'M') where both have equal radius 'R'.

  1. $\frac{\sqrt{8}}{9} \cdot \frac{GmM}{R}$
  2. $\frac{\sqrt{8}}{27} \cdot \frac{GmM}{R^2}$
  3. $\frac{2\sqrt{2}}{3} \cdot \frac{GMm}{R^2}$
  4. $\frac{1}{3\sqrt{8}} \cdot \frac{GMm}{R^2}$

Answer: (b)

Solution

We know that $$F = ME = M \left( \frac{GM \sqrt{8R}}{(R^2 + (\sqrt{8}R)^2)^{3/2}} \right)$$ $$F = \frac{GMm \sqrt{8R}}{(9R^2)^{3/2}} \implies \frac{2 \sqrt{2} GmM}{(9R^2)^{3/2}}$$ $$= \frac{2 \sqrt{2} GmM}{27R^2}$$ $$F = \frac{\sqrt{8} GmM}{27R^2}$$

Question 32

Physics · Electrostatic Potential and Capacitance · Single correct

Consider the combination of 2 capacitors $C_1$ and $C_2$, with $C_2 > C_1$, when connected in parallel, the equivalent capacitance is $\frac{15}{4}$ times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, $\frac{C_2}{C_1}$. Note: NTA has dropped this question in the final official answer key.

  1. $\frac{15}{11}$
  2. $\frac{29}{15}$
  3. $\frac{15}{4}$
  4. None of the above

Answer: (d)

Solution

Given $C_1 + C_2 = \frac{15}{4} \left( \frac{C_1 C_2}{C_1 + C_2} \right)$. Then, $$4(C_1 + C_2)^2 = 15C_1 C_2$$ Expanding, $$4C_1^2 + 4C_2^2 - 7C_1 C_2 = 0$$ Let $x = \frac{C_2}{C_1}$, then $$4 + 4 \left( \frac{C_2}{C_1} \right)^2 - 7 \frac{C_2}{C_1} = 0$$ This simplifies to $$4 \left( \frac{C_2}{C_1} \right)^2 - 7 \frac{C_2}{C_1} + 4 = 0$$ The ratio $\frac{C_2}{C_1}$ has no real value. Thus, $\frac{C_2}{C_1}$ is imaginary.

Question 33

Physics · Physical World, Units and Measurements · Single correct

In a typical combustion engine the workdone by a gas molecule is given by $W = \alpha^2 \beta e^{-\frac{Bx^2}{kT}}$, where $x$ is the displacement, $k$ is the Boltzmann constant and $T$ is the temperature. If $\alpha$ and $\beta$ are constants, dimensions of $\alpha$ will be:

  1. $[M^0LT^0]$
  2. $[M^2LT^{-2}]$
  3. $[MLT^{-2}]$
  4. $[MLT^{-1}]$

Answer: (a)

Solution

Given $\($ $\frac{\beta x^2}{K T}$ $\)$ is dimensionless. So $$ KT = \beta x^2 \implies \mathrm{M}^1 \mathrm{L}^2 \mathrm{T}^{-2} $$ $$ \beta = \frac{\mathrm{M}^1 \mathrm{L}^2 \mathrm{T}^{-2}}{\mathrm{L}^2} \implies \mathrm{M}^1 \mathrm{T}^{-2} $$ $$ \mathrm{M}^1 \mathrm{L}^2 \mathrm{T}^{-2} = \alpha^2 \mathrm{M}^1 \mathrm{T}^{-2} $$ $$ \alpha^2 = \mathrm{L}^2 $$ $$ \alpha = \mathrm{L} $$ $$ \alpha = \mathrm{M}^0 \mathrm{L}^1 \mathrm{T}^0 $$

Question 34

Physics · Atoms · Single correct

If $\lambda_1$ and $\lambda_2$ are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of $\lambda_1 : \lambda_2$ is:

  1. 1 : 3
  2. 1 : 9
  3. 7 : 135
  4. 7 : 108

Answer: (c)

Solution

For Lyman series $n_1 = 1$, $n_2 = 4$ $$\frac{1}{\lambda_1} = Rz^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$ $$\frac{1}{\lambda_1} = Rz^2 \left( \frac{1}{1^2} - \frac{1}{4^2} \right)$$ $$\frac{1}{\lambda_1} = \frac{15Rz^2}{16}$$ $$\lambda_1 = \frac{16}{15Rz^2}$$ For paschen series $n_1 = 3$, $n_2 = 4$ $$\frac{1}{\lambda_2} = Rz^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right)$$ $$\frac{1}{\lambda_2} = Rz^2 \left( \frac{16 - 9}{9 \times 16} \right)$$ $$\frac{1}{\lambda_2} = Rz^2 \left( \frac{7}{9 \times 16} \right)$$ $$\lambda_2 = \frac{9 \times 16}{7Rz^2}$$ So, $$\frac{\lambda_1}{\lambda_2} = \frac{\frac{16}{15Rz^2}}{\frac{9 \times 16}{7Rz^2}}$$ $$= \frac{16 \times 7}{15 \times 9 \times 16}$$ $$= \frac{7}{135}$$

Question 35

Physics · Ray Optics and Optical Instruments · Single correct

A short straight object of height 100 $\mathrm{\ cm}$ lies before the central axis of a spherical mirror whose focal length has absolute value $|f| = 40 \mathrm{\ cm}$. The image of object produced by the mirror is of height 25 $\mathrm{\ cm}$ and has the same orientation of the object. One may conclude from the information:

  1. Image is real, same side of concave mirror.
  2. Image is virtual, opposite side of convex mirror.
  3. Image is virtual, opposite side of concave mirror.
  4. Image is real, same side of convex mirror.

Answer: (b)

Solution

Same orientation so image is virtual. It is combination of real object and virtual image using height it is possible only from convex mirror.

Question 36

Physics · Oscillations · Single correct

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R/2) from the earth's centre, where 'R' is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period:

  1. $2\pi \sqrt{\frac{R}{g}}$
  2. $\frac{1}{2\pi} \sqrt{\frac{g}{R}}$
  3. $\frac{2\pi R}{g}$
  4. $\frac{g}{2\pi R}$

Answer: (a)

Solution

Given $\cos \theta = \frac{x}{d}$. If displaced from equilibrium position, $$F_{restoring} = \left( \frac{GMmd}{R^3} \right) \cos \theta$$ $$F_{Res.} = \frac{GMmd}{R^3} \cdot \frac{x}{d} = \frac{GMmx}{R^3}$$ $$a_R = \frac{GMx}{R^3} GMe = gR^2$$ $$T = 2\pi \sqrt{\frac{x}{a}}$$ $$T = 2\pi \sqrt{\frac{x}{GMx}} \sqrt{\frac{R^3}{gR^2}}$$ $$T = 2\pi \sqrt{\frac{R}{g}}$$

Question 37

Physics · Alternating Current · Single correct

An alternating current is given by the equation $i = i_1 \sin \omega t + i_2 \cos \omega t$. The rms current will be:

  1. $\frac{1}{2} \left( i_1^2 + i_2^2 \right)^{\frac{1}{2}}$
  2. $\frac{1}{\sqrt{2}} \left( i_1^2 + i_2^2 \right)^{\frac{1}{2}}$
  3. $\frac{1}{\sqrt{2}} \left( i_1 + i_2 \right)^2$
  4. $\frac{1}{\sqrt{2}} \left( i_1 + i_2 \right)$

Answer: (b)

Solution

Given $$I_0 = \sqrt{I_1^2 + I_2^2 + 2I_1I_2 \cos \theta}$$ $$I_0 = \sqrt{I_1^2 + I_2^2 + 2I_1I_2 \cos 90^\circ}$$ $$I_0 = \sqrt{I_1^2 + I_2^2 + 2I_1I_2(0)} \Rightarrow \sqrt{I_1^2 + I_2^2}$$ We, know that So, $$I_{rms} = \frac{I_0}{\sqrt{2}}$$ $$I_{rms} = \frac{\sqrt{I_1^2 + I_2^2}}{\sqrt{2}}$$

Question 38

Physics · Mechanical Properties of Solids · Single correct

The normal density of a material is $\rho$ and its bulk modulus of elasticity is $K$. The magnitude of increase in density of material, when a pressure $P$ is applied uniformly on all sides, will be:

  1. $\frac{\rho K}{P}$
  2. $\frac{K}{\rho P}$
  3. $\frac{P K}{\rho}$
  4. $\frac{\rho P}{K}$

Answer: (d)

Solution

Bulk modulus $K = \frac{-\Delta P}{\Delta V / V} = \frac{-\Delta p V}{\Delta V}$. We know, $\rho = \frac{M}{V}$. So, $\($ $\frac{-\Delta \rho}{\rho}$ = $\frac{\Delta V}{V}$ $\)$. Therefore, $$ K = \frac{-\Delta P}{\left( \frac{-\Delta \rho}{\rho} \right)} = \frac{\rho \Delta P}{\Delta \rho} $$. Thus, $$ \Delta \rho = \frac{\rho \Delta P}{K} $$ and $$ \Delta \rho = \frac{\rho P}{K} $$.

Question 39

Physics · Gravitation · Single correct

A particle is moving with uniform speed along the circumference of a circle of radius $R$ under the action of a central fictitious force $F$ which is inversely proportional to $R^3$. Its time period of revolution will be given by:

  1. $T \propto R^{\frac{5}{2}}$
  2. $T \propto R^2$
  3. $T \propto R^{\frac{4}{3}}$
  4. $T \propto R^{\frac{3}{2}}$

Answer: (b)

Solution

Given $F \propto \frac{1}{R^3}$. Therefore, $F = \frac{K}{R^3}$. We have $\frac{mv^2}{R} = \frac{K}{R^3}$. This implies $m(\omega R)^2 = \frac{K}{R^2}$. Thus, $m\omega^2 R^2 = \frac{K}{R^2}$. Therefore, $\omega^2 = \frac{K}{m} \left( \frac{1}{R^4} \right)$. Now, $\left( \frac{2\pi}{T} \right)^2 \propto \frac{1}{R^4}$. This gives $\frac{4\pi^2}{T^2} \propto \frac{1}{R^4}$. Hence, $T \propto R^2$.

Question 40

Physics · Gravitation · Single correct

A planet revolving in elliptical orbit has: A. a constant velocity of revolution. B. has the least velocity when it is nearest to the sun. C. its areal velocity is directly proportional to its velocity. D. areal velocity is inversely proportional to its velocity. E. to follow a trajectory such that the areal velocity is constant. Choose the correct answer from the options given below:

  1. A only
  2. E only
  3. D only
  4. C only

Answer: (b)

Solution

The rate of change of vector area with respect to time is given by the equation: $$\frac{\overrightarrow{\mathrm{dA}}}{\mathrm{dt}} = \frac{\overrightarrow{\mathrm{L}}}{2 \, \mathrm{m}}.$$

Question 41

Physics · Work, Energy and Power · Single correct

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Body 'P' having mass M moving with speed 'u' has head-on collision elastically with another body 'Q' having mass 'm' initially at rest. If $m < M$. Reason R: During elastic collision, the momentum and kinetic energy are both conserved. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. A is correct but R is not correct.
  2. Both A and R are correct but R is NOT the correct explanation of A.
  3. A is not correct but R is correct.
  4. Both A and R are correct and R is the correct explanation of A.

Answer: (d)

Solution

Given $m \ll M$. The coefficient of restitution $e$ is given by: $$e = \frac{v_2 - V_1}{u_1 - u_2}$$ For an elastic collision, $e = 1$. Thus, $$1 = \frac{v_2 - u}{u - 0}$$ Solving for $v_2$: $$u = v_2 - u$$ Therefore, $$v_2 = 2u$$ In an elastic collision, kinetic energy and momentum are conserved.

Question 42

Physics · System of Particles and Rotational Motion · Single correct

Four identical solid spheres each of mass 'm' and radius 'a' are placed with their centres on the four corners of a square of side 'b'. The moment of inertia of the system about one side of square where the axis of rotation is parallel to the plane of the square is:

  1. $\frac{4}{5}ma^2$
  2. $\frac{8}{5}ma^2 + mb^2$
  3. $\frac{4}{5}ma^2 + 2mb^2$
  4. $\frac{8}{5}ma^2 + 2mb^2$

Answer: (d)

Solution

The moment of inertia is calculated as follows: $$I = \frac{2}{5}ma^2 + \frac{2}{5}ma^2 + \left[ \frac{2}{5}ma^2 + mb^2 \right] + \frac{2}{5}ma^2 + mb^2$$ Simplifying, we have: $$I = 4 \times \frac{2}{5}ma^2 + 2mb^2$$ This results in: $$= \frac{8}{5}ma^2 + 2mb^2$$

Question 43

Physics · Wave Optics · Single correct

In a Young's double slit experiment two slits are separated by 2 mm and the screen is placed one meter away. When a light of wavelength 500 nm is used, the fringe separation will be:

  1. 0.75 mm
  2. 0.50 mm
  3. 1 mm
  4. 0.25 mm

Answer: (d)

Solution

Fringe width ($\beta$) is given by $\frac{\lambda D}{d}$. $d = 2 \times 10^{-3} \, \mathrm{m}$ $\lambda = 500 \times 10^{-9} \, \mathrm{m}$ $D = 1 \, \mathrm{m}$ Now, $$\beta = \frac{500 \times 10^{-9} \times 1}{2 \times 10^{-3}}$$ $$\beta = \frac{5}{2} \times 10^{-4}$$ $$\beta = 2.5 \times 10^{-4}$$ $$\beta = 0.25 \, \mathrm{mm}$$

Question 44

Physics · Electric Charges and Fields · Single correct

Find the electric field at point P (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length L carrying a charge Q. The distance of the point P from the centre of the rod is $a = \frac{\sqrt{3}}{2} L$

  1. $\frac{Q}{2\sqrt{3}\pi\varepsilon_0 L^2}$
  2. $\frac{\sqrt{3}Q}{4\pi\varepsilon_0 L^2}$
  3. $\frac{Q}{3\pi\varepsilon_0 L^2}$
  4. $\frac{Q}{4\pi\varepsilon_0 L^2}$

Answer: (a)

Solution

Given the diagram, we have: $$\tan \theta = \frac{L/2}{\frac{\sqrt{3}}{2}L} \Rightarrow \frac{1}{\sqrt{3}}$$ Thus, $\theta = 30^\circ$. The net electric field $E_{net}$ is given by: $$E_{net} = \frac{KQ}{\frac{\sqrt{3}}{2}L} (\sin 30^\circ + \sin 30^\circ) \Rightarrow \frac{2KQ}{\sqrt{3}L^2} \left( \frac{1}{2} + \frac{1}{2} \right)$$ Simplifying further: $$E_{net} = \frac{1}{4\pi \varepsilon_0} \frac{2Q}{\sqrt{3}L^2}$$ Finally, we have: $$E_{net} = \frac{Q}{2\sqrt{3}\pi \varepsilon_0 L^2}$$

Question 45

Physics · Oscillations · Single correct

If two similar springs each of spring constant $K_1$ are joined in series, the new spring constant and time period would be changed by a factor:

  1. $\frac{1}{2}, \sqrt{2}$
  2. $\frac{1}{4}, 2\sqrt{2}$
  3. $\frac{1}{2}, 2\sqrt{2}$
  4. $\frac{1}{4}, \sqrt{2}$

Answer: (a)

Solution

For the first system, the time period is given by the formula: $$T = 2\pi \sqrt{\frac{M}{K_1}}.$$ For the second system, the effective spring constant is the sum of the two springs: $$K_{eff} = K_1 + K_1 = 2K_1.$$ The time period for the second system is: $$T = 2\pi \sqrt{\frac{m}{K_{eff}}} = 2\pi \sqrt{\frac{m}{2K_1}}.$$

Question 46

Physics · Thermal Properties of Matter · Single correct

The temperature $\theta$ at the junction of two insulating sheets, having thermal resistances $R_1$ and $R_2$ as well as top and bottom temperatures $\theta_1$ and $\theta_2$ (as shown in figure) is given by:

  1. $\frac{\theta_1 R_2 + \theta_2 R_1}{R_1 + R_2}$
  2. $\frac{\theta_2 - \theta_1}{R_2 - R_1}$
  3. $\frac{\theta_2 R_2 - \theta_1 R_1}{R_2 - R_1}$
  4. $\frac{\theta_1 R_1 + \theta_2 R_2}{R_1 + R_2}$

Answer: (a)

Solution

Temperature at the junction is $\theta$. So using the formula $$\frac{T_2 - T}{R_1} = \frac{T - T_1}{R_2}$$ $$\frac{\theta_2 - \theta}{R_2} = \frac{\theta - \theta_1}{R_1}$$ $$R_1 \left( \theta_2 - \theta \right) = R_2 \left( \theta - \theta_1 \right)$$ $$R_1 \theta_2 - R_1 \theta = R_2 \theta - R_2 \theta_1$$ $$R_1 \theta + R_2 \theta = R_1 \theta_2 + R_2 \theta_1$$ $$\theta = \frac{R_1 \theta_2 + R_2 \theta_1}{R_1 + R_2}$$

Question 47

Physics · Experimental Physics · Single correct

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: An electron microscope can achieve better resolving power than an optical microscope. Reason R: The de Broglie's wavelength of the electrons emitted from an electron gun is much less than wavelength of visible light. In the light of the above statements, choose the correct answer from the options given below:

  1. A is true but R is false.
  2. Both A and R are true but R is NOT the correct explanation of A.
  3. Both A and R are true and R is the correct explanation of A.
  4. A is false but R is true.

Answer: (c)

Solution

Resolution limit ($\Delta \theta$) is given by $$\Delta \theta = \frac{1.22 \lambda}{d}$$ Resolution power is given by $$Resolution power = \frac{1}{Resolution limit}$$ If $\lambda$ decreases, $\Delta \theta$ decreases. If $\Delta \theta$ decreases, power increases.

Question 48

Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct

LED is constructed from Ga-As-P semiconducting material. The energy gap of this LED is 1.9eV. Calculate the wavelength of light emitted and its colour. $\left[\text{h} = 6.63 \times 10^{-34}\,\text{Js and c} = 3 \times 10^{8}\,\text{ms}^{-1}\right]$

  1. 654$\mathrm{nm}$ and red colour
  2. 1046$\mathrm{nm}$ and blue colour
  3. 1046$\mathrm{nm}$ and red colour
  4. 654$\mathrm{nm}$ and orange colour

Answer: (a)

Solution

We know that $E = \frac{hc}{\lambda}$. $$\frac{hc}{E} \Rightarrow \frac{1240 (in eV)}{E (in eV)}$$ $$= \frac{1240}{1.9}$$ $$= 652.63 \, nm \approx 654 \, nm$$ Wavelength of red light is 620 nm to 750 nm. So, answer is 1.

Question 49

Physics · Mechanical Properties of Fluids · Single correct

A large number of water drops, each of radius $r$, combine to have a drop of radius $R$. If the surface tension is $T$ and mechanical equivalent of heat is $J$, the rise in heat energy per unit volume will be:

  1. $\frac{2T}{rJ}$
  2. $\frac{3T}{rJ}$
  3. $\frac{2T}{J} \left( \frac{1}{r} - \frac{1}{R} \right)$
  4. $\frac{3T}{J} \left( \frac{1}{r} - \frac{1}{R} \right)$

Answer: (d)

Solution

R is the radius of the bigger drop. r is the radius of n water drops. Water drops are combined to make a bigger drop. So, Volume of n drops = volume of bigger drop $$n \left( \frac{4}{3} \pi r^3 \right) = \frac{4}{3} \pi R^3$$ $$R = rn^{1/3} \Rightarrow n = \left( \frac{R}{r} \right)^3$$ $$\Delta U = T (Change in surface area) \Delta U = T \left( n 4 \pi r^2 - 4 \pi R^2 \right)$$ $$\Delta U = 4 \pi T \left[ \left( \frac{R}{r} \right)^3 r^2 - R^2 \right] \Rightarrow \frac{4 \pi T \left( R^3/r - R^2 \right)}{J}$$ $$\frac{\Delta U}{V} = \frac{4 \pi T \left( R^3/r - R^2 \right)}{J \times \frac{4}{3} \pi R^3} = \frac{3\, T}{J} \left[ \frac{1}{r} - \frac{1}{R} \right]$$

Question 50

Physics · Current Electricity · Single correct

Five equal resistances are connected in a network as shown in figure. The net resistance between the points $A$ and $B$ is:

  1. $\frac{3R}{2}$
  2. $\frac{R}{2}$
  3. $R$
  4. $2R$

Answer: (c)

Solution

It is a balanced Wheatstone bridge. So, we know that $$R_1 R_4 = R_2 R_3$$ $$\frac{R_1}{R_2} = \frac{R_3}{R_4}$$ The equivalent resistance is given by $$R_{eq} = \frac{2R \times 2R}{2R + 2R}$$ $$= \frac{4R^2}{4R} \Rightarrow R$$

Question 51

Physics · Laws of Motion · Numerical

A person standing on a spring balance inside a stationary lift measures 60 kg. The weight of that person, if the lift descends with a uniform downward acceleration of $1.8\,\mathrm{m\,s^{-2}}$, will be $\_\_\_\_\,\mathrm{N}$. [Given $g = 10\,\mathrm{m\,s^{-2}}$]

Answer: 492

Solution

Given the forces acting on the man, we have the equation: $$Mg - N = Ma$$ Solving for the normal force $N$, we get: $$N = M(g - a)$$ Substituting the given values: $$N = 60(10 - 1.8)$$ Calculating further: $$N = 60 \times 8.2 = 492 \, \mathrm{N}$$ Thus, the normal force is $$N = 492$$

Question 52

Physics · Electric Charges and Fields · Numerical

In an electrical circuit, a battery is connected to pass $20 \, \mathrm{C}$ of charge through it in a certain given time. The potential difference between two plates of the battery is maintained at $15 \, \mathrm{V}$. The work done by the battery is ____ J.

Answer: 300

Solution

Charge flown $(Q) = 20 \, \mathrm{c}$ Potential difference $(V) = 15 \, \mathrm{V}$ Work done $(w) = Q \cdot V$ $$= 20 \times 15 = 300 \, \mathrm{J}$$ $w = 300 \, \mathrm{J}$

Question 53

Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Numerical

The circuit contains two diodes each with a forward resistance of 50$\Omega$ and with infinite reverse resistance. If the battery voltage is 6 $\mathrm{V}$, the current through the 120$\Omega$ resistance is ____ $\mathrm{mA}$

Answer: 20

Solution

$D_2$ is reverse bias so current does not flow through $D_2$ $D_1$ is forward bias. $I = \dfrac{6}{300} \Rightarrow 0.02\,\text{A}$ $= 20\,\text{mA}$

Question 54

Physics · Electromagnetic Waves · Numerical

A radiation is emitted by 1000 W bulb and it generates an electric field and magnetic field at $P_t$ placed at a distance of 2 m. The efficiency of the bulb is 1.25$\%$. The value of peak electric field at P is $x \times 10^{-1} \, \mathrm{V/m}$. Value of $x$ is ____. (Rounded-off to the nearest integer) [Take $\varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C^2 \, N^{-1} \, m^{-2}}$, $c = 3 \times 10^8 \, \mathrm{m/s}$]

Answer: 137

Solution

Intensity of electromagnetic wave is, $$I = \frac{1}{2} C \varepsilon_0 E_0^2 = \frac{P}{4 \pi r^2}$$ $$\frac{1}{2} \times 4 \pi \varepsilon_0 \times C \times E_0^2 = \frac{P}{r^2}$$ $$\frac{1}{2} \times \frac{3 \times 10^5 \times E_0}{9 \times 10^9} = \frac{1000 \times 1.25}{(2)^2} \times \frac{1}{100}$$ $$E_0^2 = \frac{60 \times 1000 \times 1.25}{4 \times 100} = \frac{125 \times 3}{2}$$ $$E_0^2 = \frac{375}{2} = 187.5$$ $$E_0 = 13.69$$ $$E_0 \approx 137 \times 10^{-1} \, \mathrm{v/m}$$

Question 55

Physics · Motion in a Plane · Numerical

A boy pushes a box of mass 2 kg with a force $\vec{F} = (20\hat{i} + 10\hat{j}) \mathrm{N}$ on a frictionless surface. If the box was initially at rest, then ____ m is displacement along the $x$-axis after 10 s.

Answer: 500

Solution

Given $\mathbf{F} = 20\hat{i} + 10\hat{j}$. $F_x = 20 \, \mathrm{N}$ $F_y = 10 \, \mathrm{N}$ $a_x = \frac{F_x}{M} = \frac{20}{2} = 10 \, \mathrm{m/s^2}$ $a_y = \frac{F_y}{M} = \frac{10}{2} = 5 \, \mathrm{m/s^2}$ displacement on $x$ axis is $$S_x = u_x t + \frac{1}{2} a_x t^2$$ $$S = 0 \times 10 + \frac{1}{2} \times 10 \times (10)^2$$ $$S = 500 \, \mathrm{m}$$

Question 56

Physics · Laws of Motion · Numerical

As shown in the figure, a block of mass $\sqrt{3} \, \mathrm{kg}$ is kept on a horizontal rough surface of coefficient of friction $\frac{1}{3\sqrt{3}}$. The critical force to be applied on the vertical surface as shown at an angle $60^\circ$ with horizontal such that it does not move, will be $3x$. The value of $x$ will be $\left[ g = 10 \, \mathrm{m/s^2}; \sin 60^\circ = \frac{\sqrt{3}}{2}; \cos 60^\circ = \frac{1}{2} \right]$ (round off to nearest integer)

Answer: 3

Solution

Given the forces acting on the block, we have: $$N = Mg + F \sin 60^\circ$$ Substituting the values, we get: $$N = \sqrt{3}g + \frac{F \sqrt{3}}{2}$$ For no slipping, the frictional force is given by: $$F \cos 60^\circ = Friction$$ The frictional force is also given by: $$f = \mu N = \frac{1}{3\sqrt{3}} \left( \sqrt{3}g + \frac{F \sqrt{3}}{2} \right)$$ Equating the two expressions for friction: $$\frac{F}{2} = \frac{g}{3} + \frac{F}{6}$$ Simplifying, we have: $$\frac{F}{2} - \frac{F}{6} = \frac{g}{3}$$ Solving for $F$: $$\frac{6F - 2F}{12} = \frac{g}{3}$$ This gives: $$4F = 4g$$ Therefore, $F = 10$. Since $F = 3x$, we have: $$x = \frac{F}{3} = \frac{10}{3} = 3.33$$ Thus, $x = 3.33$.

Question 57

Physics · Kinetic Theory · Numerical

A container is divided into two chambers by a partition. The volume of first chamber is 4.5 litre and second chamber is 5.5 litre. The first chamber contain 3.0 moles of gas at pressure 2.0 atm and second chamber contain 4.0 moles of gas at pressure 3.0 atm. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is $x \times 10^{-1}$ atm. Value of $x$ is

Answer: 2.5

Solution

By energy conservation, $\frac{3}{2}n_{1}RT_{1}$ $+$ $\frac{3}{2}n_{2}RT_{2}$ = $\frac{3}{2}(n_{1}+n_{2})RT$ Using $PV=nRT$ $P=\frac{P_{1}V_{1}+P_{2}V_{2}}{V_{1}+V_{2}}$ $=\frac{2\times4.5+3\times5.5}{4.5+5.5}$ $=\frac{25.5}{10}$ $\approx25\times10^{-1}$ atm

Question 58

Physics · Waves · Numerical

The mass per unit length of a uniform wire is 0.135 g/cm. A transverse wave of the form $$y = -0.21 \sin(x + 30t)$$ is produced in it, where $x$ is in meter and $t$ is in second. Then, the expected value of tension in the wire is $x \times 10^{-2}$ N. Value of $x$ is ____. (Round-off to the nearest integer)

Answer: 1215

Solution

Given $y = -0.21 \sin(x + 30t)$. The velocity $v$ is given by $v = \frac{\omega}{K} = \frac{30}{1} = 30 \, \mathrm{m/s}$. The velocity $v$ can also be expressed as $v = \sqrt{\frac{T}{\mu}}$. The tension $T$ is calculated as $T = v^2 \times \mu$. Substituting the values, $T = (30)^2 \times 0.135 \times 10^{-1}$. Given $\mu = 0.135 \, \mathrm{gm/cm}$, convert to $\mu = 0.135 \times \frac{10^{-3} \, \mathrm{kg}}{10^{-2} \, \mathrm{m}} \, \mathrm{kg/m}$. Thus, $T = 900 \times 0.135 \times 10^{-1}$. This gives $T = 12.15 \, \mathrm{N}$. Therefore, $T = 1215 \times 10^{-2} \, \mathrm{N}$. Finally, $X = 1215$.

Question 59

Physics · Alternating Current · Numerical

In a series LCR resonant circuit, the quality factor is measured as 100. If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be (round off to nearest integer)

Answer: 283

Solution

Quality factor = $\frac{X_L}{R}$ = $\frac{\omega L}{R}$ Q = $\frac{1}{\sqrt{LC}}$ $\frac{L}{R}$ Q = $\left$( $\frac{1}{\sqrt{C}}$ $\right$) $\frac{\sqrt{L}}{R}$ Q = $\frac{X_L}{R}$ = $\frac{\omega L}{R}$ = $\frac{1}{\sqrt{LC}}$ $\frac{L}{R}$ = $\frac{1}{R}$ $\frac{\sqrt{L}}{\sqrt{C}}$ Q' = $\frac{\sqrt{2} L}{\left( \frac{R}{2} \right) \sqrt{C}}$ = 2 $\sqrt{2}$ Q Q' = 282.84

Question 60

Physics · Communication Systems · Numerical

The maximum and minimum amplitude of an amplitude modulated wave is 16 V and 8 V respectively. The modulation index for this amplitude modulated wave is $x \times 10^{-2}$. The value of $x$ is

Answer: 33

Solution

$A_m=\dfrac{A_{\max}-A_{\min}}{2}$ $A_C=\dfrac{A_{\max}+A_{\min}}{2}$ $A_{\max}=16\,V$ $A_{\min}=8\,V$ Modulation index $(m_i)$ $=\dfrac{A_m}{A_C}$ $=\dfrac{\dfrac{A_{\max}-A_{\min}}{2}} {\dfrac{A_{\max}+A_{\min}}{2}}$ $=\dfrac{A_{\max}-A_{\min}} {A_{\max}+A_{\min}}$ $m_i=\dfrac{16-8}{16+8}$ $=\dfrac{8}{24}$ $=\dfrac13$ $=0.33$ $m_i=33\times10^{-2}$ $X=33$

Chemistry

Question 61

Chemistry · Polymers · Single correct

The structure of Neoprene is :

Answer: (b)

Solution

Nerprene

Question 62

Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct

Find $A$, $B$ and $C$ in the following reaction: $$\mathrm{NH}_3 + A + \mathrm{CO}_2 \rightarrow (\mathrm{NH}_4)_2\mathrm{CO}_3$$ $$(\mathrm{NH}_4)_2\mathrm{CO}_3 + \mathrm{H}_2\mathrm{O} + B \rightarrow \mathrm{NH}_4\mathrm{HCO}_3$$ $$\mathrm{NH}_4\mathrm{HCO}_3 + \mathrm{NaCl} \rightarrow \mathrm{NH}_4\mathrm{Cl} + C$$

  1. $A - \mathrm{H}_2\mathrm{O}$; $B - \mathrm{CO}_2$; $C - \mathrm{NaHCO}_3$
  2. $A - \mathrm{H}_2\mathrm{O}$; $B - \mathrm{O}_2$; $C - \mathrm{Na}_2\mathrm{CO}_3$
  3. $A - \mathrm{O}_2$; $B - \mathrm{CO}_2$; $C - \mathrm{Na}_2\mathrm{CO}_3$
  4. $A - \mathrm{H}_2\mathrm{O}$; $B - \mathrm{O}_2$; $C - \mathrm{NaHCO}_3$

Answer: (a)

Solution

Option (1) is $\mathrm{NH_3 + H_2O + CO_2 \rightarrow (NH_4)_2CO_3}$. Option (2) is $\mathrm{(NH_4)_2CO_3 + H_2O + CO_2 \rightarrow NH_4HCO_3}$. Option (3) is $\mathrm{NH_4HCO_3 + NaCl \rightarrow NH_4Cl + NaHCO_3}$.

Question 63

Chemistry · Environmental Chemistry · Single correct

The presence of ozone in troposphere:

  1. generates photochemical smog
  2. Protects us from the UV radiation
  3. Protects us from the X-ray radiation
  4. Protects us from greenhouse effect

Answer: (b)

Solution

The presence of ozone in the troposphere protects the earth from ultraviolet rays.

Question 64

Chemistry · Classification of Elements and Periodicity in Properties · Single correct

Match List-I with List-II \begin{tabular}{|c|c|c|c|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{Electronic configuration of elements} & \multicolumn{2}{c|}{$\Delta_iH$ in kJ mol$^{-1}$} \\ \hline (a) & $1s^22s^2$ & (i) & 801 \\ \hline (b) & $1s^22s^22p^4$ & (ii) & 899 \\ \hline (c) & $1s^22s^22p^3$ & (iii) & 1314 \\ \hline (d) & $1s^22s^22p^1$ & (iv) & 1402 \\ \hline \end{tabular}

  1. (a)- (ii), (b) - (iii), $(c)$ - (iv), (d) - (i)
  2. (a) - (iv), (b) - (i), $(c)$ - (ii), (d) - (iii)
  3. (a) - (i), (b) - (iv), $(c)$ - (iii), (d) - (ii)
  4. (a) - (i), (b) - (iii), $(c)$ - (iv), (d) - (ii)

Answer: (a)

Solution

Order of I.E. in second period $$\mathrm{Li} < \mathrm{B} < \mathrm{Be} < \mathrm{C} < \mathrm{O} < \mathrm{N} < \mathrm{F} < \mathrm{Ne}$$

Question 65

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 A: Dipole-dipole interactions are the only non-covalent interactions, resulting in hydrogen bond formation Reason R: Fluorine is the most electronegative element and hydrogen bonds in HF are symmetrical. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. A is false but R is true
  2. Both A and R are true and R is the correct explanation of A
  3. A is true but R is false
  4. Both A and R are true and R is not the correct explanation of A

Answer: (c)

Solution

Fact

Question 66

Chemistry · Hydrogen · Single correct

Statement about heavy water are given below A. Heavy water is used in exchange reactions for the study of reaction mechanisms B. Heavy water is prepared by exhaustive electrolysis of water C. Heavy water has higher boiling point than ordinary water D. Viscosity of $\mathrm{H_2O}$ is greater than $\mathrm{D_2O}$

  1. A and B only
  2. A and D only
  3. A, B and C only
  4. A and C only

Answer: (c)

Solution

Fact

Question 67

Chemistry · Structure of Atom · Single correct

The orbital having two radial as well as two angular nodes is

  1. 5d
  2. 4f
  3. 3p
  4. 4d

Answer: (a)

Solution

Given the formula for radial nodes, $R \cdot N = n - \ell - 1$, we can calculate the number of radial nodes for each orbital. For the $5d$ orbital, $n = 5$ and $\ell = 2$. Thus, $$R \cdot N = 5 - 2 - 1 = 2.$$ For the $4f$ orbital, $n = 4$ and $\ell = 3$. Thus, $$R \cdot N = 4 - 3 - 1 = 0.$$ For the $3p$ orbital, $n = 3$ and $\ell = 1$. Thus, $$R \cdot N = 3 - 1 - 1 = 1.$$ For the $4d$ orbital, $n = 4$ and $\ell = 2$. Thus, $$R \cdot N = 4 - 2 - 1 = 1.$$

Question 68

Chemistry · General Principles and Processes of Isolation of Elements · Single correct

Match List-I with List-II \begin{tabular}{|l|l|} \hline \textbf{List-I} & \textbf{List-II} \\ (Ore) & (Element Present) \\ \hline (a) Kernite & (i) Tin \\ \hline (b) Cassiterite & (ii) Boron \\ \hline (c) Calamine & (iii) Fluorine \\ \hline (d) Cryolite & (iv) Zinc \\ \hline \end{tabular} Choose the most appropriate answer from the option given below :

  1. (a) - (ii), (b) - (iv), ($c$) - (i), (d) - (iii)
  2. (a) - (ii), (b) - (i), ($c$) - (iv), (d) - (iii)
  3. (a) - (i), (b) - (iii), ($c$) - (iv), (d) - (ii)
  4. (a) - (iii), (b) - (i), ($c$) - (ii), (d) - (iv)

Answer: (b)

Solution

Fact

Question 69

Chemistry · Alcohols, Phenols and Ethers · Single correct

Identify the major products $A$ and $B$ respectively in the following reaction of phenol:

Answer: (a)

Solution

The reaction involves the bromination of phenol in carbon disulfide at 273 K, resulting in the formation of bromophenol as the major product. The reverse reaction involves the use of chloroform and sodium hydroxide in water to regenerate phenol.

Question 70

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

Given below are two statements: Statement I: A mixture of chloroform and aniline can be separated by simple distillation Statement II: When separating aniline from a mixture of aniline and water by steam distillation aniline boils below its boiling point In the light of the above statements, choose the most appropriate answer from the options given below

  1. Statement I is true, statement II is false
  2. Both Statement I and Statement II are true
  3. Both Statement I and Statement II are false
  4. Statement I is false, Statement II is true

Answer: (b)

Solution

A suitable method for separating a mixture of aniline and chloroform would be steam distillation. Steam distillation is the process used to separate aromatic compounds from a mixture because of their temperature sensitivity. Therefore, steam distillation is an ideal method for their separation.

Question 71

Chemistry · Haloalkanes and Haloarenes · Single correct

For the given reaction:

  1. CH_3CH_2CH_2NH_2

Answer: (d)

Solution

Question 72

Chemistry · The d-and f-Block Elements · Single correct

On treating a compound with warm dil. $\mathrm{H_2SO_4}$, gas $X$ is evolved which turns $\mathrm{K_2Cr_2O_7}$ paper acidified with dil. $\mathrm{H_2SO_4}$ to a green compound $Y$. $X$ and $Y$ respectively are:

  1. $X = \mathrm{SO_2}, Y = \mathrm{Cr_2(SO_4)_3}$
  2. $X = \mathrm{SO_2}, Y = \mathrm{Cr_2O_3}$
  3. $X = \mathrm{SO_3}, Y = \mathrm{Cr_2O_3}$
  4. $X = \mathrm{SO_3}, Y = \mathrm{Cr_2(SO_4)_3}$

Answer: (a)

Solution

The reaction is given as: $$\mathrm{SO_2 + K_2Cr_2O_7 + H_2SO_4 \rightarrow Cr_2(SO_4)_3 + K_2SO_4 + H_2O}$$

Question 73

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

Which of the following is 'a' FALSE statement?

  1. Carius tube used in the estimation of sulphur in an organic compound
  2. Kjeldahl's method is used for the estimation of nitrogen in an organic compound
  3. Phosphoric acid produced on oxidation of phosphorus present in an organic compound is precipitated as $\mathrm{Mg}_2\mathrm{P}_2\mathrm{O}_7$ by adding magnesia mixture
  4. Carius method is used for the estimation of nitrogen in an organic compound

Answer: (d)

Solution

Fact

Question 74

Chemistry · Biomolecules · Single correct

Which of the following vitamin is helpful in delaying the blood clotting?

  1. Vitamin B
  2. Vitamin C
  3. Vitamin K
  4. Vitamin E

Answer: (c)

Solution

Vitamin K is used by the body to help blood clot.

Question 75

Chemistry · Alcohols, Phenols and Ethers · Single correct

B reacts with Hydroxyl amine but does not give Tollen's test. Identify A and B.

  1. 1, 1-Dichlorobutane and 2-Butanone
  2. 2, 2- Dichlorobutane and Butan-2-one
  3. 2,2 - Dichlorobutane and Butanal
  4. 1,1 - Dichlorobutane and Butanal

Answer: (b)

Solution

Compound 'B' does not give Tollen's test due to the presence of a ketonic group but reacts with hydroxyl amine.

Question 76

Chemistry · The p-Block Elements (Group-13 and 14) · Single correct

Compound A used as a strong oxidizing agent is amphoteric in nature. It is the part of lead storage batteries. Compound A is :

  1. Pb_3O_4
  2. PbO_2
  3. PbSO_4
  4. PbO

Answer: (b)

Solution

In lead storage batteries $\mathrm{PbO_2}$ is used. In this O.S. of Pb is $+4$ so it is always reduced and behaves as oxidizing agent.

Question 77

Chemistry · The d-and f-Block Elements · Single correct

Which one of the following lanthanoids does not form $\mathrm{MO}_2$ ? [M is lanthanoid metal]

  1. Nd
  2. Yb
  3. Dy
  4. Pr

Answer: (b)

Solution

Fact

Question 78

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

Given below are two statement : Statement I : o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding Statement II : o-Nitrophenol has high melting point due to hydrogen bonding. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. Both Statement I and Statement II are false
  2. Statement I is false but Statement II is true
  3. Both Statement I and Statement II are true
  4. Statement I is true but Statement II is false

Answer: (d)

Solution

o-Nitrophenol is steam volatile due to intramolecular hydrogen H-bonding, but m-Nitrophenol has more melting point due to its symmetry.

Question 79

Chemistry · Hydrocarbons · Single correct

For the given reaction: What is 'A'?

Answer: (d)

Solution

It is benzylic substitution reaction.

Question 80

Chemistry · Amines · Single correct

An amine on reaction with benzenesulphonyl chloride produces a compound insoluble in alkaline solution. This amine can be prepared by ammonolysis of ethyl chloride. The correct structure of amine is :

  1. $\mathrm{CH_3CH_2CH_2\overset{H}{N}-CH_2CH_3}$
  2. $\mathrm{CH_3CH_2CH_2NHCH_3}$
  3. $\mathrm{CH_3CH_2NH_2}$

Answer: (a)

Solution

A primary amine $\mathrm{R-NH_2}$ reacts with $\mathrm{C_6H_5SO_2Cl}$ to form $\mathrm{R-NH-SO_2-C_6H_5}$, which is soluble in alkalines. A secondary amine $\mathrm{R-NH-R}$ reacts with $\mathrm{C_6H_5SO_2Cl}$ to form $\mathrm{R-N(SO_2-C_6H_5)-R}$, which is insoluble in alkalines. According to the question, the amine should be a secondary amine, in which one of the alkyl groups should be ethyl, because it can be formed by ammonolysis of ethyl chloride.

Question 81

Chemistry · Thermodynamics · Fill in the blank

For a chemical reaction $A+B\rightleftharpoons C+D$ $\left(\Delta_rH^\Theta=80\ \mathrm{kJ\,mol^{-1}}\right)$ the entropy change $\Delta_rS^\Theta$ depends on the temperature $T$ (in K) as $\Delta_rS^\Theta=2T\ \left(J\,K^{-1}\,mol^{-1}\right)$. Minimum temperature at which it will become spontaneous is __ K.

Answer: 200

Solution

Given $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$. To make the process spontaneous, $\Delta G^\circ \frac{\Delta H^\circ}{\Delta S^\circ}$. Given $T > \frac{80000}{2T}$, we have $2T^2 > 80000$. Thus, $T^2 > 40000$ and $T > 200$. The minimum temperature to make it spontaneous is 200 $\mathrm{\ K}$.

Question 82

Chemistry · Some Basic Concepts of Chemistry · Numerical

The number of significant figures in $50000.020 \times 10^{-3}$ is

Answer: 7

Solution

Given $50000.020 \times 10^{-3}$. Number of significant figures $= 7$.

Question 83

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

An exothermic reaction $X \rightarrow Y$ has an activation energy $30 \, \mathrm{kJ \, mol^{-1}}$. If energy change $\Delta E$ during the reaction is $-20 \, \mathrm{kJ}$, then the activation energy for the reverse reaction in kJ is

Answer: 50

Solution

Given $\Delta H = E_{a,f} - E_{a,b}$. $$-20 = 30 - E_{a,b}$$ Solving for $E_{a,b}$, we get: $$E_{a,b} = 50 \, \mathrm{kJ/mole}$$

Question 84

Chemistry · Electrochemistry · Numerical

Consider the following reaction $\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{+2} + 4\text{H}_2\text{O}, \quad \text{E}^\circ = 1.51\text{ V}$ The quantity of electricity required in Faraday to reduce five moles of $\text{MnO}_4^-$ is

Answer: 25

Solution

Given the reaction: $$\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{+2} + 4H_2O}$$ 1 mole of $\mathrm{MnO_4^-}$ requires 5 faraday charge. 5 moles of $\mathrm{MnO_4^-}$ will require 25 faraday charge.

Question 85

Chemistry · States of Matter · Numerical

A certain gas obeys $P \left(V_m - b\right) = RT$. The value of $$\left(\frac{\partial Z}{\partial P}\right)_T$$ is $\frac{xb}{RT}$. The value of $x$ is

Answer: 1

Solution

Given the equation $P(v - b) = RT$, we can rewrite it as $PV - Pb = RT$. Dividing through by $RT$, we have $$\frac{PV}{RT} - \frac{Pb}{RT} = 1.$$ Therefore, $$Z = 1 + \frac{PV}{RT}.$$ Differentiating $Z$ with respect to $P$, we get $$\frac{dz}{dp} = 0 + \frac{b}{RT}.$$ This implies $$\frac{b}{RT} = \frac{xb}{RT}$$ which leads to $x = 1$.

Question 86

Chemistry · Equilibrium · Numerical

A homogeneous ideal gaseous reaction $\mathrm{AB}_2{(g)} \rightleftharpoons \mathrm{A}_{(g)} + 2 \mathrm{B}_{(g)}$ is carried out in a 25 litre flask at $27^\circ \mathrm{C}$. The initial amount of $\mathrm{AB}_2$ was 1 mole and the equilibrium pressure was 1.9 atm. The value of $K_p$ is $x \times 10^{-2}$. The value of $x$ is ____ $\left[ R = 0.08206 \mathrm{dm}^3 \mathrm{atm} \mathrm{K}^{-1} \mathrm{mol}^{-1} \right]$

Answer: 74

Solution

The reaction is given as: $$\mathrm{AB_2}_{(g)} \rightarrow \mathrm{A}_{(g)} + 2 \mathrm{B}_{(g)}$$ Initially, the concentrations are: $$initial: 1-x x 2x$$ At equilibrium, the concentrations are: $$at eq.: \frac{1}{1+2x} \frac{1}{1.9}$$ By the ratio of pressure and mole: $$\frac{1}{1+2x} = \frac{0.985}{1.9}$$ Solving for $x$: $$1.9 = 0.985 + 1.9x$$ $$0.915 = 1.9x$$ $$\frac{0.915}{1.9} = x$$ The equilibrium constant $K_p$ is given by: $$K_p = \frac{4x^2 \cdot x}{(1-x)} \left[ \frac{P_{total}}{n_{total}} \right]^2$$ This simplifies to: $$\Rightarrow \frac{4x^3}{1-x} \left( \frac{RT}{V} \right)^2$$ On substituting the values, $K_p = 74 \times 10^{-2}$.

Question 87

Chemistry · Redox Reactions · Fill in the blank

Dichromate ion is treated with base, the oxidation number of Cr in the product formed is :

Answer: 6

Solution

The given reaction is: $$\mathrm{Cr_2O_7^{2-} }+ 2OH^- = 2CrO_4^{2-} + H_2O$$ The coefficient of $\mathrm{Cr_2O_7^{2-}}$ is 2. Solving the equation: $$x + (-2 \times 4) = -2$$ Therefore, $$x = 6$$

Question 88

Chemistry · Solutions · Numerical

$224\ \mathrm{mL}$ of $\mathrm{SO_2}(g)$ at $298\ \mathrm{K}$ and $1\ \mathrm{atm}$ is passed through $100\ \mathrm{mL}$ of $0.1\,\mathrm{M}\ \mathrm{NaOH}$ solution. The non-volatile solute produced is dissolved in $36\ \mathrm{g}$ of water. The lowering of vapour pressure of solution (assuming the solution is dilute) $(P^\circ_{\mathrm{H_2O}}=24\ \mathrm{mm\ of\ Hg})$ is $x\times10^{-2}\ \mathrm{mm\ of\ Hg}$, the value of $x$ is (round off to nearest integer).

Answer: 0

Solution

The balanced equation is $\mathrm{SO_2+2NaOH\rightarrow Na_2SO_3+H_2O}$ moles of NaOH $=$ molarity $\times$ volume (in litre) $=0.1\times0.1$ $=0.01$ moles Here NaOH is limiting Reagent. $2$ mole NaOH $\rightarrow1$ mole $\mathrm{Na_2SO_3}$ $0.01$ mole NaOH $\rightarrow\frac{1}{2}\times0.01$ mole $\mathrm{Na_2SO_3}$ Moles of $\mathrm{Na_2SO_3}\rightarrow0.005$ mole $\mathrm{Na_2SO_3\rightarrow2Na^++SO_3^{2-}}$ $i=3$ Moles of $\mathrm{H_2O}=\frac{36}{18}=2$ moles According to RLVP - $\frac{P_A^0-P_A}{P_A^0}=iX_B$ $\frac{P_A^0-P_A}{P_A^0}=\frac{i\,n_B}{i\,n_B+n_A}\;(i\,n_B\simeq0)$ $n_B\ll n_A$ $\{\,n_A+n_B\simeq n_A\,\}$ $\frac{P_A^0-P_A}{P_A^0}=i\times\frac{n_B}{n_A}$ $\frac{24-P_A}{24}=3\times\frac{0.005}{2}$ $\Rightarrow24-P_A=0.18$ Lowering in pressure $=0.18$ mm of Hg Lowering in pressure $=18\times10^{-2}$ mm of Hg $X=18$

Question 89

Chemistry · Surface Chemistry · Fill in the blank

3.12 g of oxygen is adsorbed on 1.2 g of platinum metal. The value of oxygen adsorbed per gram of the adsorbent at 1 atm and 300 K in L is __ $\left[R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}\right]$

Answer: 2

Solution

Moles of $\mathrm{O_2} = \frac{3.12}{32} = 0.0975$ Volume of $\mathrm{O_2} = \frac{nRT}{p} = \frac{0.0975 \times 0.082 \times 300}{1}$ $$= 2.3985 \, \mathrm{L} \simeq 2.4 \, \mathrm{L}$$ Volume of $\mathrm{O_2}$ absorbed per gm of pt $= \frac{2.4}{1.2} = 2$

Question 90

Chemistry · Co-ordination Compounds · Numerical

Number of bridging CO ligands in $[\mathrm{Mn}_2(\mathrm{CO})_{10}]$ is

Answer: 0

Solution

The structure shown is a manganese carbonyl complex. It consists of two manganese (Mn) atoms each bonded to five carbon monoxide (CO) ligands. The Mn atoms are also bonded to each other.