JEE Main 6 April 2024 Shift 2 question paper with solutions

JEE Main 6 April 2024 Shift 2: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.

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Maths

Question 1

Maths · Properties of Triangles · Single correct

Let ABC be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle ABC and the same process is repeated infinitely many times. If P is the sum of perimeters and Q is the sum of areas of all the triangles formed in this process, then :

  1. $P^2 = 6\sqrt{3}Q$
  2. $P^2 = 36\sqrt{3}Q$
  3. $P = 36\sqrt{3}Q^2$
  4. $P^2 = 72\sqrt{3}Q$

Answer: (b)

Solution

Area of first $\Delta = \frac{\sqrt{3}a^2}{4}$. Area of second $\Delta = \frac{\sqrt{3}a^2}{4} \cdot \frac{a^2}{4} = \frac{\sqrt{3}a^2}{16}$. Area of third $\Delta = \frac{\sqrt{3}a^2}{64}$. Sum of area $= \frac{\sqrt{3}a^2}{4} \left( 1 + \frac{1}{4} + \frac{1}{16} \cdots \right)$. $Q = \frac{\sqrt{3}a^2}{4} \cdot \frac{1}{\frac{3}{4}} = \frac{a^2}{\sqrt{3}}$. Perimeter of 1st $\Delta = 3a$. Perimeter of 2nd $\Delta = \frac{3a}{2}$. Perimeter of 3rd $\Delta = \frac{3a}{4}$. $P = 3a \left( 1 + \frac{1}{2} + \frac{1}{4} + \cdots \right)$. $P = 3a \cdot 2 = 6a$. $a = \frac{P}{6}$. $Q = \frac{1}{\sqrt{3}} \cdot \frac{P^2}{36}$. $P^2 = 36\sqrt{3}Q$.

Question 2

Maths · Relations and Functions · Single correct

Let A = $\{1, 2, 3, 4, 5\}$. Let R be a relation on A defined by $xRy$ if and only if $4x \leq 5y$. Let $m$ be the number of elements in R and $n$ be the minimum number of elements from $A \times A$ that are required to be added to R to make it a symmetric relation. Then $m + n$ is equal to:

  1. 25
  2. 24
  3. 26
  4. 23

Answer: (a)

Solution

Given: $4x \leq 5y$ then $$R = \{(1,1), (1,2), (1,3), (1,4), (1,5), (2,2), (2,3), (2,4), (2,5), (3,3), (3,4), (3,5), (4,4), (4,5), (5,4), (5,5)\}$$ i.e. 16 elements. i.e. $m = 16$ Now to make $R$ a symmetric relation add $$\{(2,1), (3,2), (4,3), (3,1), (4,2), (5,3), (4,1), (5,2), (5,1)\}$$ i.e. $n = 9$ So $m + n = 25$

Question 3

Maths · Probability · Single correct

If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly two addresses is:

  1. $\frac{18}{25}$
  2. $\frac{12}{25}$
  3. $\frac{6}{25}$
  4. $\frac{4}{25}$

Answer: (b)

Solution

Total method $= 5^3$. Favorable $= \binom{5}{2} (2^3 - 2) = 60$. Probability $= \frac{60}{125} = \frac{12}{25}$.

Question 4

Maths · Differential Equations · Single correct

Suppose the solution of the differential equation $\frac{dy}{dx} = \frac{(2+\alpha)x - \beta y + 2}{\beta x - 2\alpha y - (\beta \gamma - 4\alpha)}$ represents a circle passing through origin. Then the radius of this circle is:

  1. 2
  2. $\sqrt{17}$
  3. $\frac{1}{2}$
  4. $\frac{\sqrt{17}}{2}$

Answer: (d)

Solution

Given $\($ $\frac{dy}{dx}$ = $\frac{(2 + \alpha)x - \beta y + 2}{\beta x - y(2\alpha + \beta) + 4\alpha}$ $\)$ $\($ $\beta$ x dy - (2$\alpha$ + $\beta$) y dy + 4$\alpha$ dy = (2 + $\alpha$) x dx - $\beta$ y dx + 2 dx $\)$ $\($ $\beta$ (x dy + y dx) - (2$\alpha$ + $\beta$) y dy + 4$\alpha$ dy = (2 + $\alpha$) x dx + 2 dx $\)$ $\($ $\beta$ xy - $\frac{(2\alpha + \beta) y^2}{2}$ + 4$\alpha$ y = $\frac{(2 + \alpha) x^2}{2}$ $\)$ $\($ $\Rightarrow$ $\beta$ = 0 $\)$ for this to be a circle $\($ (2 + $\alpha$) $\frac{x^2}{2}$ + $\alpha$ y^2 + 2x - 4$\alpha$ y = 0 $\)$ Coefficient of $\($ x^2 = y^2 $\)$ $\($ 2 + a = 2a $\)$ $\($ $\Rightarrow$ $\alpha$ = 2 $\)$ i.e. $\($ 2x^2 + 2y^2 + 2x - 8y = 0 $\)$ $\($ x^2 + y^2 + x - 4y = 0 $\)$ $\($ rd = $\sqrt{\frac{1}{4} + 4}$ = $\frac{\sqrt{17}}{2}$ $\)$

Question 5

Maths · Conic Sections · Single correct

If the locus of the point, whose distances from the point $(2,1)$ and $(1,3)$ are in the ratio $5:4$, is $$ax^2 + by^2 + cxy + dx + ey + 170 = 0,$$ then the value of $a^2 + 2b + 3c + 4d + e$ is equal to:

  1. 37
  2. 437
  3. -27
  4. 5

Answer: (a)

Solution

Let $P(x, y)$ $$\frac{(x - 2)^2 + (y - 1)^2}{(x - 1)^2 + (y - 3)^2} = \frac{25}{16}$$ $$9x^2 + 9y^2 + 14x - 118y + 170 = 0$$ $$a^2 + 2b + 3c + 4d + e$$ $$= 81 + 18 + 0 + 56 - 118$$ $$= 155 - 118$$ $$= 37$$

Question 6

Maths · Sequences and Series · Single correct

$\lim_{n \to \infty}$ $\frac{(1^2-1)(n-1)+(2^2-2)(n-2)+\cdots+((n-1)^2-(n-1)) \cdot 1}{(1^3+2^3+\cdots+n^3)-(1^2+2^2+\cdots+n^2)}$ is equal to :

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

Answer: (b)

Solution

The given limit is $$\lim_{n \to \infty} \frac{\sum_{r=1}^{n-1} \left( r^2 - r \right) \left( n - r \right)}{\sum_{r=1}^{n} r^3 - \sum_{r=1}^{n} r^2}$$ which simplifies to $$\lim_{n \to \infty} \frac{\sum_{r=1}^{n-1} \left( -r^3 + r^2(n+1) - nr \right)}{\frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}}.$$ Further simplification gives $$\lim_{n \to \infty} \frac{\frac{((n-1)n)^2}{2} + \frac{(n+1)(n-1)n(2n-1)}{6} - \frac{n^2(n-1)}{2}}{\frac{n(n+1)}{2} \left( \frac{n(n+1)}{2} - \frac{2n+1}{3} \right)}.$$ This reduces to $$\lim_{n \to \infty} \frac{n(n-1)}{2} \left( \frac{-n(n-1)}{2} + \frac{(n+1)(2n-1)}{3} - n \right).$$ Finally, we have $$\lim_{n \to \infty} \frac{n(n+1)}{2} \left( \frac{3n^2 + 3n - 4n - 2}{6} \right).$$ Simplifying further, $$\lim_{n \to \infty} \frac{(n-1) \left( -3n^2 + 3n + 2 \left( 2n^2 + n - 1 \right) - 6 \right)}{(n+1) \left( 3n^2 - n - 2 \right)}.$$ This becomes $$\lim_{n \to \infty} \frac{(n-1) \left( n^2 + 5n - 8 \right)}{(n+1) \left( 3n^2 - n - 2 \right)} = \frac{1}{3}.$$

Question 7

Maths · Binomial Theorem · Single correct

Let $0 \leq r \leq n$. If $\binom{n+1}{r+1} : \binom{n}{r} : \binom{n-1}{r-1} = 55 : 35 : 21$, then $2n + 5r$ is equal to:

  1. 50
  2. 62
  3. 55
  4. 60

Answer: (a)

Solution

${}^{\,n+1}C_r$ $\div$ ${}^{\,n}C_r$ $=\dfrac{55}{35}$ $\dfrac{(n+1)!}{r!(n-r+1)!}$ $\times$ $\dfrac{r!(n-r)!}{n!}$ $=\dfrac{11}{7}$ $\dfrac{n+1}{r+1}$ $=\dfrac{11}{7}$ $7n=4+11r$ ${}^{\,n}C_r$ $\div$ ${}^{\,n-1}C_{r-1}$ $=\dfrac{35}{21}$ $\dfrac{n!}{r!(n-r)!}$ $\times$ $\dfrac{(r-1)!(n-r)!}{(n-1)!}$ $=\dfrac53$ $\dfrac nr$ $=\dfrac53$ $3n=5r$ By solving $r=6$ $n=10$ $2n+5r$ $=50$

Question 8

Maths · Sequences and Series · Single correct

A software company sets up $m$ number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of $m$ is equal to:

  1. 150
  2. 180
  3. 160
  4. 125

Answer: (a)

Solution

Given the equation: $$17m = m + (m - 4) + (m - 4 \times 2) + \ldots + (m - 4 \times 24)$$ Simplifying, we have: $$17m = 25m - 4(1 + 2 + \ldots + 24)$$ The sum of the series is calculated as: $$8m = \frac{4 \cdot 24 \cdot 25}{2} = 150$$

Question 9

Maths · Complex Numbers and Quadratic Equations · Single correct

If $z_1, z_2$ are two distinct complex number such that $$\left| \frac{z_1 - 2z_2}{\frac{1}{2} - z_1 z_2} \right| = 2,$$ then

  1. $z_1$ lies on a circle of radius $\frac{1}{2}$ and $z_2$ lies on a circle of radius $1$.
  2. both $z_1$ and $z_2$ lie on the same circle.
  3. either $z_1$ lies on a circle of radius $\frac{1}{2}$ or $z_2$ lies on a circle of radius $1$.
  4. either $z_1$ lies on a circle of radius $1$ or $z_2$ lies on a circle of radius $\frac{1}{2}$.

Answer: (d)

Solution

Given the expression: $$\frac{z_1 - 2z_2}{\frac{1}{2} - z_1 \overline{z_2}} \times \frac{\overline{z_1} - 2\overline{z_2}}{\frac{1}{2} - \overline{z_1} z_2} = 4$$ We have: $$|z_1|^2 2z_1 \overline{z_2} - 2 \overline{z_1} z_2 + 4|z_2|^2$$ $$= 4 \left( \frac{1}{4} - \frac{\overline{z_1} z_2}{2} - \frac{z_1 \overline{z_2}}{2} + |z_1|^2 |z_2|^2 \right)$$ $$z_1 \overline{z_1} + 2z_2 \cdot 2\overline{z_2} - z_1 \overline{z_1} 2z_2 2\overline{z_2} - 1 = 0$$ Therefore: $$(z, \overline{z_1} - 1)(1 - 2z_2 \cdot 2\overline{z_2}) = 0$$ $$(|z_1|^2 - 1) \left( |2z_2|^2 - 1 \right) = 0$$

Question 10

Maths · Applications of Derivatives · Single correct

If the function $f(x) = \left(\frac{1}{x}\right)^{2x}$; $x > 0$ attains the maximum value at $x = \frac{1}{e}$ then:

  1. $e^{\pi} < \pi^{e}$
  2. $e^{\pi} > \pi^{e}$
  3. $(2e)^{\pi} > \pi^{(2e)}$
  4. $e^{2\pi} < (2\pi)^{e}$

Answer: (b)

Solution

Let $y = \left( \frac{1}{x} \right)^{2x}$. Taking the natural logarithm on both sides, we have $\ln y = 2x \ln \left( \frac{1}{x} \right)$. Simplifying, $\ln y = -2x \ln x$. Differentiating both sides with respect to $x$, we get $\frac{1}{y} \frac{dy}{dx} = -2(1 + \ln x)$. For $x > \frac{1}{e}$, $f^n$ is decreasing. Therefore, $e \left( \frac{1}{\pi} \right)^{2\pi}$, which implies $e^\pi > \pi^e$.

Question 11

Maths · Vector Algebra · Single correct

Let $$\vec{a}=6\hat{i}+\hat{j}-\hat{k}\quad\text{and}\quad\vec{b}=\hat{i}+\hat{j}.$$ If $\vec{c}$ is a vector such that $$|\vec{c}|\ge6,\quad \vec{a}\cdot\vec{c}=6|\vec{c}|,\quad |\vec{c}-\vec{a}|=2\sqrt{2},$$ and the angle between $(\vec{a}\times\vec{b})$ and $\vec{c}$ is $60^\circ$, then $$|(\vec{a}\times\vec{b})\times\vec{c}|$$ is equal to:

  1. $\frac{9}{2}$(6 - $\sqrt{6}$)
  2. $\frac{3}{2}$$\sqrt{6}$
  3. $\frac{9}{2}$(6 + $\sqrt{6}$)
  4. $\frac{3}{2}$$\sqrt{3}$

Answer: (c)

Solution

Given $|(\vec{a} \times \vec{b} \times \vec{c})| = |\vec{a} \times \vec{b}||\vec{c}| \frac{\sqrt{3}}{2}$. $|\vec{c} - \vec{a}| = 2\sqrt{2}$ $|c|^2 + |a|^2 - 2 \vec{c} \cdot \vec{a} = 8$ $|z|^2 + 38 - 12|z| = 8$ $|z|^2 - 12|z| + 30 = 0$ $|z| = \frac{12 \pm \sqrt{144 - 120}}{2}$ $= \frac{12 \pm 2\sqrt{6}}{2}$ $|z| = 6 + \sqrt{6}$ $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 1 & -1 \\ 1 & 1 & 0 \end{vmatrix}$ $= \hat{i} - \hat{j} + 5\hat{k}$ $|\vec{a} \times \vec{b}| = \sqrt{27}$ $|(\vec{a} \times \vec{b}) \times z| = \sqrt{27}(6 + \sqrt{6}) \frac{\sqrt{3}}{2}$ $= \frac{9}{2}(6 + \sqrt{6})$

Question 12

Maths · Permutations and Combinations · Single correct

If all the words with or without meaning made using all the letters of the word "NAGPUR" are arranged as in a dictionary, then the word at $315^{th}$ position in this arrangement is :

  1. NRAGUP
  2. NRAPUG
  3. NRAPGU
  4. NRAGPU

Answer: (c)

Solution

NAGPUR A $\rightarrow$ 5! = 120 G $\Rightarrow$ 5! = 120 240 NA $\Rightarrow$ 4! = 24 264 NG $\Rightarrow$ 4! = 24 288 NP $\Rightarrow$ 4! = 24 312 NRAGPU = 1 313 NRAGUP 314 NRAPGU 315

Question 13

Maths · Continuity and Differentiability · Single correct

Suppose for a differentiable function $h$, $h(0) = 0$, $h(1) = 1$ and $h'(0) = h'(1) = 2$. If $g(x) = h(e^x) e^{h(x)}$, then $g'(0)$ is equal to:

  1. 5
  2. 4
  3. 8
  4. 3

Answer: (b)

Solution

Given $g(x) = h(e^x) \cdot e^{h(x)}$. Differentiating, we have: $$g'(x) = h(e^x) \cdot e^{h(x)} \cdot h'(x) + e^{h(x)} h'(e^x) \cdot e^x$$ Evaluating at $x = 0$: $$g'(0) = h(1)e^{h(0)}h'(0) + e^{h(0)}h'(1)$$ Simplifying, we get: $$= 2 + 2 = 4$$

Question 14

Maths · Three Dimensional Geometry · Single correct

Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the image of the point $\mathrm{Q}(3, -3, 1)$ in the line $\frac{x-0}{1} = \frac{y-3}{1} = \frac{z-1}{-1}$ and $\mathrm{R}$ be the point $(2, 5, -1)$. If the area of the triangle $\mathrm{PQR}$ is $\lambda$ and $\lambda^2 = 14K$, then $K$ is equal to:

  1. 36
  2. 81
  3. 72
  4. 18

Answer: (b)

Solution

RQ = $\sqrt{1 + 64 + 4}$ = $\sqrt{69}$ $\newline$ $\overrightarrow{RQ}$ = $\hat{i}$ - 8$\hat{j}$ + 2$\hat{k}$ $\newline$ $\overrightarrow{RS}$ = $\hat{i}$ + $\hat{j}$ - $\hat{k}$ $\newline$ $\cos$ $\theta$ = $\frac{\overrightarrow{RQ} \cdot \overrightarrow{RS}}{\left| \overrightarrow{RQ} \right| \left| \overrightarrow{RS} \right|}$ = $\frac{1 - 8 - 2}{\sqrt{69} \sqrt{3}}$ = $\frac{9}{3\sqrt{23}}$ $\newline$ $\cos$ $\theta$ = $\frac{3}{\sqrt{23}}$ = $\frac{\mathrm{RS}}{\mathrm{RQ}}$ = $\frac{\mathrm{RS}}{\sqrt{69}}$ $\newline$ $\mathrm{RS}$ = 3$\sqrt{3}$ $\newline$ $\sin$ $\theta$ = $\frac{\sqrt{14}}{\sqrt{23}}$ = $\frac{\mathrm{QS}}{\sqrt{69}}$ $\newline$ $\mathrm{QS}$ = $\sqrt{42}$ $\newline$ area = $\frac{1}{2}$ $\cdot$ 2 $\mathrm{QS}$ $\cdot$ $\mathrm{RS}$ = $\sqrt{42}$ $\cdot$ 3$\sqrt{3}$ $\newline$ $\lambda$ = $\sqrt{\frac{9}{14}}$ $\newline$ $\lambda$^2 = 81.14 = 14k $\newline$ k = 81

Question 15

Maths · Properties of Triangles · Single correct

If P(6, 1) be the orthocentre of the triangle whose vertices are A(5, -2), B(8, 3) and C(h, k), then the point C lies on the circle:

  1. $x^2 + y^2 - 61 = 0$
  2. $x^2 + y^2 - 52 = 0$
  3. $x^2 + y^2 - 65 = 0$
  4. $x^2 + y^2 - 74 = 0$

Answer: (c)

Solution

Slope of $AD = 3$. Slope of $BC = -\frac{1}{3}$. Equation of $BC$ is $3y + x - 17 = 0$. Slope of $BE = 1$. Slope of $AC = -1$. Equation of $AC$ is $x + y - 3 = 0$. Point $C$ is $(-4, 7)$.

Question 16

Maths · Relations and Functions · Single correct

Let $f(x) = \frac{1}{7 - \sin 5x}$ be a function defined on $\mathbb{R}$. Then the range of the function $f(x)$ is equal to ;

  1. $\left[ \frac{1}{7}, \frac{1}{6} \right]$
  2. $\left[ \frac{1}{8}, \frac{1}{5} \right]$
  3. $\left[ \frac{1}{7}, \frac{1}{5} \right]$
  4. $\left[ \frac{1}{8}, \frac{1}{6} \right]$

Answer: (d)

Solution

Given $\sin 5x \in [-1, 1]$. Therefore, $-\sin 5x \in [-1, 1]$. Adding 7 to each part, we get $7 - \sin 5x \in [6, 8]$. Taking the reciprocal, we have $$\frac{1}{7 - \sin 5x} \in \left[ \frac{1}{8}, \frac{1}{6} \right].$$

Question 17

Maths · Vector Algebra · Single correct

Let $\vec{a}$ = 2$\hat{i}$ + $\hat{j}$ - $\hat{k}$, $\vec{b}$ = (($\vec{a}$ $\times$ ($\hat{i}$ + $\hat{j}$)) $\times$ $\hat{i}$) $\times$ $\hat{i}$ . Then the square of the projection of $\vec{a}$ on $\vec{b}$ is :

  1. $\frac{1}{3}$
  2. $\frac{2}{3}$
  3. 2
  4. $\frac{1}{5}$

Answer: (c)

Solution

Given $\vec{a} \times (\hat{i} + \hat{j}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 1 & 0 \end{vmatrix}$. This simplifies to $\hat{i} - \hat{j} + \hat{k}$. Then, $(\vec{a} \times (\hat{i} \times \hat{j})) \times \hat{i} = \hat{k} + \hat{j}$. Further, $((\vec{a} \times (\hat{i} \times \hat{j})) \times \hat{i}) \times \hat{i} = \hat{j} - \hat{k}$. The projection of $\vec{a}$ on $\hat{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. This equals $\frac{1 + 1}{\sqrt{2}} = \sqrt{2}$.

Question 18

Maths · Applications of Integrals · Single correct

If the area of the region \[ \left\{ (x,y): \frac{a}{x^2}\le y\le\frac{1}{x},\; 1\le x\le2,\; 0<a<1 \right\} \] is \[ (\log_e2)-\frac{1}{7}, \] then the value of $7a-3$ is equal to:

  1. 0
  2. 2
  3. -1
  4. 1

Answer: (c)

Solution

The area is given by the integral $$\int_1^2 \left( \frac{1}{x} - \frac{a}{x^2} \right) \, dx$$. Evaluating the integral, we have: $$\left[ \ln x + \frac{a}{x} \right]_1^2$$. This simplifies to $$\ln 2 + \frac{a}{2} - a = \log_e 2 - \frac{1}{7}$$. Solving for $a$, we get: $$\frac{-a}{2} = -\frac{1}{7}$$. Therefore, $$a = \frac{2}{7}$$. Multiplying both sides by 7 gives: $$7a = 2$$. Finally, $$7a - 3 = -1$$.

Question 19

Maths · Trigonometric Functions · Single correct

If $\int \dfrac{1}{a^2\sin^2 x + b^2\cos^2 x}\, dx = \dfrac{1}{12}\tan^{-1}(3\tan x) + \text{constant}$, then the maximum value of $a\sin x + b\cos x$, is :

  1. $\sqrt{40}$
  2. $\sqrt{41}$
  3. $\sqrt{39}$
  4. $\sqrt{42}$

Answer: (a)

Solution

Given $$\int \frac{\sec^2 x \, dx}{a^2 \tan^2 x + b^2}$$. Let $$\tan x = t$$, then $$\sec^2 x \, dx = dt$$. The integral becomes $$\int \frac{dt}{a^2 t^2 + b^2}$$. This can be rewritten as $$\frac{1}{a^2} \int \frac{dt}{t^2 + \left( \frac{b}{a} \right)^2}$$. The solution is $$\frac{1}{a^2} \cdot \frac{1}{\frac{b}{a}} \tan^{-1} \left( \frac{t}{\frac{b}{a}} \right) + c$$. Substituting back, we have $$\frac{1}{ab} \tan^{-1} \left( \frac{a}{b} \tan x \right) + c$$. On comparing, $$\frac{a}{b} = 3$$. Given $$ab = 12$$, we find $$a = 6, b = 2$$. The maximum value of $$6 \sin x + 2 \cos x$$ is $$\sqrt{40}$$.

Question 20

Maths · Matrices · Single correct

If $A$ is a square matrix of order 3 such that $\det(A) = 3$ and $\det(\mathrm{adj}(-4 \mathrm{adj}(-3 \mathrm{adj}(3 \mathrm{adj}((2 \, A)^{-1})))) = 2^m 3^n$, then $m + 2n$ is equal to:

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

Answer: (d)

Solution

Given $|A| = 3$. $$|adj(-4 adj(-3 adj(3 adj((2A)^{-1}))))|$$ $$= |-4 adj(-3 adj(3 adj(2A)^{-1}))|^2$$ $$= 4^6 |adj(-3 adj(3 adj(2A)^{-1}))|^2$$ $$= 2^{12} \cdot 3^{12} |3 adj(2A)^{-1}|^8$$ $$= 2^{12} \cdot 3^{12} \cdot 3^{24} |adj(2A)^{-1}|^8$$ $$= 2^{12} \cdot 3^{36} |(2A)^{-1}|^{16}$$ $$= 2^{12} \cdot 3^{36} \frac{1}{|2A|^{16}}$$ $$= 2^{12} \cdot 3^{36} \frac{1}{2^{48} |A|^{16}}$$ $$= 2^{12} \cdot 3^{36} \frac{1}{2^{48} \cdot 3^{16}}$$ $$= \frac{3^{20}}{2^{36}} = 2^{-36} \cdot 3^{20}$$ Let $m = -36$ and $n = 20$. Then $m + 2n = 4$.

Question 21

Maths · Continuity and Differentiability · Numerical

Let [t] denote the greatest integer less than or equal to t. Let $f : [0, \infty) \to \mathbb{R}$ be a function defined by $$f(x) = \left[ \frac{x}{2} + 3 \right] - \left[ \sqrt{x} \right].$$ Let $S$ be the set of all points in the interval $[0, 8]$ at which $f$ is not continuous. Then $$\sum_{a \in S} a$$ is equal to .

Answer: 17

Solution

The function $\left[ \frac{x}{2} + 3 \right]$ is discontinuous at $x = 2, 4, 6, 8$. The function $\sqrt{x}$ is discontinuous at $x = 1, 4$. The function $F(x)$ is discontinuous at $x = 1, 2, 6, 8$. The sum $\sum a = 1 + 2 + 6 + 8 = 17$.

Question 22

Maths · Conic Sections · Numerical

The length of the latus rectum and directrices of a hyperbola with eccentricity $e$ are 9 and $x = \pm \frac{4}{\sqrt{13}}$, respectively. Let the line $y - \sqrt{3}x + \sqrt{3} = 0$ touch this hyperbola at $(x_0, y_0)$. If $m$ is the product of the focal distances of the point $(x_0, y_0)$, then $4e^2 + m$ is equal to

Answer: 61

Solution

Given $\frac{2b^2}{a} = 9$ and $\frac{a}{e} = \pm \frac{4}{\sqrt{3}}$. Equation of tangent $y - \sqrt{3}x + \sqrt{3} = 0$ by equation of tangent. Let slope $= S = \sqrt{3}$. Constant $= -\sqrt{3}$. By condition of tangency $$6 = 6a^2 - 9a$$ $$\Rightarrow a = 2, \ b^2 = 9$$ Equation of Hyperbola is $$\frac{x^2}{4} - \frac{y^2}{9} = 1$$ and for tangent. Point of contact is $(4, 3\sqrt{3}) = (x_0, y_0)$. Now $e = \sqrt{1 + \frac{9}{4}} = \frac{\sqrt{13}}{2}$. Again product of focal distances $$m = (x_0 e + a)(x_0 e - a)$$ $$m + 4e^2 = 20e^2 - a^2$$ $$= 20 \times \frac{13}{4} - 4 = 61$$ (There is a printing mistake in the equation of directrix $x = \pm \frac{4}{\sqrt{3}}$. Corrected equation is $x = \pm \frac{4}{\sqrt{13}}$ for directrix, as eccentricity must be greater than one, so question must be bonus)

Question 23

Maths · Sequences and Series · Numerical

If $S(x) = (1 + x) + 2(1 + x)^2 + 3(1 + x)^3 + \cdots + 60(1 + x)^{60}, x \neq 0$, and $(60)^2 S(60) = a(b)^b + b$, where $a, b \in \mathbb{N}$, then $(a + b)$ equal to

Answer: 3660

Solution

$S(x) = (1+x) + 2(1+x)^2 + 3(1+x)^3 + \ldots + 60(1+x)^{60}$ $(1+x)S = (1+x)^2 + \ldots + 59(1+x)^{60} + 60(1+x)^{61}$ $$-xS = \frac{(1+x)(1+x)^{60} - 1}{x} - 60(1+x)^{61}$$ Put $x = 60$ $$-60S = \frac{61\left((61)^{60} - 1\right)}{60} - 60(61)^{61}$$ On solving $= 3660$

Question 24

Maths · Integrals · Numerical

Let [t] denote the largest integer less than or equal to t. If $$\int_0^3 \left( \left[ x^2 \right] + \left[ \frac{x^2}{2} \right] \right) \, dx = a + b\sqrt{2} - \sqrt{3} - \sqrt{5} + c\sqrt{6} - \sqrt{7},$$ where $a, b, c \in \mathbb{Z}$, then $a + b + c$ is equal to

Answer: 23

Solution

The integral is given by: $$\int_0^3 [x^2] \, dx + \int_0^3 \left[ \frac{x^2}{2} \right] \, dx$$ which simplifies to: $$= \int_0^1 0 \, dx + \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx$$ $$+ \int_{\sqrt{3}}^2 3 \, dx + \int_2^{\sqrt{5}} 4 \, dx + \int_{\sqrt{5}}^{\sqrt{6}} 5 \, dx$$ $$+ \int_{\sqrt{6}}^{\sqrt{7}} 6 \, dx + \int_{\sqrt{7}}^{\sqrt{8}} 7 \, dx + \int_{\sqrt{8}}^3 8 \, dx$$ $$+ \int_0^{\sqrt{2}} 0 \, dx + \int_{\sqrt{2}}^2 1 \, dx$$ $$+ \int_2^{\sqrt{6}} 2 \, dx + \int_{\sqrt{6}}^{\sqrt{8}} 3 \, dx + \int_{\sqrt{8}}^3 4 \, dx$$ This results in: $$= 31 - 6\sqrt{2} - \sqrt{3} - \sqrt{5} - 2\sqrt{6} - \sqrt{7}$$ Given values are: $a = 31$, $b = -6$, $c = -2$. Therefore, $a + b + c = 31 - 6 - 2 = 23$.

Question 25

Maths · Probability · Numerical

From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable X denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of X is $\frac{m}{n}$, where $\mathrm{gcd}(m, n) = 1$, then $n - m$ is equal to

Answer: 71

Solution

a = 1 - $\frac{{^3C_5}}{{^{12}C_5}}$ b = 3 $\cdot$ $\frac{{^9C_4}}{{^{12}C_5}}$ c = 3 $\cdot$ $\frac{{^9C_3}}{{^{12}C_5}}$ d = 1 $\cdot$ $\frac{{^9C_2}}{{^{12}C_5}}$ u = 0 $\cdot$ a + 1 $\cdot$ b + 2 $\cdot$ c + 3 $\cdot$ d = 1.25 $\sigma$^2 = 0 $\cdot$ a + 1 $\cdot$ b + 4 $\cdot$ c + 9 $\cdot$ d - u^2 $\sigma$^2 = $\frac{105}{176}$ Ans. 176 - 105 = 71

Question 26

Maths · Properties of Triangles · Numerical

In a triangle ABC, BC = 7, AC = 8, AB = $\alpha$ $\in$ $\mathbb{N}$ and $\cos$ A = $\frac{2}{3}$. If 49 $\cos$(3C) + 42 = $\frac{m}{n}$, where gcd(m, n) = 1, then m + n is equal to

Answer: 39

Solution

Given $$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$ $$\frac{2}{3} = \frac{8^2 + c^2 - 7^2}{2 \times 8 \times c}$$ $$C = 9$$ $$\cos C = \frac{7^2 + 8^2 - 9^2}{2 \times 7 \times 8} = \frac{2}{7}$$ $$49 \cos 3C + 42$$ $$49 \left( 4 \cos^3 C - 3 \cos C \right) + 42$$ $$49 \left( 4 \left( \frac{2}{7} \right)^3 - 3 \left( \frac{2}{7} \right) \right) + 42$$ $$= \frac{32}{7}$$ $$m + n = 32 + 7 = 39$$

Question 27

Maths · Three Dimensional Geometry · Numerical

If the shortest distance between the lines $\frac{x-\lambda}{3} = \frac{y-2}{-1} = \frac{z-1}{1}$ and $\frac{x+2}{-3} = \frac{y+5}{2} = \frac{z-4}{4}$ is $\frac{44}{\sqrt{30}}$, then the largest possible value of $|\lambda|$ is equal to _______.

Answer: 43

Solution

Given $\vec{a}_1 = \lambda \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{a}_2 = -2 \hat{i} - 5 \hat{j} + 4 \hat{k}$. The vector $\vec{p} = -3 \hat{i} - \hat{j} + \hat{k}$ and $\vec{q} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}$. The equation $(\lambda + 2) \hat{i} + 7 \hat{j} - 3 \hat{k} = \vec{a}_1 - \vec{a}_2$. The cross product $\vec{p} \times \vec{q} = -6 \hat{i} - 15 \hat{j} + 3 \hat{k}$. $$\frac{44}{\sqrt{30}} = \frac{| -6 \lambda - 12 - 105 - 9 |}{\sqrt{(-6)^2 + (-15)^2 + 3^2}}$$ Simplifying, $$\frac{44}{\sqrt{30}} = \frac{|6 \lambda + 126|}{3 \sqrt{30}}$$ Thus, $$132 = |6 \lambda + 126|$$ Solving for $\lambda$, $$\lambda = 1, \lambda = -43$$ Therefore, $|\lambda| = 43$.

Question 28

Maths · Complex Numbers and Quadratic Equations · Numerical

Let $\alpha$, $\beta$ be roots of $x^2 + \sqrt{2}x - 8 = 0$. If $U_n = \alpha^n + \beta^n$, then $\frac{U_{10} + \sqrt{2}U_9}{2U_8}$ is equal to

Answer: 4

Solution

The given expression is: $$\frac{\alpha^{10} + \beta^{10} + \sqrt{2}\left(\alpha^9 + \beta^9\right)}{2\left(\alpha^8 + \beta^8\right)}$$ Simplifying the expression: $$\frac{\alpha^8\left(\alpha^2 + \sqrt{2}\alpha\right) + \beta^8\left(\beta^2 + \sqrt{2}\beta\right)}{2\left(\alpha^8 + \beta^8\right)}$$ Further simplification gives: $$\frac{8\alpha^8 + 8\beta^8}{2\left(\alpha^8 + \beta^8\right)} = 4$$

Question 29

Maths · Determinants · Numerical

If the system of equations $$2x + 7y + \lambda z = 3$$ $$3x + 2y + 5z = 4$$ $$x + \mu y + 32z = -1$$ has infinitely many solutions, then $(\lambda - \mu)$ is equal to

Answer: 38

Solution

Given $D = D_1 = D_2 = D_3 = 0$. $$D_3 = \begin{vmatrix} 2 & 7 & 3 \\ 3 & 2 & 4 \\ 1 & \mu & -1 \end{vmatrix} = 0 \Rightarrow \mu = -39$$ $$D = \begin{vmatrix} 2 & 7 & \lambda \\ 3 & 2 & 5 \\ 1 & -39 & 32 \end{vmatrix} = 0 \Rightarrow \lambda = -1$$ $\lambda - \mu = 38$

Question 30

Maths · Differential Equations · Numerical

If the solution $y(x)$ of the given differential equation $(e^y + 1) \cos x \, dx + e^y \sin x \, dy = 0$ passes through the point $\left( \frac{\pi}{2}, 0 \right)$, then the value of $e^{y\left( \frac{\pi}{6} \right)}$ is equal to

Answer: 3

Solution

$(e^y+1)\cos x\,dx + e^y\sin x\,dy = 0$ $\Rightarrow\ d\!\left((e^y+1)\sin x\right)=0$ $\therefore\ (e^y+1)\sin x=c$ It passes through $\left(\frac{\pi}{6},0\right)$ $\Rightarrow\ c=1$ Now, $x=\frac{\pi}{6}$ $\Rightarrow\ e^y=3$

Physics

Question 31

Physics · Atoms · Single correct

The longest wavelength associated with Paschen series is : (Given $R_H = 1.097 \times 10^7$ SI unit)

  1. 3.646 $\times$ $10^{-6}$m
  2. 1.876 $\times$ $10^{-6}$m
  3. 2.973 $\times$ $10^{-6}$m
  4. 1.094 $\times$ $10^{-6}$m

Answer: (b)

Solution

For longest wavelength in Paschen's series: $$\frac{1}{\lambda} = \mathrm{R} \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$$ For longest $n_1 = 3$ $n_2 = 4$ $$\frac{1}{\lambda} = \mathrm{R} \left[ \frac{1}{(3)^2} - \frac{1}{(4)^2} \right]$$ $$\frac{1}{\lambda} = \mathrm{R} \left[ \frac{1}{9} - \frac{1}{16} \right]$$ $$\frac{1}{\lambda} = \mathrm{R} \left[ \frac{16 - 9}{16 \times 9} \right]$$ $$\Rightarrow \lambda = \frac{16 \times 9}{7 \mathrm{R}} = \frac{16 \times 9}{7 \times 1.097 \times 10^7}$$ $$\lambda = 1.876 \times 10^{-6} \, \mathrm{m}$$

Question 32

Physics · Thermodynamics · Single correct

A total of 48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by $2^\circ \mathrm{C}$. The work done by the gas is: Given, $R = 8.3 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1}$.

  1. 24.9 $\mathrm{J}$
  2. 72.9 $\mathrm{J}$
  3. 48 $\mathrm{J}$
  4. 23.1 $\mathrm{J}$

Answer: (d)

Solution

1st law of thermodynamics $$\Delta Q = \Delta U + W$$ $$\Rightarrow +48 = nC_v \Delta T + W$$ $$\Rightarrow 48 = (1) \left( \frac{3R}{2} \right) (2) + W$$ $$\Rightarrow W = 48 - 3 \times R$$ $$\Rightarrow W = 48 - 3 \times (8.3)$$ $$\Rightarrow W = 23.1 \, Joule$$

Question 33

Physics · Ray Optics and Optical Instruments · Single correct

In finding out refractive index of glass slab the following observations were made through travelling microscope 50 vernier scale division = 49MSD; 20 divisions on main scale in each cm For mark on paper MSR = 8.45 $\mathrm{cm}$, VC = 26 For mark on paper seen through slab MSR = 7.12 $\mathrm{cm}$, VC = 41 For powder particle on the top surface of the glass slab MSR = 4.05 $\mathrm{cm}$, VC = 1 (MSR = Main Scale Reading, VC = Vernier Coincidence) Refractive index of the glass slab is :

  1. 1.52
  2. 1.35
  3. 1.42
  4. 1.24

Answer: (c)

Solution

1 MSD $=\frac{1\,\mathrm{cm}}{20}=0.05\,\mathrm{cm}$ 1 VSD $=\frac{49}{50}\,\mathrm{MSD}=\frac{49}{50}\times0.05\,\mathrm{cm}=0.049\,\mathrm{cm}$ LC $=1\,\mathrm{MSD}-1\,\mathrm{VSD}=0.001\,\mathrm{cm}$ For mark on paper, $L_1=8.45\,\mathrm{cm}+26\times0.001\,\mathrm{cm}=84.76\,\mathrm{mm}$ For mark on paper through slab, $L_2=7.12\,\mathrm{cm}+41\times0.001\,\mathrm{cm}=71.61\,\mathrm{mm}$ For powder particle on top surface, $ZE=4.05\,\mathrm{cm}+1\times0.001\,\mathrm{cm}=40.51\,\mathrm{mm}$ $\therefore$ actual $L_1=84.76-40.51=44.25\,\mathrm{mm}$ actual $L_2=71.61-40.51=31.10\,\mathrm{mm}$ $L_2=\frac{L_1}{\mu}$ $\Rightarrow \mu=\frac{L_1}{L_2}=\frac{44.25}{31.10}=1.42$

Question 34

Physics · Electromagnetic Waves · Single correct

In the given electromagnetic wave $E_y = 600 \sin(\omega t - kx) \mathrm{Vm}^{-1}$, intensity of the associated light beam is (in $\mathrm{W/m}^2$): (Given $\epsilon_0 = 9 \times 10^{-12} \mathrm{C}^2 \mathrm{N}^{-1} \mathrm{m}^{-2}$)

  1. 243
  2. 729
  3. 972
  4. 486

Answer: (d)

Solution

Intensity $$= \frac{1}{2} \varepsilon_0 E_0^2 c$$ $$= \frac{1}{2} \times 9 \times 10^{-12} \times (600)^2 \times 3 \times 10^8$$ $$= \frac{9}{2} \times 36 \times 3 = 486 \, \mathrm{w/m^2}$$

Question 35

Physics · Gravitation · Single correct

Assuming the earth to be a sphere of uniform mass density, a body weighed 300 $\mathrm{N}$ on the surface of earth. How much it would weigh at R/4 depth under surface of earth ?

  1. 75 $\mathrm{N}$
  2. 300 $\mathrm{N}$
  3. 375 $\mathrm{N}$
  4. 225 $\mathrm{N}$

Answer: (d)

Solution

At surface: $mg = 300 \, \mathrm{N}$ $$m = \frac{300}{g_s}$$ At Depth $\frac{R}{4}$: $g_d = g_s \left[ 1 - \frac{d}{R} \right]$ $$g_d = g_s \left[ 1 - \frac{R}{4R} \right]$$ $$g_d = \frac{3}{4} g_s$$ Weight at depth $= m \times g_d$ $$= m \times \frac{3}{4} g_s$$ $$= \frac{3}{4} \times 300$$ $$= 225 \, \mathrm{N}$$

Question 36

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

The acceptor level of a p-type semiconductor is $6\,\mathrm{eV}$. The maximum wavelength of light which can create a hole would be : Given $\mathrm{hc} = 1242\,\mathrm{eVnm}$.

  1. 414 nm
  2. 103.5 nm
  3. 207 nm
  4. 407 nm

Answer: (c)

Solution

Energy = $\frac{hc}{\lambda}$ $\newline$ E = $\frac{1240}{\lambda (\mathrm{nm})}$ $\mathrm{eV}$ $\newline$ 6 = $\frac{1240}{\lambda (\mathrm{nm})}$ $\newline$ $\lambda$ = $\frac{1240}{6}$ = 207 $\mathrm{nm}$

Question 37

Physics · Laws of Motion · Single correct

A car of 800 kg is taking turn on a banked road of radius 300 m and angle of banking $30^\circ$. If coefficient of static friction is 0.2 then the maximum speed with which car can negotiate the turn safely: $\left( g = 10 \, \mathrm{m/s^2}, \sqrt{3} = 1.73 \right)$

  1. 264 $\mathrm{m/s}$
  2. 51.4 $\mathrm{m/s}$
  3. 70.4 $\mathrm{m/s}$
  4. 102.8 $\mathrm{m/s}$

Answer: (b)

Solution

Given $m = 800 \, \mathrm{kg}$, $r = 300 \, \mathrm{m}$, $\theta = 30^\circ$, $\mu_s = 0.2$. The maximum velocity $V_{\max}$ is given by: $$V_{\max} = \sqrt{Rg \left[ \frac{\tan \theta + \mu}{1 - \mu \tan \theta} \right]}$$ Substituting the values: $$= \sqrt{300 \times g \times \left[ \frac{\tan 30^\circ + 0.2}{1 - 0.2 \times \tan 30^\circ} \right]}$$ Calculating further: $$= \sqrt{300 \times 10 \times \left[ \frac{0.57 + 0.2}{1 - 0.2 \times 0.57} \right]}$$ Finally, we find: $$V_{\max} = 51.4 \, \mathrm{m/s}$$

Question 38

Physics · Electric Charges and Fields · Single correct

Two identical conducting spheres $P$ and $S$ with charge $Q$ on each, repel each other with a force $16 \, \mathrm{N}$. A third identical uncharged conducting sphere $R$ is successively brought in contact with the two spheres. The new force of repulsion between $P$ and $S$ is:

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

Answer: (b)

Solution

Given $F_{PS} \propto Q^2$ and $F_{PS} = 16 \, \mathrm{N}$. Now if $P$ and $R$ are brought in contact, then the charges are $\frac{Q}{2}$, $Q$, and $\frac{Q}{2}$ for $P$, $S$, and $R$ respectively. Now if $S$ and $R$ are brought in contact, then the charges become $\frac{Q}{2}$, $\frac{3Q}{4}$, and $\frac{3Q}{4}$ for $P$, $S$, and $R$ respectively. The new force between $P$ and $S$ is: $$F_{PS} \propto \frac{Q}{2} \times \frac{3Q}{4}$$ $$F_{PS} \propto \frac{3Q^2}{8} = \frac{3}{8} \times 16 = 6 \, \mathrm{N}$$

Question 39

Physics · Electromagnetic Induction · Single correct

In a coil, the current changes from $-2 \, \mathrm{A}$ to $+2 \, \mathrm{A}$ in $0.2 \, \mathrm{s}$ and induces an emf of $0.1 \, \mathrm{V}$. The self inductance of the coil is:

  1. 4$\mathrm{mH}$
  2. 1$\mathrm{mH}$
  3. 5$\mathrm{mH}$
  4. 2.5$\mathrm{mH}$

Answer: (c)

Solution

(Emf)_{induced} = -L $\frac{di}{dt}$ In magnitude form, $$|Emf_{ind}| = \left| (-)L \frac{di}{dt} \right|$$ $$\Rightarrow 0.1 = \frac{(L)[+2 - (-2)]}{0.2}$$ $$\Rightarrow L = \frac{0.1 \times 0.2}{4} = 5 \, mH$$

Question 40

Physics · Ray Optics and Optical Instruments · Single correct

For the thin convex lens, the radii of curvature are at 15 cm and 30 cm respectively. The focal length the lens is 20 cm. The refractive index of the material is :

  1. 1.2
  2. 1.8
  3. 1.5
  4. 1.4

Answer: (c)

Solution

Given $\($ $\frac{1}{f}$ = $\left$( $\frac{\mu_{lens}}{\mu_{air}}$ - 1 $\right$) $\left$( $\frac{1}{R_1}$ - $\frac{1}{R_2}$ $\right$) $\)$. Therefore, $\($ $\frac{1}{+20}$ = $\left$( $\frac{\mu}{1}$ - 1 $\right$) $\left$( $\frac{1}{+15}$ - $\frac{1}{-30}$ $\right$) $\)$. This simplifies to $\($ $\frac{1}{20}$ = ($\mu$ - 1) $\left$( $\frac{3}{30}$ $\right$) $\)$. Solving for $\($ $\mu$ $\)$, we have $\($ $\mu$ - 1 = $\frac{1}{2}$ $\)$. Thus, $\($ $\mu$ = 1 + $\frac{1}{2}$ = $\frac{3}{2}$ = 1.5 $\)$.

Question 41

Physics · Kinetic Theory · Single correct

Energy of 10 non rigid diatomic molecules at temperature $T$ is :

  1. $70 \, K_B T$
  2. $35 \, K_B T$
  3. $\frac{7}{2} RT$
  4. $35 RT$

Answer: (b)

Solution

Degree of freedom $(f) = 5 + 2(3N - 5)$ $$f = 5 + 2(3 \times 2 - 1) = 7$$ Energy of one molecule $= \frac{f}{2} K_B T$ Energy of 10 molecules $$= 10 \left( \frac{f}{2} K_B T \right) = 10 \left( \frac{7}{2} K_B T \right) = 35 \, K_B T$$

Question 42

Physics · Laws of Motion · Single correct

A body of weight 200 $\mathrm{N}$ is suspended from a tree branch through a chain of mass 10 $\mathrm{kg}$. The branch pulls the chain by a force equal to (if $g = 10 \, \mathrm{m/s^2}$):

  1. 100 $\mathrm{N}$
  2. 200 $\mathrm{N}$
  3. 300 $\mathrm{N}$
  4. 150 $\mathrm{N}$

Answer: (c)

Solution

Chain block system is in equilibrium so $$T = 200 + 100 = 300 \, \mathrm{N}.$$

Question 43

Physics · Dual Nature of Radiation and Matter · Single correct

When UV light of wavelength 300 nm is incident on the metal surface having work function 2.13 eV, electron emission takes place. The stopping potential is: (Given $hc = 1240 \, \mathrm{eV} \, \mathrm{nm}$ )

  1. 1.5 V
  2. 4.1 V
  3. 2 V
  4. 4 V

Answer: (c)

Solution

$\dfrac{hc}{\lambda} - \phi = e \cdot V_s$ $\Rightarrow \dfrac{1240}{300}\,\mathrm{eV} - 2.13\,\mathrm{eV} = eV_s$ $\Rightarrow 4.13\,\mathrm{eV} - 2.13\,\mathrm{eV} = eV_s$ $\Rightarrow$ So, $V_s = 2\,\mathrm{volt}$

Question 44

Physics · Current Electricity · Single correct

The number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V is : ( Given $e = 1.6 \times 10^{-19} \mathrm{C}$)

  1. $6.25 \times 10^{17}$
  2. $1.25 \times 10^{19}$
  3. $6.25 \times 10^{18}$
  4. $31.25 \times 10^{17}$

Answer: (d)

Solution

Power $(P)=V.I$ $\Rightarrow\ 110=(220)(I)$ $\Rightarrow\ I=0.5\ A$ Now, $I=\dfrac{ne}{t}$ $\Rightarrow\ 0.5$ $=\left(\dfrac{n}{t}\right) \left(1.6\times10^{-19}\right)$ $\Rightarrow\ \dfrac{n}{t}$ $=\dfrac{0.5}{1.6\times10^{-19}}$ $\Rightarrow\ \dfrac{n}{t}$ $=31.25\times10^{17}$

Question 45

Physics · System of Particles and Rotational Motion · Single correct

When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be:

  1. 6$\%$
  2. 600$\%$
  3. 60$\%$
  4. 500$\%$

Answer: (d)

Solution

Kinetic energy $(K) = \frac{P^2}{2m}$. Therefore, $P = \sqrt{2mK}$. If $K_f = 36K_i$, then $P_f = 6P_i$. The percentage increase in momentum is given by: $$\frac{P_f - P_i}{P_i} \times 100\%$$ Substituting the values, we get: $$\frac{6P_i - P_i}{P_i} \times 100\%$$ This simplifies to: $$= 500\%$$

Question 46

Physics · Mechanical Properties of Fluids · Single correct

Pressure inside a soap bubble is greater than the pressure outside by an amount: (given: $R =$ Radius of bubble $S =$ Surface tension of bubble)

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

Answer: (d)

Solution

There are two liquid-air surfaces in bubble so $$\Delta P = 2 \left( \frac{2S}{R} \right) = \frac{4S}{R}$$

Question 47

Physics · Moving Charges and Magnetism · Single correct

Match List-I with List-II : Choose the correct answer from the options given below :

  1. (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
  2. (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
  3. (A)-(III), (B)-(IV), (C)-( I), (D)-(II)
  4. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

Answer: (d)

Solution

(A) Graph between Magnetic susceptibility and magnetising field is a horizontal line. (B) Magnetic field due to a current carrying wire for $x a$ is given by: $$B = \frac{\mu_0 i}{2 \pi a}$$ The graph is a decreasing curve. (D) Magnetic field inside solenoid varies as a constant value followed by a drop.

Question 48

Physics · Physical World, Units and Measurements · Single correct

In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its $4^{th}$ division coincides exactly with a certain division on main scale. If 50 vernier scale divisions equal to 49 main scale divisions and zero error in the instrument is $0.04 \, \mathrm{mm}$ then how many main scale divisions are there in 1 cm?

  1. 10
  2. 5
  3. 20
  4. 40

Answer: (c)

Solution

50VSD = 49MSD 1VSD = $\frac{49}{50}$ MSD LC = 1MSD - 1VSD LC = 1MSD $\cdot$ $\frac{49}{50}$ MSD = $\frac{MSD}{50}$ $\ldots$ $\ldots$ $\ldots$ $\ldots$ $\ldots$ $\ldots$ (1) Now, $\;$ | zero error | = | 0 - 4(L.C) | 0.04 = | - 4(L.C) | $\Rightarrow$ L.C = 0.01 mm 0.04 = | - 4(L.C) | $\Rightarrow$ L.C $\ldots$ $\ldots$ $\ldots$ (2) = 0.01 mm Using (1) and (2) $\frac{MSD}{50}$ = 0.01 MSD = 0.5 mm MSD = $\frac{1 cm}{N}$ N = $\frac{1 cm}{0.5 mm}$ = $\frac{100 mm}{5 mm}$ N = 20

Question 49

Physics · Physical World, Units and Measurements · Single correct

Given below are two statements: Statement (I): Dimensions of specific heat is $[L^2 \, T^{-2} \, K^{-1}]$. Statement (II): Dimensions of gas constant is $[M \, L^2 \, T^{-1} \, K^{-1}]$. 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 correct
  2. Statement (I) is correct but statement (II) is incorrect
  3. Both statement (I) and statement (II) are incorrect
  4. Statement (I) is incorrect but statement (II) is correct

Answer: (b)

Solution

Given $\Delta Q = mS \Delta T$, we have $s = \frac{\Delta Q}{m \Delta T}$. The dimensional formula for $[s]$ is $$[s] = \frac{[ML^2 \ T^{-2}]}{[MK]}$$ which simplifies to $$[s] = [L^2 \ T^{-2} \ K^{-1}].$$ Statement-I is correct. For the ideal gas equation $PV = nRT$, we have $R = \frac{PV}{nT}$. The dimensional formula for $[R]$ is $$[R] = \frac{[ML^{-1} \ T^{-2}] [L^3]}{[mol] [K]}$$ which simplifies to $$[R] = [ML^2 \ T^{-2} \ mol^{-1} \ K^{-1}].$$ Statement-II is incorrect.

Question 50

Physics · Motion in a Straight Line · Single correct

A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in $t_1$. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in $t_2$. Time required to reach the ground, if it is dropped from the top of the tower, is:

  1. $\sqrt{t_1 t_2}$
  2. $\sqrt{t_1 - t_2}$
  3. $\sqrt{\frac{t_1}{t_2}}$
  4. $\sqrt{t_1 + t_2}$

Answer: (a)

Solution

Question 51

Physics · Atoms · Fill in the blank

In Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at 10.2 $\mathrm{V}$. The wavelength of light emitted by hydrogen atom when excited to the first excitation level is ________ nm. (Given $hc = 1245 \, \mathrm{eVnm}$, $e = 1.6 \times 10^{-19} \, \mathrm{C}$).

Answer: 122

Solution

Given $10.2 \, \mathrm{eV} = \frac{hc}{\lambda}$. $$\lambda = \frac{1245 \, \mathrm{eV} \cdot \mathrm{nm}}{10.2 \, \mathrm{eV}} = 122.06 \, \mathrm{nm}$$

Question 52

Physics · Alternating Current · Numerical

For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is $25\,\mathrm{nF}$. If resistance of $200\,\Omega$ and $100\,\mathrm{mH}$ inductor is being used in the given circuit. The frequency of ac source is ______ $\times 10^3 \, \mathrm{Hz}$. (given $\pi^2 = 10$)

Answer: 10

Solution

Given the formula for the resonant frequency $f_0$: $$f_0 = \frac{1}{2\pi \sqrt{L \times C}}$$ Substitute the given values: $$f_0 = \frac{1}{2\pi \sqrt{100 \times 10^{-3} \times 25 \times 10^{-9}}}$$ Simplify the expression: $$= \frac{1}{2\pi \sqrt{25 \times 10^{-10}}}$$ Further simplification gives: $$= \frac{1}{2\pi \times 5 \times 10^5 \times \sqrt{10} \, \mathrm{Hz}}$$ Calculate the final value: $$= \frac{100}{10} \times 10^3 \, \mathrm{Hz}$$ Thus, $$f_0 = 10 \times 10^3 \, \mathrm{Hz}$$

Question 53

Physics · Motion in a Straight Line · Numerical

A particle moves in a straight line so that its displacement $x$ at any time $t$ is given by $x^2 = 1 + t^2$. Its acceleration at any time $t$ is $x^{-n}$ where $n = \ldots$

Answer: 3

Solution

Question 54

Physics · System of Particles and Rotational Motion · Numerical

Three balls of masses 2 kg, 4 kg and 6 kg respectively are arranged at centre of the edges of an equilateral triangle of side 2 m. The moment of inertia of the system about an axis through the centroid and perpendicular to the plane of triangle, will be _____ $kgm^2$.

Answer: 4

Solution

Moment of inertia about C and perpendicular to the plane is: $$I = r^2 [2 + 4 + 6]$$ $$= \frac{1}{3} \times 12$$ $$I = 4 \, \mathrm{kg} - \mathrm{m}^2$$

Question 55

Physics · Moving Charges and Magnetism · Numerical

A coil having 100 turns, area of $5 \times 10^{-3} \, \mathrm{m}^2$, carrying current of $1 \, \mathrm{mA}$ is placed in uniform magnetic field of $0.20 \, \mathrm{T}$ such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through $90^\circ$ is _______ $\mu \mathrm{J}$.

Answer: 100

Solution

Given $W = \Delta U = U_f - U_i$. $$W = (-\vec{\mu} \cdot \vec{B})_f - (-\vec{\mu} \cdot \vec{B})_i$$ $$= 0 + (\vec{\mu} \cdot \vec{B})_i$$ $$= (100 \times 5 \times 10^{-3} \times 1 \times 10^{-3}) \times 0.2 \, \mathrm{J}$$ $$= 1 \times 10^{-4} \, \mathrm{J} = 100 \, \mu \mathrm{J}$$

Question 56

Physics · Current Electricity · Numerical

In the given figure an ammeter A consists of a 240 $\Omega$ coil connected in parallel to a 10 $\Omega$ shunt. The reading of the ammeter is ____ mA

Answer: 160

Solution

Req = 140.4 + $\frac{240 \times 10}{240 + 10}$ = 140.4 + $\frac{2400}{250}$ Req. = 150 $\Omega$ $\therefore$ Current in ammeter = $\frac{24}{150}$ = 160 mA

Question 57

Physics · Mechanical Properties of Solids · Numerical

A wire of cross sectional area $A$, modulus of elasticity $2 \times 10^{11} \, \mathrm{Nm}^{-2}$ and length $2 \, \mathrm{m}$ is stretched between two vertical rigid supports. When a mass of $2 \, \mathrm{kg}$ is suspended at the middle it sags lower from its original position making angle $\theta = \frac{1}{100}$ radian on the points of support. The value of $A$ is $\times 10^{-4} \, \mathrm{m}^2$ (consider $x \ll L$).(given: g= 10 m/s^2)

Answer: 1

Solution

In vertical direction $$2T \sin \theta = 20$$ using small angle approximation $\sin \theta = \theta$ $$\theta = \frac{1}{100}$$ $$\therefore T = \frac{10}{\theta}$$ $$T = 1000 \, \mathrm{N}$$ Change in length $\Delta L = 2 \sqrt{x^2 + L^2} - 2L$ $$= 2L \left[ 1 + \frac{x^2}{2L^2} - 1 \right]$$ $$\Delta L = \frac{x^2}{L}$$ $$\therefore Modulus of elasticity = \frac{stress}{strain}$$ $$2 \times 10^{11} = \frac{10^3}{A \times \frac{x^2}{L}} \times 2L$$ $$\therefore A = 1 \times 10^{-4} \, \mathrm{m^2}$$

Question 58

Physics · Wave Optics · Numerical

Two coherent monochromatic light beams of intensities $I$ and $4I$ are superimposed. The difference between maximum and minimum possible intensities in the resulting beam is $xI$. The value of $x$ is ___.

Answer: 8

Solution

Given $$I_{max} = (\sqrt{I} + \sqrt{4I})^2 = 9I$$ $$I_{min} = (\sqrt{4I} - \sqrt{I})^2 = I$$ Therefore, $$I_{max} - I_{min} = 8I$$

Question 59

Physics · Waves · Numerical

Two open organ pipes of lengths 60 cm and 90 cm resonate at $6^{th}$ and $5^{th}$ harmonics respectively. The difference of frequencies for the given modes is _______ Hz. (Velocity of sound in air = $333 \, \mathrm{m/s}$)

Answer: 740

Solution

The difference in frequency in open organ pipe is given by: $$f = \frac{nv}{2L}$$ The change in frequency is: $$\Delta f = \frac{6v}{2 \times 0.6} - \frac{5v}{2 \times 0.9}$$ Given $v = 333 \, \mathrm{m/s}$, we find: $$\Delta f = 740 \, \mathrm{Hz}$$

Question 60

Physics · Electrostatic Potential and Capacitance · Numerical

A capacitor of $10 \, \mu \mathrm{F}$ capacitance whose plates are separated by $10 \, \mathrm{mm}$ through air and each plate has area $4 \, \mathrm{cm}^2$ is now filled equally with two dielectric media of $K_1 = 2$, $K_2 = 3$ respectively as shown in figure. If new force between the plates is $8 \, \mathrm{N}$. The supply voltage is ________ $\times 10^{-4} \, \mathrm{V}$.

Answer: 93

Solution

Given the circuit, the equivalent capacitance is $C_{eq} = C_1 + C_2$. For $C_1$, we have: $$C_1 = \frac{2 \varepsilon_0 \, A}{2 \times d} = 10 \mu \mathrm{F}$$ For $C_2$, we have: $$C_2 = \frac{3 \varepsilon_0 \, A}{2 \, d} = 15 \mu \mathrm{F}$$ Thus, the equivalent capacitance is: $$C_{eq} = 25 \mu \mathrm{F}$$ Now the charge on $C_1$ is $10 \, V \mu \mathrm{C}$ and on $C_2$ is $1.5 \, V \mu \mathrm{C}$. Now, the force between the plates is given by: $$F = \frac{Q^2}{2 A \varepsilon_0}$$ Calculating for $100 \, V^2 \times 10^{-12}$ and $225 \, V^2 \times 10^{-12}$: $$\frac{100 \, V^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \varepsilon_0} + \frac{225 \, V^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \times \varepsilon_0} = 8$$ Thus: $$325 \, V^2 = 8 \times 4 \times 10^{-4} \times 8.85$$ Solving for $V^2$: $$V^2 = \frac{32 \times 8.85 \times 10^{-4}}{325}$$ Therefore: $$\therefore \, V = \sqrt{\frac{283.2 \times 10^{-4}}{325}}$$ Finally, we find: $$V = 0.93 \times 10^{-2}$$

Chemistry

Question 61

Chemistry · Haloalkanes and Haloarenes · Single correct

The correct arrangement for decreasing order of electrophilic substitution for above compounds is :

  1. (III) > (I) > (II) > (IV)
  2. (IV) > (I) > (II) > (III)
  3. (III) > (IV) > (II) > (I)
  4. (II) > (IV) > (III) > (I)

Answer: (a)

Solution

The order of stability based on the substituents is as follows: $$OMe (+M) > CH_3 (+H/+I) > CF_3 (-I)$$.

Question 62

Chemistry · Solutions · Single correct

Molality ( m ) of 3M aqueous solution of NaCl is : (Given : Density of solution = 1.25 $\mathrm{g \, mL^{-1}}$, Molar mass in gmol^{-1} : Na = 23, Cl = 35.5)

  1. 1.9 m
  2. 3.85 m
  3. 2.79 m
  4. 2.90 m

Answer: (c)

Solution

3 moles are present in 1 litre solution $$molality = \frac{3 \times 1000}{1.25 \times 1000 - [3 \times 58.5]} = 2.79 \, \mathrm{m}$$

Question 63

Chemistry · Biomolecules · Single correct

The incorrect statements regarding enzymes are : (A) Enzymes are biocatalysts. (B) Enzymes are non-specific and can catalyse different kinds of reactions. $(C)$ Most Enzymes are globular proteins. (D) Enzyme - oxidase catalyses the hydrolysis of maltose into glucose.

  1. (B), $(C)$ and (D)
  2. (B) and (D)
  3. (A), (B) and $(C)$
  4. (B) and $(C)$

Answer: (b)

Solution

Direct NCERT Based

Question 64

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Consider the above chemical reaction. Product "A" is:

Answer: (a)

Solution

The reaction begins with the given alkyl chloride. In the presence of $\mathrm{NaOH/H_2O}$, an $\mathrm{S_N1}$ reaction occurs, leading to the formation of a carbocation. A 1,2-hydride shift takes place, resulting in a more stable carbocation. Finally, the hydroxide ion attacks, forming the major product, which is an alcohol.

Question 65

Chemistry · Analytical Chemistry · Single correct

During the detection of acidic radical present in a salt, a student gets a pale yellow precipitate soluble with difficulty in $\mathrm{NH_4OH}$ solution when sodium carbonate extract was first acidified with dil. $\mathrm{HNO_3}$ and then $\mathrm{AgNO_3}$ solution was added. This indicates presence of:

  1. $\mathrm{Br^-}$
  2. $\mathrm{I^-}$
  3. $\mathrm{Cl^-}$
  4. $\mathrm{CO_3^{2-}}$

Answer: (a)

Solution

Ag^+ + I^- $\rightarrow$ $\mathrm{AgI}$ Yellow ppt. Ag^+ + $\mathrm{Cl}$^- $\rightarrow$ $\mathrm{AgCl}$ White ppt Ag^+ + $\mathrm{Br}$^- $\rightarrow$ $\mathrm{AgBr}$ Pale yellow ppt

Question 66

Chemistry · Electrochemistry · Single correct

How can an electrochemical cell be converted into an electrolytic cell?

  1. Applying an external opposite potential lower than $E^0$ cell.
  2. Reversing the flow of ions in salt bridge.
  3. Applying an external opposite potential greater than $E^0$ cell.
  4. Exchanging the electrodes at anode and cathode.

Answer: (c)

Solution

Applied external potential should be greater than $E_{cell}^0$ in opposite direction.

Question 67

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

Arrange the following elements in the increasing order of number of unpaired electrons in it. (A) Sc (B) Cr ( C ) V (D) Ti (E) Mn Choose the correct answer from the options given below :

  1. $< (D) < (C) < (B) < (E)$
  2. ( C ) $< (E) < (B) < (A) < (D)$
  3. $< (D) < (C) < (E) < (B)$
  4. $< (C) < (D) < (E) < (A)$

Answer: (c)

Solution

Unpaired electron Sc : $[\mathrm{Ar}]\,4s^2\,3d^1$ \quad $1$ Cr : $[\mathrm{Ar}]\,4s^1\,3d^5$ \quad $6$ V : $[\mathrm{Ar}]\,4s^2\,3d^3$ \quad $3$ Ti : $[\mathrm{Ar}]\,4s^2\,3d^2$ \quad $2$ Mn : $[\mathrm{Ar}]\,4s^2\,3d^5$ \quad $5$

Question 68

Chemistry · The s-Block Elements · Single correct

Match List-I with List-II. \begin{tabular}{|c|c|c|c|} \hline \multicolumn{2}{|c|}{\textbf{List-I}} & \multicolumn{2}{c|}{\textbf{List-II}} \\ \multicolumn{2}{|c|}{\textbf{Reaction}} & \multicolumn{2}{c|}{\textbf{Type of redox reaction}} \\ \hline (A) & Li & (I) & 589.2 \\ \hline (B) & Na & (II) & 455.5 \\ \hline (C) & Rb & (III) & 670.8 \\ \hline (D) & Cs & (IV) & 780.0 \\ \hline \end{tabular} Choose the correct answer from the options given below:

  1. (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
  2. (A)-(I), (B)-(IV), (C)-(III), (D)-(II)
  3. (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
  4. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

Answer: (d)

Solution

Fact Based

Question 69

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

The major products formed: A and B respectively are:

Answer: (a)

Solution

The reaction sequence involves two steps. In the first step, anisole undergoes nitration using $\mathrm{HNO_3}$ and $\mathrm{H_2SO_4}$ to form a nitro compound (A). In the second step, compound (A) is treated with excess $\mathrm{Br_2}$ in the presence of iron (Fe) to form compound (B) with bromine substitutions.

Question 70

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

The incorrect statement regarding the geometrical isomers of 2-butene is :

  1. cis-2-butene and trans-2-butene are not interconvertible at room temperature.
  2. cis-2-butene and trans-2-butene are stereoisomers.
  3. cis-2-butene has less dipole moment than trans-2-butene.
  4. trans-2-butene is more stable than cis-2-butene.

Answer: (c)

Solution

The image shows two isomers of but-2-ene. The first isomer is Cis-but-2-ene, which is polar. The second isomer is Trans-but-2-ene, which is non-polar.

Question 71

Chemistry · Co-ordination Compounds · Single correct

Given below are two statements : Statement I : $\mathrm{PF}_5$ and $\mathrm{BrF}_5$ both exhibit $sp^3d$ hybridisation. Statement II : Both $\mathrm{SF}_6$ and $[\mathrm{Co(NH}_3)_6]^{3+}$ exhibit $sp^3d^2$ hybridisation. In the light of the above statements, choose the correct answer from the options given below :

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

Answer: (d)

Solution

PF_5 has hybridisation $sp^3d$. BrF_5 has hybridisation $sp^3d^2$. SF_6 has hybridisation $sp^3d^2$. [$\mathrm{Co(NH_3)_6}$]^{+3}$ has hybridisation $d^2sp^3$. Both Statement (1) and (2) are false.

Question 72

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

The number of ions from the following that are expected to behave as oxidising agent is : $\mathrm{Sn}^{4+}, \mathrm{Sn}^{2+}, \mathrm{Pb}^{2+}, \mathrm{Tl}^{3+}, \mathrm{Pb}^{4+}, \mathrm{Tl}^{+}$

  1. 3
  2. 2
  3. 1
  4. 4

Answer: (b)

Solution

Due to inert pair effect; $\mathrm{Tl^{+3}}$ and $\mathrm{Pb^{+4}}$ can behave as oxidising agents.

Question 73

Chemistry · Amines · Single correct

Identify the product (A) in the following reaction.

Answer: (d)

Solution

The given reaction sequence involves two steps. First, aniline $\mathrm{(NH_2)}$ is converted to a diazonium salt using $\mathrm{NaNO_2}$ and $\mathrm{HCl}$. This diazonium salt is then reacted with $\mathrm{Cu_2Cl_2}$ to form chlorobenzene. Second, chlorobenzene is treated with $\mathrm{NaOH}$ at $623 \, \mathrm{K}$ and $300 \, \mathrm{atm}$ to form phenol. Thus, the final product (A) is phenol.

Question 74

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

The correct statement among the following, for a "chromatography" purification method is :

  1. Organic compounds run faster than solvent in the thin layer chromatographic plate.
  2. $R_f$ is an integral value.
  3. $R_f$ of a polar compound is smaller than that of a non-polar compound.
  4. Non-polar compounds are retained at top and polar compounds come down in column chromatography.

Answer: (c)

Solution

Non polar compounds are having higher value of $R_f$ than polar compound.

Question 75

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

Evaluate the following statements related to group 14 elements for their correctness. (A) Covalent radius decreases down the group from C to Pb in a regular manner. (B) Electronegativity decreases from C to Pb down the group gradually. (C) Maximum covalence of C is 4 whereas other elements can expand their covalence due to presence of d orbitals. (D) Heavier elements do not form $p\pi - p\pi$ bonds. (E) Carbon can exhibit negative oxidation states. Choose the correct answer from the options given below :

  1. (A), (B) and (C) Only
  2. (C), (D) and (E) Only
  3. (C) and (D) Only
  4. (A) and (B) Only

Answer: (b)

Solution

(A) Down the group; radius increases (B) EN does not decrease gradually from C to Pb. (C) Correct. (D) Correct. (E) Range of oxidation state of carbon; -4 to +4

Question 76

Chemistry · Redox Reactions · Single correct

Match List - I with List - II. \begin{tabular}{|l|l|} \hline \textbf{List - I} & \textbf{List - II} \\ Reaction & Type of redox reaction \\ \hline (A) $\mathrm{N_2(g) + O_2(g)}$ & (I) Decomposition \\ $\mathrm{\rightarrow 2NO_{(g)}}$ & \\ \hline (B) $\mathrm{2Pb(NO_3)_2(s)}$ & (II) Displacement \\ $\mathrm{\rightarrow 2PbO_{(s)} + 4NO_2(g) + O_2(g)}$ & \\ \hline (C) $\mathrm{2Na_{(s)} + 2H_2O_{(l)}}$ & (III) Disproportionation \\ $\mathrm{\rightarrow 2NaOH_{(aq.)} + H_2(g)}$ & \\ \hline (D) $\mathrm{2NO_2(g) + 2OH^-_{(aq.)}}$ & (IV) Combination \\ $\mathrm{\rightarrow NO_2^-{}_{(aq.)} + NO_3^-{}_{(aq.)} + H_2O_{(l)}}$ & \\ \hline \end{tabular} Choose the correct answer from the options given below:

  1. (A)-(II), (B)-(III), ($C$)-(IV), (D)-(I)
  2. (A)-(III), (B)-(II), ($C$)-(I), (D)-(IV)
  3. (A)-(IV), (B)-(I), ($C$)-(II), (D)-(III)
  4. (A)-(I), (B)-(II), ($C$)-(III), (D)-(IV)

Answer: (c)

Solution

A $\rightarrow$ (IV). B $\rightarrow$ (I). C $\rightarrow$ (II). D $\rightarrow$ (III).

Question 77

Chemistry · Alcohols, Phenols and Ethers · Single correct

Consider the given reaction, identify the major product $P$. $$CH_3 - COOH \xrightarrow{(i) LiAlH_4 \ (ii) PCC \ (iii) HCN/\overline{OH} \ (iv) H_2O/\overline{OH}, \Delta} ''P''$$

  1. CH_3 - CH_2 - CH_2 - OH

Answer: (c)

Solution

The reaction sequence is as follows: 1. The carboxylic acid $\mathrm{CH_3COOH}$ is reduced by $\mathrm{LiAlH_4}$ to form the primary alcohol $\mathrm{CH_3CH_2OH}$. 2. The primary alcohol is oxidized by PCC to form the aldehyde $\mathrm{CH_3CHO}$. 3. The aldehyde undergoes cyanohydrin formation with $\mathrm{HCN/OH^-}$ to form $\mathrm{CH_3CH(OH)CN}$. 4. The cyanohydrin is hydrolyzed in the presence of $\mathrm{H_2O/OH^-}$ and heat $\Delta$ to form the hydroxy acid $\mathrm{CH_3CH(OH)COOH}$.

Question 78

Chemistry · Co-ordination Compounds · Single correct

The correct IUPAC name of $[PtBr_2(PMe_3)_2]$ is:

  1. dibromodi(trimethylphosphine)platinum(II)
  2. bis(trimethylphosphine)dibromoplatinum(II)
  3. dibromobis(trimethylphosphine)platinum(II)
  4. bis[bromo(trimethylphosphine)]platinum(II)

Answer: (c)

Solution

Dibromo bis(trimethylphosphine) platinum (II)

Question 79

Chemistry · Redox Reactions · Single correct

Match List - I with List - II. \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{List - I (Reaction)} & \multicolumn{2}{c|}{List - II (Type of redox reaction)} \\ \hline (A) & $\mathrm{TiCl_4}$ & (I) & $e^2,\ t_2^0$ \\ \hline (B) & $[\mathrm{FeO_4}]^{2-}$ & (II) & $e^4,\ t_2^3$ \\ \hline (C) & $[\mathrm{FeCl_4}]^{-}$ & (III) & $e^0,\ t_2^0$ \\ \hline (D) & $[\mathrm{CoCl_4}]^{2-}$ & (IV) & $e^2,\ t_2^3$ \\ \hline \end{tabular} Choose the correct answer from the options given below:

  1. (1) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
  2. (2) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
  3. (3) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  4. (4) (A)-(I), (B)-(III), (C)-(IV), (D)-(II)

Answer: (c)

Solution

Question 80

Chemistry · Equilibrium · Single correct

The ratio $\frac{K_P}{K_C}$ for the reaction : CO(g) + ½ O₂(g) ⇌ CO₂(g) is :

  1. $\frac{1}{\sqrt{RT}}$
  2. $(RT)^{1/2}$
  3. $RT$
  4. $1$

Answer: (a)

Solution

The reaction is given by: $$\mathrm{CO(g) + \frac{1}{2}O_2(g) \rightleftharpoons CO_2(g)}$$ The change in moles of gas, $\Delta n_g$, is calculated as: $$\Delta n_g = 1 - \left(1 + \frac{1}{2}\right) = -\frac{1}{2}$$ The relationship between $K_p$ and $K_C$ is given by: $$\frac{K_p}{K_C} = (RT)^{\Delta n_g} = \frac{1}{\sqrt{RT}}$$

Question 81

Chemistry · Amines · Numerical

An amine (X) is prepared by ammonolysis of benzyl chloride. On adding p-toluenesulphonyl chloride to it the solution remains clear. Molar mass of the amine (X) formed is ______ gmol$^{-1}$. (Given molar mass in gmol$^{-1}$ C : 12, H : 1, O : 16, N : 14)

Answer: 287

Solution

The reaction involves benzyl chloride reacting with ammonia to form a tertiary amine. The product is a tertiary amine with the structure PhCH_2-N(CH_2Ph)_2. The molar mass of the tertiary amine (X) is given as 287 $\mathrm{g \, mol^{-1}}$.

Question 82

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Numerical

Consider the following reactions \[ \mathrm{NiS + HNO_3 + HCl \rightarrow A + NO + S + H_2O} \] The number of protons that do not involve in hydrogen bonding in the product $B$ is \(\underline{\hspace{1cm}}\).

Solution

Question 83

Chemistry · Solutions · Numerical

When $x \times 10^{-2}$ $\mathrm{mL}$ $\mathrm{methanol}$ ($\mathrm{molar \ mass}$ = 32 $\mathrm{g}$; $\mathrm{density}$ = 0.792 $\mathrm{g/cm^3}$) $\mathrm{is \ added \ to}$ 100 $\mathrm{mL \ water}$ ($\mathrm{density}$ = 1 $\mathrm{g/cm^3}$), $\mathrm{the \ following \ diagram \ is \ obtained.}$ $x =$ ($\mathrm{nearest \ integer}$). [$\mathrm{Given : \ Molal \ freezing \ point \ depression \ constant \ of \ water \ at \ 273.15 \ K \ is \ 1.86 \ K \ kg \ mol}^{-1}$]

Answer: 543

Solution

Given $\Delta T_f = 273.15 - 270.65 = 2.5 \, \mathrm{K}$. Using $\Delta T_f = K_f m$, we have $2.5 = 1.86 \times \frac{n}{0.1}$. This implies $n = 0.1344$ moles. Therefore, $w = 0.1344 \times 32 = 4.3 \, \mathrm{g}$. The volume is calculated as $\frac{4.3}{0.792} = 5.43 \, \mathrm{ml} = 5.43 \times 10^{-2} \, \mathrm{ml}$.

Question 84

Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Fill in the blank

The ratio of number of oxygen atoms to bromine atoms in the product Q is ___ $\times$ 10^{-1}.

Answer: 15

Solution

The reaction begins with the nitration of ethoxybenzene using $\mathrm{HNO_3}$ and $\mathrm{H_2SO_4}$, resulting in the formation of nitroethoxybenzene. This is followed by bromination using $2\mathrm{Br_2}$ and $\mathrm{Fe}$, which adds bromine atoms to the aromatic ring, resulting in the final product with two bromine atoms and one nitro group.

Question 85

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Numerical

Number of carbocations from the following that are not stabilized by hyperconjugation is _______

Answer: 5

Solution

Question 86

Chemistry · Thermodynamics · Fill in the blank

For the reaction at 298 K, \[ 2\,\mathrm{A}+\mathrm{B}\rightarrow \mathrm{C} \] \[ \Delta H = 400\,\mathrm{kJ\,mol^{-1}} \] and \[ \Delta S = 0.2\,\mathrm{kJ\,mol^{-1}\,K^{-1}}. \] The reaction will become spontaneous above \[ \underline{\hspace{2cm}}\ \mathrm{K}. \]

Answer: 2000

Solution

Given $\Delta G = 0$. $$T = \frac{\Delta H}{\Delta S} = \frac{400}{0.2} = 2000 \, \mathrm{K}$$

Question 87

Chemistry · The d-and f-Block Elements · Numerical

Total number of species from the following with central atom utilising $sp^2$ hybrid orbitals for bonding is _____. NH$_3$, SO$_2$, SiO$_2$, BeCl$_2$, C$_2$H$_2$, C$_2$H$_4$, BCl$_3$, HCHO, C$_6$H$_6$, BF$_3$, C$_2$H$_4$Cl$_2$

Answer: 6

Solution

Central atom utilizing $sp^2$ hybrid orbitals. $\mathrm{SO_2}$, $\mathrm{C_2H_4}$, $\mathrm{BCl_3}$, $\mathrm{HCHO}$, $\mathrm{C_6H_6}$, $\mathrm{BF_3}$

Question 88

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

Consider the two different first order reactions given below $$\mathrm{A + B \rightarrow C \ (Reaction \ 1)}$$ $$\mathrm{P \rightarrow Q \ (Reaction \ 2)}$$ The ratio of the half life of Reaction 1 : Reaction 2 is 5 : 2. If $t_1$ and $t_2$ represent the time taken to complete $$\frac{2}{3}^{\mathrm{rd}}$$ and $$\frac{4}{5}^{\mathrm{th}}$$ of Reaction 1 and Reaction 2, respectively, then the value of the ratio $t_1 : t_2$ is ______ $\times 10^{-1}$ (nearest integer). [Given : $\log_{10}(3) = 0.477$ and $\log_{10}(5) = 0.699$]

Answer: 17

Solution

Given $\($ $\frac{(t_{1/2})_1}{(t_{1/2})_{II}}$ = $\frac{K_2}{K_1}$ = $\frac{5}{2}$ $\)$. Therefore, $\($ K_1 t_1 = $\ln$ $\frac{1}{1 - \frac{2}{3}}$ = $\ln$ 3 $\)$. $\($ K_2 t_2 = $\ln$ $\frac{1}{1 - \frac{4}{5}}$ = $\ln$ 5 $\)$. Thus, $\($ $\frac{K_1}{K_2}$ $\times$ $\frac{t_1}{t_2}$ = $\frac{0.477}{0.699}$ $\)$. $\($ $\Rightarrow$ $\frac{t_1}{t_2}$ = $\frac{0.477}{0.699}$ $\times$ $\frac{5}{2}$ = 1.7 = 17 $\times$ 10^{-1} $\)$

Question 89

Chemistry · Structure of Atom · Numerical

For hydrogen atom, energy of an electron in first excited state is $-3.4 \, \mathrm{eV}$, K. E. of the same electron of hydrogen atom is $x \, \mathrm{eV}$. Value of $x$ is ____ $\times$ $10^{-1}$ \, $\mathrm{eV}$. (Nearest integer)

Answer: 34

Solution

Kinetic energy of electron is, $$\mathrm{KE} = \frac{13.6Z^2}{n^2} \, \mathrm{eV}$$ For the first excited state of the hydrogen atom, $n = 2$ and $Z = 1$. Therefore, $\mathrm{KE} = \frac{13.6}{2^2} = 3.4 \, \mathrm{eV}$

Question 90

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Numerical

Among $\mathrm{VO}_2^+$, $\mathrm{MnO}_4^-$ and $\mathrm{Cr}_2\mathrm{O}_7^{2-}$, the spin-only magnetic moment value of the species with least oxidising ability is ______ BM (Nearest integer). (Given atomic member $\mathrm{V} = 23$, $\mathrm{Mn} = 25$, $\mathrm{Cr} = 24$)

Answer: 0

Solution

For 3d transition series; Oxidising power: $\mathrm{V}^{+5} < \mathrm{Cr}^{+6} < \mathrm{Mn}^{+7}$ $\mathrm{V}^{+5}$: $[\mathrm{Ar}] 4s^0 3d^0$ Number of unpaired electron $= 0$ $\mu = 0$