JEE Main 5 April 2024 Shift 1 question paper with solutions
JEE Main 5 April 2024 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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Maths
Question 1
Maths · Three Dimensional Geometry · Single correct
Let d be the distance of the point of intersection of the lines $\frac{x+6}{3} = \frac{y}{2} = \frac{z+1}{1}$ and $\frac{x-7}{4} = \frac{y-9}{3} = \frac{z-4}{2}$ from the point $(7, 8, 9)$. Then $d^2 + 6$ is equal to:
Maths · Applications of Derivatives · Single correct
Let a rectangle $ABCD$ of sides $2$ and $4$ be inscribed in another rectangle $PQRS$ such that the vertices of the rectangle $ABCD$ lie on the sides of the rectangle $PQRS$. Let $a$ and $b$ be the sides of the rectangle $PQRS$ when its area is maximum. Then $(a + b)^2$ is equal to:
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let two straight lines drawn from the origin O intersect the line $3x + 4y = 12$ at the points P and Q such that $\triangle OPQ$ is an isosceles triangle and $\angle POQ = 90^\circ$. If $l = OP^2 + PQ^2 + QO^2$, then the greatest integer less than or equal to $l$ is:
If $y = y(x)$ is the solution of the differential equation $\frac{dy}{dx} + 2y = \sin(2x)$, $y(0) = \frac{3}{4}$, then $y\left(\frac{\pi}{8}\right)$ is equal to:
$e^{\pi/8}$
$e^{\pi/4}$
$e^{-\pi/4}$
$e^{-\pi/8}$
Answer: (c)
Solution
Given $\dfrac{dy}{dx}+2y=\sin 2x,\quad y(0)=\dfrac{3}{4}$. The integrating factor is $\mathrm{I.F.}=e^{\int 2\,dx}=e^{2x}$. Multiply through by the integrating factor: $y\cdot e^{2x}=\int e^{2x}\sin 2x\,dx$. Using integration by parts, $y\cdot e^{2x}=\dfrac{e^{2x}(2\sin 2x-2\cos 2x)}{4+4}+C$. Given $x=0,\ y=\dfrac{3}{4}$, $\dfrac{3}{4}=\dfrac{-1}{4}+C$. Solving for $C$, $C=1$. Thus, $y=\dfrac{2\sin 2x-2\cos 2x}{8}+e^{-2x}$. For $x=\dfrac{\pi}{8}$, $y=\dfrac{1}{8}\left(2\sin\frac{\pi}{4}-2\cos\frac{\pi}{4}\right)+e^{-2\left(\frac{\pi}{8}\right)}$. Therefore, $y=0+e^{-\frac{\pi}{4}}$. $\therefore\ y=e^{-\frac{\pi}{4}}$.
Question 5
Maths · Applications of Derivatives · Single correct
For the function $$f(x) = \sin x + 3x - \frac{2}{\pi}(x^2 + x), where x \in \left[0, \frac{\pi}{2}\right],$$ consider the following two statements: (I) $f$ is increasing in $\left(0, \frac{\pi}{2}\right)$. (II) $f'$ is decreasing in $\left(0, \frac{\pi}{2}\right)$. Between the above two statements,
only (II) is true.
only (I) is true.
neither (I) nor (II) is true.
both (I) and (II) are true
Answer: (d)
Solution
Given $f(x) = \sin x + 3x - \frac{2}{\pi}(x^2 + x)$ for $x \in \left[0, \frac{\pi}{2}\right]$. The derivative is $f'(x) = \cos x + 3 - \frac{2}{\pi}(2x + 1) > 0$, indicating $f(x)$ is increasing. At $x = 0$, $f'(x) = -\sin x + 0 - \frac{\pi}{2}(2)$. This simplifies to $-\sin x - \frac{4}{\pi} -\frac{2}{\pi} \left(2x + 1\right) > -\frac{2}{\pi} \left(\pi + 1\right)$$ $$3_{+ve} - \frac{2}{\pi} > 3 - \frac{2}{\pi} (2x + 1) > 3 - \frac{2}{\pi} (\pi_{+ve} + 1)$$
Question 6
Maths · Determinants · Single correct
If the system of equations $$11x + y + \lambda z = -5$$ $$2x + 3y + 5z = 3$$ $$8x - 19y - 39z = \mu$$ has infinitely many solutions, then $\lambda^4 - \mu$ is equal to:
Maths · Permutations and Combinations · Single correct
Let $A = \{1, 3, 7, 9, 11\}$ and $B = \{2, 4, 5, 7, 8, 10, 12\}$. Then the total number of one-one maps $f : A \rightarrow B$, such that $f(1) + f(3) = 14$, is :
The integral $\int_0^{\pi/4} \frac{136 \sin x}{3 \sin x + 5 \cos x} \, dx$ is equal to:
$3\pi - 50 \log_e 2 + 20 \log_e 5$
$3\pi - 25 \log_e 2 + 10 \log_e 5$
$3\pi - 10 \log_e (2\sqrt{2}) + 10 \log_e 5$
$3\pi - 30 \log_e 2 + 20 \log_e 5$
Answer: (a)
Solution
Given $$I = \int_0^{\pi/4} \frac{136 \sin x}{3 \sin x + 5 \cos x} \, dx$$ Assume $$136 \sin x = A(3 \sin x + 5 \cos x) + B(3 \cos x - 5 \sin x)$$ This gives the equations: $$136 = 3A - 5B \ldots (i)$$ $$0 = 5A + 3B \ldots (ii)$$ From equation (ii): $$3B = -5A \Rightarrow B = -\frac{5}{3}A$$ Substitute in equation (i): $$136 = 3A - 5 \left(-\frac{5}{3}A\right)$$ $$136 = 3A + \frac{25}{3}A$$ Combine terms: $$136 = \frac{34}{3}A$$ Solve for $A$: $$A = \frac{136 \times 3}{34} = 12$$ Then $B$ is: $$B = -\frac{5}{3}(12) = -20$$ Now, substitute back into the integral: $$I = \int_0^{\pi/4} \frac{A(3 \sin x + 5 \cos x)}{3 \sin x + 5 \cos x} + \int_0^{\pi/4} \frac{B(3 \cos x - 5 \sin x)}{3 \sin x + 5 \cos x}$$ This simplifies to: $$= A(x)\bigg|_0^{\pi/4} + B[\ln(3 \sin x + 5 \cos x)]\bigg|_0^{\pi/4} - \ln(0 + 5)$$ Substitute the values of $A$ and $B$: $$= 12 \left(\frac{\pi}{4}\right) - 20 \ln \left(\frac{3}{\sqrt{2}} + \frac{5}{\sqrt{2}}\right) - \ln(5)$$ Simplify: $$= 3\pi - 20 \ln 4\sqrt{2} + 20 \ln 5$$ $$= 3\pi - 20 \times \frac{5}{2} \ln 2 + 20 \ln 5$$ $$= 3\pi - 50 \ln 2 + 20 \ln 5$$
Question 10
Maths · Probability · Single correct
The coefficients $a, b, c$ in the quadratic equation $ax^2 + bx + c = 0$ are chosen from the set $\{$1, 2, 3, 4, 5, 6, 7, 8$\}$. The probability of this equation having repeated roots is:
$\frac{1}{128}$
$\frac{1}{64}$
$\frac{3}{256}$
$\frac{3}{128}$
Answer: (b)
Solution
Given the quadratic equation $ax^2 + bx + c = 0$ where $a, b, c \in \{1, 2, 3, 4, 5, 6, 7, 8\}$. For repeated roots, the discriminant $D = 0$. This implies $b^2 - 4ac = 0$ which leads to $b^2 = 4ac$. The probability is calculated as $$Prob = \frac{8}{8 \times 8 \times 8} = \frac{1}{64}$$ which implies the solutions for $(a, b, c)$ are: $(1, 2, 1); (2, 4, 2); (1, 4, 4); (4, 4, 1); (3, 6, 3); (2, 8, 8); (8, 8, 2); (4, 8, 4)$. There are 8 cases.
Question 11
Maths · Matrices · Single correct
Let A and B be two square matrices of order 3 such that $|A| = 3$ and $|B| = 2$. Then $$\left| A^T A (adj(2 \, A))^{-1} (adj(4 \, B))(adj(AB))^{-1} AA^T \right|$$ is equal to:
Let a circle $C$ of radius $1$ and closer to the origin be such that the lines passing through the point $(3, 2)$ and parallel to the coordinate axes touch it. Then the shortest distance of the circle $C$ from the point $(5, 5)$ is :
$2\sqrt{2}$
$4\sqrt{2}$
$4$
$5$
Answer: (c)
Solution
Coordinates of the centre will be $(2, 1)$. Equation of circle will be $$ (x - 2)^2 + (y - 1)^2 = 1 $$ $$ QC = \sqrt{(5 - 2)^2 + (5 - 1)^2} $$ $$ QC = 5 $$ shortest distance $$ = RQ = CQ - CR $$ $$ = 5 - 1 $$ $$ = 4 $$
Question 13
Maths · Conic Sections · Single correct
Let the line $2x + 3y - k = 0$, $k > 0$, intersect the $x$-axis and $y$-axis at the points $A$ and $B$, respectively. If the equation of the circle having the line segment $AB$ as a diameter is $x^2 + y^2 - 3x - 2y = 0$ and the length of the latus rectum of the ellipse $x^2 + 9y^2 = k^2$ is $\frac{m}{n}$, where $m$ and $n$ are coprime, then $2m + n$ is equal to
11
10
12
13
Answer: (a)
Solution
Centre of the circle = $\left$( $\frac{3}{2}$, 1 $\right$) Equation of diameter = 2x + 3y - k = 0 2 $\left$( $\frac{3}{2}$ $\right$) + 3(1) - k = 0 $\Rightarrow$ k = 6 Now, Equation of ellipse becomes $$x^2 + 9y^2 = 36$$ $$\frac{x^2}{6^2} + \frac{y^2}{2^2} = 1$$ length of $LR = \frac{2b^2}{a} = \frac{2 \cdot 2^2}{6} = \frac{8}{6} = \frac{m}{n}$ $\therefore 2m + n = 2(4) + 3 = 11$
Question 14
Maths · Complex Numbers and Quadratic Equations · Single correct
Consider the following two statements : Statement I : For any two non-zero complex numbers $z_1, z_2$, $$\left( |z_1| + |z_2| \right) \left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right| \leq 2 \left( |z_1| + |z_2| \right)$$, and Statement II : If $x, y, z$ are three distinct complex numbers and $a, b, c$ are three positive real numbers such that $$\frac{a}{|y-z|} = \frac{b}{|z-x|} = \frac{c}{|x-y|}$$, then $$\frac{a^2}{y-z} + \frac{b^2}{z-x} + \frac{c^2}{x-y} = 1$$. Between the above two statements,
Statement I is correct but Statement II is incorrect.
both Statement I and Statement II are correct.
both Statement I and Statement II are incorrect.
Statement I is incorrect but Statement II is correct.
If $\frac{1}{\sqrt{1+\sqrt{2}}} + \frac{1}{\sqrt{2+\sqrt{3}}} + \cdots + \frac{1}{\sqrt{99+\sqrt{100}}} = m$ and $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{99 \cdot 100} = n$, then the point $(m, n)$ lies on the line
$11(x - 1) - 100(y - 2) = 0$
$11x - 100y = 0$
$11(x - 2) - 100(y - 1) = 0$
$11(x - 1) - 100y = 0$
Answer: (b)
Solution
The expression is given by $$\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \cdots + \frac{1}{\sqrt{99} + \sqrt{100}} = m$$ which simplifies to $$\frac{1}{\sqrt{1} - \sqrt{2}} + \frac{1}{\sqrt{2} - \sqrt{3}} + \cdots + \frac{1}{\sqrt{99} - \sqrt{100}} = m$$ Simplifying further, we have $$\sqrt{100} - 1 = m \Rightarrow m = 9$$ The next expression is $$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{99 \cdot 100} = n$$ which simplifies to $$1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} \cdots \frac{1}{99} - \frac{1}{100} = n$$ Thus, $$1 - \frac{1}{100} = n$$ which gives $$\frac{99}{100} = n$$ Therefore, $$(m, n) = \left(9, \frac{99}{100}\right)$$ This implies $$\Rightarrow 11(9) - 100\left(\frac{99}{100}\right)$$ which simplifies to $$= 99 - 99 = 0$$ The answer is $11x - 100y = 0$.
Question 17
Maths · Relations and Functions · Single correct
Let $f(x) = x^5 + 2x^3 + 3x + 1, x \in \mathbb{R}$, and $g(x)$ be a function such that $g(f(x)) = x$ for all $x \in \mathbb{R}$. Then $\frac{g(7)}{g'(7)}$ is equal to:
14
42
7
1
Answer: (a)
Solution
Given $f(x) = x^5 + 2x^3 + 3x + 1$. The derivative is $f'(x) = 5x^4 + 6x^2 + 3$. Evaluating at $x = 1$, we have $f'(1) = 5 + 6 + 3 = 14$. Given $g(f(x)) = x$, we have $g'(f(x)) f'(x) = 1$. For $f(x) = 7$, we solve $x^5 + 2x^3 + 3x + 1 = 7$, which gives $x = 1$. Thus, $g'(7) f'(1) = 1 \Rightarrow g'(7) = \frac{1}{f'(1)} = \frac{1}{14}$. With $x = 1$, $f(x) = 7 \Rightarrow g(7) = 1$. Therefore, $\frac{g(7)}{g'(7)} = \frac{1}{1/14} = 14$.
Question 18
Maths · Three Dimensional Geometry · Single correct
If $A(1, -1, 2)$, $B(5, 7, -6)$, $C(3, 4, -10)$ and $D(-1, -4, -2)$ are the vertices of a quadrilateral $ABCD$, then its area is:
The value of $\int_{-\pi}^{\pi} \frac{2y(1+\sin y)}{1+\cos^2 y} \, dy$ is :
$2\pi^2$
$\frac{\pi^2}{2}$
$\frac{\pi}{2}$
$\pi^2$
Answer: (d)
Solution
Given $$\int_{-\pi}^{\pi} \frac{2y(1+\sin y)}{1+\cos^2 y} \, dy$$ we can split it as $$\int_{-\pi}^{\pi} \frac{2y}{1+\cos^2 y} \, dy + \int_{-\pi}^{\pi} \frac{2y \sin y}{1+\cos^2 y} \, dy$$ where the first integral is odd and the second is even. This simplifies to $$0 + 2 \cdot 2 \int_0^{\pi} y \left( \frac{\sin y}{1+\cos^2 y} \right) \, dy$$. Let $$I = 4 \int_0^{\pi} \frac{y \sin y}{1+\cos^2 y} \, dy$$. Then $$I = 4 \int_0^{\pi} \frac{(\pi - y) \sin y}{1+\cos^2 y} \, dy$$. Adding these, $$2I = 4 \int_0^{\pi} \frac{\pi \sin y}{1+\cos^2 y} \, dy$$. Therefore, $$I = 2\pi \int_0^{\pi} \frac{\sin y}{1+\cos^2 y} \, dy$$. This evaluates to $$2\pi (- \tan^{-1}(\cos y)) \bigg|_0^{\pi}$$ which simplifies to $$-2\pi \left[ \left( -\frac{\pi}{4} \right) - \left( \frac{\pi}{4} \right) \right]$$. Thus, $$= -2\pi \left[ -\frac{2\pi}{4} \right] = \pi^2$$.
Question 20
Maths · Three Dimensional Geometry · Single correct
If the line $\frac{2-x}{3} = \frac{3y-2}{4\lambda+1} = 4 - z$ makes a right angle with the line $\frac{x+3}{3\mu} = \frac{1-2y}{6} = \frac{5-z}{7}$, then $4\lambda + 9\mu$ is equal to:
4
13
5
6
Answer: (d)
Solution
Given the equations: $$\frac{2-x}{3} = \frac{3y-2}{4\lambda+1} = 4-z \ldots (1)$$ $$\frac{x-2}{(-3)} = \frac{y-\frac{2}{3}}{\left(4\lambda+\frac{1}{3}\right)} = \frac{z-4}{(-1)}$$ From equation (2): $$\frac{x+3}{3\mu} = \frac{1-2y}{6} = \frac{5-z}{7} \ldots (2)$$ We have: $$\frac{x+3}{3\mu} = \frac{y-\frac{1}{2}}{(-3)} = \frac{z-5}{(-7)}$$ For a right angle, $$(-3)(3\mu) + \left(\frac{4\lambda+1}{3}\right)(-3) + (-1)(-7) = 0$$ Simplifying, $$-9\mu - 4\lambda - 1 + 7 = 0$$ Thus, $$4\lambda + 9\mu = 6$$
Question 21
Maths · Probability · Numerical
From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the variance of $X$ is $\sigma^2$, then $96\sigma^2$ is equal to .
If $S = \{ a \in \mathbb{R} : |2a - 1| = 3[a] + 2\{a\} \}$, where $[t]$ denotes the greatest integer less than or equal to $t$ and $\{t\}$ represents the fractional part of $t$, then $72 \sum_{a \in S} a$ is equal to
Maths · Continuity and Differentiability · Numerical
Let $f$ be a differentiable function in the interval $(0, \infty)$ such that $f(1) = 1$ and $\lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t-x} = 1$ for each $x > 0$. Then $2f(2) + 3f(3)$ is equal to
Let $a_1, a_2, a_3, \ldots$ be in an arithmetic progression of positive terms. Let $A_k = a_1^2 - a_2^2 + a_3^2 - a_4^2 + \ldots + a_{2k-1}^2 - a_{2k}^2$. If $A_3 = -153$, $A_5 = -435$ and $a_1^2 + a_2^2 + a_3^2 = 66$, then $a_{17} - A_7$ is equal to
Maths · Complex Numbers and Quadratic Equations · Fill in the blank
The number of distinct real roots of the equation $|x||x + 2| - 5|x + 1| - 1 = 0$ is
Answer: 3
Solution
Case-1 $x \geq 0$ $$x^2 + 2x - 5x - 5 - 1 = 0$$ $$x^2 - 3x - 6 = 0$$ $$x = \frac{3 \pm \sqrt{9 + 24}}{2} = \frac{3 \pm \sqrt{33}}{2}$$ One positive root Case-2 $-1 \leq x < 0$ $$-x^2 - 2x - 5x - 5 - 1 = 0$$ $$x^2 + 7x + 6 = 0$$ $$(x + 6)(x + 1) = 0$$ $x = -1$ one root in range Case-3 $-2 \leq x < -1$ $$x^2 - 2x + 5x + 5 - 1 = 0$$ $$x^2 - 3x - 4 = 0$$ $$(x - 4)(x + 1) = 0$$ No root in range Case-4 $x < -2$ $$x^2 + 7x + 4 = 0$$ $$x = \frac{-7 \pm \sqrt{49 - 16}}{2} = \frac{7 \pm \sqrt{33}}{2}$$ one root in range Total number of distinct roots are 3
Question 30
Maths · Conic Sections · Numerical
Suppose $AB$ is a focal chord of the parabola $y^2 = 12x$ of length $l$ and slope $m < \sqrt{3}$. If the distance of the chord $AB$ from the origin is $d$, then $l \, d^2$ is equal to
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
Following gates section is connected in a complete suitable circuit. For which of the following combination, bulb will glow (ON):
A = 0, B = 0, C = 0, D = 1
A = 0, B = 1, C = 1, D = 1
A = 1, B = 0, C = 0, D = 0
A = 1, B = 1, C = 1, D = 0
Answer: (c)
Solution
Bulb will glow if bulb have potential drop on it. One end of bulb must be at high (1) and other must be at low (0). Option (3) satisfy this condition.
Question 33
Physics · Physical World, Units and Measurements · Single correct
If $G$ be the gravitational constant and $u$ be the energy density then which of the following quantity have the dimensions as that of the $\sqrt{uG}$:
Physics · Mechanical Properties of Fluids · Single correct
Given below are two statements: Statement I: When a capillary tube is dipped into a liquid, the liquid neither rises nor falls in the capillary. The contact angle may be $0^\circ$. Statement II: The contact angle between a solid and a liquid is a property of the material of the solid and liquid as well. In the light of the above statements, choose the correct answer from the options given below.
Both Statement I and Statement II are false
Both Statement I and Statement II are true
Statement I is false but Statement II is true
Statement I is true and Statement II is false
Answer: (c)
Solution
Capillary rise $$h = \frac{2 \, T \, \cos \theta}{\rho g r};$$ If $\theta = 0^\circ$ then rise is non-zero. Therefore, Statement-1 is incorrect.
Question 35
Physics · Dual Nature of Radiation and Matter · Single correct
Given below are two statements: Statement I: Figure shows the variation of stopping potential with frequency $(\nu)$ for the two photosensitive materials $M_1$ and $M_2$. The slope gives value of $\frac{h}{e}$, where $h$ is Planck's constant, $e$ is the charge of electron. Statement II: $M_2$ will emit photoelectrons of greater kinetic energy for the incident radiation having same frequency. In the light of the above statements, choose the most appropriate answer from the options given below.
Both Statement I and Statement II are correct
Statement I is incorrect but Statement II is correct
Both Statement I and Statement II are incorrect
Statement I is correct and Statement II is incorrect
Answer: (d)
Solution
Given $eV_0 = h\nu - \phi$. $$V_0 = \frac{h}{e} \nu - \frac{\phi}{e}$$ $M_2$ material has higher work function, so statement-(II) is incorrect.
Question 36
Physics · Mathematics in Physics · Single correct
The angle between vector $\vec{Q}$ and the resultant of $(2\vec{Q} + 2\vec{P})$ and $(2\vec{Q} - 2\vec{P})$ is:
Physics · Moving Charges and Magnetism · Single correct
In a co-axial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero:
outside the cable
inside the outer conductor
inside the inner conductor
in between the two conductors
Answer: (a)
Solution
The integral of the magnetic field $\vec{B}$ over a closed loop $\oint \vec{B} \cdot d\vec{\ell}$ is equal to $\mu_0 I_{enc} = 0$. Therefore, $B = 0$ outside the cable.
Question 39
Physics · Atoms · Single correct
An electron rotates in a circle around a nucleus having positive charge Ze. Correct relation between total energy (E) of electron to its potential energy (U) is :
E = U
2E = U
2E = 3U
E = 2U
Answer: (b)
Solution
Given the force equation, $$F = \frac{k(Ze)(e)}{r^2} = \frac{mv^2}{r}$$ we can find the kinetic energy as $$KE = \frac{1}{2} mv^2 = \frac{1}{2} \frac{K(Ze)(e)}{r}$$ The potential energy is $$PE = -\frac{K(Ze)(e)}{r}$$ The total energy is $$TE = \frac{K(Ze)(e)}{2r} - \frac{K(Ze)(e)}{r} = -\frac{K(Ze)(e)}{2r}$$ Therefore, $$TE = \frac{PE}{2}$$ and $$2TE = PE$$
Question 40
Physics · Kinetic Theory · Single correct
If the collision frequency of hydrogen molecules in a closed chamber at 27°C is $Z$, then the collision frequency of the same system at 127°C is:
Physics · System of Particles and Rotational Motion · Single correct
Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of Inertia about their diameter axis AB as shown in figure is $\sqrt{\frac{8}{x}}$. The value of $x$ is :
51
34
17
67
Answer: (d)
Solution
Given $I_{sphere} = \frac{2}{3} MR^2 = M k_1^2$. $I_{cylinder} = \frac{1}{12} M \left(4R^2\right) + \frac{1}{4} MR^2 + M(2R)^2$ $$= \frac{67}{12} MR^2 = M k_2^2$$ $$\frac{k_1}{k_2} = \sqrt{\frac{\frac{2}{3} \cdot 12}{67}} = \sqrt{\frac{8}{67}}$$
Question 42
Physics · Electromagnetic Induction · Single correct
Two conducting circular loops A and B are placed in the same plane with their centers coinciding as shown in figure. The mutual inductance between them is :
$\frac{\mu_0 \pi b^2}{2a}$
$\frac{\mu_0}{2\pi} \cdot \frac{b^2}{a}$
$\frac{\mu_0}{2\pi} \cdot \frac{a^2}{b}$
$\frac{\mu_0 \pi a^2}{2b}$
Answer: (d)
Solution
Given $\phi = Mi = BA$. Therefore, Mi = $\frac{\mu_0 i}{2 \, b}$ $\pi$ a^2$. Thus, $M = $\frac{\mu_0 \pi a^2}{2 \, b}$$.
Question 43
Physics · Gravitation · Single correct
Match List-I with List-II. \begin{tabular}{|c|l|c|c|} \hline & \textbf{List-I} & & \textbf{List-II} \\ \hline (A) & Kinetic energy of planet & (I) & $\displaystyle -\frac{GMm}{a}$ \\ \hline (B) & Gravitational potential energy of Sun-planet system & (II) & $\displaystyle \frac{GMm}{2a}$ \\ \hline (C) & Total mechanical energy of planet & (III) & $\displaystyle \frac{Gm}{r}$ \\ \hline (D) & Escape energy at surface of planet for unit mass object & (IV) & $\displaystyle \frac{GMm}{2a}$ \\ \hline \end{tabular} where $a$ = radius of planet orbit, $r$ = radius of planet, $M$ = mass of Sun, $m$ = mass of planet. Choose the correct answer from the options given below:
(A) - II, (B) - I, (C) - IV, (D) - III
(A) - III, (B) - IV, (C) - I, (D) - II
(A) - I, (B) - IV, (C) - II, (D) - III
(A) - I, (B) - II, (C) - III, (D) - IV
Answer: (a)
Solution
Question 44
Physics · Laws of Motion · Single correct
A wooden block of mass 5 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on the top of the block, the floor yields and the block and the cylinder together go down with an acceleration of 0.1 $\,$ $\mathrm{ms^{-2}}$. The action force of the system on the floor is equal to:
196 $\mathrm{N}$
291 $\mathrm{N}$
294 $\mathrm{N}$
297 $\mathrm{N}$
Answer: (b)
Solution
Taking $g = 9.8 \, \mathrm{m/s^2}$. The total weight is $30 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} = 294 \, \mathrm{N}$. Using the equation: $$294 - N = 30 \times 0.1$$ Solving for $N$: $$N = 291$$
Question 45
Physics · Gravitation · Single correct
A simple pendulum doing small oscillations at a place $R$ height above earth surface has time period of $T_1 = 4 \, \mathrm{s}$. $T_2$ would be it's time period if it is brought to a point which is at a height $2R$ from earth surface. Choose the correct relation $[R = radius of earth]$:
A body of mass 50 kg is lifted to a height of 20 m from the ground in the two different ways as shown in the figures. The ratio of work done against the gravity in both the respective cases, will be:
1 : 2
$\sqrt{3} : 2$
2 : 1
1 : 1
Answer: (d)
Solution
Work done by gravity is independent of path. It depends only on vertical displacement so work done in both cases will be same.
Question 47
Physics · Physical World, Units and Measurements · Single correct
Time periods of oscillation of the same simple pendulum measured using four different measuring clocks were recorded as 4.62 $\mathrm{s}$, 4.632 $\mathrm{s}$, 4.6 $\mathrm{s}$ and 4.64 $\mathrm{s}$. The arithmetic mean of these readings in correct significant figure is :
5 $\mathrm{s}$
4.623 $\mathrm{s}$
4.6 $\mathrm{s}$
4.62 $\mathrm{s}$
Answer: (c)
Solution
Sum of number by considering significant digits. $$sum = 4.6 + 4.6 + 4.6 + 4.6 = 18.4$$ Arithmetic Mean = $$\frac{sum}{4} = \frac{18.4}{4} = 4.6$$
Question 48
Physics · Thermodynamics · Single correct
The heat absorbed by a system in going through the given cyclic process is :
19.6 J
61.6 J
616 J
431.2 J
Answer: (b)
Solution
Given $\Delta U = 0$ (Cyclic process). $\Delta Q = W =$ area of $P - V$ curve. $$= \pi \times (140 \times 10^3 \, \mathrm{Pa}) \times (140 \times 10^{-6} \, \mathrm{m^3})$$ $\Delta Q = 61.6 \, \mathrm{J}$
Question 49
Physics · Current Electricity · Single correct
In the given figure $R_1 = 10\,\Omega$, $R_2 = 8\,\Omega$, $R_3 = 4\,\Omega$ and $R_4 = 8\,\Omega$. Battery is ideal with emf $12\,\mathrm{V}$. Equivalent resistant of the circuit and current supplied by battery are respectively:
$10.5\,\Omega$ and $1.14\,\mathrm{A}$
$12\,\Omega$ and $1\,\mathrm{A}$
$10.5\,\Omega$ and $1\,\mathrm{A}$
$12\,\Omega$ and $11.4\,\mathrm{A}$
Answer: (b)
Solution
Here $R_2$, $R_3$, $R_4$ are in parallel $$\frac{1}{R_{234}} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}$$ $$R_{234} = 2\, \Omega$$ $R_{234}$ is in series with $R_1$ so $$R_{eq} = R_{234} + R_1 = 2 + 10 = 12\, \Omega$$ $$i = \frac{12}{12} = 1\, Amp$$
Question 50
Physics · Electromagnetic Waves · Single correct
An alternating voltage of amplitude 40 $\mathrm{V}$ and frequency 4 $\mathrm{kHz}$ is applied directly across the capacitor of 12 $\mu$$\mathrm{F}$. The maximum displacement current between the plates of the capacitor is nearly:
10 $\mathrm{A}$
12 $\mathrm{A}$
8 $\mathrm{A}$
13 $\mathrm{A}$
Answer: (b)
Solution
Displacement current is same as conduction current in capacitor. $$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$$ $$= \frac{1}{2\pi \times 4 \times 10^3 \times 12 \times 10^{-6}} = 3.317 \, \Omega$$ $$I = \frac{V}{X_C} = \frac{40}{3.317} = 12 \, \mathrm{A}$$
Question 51
Physics · Wave Optics · Numerical
In Young's double slit experiment, carried out with light of wavelength $5000 \, \mathrm{\AA}$, the distance between the slits is $0.3 \, \mathrm{mm}$ and the screen is at $200 \, \mathrm{cm}$ from the slits. The central maximum is at $x = 0 \, \mathrm{cm}$. The value of $x$ for third maxima is _____ mm.
Answer: 10
Solution
Given $\($ $\beta$ = $\frac{\lambda D}{d}$ = $\frac{5 \times 10^{-7} \times 2}{3 \times 10^{-4}}$ = $\frac{10 \times 10^{-3}}{3}$ $\)$ m. For 3rd maxima $\($ y_3 = 3 $\beta$ = 10 $\times$ $10^{-3} )$ m = 10 mm.
Question 52
Physics · Moving Charges and Magnetism · Numerical
A 2 A current carrying straight metal wire of resistance 1 $\Omega$, resistivity $2 \times 10^{-6} \, \Omega \mathrm{m}$, area of cross-section $10 \, \mathrm{mm}^2$ and mass $500 \, \mathrm{g}$ is suspended horizontally in mid air by applying a uniform magnetic field $\vec{B}$. The magnitude of $B$ is _____ $\times 10^{-1} \, \mathrm{T}$ (given, $g = 10 \, \mathrm{m/s}^2$).
Answer: 5
Solution
Given $R = \frac{\rho \ell}{A}$, we have $$\frac{2 \times 10^{-6} \times \ell}{10^{-5}} = 1 \Rightarrow \ell = 5$$ $$mg = Bi \ell$$ $$B = \frac{mg}{i \ell} = \frac{5}{2 \times 5} = 0.5 = 5 \times 10^{-1} Tesla$$
Question 53
Physics · Electrostatic Potential and Capacitance · Numerical
The electric field between the two parallel plates of a capacitor of $1.5 \, \mu \mathrm{F}$ capacitance drops to one third of its initial value in $6.6 \, \mu \mathrm{s}$ when the plates are connected by a thin wire. The resistance of this wire is _____ $\Omega$. (Given, $\log 3 = 1.1$)
Answer: 4
Solution
Given $E = \frac{E_0}{3}$, it follows that $V = \frac{V_0}{3}$. We have $\frac{V_0}{3} = V_0 e^{-\frac{t}{\tau}}$. Solving for $t$, we get $t = \tau \ln 3$. Given $6.6 \times 10^{-6} = R (1.5 \times 10^{-6}) (1.1)$, we find $R = \frac{6}{1.5} = 4 \, \Omega$.
Question 54
Physics · Laws of Motion · Numerical
Three blocks $M_1, M_2, M_3$ having masses $4 \, \mathrm{kg}, 6 \, \mathrm{kg}$ and $10 \, \mathrm{kg}$ respectively are hanging from a smooth pully using rope $1, 2$ and $3$ as shown in figure. The tension in the rope $1, T_1$ when they are moving upward with acceleration of $2 \, \mathrm{ms^{-2}}$ is $\mathrm{N}$ ( if $g = 10 \, \mathrm{m/s^2}$ ).
Physics · Mechanical Properties of Solids · Numerical
The density and breaking stress of a wire are $6 \times 10^4 \, \mathrm{kg/m^3}$ and $1.2 \times 10^8 \, \mathrm{N/m^2}$ respectively. The wire is suspended from a rigid support on a planet where acceleration due to gravity $g$ is $\frac{1}{3}$ of the value on the surface of earth. The maximum length of the wire with breaking is _____ m (take, $g = 10 \, \mathrm{m/s^2}$).
Answer: 600
Solution
Given $T = mg$. The stress $\sigma = \frac{T}{A} = \frac{mg}{A}$. The expression becomes $\frac{(\sigma A \ell) g}{A}$. Thus, $\ell = \frac{\sigma}{\rho g} = \frac{1.2 \times 10^8 \times 3}{6 \times 10^4 \times 10} = 600$.
Question 56
Physics · Motion in a Straight Line · Numerical
A body moves on a frictionless plane starting from rest. If $S_n$ is distance moved between $t = n - 1$ and $t = n$ and $S_{n-1}$ is distance moved between $t = n - 2$ and $t = n - 1$, then the ratio $\frac{S_{n-1}}{S_n}$ is $\left(1 - \frac{2}{x}\right)$ for $n = 10$. The value of $x$ is $\ldots$
Answer: 19
Solution
Question 57
Physics · Nuclei · Numerical
If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is _____ $\times 10^{-2} \, \mathrm{MeV}$. (Given $1 \, \mathrm{u} = 931 \, \mathrm{MeV}/c^2$, atomic mass of helium $= 4.002603 \, \mathrm{u}$ )
Answer: 727
Solution
Reaction: $$^4_2\mathrm{He} \longrightarrow ^{12}_6\mathrm{C} + \gamma rays$$ Mass defect $= \Delta m = (3 \, m_{\mathrm{He}} - m_{\mathrm{C}})$ $$= (3 \times 4.002603 - 12) = 0.007890 \, u$$ Energy released $$= 931 \Delta m \, MeV$$ $$= 7.27 \, MeV = 727 \times 10^{-2} \, MeV$$
Question 58
Physics · Alternating Current · Numerical
An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of $20\mu F$ is _____ V.
In the experiment to determine the galvanometer resistance by half-deflection method, the plot of $1/\theta$ vs the resistance $(R)$ of the resistance box is shown in the figure. The figure of merit of the galvanometer is ____ $\times 10^{-1}$ $\mathrm{A/ division}$. [The source has emf 2V]
Answer: 5
Solution
Given $i = K \theta$. $$\frac{2}{G + R} = K \theta$$ Therefore, $$\frac{1}{\theta} = \frac{(G + R)K}{2} = R \left( \frac{K}{2} \right) + \frac{KG}{2}$$ The slope is $$\frac{K}{2} = \frac{1}{4} \Rightarrow K = 0.5 = 5 \times 10^{-1} \, \mathrm{A}$$
Question 60
Physics · Electrostatic Potential and Capacitance · Numerical
Three capacitors of capacitances $25\mu F$, $30\mu F$ and $45\mu F$ are connected in parallel to a supply of $100 \, V$. Energy stored in the above combination is $E$. When these capacitors are connected in series to the same supply, the stored energy is $\frac{9}{x}E$. The value of $x$ is _____.
Answer: 86
Solution
In parallel combination: Potential difference is same across all Energy $= \frac{1}{2} (C_1 + C_2 + C_3) V^2$ $$= \frac{1}{2} (25 + 30 + 45) \times (100)^2 \times 10^{-6} = 0.5 = E$$ In series combination: Charge is same on all. $$\frac{1}{C_{equ}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{25} + \frac{1}{30} + \frac{1}{45}$$ $$\frac{1}{C_{equ}} = \frac{(18 + 15 + 10)}{450} = \frac{43}{450} \Rightarrow C_{equ} = \frac{450}{43}$$ Energy $= \frac{Q^2}{2C_1} + \frac{Q^2}{2C_2} + \frac{Q^2}{2C_3}$ $$= \frac{Q^2}{2} \left[ \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \right]$$ $$(V \times C_{equ})^2 \times \frac{1}{C_{equ}} = \frac{V^2 C_{equ}}{2}$$ $$\frac{(100)^2}{2} \times \frac{450}{43} \times 9 \times 10^{-6}$$ $$\Rightarrow \frac{4.5}{86} = \frac{9}{x} = E = \frac{9}{x} \times 0.5 \Rightarrow x = 86$$
Chemistry
Question 61
Chemistry · Some Basic Concepts of Chemistry · Single correct
The incorrect postulates of the Dalton's atomic theory are : (A) Atoms of different elements differ in mass. (B) Matter consists of divisible atoms. ($C$) Compounds are formed when atoms of different element combine in a fixed ratio. (D) All the atoms of given element have different properties including mass. (E) Chemical reactions involve reorganisation of atoms. Choose the correct answer from the options given below :
($C$), (D), (E) only
(B), (D) only
(A), (B), (D) only
(B), (D), (E) only
Answer: (b)
Solution
b,d
Question 62
Chemistry · Equilibrium · Single correct
The following reaction occurs in the Blast furnace where iron ore is reduced to iron metal $Fe_2O_3(s)+3CO(g)⇌Fe(l)+3CO_2(g)$ Using the Le-chatelier's principle, predict which one of the following will not disturb the equilibrium.
Addition of $\mathrm{CO_2}$
Removal of $\mathrm{CO_2}$
Addition of $\mathrm{Fe_2O_3}$
Removal of $\mathrm{CO}$
Answer: (c)
Solution
When solid added no effect on equilibrium.
Question 63
Chemistry · Alcohols, Phenols and Ethers · Single correct
Identify compound (Z) in the following reaction sequence.
Answer: (a)
Solution
The reaction starts with chlorobenzene reacting with $\mathrm{NaOH}$ at $623 \, \mathrm{K}$ and $300 \, \mathrm{atm}$ to form sodium phenoxide $(\mathrm{X})$. This is then treated with $\mathrm{HCl}$ to form phenol $(\mathrm{Y})$. Finally, phenol undergoes nitration with concentrated $\mathrm{HNO_3}$ to form 2,4,6-trinitrophenol $(\mathrm{Z})$.
Question 64
Chemistry · Thermodynamics · Single correct
Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason ($R$) Assertion (A) : Enthalpy of neutralisation of strong monobasic acid with strong monoacidic base is always $-57 \, \mathrm{kJ \, mol^{-1}}$. Reason $(R)$ : Enthalpy of neutralisation is the amount of heat liberated when one mole of $\mathrm{H^+}$ ions furnished by acid combine with one mole of $\mathrm{OH^-}$ ions furnished by base to form one mole of water. In the light of the above statements, choose the correct answer from the options given below.
(A) is true but ($R$) is false
Both (A) and ($R$) are true but ($R$) is not the correct explanation of (A)
Both (A) and ($R$) are true and ($R$) is the correct explanation of (A)
(A) is false but ($R$) is true
Answer: (c)
Solution
Enthalpy of neutralization of SA & SB is always $-57 \, \mathrm{kJ/mol}$ because strong monoacid gives one mole of $\mathrm{H}^+$ and strong mono base gives one mole of $\mathrm{OH}^-$ which form one mole of water.
Question 65
Chemistry · Classification of Elements and Periodicity in Properties · Multiple correct
The statement(s) that are correct about the species $\mathrm{O^{2-}}$, $\mathrm{F^-}$, $\mathrm{Na^+}$ and $\mathrm{Mg^{2+}}$. (A) All are isoelectronic (B) All have the same nuclear charge ($C$) $\mathrm{O^{2-}}$ has the largest ionic radii (D) $\mathrm{Mg^{2+}}$ has the smallest ionic radii Choose the most appropriate answer from the options given below:
(1) (B), (C) and (D) only
(2) (C) and (D) only
(3) (A), (C) and (D) only
(4) (A), (B), (C) and (D)
Answer: (c)
Solution
For $\mathrm{O}^{-2}$, $\mathrm{F}^{-}$, $\mathrm{Na}^{+}$, $\mathrm{Mg}^{+2}$, the number of electrons is 10 for each. The ionic radius order is $\mathrm{O}^{-2} > \mathrm{F}^{-} > \mathrm{Na}^{+} > \mathrm{Mg}^{+2}$. The effective nuclear charge (Zeff) order is $\mathrm{O}^{-2} < \mathrm{F}^{-} < \mathrm{Na}^{+} < \mathrm{Mg}^{+2}$.
Question 66
Chemistry · Alcohols, Phenols and Ethers · Single correct
For the compounds: The increasing order of boiling point is: Choose the correct answer from the options given below:
< $(C)$ < (A) < (B)
< (A) < $(C)$ < (D)
< (B) < $(C)$ < (D)
< (A) < (D) < $(C)$
Answer: (b)
Solution
Question 67
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Given below are two statements: Statement I: In group 13, the stability of +1 oxidation state increases down the group. Statement II: The atomic size of gallium is greater than that of aluminium. In the light of the above statements, choose the most appropriate answer from the options given below:
Both Statement I and Statement II are correct
Statement I is incorrect but Statement II is correct
Both Statement I and Statement II are incorrect
Statement I is correct but Statement II is incorrect
Answer: (d)
Solution
Statement I: Number of $d$ and $f$ electrons increases down the group and due to poor shielding of $d$ and $f$ electrons, stability of lower oxidation states increases down the group. Statement II: The atomic size of aluminium is greater than that of gallium.
Question 68
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Number of $\sigma$ and $\pi$ bonds present in ethylene molecule is respectively:
4 and 1
5 and 2
3 and 1
5 and 1
Answer: (d)
Solution
Ethyene is it has 5 $\sigma$ bonds and 1 $\pi$ bond.
Question 69
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Identify 'A' in the following reaction:
Answer: (d)
Solution
The reaction shown is a Wolff-Kishner reduction. The carbonyl group is reduced to a methylene group using hydrazine $\mathrm{(N_2H_4)}$ and ethylene glycol with potassium hydroxide $\mathrm{(KOH)}$. The product is an alkane.
Question 70
Chemistry · Electrochemistry · Single correct
The reaction at cathode in the cells commonly used in clocks involves.
reduction of Mn from +7 to +2
reduction of Mn from +4 to +3
oxidation of Mn from +3 to +4
oxidation of Mn from +2 to +7
Answer: (b)
Solution
In the cathode reaction manganese (Mn) is reduced from the $+4$ oxidation state to the $+3$ state.
Question 71
Chemistry · Co-ordination Compounds · Single correct
Which one of the following complexes will exhibit the least paramagnetic behaviour? [Atomic number, Cr = 24, Mn = 25, Fe = 26, Co = 27]
$[\mathrm{Cr(H_2O)_6}]^{2+}$
$[\mathrm{Fe(H_2O)_6}]^{2+}$
$[\mathrm{Co(H_2O)_6}]^{2+}$
$[\mathrm{Mn(H_2O)_6}]^{2+}$
Answer: (c)
Solution
The table shows the number of unpaired electrons and the magnetic moment $\mu = \sqrt{n(n+2)} \, \mathrm{B.M.}$ for different complexes. For $[\mathrm{Co(H_2O)_6}]^{2+}$, the number of unpaired electrons is 3, and $\mu = 3.87$. For $[\mathrm{Fe(H_2O)_6}]^{2+}$, the number of unpaired electrons is 4, and $\mu = 4.89$. For $[\mathrm{Mn(H_2O)_6}]^{2+}$, the number of unpaired electrons is 5, and $\mu = 5.92$. For $[\mathrm{Cr(H_2O)_6}]^{2+}$, the number of unpaired electrons is 4, and $\mu = 4.89$. The complex with the least paramagnetic behavior is $[\mathrm{Co(H_2O)_6}]^{2+}$.
Question 72
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason $(R)$. Assertion (A) : Cis form of alkene is found to be more polar than the trans form. Reason $(R)$: Dipole moment of trans isomer of 2-butene is zero. In the light of the above statements, choose the correct answer from the options given below :
(A) is false but $(R)$ is true
Both (A) and $(R)$ are true and $(R)$ is the correct explanation of (A)
(A) is true but $(R)$ is false
Both (A) and $(R)$ are true but $(R)$ is NOT the correct explanation of (A)
Answer: (b)
Solution
Dipole moment is a vector quantity and for compound net dipole moment is the vector sum of all dipoles hence dipole moment of cis form is greater than trans form.
Question 73
Chemistry · Hydrocarbons · Single correct
Given below are two statements: In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I is correct but Statement II is incorrect
Statement I is incorrect but Statement II is correct
Both Statement I and Statement II are correct
Both Statement I and Statement II are incorrect
Answer: (a)
Solution
In nitration of benzene concentrated $\mathrm{H_2SO_4}$ and $\mathrm{HNO_3}$ is used as reagent which generates electrophile $\mathrm{N_2}$ in following steps: $$\mathrm{H_2SO_4 + HNO_3 \rightleftharpoons HSO_4^- + H-O-NO_2^+}$$ $$\mathrm{HSO_4^- + H_2O + NO_2^+}$$ Lewis acids can promote the formation of electrophiles not Lewis base.
Question 74
Chemistry · Co-ordination Compounds · Single correct
The correct order of ligands arranged in increasing field strength.
Experimental order $\mathrm{Br}^- < \mathrm{F}^- < \mathrm{H_2O} < \mathrm{NH_3}$
Question 75
Chemistry · Biomolecules · Single correct
Which of the following gives a positive test with ninhydrin?
Starch
Egg albumin
Polyvinyl chloride
Cellulose
Answer: (b)
Solution
Ninhydrin test is a test of amino acids. Egg albumin contains protein which is a natural polymer of amino acids which will show positive ninhydrin test.
Question 76
Chemistry · The d-and f-Block Elements · Single correct
The metal that shows highest and maximum number of oxidation state is :
Fe
Mn
Co
Ti
Answer: (b)
Solution
Mn shows highest oxidation state $\left( \mathrm{Mn}^{+7} \right)$ in 3 d series metals.
Question 77
Chemistry · Some Basic Concepts of Chemistry · Single correct
An organic compound has 42.1$\%$ carbon, 6.4$\%$ hydrogen and remainder is oxygen. If its molecular weight is 342, then its molecular formula is :
C_{11}H_{18}O_{12}
C_{12}H_{20}O_{12}
C_{12}H_{22}O_{11}
C_{14}H_{20}O_{10}
Answer: (c)
Solution
Only $\mathrm{C_{12}H_{22}O_{11}}$ has $42.1\%$ carbon, $6.4\%$ hydrogen, and $51.5\%$ oxygen.
Question 78
Chemistry · Alcohols, Phenols and Ethers · Single correct
Given below are two statement: Statements I : Bromination of phenol in solvent with low polarity such as CHCl$_3$ or CS$_2$ requires Lewis acid catalyst. Statements II : The Lewis acid catalyst polarises the bromine to generate Br$^+$. In the light of the above statements, choose the correct answer from the options given below :
Both Statement I and Statement II are true
Statement I is true but Statement II is false
Statement I is false but Statement II is true
Both Statement I and Statement II are false
Answer: (c)
Solution
Phenol is a highly activated compound which can undergo bromination directly with Bromine without any Lewis acid.
Question 79
Chemistry · Electrochemistry · Single correct
Molar ionic conductivities of divalent cation and anion are $57 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$ and $73 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$ respectively. The molar conductivity of solution of an electrolyte with the above cation and anion will be :
Given $\Lambda_C^{+2} = 57 \, \mathrm{Scm^2 \, mol^{-1}}$ and $\Lambda_A^{+2} = 73 \, \mathrm{Scm^2 \, mol^{-1}}$. The solution is calculated as follows: $$\Lambda_{Solution} = \Lambda_C^{+2} + \Lambda_A^{-2}$$ $$= 57 + 73 = 130$$
Question 80
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
The number of neutrons present in the more abundant isotope of boron is '$x$'. Amorphous boron upon heating with air forms a product, in which the oxidation state of boron is '$y$'. The value of $x + y$ is _______
3
9
4
6
Answer: (b)
Solution
More abundant isotope = $\mathrm{B^{11}}$ [Number of neutrons = 6] $x = 6$ $\mathrm{B + O_2 \rightarrow B_2O_3}$ Oxidation state of B in $\mathrm{B_2O_3} = +3$ So, $y = 3$ Hence $x + y = 9$
Question 81
Chemistry · Structure of Atom · Numerical
The value of Rydberg constant ($R_H$) is $2.18 \times 10^{-18} \, \mathrm{J}$. The velocity of electron having mass $9.1 \times 10^{-31} \, \mathrm{kg}$ in Bohr's first orbit of hydrogen atom = \text{_____} $\times 10^5 \, \mathrm{ms}^{-1}$ (nearest integer).
Answer: 22
Solution
Given $$V = 2.18 \times 10^6 \times \frac{Z}{n}$$ Substituting the values, we have $$= 21.8 \times 10^5 \times \frac{1}{1} \approx 22 \times 10^5 (nearest)$$
Question 82
Chemistry · Analytical Chemistry · Numerical
In a borax bead test under hot condition, a metal salt (one from the given) is heated at point B of the flame, resulted in green colour salt bead. The spin-only magnetic moment value of the salt is _______ BM (Nearest integer) [Given atomic number of Cu = 29, Ni = 28, Mn = 25, Fe = 26]
Answer: 6
Solution
Fe^{+3} will give green coloured bead when heated at point B. Number of unpaired e^{-} in Fe^{+3} = 5. $$\mu = 5.92$$ Nearest integer = 6
Question 83
Chemistry · Thermodynamics · Fill in the blank
The heat of combustion of solid benzoic acid at constant volume is $-321.30 \, \mathrm{kJ}$ at $27^\circ \mathrm{C}$. The heat of combustion at constant pressure is $(-321.30 - xR) \, \mathrm{kJ}$, the value of $x$ is \text{_____}.
Answer: 150
Solution
For the reaction: $C_6H_5COOH$(S) + $\frac{15}{2}$ $O_2$(g) $\rightarrow$ 7 $CO_2$(g) + 3$H_2O$($\ell$) The change in enthalpy is given by: $$\Delta H = \Delta U + \Delta n_g RT$$ Substituting the values: $$= -321.30 - \frac{1}{2} \frac{R}{100} \times 300$$ $$= (-321.30 - 150R) kJ$$
Question 84
Chemistry · Alcohols, Phenols and Ethers · Numerical
Consider the given chemical reaction sequence: Total sum of oxygen atoms in Product A and Product B are _______
Answer: 14
Solution
Picric acid is prepared by treating phenol first with concentrated sulphuric acid which converts it to phenol-2,4-disulphonic acid and then with concentrated nitric acid to get 2, 4, 6 trinitrophenol.
Question 85
Chemistry · The d-and f-Block Elements · Numerical
The spin-only magnetic moment value of the ion among $\mathrm{Ti}^{2+}$, $\mathrm{V}^{2+}$, $\mathrm{Co}^{3+}$ and $\mathrm{Cr}^{2+}$, that acts as strong oxidising agent in aqueous solution is ______ BM (Near integer). (Given atomic numbers : Ti : 22, V : 23, Cr : 24, Co : 27)
Answer: 5
Solution
Strong oxidising agent is $\mathrm{Co^{+3}}$. The number of unpaired electrons in $\mathrm{Co^{+3}} \left[ 3d^6 \right]$ is $4$. Hence $\mu = \sqrt{n(n+2)} = \sqrt{24} \mathrm{BM}$. Nearest integer is $5$.
Question 86
Chemistry · Chemical Kinetics and Nuclear Chemistry · Fill in the blank
During Kinetic study of reaction $2A + B \rightarrow C + D$, the following results were obtained: \begin{tabular}{|c|c|c|c|} \hline & A $[\mathrm{M}]$ & B $[\mathrm{M}]$ & Initial rate of formation of D \\ \hline I & 0.1 & 0.1 & $6.0 \times 10^{-3}$ \\ \hline II & 0.3 & 0.2 & $7.20 \times 10^{-2}$ \\ \hline III & 0.3 & 0.4 & $2.88 \times 10^{-1}$ \\ \hline IV & 0.4 & 0.1 & $2.40 \times 10^{-2}$ \\ \hline \end{tabular} Based on above data, overall order of the reaction is \_\_\_\_\_
Answer: 3
Solution
Given $r = K[A]^x[B]^y$. (I) $6 \times 10^{-3} = K[0.1]^x[0.1]^y$ (IV) $2.4 \times 10^{-2} = K[0.4]^x[0.1]^y$ Dividing (IV) by (I): $$4 = (4)^x$$ Thus, $x = 1$. Now, $r = K[A]^x[B]^y$. (III) $2.88 \times 10^{-1} = K[0.3]^x[0.4]^y$ (II) $7.2 \times 10^{-2} = K[0.3]^x[0.2]^y$ Dividing (III) by (II): $$4 = 2^y$$ Thus, $y = 2$. Overall order $= x + y = 1 + 2 = 3$
Question 87
Chemistry · Solutions · Numerical
An artificial cell is made by encapsulating $0.2\mathrm{M}$ glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a $0.05\mathrm{M}$ solution of $\mathrm{NaCl}$ at $300\mathrm{K}$ is ______ $\times 10^{-1}$ bar. (nearest integer). [Given : $R = 0.083\ \mathrm{Lbar}\ mol^{-1}\ K^{-1}$ Assume complete dissociation of $\mathrm{NaCl}$.
Answer: 25
Solution
NaCl dissociates into $\mathrm{Na^+}$ and $\mathrm{Cl^-}$. The concentrations are $0.05 \, \mathrm{M}$ for both ions. The total concentration $C_1$ is $0.05 + 0.05 = 0.1 \, \mathrm{M}$ (NaCl). The concentration $C_2$ is $0.2 \, \mathrm{M}$ (glucose). The osmotic pressure $\pi$ is given by $\pi = (C_2 - C_1) RT$. Substituting the values, we have $$\pi = (0.2 - 0.1) \times 0.083 \times 300$$ which equals $2.49 \, \mathrm{bar}$ or $24.9 \times 10^{-1} \, \mathrm{bar}$.
Question 88
Chemistry · Haloalkanes and Haloarenes · Numerical
The number of halobenzenes from the following that can be prepared by Sandmeyer's reaction is _______
Answer: 2
Solution
In Sandmeyer reaction only bromobenzene and chlorobenzene are prepared.
Question 89
Chemistry · Chemical Bonding and Molecular Structure · Numerical
In the lewis dot structure for $\mathrm{NO}_2^-$, total number of valence electrons around nitrogen is
Answer: 8
Solution
Number of valence $e^-$ around N-atom $= 8$
Question 90
Chemistry · Amines · Numerical
9.3 $\mathrm{g}$ of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product 'P'. The mass of product 'P' obtained is 26.4 $\mathrm{g}$. The percentage yield is \text{_____}\%.
Answer: 80
Solution
93 g of aniline produces 330 g of 2, 4, 6-tribromoaniline. Hence 9.3 g of aniline should produce 33 g of 2, 4, 6-tribromoaniline. Hence percentage yield $$\frac{26.4 \times 100}{33} = 80\%$$