JEE-Main-2025-Question-Paper-with-Solution-23-Jan-Shift-1

Question 1

1. The value of $ \displaystyle \int_{e^2}^{e^4} \frac{1}{x}
\left( \dfrac{e^{\left( (\log_e x)^2 + 1 \right)^{-1}}}
{e^{\left( (\log_e x)^2 + 1 \right)^{-1}} + e^{\left( (6 – \log_e x)^2 + 1 \right)^{-1}}} \right) dx $ is

(1) $ \log_e 2 $

(2) $ 2 $

(3) $ 1 $

(4) $ e^2 $

▶️ Answer/Explanation

Answer: (3)

Solution:

$\text{Ans. } (3)$

$\text{Sol. } \text{Let } \ln x = t \Rightarrow \frac{dx}{x} = dt$

$ I = \displaystyle\int_{2}^{4} \frac{e^{\frac{1}{1 + t^2}}}{e^{\frac{1}{1 + t^2}} + e^{\frac{1}{1 + (6 – t)^2}}} \, dt $

$ I = \displaystyle\int_{2}^{4} \frac{1}{1 + e^{\frac{1}{1 + (6 – t)^2} – \frac{1}{1 + t^2}}} \, dt $

$ I = \displaystyle\int_{2}^{4} \frac{1}{e^{\frac{1}{1 + (6 – t)^2}} + e^{\frac{1}{1 + t^2}}} \, dt $

$\text{Now, let’s replace } t \text{ by } (6 – t) \text{ in the integral.}$

$\therefore 2I = \displaystyle\int_{2}^{4} dt = (t)_{2}^{4} = 4 – 2 = 2$

$\Rightarrow I = 1$

Question 2

Let $ I(x) = \int \frac{dx}{(x – 11)^{\frac{11}{13}} (x + 15)^{\frac{15}{13}}} $.

If $ I(37) – I(24) = \frac{1}{4} \left( \frac{1}{b^{\frac{11}{13}}} – \frac{1}{c^{\frac{11}{13}}} \right), \, b, c \in \mathbb{N} $,
then $ 3(b + c) $ is equal to

(1) \( 40 \

(2) $ 39 $

(3) $ 22 $

(4) $ 26 $

▶️ Answer/Explanation

$\text{Ans. } (2)$

$\text{Sol. } I(x) = \int \frac{dx}{(x – 11)^{\frac{11}{13}} (x + 15)^{\frac{15}{13}}}$

$\text{Put } \frac{x – 11}{x + 15} = t \Rightarrow \frac{26}{(x + 15)^2} dx = dt$

$\Rightarrow I(x) = \frac{1}{26} \int \frac{dt}{t^{\frac{11}{13}}} = \frac{1}{26} \cdot \frac{t^{\frac{2}{13}}}{\frac{2}{13}}$

$\therefore I(x) = \frac{1}{4} \left( \frac{x – 11}{x + 15} \right)^{\frac{2}{13}} + C$

$I(37) – I(24) = \frac{1}{4} \left[ \left( \frac{26}{52} \right)^{\frac{2}{13}} – \left( \frac{13}{39} \right)^{\frac{2}{13}} \right]$

$= \frac{1}{4} \left( \frac{1}{2^{\frac{2}{13}}} – \frac{1}{3^{\frac{2}{13}}} \right)$

$\therefore b = 4, \, c = 9$

$\Rightarrow 3(b + c) = 3(4 + 9) = 39$

Question 3

If the function

$
f(x) =
\begin{cases}
\dfrac{2}{x} \left[ \sin{(k_1 + 1)x} + \sin{(k_2 – 1)x} \right], & x < 0 \\(6pt]
4, & x = 0 \\(6pt]
\dfrac{2}{x} \log_e{\left( \dfrac{2 + k_1 x}{2 + k_2 x} \right)}, & x > 0
\end{cases}
$

is continuous at $ x = 0 $, then $ k_1^2 + k_2^2 $ is equal to

(1) $ 8 $

(2) $ 20 $

(3) $ 5 $

(4) $ 10 $

▶️ Answer/Explanation

$\text{Ans. } (4)$

$\text{Sol. }$

\(
\lim_{x \to 0^-} \frac{2}{x} \left[ \sin{(k_1 + 1)x} + \sin{(k_2 – 1)x} \right] = 4
\(

$\Rightarrow 2(k_1 + 1) + 2(k_2 – 1) = 4$

$\Rightarrow k_1 + k_2 = 2$

\(
\lim_{x \to 0^+} \frac{2}{x} \ln{\left( \frac{2 + k_1 x}{2 + k_2 x} \right)} = 4
\(

$\Rightarrow \lim_{x \to 0} 2 \ln{\left( \frac{1 + \frac{k_1 x}{2}}{1 + \frac{k_2 x}{2}} \right)} \cdot \frac{1}{x} = 4$

Using $\ln(1 + a) \approx a$ for small $a$,

$\Rightarrow 2 \times \frac{1}{x} \times \left( \frac{k_1 x}{2} – \frac{k_2 x}{2} \right) = 4$

$\Rightarrow k_1 – k_2 = 2$

Now, solving the two equations:

\(
\begin{cases}
k_1 + k_2 = 2 \\
k_1 – k_2 = 2
\end{cases}
\Rightarrow
k_1 = 2, \, k_2 = 0
\(

$\therefore k_1^2 + k_2^2 = 2^2 + 0^2 = 4$

However, checking with correct substitution in the logarithmic condition gives
$ k_1 – k_2 = 4 $, not $2$, hence:

\(
\begin{cases}
k_1 + k_2 = 2 \\
k_1 – k_2 = 4
\end{cases}
\Rightarrow
k_1 = 3, \, k_2 = -1
\(

$\therefore k_1^2 + k_2^2 = 3^2 + (-1)^2 = 9 + 1 = 10$

Question 4

4. If the line $ 3x – 2y + 12 = 0 $ intersects the parabola $ 4y = 3x^2 $ at the points A and B, then at the vertex of the parabola, the line segment AB subtends an angle equal to

(1) $ \tan^{-1}\!\left(\frac{1}{9}\right) $ \quad
(2) $ \pi – \tan^{-1}\!\left(\frac{3}{2}\right) $ \quad
(3) $ \tan^{-1}\!\left(\frac{1}{4}\right) $ \quad
(4) $ \tan^{-1}\!\left(\frac{1}{9/7}\right) $

▶️ Answer/Explanation

Ans. (4)

Sol.

$ 3x – 2y + 12 = 0 $
$ 4y = 3x^2 $

$ \therefore 2(3x + 12) = 3x^2 $

$ \Rightarrow x^2 – 2x – 8 = 0 $

$ \Rightarrow x = -2, 4 $

$ m_{OA} = -\frac{3}{2}, \; m_{OB} = 3 $

$ \tan\theta = \left| \frac{3 – (-\frac{3}{2})}{1 + (-\frac{3}{2})(3)} \right| = \frac{9}{7} $

$ \therefore \theta = \tan^{-1}\!\left(\frac{9}{7}\right) $

Question 5

5. Let a curve $ y = f(x) $ pass through the points $ (0, 5) $ and $ (\log_e 2, k) $. If the curve satisfies the differential equation
$ 2(3 + y)e^{2x}dx – (7 + e^{2x})dy = 0 $, then $ k $ is equal to

(1) $16 $

(2) $8$

(3) $32 $

(4) $4$

▶️ Answer/Explanation

Ans. (2)

Sol.
$ 2(3 + y)e^{2x}dx = (7 + e^{2x})dy $

$ \Rightarrow \frac{dy}{dx} = \frac{2(3 + y)e^{2x}}{7 + e^{2x}} $

$ \Rightarrow \frac{dy}{dx} – \frac{2e^{2x}}{7 + e^{2x}}y = \frac{6e^{2x}}{7 + e^{2x}} $

Integrating factor $ I.F. = e^{-\int \frac{2e^{2x}}{7 + e^{2x}}dx} = e^{-\ln(7 + e^{2x})} = \frac{1}{7 + e^{2x}} $

$ \therefore y \cdot \frac{1}{7 + e^{2x}} = \int \frac{6e^{2x}}{(7 + e^{2x})^2} dx $

Put $ 7 + e^{2x} = t \Rightarrow dt = 2e^{2x} dx $

$ \Rightarrow \int \frac{6e^{2x}}{(7 + e^{2x})^2} dx = 3 \int \frac{dt}{t^2} = -\frac{3}{t} + C $

$ \therefore y = -3(7 + e^{2x}) + C(7 + e^{2x}) $

Using $ (0, 5) \Rightarrow 5 = -3(8) + 8C \Rightarrow 5 = -24 + 8C \Rightarrow C = \frac{29}{8} $

$ \therefore y = -3 + 7 + e^{2x} = e^{2x} + 4 $

$ \therefore k = 8 $

Question 6

Let $f(x) = \log_e x$ and $g(x) = \frac{x^4 – 2x^3 + 3x^2 – 2x + 2}{2x^2 – 2x + 1}$. Then the domain of $fog$ is

(1) $\mathbb{R}$
(2) $(0, \infty)$
(3) $[0, \infty)$
(4) $[1, \infty)$

▶️ Answer/Explanation

Ans. (1)

Sol. $f(x) = \ln x$
$g(x) = \frac{x^4 – 2x^3 + 3x^2 – 2x + 2}{2x^2 – 2x + 1}$
$D_g \in \mathbb{R}$
$D_f \in (0, \infty)$
For $D_{fog} \Rightarrow g(x) > 0$
$\Rightarrow x^4 – 2x^3 + 3x^2 – 2x + 2 > 0$
Clearly $x < 0$ satisfies which are included in option (1) only.

Question 7

Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC divides the arc AC such that
$ \frac{\text{length of arc AB}}{\text{length of arc BC}} = \frac{1}{5} $,
and $ \overrightarrow{OC} = \alpha \overrightarrow{OA} + \beta \overrightarrow{OB} $,
then $ (2\alpha – \beta) = $

(1) $ 2\sqrt{3} – 3 $

(2) $ 2\sqrt{3} $

(3) $ 5\sqrt{3} $

(4) $ 2 + \sqrt{3} $

▶️ Answer/Explanation

Ans. (1)

Sol.

Let $ \angle AOB = 15^\circ $, $ \angle BOC = 75^\circ $.

Using vector relations:

$ \overrightarrow{OC} = \alpha \overrightarrow{OA} + \beta \overrightarrow{OB} $

$ \overrightarrow{a} \cdot \overrightarrow{c} = \alpha a^2 + \beta (\overrightarrow{a} \cdot \overrightarrow{b}) \Rightarrow 0 = \alpha + \beta \cos 15^\circ $ …(1)

$ \overrightarrow{b} \cdot \overrightarrow{c} = \alpha (\overrightarrow{b} \cdot \overrightarrow{a}) + \beta b^2 \Rightarrow \cos 75^\circ = \alpha \cos 15^\circ + \beta $ …(2)

From (1) and (2),

$ \cos 75^\circ = -\beta \cos^2 15^\circ + \beta $

$ \Rightarrow \beta = \frac{\sin 15^\circ}{\sin 15^\circ \cos 15^\circ} = 2 – \sqrt{3} $

$ \alpha = \frac{3 – \sqrt{3}}{2} $

$ \therefore (2\alpha – \beta) = 2\sqrt{3} – 3 $

Question 8

If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to

(1) -1200
(2) -1080
(3) -1020
(4) -120

▶️ Answer/Explanation

Ans. (2)

Sol. $a = 3$
$S_4 = \frac{1}{5} (S_8 – S_4)$
$\Rightarrow 5S_4 = S_8 – S_4$
$\Rightarrow 6S_4 = S_8$
$\Rightarrow \frac{4}{2} [2 \cdot 3 + (4-1)d] = \frac{8}{2} [2 \cdot 3 + (8-1)d]$
$\Rightarrow 12(6 + 3d) = 4(6 + 7d)$
$\Rightarrow 18 + 9d = 6 + 7d$
$\Rightarrow d = -6$
$S_{20} = \frac{20}{2} [2 \cdot 3 + (20-1)(-6)]$
$= 10 [6 – 114]$
$= -1080$

Question 9

9. Let P be the foot of the perpendicular from the point Q(10, –3, –1) on the line
$ \frac{x – 3}{7} = \frac{y – 2}{-1} = \frac{z – 1}{-2} $.
Then the area of the right angled triangle PQR, where R is the point (3, –2, 1), is

(1) $ 9\sqrt{15} $

(2) $30 $

(3) $ 8\sqrt{15} $

(4) $ 3\sqrt{30} $

▶️ Answer/Explanation

Ans. (4)

Sol.

Let point on line be $ P(7\lambda + 3, -\lambda + 2, -2\lambda + 1) $.

Direction ratios of $ QP \Rightarrow (7\lambda – 7, -\lambda + 5, -2\lambda – 2) $

Since $ QP \perp \text{line} \Rightarrow (7\lambda – 7)(7) + (-\lambda + 5)(-1) + (-2\lambda – 2)(-2) = 0 $

$ 54\lambda – 54 = 0 \Rightarrow \lambda = 1 $

$ \therefore P = (10, 1, -3) $

$ \overrightarrow{PQ} = 4\hat{j} + 2\hat{k} $, $ \overrightarrow{PR} = -7\hat{i} – 3\hat{j} + 4\hat{k} $

Area $ = \frac{1}{2} |\overrightarrow{PQ} \times \overrightarrow{PR}| = \frac{1}{2} \sqrt{(0)^2 + (4)^2 + (2)^2} = 3\sqrt{30} $

Question 10

10. Let $ \frac{z – i}{2z + i} = 3 $, $ z \in \mathbb{C} $, be the equation of a circle with center at C. If the area of the triangle whose vertices are at the points (0, 0), C and (α, 0) is 11 square units, then $ \alpha^2 $ equals

(1) $100$

(2) $50$

(3) $ \frac{121}{25} $

(4) $ \frac{81}{25} $

▶️ Answer/Explanation

Ans. (1)

Sol.
$ \frac{z – i}{2z + i} = 3 \Rightarrow |z – i| = \frac{2}{3}|z + \frac{i}{2}| $

$ 3|x – iy – i| = 2|x – iy + \frac{i}{2}| $

$ 9(x^2 + (y + 1)^2) = 4(x^2 + (y – \frac{1}{3})^2) $

$ 9x^2 + 9y^2 + 18y + 9 = 4x^2 + 4y^2 – \frac{8y}{3} + \frac{4}{9} $

$ 5x^2 + 5y^2 + 22y + 8 = 0 \Rightarrow x^2 + y^2 + \frac{22}{5}y + \frac{8}{5} = 0 $

Center $ C(0, -\frac{11}{5}) $

Area $ = \frac{1}{2} \alpha \times \frac{11}{5} = 11 \Rightarrow \alpha = 10 \Rightarrow \alpha^2 = 100 $

Question 11

Let $R = \{(1, 2), (2, 3), (3,3)\}$ be a relation defined on the set $\{1, 2, 3, 4\}$. Then the minimum number of elements, needed to be added in R so the R becomes an equivalence relation, is :

(1) 10
(2) 8
(3) 9
(4) 7

▶️ Answer/Explanation

Ans. (4)

Sol. $A = \{1, 2, 3, 4\}$
For relation to be reflexive
$R = \{(1,2), (2, 3), (3,3)\}$
Minimum elements added will be
$(1,1), (2,2), (4,4) (2,1) (3,2) (3,2) (3,1) (1,3)$
$\therefore$ Minimum number of elements = 7
Option : (4)

Question 12

The number of words, which can be formed using all the letters of the word “DAUGHTER”, so that all the vowels never come together, is

(1) 34000
(2) 37000
(3) 36000
(4) 35000

▶️ Answer/Explanation

Ans. (3)

Sol. DAUGHTER
Total words = $8!$
Total words in which vowels are together = $6! \times 3!$
words in which all vowels are not together
$= 8! – 6! \times 3!$
$= 6! [56 – 6]$
$= 720 \times 50$
$= 36000$
Ans.(3)

Question 13

Let the area of a $\Delta PQR$ with vertices $P(5, 4)$, $Q(-2, 4)$ and $R(a, b)$ be 35 square units. If its orthocenter and centroid are $O\left(\frac{14}{3}, 5\right)$ and $C(c, d)$ respectively, then $c + 2d$ is equal to

(1) $\frac{7}{3}$
(2) 3
(3) 2
(4) $\frac{8}{3}$

▶️ Answer/Explanation

Ans. (2)

Sol.

Equation of lines $QR = 5x + 2y + 2 = 0$
Equation of lines $PR = 10x – 3y – 38 = 0$
$\therefore$ Point $R (2, -6)$
Centroid = $\left(\frac{5-2+2}{3}, \frac{4+4-6}{3}\right)$
$= \left(\frac{5}{3}, \frac{2}{3}\right)$
$c + 2d = \frac{5}{3} + 2 \cdot \frac{2}{3} = 3$

Question 14

14. If $ \frac{\pi}{3} \le x \le \frac{\pi}{2} $, then
$ \cos^{-1}\!\left( \frac{12}{13}\cos x + \frac{5}{13}\sin x \right) $ is equal to

(1) $ x – \tan^{-1}\!\frac{4}{3} $

(2) $ x – \tan^{-1}\!\frac{5}{12} $

(3) $ x + \tan^{-1}\!\frac{4}{5} $

(4) $ x + \tan^{-1}\!\frac{5}{12} $

▶️ Answer/Explanation

Ans. (2)

Sol.
For $ \frac{\pi}{3} \le x \le \frac{\pi}{2} $,

$ \cos^{-1}\!\left( \frac{12}{13}\cos x + \frac{5}{13}\sin x \right) = \cos^{-1}\!\left( \cos x \cos \alpha + \sin x \sin \alpha \right) $

$ = \cos^{-1}(\cos(x – \alpha)) = x – \alpha $

$ \text{where } \alpha = \tan^{-1}\!\frac{5}{12} $

$ \therefore \text{ Required value } = x – \tan^{-1}\!\frac{5}{12} $

Question 15

The value of $(\sin 70^\circ)(\cot 10^\circ \cot 70^\circ – 1)$ is

(1) 1
(2) 0
(3) $\frac{3}{2}$
(4) $\frac{2}{3}$

▶️ Answer/Explanation

Ans. (1)

Sol. $\sin 70^\circ (\cot 10^\circ \cot 70^\circ – 1)$
$\Rightarrow \frac{\cos 80^\circ}{\sin 10^\circ} = 1$

Question 16

Marks obtains by all the students of class 12 are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18, then the total number of students is

(1) 48
(2) 44
(3) 40
(4) 52

▶️ Answer/Explanation

Ans. (2)

Sol.

Median $ = l + \frac{\left(\frac{N}{2} – F\right)}{f} \times h $

$ 14 = 12 + \frac{\left(\frac{N}{2} – 18\right)}{12} \times 6 $

$ \Rightarrow 2 = \frac{N – 36}{4} \Rightarrow N – 36 = 8 \Rightarrow N = 44 $

Question 17

Let the position vectors of the vertices A, B and C of a tetrahedron ABCD be $\hat{i} + 2\hat{j} + \hat{k}$, $\hat{i} + 3\hat{j} – 2\hat{k}$ and $2\hat{i} – \hat{j} + \hat{k}$ respectively. The altitude from the vertex D to the opposite face ABC meets the median line segment through A of the triangle ABC at the point E. If the length of AD is $\frac{\sqrt{110}}{3}$ and the volume of the tetrahedron is $\frac{\sqrt{805}}{6\sqrt{2}}$, then the position vector of E is

(1) $\frac{1}{2} (\hat{i} + 4\hat{j} + 7\hat{k})$
(2) $\frac{1}{12} (7\hat{i} + 4\hat{j} – 3\hat{k})$
(3) $\frac{1}{6} (12\hat{i} + 12\hat{j} + \hat{k})$
(4) $\frac{1}{6} (7\hat{i} + 12\hat{j} + \hat{k})$

▶️ Answer/Explanation

Ans. (4)

Sol.

$\text{Area of }\triangle ABC = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}$

$= \frac{1}{2} |\hat{i} + 3\hat{j} + \hat{k}| = \frac{1}{2} \sqrt{35}$

$\text{volume of tetrahedron}$

$= \frac{1}{3} \times \text{Base area} \times h = \frac{\sqrt{805}}{6\sqrt{2}}$

$\frac{1}{3} \times \frac{1}{2} \sqrt{35} \times h = \frac{\sqrt{805}}{6\sqrt{2}}$

$h = \frac{\sqrt{23}}{\sqrt{2}}$

$\text{AE}^2 = \text{AD}^2 – \text{DE}^2 = \frac{13}{18} \quad : \quad \text{AE} = \frac{\sqrt{13}}{\sqrt{18}}$

$\overrightarrow{AE} = |\overrightarrow{AE}| \left( \frac{\hat{i} – 5\hat{k}}{\sqrt{26}} \right)$

$= \frac{\sqrt{13}}{\sqrt{18}} \cdot \frac{\hat{i} – 5\hat{k}}{\sqrt{26}}$

$= \frac{\sqrt{13} (\hat{i} – 5\hat{k})}{\sqrt{18} \cdot \sqrt{26}} = \frac{\hat{i} – 5\hat{k}}{6}$

$\text{P.V. of E} = \frac{\hat{i} – 5\hat{k}}{6} + \hat{i} + 2\hat{j} + \hat{k} = \frac{1}{6} (7\hat{i} + 12\hat{j} + \hat{k})$

Question 18

If A, B and $(\text{adj}(A^{-1}) + \text{adj}(B^{-1}))$ are non-singular matrices of same order, then the inverse of $A(\text{adj}(A^{-1}) + \text{adj}(B^{-1}))^{-1}B$, is equal to

(1) $AB^{-1} + A^{-1}B$
(2) $\text{adj}(B^{-1}) + \text{adj}(A^{-1})$
(3) $AB^{-1} (\text{adj}(B) + \text{adj}(A))$
(4) $\frac{1}{AB} (A^{-1}B + BA^{-1})$

▶️ Answer/Explanation

Ans. (3)

Sol.

$\left[ A(\text{adj}(A^{-1}) + \text{adj}(B^{-1})) \right] B^{-1}$

$B^{-1}(\text{adj}(A^{-1}) + \text{adj}(B^{-1})) A^{-1}$

$B^{-1}\text{adj}(A^{-1}) A^{-1} + B^{-1}(\text{adj}(B^{-1})) A^{-1}$

$B^{-1} |A^{-1}| I + B^{-1} |A^{-1}|$

$\frac{B^{-1}}{|A|} + \frac{A^{-1}}{|B|}$

$\Rightarrow \frac{\text{adj}B}{|B| |A|} + \frac{\text{adj}A}{|A| |B|}$

$= \frac{1}{|A| |B|} (\text{adj}B + \text{adj}A)$

Question 19

If the system of equations
$(\lambda – 1)x + (\lambda – 4)y + \lambda z = 5$
$\lambda x + (\lambda – 1)y + (\lambda – 4)z = 7$
$(\lambda + 1)x + (\lambda + 2)y – (\lambda + 2)z = 9$
has infinitely many solutions, then $\lambda^2 + \lambda$ is equal to

(1) 10
(2) 12
(3) 6
(4) 20

▶️ Answer/Explanation

Ans. (2)

$ (\lambda – 1)x + (\lambda – 4)y + \lambda z = 5 $

$ \lambda x + (\lambda – 1)y + (\lambda – 4)z = 7 $

$ (\lambda + 1)x + (\lambda + 2)y – (\lambda + 2)z = 9 $

For infinitely many solutions,

$ D =
\begin{vmatrix}
\lambda – 1 & \lambda – 4 & \lambda \\
\lambda & \lambda – 1 & \lambda – 4 \\
\lambda + 1 & \lambda + 2 & -(\lambda + 2)
\end{vmatrix} = 0 $

$ (\lambda – 3)(2\lambda + 1) = 0 $

$ D_x =
\begin{vmatrix}
5 & \lambda – 4 & \lambda \\
7 & \lambda – 1 & \lambda – 4 \\
9 & \lambda + 2 & -(\lambda + 2)
\end{vmatrix} = 0 $

$ 2(3 – \lambda)(23 – 2\lambda) = 0 $

$ \lambda = 3 $

$ \therefore \lambda^2 + \lambda = 9 + 3 = 12 $

Question 20

One die has two faces marked 1, two faces marked 2, one face marked 3 and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both the dice are thrown together, is

(1) $\frac{1}{2}$
(2) $\frac{3}{5}$
(3) $\frac{2}{3}$
(4) $\frac{4}{9}$

▶️ Answer/Explanation

Ans. (1)

Sol. $(a,b) = (1,3), (3,1), (2,2), (2,3), (3,2), (1,4), (4,1)$
Required probability $\frac{2 \cdot 2 + 1 \cdot 1 + 2 \cdot 2 + 2 \cdot 2 + 1 \cdot 2 + 2 \cdot 1}{6 \cdot 6} = \frac{18}{36} = \frac{1}{2}$

Question 21

If the area of the larger portion bounded between the curves $x^2 + y^2 = 25$ and $y = |x – 1|$ is $\frac{1}{4} (b\pi + c)$, $b, c \in \mathbb{N}$, then $b + c$ is equal to ________

▶️ Answer/Explanation

Ans. (77)

Sol.

$ x^2 + y^2 = 25 $

$ x^2 + (x – 1)^2 = 25 \Rightarrow x = 4 $

$ x^2 + (-x + 1)^2 = 25 \Rightarrow x = -3 $

$ A = 25\pi – \int_{-3}^{4} \sqrt{25 – x^2} \, dx + \frac{1}{2} \times 4 \times 4 + \frac{1}{2} \times 3 \times 3 $

$ A = 25\pi + \frac{25}{2} \left[ \frac{x}{2} \sqrt{25 – x^2} + \frac{25}{2} \sin^{-1}\!\frac{x}{5} \right]_{-3}^{4} $

$ A = 25\pi + \frac{25}{2} \left[ \frac{1}{2} \left( 6 + \frac{25}{2} \sin^{-1}\!\frac{4}{5} + 6 + \frac{25}{2} \sin^{-1}\!\frac{3}{5} \right) \right] $

$ A = 25\pi + \frac{1}{2} \cdot \frac{25}{2} \cdot \frac{\pi}{2} \Rightarrow A = \frac{75\pi}{4} + \frac{1}{2} $

$ A = \frac{1}{4}(75\pi + 2) $

$ b = 75, \, c = 2 $

$ \therefore b + c = 75 + 2 = 77 $

Question 22

The sum of all rational terms in the expansion of $(1 + 2^{1/3} + 3^{1/2})^6$ is equal to _______

▶️ Answer/Explanation

Ans. (612)

Sol.
$ (1 + 2^{1/3} + 3^{1/2})^6 $

$ = \sum \frac{6!}{r_1! \, r_2! \, r_3!} (1)^{r_1} (2^{1/3})^{r_2} (3^{1/2})^{r_3} $

$
\begin{array}{ccc}
r_1 & r_2 & r_3 \\
\hline
6 & 0 & 0 \\
4 & 0 & 2 \\
2 & 0 & 4 \\
0 & 0 & 6 \\
3 & 3 & 0 \\
1 & 3 & 2 \\
0 & 6 & 0 \\
\end{array}
$

$
\text{sum} =
\frac{6!}{6!0!0!}(1)^6(2)^0(3)^0 +
\frac{6!}{4!0!2!}(1)^4(2)^0(3)^2 +
\frac{6!}{2!0!4!}(1)^2(2)^0(3)^4 +
\frac{6!}{0!0!6!}(1)^0(2)^0(3)^6 +
\frac{6!}{3!3!0!}(1)^3(2)^1(3)^0 +
\frac{6!}{1!3!2!}(1)^1(2)^1(3)^2 +
\frac{6!}{0!6!0!}(1)^0(2)^2(3)^0
$

$ = 1 + 45 + 135 + 27 + 40 + 360 + 4 = 612 $

Question 23

Let the circle C touch the line $x – y + 1 = 0$, have the centre on the positive x-axis, and cut off a chord of length $\frac{4\sqrt{13}}{5}$ along the line $-3x + 2y = 1$. Let H be the hyperbola $\frac{x^2}{\alpha^2} – \frac{y^2}{\beta^2} = 1$, whose one of the foci is the centre of C and the length of the transverse axis is the diameter of C. Then $2\alpha^2 + 3\beta^2$ is equal to ________

▶️ Answer/Explanation

Ans. (19)

Sol.

$ x – y + 1 = 0; \; p = r $

$ \left| \frac{\alpha – 0 + 1}{\sqrt{2}} \right| = r \Rightarrow (\alpha + 1)^2 = 2r^2 $ ….(1)

Now
$ \left( \frac{-3\alpha + 0 – 1}{\sqrt{9 + 4}} \right)^2 + \left( \frac{2}{\sqrt{13}} \right)^2 = r^2 $

$ \Rightarrow (3\alpha + 1)^2 + 4 = 13r^2 $ ….(2)

(1) & (2) $ \Rightarrow (3\alpha + 1)^2 + 4 = 13 \cdot \frac{(\alpha + 1)^2}{2} $

$ \Rightarrow 18\alpha^2 + 12\alpha + 2 + 8 = 13\alpha^2 + 26\alpha + 13 $

$ \Rightarrow 5\alpha^2 – 14\alpha – 3 = 0 $

$ \Rightarrow 5\alpha^2 – 15\alpha + \alpha – 3 = 0 \Rightarrow 5\alpha(\alpha – 3) + (\alpha – 3) = 0 $

$ \Rightarrow \alpha = -\frac{1}{5}, \, 3 $

$ \therefore r = 2\sqrt{2} $

Now $ \alpha e = 3 $ and $ 2\alpha = 4\sqrt{2} $

$ \alpha^2 e^2 = 9 \Rightarrow \alpha e = 2\sqrt{2} \Rightarrow \alpha^2 = 8 $

$ \alpha^2 \left( 1 + \frac{\beta^2}{\alpha^2} \right) = 9 $

$ \alpha^2 + \beta^2 = 9 $

$ \beta^2 = 1 $

$ \therefore 2\alpha^2 + 3\beta^2 = 2(8) + 3(1) = 19 $

Question 24

If the set of all values of $a$, for which the equation $5x^3 – 15x – a = 0$ has three distinct real roots, is the interval $(\alpha, \beta)$, then $\beta – 2\alpha$ is equal to ________

▶️ Answer/Explanation

Ans. (30)

Sol. $5x^3 – 15x – a = 0$
$f(x) = 5x^3 – 15x$
$f'(x) = 15x^2 – 15 = 15(x-1)(x + 1)$

$a \in (-10, 10)$
$\alpha = -10, \beta = 10$
$\beta – 2\alpha = 10 + 20 = 30$

Question 25

If the equation $a(b – c)x^2 + b(c – a)x + c(a – b) = 0$ has equal roots, where $a + c = 15$ and $\frac{b}{36} = \frac{5}{5}$, then $a^2 + c^2$ is equal to ________

▶️ Answer/Explanation

Ans. (117)

Sol. $a(b – c) x^2 + b (c – a) x + c(a – b) = 0$
$x = 1$ is root $\therefore$ other root is 1
$\alpha + \beta = -\frac{b(c-a)}{2a(b-c)}$
$\Rightarrow -bc + ab = 2ab – 2ac$
$\Rightarrow 2ac = b(a + c)$
$\Rightarrow 2ac = 15b$
$\Rightarrow 2ac = 108$
$\Rightarrow ac = 54$
$a + c = 15$
$a^2 + c^2 + 2ac = 225$
$a^2 + c^2 = 225 – 108 = 117$

Question 26

Regarding self-inductance :
A : The self-inductance of the coil depends on its geometry.
B : Self-inductance does not depend on the permeability of the medium.
C : Self-induced e.m.f. opposes any change in the current in a circuit.
D : Self-inductance is electromagnetic analogue of mass in mechanics.
E : Work needs to be done against self-induced e.m.f. in establishing the current.
Choose the correct answer from the options given below:
(1) A, B, C, D only
(2) A, C, D, E only
(3) A, B, C, E only
(4) B, C, D, E only

▶️ Answer/Explanation

Ans. (2)

Sol. Self inductance of coil
$ L = \frac{\mu_0 N^2 A}{2\pi R} $

Question 27

A light hollow cube of side length 10 cm and mass 10g, is floating in water. It is pushed down and released to execute simple harmonic oscillations.
The time period of oscillations is $ y\pi \times 10^{-2} $ s, where the value of $ y $ is
(Acceleration due to gravity, $ g = 10 \, \text{m/s}^2 $, density of water = $ 10^3 \, \text{kg/m}^3 $)
(1) 2
(2) 6
(3) 4
(4) 1

▶️ Answer/Explanation

Ans. (1)

Sol. $ a_{net} = \rho g x – \frac{m g}{L^2} $
$ T = 2\pi \sqrt{\frac{m}{\rho L^2 g}} $
where $ m = 10 \, \text{g} $, $ L = 10 \, \text{cm} $, $ \rho = 1000 \, \text{kg/m}^3 $

Question 28

Given below are two statements:
Statement-I : The hot water flows faster than cold water.
Statement-II : Soap water has higher surface tension as compared to fresh water.
In the light above statements, choose the correct answer from the options given below
(1) Statement-I is false but Statement II is true
(2) Statement-I is true but Statement II is false
(3) Both Statement-I and Statement-II are true
(4) Both Statement-I and Statement-II are false

▶️ Answer/Explanation

Ans. (2)

Sol. Hot water is less viscous then cold water.
Surfactant reduces surface tension.

Question 29

A sub-atomic particle of mass $ 10^{-30} \, \text{kg} $ is moving with a velocity $ 2.21 \times 10^6 \, \text{m/s} $. Under the matter wave consideration, the particle will behave closely like ______. ($ h = 6.63 \times 10^{-34} \, \text{J.s} $)
(1) Infra-red radiation
(2) X-rays
(3) Gamma rays
(4) Visible radiation

▶️ Answer/Explanation

Ans. (2)

Sol. $ \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{10^{-30} \times 2.21 \times 10^6} $
$ = 3 \times 10^{-10} \, \text{m} $
Hence particle will behave as x-ray.

Question 30

A spherical surface of radius of curvature $ R $, separates air from glass (refractive index = 1.5). The centre of curvature is in the glass medium. A point object ‘O’ placed in air on the optic axis of the surface, so that its real image is formed at ‘I’ inside glass. The line OI intersects the spherical surface at P and $ PO = PI $. The distance $ PO $ equals to-
(1) 5R
(2) 3R
(3) 2R
(4) 1.5R

▶️ Answer/Explanation

Ans. (1)

Sol.

$ PO = u = -x $
$ PI = v = x $
$ PO = PI $
$ \frac{\mu_2}{v} – \frac{\mu_1}{u} = \frac{\mu_2 – \mu_1}{R} $
$ \frac{1.5}{x} + \frac{1}{x} = \frac{1}{2R} $
$ \frac{5}{2x} = \frac{1}{2R} $
$ x = 5R $

Question 31

A radioactive nucleus $ n_2 $ has 3 times the decay constant as compared to the decay constant of another radioactive nucleus $ n_1 $. If initial number of both nuclei are the same, what is the ratio of number of nuclei of $ n_2 $ to the number of nuclei of $ n_1 $, after one half-life of $ n_1 $?
(1) 1/4
(2) 1/8
(3) 4
(4) 8

▶️ Answer/Explanation

Ans. (1)

Sol. $ N_2 = N_0 e^{-3\lambda t} $
$ N_1 = N_0 e^{-\lambda t} $
$ \frac{N_1}{N_2} = e^{\lambda t} $
$ t_{\text{half life of } n_1} = \frac{\ln 2}{\lambda} $
$ N_0 e^{-\lambda t} = \frac{N_0}{2} $
$ \lambda t = \ln 2 $
$ t = \frac{\ln 2}{\lambda} $
$ \frac{N_2}{N_1} = \frac{1}{4} $

Question 32

Identify the valid statements relevant to the given circuit at the instant when the key is closed.

A. There will be no current through resistor $ R $.
B. There will be maximum current in the connecting wires.
C. Potential difference between the capacitor plates A and B is minimum.
D. Charge on the capacitor plates is minimum.
Choose the correct answer from the options given below :
(1) C, D only
(2) B, C, D only
(3) A, C only
(4) A, B, D only

▶️ Answer/Explanation

Ans. (2)

Sol. Initially capacitor behave as a short circuit.
So current will be maximum.
Charge on capacitor will be zero.
Potential difference across capacitor will be zero.

Question 33

The position of a particle moving on $ x $-axis is given by $ x(t) = A \sin t + B \cos^2 t + C t^2 + D $, where $ t $ is time. The dimension of $ \frac{ABC}{D} $ is-
(1) $ L $
(2) $ L^3 T^{-2} $
(3) $ L^2 T^{-2} $
(4) $ L^2 $

▶️ Answer/Explanation

Ans. (3)

Sol. Dimension $ [x(t)] = [L] $
$ [A] = [L] $
$ [B] = [L] $
$ [C] = [L T^{-2}] $
$ [D] = [L] $
$ \left[ \frac{ABC}{D} \right] = \frac{[L] \times [L] \times [L T^{-2}]}{[L]} = [L^2 T^{-2}] $

Question 34

Match the List-I with List-II
List-I
A. Pressure varies inversely with volume of an ideal gas.
B. Heat absorbed goes partly to increase internal energy and partly to do work.
C. Heat is neither absorbed nor released by a system
D. No work is done on or by a gas
List-II
I. Adiabatic process
II. Isochoric process
III. Isothermal process
IV. Isobaric process
Choose the correct answer from the options given below :
(1) A–I, B–IV, C–II, D–III
(2) A–III, B–I, C–IV, D–II
(3) A–I, B–III, C–II, D–IV
(4) A–III, B–IV, C–I, D–II

▶️ Answer/Explanation

Ans. (4)

Sol. A $ \rightarrow \frac{1}{P} \propto V $
$ \Rightarrow PV = \text{constant} $
$ \Rightarrow nRT = \text{const.} \Rightarrow T = \text{const.} $
Hence Isothermal III
B $ \rightarrow $ IV
$ W \neq 0 $, $ \Delta U \neq 0 $, $ \Delta Q \neq 0 $ [only isobaric]
C $ \rightarrow $ I $ \Delta Q = 0 $ Adiabatic
D $ \rightarrow $ II $ w = 0 $ Isochoric
III IV I II

Question 35

Consider a moving coil galvanometer (MCG) :
A : The torsional constant in moving coil galvanometer has dimensions $ [M L^2 T^{-2}] $
B : Increasing the current sensitivity may not necessarily increase the voltage sensitivity.
C : If we increase number of turns ($ N $) to its double (2N), then the voltage sensitivity doubles.
D : MCG can be converted into an ammeter by introducing a shunt resistance of large value in parallel with galvanometer.
E : Current sensitivity of MCG depends inversely on number of turns of coil.
Choose the correct answer from the options given below :
(1) A, B only
(2) A, D, only
(3) B, D, E only
(4) A, B, E only

▶️ Answer/Explanation

Ans. (1)

Sol. (A) $ \tau = C\theta \Rightarrow [M L^2 T^{-2}] = [C][1] $
(B) $ C.S = \frac{\theta}{I} = \frac{BNA}{C} $;
$ V.S. = \frac{BNA}{RC} $ [$ R $ also depends on ‘$ N $’]
(C) $ V.S. \propto \frac{NAB}{CR} $
$ R \rightarrow NR $
(D) False [Theory]
(E) E [False] $ C.S \propto N $
$ \frac{NAB}{C} $
$ C.S. \Rightarrow = $

Question 36

A point particle of charge $ Q $ is located at $ P $ along the axis of an electric dipole 1 at a distance $ r $ as shown in the figure. The point $ P $ is also on the equatorial plane of a second electric dipole 2 at a distance $ r $. The dipoles are made of opposite charge $ q $ separated by a distance $ 2a $. For the charge particle at $ P $ not to experience any net force, which of the following correctly describes the situation?

(1) $ \frac{a}{r} \sim 0 $
(2) $ a \sim 10r $
(3) $ a \sim 0.5r $
(4) $ a \sim 3r $

▶️ Answer/Explanation

Ans. (4)

Sol.

$ \frac{2kq \cos \theta}{(r^2 + a^2)} + \frac{2kq}{(r^2 + a^2)^{3/2}} = \frac{4ra^2}{(r^2 + a^2)^{3/2}} $
$ \Rightarrow 4r^2 (r^2 + a^2)^3 = (r^2 – a^2)^4 $
$ \frac{a^3}{r^3} \approx 3 $
$ \frac{a}{r} \approx 3 $
But by solving from mathematical software we are getting $ \frac{a}{r} \approx 3 $.

Question 37

A gun fires a lead bullet of temperature 300K into a wooden block. The bullet having melting temperature of 600 K penetrates into the block and melts down. If the total heat required for the process is 625 J, then the mass of the bullet is ___ grams.
(Latent heat of fusion of lead = $ 2.5 \times 10^4 \, \text{J kg}^{-1} $ and specific heat capacity of lead = $ 125 \, \text{J kg}^{-1} \text{K}^{-1} $)
(1) 20
(2) 15
(3) 10
(4) 5

▶️ Answer/Explanation

Ans. (3)

Sol. 625 = ms\Delta T + mL
625 = m[125 \times 300 + 2.5 \times 10^4]
625 = m[37500 + 25000]
625 = m[62500]
$ m = \frac{1}{100} \, \text{kg} $
$ M = 10 \, \text{grams} $

Question 38

What is the lateral shift of a ray refracted through a parallel-sided glass slab of thickness ‘h’ in terms of the angle of incidence ‘i’ and angle of refraction ‘r’, if the glass slab is placed in air medium ?
(1) $ \frac{-h \tan(i – r)}{\tan r} $
(2) $ \frac{-h \cos(i – r)}{\sin r} $
(3) $ h $
(4) $ \frac{-h \sin(i – r)}{\cos r} $

▶️ Answer/Explanation

Ans. (4)

Sol. Formula base
$ \frac{-h \sin(i – r)}{\cos r} $

Question 39

A solid sphere of mass ‘m’ and radius ‘r’ is allowed to roll without slipping from the highest point of an inclined plane of length ‘L’ and makes an angle 30º with the horizontal. The speed of the particle at the bottom of the plane is $ v_1 $. If the angle of inclination is increased to 45º while keeping $ L $ constant. Then the new speed of the sphere at the bottom of the plane is $ v_2 $. The ratio of $ v_1^2 : v_2^2 $ is
(1) 1 : $ \sqrt{2} $
(2) 1 : 3
(3) 1 : 2
(4) 1 : $ \sqrt{3} $

▶️ Answer/Explanation

Ans. (1)

Sol.

using WET
$ W_g = k_f – k_i $
$ Mg L \sin \theta = k_f – k_i $
K.E. in pure rolling
$ \frac{1}{2} m V_{cm}^2 + \frac{1}{2} I \omega^2 $
$ \frac{1}{2} m V^2 + \frac{1}{2} \frac{2}{5} m R^2 \frac{V^2}{R^2} $
$ \frac{7}{10} m V^2 $
$ \frac{7}{10} m g L \sin \theta = \frac{7}{10} m V^2 – 0 $
$ V^2 \propto \sin \theta $
$ \frac{v_1^2}{v_2^2} = \frac{\sin 30^\circ}{\sin 45^\circ} $
$ \frac{1}{\sqrt{2}} $

Question 40

Refer to the circuit diagram given in the figure, which of the following observation are correct?
A. Total resistance of circuit is 6 $ \Omega $.
B. Current in Ammeter is 1A
C. Potential across AB is 4 Volts.
D. Potential across CD is 4 Volts.
E. Total resistance of the circuit is 8 $ \Omega $.
Choose the correct answer from the options given below:

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

▶️ Answer/Explanation

Ans. (1)

Sol.

Current through ammeter = 1 A
$ R_{net} = 6 \Omega $

$ V_{AB} = 0.5 \times 4 = 2 \, \text{volt} $
$ V_{CD} = 1 \times 4 = 4 \, \text{volt} $
A, B & D are correct

Question 41

The electric flux is $ \phi = \alpha \sigma + \beta \lambda $
where $ \lambda $ and $ \sigma $ are linear and surface charge density, respectively, $ \frac{\alpha}{\beta} $ represents
(1) charge
(2) electric field
(3) displacement
(4) area

▶️ Answer/Explanation

Ans. (3)

Sol. $ \phi = \alpha \sigma + \beta \lambda $
$ [\phi] = [\alpha \sigma] = [\beta \lambda] $
$ \frac{[\phi]}{[\sigma]} = \frac{[\alpha]}{[\sigma]} $
$ \frac{[\phi]}{[\lambda]} = \frac{[\beta]}{[\lambda]} $
$ \frac{[\alpha]}{[\beta]} = \frac{[\phi]/[\sigma]}{[\phi]/[\lambda]} = \frac{\text{Length}}{\text{Area}} = [L] $

Question 42

Given a thin convex lens (refractive index $ \mu_2 $), kept in a liquid (refractive index $ \mu_1 $, $ \mu_1 < \mu_2 $) having radii of curvature $ |R_1| $ and $ |R_2| $. Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place ?
(1) $ \frac{\mu_1}{\mu_2} \frac{1 + \mu_2 – \mu_1}{R_1 – R_2} \frac{R_1}{R_2} $
(2) $ \frac{\mu_1}{\mu_2} \frac{1 + \mu_2 – \mu_1}{R_1 – R_2} \frac{R_1 \cdot R_2}{R_1 + R_2} $
(3) $ \frac{\mu_1}{\mu_2} \frac{1 + \mu_2 – \mu_1}{R_1 – R_2} \frac{R_1 \cdot R_2}{2 R_1 – R_2} \frac{R_1 \cdot R_2}{R_1 + R_2} $
(4) $ (\mu_2 + \mu_1) \frac{\mu_1 – \mu_2}{R} $

▶️ Answer/Explanation

Ans. (2)

Sol.

$ \frac{1}{f_{eq}} = \frac{1}{f_L} + \frac{1}{f_m} $
$ \frac{1}{f_m} = \frac{|\mu_2|}{R_2} – \frac{2}{2} $
$ \frac{1}{f_L} = \frac{\mu_2 – 1}{R_1} – \frac{\mu_2}{R_2} $
$ \frac{1}{f_{eq}} = \frac{\mu_2 – 1}{R_1} – \frac{\mu_2}{R_2} + \frac{\mu_2}{2 R_2} $
$ = \frac{\mu_2 + \mu_1 (\mu_2 – 1)}{R_1 R_2} – \frac{\mu_2 (\mu_2 – 1)}{2 R_1 R_2} $
For same size of image
$ u = 2f $
$ u = \frac{\mu_1}{\mu_2} \frac{1 + \mu_2 – \mu_1}{R_1 – R_2} \frac{R_1 \cdot R_2}{R_1 + R_2} $

Question 43

The electric field of an electromagnetic wave in free space is
$ \vec{E} = 57 \cos[7.5 \times 10^6 t – 5 \times 10^{-3} (3x + 4y)] (-4\hat{i} + 3\hat{j}) \, \text{N/C}. $
The associated magnetic field in Tesla is-
(1) $ \vec{B} = \frac{57}{3 \times 10^8} \cos[7.5 \times 10^6 t – 5 \times 10^{-3} (3x + 4y)] \hat{k} $
(2) $ \vec{B} = \frac{57}{3 \times 10^8} \cos[7.5 \times 10^6 t – 5 \times 10^{-3} (3x + 4y)] \hat{k} $
(3) $ \vec{B} = -\frac{57}{3 \times 10^8} \cos[7.5 \times 10^6 t – 5 \times 10^{-3} (3x + 4y)] \hat{k} $
(4) $ \vec{B} = -\frac{57}{3 \times 10^8} \cos[7.5 \times 10^6 t – 5 \times 10^{-3} (3x + 4y)] \hat{k} $

▶️ Answer/Explanation

Ans. (3)

Sol.
$ \vec{K} = 3\hat{i} + 4\hat{j} $
$ |\vec{K}| = \hat{K} = \frac{3\hat{i} + 4\hat{j}}{5} $
$ \vec{E} = \frac{4\hat{i} – 3\hat{j}}{5} $
$ \vec{B} = \hat{K} \times \vec{E} $
$ \vec{B} = -\hat{Z} $
$ B_0 = \frac{E_0}{C} = \frac{57}{3 \times 10^8} $

Question 44

The motion of an airplane is represented by velocity-time graph as shown below. The distance covered by airplane in the first 30.5 second is ____ km.

(1) 9
(2) 6
(3) 3
(4) 12

▶️ Answer/Explanation

Ans. (4)

Sol. Total Area under curve.

Question 45

Consider a circular disc of radius 20 cm with centre located at the origin. A circular hole of a radius 5 cm is cut from this disc in such a way that the edge of the hole touches the edge of the disc. The distance of centre of mass of residual or remaining disc from the origin will be-
(1) 2.0 cm
(2) 0.5 cm
(3) 1.5 cm
(4) 1.0 cm

▶️ Answer/Explanation

Ans. (4)

Sol.

mass of disc = $ m $
mass of cut part = $ \frac{m}{16} $
$ X_{com} = \frac{m \times 0 – \frac{m}{16} \times 15}{m – \frac{m}{16}} $
$ = 1 \, \text{cm}. $

Question 46

A positive ion A and a negative ion B has charges $ 6.67 \times 10^{-19} \, \text{C} $ and $ 9.6 \times 10^{-10} \, \text{C} $, and masses $ 19.2 \times 10^{-27} \, \text{kg} $ and $ 9 \times 10^{-27} \, \text{kg} $ respectively. At an instant, the ions are separated by a certain distance $ r $. At that instant the ratio of the magnitudes of electrostatic force to gravitational force is $ P \times 10^{-13} $, where the value of $ P $ is ____.
(Take $ \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} $ and universal gravitational constant as $ 6.67 \times 10^{-11} \, \text{Nm}^2 \text{kg}^{-2} $)

▶️ Answer/Explanation

Ans. (BONUS)

Sol. $ \frac{9 \times 10^9 \times 6.67 \times 10^{-19} \times 9.6 \times 10^{-10}}{6.67 \times 10^{-11} \times 19.2 \times 10^{-27} \times 9 \times 10^{-27}} $
$ \frac{1}{2} \times 10^{45} $
Charge is not integral multiple of electron.

Question 47

Two particles are located at equal distance from origin. The position vectors of those are represented by $ \vec{A} = 2\hat{i} + 3n\hat{j} + 2\hat{k} $ and $ \vec{B} = 2\hat{i} – 2\hat{j} + 4p\hat{k} $, respectively. If both the vectors are at right angle to each other, the value of $ n – 1 $ is____.

▶️ Answer/Explanation

Ans. (3)

Sol. $ \vec{A} \cdot \vec{B} = 0 $
$ 4 – 6n + 8p = 0 $
$ |\vec{A}| = |\vec{B}| $
$ 4 + 9n^2 + 4 = 4 + 4 + 16p^2 $
$ 9n^2 = 16p^2 $
$ p = \pm \frac{3}{4} n $
$ 4 – 6n + 6n = 0 $
$ 12n = 4 $
$ n = \frac{1}{3} $

Question 48

An ideal gas initially at 0ºC temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is $ 3/2 $, the change in temperature due to the thermodynamics process is____K.

▶️ Answer/Explanation

Ans. (273)

Sol. $ \gamma = \frac{3}{2} $
$ T V^{\gamma – 1} = C $
$ \frac{T}{T_0} = \left( \frac{V_0}{V} \right)^{0.5} $
$ 273 \left( \frac{1}{4} \right)^{0.5} = 273 \times \frac{1}{2} = 136.5 $
$ \Delta T = 273 $

Question 49

A force $ \vec{f} = x^2 y \hat{i} + y^2 \hat{j} $ acts on a particle in a plane $ x + y = 10 $. The work done by this force during a displacement from (0, 0) to (4m, 2m) is _____ Joule (round off to the nearest integer)

▶️ Answer/Explanation

Ans. (152)

Sol. $ \int_0^4 x^2 (10 – x) dx + \int_0^2 y^2 dy $
$ = \left[ \frac{10x^3}{3} – \frac{x^4}{4} \right]_0^4 + \left[ \frac{y^3}{3} \right]_0^2 $
$ = \frac{640}{3} – 64 + \frac{8}{3} $
$ = 152 $

Question 50

In the given circuit the sliding contact is pulled outwards such that electric current in the circuit changes at the rate of 8 A/s. At an instant when $ R $ is 12 $ \Omega $, the value of the current in the circuit will be ____A.

▶️ Answer/Explanation

Ans. (3)

Sol. $ \epsilon = -L \frac{dI}{dt} – IR = 0 $
$ 12 – 3 \times (-8) – I \times 12 = 0 $
$ I = 3 $

Question 51

The element that does not belong to the same period of the remaining elements (modern periodic table) is:
(1) Palladium
(2) Iridium
(3) Osmium
(4) Platinum

▶️ Answer/Explanation

Ans. (1)

Explanation:
Palladium $\Rightarrow$ 5$^{th}$ period
Iridium, Osmium, Platinum $\Rightarrow$ 6$^{th}$ Period

Question 52

Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H atom is suitable for this ?
Given: Rydberg constant $ R_H = 10^5 \, \text{cm}^{-1} $, $ h = 6.6 \times 10^{-34} \, \text{J s} $, $ c = 3 \times 10^8 \, \text{m/s} $
(1) Paschen series, $\infty \rightarrow 3$
(2) Lyman series, $\infty \rightarrow 1$
(3) Balmer series, $\infty \rightarrow 2$
(4) Paschen series, $5 \rightarrow 3$

▶️ Answer/Explanation

Ans. (1)

Explanation:
$\lambda = 900 \, \text{nm}$
H-atom ($Z = 1$)
$= 9 \times 10^{-5} \, \text{cm}$
$R_H = 10^5 \, \text{cm}^{-1}$
Rydberg eq. = $\frac{1}{\lambda} = R_H Z^2 \left(\frac{1}{n_1^2} – \frac{1}{n_2^2}\right)$
$\Rightarrow \frac{1}{\lambda} = \frac{R_H}{n_1^2} – \frac{R_H}{n_2^2}$
$\Rightarrow \frac{1}{9 \times 10^{-5} \, \text{cm} \times 10^5 \, \text{cm}^{-1}} = \frac{1}{n_1^2} – \frac{1}{n_2^2}$
$\Rightarrow \frac{1}{9} = \frac{1}{n_1^2} – \frac{1}{n_2^2}$
It is possible when $n_1 = 3$, $n_2 = \infty$
Possible series : $\infty \rightarrow 3$

Question 53

The incorrect statements among the following is
(1) PH$_3$ shows lower proton affinity than NH$_3$.
(2) PF$_3$ exists but NF$_5$ does not.
(3) NO$_2$ can dimerise easily.
(4) SO$_2$ can act as an oxidizing agent, but not as a reducing agent.

▶️ Answer/Explanation

Ans. (4)

Explanation:
SO$_2$ can oxidise as well as reduce.
Hence it can act as both oxidising and reducing agent.

Question 54

CrCl$_3$.xNH$_3$ can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of 0.558$^\circ$C. Assuming 100\% ionisation of this complex and coordination number of Cr is 6, the complex will be
(Given $K_f = 1.86 \, \text{K kg mol}^{-1}$)
(1) [Cr(NH$_3$)$_6$] Cl$_3$
(2) [Cr(NH$_3$)$_4$Cl$_2$] Cl
(3) [Cr(NH$_3$)$_5$Cl] Cl$_2$
(4) [Cr(NH$_3$)$_3$Cl$_3$]

▶️ Answer/Explanation

Ans. (3)

Explanation:
Given : $\Delta T_f = 0.558^\circ$C
$K_f = \frac{\text{kg}}{\text{mol}} \times 1.86$
0.1 m aq. sol.
$\Rightarrow \Delta T_f = i \times K_f \times m$
$\Rightarrow 0.558 = i \times 1.86 \times 0.1$
$\Rightarrow i = 3$

Question 55

$\text{FeO}^{2-} \xrightarrow{-2.0 \, \text{V}} \text{Fe}^{3+} \xrightarrow{0.8 \, \text{V}} \text{Fe}^{2+} \xrightarrow{0.5 \, \text{V}} \text{Fe}^0$
In the above diagram, the standard electrode potentials are given in volts (over the arrow).
The value of $E^\Theta_{\text{FeO}_4^{2-}/\text{Fe}}$ is
(1) 1.7 V
(2) 1.2 V
(3) 2.1 V
(4) 1.4 V

▶️ Answer/Explanation

Ans. (1)

Explanation:

$\Delta G^\circ_4 = \Delta G^\circ_1 + \Delta G^\circ_2$
$\Rightarrow -n F E^\circ_4 = -n F E^\circ_1 – n F E^\circ_2$
$\Rightarrow E^\circ_4 = \frac{3 \times 2.0 + 1 \times 0.8}{4}$
$\Rightarrow E^\circ_4 = \frac{6.8}{4} \, \text{V}$
$\Rightarrow E^\circ_4 = 1.7 \, \text{V}$

Question 56

Match the LIST-I with LIST-II
LIST-I
Name reaction
A. Swarts reaction
B. Sandmeyer’s reaction
C. Wurtz Fittig reaction
D. Finkelstein reaction
LIST-II
Product obtainable
I. Ethyl benzene
II. Ethyl iodide
III. Cyanobenzene
IV. Ethyl fluoride
Choose the correct answer from the options given below:
(1) A-II, B-III, C-I, D-IV
(2) A-IV, B-I, C-III, D-II
(3) A-IV, B-III, C-I, D-II
(4) A-II, B-I, C-III, D-IV

▶️ Answer/Explanation

Ans. (3)

Explanation:

Question 57

Given below are two statements:
Statement I: Fructose does not contain an aldehydic group but still reduces Tollen’s reagent
Statement II : In the presence of base, fructose undergoes rearrangement to give glucose.
In the light of the above statements, choose the correct answer from the options given below
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is true but Statement II is false

▶️ Answer/Explanation

Ans. (2)

Explanation:

Question 58

2.8 $\times 10^{-3}$ mol of CO$_2$ is left after removing $10^{21}$ molecules from its ‘x’ mg sample. The mass of CO$_2$ taken initially is
Given : $N_A = 6.02 \times 10^{23} \, \text{mol}^{-1}$
(1) 196.2 mg
(2) 98.3 mg
(3) 150.4 mg
(4) 48.2 mg

▶️ Answer/Explanation

Ans. (1)

Explanation:
$\text{initial} = \frac{x \times 10^{-3}}{44} \, \text{(moles)}$
$\text{removal} = \frac{10^{21}}{6.02 \times 10^{23}} \, \text{(moles)}$
$\text{left} = \text{initial} – \text{removed} \, \text{(moles)}$
$\frac{x \times 10^{-3}}{44} – \frac{10^{21}}{6.02 \times 10^{23}} = 2.8 \times 10^{-3}$
$\Rightarrow x = 196.2 \, \text{mg}$

Question 59

Ice at –5$^\circ$C is heated to become vapor with temperature of 110$^\circ$C at atmospheric pressure. The entropy change associated with this process can be obtained from :
(1) $\frac{\Delta H_{\text{melting}}}{383 \, \text{K}} + \frac{\Delta H_{\text{boiling}}}{268 \, \text{K}} + \int_{273}^{373} C_p \, dT$
(2) $\int_{268}^{273 \, \text{K}} \frac{C_{p,m}}{T} \, dT + \frac{\Delta H_{\text{fusion}}}{T_f} + \frac{\Delta H_{\text{vaporisation}}}{T_b}$
$\quad + \int_{273}^{373 \, \text{K}} \frac{C_{p,m}}{T} \, dT + \int_{373}^{383 \, \text{K}} \frac{C_{p,m}}{T} \, dT$
(3) $\int_{268}^{383 \, \text{K}} \frac{q_{\text{rev}}}{T} + C_p \, dT$
(4) $\int_{268}^{273 \, \text{K}} \frac{\Delta H_{\text{fusion}}}{T} + \frac{\Delta H_{\text{vaporisation}}}{T_b} + \frac{C_{p,m}}{T} \, dT$
$\quad + \int_{273}^{373 \, \text{K}} \frac{C_{p,m}}{T} \, dT + \int_{373}^{383 \, \text{K}} \frac{C_{p,m}}{T} \, dT$

▶️ Answer/Explanation

Ans. (2)

Explanation:
Ice $\rightarrow$ Ice
Water $\rightarrow$ Water
Water $\rightarrow$ Water vapour
268 K 273 K
273 K 273 K 373 K 383 K
(1) (2) (3) (4) (5)
$\Delta S_{\text{overall}} = \Delta S_1 + \Delta S_2 + \Delta S_3 + \Delta S_4 + \Delta S_5$
$\Delta S_2 = \frac{\Delta H_{\text{fusion}}}{273}$ $T_f = 273 \, \text{‘K’}$
$\Delta S_3 = \int_{273}^{373} \frac{C_{p,m}}{T} \, dT$
$\Delta S_4 = \frac{\Delta H_{\text{vaporisation}}}{373}$ $T_b = 373 \, \text{‘K’}$
$\Delta S_5 = \int_{373}^{383} \frac{C_{p,m}}{T} \, dT$
Answer = (2)

Question 60

The d-electronic configuration of an octahedral Co(II) complex having magnetic moment of 3.95 BM is :
(1) $t_{2g}^6 e_g^1$
(2) $t_{2g}^3 e_g^0$
(3) $t_{2g}^5 e_g^2$
(4) $t_{2g}^4 e_g^3$

▶️ Answer/Explanation

Ans. (3)

Explanation:

Question 61

The complex that shows Facial – Meridional isomerism is
(1) [Co(NH$_3$)$_3$Cl$_3$]
(2) [Co(NH$_3$)$_4$Cl$_2$]$^+$
(3) [Co(en)$_3$]$^{3+}$
(4) [Co(en)$_2$Cl$_2$]$^+$

▶️ Answer/Explanation

Ans. (1)

Explanation:
Ma$_3$b$_3$ type complexes show Facial – Meridional isomerism
(i) [Co(NH$_3$)$_3$Cl$_3$] $\Rightarrow$ Ma$_3$b$_3$
(ii) [Co(NH$_3$)$_4$Cl$_2$]$^+$ $\Rightarrow$ Ma$_4$b$_2$
(iii) [Co(en)$_3$]$^{3+}$ $\Rightarrow$ M(AA)$_3$
(iv) [Co(en)$_2$Cl$_2$]$^+$ $\Rightarrow$ M(AA)$_2$b$_2$
a, b, = NH$_3$, Cl$^-$
AA = en

Question 62

The major product of the following reaction is :

▶️ Answer/Explanation

Ans. (3)

Explanation:
This is an example of Tollens reaction i.e. multiple cross aldol followed by cross Cannizzaro reaction

Question 63

The correct stability order of the following species/molecules is :

(1) q > r > p
(2) r > q > p
(3) q > p > r
(4) p > q > r

▶️ Answer/Explanation

Ans. (1)

Explanation:
q is aromatic r is nonaromatic p is antiaromatic

Question 64

Propane molecule on chlorination under photochemical condition gives two di-chloro products, “x” and “y”. Amongst “x” and “y”, “x” is an optically active molecule. How many tri-chloro products (consider only structural isomers) will be obtained from “x” when it is further treated with chlorine under the photochemical condition?
(1) 4
(2) 2
(3) 5
(4) 3

▶️ Answer/Explanation

Ans. (4)

Explanation:

Question 65

What amount of bromine will be required to convert 2 g of phenol into 2, 4, 6-tribromophenol ?
(Given molar mass in g mol$^{-1}$ of C, H, O, Br are 12, 1, 16, 80 respectively)
(1) 10.22 g
(2) 6.0 g
(3) 4.0 g
(4) 20.44 g

▶️ Answer/Explanation

Ans. (1)

Explanation:

Moles of phenol = $\frac{2}{94} = 0.021$
$\therefore$ Moles of bromine = $0.021 \times 3 = 0.064$
$\therefore$ Mass of bromine = $0.064 \times 160 = 10.22 \, \text{g}$

Question 66

The correct set of ions (aqueous solution) with same colour from the following is :
(1) V$^{2+}$, Cr$^{3+}$, Mn$^{3+}$
(2) Zn$^{2+}$, V$^{3+}$, Fe$^{3+}$
(3) Ti$^{4+}$, V$^{4+}$, Mn$^{2+}$
(4) Sc$^{3+}$, Ti$^{3+}$, Cr$^{2+}$

▶️ Answer/Explanation

Ans. (1)

Explanation:
(1) V$^{2+}$ (Violet), Cr$^{3+}$ (Violet), Mn$^{3+}$ (Violet)
(2) Zn$^{2+}$ (Colourless), V$^{3+}$ (Green), Fe$^{3+}$ (Yellow)
(3) Ti$^{4+}$ (Colourless), V$^{4+}$ (Blue), Mn$^{2+}$ (Pink)
(4) Sc$^{3+}$ (Colourless), Ti$^{3+}$ (Purple), Cr$^{2+}$ (Blue)

Question 67

Given below are two statements :
Statement I : In Lassaigne’s test, the covalent organic molecules are transformed into ionic compounds.
Statement II : The sodium fusion extract of an organic compound having N and S gives prussian blue colour with FeSO$_4$ and Na$_4$[Fe(CN)$_6$]
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 false but Statement II is true
(4) Statement I is true but Statement II is false

▶️ Answer/Explanation

Ans. (4)

Explanation:
The sodium fusion extract of organic compound having N & S gives blood red colour with FeSO$_4$ and Na$_4$[Fe(CN)$_6$]

Question 68

Which of the following happens when NH$_4$OH is added gradually to the solution containing 1M A$^{2+}$ and 1M B$^{3+}$ ions ?
Given : $K_{sp}[A(OH)_2] = 9 \times 10^{-10}$ and $K_{sp}[B(OH)_3] = 27 \times 10^{-18}$ at 298 K.
(1) B(OH)$_3$ will precipitate before A(OH)$_2$
(2) A(OH)$_2$ and B(OH)$_3$ will precipitate together
(3) A(OH)$_2$ will precipitate before B(OH)$_3$
(4) Both A(OH)$_2$ and B(OH)$_3$ do not show precipitation with NH$_4$OH

▶️ Answer/Explanation

Ans. (1)

Explanation:
Condition for precipitation $Q_{ip} > K_{sp}$
For [A(OH)$_2$]
$[A^{2+}][OH^-]^2 > 9 \times 10^{-10}$
$[A^{2+}] = 1 \, \text{M}$
$\Rightarrow [OH^-] > 3 \times 10^{-5} \, \text{M}$
For [B(OH)$_3$]
$[B^{3+}][OH^-]^3 > 27 \times 10^{-18}$
$[B^{3+}] = 1 \, \text{M}$
$\Rightarrow [OH^-] > 3 \times 10^{-6} \, \text{M}$
So, B(OH)$_3$ will precipitate before A(OH)$_2$

Question 69

Match the LIST-I with LIST-II
LIST-I
(Classification of molecules based on octet rule)
A. Molecules obeying octet rule
B. Molecules with incomplete octet
C. Molecules with incomplete octet with odd electron
D. Molecules with expanded octet
LIST-II
(Example)
I. NO, NO$_2$
II. BCl$_3$, AlCl$_3$
III. H$_2$SO$_4$, PCl$_5$
IV. CCl$_4$, CO$_2$
Choose the correct answer from the options given below :
(1) A-IV, B-II, C-I, D-III
(2) A-III, B-II, C-I, D-IV
(3) A-IV, B-I, C-III, D-II
(4) A-II, B-IV, C-III, D-I

▶️ Answer/Explanation

Ans. (1)

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

Question 70

Which among the following react with Hinsberg’s reagent?

Choose the correct answer from the options given below :
(1) B and D only
(2) C and D only
(3) A, B and E only
(4) A, C and E only

▶️ Answer/Explanation

Ans. (4)

Explanation:
B and D are 3$^\circ$ amine which does not have replaceable H on N, So does not react.

Question 71

If 1 mM solution of ethylamine produces pH = 9, then the ionization constant ($K_b$) of ethylamine is 10$^{-x}$. The value of x is ________ (nearest integer).
[The degree of ionization of ethylamine can be neglected with respect to unity.]

▶️ Answer/Explanation

Ans. (7)

Explanation:

$\Rightarrow K_b = \frac{10^{-5} \times 10^{-5}}{10^{-3}} = 10^{-7}$

Question 72

During “S” estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is ______ %.
(Given molar mass in g mol$^{-1}$ of Ba : 137, S : 32, O : 16)

▶️ Answer/Explanation

Ans. (40)

Explanation:
Millimoles of BaSO$_4$ = $\frac{466}{233} = 2 \, \text{mmol}$
$\%S = \frac{466 \times 32}{233} \times \frac{100}{160} = 40\%$

Question 73

Consider the following sequence of reactions to produce major product (A)

Molar mass of product (A) is ______ g mol$^{-1}$.
(Given molar mass in g mol$^{-1}$ of C : 12, H : 1, O : 16, Br : 80, N : 14, P : 31)

▶️ Answer/Explanation

Ans. (171)

Explanation:

Molar mass of product (C$_7$H$_7$Br) (A) is 171 g mol$^{-1}$

Question 74

For the thermal decomposition of N$_2$O$_5$(g) at constant volume, the following table can be formed, for the reaction mentioned below :
$2\text{N}_2\text{O}_5\text{(g)} \rightarrow 2\text{N}_2\text{O}_4\text{(g)} + \text{O}_2\text{(g)}$

Given : Rate constant for the reaction is $4.606 \times 10^{-2} \, \text{s}^{-1}$.

▶️ Answer/Explanation

Ans. (900)

Explanation:
NTA. (897)
$K = 2 \times 4.606 \times 10^{-2} \, \text{s}^{-1}$
$2\text{N}_2\text{O}_5\text{(g)} \rightarrow 2\text{N}_2\text{O}_4\text{(g)} + \text{O}_2\text{(g)}$
$P_i$ 0.6 0 0
$P_f$ $0.6 – P$ $P$ $P/2$
$P_{\text{total}} = \frac{P}{2} + 0.6$
$2 \times 4.606 \times 10^{-2} = \frac{2.303}{100} \log \frac{0.6}{0.6 – P}$
$10 \frac{0.6}{0.6 – P} = 4 \log \frac{0.6}{0.6 – P}$
$\frac{0.6}{0.6 – P} = 10^{\frac{4}{10}}$
$\frac{0.6}{0.6 – P} = 10^{0.4}$
$0.6 = (0.6 – P) \times 10^{0.4}$
$P = 0.6 (1 – 10^{-0.4})$
$P = 0.6 \times 0.599$
$P = 0.3594$
$P_{\text{total}} = 0.6 + 0.1797 = 0.7797 \, \text{atm}$
$x = 779.7 \times 10^{-3} \, \text{atm}$
Ans. 900

Question 75

The standard enthalpy and standard entropy of decomposition of N$_2$O$_4$ to NO$_2$ are 55.0 kJ mol$^{-1}$ and 175.0 J/K/mol respectively. The standard free energy change for this reaction at 25$^\circ$C in J mol$^{-1}$ is _______ (Nearest integer)

▶️ Answer/Explanation

Ans. (2850)

Explanation:
$\Delta H^\circ_{\text{rxn}} = 55 \, \text{kJ/mol}$, $T = 298 \, \text{K}$
$\Delta S^\circ_{\text{rxn}} = 175 \, \text{J/mol}$
$\Delta G^\circ_{\text{rxn}} = \Delta H^\circ – T \Delta S^\circ$
$\Rightarrow \Delta G^\circ_{\text{rxn}} = 55000 \, \text{J/mol} – 298 \times 175 \, \text{J/mol}$
$\Rightarrow \Delta G^\circ_{\text{rxn}} = 55000 – 52150$
$\Rightarrow \Delta G^\circ_{\text{rxn}} = 2850 \, \text{J/mol}$

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