JEE Main 5 April 2026 Shift 2 question paper with solutions

JEE Main 5 April 2026 Shift 2: all 75 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.

Register free to attempt this shift as a timed 180-minute test with instant scoring and chapter-wise analysis.

Maths

Question 1

Maths · Complex Numbers and Quadratic Equations · Single correct

Let $\alpha$, $\beta$ be the roots of the equation $x^2 - x + p = 0$ and $\gamma$, $\delta$ be the roots the equation $x^2 - 4x + q = 0$; $p, q \in \mathbb{Z}$. If $\alpha$, $\beta$, $\gamma$, $\delta$ are in G.P., then $|p+q|$ equals:

  1. 16
  2. 32
  3. 34
  4. 38

Answer: (c)

Solution

Let $\alpha = a$, $\beta = ar$, $\gamma = ar^2$, $\delta = ar^3$. $a + ar = 1$ $ar^2 + ar^3 = 4$ $$\Rightarrow ar^2(1 + r) = 4$$ $$\Rightarrow r = 2, \ a = \frac{1}{3} (reject as p \in \mathbb{Z})$$ or $r = -2$, $a = -1$ Now $|P+q| = |a(ar) + ar^2 - ar^3|$ $$= |a^2 r(1 + r^4)|$$ $$|1 \times -2 \times 17| = 34$$

Question 2

Maths · Complex Numbers and Quadratic Equations · Single correct

Let $z_1, z_2 \in \mathbb{C}$ be the distinct solutions of the equation $z^2 + 4z - (1 + 12i) = 0$. Then $|z_1|^2 + |z_2|^2$ is equal to:

  1. 18
  2. 22
  3. 29
  4. 34

Answer: (d)

Solution

Given $Z^2 + 4Z - (1 + 12i) = 0$. Therefore, $$Z = \frac{-4 \pm \sqrt{16 + 4(1 + 12i)}}{2}$$ This simplifies to $$Z = -2 \pm \sqrt{5 + 12i}$$ Further simplifying, $$Z = -2 \pm (3 + 2i)$$ Thus, $$Z = 1 + 2i, -5 - 2i$$ Therefore, $$|Z_1|^2 + |Z_2|^2$$ is $$= 5 + 29$$ which equals $$= 34$$

Question 3

Maths · Determinants · Single correct

If $f:\mathbb{N}\to\mathbb{Z}$ is defined by $$ f(n)=\left| \begin{array}{ccc} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{array} \right|, \qquad k\in\mathbb{N}, $$ and $\sum_{n=1}^{k}f(n)=98$, then $k$ is equal to:

  1. 3
  2. 4
  3. 5
  4. 6

Answer: (a)

Solution

Given $$f(n) = \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}$$ We have $$\sum_{n=1}^{k} f(n) = \begin{vmatrix} \frac{k(k+1)}{2} & -1 & -5 \\ -2 \frac{k(k+1)(2k+1)}{6} & 3(2k+1) & 2k+1 \\ -3 \frac{k^2(k+1)^2}{4} & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix} = 98$$ Thus, $$\frac{k(k+1)(2k+1)}{2} \cdot \frac{7}{3} = 98$$ This implies $$k(k+1)(2k+1) = 84$$ Solving gives $$k = 3$$

Question 4

Maths · Matrices · Single correct

Let $\mathrm{M}$ be a $3\times3$ matrix such that $\mathrm{M}\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}1\\2\\3\end{pmatrix}$, $\mathrm{M}\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\1\\0\end{pmatrix}$ and $\mathrm{M}\begin{pmatrix}0\\0\\1\end{pmatrix}=\begin{pmatrix}-1\\1\\1\end{pmatrix}$. If $\mathrm{M}(x,y,z)^{T}$ $=$ $\begin{pmatrix}1\\7\\11\end{pmatrix}$, then $x+y+z$ equals:

  1. 4
  2. 5
  3. 7
  4. 11

Answer: (c)

Solution

Let $M = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}$. Then $$M \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \Rightarrow a_1 = 1, b_1 = 2, c_1 = 3$$ $$M \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} \Rightarrow a_2 = 0, b_2 = 1, c_2 = 2$$ $$\& M \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \Rightarrow a_3 = -1, b_3 = 1, c_3 = 1$$ Now $$M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix}$$ $$\Rightarrow x + 0y - z = 1$$ $$2x + y + z = 7$$ $$3x + 2y + z = 11$$ On solving $x = 2, y = 2, z = 1$. Therefore, $x + y + z = 5$

Question 5

Maths · Sequences and Series · Single correct

If the sum of the first 10 terms of the series $$\frac{1}{1+1^4 \times 4} + \frac{2}{1+2^4 \times 4} + \frac{3}{1+3^4 \times 4} + \frac{4}{1+4^4 \times 4} + \ldots$$ is $\frac{m}{n}$, gcd (m, n) = 1, then m + n is equal to :

  1. 256
  2. 264
  3. 276
  4. 284

Answer: (c)

Solution

Given $T_r = \frac{r}{1 + 4r^4} = \frac{r}{4r^4 + 4r^2 + 1 - 4r^2} = \frac{r}{(2r^2 + 1)^2 - (2r)^2}$. This simplifies to $\frac{r}{(2r^2 + 2r + 1)(2r^2 - 2r + 1)}$. $S_{10} = \frac{1}{4} \left( 1 - \frac{1}{221} \right) = \frac{55}{221} = \frac{m}{n}$. Therefore, $m + n = 276$.

Question 6

Maths · Sequences and Series · Single correct

Let $A_1, A_2, A_3, \ldots, A_{39}$ be 39 arithmetic means between the numbers 59 and 159. Then the mean of $A_{25}, A_{28}, A_{31}$ and $A_{36}$ is equal to:

  1. 129
  2. 136
  3. 131.50
  4. 134

Answer: (d)

Solution

Given the sequence $59, A_1, A_2, A_3, \ldots, A_{39}, 159$, we find the common difference $d$ as follows: $$d = \frac{159 - 59}{39 + 1} = \frac{5}{2}.$$ Now, calculate the terms: $$A_{25} = 59 + 25d = 121.5,$$ $$A_{28} = 59 + 28d = 129,$$ $$A_{31} = 59 + 31d = 136.5,$$ $$A_{36} = 59 + 36d = 149.$$ The mean of $A_{25}, A_{28}, A_{31}, A_{36}$ is: $$\frac{121.5 + 129 + 136.5 + 149}{4} = \frac{536}{4} = 134.$$

Question 7

Maths · Binomial Theorem · Single correct

The coefficient of $x^2$ in the expansion of $$\left( 2x^2 + \frac{1}{x} \right)^{10}, x \neq 0,$$ is:

  1. 3240
  2. 3360
  3. 3480
  4. 3600

Answer: (c)

Solution

The term $T_{r+1} = \binom{10}{r} (2x^2)^{10-r} (1/x)^r$ is given by $$= \binom{10}{r} 2^{10-r} x^{20-2r-r}$$ $$= \binom{10}{r} 2^{10-r} x^{20-3r}$$ Solving $20 - 3r = 2$ gives $$r = 6$$ So the required coefficient is $\binom{10}{6} 2^4$

Question 8

Maths · Probability · Single correct

The probabilities that players A and B of a team are selected for the captaincy for a tournament are 0.6 and 0.4, respectively. If A is selected the captain, the probability that the team wins the tournament is 0.8 and if B is selected the captain, the probability that the team wins the tournament is 0.7. Then the probability, that the team wins the tournament, is:

  1. 0.74
  2. 0.76
  3. 0.72
  4. 0.78

Answer: (b)

Solution

P(Win) = P(A $\cap$ Win) + P(B $\cap$ Win) = P(A) $\cdot$ P$\left(\frac{Win}{A}\right)$ + P(B) $\cdot$ P$\left(\frac{Win}{B}\right)$ = (0.6)(0.8) + (0.4)(0.7) = 0.48 + 0.28 = 0.76

Question 9

Maths · Permutations and Combinations · Single correct

A box contains 5 blue, 6 yellow and 4 red balls. The number of ways, of drawing 8 balls containing at least two balls of each colour, is:

  1. 4100
  2. 4140
  3. 4230
  4. 4290

Answer: (a)

Solution

But in this question, it is not mentioned that balls are distinct/alike. Blue – 5, Yellow – 6, Red – 4. Reqd. No. of ways $$= \binom{5}{2} \times \binom{6}{2} \times \binom{4}{4} + \binom{5}{2} \times \binom{6}{3} \times \binom{4}{3} + \binom{5}{2} \times \binom{6}{4} \times \binom{4}{2}$$ $$+ \binom{5}{3} \times \binom{6}{2} \times \binom{4}{3} + \binom{5}{3} \times \binom{6}{3} \times \binom{4}{2} + \binom{5}{4} \times \binom{6}{2} \times \binom{4}{2}$$ $$= 10 \times 15 \times 1 + 10 \times 20 \times 4 + 10 \times 15 \times 6 + 10 \times 15 \times 4 + 10 \times 20 \times 6 + 5 \times 15 \times 6$$ $$= 150 + 800 + 900 + 600 + 1200 + 450$$ $$= 4100$$

Question 10

Maths · Statistics · Single correct

A variable $X$ takes values $0,0,2,6,12,20,\ldots,n(n-1)$ with frequencies ${}^{n}C_{0},{}^{n}C_{1},{}^{n}C_{2},{}^{n}C_{3},{}^{n}C_{4},{}^{n}C_{5},\ldots,{}^{n}C_{n}$, respectively. If the mean of this data is $60$, then its median is:

  1. 56
  2. 42
  3. 72
  4. 90

Answer: (a)

Solution

Given $$ \overline{x} = \frac{0 \cdot \binom{n}{0} + 0 \cdot \binom{n}{1} + 2 \cdot \binom{n}{2} + \ldots + n(n-1) \cdot \binom{n}{n}}{\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n}} $$ $$ = \frac{\sum_{r=0}^{n} r(r-1) \binom{n}{r}}{2^n} $$ $$ = \frac{\sum_{r=1}^{n} n(r-1)^{n-1} \binom{n-1}{r-1}}{2^n} = \frac{\sum_{r=2}^{n} n(n-1)^{n-2} \binom{n-1}{r-2}}{2^n} $$ $$ = \frac{n(n-1)2^{n-2}}{2^n} = \frac{n(n-1)}{4} $$ Thus, $$ \frac{n(n-1)}{4} = 60 \Rightarrow n^2 - n - 240 = 0 $$ Solving, $$ n = 16, -15 (Reject) $$ Then, $$ N = \sum f_i = \binom{16}{0} + \binom{16}{1} + \ldots + \binom{16}{16} = 2^{16} $$ Since $N = 2^{16}$ is even, $$ Median = \frac{\left( \frac{N}{2} \right)^{th} + \left( \frac{N}{2} + 1 \right)^{th}}{2} term $$ Calculating, $$ \frac{N}{2} = 2^{15} = 32768 $$ The table of $x_i$ and $f_i$ is as follows: $$\begin{array}{ccc} x_i & f_i & CF \\ 0 & \binom{16}{0} & 1 \\ 0 & \binom{16}{1} & 16 \\ 2 & \binom{16}{2} & 137 \\ 6 & \binom{16}{3} & 697 \\ 12 & \binom{16}{4} & 2517 \\ 20 & \binom{16}{5} & 6885 \\ 30 & \binom{16}{6} & 14893 \\ 42 & \binom{16}{7} & 26333 \\ 56 & \binom{16}{8} & 39203 \rightarrow \frac{N}{2} \& \left( \frac{N}{2} + 1 \right) lies \\ 72 & \binom{16}{9} & \frac{n(n-1)2^{n-2}}{2^n} \\ \end{array}$$ Thus, $$ \Rightarrow median = \frac{56 + 56}{2} = 56 $$

Question 11

Maths · Conic Sections · Single correct

Let the point P be the vertex of the parabola $y = x^2 - 6x + 12$. If a line passing through the point P intersects the circle $x^2 + y^2 - 2x - 4y + 3 = 0$ at the points R and S, then the maximum value of $(PR + PS)^2$ is:

  1. 10
  2. 20
  3. 25
  4. 5

Answer: (b)

Solution

Parabola $y = x^2 - 6x + 12 = (x-3)^2 + 3$. $y - 3 = (x-3)^2 \Rightarrow$ Vertex $P(3, 3)$. $x^2 + y^2 - 2x - 4y + 3 = 0$. $C(1, 2); \; r = \sqrt{2}$. For max. of $(PR + PS)^2$, RS should be diameter. $$(PR + PS)^2_{\max} = (PC - r + PC + r)^2$$ $$= (2PC)^2 = 4(PC)^2$$ $$= 4 \times 5 = 20$$

Question 12

Maths · Conic Sections · Single correct

Let the directrix of the parabola P : $y^2 = 8x$, cut x-axis at the point A. Let B $(\alpha, \beta)$, $\alpha > 1$, be a point on P such that the slope of AB is $3/5$. If BC is a focal chord of P, then six times the area of $\Delta ABC$ is:

  1. 80
  2. 160
  3. 174
  4. 192

Answer: (b)

Solution

Let $B(2t^2, 4t) \Rightarrow C \left( \frac{2}{t^2}, \frac{-4}{t} \right); A(-2,0)$ given $m_{AB} = \frac{4t}{2t^2 + 2} = \frac{3}{5}$. $$20t = 6t^2 + 6$$ $$3t^2 - 10t + 3 = 0$$ $$t = 3, \frac{1}{3} but t > 1 (given)$$ $$\Rightarrow P(18,12), Q \left( \frac{2}{9}, \frac{-4}{3} \right)$$ $$A = \frac{1}{2} \begin{vmatrix} -2 & 0 & 1 \\ 18 & 12 & 1 \\ \frac{2}{9} & \frac{-4}{3} & 1 \end{vmatrix} = \frac{80}{3}$$ $$\Rightarrow 6A = 160$$

Question 13

Maths · Conic Sections · Single correct

Let the eccentricity $e$ of a hyperbola satisfy the equation $6e^2 - 11e + 3 = 0$. If the foci of the hyperbola are $(3, 5) (3, -4)$, then the length of its latus rectum is:

  1. $\frac{11}{3}$
  2. $\frac{17}{3}$
  3. $\frac{15}{2}$
  4. $\frac{17}{2}$

Answer: (c)

Solution

Given $S_1(3,5)$ and $S_2(3,-4)$. Therefore, $S_1 S_2 = 2ae$. $$9 = 2ae ...(1)$$ And $6e^2 - 11e + 3 = 0$. $$e = \frac{3}{2}, e = \frac{1}{3} < 1 (rejected)$$ From (1); $a = 3$. Therefore, $$LR = 2 \cdot \frac{b^2}{a} = 2a(e^2 - 1) = 2 \cdot 3 \left( \frac{9}{4} - 1 \right) = \frac{15}{2}$$

Question 14

Maths · Three Dimensional Geometry · Single correct

Let a triangle PQR be such that P and Q lie on the line $\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}$ and are at a distance of 6 units from R (1, 2, 3). If $(\alpha, \beta, \gamma)$ is the centroid of $\Delta PQR$, then $\alpha + \beta + \gamma$ is equal to:

  1. 4
  2. 5
  3. 6
  4. 8

Answer: (c)

Solution

Let the point P be $(8\lambda - 3, 2\lambda + 4, 2\lambda - 1)$. PR = 6 implies $\mathrm{PR}^2 = 36$. $$(8\lambda - 4)^2 + (2\lambda + 2)^2 + (2\lambda - 4)^2 = 36$$ $$(4\lambda - 2)^2 + (\lambda + 1)^2 + (\lambda - 2)^2 = 9$$ $\lambda^2 - \lambda = 0 \implies \lambda = 0, 1 For \lambda$ = 0$, $P(-3, 4, -1)$. For \lambda$ = 1$, $P(5, 6, 1)$. R(1, 2, 3) G(1, 4, 1) \alpha$ + $\beta$ + $\gamma$ = 6$

Question 15

Maths · Three Dimensional Geometry · Single correct

If the distance of the point (a, 2, 5) from the image of the point (1, 2, 7) in the line $\frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2}$ is 4, then the sum of all possible values of a is equal to:

  1. 11
  2. 9
  3. 6
  4. 4

Answer: (c)

Solution

PQ $\perp$ L $\Rightarrow$ ($\alpha$ - 1) + ($\beta$ - 2) + 2($\gamma$ - 7) = 0 $\Rightarrow \alpha$ + $\beta$ + 2$\gamma$ = 17 $\ldots$ (1) M is the midpoint of PQ which will satisfy L $$\frac{\alpha + 1}{2} = \frac{\beta + 2}{2} - 1 = \frac{\gamma + 7}{2} - 2$$ $$\Rightarrow \frac{\alpha + 1}{2} = \frac{\beta}{2} = \frac{\gamma + 3}{4}$$ $$\Rightarrow \alpha + 1 = \beta \ldots (2)$$ and $2\beta = \gamma + 3 \ldots (3)$ $\Rightarrow \alpha = 3, \beta = 4, \gamma = 5$ Distance from $(a, 2, 5)$ is $= \sqrt{(a - 3)^2 + 4 + 0} = 4$ $(a - 3)^2 + 4 = 16 \Rightarrow a^2 - 6a - 3 = 0 \Rightarrow sum of values of a = 6$

Question 16

Maths · Vector Algebra · Single correct

Let O be the origin, $\overrightarrow{OP} = \vec{a}$ and $\overrightarrow{OQ} = \vec{b}$. If R is the point on $\overrightarrow{OP}$ such that $\overrightarrow{OR} = \frac{5}{6}\overrightarrow{OP}$, and M is the point such that $\overrightarrow{OQ} = 5\overrightarrow{RM}$, then $\overrightarrow{PM}$ is equal to:

  1. $\frac{1}{5}(\vec{a} - 4\vec{b})$
  2. $\frac{1}{5}(\vec{b} - 4\vec{a})$
  3. $\frac{1}{5}(-\vec{a} + 4\vec{b})$
  4. $\frac{1}{5}(-\vec{b} + 4\vec{a})$

Answer: (b)

Solution

Given $\overrightarrow{\mathrm{OR}} = \frac{\overrightarrow{\mathrm{OP}}}{5} = \frac{\vec{a}}{5}$ $\overrightarrow{\mathrm{RM}} = \frac{\overrightarrow{\mathrm{OQ}}}{5} = \frac{\vec{b}}{5}$ $\overrightarrow{\mathrm{OM}} - \overrightarrow{\mathrm{OR}} = \frac{\vec{b}}{5}$ $\overrightarrow{\mathrm{OM}} = \frac{\vec{a} + \vec{b}}{5}$ $\overrightarrow{\mathrm{PM}} = \overrightarrow{\mathrm{OM}} - \overrightarrow{\mathrm{OP}}$ $\frac{\vec{a} + \vec{b}}{5} - \vec{a}$ $= \frac{\vec{b} - 4\vec{a}}{5}$

Question 17

Maths · Limits and Derivatives · Single correct

Let $f(x) = \lim_{y \to 0} \frac{(1 - \cos(xy)) \tan(xy)}{y^3}$. Then the number of solutions of the equation $f(x) = \sin x$, $x \in \mathbb{R}$ is:

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

Answer: (c)

Solution

Given $$f(x) = \lim_{y \to 0} \frac{1 - \cos(xy)}{(xy)^2} \cdot \frac{\tan xy}{xy} \frac{(xy)^3}{y^3}$$ $$f(x) = \frac{x^3}{2}$$ $$f(x) = \sin x$$ $$\frac{x^3}{2} = \sin x$$ 3 points of intersection

Question 18

Maths · Sequences and Series · Single correct

Let $\left(2^{1-a} + 2^{1+a}\right)$, $f(a)$, $\left(3^a + 3^{-a}\right)$ be a A. P. and $\alpha$ be the minimum value of $f(a)$. Then the value of the integral $$\int_{\log_e (\alpha - 1)}^{\log_e (\alpha)} \frac{dx}{(e^{2x} - e^{-2x})}$$ is:

  1. $\frac{1}{2} \log_e \left(\frac{4}{3}\right)$
  2. $\frac{1}{4} \log_e \left(\frac{4}{3}\right)$
  3. $\frac{1}{2} \log_e \left(\frac{8}{5}\right)$
  4. $\frac{1}{4} \log_e \left(\frac{8}{5}\right)$

Answer: (b)

Solution

Given $$f(a) = \frac{1}{2} \left( 2(2^a + 2^{-a}) + (3^a + 3^{-a}) \right)$$ Calculate $$a = \frac{1}{2} (2 \times 2 + 2) = 3$$ Now, $$I = \int_{\ln(3-1)}^{\ln 3} \frac{e^{2x}}{e^{4x} - 1} \, dx$$ Let $e^{2x} = t$. Then, $$I = \frac{1}{2} \int_{4}^{9} \frac{dt}{t^2 - 1}$$ This becomes $$I = \frac{1}{4} \left( \ln \left( \frac{t-1}{t+1} \right) \right)_{4}^{9}$$ Finally, $$I = \frac{1}{4} \ln \left( \frac{4}{3} \right)$$

Question 19

Maths · Differential Equations · Single correct

Let $f : [1, \infty) \to \mathbb{R}$ be a differentiable function defined as $f(x) = \int_{1}^{x} f(t) \, dt + (1-x)(\log_e x - 1) + e$. Then the value fo $f(f(1))$ is:

  1. (1 + e^e)
  2. (1 + e)
  3. (1 + e + e^e)
  4. 1 + 2e

Answer: (a)

Solution

Given $$f(1) = \int_1^1 f(t) \, dt + (1-1)(\ln 1 - 1) + e$$ We have $$f(1) = e$$ Now, $$f(x) = \int_1^x f(t) \, dt + (1-x)(\ln x - 1) + e$$ Differentiating with respect to $x$: $$f'(x) = f(x) + \frac{1}{x} - \ln x$$ The integrating factor (I.F.) is $$e^{\int -1 \, dx} = e^{-x}$$ Thus, $$f(x) \, e^{-x} = \int e^{-x} \left( \frac{1}{x} - \ln x \right) \, dx$$ This gives $$f(x) \, e^{-x} = e^{-x} \ln x + C$$ Putting $x = 1$, $y = e$, we find $$\Rightarrow C = 1$$ Therefore, $$f(x) e^{-x} = e^{-x} \ln x + 1$$ So, $$f(x) = e^x + \ln x$$ Finally, $$f(e) = e^e + 1$$

Question 20

Maths · Continuity and Differentiability · Single correct

Let $f(x)$ and $g(x)$ be twice differentiable functions satisfying $f''(x) = g''(x)$ for all $x \in \mathbb{R}$, $f'(1) = 2g'(1) = 4$ and $g(2) = 3f(2) = 9$. Then $f(25) - g(25)$ is equal to:

  1. 20
  2. 40
  3. -20
  4. -40

Answer: (b)

Solution

Given $f''(x) = g''(x)$. Integrate to get $f'(x) = g'(x) + C_1$. Put $x = 1$. Then $f'(1) = g'(1) + C_1$. Since $4 = 2 + C_1$, we have $C_1 = 2$. Therefore, $f'(x) = g'(x) + 2$. Integrate again to get $f(x) = g(x) + 2x + C_2$. Put $x = 2$. Then $f(2) = g(2) + 4 + C_2$. Since $3 = 9 + 4 + C_2$, we have $C_2 = -10$. Thus, $f(x) = g(x) + 2x - 10$. Put $x = 25$. Then $f(x) - g(25) = 40$.

Question 21

Maths · Relations and Functions · Numerical

Let $A=\{1,4,7\}$ and $B=\{2,3,8\}$. Then the number of elements in the relation $R=\{((a_1,b_1),(a_2,b_2))\in((A\times B)\times(A\times B)):a_1+b_2\text{ divides }a_2+b_1\}$ is $\ldots$

Answer: 18

Solution

These are possible sums of $a_1 + b_j$. Now $b_1 + a_2$ divides $a_1 + b_2$. So, $$\begin{array}{ccc} a_1 + b_2 & a_2 + b_1 & No. of pairs \\ 15 & 3, 15 & 2 \\ 12 & 3, 4, 6, 12 & 4 \\ 10 & 10 & 1 \\ 9 & 3, 9 & 6 \\ 7 & 7 & 1 \\ 6 & 3, 6 & 2 \\ 4 & 4 & 1 \\ 3 & 3 & 1 \\ \end{array}$$ Number of relations = 18

Question 22

Maths · Straight Lines and Pair of Straight Lines · Numerical

From the point $(-1, -1)$, two rays are sent making angles of $45^\circ$ with the line $x + y = 0$. These rays get reflected from the mirror $x + 2y = 1$. If the equations of the reflected rays are $ax + by = 9$ and $cx + dy = 7$, $a, b, c, d \in \mathbb{Z}$, then the value of $ad + bc$ is $\ldots$.

Answer: 7

Solution

Given $\tan 45^\circ = \left| \frac{m+1}{1-m} \right|$. Therefore, $\pm 1 = \frac{m+1}{1-m}$. This implies $m = 0$ and one line is perpendicular to the x-axis. $L_1: x = -1$, $L_2: y = -1$. Now taking the mirror image of $x + 1 = 0$ in $x + 2y = 1$. $A': \frac{x+1}{1} = \frac{y+1}{2} = \frac{-2(-1-2-1)}{1^2+2^2}$. Thus, $x = \frac{3}{5}$, $y = \frac{11}{5}$. $A'\left( \frac{3}{5}, \frac{11}{5} \right)$. The line is $3x - 4y + 7 = 0$. Now taking the mirror image of $y + 1 = 0$ in $x + 2y = 1$. $A'': \frac{x+1}{1} = \frac{y+1}{2} = \frac{-2(-1-2-1)}{1^2+2^2}$. Thus, $x = \frac{3}{5}$, $y = \frac{11}{5}$. $A''\left( \frac{3}{5}, \frac{11}{5} \right)$. The line is $\left( \frac{11}{5} + 1 \right) = \left( \frac{3}{5} - 3 \right)(x - 3)$. $4x + 3y = 9$. Comparing $4x + 3y = 9$ with $ax + by = 9$, we have $a = 4$, $b = 3$. By line $-3x + 4y = 7$. Comparing with $cx + dy = 7$, we have $c = -3$, $d = 4$. Therefore, $ad + bc = 4 \times 4 + 3 \times (-3) = 16 - 9 = 7$.

Question 23

Maths · Trigonometric Functions · Numerical

If $S = \left\{ \theta \in [-\pi, \pi] : \cos \theta \cos \frac{5 \theta}{2} = \cos 7 \theta \cos \frac{7 \theta}{2} \right\}$, then $n(S)$ is equal to $\ldots$

Answer: 19

Solution

Given $\cos \theta \cos \frac{5\theta}{2} = \cos 7\theta \cos \frac{7\theta}{2}$. $$\cos \frac{7\theta}{2} + \cos \frac{3\theta}{2} = \cos \frac{21\theta}{2} + \cos \frac{7\theta}{2}$$ $$\cos \frac{21\theta}{2} - \cos \frac{3\theta}{2} = 0$$ $$-2 \sin 6\theta \sin \frac{9\theta}{2} = 0$$ $$\sin 6\theta = 0$$ $$\theta = 0, \pm \frac{\pi}{6}, \pm \frac{2\pi}{6}, \ldots \pm \frac{5\pi}{6}, \pm \pi (13 solutions)$$ $$\sin \frac{9\theta}{2} = 0$$ $$\theta = \pm \frac{2\pi}{9}, \pm \frac{4\pi}{9}, \pm \frac{8\pi}{9} (6 more solutions)$$ Total = 19 solutions

Question 24

Maths · Applications of Integrals · Numerical

Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that $f(x) + 3f\left(\frac{\pi}{2} - x\right) = \sin x$, $x \in \mathbb{R}$. Let maximum value of $f$ on $\mathbb{R}$ be $\alpha$. If the area of the region bounded by the curves $g(x) = x^2$ and $h(x) = \beta x^3$, $\beta > 0$, is $\alpha^2$, then $30\beta^3$ is equal to $\ldots$

Answer: 16

Solution

Given $f(x) + 3f\left(\frac{\pi}{2} - x\right) = \sin x \ldots$ (1). Put $x \to \frac{\pi}{2} - x$. Therefore, $f\left(\frac{\pi}{2} - x\right) + 3f(x) = \cos x \ldots$ (2). From (1) and (2), $f(x) = \frac{1}{8}(3 \cos x - \sin x)$. The maximum value of $f$ is $f_{\max} = \frac{\sqrt{10}}{8}$. The curves $y = g(x)$ and $y = h(x)$ intersect as shown in the figure. $g(x) = x^2$ and $h(x) = \beta x^3$. The area bounded $= \Delta = \left| \int_{0}^{\frac{1}{\beta}} (\beta x^3 - x^2) \, dx \right| = \frac{1}{12\beta^3} = \alpha^2$ (given). Therefore, $30\beta^3 = 16$.

Question 25

Maths · Differential Equations · Numerical

Let $y = y(x)$ be the solution of the differential equation $(\tan x)^{1/2} dy = (\sec^3 x - (\tan x)^{3/2} y) dx$, $0 < x < \frac{\pi}{2}, y\left(\frac{\pi}{4}\right) = \frac{6\sqrt{2}}{5}$. If $y\left(\frac{\pi}{3}\right) = \frac{4}{5} \alpha$, then $\alpha^4$ equals $\ldots$

Answer: 48

Solution

Given $\frac{dy}{dx} + y \tan x = \frac{\sec^3 x}{\sqrt{\tan x}}$ IF $= e^{\int \tan x \, dx} = \sec x$ $y \sec x = \int \frac{\sec^4 x}{\sqrt{\tan x}} \, dx$ $y \sec x = \int \frac{(1 + \tan^2 x)}{\sqrt{\tan x}} \sec^2 x \, dx$ Let $\tan x = t$ $y \sec x = 2 \sqrt{\tan x} + \frac{2}{5} (\tan x)^{5/2} + c$ $y \left( \frac{\pi}{4} \right) = \frac{6 \sqrt{2}}{5}$ $\frac{6 \sqrt{2}}{5} \times \sqrt{2} = 2 + \frac{2}{5} + c \Rightarrow \frac{12}{5} = \frac{12}{5} + c$ $\Rightarrow c = 0$ $y(\pi/3) = \frac{8}{5} \cdot 3^{1/4} \Rightarrow \alpha = 2 \times 3^{3/4}$ $\therefore \alpha^4 = 48$

Physics

Question 26

Physics · Physical World, Units and Measurements · Single correct

26. Match List-I with List-II. where $h$ (Planck's constant), $G$ (gravitational constant) and $c$ (speed of light in vacuum) as fundamental units. Choose the correct answer from the options given below:

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

Answer: b

Solution

[h] = [ML^2T^{-1}] [c] = [LT^{-1}] and [G] = [M^{-1}L^3T^{-2}] then $$\left[ \sqrt{\frac{hc}{G}} \right] = \left[ \frac{ML^2T^{-1}LT^{-1}}{M^{-1}L^3T^{-2}} \right]^{1/2} = [M]$$ $$\left[ \sqrt{\frac{Gh}{c^5}} \right] = \left[ \frac{M^{-1}L^3T^{-2}ML^2T^{-1}}{L^5T^{-5}} \right]^{1/2} = [T]$$ $$\left[ \sqrt{\frac{k^2L^2c^3}{Gh}} \right] = \left[ \frac{k^2L^2L^3T^{-3}}{M^{-1}L^3T^{-2}ML^2T^{-1}} \right]^{1/2} = [k]$$ $$\left[ \sqrt{\frac{Gh}{c^3}} \right] = \left[ \frac{M^{-1}L^3T^{-2}ML^2T^{-1}}{L^3T^{-3}} \right]^{1/2} = [L]$$ A $\rightarrow$ IV, B $\rightarrow$ II, C $\rightarrow$ I, D $\rightarrow$ III

Question 27

Physics · Current Electricity · Single correct

In an experiment to determine the resistance of a given wire using Ohm's law, the voltmeter and ammeter readings are noted as 10 V and 5 A, respectively. The least counts of voltmeter and ammeter are $500\,\mathrm{mV}$ and $200\,\mathrm{mA}$, respectively. The estimated error in the resistance measurement is $\ldots \Omega$

  1. 0.25
  2. 2
  3. 2.5
  4. 0.18

Answer: (d)

Solution

Given $V = IR$. $R = \frac{10}{5} = 2 \, \Omega$ $$\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$$ $$\frac{\Delta R}{R} = \frac{500 \times 10^{-3}}{10} + \frac{200 \times 10^{-3}}{5}$$ $$\frac{\Delta R}{2} = 0.05 + 0.04$$ $$\Delta R = 0.09 \times 2 = 0.18 \, \Omega$$

Question 28

Physics · Laws of Motion · Single correct

A mass of $1\,\mathrm{kg}$ kept in a inclined plane with $30^\circ$ inclination with respect to horizontal plane and it is at rest initially. Then the whole assembly is moved up with constant velocity of $4 \, \mathrm{m/s}$. The work done by the frictional force in time $2 \, \mathrm{s}$ is $\ldots$ J. (Take $g = 10 \, \mathrm{m/s^2}$)

  1. 20
  2. 25
  3. 30
  4. 10

Answer: (a)

Solution

Given $m = 1 \, \mathrm{kg}$ and a constant velocity of $4 \, \mathrm{m/s}$. The angle of the incline is $30^\circ$. The force $f$ is given by $f = mg \sin 30^\circ = 5 \, \mathrm{N}$. The work done in $2 \, \mathrm{sec}$ will be calculated as follows. The displacement $d$ is $d = 4 \times 2 = 8 \, \mathrm{m}$. The work $W$ is given by $W = fd \cos 60^\circ$. Substituting the values, $W = (5)(8) \cos 60^\circ$. Therefore, $W = 20 \, \mathrm{joule}$.

Question 29

Physics · Motion in a Straight Line · Single correct

The velocity $(v)$ versus time $(t)$ plot of a particle is shown in the figure, for a time interval of $40 \, \mathrm{s}$. The total distance travelled by the particle and the average velocity during this period are, respectively

  1. $25 \, \mathrm{m}$ and zero
  2. $50 \, \mathrm{m}$ and zero
  3. $100 \, \mathrm{m}$ and zero
  4. $100 \, \mathrm{m}$ and $2.5 \, \mathrm{m/s}$

Answer: (c)

Solution

Total distance travel $$d = \frac{1}{2}(20)(5) + \frac{1}{2}(20) \times 5$$ $$d = 100 \, \mathrm{m}$$ Average velocity = $\frac{Total displacement}{Total time}$ $$\langle v \rangle = \frac{0}{40} = 0$$

Question 30

Physics · System of Particles and Rotational Motion · Single correct

A wheel initially at rest is subjected to a uniform angular acceleration about its axis. In the first 2 s it rotates through an angle $\theta_1$ and in the next 2 s it rotates through an angle $\theta_2$. The ratio $\frac{\theta_2}{\theta_1}$ is $\ldots$.

  1. 6
  2. 3
  3. 4
  4. $\frac{1}{3}$

Answer: (b)

Solution

Let $\alpha$ be angular acceleration. In first 2 sec $$\theta_1 = \frac{1}{2} \cdot \alpha (2)^2 = 2\alpha$$ Angular covered in 4 sec $$\theta' = \frac{1}{2} (\alpha)(4)^2 = 8\alpha$$ Angle in next 2 sec $$\theta_2 = \theta' - \theta_1 = 8\alpha - 2\alpha$$ $$\theta_2 = 6\alpha$$ $$\frac{\theta_2}{\theta_1} = \frac{6\alpha}{2\alpha} = 3$$

Question 31

Physics · System of Particles and Rotational Motion · Single correct

An object of uniform density rolls up the curved path with the initial velocity $v_0$ as shown in the figure. If the maximum height attained by an object is $\frac{7v_0^2}{10g}$ ($g$ = acceleration due to gravity), the object is a $\ldots$.

  1. solid cylinder
  2. ring
  3. disc
  4. solid sphere

Answer: (d)

Solution

Given $h = \frac{7v_0^2}{10g}$. Applying work energy theorem: $$-mgh = 0 - \frac{1}{2} mv^2 - \frac{1}{2} I \omega^2$$ Substituting the value of $h$: $$-mg \left( \frac{7v_0^2}{10g} \right) = -\frac{1}{2} mv_0^2 - \frac{1}{2} \frac{Iv_0^2}{2R^2}$$ Simplifying: $$\frac{7mv_0^2}{10} = \frac{mv_0^2}{2} + \frac{Iv_0^2}{2R^2}$$ Rearranging terms: $$\frac{mv_0^2}{5} = \frac{Iv_0^2}{2R^2}$$ Solving for $I$: $$I = \frac{2mR^2}{5}$$

Question 32

Physics · Gravitation · Single correct

A body of mass $m$ is taken from the surface of earth to a height equal to twice the radius of earth ($R_e$). The increase in potential energy will be $\ldots$. (g is acceleration due to gravity at the surface of earth)

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

Answer: (d)

Solution

Given $$U_1 = \frac{-GMm}{R_e}$$ $$U_2 = \frac{-GMm}{R_e + 2R_e} = \frac{-GMm}{3R_e}$$ $$\Delta U = U_2 - U_1 = \frac{-GMm}{3R_e} + \frac{GMm}{R_e}$$ $$\Delta U = \frac{2}{3} \frac{GMm}{R_e} = \frac{2}{3} \left[ \frac{GM}{R_e^2} \right] (mR_e)$$ $$\Delta U = \frac{2}{3} mgR_e$$

Question 33

Physics · Mechanical Properties of Fluids · Single correct

Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is $\ldots$. (S is the surface tension of mercury).

  1. 8$\pi$ r^2 S
  2. 16$\pi$ r^2 S
  3. 64$\pi$ r^2 S
  4. 4$\pi$ r^2 S

Answer: (b)

Solution

Volume conservation $$ (8) \left[ \frac{4}{3} \pi r^3 \right] = \frac{4}{3} \pi R^3 $$ $R = 2r$ $U_i = (8)(4 \pi r^2)(s)$; $s =$ surface tension $U_i = 32 \pi r^2 s$ $U_f = (4 \pi R^2)(s)$ $U_f = 4 \pi (2r)^2 (s) = 16 \pi r^2 s$ Energy released $= U_i - U_f$ Energy released $= 16 \pi r^2 s$

Question 34

Physics · Thermodynamics · Single correct

An ideal gas at pressure $P$ and temperature $T$ is expanding such that $PT^3 = constant$. The coefficient of volume expansion of the gas is____.

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

Answer: (c)

Solution

PT^3 = constant (C) $\frac{nRT}{V} \cdot$ T^3 = C V = $\frac{nR}{C} \cdot$ T^4 Volume expansion coeff. $\gamma$ = $\frac{1}{V} \frac{dV}{dT} \gamma$ = $\frac{C}{nRT^4} \cdot \frac{nR}{C} \cdot$ 4T^3 = $\frac{4}{T}$ Ans. $\gamma$ = $\frac{4}{T}$

Question 35

Physics · Oscillations · Single correct

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

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

Answer: (a)

Solution

(A) $\sin^2 \omega t = \frac{1}{2} - \frac{1}{2} \cos 2\omega t$ it represents SHM motion having period $= \frac{\pi}{\omega}$ (B) $\sin^3 2\omega t = \frac{3}{4} \sin 2\omega t - \frac{1}{4} \sin 6\omega t$ common period $\frac{\pi}{\omega}$ but not SHM. (C) $\sin \omega t + \cos \pi \omega t \rightarrow$ no common period hence it is a nonperiodic function (D) $\cos \omega t + \cos 2\omega t$ common period $\frac{2\pi}{\omega}$ but it do not represent SHM. A $\rightarrow$ III, B $\rightarrow$ I, C $\rightarrow$ IV, D $\rightarrow$ II

Question 36

Physics · Electromagnetic Induction · Single correct

A metal rod of length $L$ rotates about on end at origin with a uniform angular velocity $\omega$. The magnetic field radially falls off as $B(r) = B_o \, e^{-\lambda r}$; $\lambda$ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is:

  1. $B_o \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{1}{\lambda^2} + \frac{L}{\lambda} \right) \right]$
  2. $B_o \omega \left[ \frac{1}{\lambda^2} + e^{-\lambda L} \left( \frac{1}{\lambda^2} + \frac{L}{\lambda} \right) \right]$
  3. $B_o \omega \left[ \frac{4}{\lambda^2} - e^{-2\lambda L} \left( \frac{1}{\lambda^2} + \frac{2L}{\lambda} \right) \right]$
  4. $B_o \omega \left[ \frac{3}{\lambda^2} - e^{-3\lambda L} \left( \frac{3}{\lambda^2} + \frac{L}{\lambda} \right) \right]$

Answer: (a)

Solution

Given $$e = \int_0^L B(\omega r) \, dr = B_0 \omega \int_0^L e^{-\lambda r} \, rdr$$ This simplifies to $$= B_0 \omega \left[ r \int e^{-\lambda r} \, dr - \int 1 \cdot \frac{e^{-\lambda r}}{-\lambda} \, dr \right]_0^L$$ Evaluating the integrals, we have $$= B_0 \omega \left[ \frac{re^{-\lambda r}}{-\lambda} - \frac{e^{-\lambda r}}{\lambda^2} \right]_0^L$$ Substituting the limits, $$= B_0 \omega \left[ \left( 0 + \frac{1}{\lambda^2} \right) - \left( \frac{Le^{-\lambda L}}{\lambda} + \frac{e^{-\lambda L}}{\lambda^2} \right) \right]$$ Finally, we get $$e = B_0 \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{1}{\lambda^2} + \frac{L}{\lambda} \right) \right]$$

Question 37

Physics · Electrostatic Potential and Capacitance · Single correct

Under steady state condition, the potential difference across the capacitor in the circuit is $\_\_\_\_$ V.

  1. 0.5
  2. 1.5
  3. 0
  4. 2

Answer: (a)

Solution

The current $i$ is calculated as $$i = \frac{2}{6+2} = \frac{1}{4} \, \mathrm{A}$$ The voltage $v_c$ is equal to $v_{2\Omega}$, which is $$v_c = v_{2\Omega} = i \times 2 = \frac{1}{2} \, \mathrm{V} = 0.5 \, \mathrm{V}$$

Question 38

Physics · Moving Charges and Magnetism · Single correct

A particle of charge $q$ and mass $m$ is projected from origin with an initial velocity $\vec{v} = \left( \frac{v_0}{\sqrt{2}} \hat{x} + \frac{v_0}{\sqrt{2}} \hat{y} \right)$. There exists a uniform magnetic field $\vec{B} = B_0 \hat{z}$ and a space varying electric field $\vec{E} = E_0 e^{-\lambda x} \hat{x}$ within the region $0 \leq x \leq L$. After travelling a distance such that x-coordinate has changed from $x = 0$ to $x = L$, the change in the kinetic energy is_____

  1. $\frac{qE_0}{\lambda} [1 - e^{-\lambda L}]$
  2. $\left( \frac{v_0 q B_0}{2 \lambda} \right) [2 - e^{-2 \lambda L}]$
  3. $\frac{qE_0}{\lambda} [1 + e^{-\lambda L}]$
  4. $q \left[ \frac{E_0 + v_0 B_0}{\lambda} \right] [1 - e^{-\lambda L/2}]$

Answer: (a)

Solution

The change in kinetic energy $\Delta K$ is equal to the work done by the electric field $W_e$ plus the work done by the magnetic field $W_m$. Since $W_m = 0$, we have: $$\Delta K = W_e + W_m = W_e + 0$$ The work done by the electric field is given by: $$= \int_0^L qE \, dx = qE_0 \int_0^L e^{-\lambda x} \, dx$$ Evaluating the integral, we get: $$= \frac{qE_0}{\lambda} \left( 1 - e^{-\lambda L} \right)$$

Question 39

Physics · Electromagnetic Waves · Single correct

Given below are two statements: one is labelled as Assertion (A): and the other is labelled as Reason (R). Assertion (A): The electromagnetic wave exerts pressure on the surface on which they are allowed to fall. Reason (R): There is no mass associated with the electromagnetic waves. In the light of the above statements, choose the correct answer from the options given below:

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

Answer: (b)

Solution

Assertion and reason both are correct but not related.

Question 40

Physics · Ray Optics and Optical Instruments · Single correct

A thin convex lens and a thin concave lens are kept in contact and are co-axial. Which of the following statements is correct for this combination of two lenses?

  1. behaves as concave lens if $|f_{convex}| > |f_{concave}|$
  2. behaves as concave lens if $|f_{convex}| < |f_{concave}|$
  3. behaves as convex lens if $|f_{convex}| > |f_{concave}|$
  4. Focal length of the lens system will change if the position of two lenses are interchanged

Answer: (a)

Solution

1 $\rightarrow$ convex lens 2 $\rightarrow$ concave lens $$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{-f_2} : \frac{1}{f_1} - \frac{1}{f_2}$$ for $f_1 > f_2 \therefore f < 0$ $\therefore$ concave lens.

Question 41

Physics · Ray Optics and Optical Instruments · Single correct

An object AB is placed $15\,\mathrm{cm}$ on the left of a convex lens P of focal length $10 \, \mathrm{cm}$. Another convex lens Q is now placed $15 \, \mathrm{cm}$ right of lens P. If the focal length of lens Q is $15 \, \mathrm{cm}$, the final image is $\ldots$

  1. virtual, formed at $7.5\,\mathrm{cm}$ right of lens Q, with a size bigger than that of AB
  2. real, formed at $7.5\,\mathrm{cm}$ right of lens Q, with a size same than that of AB
  3. formed at infinity.
  4. real, formed at $7\,\mathrm{cm}$ right of lens Q, with a size smaller than that of AB

Answer: (b)

Solution

For P, $\frac{1}{V} - \frac{1}{-15} = \frac{1}{10}$. Therefore, $v = +30 \, \mathrm{cm} = +15 \, \mathrm{cm}$ right of Q. For Q, $\frac{1}{V'} - \frac{1}{15} = \frac{1}{15}$. $V' = 7.5 \, \mathrm{cm}$. $m = m_1 m_2 = \frac{30}{-15} \times \frac{7.5}{15} = -1$.

Question 42

Physics · Wave Optics · Single correct

The maximum intensity in a Young's double slit experiment is $I_0$. Distance between the slits $(d)$ is $5\lambda$, where $\lambda$ is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at $D = 10 \, d$ is $\ldots$

  1. $\frac{I_0}{4}$
  2. $\frac{I_0}{2}$
  3. $I_0$
  4. $\frac{3I_0}{4}$

Answer: (b)

Solution

At P, $y = \frac{d}{2}$. $$\Delta Q = \frac{2 \pi \, y d}{\lambda \, D} = \frac{2 \pi}{\lambda} \frac{d}{2 \, D} = \frac{\pi d^2}{\lambda D}$$ $$\Delta Q = \frac{\pi \times 25 \lambda^2}{\lambda \times 10 d} = \frac{5 \pi \lambda}{2 \times 5 \lambda} = \frac{\pi}{2}$$ $$I = I_0 \cos^2 \left( \frac{\Delta Q}{2} \right) = I_0 \cos^2 \left( \frac{\pi}{2 \times 2} \right) = \frac{I_0}{2}$$

Question 43

Physics · Dual Nature of Radiation and Matter · Single correct

An electron is travelling with a velocity $v$ in free space and when it enters a medium, its velocity is reduced by 20$\%$. The de Broglie wavelength of electron in the medium is $\alpha \lambda_0$, where $\lambda_0$ is its de Broglie wavelength in free space. The value of $\alpha$ is

  1. 1.20
  2. 1.0
  3. 1.25
  4. 0.75

Answer: (c)

Solution

Given $\lambda_0 = \frac{h}{mv_0}$. $$\lambda = \frac{h}{mv} = \frac{h}{m(0.8v_0)} = \frac{1}{0.8} \frac{h}{mv_0} = 1.25 \lambda_0$$ Therefore, $\alpha = 1.25$

Question 44

Physics · Nuclei · Single correct

Assuming the experimental mass of $^{12}_{6}\mathrm{C}$ as $12\,\mathrm{u}$, the mass defect of $^{12}_{6}\mathrm{C}$ atom is $\ldots\,\mathrm{MeV}/c^2$. (Mass of proton $= 1.00727\,\mathrm{u}$, mass of neutron $= 1.00866\,\mathrm{u}$, $1\,\mathrm{u} = 931.5\,\mathrm{MeV}/c^2$ and $c$ is the speed of the light in vacuum.)

  1. 127.5
  2. 89.03
  3. 272.0
  4. 92.0

Answer: (b)

Solution

The change in mass $\Delta m$ is given by the equation: $$\Delta m = (6m_p + 6m_n) - m_C$$ Substituting the values, we have: $$= 6(1.00727 + 1.00866) - 12$$ This simplifies to: $$= 0.09558 \, amu = 89.03 \, MeV/C^2$$

Question 45

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

In a semiconductor p-n diode, the doping concentrations on p-side and n-side are $10^{15}$ atoms/cm$^3$ and $10^{18}$ atoms/cm$^3$, respectively. Which one of the following statement is true?

  1. Widths of depletion region on either side of the interface are equal
  2. The depletion region width is more on p-side compared to that in n-side
  3. The depletion region width is more on n-side compared to that in p-side
  4. No depletion region forms because of unequal doping concentrations on p and n-sides

Answer: (b)

Solution

Given $N_p x_p = N_n x_n$. Since $N_p x_n$ in the depletion region.

Question 46

Physics · Mechanical Properties of Solids · Numerical

A copper wire of length 3 m is stretched by 3 mm by applying an external force. The volume of the wire is $600 \times 10^{-6} \, \mathrm{m}^3$. The elastic potential energy stored in the wire in stretched condition would be $\ldots$ J. (Given Young modulus of copper = $1.1 \times 10^{11} \, \mathrm{N/m}^2$)

Answer: 33

Solution

The energy stored, $U$, is given by the formula: $$U = \frac{1}{2} y (strain)^2 \times volume$$ Substituting the given values: $$= \frac{1}{2} \times 1.1 \times 10^{11} \times \left( \frac{3 \times 10^{-3}}{3} \right)^2 \times 600 \times 10^{-6}$$ Calculating the result: $$U = 33 \, J$$

Question 47

Physics · Thermal Properties of Matter · Numerical

The heat extracted out of $x$ gram of water initially at $50^\circ \mathrm{C}$ to cool it down to $0^\circ \mathrm{C}$ is sufficient to evaporate $(1000 - x)$ gram of water also initially at $50^\circ \mathrm{C}$. The value of $x$ (closest integer) is $\ldots$. (Take latent heat of water $2256 \, \mathrm{kJ/kg}$. $\mathrm{K}$, specific heat capacity of water $4200 \, \mathrm{J/kg}$. $\mathrm{K}$)

Answer: 922

Solution

Given the equation $x \times 4200 \times 50 = (1000 - x) \times 4200 \times 50 + (1000 - x) \times 2256 \times 10^3$. Simplifying, we have: $$x = 1000 - x + (1000 - x) \times 10.74$$ Solving for $x$, we find: $$x = 921.52$$

Question 48

Physics · Alternating Current · Fill in the blank

A series LCR circuit with $R = 20\,\Omega$, $L = 1.6\,\mathrm{H}$ and $C = 40\,\mu\mathrm{F}$ is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is $\ldots\,\Omega$.

Answer: 200

Solution

For resonance $$\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1.6 \times 40 \times 10^{-6}}} = \frac{10^3}{8} = 125$$ $$X_L = \omega L = 125 \times 1.6 = 200 \, \Omega$$

Question 49

Physics · Current Electricity · Fill in the blank

When an external resistance of $5\,\Omega$ is connected across terminals of a cell, a current of $0.25\,\mathrm{A}$ flows through it. When the $5\,\Omega$ resistor is replaced by a $2\,\Omega$ resistor, a current of $0.5\,\mathrm{A}$ flows through it. The internal resistance of the cell is $\ldots\,\Omega$.

Answer: 1

Solution

The current $i$ is given by the equation $$i = \frac{\varepsilon}{R + r}.$$ Substituting the given values, we have $$0.25 = \frac{\varepsilon}{5 + r}$$ and $$0.5 = \frac{\varepsilon}{2 + r}.$$ On solving, $r = 1 \, \Omega$.

Question 50

Physics · Electromagnetic Induction · Numerical

A circular loop of radius $20\,\mathrm{cm}$ and resistance 2 $\Omega$ is placed in a time varying magnetic field $\mathbf{B} = (2t^2 + 2t + 3) \, \mathrm{T}$. At $t = 0$, for the plane of the loop being perpendicular to the magnetic field and, the induced current in the loop at $t = 3 \, \mathrm{s}$ is $\frac{\alpha}{50} \, \mathrm{A}$. The value of $\alpha$ is $\ldots$ (Take $\pi = 22/7$)

Answer: 44

Solution

Given $\phi = B \cdot \pi r^2$. $\varepsilon = \frac{d\phi}{dt} = \pi r^2 \frac{dB}{dt}$. $i = \frac{\varepsilon}{R} = \frac{\pi r^2}{R} \cdot \left( \frac{dB}{dt} \right) = \frac{\pi}{2} \left( \frac{1}{5} \right)^2 (4t + 2)$. For $t = 3$, $$i = \frac{11}{7} \times \frac{1}{25} \times 14$$ $$i = \frac{44}{50}$$

Chemistry

Question 51

Chemistry · Some Basic Concepts of Chemistry · Single correct

What volume of hydrogen gas at STP would be liberated by action of $50\,\mathrm{mL}$ of $\mathrm{H_2SO_4}$ of 50$\%$ purity (density = $1.3\,\mathrm{g \, mL^{-1})}$ on $20\,\mathrm{g}$ of zinc? Given: Molar mass of H, O, S, Zn are 1, 16, 32, $65\,\mathrm{g \, mol^{-1}}$ respectively.

  1. $5.824\,\mathrm{L}$
  2. $7.428\,\mathrm{L}$
  3. $6.892\,\mathrm{L}$
  4. $8.375\,\mathrm{L}$

Answer: (c)

Solution

Mass of $\mathrm{H_2SO_4} = \frac{50 \times 1.3}{98} \times \frac{50}{100} = \frac{32.5}{98}$. $\mathrm{Zn(s) + H_2SO_4(aq) \rightarrow ZnSO_4(aq) + H_2(g)}$ $$\frac{20}{65} \frac{32.5}{98}$$ Moles of $\mathrm{H_2}$ formed $= \frac{20}{65}$. Volume of $\mathrm{H_2} = \frac{20}{65} \times 22.7 = 6.9 \, \mathrm{L}$.

Question 52

Chemistry · Structure of Atom · Single correct

Which of the following statement(s) is/are true? (A) If two orbitals have the same value of $(n + l)$, the orbital with lower value of $n$ will have lower energy. (B) Energies of the orbitals in the same subshell increase with increase in atomic number. (C) The size of $2p_x$ orbital is less than the size of $3p_x$ orbital. (D) Among $5f, 6s, 4d, 5p$ and $5d$ orbitals, none of the orbitals have 2 radial nodes. Choose the correct answer from the options given below:

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

Answer: (b)

Solution

(a) If two orbitals are having same value of $n + \ell$ then the orbital having lower value of $n$ will have lower energy. (b) Energies of the orbitals in the same subshell decreases with increase in atomic number. (c) Size of $2p_x < 3p_x$ (d) Orbital Radial node $(n - \ell - 1)$ 5f $\hspace{1cm}$ 1 6s $\hspace{1cm}$ 5 4d $\hspace{1cm}$ 1 5p $\hspace{1cm}$ 3 5d $\hspace{1cm}$ 2

Question 53

Chemistry · Chemical Bonding and Molecular Structure · Single correct

The covalent radii of atoms A and B are $r_A$ and $r_B$, respectively. The covalent bond length and total length of AB molecule are respectively:

  1. $(r_A + r_B), 2(r_A + r_B)$
  2. $\frac{1}{2}(r_A + r_B), (r_A + r_B)$
  3. $(r_A + r_B), (r_A + r_B)$
  4. $2(r_A + r_B), \frac{1}{2}(r_A + r_B)$

Answer: (a)

Solution

According to NCERT, bond length is $r_A + r_B$. Total length of molecule is $2(r_A + r_B)$.

Question 54

Chemistry · Thermodynamics · Single correct

Consider the following data for the reaction $$\mathrm{X_2(g) + Y_2(g) \rightleftharpoons 2XY(g)}$$ at 600 K. The $\Delta_r G^\Theta$ (in kJ mol$^{-1}$) for the reaction is:

  1. -21000
  2. -10
  3. -1000
  4. -9.012

Answer: (b)

Solution

Given $\Delta_r G = \Delta_r H - T \cdot \Delta_r S$. $\Delta_r H = (2 \times 42 - 80 - 8) = -4 \, \mathrm{kJ/mol}$. $\Delta_r S = (400 - 250 - 140) = +10 \, \mathrm{J/K \cdot mol}$. $\Delta_r G = -4000 - 600 \times (10)$. $= -10,000 \, \mathrm{J/mol} = -10 \, \mathrm{kJ/mol}$.

Question 55

Chemistry · Thermodynamics · Single correct

The correct order of molar heat capacities measured at 298 K and 1 bar is:

  1. Copper (s) > Bromine (l) > Helium (g)
  2. Bromine (l) > Copper (s) > Helium (g)
  3. Helium (g) > Bromine (l) > Copper (s)
  4. Helium (g) > Bromine (l) = Copper (s)

Answer: (b)

Solution

Generally molar heat capacity of liquids is greater than solids because liquids have greater degrees of freedom and thus can store more energy with lesser temperature rise. Also for monoatomic gases molar heat capacity is low as only translational degrees of freedom are present. Therefore, the order is: $$\mathrm{Br_2(\ell) > Cu(s) > He(g)}$$

Question 56

Chemistry · Equilibrium · Single correct

The reaction $\mathrm{A(g)\rightleftharpoons B(g)+C(g)}$ was initiated with the amount 'a' of $\mathrm{A(g)}$. At equilibrium it is found that the amount of $\mathrm{A(g)}$ remaining is $(a-x)$ at a total pressure of $p$. The equilibrium constant $K_p$ of the reaction can be calculated from the expression:

  1. $\frac{x^2}{a^2 + x^2} \times p$
  2. $\frac{x^2}{a^2 - x^2} \times p$
  3. $\frac{a + x^2}{x^2} \times p$
  4. $\frac{a - x^2}{x^2} \times p$

Answer: (b)

Solution

The reaction is given as $\mathrm{A(g) \rightleftharpoons B(g) + C(g)}$. At time $t = 0$, the concentrations are $a$, $0$, and $0$ respectively. At equilibrium $t_{eq}$, the concentrations are $a-x$, $x$, and $x$ respectively. The equilibrium constant $K_p$ is given by: $$ K_p = \frac{(x)(x)}{(a-x)} \times \left( \frac{P}{a+x} \right) $$ Simplifying, we have: $$ K_p = \frac{x^2 P}{a^2 - x^2} $$

Question 57

Chemistry · Electrochemistry · Single correct

One half cell in a voltaic cell is constructed by dipping silver rod in $\mathrm{AgNO_3}$ solution of unknown concentration, other half cell is Zn rod dipped in $1$ molar solution of $\mathrm{ZnSO_4}$. A voltage of $1.60\,\mathrm{V}$ is measured at $298\,\mathrm{K}$ for this cell. What is the concentration of $\mathrm{Ag^+}$ ions used in terms of $\log x$ ($x=[\mathrm{Ag^+}]$)? $E^\ominus_{\mathrm{Zn^{2+}/Zn}}=-0.76\,\mathrm{V}$, $E^\ominus_{\mathrm{Ag^+/Ag}}=+0.80\,\mathrm{V}$, $\dfrac{2.303RT}{F}=0.059\,\mathrm{V}$

  1. $\frac{2}{3.9}$
  2. $\frac{4}{5.9}$
  3. $\frac{2.9}{2}$
  4. $\frac{5.9}{4}$

Answer: (b)

Solution

Cell Reaction: $$\mathrm{Zn(s) + 2Ag^+(aq) \longrightarrow 2Ag(s) + Zn^{+2}(aq)}$$ $$E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log(Q)$$ $$1.6 = 1.56 - \frac{0.059}{2} \log \frac{1}{(\mathrm{Ag^+})^2}$$ $$0.04 = + 0.059 \log[\mathrm{Ag^+}]$$ $$\log[\mathrm{Ag^+}] = \frac{4}{5.9}$$

Question 58

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

Given below are two statements : Statement-I : The number of pairs among $[\mathrm{Al}_2\mathrm{O}_3, \mathrm{Cr}_2\mathrm{O}_3], [\mathrm{Cl}_2\mathrm{O}_7, \mathrm{Mn}_2\mathrm{O}_7], [\mathrm{Na}_2\mathrm{O}, \mathrm{V}_2\mathrm{O}_3]$ and $[\mathrm{CO}, \mathrm{N}_2\mathrm{O}]$ that contain oxides of same nature (acidic, basic, neutral or amphoteric) is 4. Statement-II : Among $\mathrm{Na}_2\mathrm{O}, \mathrm{Al}_2\mathrm{O}_3, \mathrm{CO}$ and $\mathrm{Cl}_2\mathrm{O}_7$, the most basic and acidic oxides are $\mathrm{Na}_2\mathrm{O}$ and $\mathrm{Cl}_2\mathrm{O}_7$, respectively. In the light of the above statements, choose the correct answer from the options given below :

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

Answer: (a)

Solution

Statement I: $\mathrm{Al_2O_3}$, $\mathrm{Cr_2O_3}$: Amphoteric $\mathrm{CO}$, $\mathrm{N_2O}$: Neutral $\mathrm{Na_2O}$, $\mathrm{V_2O_3}$: Basic $\mathrm{Cl_2O_7}$, $\mathrm{Mn_2O_7}$: Acidic Statement II: Most basic oxide: $\mathrm{Na_2O}$ Most acidic oxide: $\mathrm{Cl_2O_7}$

Question 59

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

Given below are two statements: Statement-I: Aluminium upon reaction with NaOH forms $[\mathrm{Al(OH)}_6]^{3-}$ ion. Statement-II: The geometry of $\mathrm{ICl}_4^-$, $\mathrm{ClO}_3^-$ and $\mathrm{IBr}_2^-$ is square planar, pyramidal and linear respectively. In the light of the above statements, choose the correct answer from the options given below:

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

Answer: (d)

Solution

Statement I: $\mathrm{Al}$ + $\mathrm{NaOH}$ (excess) $\rightarrow \mathrm{Na[Al(OH)_4]}$ + $\mathrm{H_2(g)}$ Statement II: $\mathrm{ClO_3^{\Theta}}$ is sp^3 and pyramidal $\mathrm{ICl_4^{\Theta}}$ is sp^3d^2 and square planar $\mathrm{IBr_2^{\Theta}}$ is sp^3d and linear.

Question 60

Chemistry · Co-ordination Compounds · Single correct

Given below are two statements: Statement-I: Presence of large number of unpaired electrons in transition metal atoms results in higher enthalpies of their atomisation. Statement-II: $d_{xy}=d_{xz}=d_{yz} In the light of the above statements, choose the correct answer from the options given below:

  1. Both Statement-I and Statement-II are correct.
  2. Both Statement-I and Statement-II are incorrect.
  3. Statement-I is correct but Statement-II is incorrect.
  4. Statement-I is incorrect but Statement-II is correct.

Answer: (a)

Solution

Statement I: More are the number of unpaired electrons, stronger is metal-metal bonding thus higher enthalpy of atomisation. Statement II: $[\mathrm{Fe(H_2O)_6}]^{3+}$ is octahedral complex. Thus, $$\left[ t_{2g} \left( d_{xy} = d_{yz} = d_{zx} \right) < e_g \left( d_{x^2-y^2} = d_{z^2} \right) \right]$$ $[\mathrm{NiCl_4}]^{2-}$ is tetrahedral complex. Thus, $$\left[ e \left( d_{x^2-y^2} = d_{z^2} \right) < t_2 \left( d_{xy} = d_{yz} = d_{zx} \right) \right]$$

Question 61

Chemistry · Co-ordination Compounds · Single correct

Identify the correct statements from the following: (A) $[\mathrm{Fe(CO)_5}]$ is the most stable complex among $[\mathrm{Fe(OH)_6}]^{3-}$, $[\mathrm{Fe(C_2O_4)_3}]^{3-}$ and $[\mathrm{Fe(SCN)_6}]^{3-}$. (B) The stability of $[\mathrm{Cu(NH_3)_4}]^{2+}$ is greater than that of $[\mathrm{Cu(en)_2}]^{2+}$. (C) The hybridization of Fe in $K_3[\mathrm{Fe(CN)_6}]$ is $d^2sp^3$. (D) $[\mathrm{Fe(NO_2)_6}]^{3-}$ exhibits linkage isomerism. (E) $\mathrm{NO_2^-}$ and $\mathrm{SCN^-}$ ligands are NOT ambidentate ligands. Choose the correct answer from the options given below:

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

Answer: (c)

Solution

(A) $[\mathrm{Fe(C_2O_4)_3}]^{3-}$ has oxalate ion as ligand which leads to chelation thus maximum stability among given complexes. (B) Stability: $[\mathrm{Cu(NH_3)_4}]^{2+} < [\mathrm{Cu(en)_2}]^{2+}$; due to higher CFSE and chelation. (C) $\mathrm{K_4[Fe(CN)_6]}$ $$\Rightarrow \mathrm{Fe^{2+} = [Ar]3d^6}$$ In presence of SFL the electronic configuration of $\mathrm{Fe^{+2} \left[t_{2g}^6, e_g^0\right]}$, thus $d^2sp^3$ hybridized. (D) $[\mathrm{Fe(NO_2)_3Cl_3}]^{3-}$ has $\mathrm{NO_2^-}$ as ambidentate ligand thus can show linkage isomerism. (E) $\mathrm{NO_2^-}$ and $\mathrm{SCN^-}$ both are ambidentate ligands.

Question 62

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

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

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

Answer: (c)

Solution

It is a theory based question.

Question 63

Chemistry · Hydrocarbons · Single correct

IUPAC names of some alkenes are given below. Find out the correct stability order. A. 2-Methylbut-2-ene B. cis-But-2-ene C. 2,3-Dimethylbut-2-ene D. Prop-1-ene Choose the correct answer from the options given below:

  1. C > A > B > D
  2. C > A > D > B
  3. B > D > A > C
  4. A > B > C > D

Answer: (a)

Solution

The stability of alkenes is proportional to the number of $\alpha$ hydrogens. The order of stability is shown as follows: $$\alpha H = 12 > \alpha H = 9 > \alpha H = 6 > \alpha H = 3.$$

Question 64

Chemistry · Hydrocarbons · Single correct

Identify the correct IUPAC name of hydrocarbon (x) containing three primary carbon atoms and with molar mass $72 \, \mathrm{g \, mol^{-1}}$.

  1. 1,1-Dimethylcyclopropane
  2. 2,2-Dimethylpropane
  3. 2-Methylbutane
  4. n-pentane

Answer: (c)

Solution

(A) (1^$\circ \mathrm{C}$ = 2) (B) (1^$\circ \mathrm{C}$ = 4) (C) (1^$\circ \mathrm{C}$ = 3) (B) (1^$\circ \mathrm{C}$ = 2) Mol. mass = 72

Question 65

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Complete the following reaction sequence and give the name of major product $P$.

  1. 2-Chloropropanoic acid
  2. 3-Chloropropanoic acid
  3. 1-Chloropropane
  4. 2-Chloropropane

Answer: (a)

Solution

The reaction starts with $\mathrm{CH_3-CH=C\equiv N}$. Under hydrolysis of cyanide with $\mathrm{OH^-}/\mathrm{H_2O}$, it forms $\mathrm{CH_3-CH_2-COO^-}$. Upon treatment with $\mathrm{H_3O^+}$, it converts to $\mathrm{CH_3-CH_2-COOH}$. In the HVZ reaction, using (i) Red P + $\mathrm{Cl_2}$ and (ii) $\mathrm{H_2O}$, it forms $\mathrm{CH_3-CH-COOH}$ with a $\mathrm{Cl}$ substituent.

Question 66

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Given below are two statements: In the light of the above statements, choose the correct answer from the options given

Answer: d

Question 67

Chemistry · Amines · Single correct

Given below are two statements: Statement-I: Heating benzamide with bromine in an ethanolic solution of sodium hydroxide will give benzylamine. Statement-II: Nitration of aniline with $\mathrm{HNO_3/H_2SO_4}$ at $288\,\mathrm{K}$ produces $m$-nitroaniline in higher amount than $o$-nitroaniline (pH adjusted). In the light of the above statements, choose the correct answer from the options given below:

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

Answer: (d)

Solution

Statement-I: The reaction of benzamide with $\mathrm{Br_2 + NaOH}$ in ethanolic solution under heat does not give benzylamine but gives aniline. Hence Statement-I is incorrect. Statement-II: Nitration of aniline with $\mathrm{HNO_3 + H_2SO_4}$ at $288 \, \mathrm{K}$ gives a mixture of products: para-nitroaniline (51$\%$), meta-nitroaniline (47$\%$), and ortho-nitroaniline (2$\%$). Statement-II is correct.

Question 68

Chemistry · Biomolecules · Single correct

Identify the incorrect statement about tertiary structure of proteins.

  1. They can be fibrous or globular in structure.
  2. The main forces that stabilize the structure are hydrogen bonding, disulphide links, van der Waals and electrostatic forces of attraction.
  3. The structure remains intact when exposed to pH changes.
  4. A linear polypeptide chain will convert to a secondary structure and then further folding of the secondary structure will convert to tertiary structure.

Answer: (c)

Solution

On changing pH, the tertiary structure of protein is disrupted and amino acids remain intact only by peptide linkage.

Question 69

Chemistry · Biomolecules · Single correct

Given below are two statements: Statement-I: are two anomers of D-(+)-glucose. Statement-II: The open chain forms of D-glucose and D-fructose contain three similar chiral carbons at $C_3$, $C_4$ and $C_5$. In the light of the above statements, choose the correct answer from the options given below:

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

Answer: (a)

Solution

α-D-Glucose and β-D-Glucose are anomers. At C-3, C-4 and C-5 are identical configuration in glucose and fructose.

Question 70

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

A paper dipped in a dil. $H_2SO_4$ solution of 'X' upon treatment with $SO_2$ gas turns into green. The compound 'X' is :

  1. KI-starch
  2. KMnO_4
  3. Pb(CH_3COO)_2
  4. K_2Cr_2O_7

Answer: (d)

Solution

SO_2 gas acts as reducing agent. $$\mathrm{K_2Cr_2O_7 + SO_2 \xrightarrow{H_2SO_4} Cr_2(SO_4)_3 + SO_4^{2-}}$$ Green solution

Question 71

Chemistry · Co-ordination Compounds · Numerical

The total number of unpaired electrons present in the $d^3$, $d^4$ (low spin) $d^5$ (high spin), $d^6$ (high spin) and $d^7$ (low spin) octahedral complex systems is $\ldots$.

Answer: 15

Solution

Given $\mathrm{d}^3 \rightarrow \left[ t_{2g}^3, e_g^0 \right]$, 3 unpaired electrons. $\mathrm{d}^4$ (low spin) $\rightarrow \left[ t_{2g}^4, e_g^0 \right]$, 2 unpaired electrons. $\mathrm{d}^5$ (high spin) $\rightarrow \left[ t_{2g}^3, e_g^2 \right]$, 5 unpaired electrons. $\mathrm{d}^6$ (high spin) $\rightarrow \left[ t_{2g}^4, e_g^2 \right]$, 4 unpaired electrons. $\mathrm{d}^7$ (low spin) $\rightarrow \left[ t_{2g}^6, e_g^1 \right]$, 1 unpaired electrons.

Question 72

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

RMgI when treated with ice cold water liberated a gas which occupied $1.4 \, \mathrm{dm^3/g}$ at STP. The gas produced is further reacted with iodine in presence of $\mathrm{HIO_3}$ to give compound (X). Compound (X) in presence of Na and dry ether produced compound (Y). Molar mass of compound (Y) is $\ldots \, \mathrm{g \, mol^{-1}}$. (Nearest integer)

Answer: 30

Solution

R Mg I + Ice cold H_2O $\rightarrow$ R--H + Mg $\begin{array}{c}$ I $\\$ OH $\end{array}$ Gas Vol. = $1.4\,\mathrm{dm^3/gm}$ at STP Hence according to this mol. Mass of gas is = 16 Hence liberated gas is = $\mathrm{CH_4}$ (Methane) $\mathrm{CH_4}$ + $\mathrm{I_2} \xrightarrow{\mathrm{HIO_3}} \mathrm{CH_3I}$ + $\mathrm{HI}$ $\boxed{X}$ $\mathrm{CH_3}$ + $\mathrm{I} \xrightarrow{Na/Dry ether} \mathrm{CH_3}$ - $\mathrm{CH_3}$ $\boxed{X} \boxed{Y}$ Molar mass = 30

Question 73

Chemistry · Solutions · Numerical

$20\,\mathrm{g}$ hemoglobin in a $1\,\mathrm{L}$ aqueous solution (A) at $300\,\mathrm{K}$ is separated from pure water by semi permeable membrane. At equilibrium the height of solution in a tube dipped in a solution (A) is found to be $80.0\,\mathrm{mm}$ higher than the tube dipped in water. The molar mass of hemoglobin is $\ldots \mathrm{kg} \mathrm{mol}^{-1}$. (Nearest integer) (Given : g = $10\,\mathrm{ms}^{-2}$, R = $8.3\,\mathrm{kPa} \mathrm{dm}^{3} \mathrm{K}^{-1} \mathrm{mol}^{-1}$, density of solution = $1000\,\mathrm{kg} \mathrm{m}^{-3}$)

Answer: 62

Solution

Osmotic pressure, $\pi = \rho g h$ $$= 1000 \times 10 \times 80 \times 10^{-3}$$ $$= 800 \, \mathrm{Pa} \ldots (1)$$ Let molar mass of haemoglobin $= M \, \mathrm{g/mol}$. Concentration of haemoglobin $= \frac{20/M}{10^{-3}} \left( \mathrm{mol/m^3} \right)$. $\pi = CRT$ (S.I. units) $$\Rightarrow \pi = \left[ \frac{20}{10^{-3} M} \right] \times 8.3 \times 300$$ $$= 800 [From (1)]$$ Solving we get: $$M = \frac{20 \times 8.3 \times 300}{0.8} \, \mathrm{g/mol}$$ $$= 62.25 \, \mathrm{kg \, mol^{-1}}$$ Ans = 62

Question 74

Chemistry · Electrochemistry · Numerical

At $298\,\mathrm{K}$, the molar conductivity of $x\%$ (w/w) MX solution (aqueous) is $123.5 \, \mathrm{S \, cm^2 \, mol^{-1}}$. The conductance of same solution is $1.9 \times 10^{-3} \, \mathrm{S}$. The value of $x$ is $\ldots \times 10^{-2}$. (Given : cell constant = $1.3 \, \mathrm{cm^{-1}}$; molar mass of MX is $75 \, \mathrm{g \, mol^{-1}}$, density of aqueous solution of MX at 298 K is $1.0 \, \mathrm{g \, mL^{-1}}$)

Answer: 15

Solution

Let molarity of MX solution = y M $$\Rightarrow \Lambda_m = c(\ell/A) \times \frac{1000}{y}$$ $$\Rightarrow 123.5 = 1.9 \times 10^{-3} \times 1.3 \times \frac{1000}{y}$$ $$\Rightarrow y = 0.02 \, \mathrm{M}$$ Let molar mass of MX = M g mol^{-1} 0.2 mol MX is dissolve in 1L solution (0.02 $\times$ 75) gm MX is dissolved in 1000 g solution $$\% \, w/w = x\% = \frac{0.02 \times 75}{1000} \times 100$$ = 0.15$\%$ Ans = 15

Question 75

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

For a reaction $\mathrm{A\rightarrow P}$ at $T\,\mathrm{K}$, the half life $(t_{1/2})$ is plotted as a function of initial concentration $[A]_0$ of A as given below: The value of x in the given figure is $\ldots \mathrm{s}$ (Nearest integer)

Answer: 90

Solution

From the plot, $t_{1/2} \propto [A]_0$ and $\propto [A_0]^{1-N}$. $1 - N = 1$ $N = 0$ (order) $$t_{1/2} = \frac{A_0}{2K}$$ $$240 = \frac{4 \times 10^{-3}}{2K} ....(1)$$ $$x = \frac{1.5 \times 10^{-3}}{2K} ....(2)$$ Divide (1) / (2) $$\frac{240}{x} = \frac{4}{1.5}$$ $$x = 90$$