JEE Main 26 July 2022 Shift 1 question paper with solutions
JEE Main 26 July 2022 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 · Continuity and Differentiability · Single correct
Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function such that $f(3x) - f(x) = x$. If $f(8) = 7$, then $f(14)$ is equal to:
Maths · Complex Numbers and Quadratic Equations · Single correct
Let O be the origin and A be the point $z_1 = 1 + 2i$. If B is the point $z_2$, $\mathrm{Re}(z_2) < 0$, such that OAB is a right angled isosceles triangle with OB as hypotenuse, then which of the following is NOT true?
If the system of linear equations. $$8x + y + 4z = -2$$ $$x + y + z = 0$$ $$\lambda x - 3y = \mu$$ has infinitely many solutions, then the distance of the point $$\left( \lambda, \mu, -\frac{1}{2} \right)$$ from the plane $$8x + y + 4z + 2 = 0$$ is :
Let A be a 2 $\times$ 2 matrix with $\det(A) = -1$ and $\det((A + I)(Adj(A) + I)) = 4$. Then the sum of the diagonal elements of A can be:
-1
2
1
-$\sqrt{2}$
Answer: (b)
Solution
Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$; $ad - bc = -1$. $$|A + I||adj A + I| = 4$$ $$\Rightarrow ad - bc + a + d + 1 = 2 or -2$$ $a + d = 2 or -2$
Question 5
Maths · Applications of Integrals · Single correct
The odd natural number $a$, such that the area of the region bounded by $y = 1$, $y = 3$, $x = 0$, $x = y^a$ is $\frac{364}{3}$, equal to:
3
5
7
9
Answer: (b)
Solution
Given $$A = \int_1^3 y^a \, dy = \left. \frac{y^{a+1}}{a+1} \right|_1^3 = \frac{364}{3}$$ which implies $$a = 5$$
Question 6
Maths · Sequences and Series · Single correct
Consider two G.Ps. 2, 2^2, 2^3, ... and 4, 4^2, 4^3,.... of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is $(2)^{\frac{25}{8}}$, then $\sum_{k=1}^{n} k(n-k)$ is equal to :
560
1540
1330
2600
Answer: (c)
Solution
Given $$\left( (2^2 \cdot \ldots \cdot 2^{60})(4^1 \cdot 4^2 \cdot \ldots \cdot 4^n) \right)^{\frac{1}{60+n}} = 2^{\frac{225}{8}}$$ We have: $$\left( 2^{30 \times 61} \cdot 4^{\frac{n(n+1)}{2}} \right)^{\frac{1}{60+n}} = 2^{\frac{225}{8}}$$ This simplifies to: $$2^{1830 + \frac{n^2 + n}{2}} = 2^{\frac{(225)(60+n)}{8}}$$ Equating the exponents: $$8n^2 - 217n + 1140 = 0$$ Solving for $$n$$ gives: $$n = 20, \frac{57}{8}$$ Calculating the sum: $$\sum_{k=1}^{n} nk - k^2 = \frac{n^2(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}$$ This results in: $$= 1330$$
Question 7
Maths · Limits and Derivatives · Single correct
If the function $$f(x) = \begin{cases} \frac{\log_e (1-x+x^2) + \log_e (1+x+x^2)}{\sec x - \cos x}, & x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) - \{0\} \\ k, & x = 0 \end{cases}$$ is continuous at $x = 0$, then $k$ is equal to:
Maths · Continuity and Differentiability · Single correct
If $f(x) = \begin{cases} x + a, & x \leq 0 \\ |x - 4|, & x > 0 \end{cases}$ and $g(x) = \begin{cases} x + 1, & x < 0 \\ (x - 4)^2 + b, & x \geq 0 \end{cases}$ are continuous on $\mathbb{R}$, then $(gof)(2) + (fog)(-2)$ is equal to:
$-10$
$10$
$8$
$-8$
Answer: (d)
Solution
Given $$f(x) = \begin{cases} x + a, & x \leq 0 \\ |x - 4|, & x > 0 \end{cases}$$ and $$g(x) = \begin{cases} x + 1, & x < 0 \\ (x - 4)^2 + b, & x \geq 0 \end{cases}$$ For continuity, $a = 4$ and $b = -15$. We have $$g(f(2)) + f(g(-2))$$ which simplifies to $$= g(2) + f(-1) = -8.$$
Question 9
Maths · Applications of Derivatives · Single correct
Let $f(x)=x^{3}-7x^{2}+10x-7,\quad x\leq 1$ $f(x)=-2x+\log_{2}(b^{2}-4),\quad x>1$ Then the set of all values of b, for which f(x) has maximum value at x = 1, is :
(-6, -2)
(2, 6)
[-6, -2) $\cup$ (2, 6]
[-$\sqrt{6}$, -2) $\cup$ (2, $\sqrt{6}$]
Answer: (c)
Solution
Given $f(1) = 3$. For $x 0$. Therefore, $f(x)$ is increasing. For $x > 1$, $f'(x) 0$$ $$b \in [-6, -2) \cup (2, 6]$$
Question 10
Maths · Limits and Derivatives · Single correct
If $a = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{2n}{n^2 + k^2}$ and $f(x) = \sqrt{\frac{1-\cos x}{1+\cos x}}, x \in (0,1)$, then :
A point P moves so that the sum of squares of its distances from the points (1, 2) and (-2, 1) is 14. Let f(x, y) = 0 be the locus of P, which intersects the x-axis at the points A, B and the y-axis at the point C, D. Then the area of the quadrilateral ACBD is equal to
$\frac{9}{2}$
$\frac{3\sqrt{17}}{2}$
$\frac{3\sqrt{17}}{4}$
9
Answer: (b)
Solution
Given $$(x-1)^2 + (y-2)^2 + (x+2)^2 + (y-1)^2 = 14$$ This implies $$x^2 + y^2 + x - 3y - 2 = 0$$ Put $x = 0$ This implies $$y^2 - 3y - 2 = 0$$ This gives $$y = \frac{3 \pm \sqrt{17}}{2}$$ Put $y = 0$ This implies $$x^2 + x - 2 = 0$$ $$(x+2)(x-1) = 0$$ Therefore, the points are $A(-2, 0)$, $B(1, 0)$, $C\left(0, \frac{3+\sqrt{17}}{2}\right)$, $D\left(0, \frac{3-\sqrt{17}}{2}\right)$. The area is $$\frac{1}{2} \cdot 3 \cdot \sqrt{17} = \frac{3\sqrt{17}}{2}$$
Question 13
Maths · Conic Sections · Single correct
Let the tangent drawn to the parabola $y^2 = 24x$ at the point $(\alpha, \beta)$ is perpendicular to the line $2x + 2y = 5$. Then the normal to the hyperbola $$\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$$ at the point $(\alpha + 4, \beta + 4)$ does NOT pass through the point:
(25, 10)
(20, 12)
(30, 8)
(15, 13)
Answer: (d)
Solution
Tangent at $(\alpha, \beta)$ has slope 1 $$\beta^2 = 24 \alpha$$ Equation of tangent $y \beta = 12 (x + \alpha)$, $\frac{12}{\beta} = 1$ $$\Rightarrow \alpha = 6, \beta = 12$$ Therefore, $(\alpha + 4, \beta + 4) = (10, 16)$ Normal at $(10, 16)$ to $$\frac{x^2}{36} - \frac{y^2}{144} = 1$$ is $$2x + 5y = 100$$
Question 14
Maths · Three Dimensional Geometry · Single correct
The length of the perpendicular from the point $(1, -2, 5)$ on the line passing through $(1, 2, 4)$ and parallel to the line $x + y - z = 0 = x - 2y + 3z - 5$ is:
$\sqrt{\frac{21}{2}}$
$\sqrt{\frac{9}{2}}$
$\sqrt{\frac{73}{2}}$
1
Answer: (a)
Solution
The direction ratios of the line are given by the determinant: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 1 & -2 & 3 \end{vmatrix} = \hat{i} - 4\hat{j} - 3\hat{k}$$ Therefore, the equation of the line is $$\vec{r} = \hat{i} + 2\hat{j} + 4\hat{k} + \lambda (\hat{i} - 4\hat{j} - 3\hat{k})$$ Let $A(1, 2, 4)$ and $P$ be $(1 + \lambda, 2 - 4\lambda, 4 - 3\lambda)$. Thus, $$\overrightarrow{PA} \cdot (\hat{i} - 4\hat{j} - 3\hat{k}) = 0$$ Solving for $\lambda$, we get $$\lambda = \frac{1}{2}$$ Therefore, $$\Rightarrow P \left( \frac{1}{2}, 2, -\frac{5}{2} \right)$$ The magnitude of $|AP|$ is $$|AP| = \frac{\sqrt{21}}{2}$$
Question 15
Maths · Vector Algebra · Single correct
Let $\vec{a} = \alpha \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - \alpha \hat{k}$, $\alpha > 0$. If the projection of $\vec{a} \times \vec{b}$ on the vector $-\hat{i} + 2 \hat{j} - 2 \hat{k}$ is 30, then $\alpha$ is equal to
$\frac{15}{2}$
8
$\frac{13}{2}$
7
Answer: (d)
Solution
Given $\vec{a} \times \vec{b} = (1 - \alpha) \hat{i} + (\alpha^2 - 2) \hat{j} + (\alpha - 2) \hat{k}$. Projection of $\vec{a} \times \vec{b}$ on $-\hat{i} + 2\hat{j} - 2\hat{k}$ is given by $$\frac{(\vec{a} \times \vec{b}) \cdot (-\hat{i} + 2\hat{j} - 2\hat{k})}{3} = 30$$ which simplifies to $$\Rightarrow 2\alpha^2 - \alpha - 91 = 0$$ Solving this equation gives $$\Rightarrow \alpha = 7, -\frac{13}{2}$$
Question 16
Maths · Probability · Single correct
The mean and variance of a binomial distribution are $\alpha$ and $\frac{\alpha}{3}$ respectively. If $\mathrm{P}(X=1) = \frac{4}{243}$, then $\mathrm{P}(X=4 or 5)$ is equal to:
$\frac{5}{9}$
$\frac{64}{81}$
$\frac{16}{27}$
$\frac{145}{243}$
Answer: (c)
Solution
Given $np = \alpha$ $\($$\ldots$$\)$(1) and $npq = \alpha / 3$ $\($$\ldots$$\)$(2). From (1) and (2), $q = 1/3$ and $p = 2/3$. $$\binom{n}{1} q^{n-1} p^1 = \frac{4}{243}$$ $$\frac{n}{3^n} = \frac{2}{243}$$ Thus, $n = 6$. The probability $P(4 or 5) = \binom{6}{4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^2 + \binom{6}{5} \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^0 = \frac{16}{27}.$
Question 17
Maths · Probability · Single correct
Let $E_1$, $E_2$, $E_3$ be three mutually exclusive events such that $\mathrm{P}(E_1) = \frac{2 + 3p}{6}$, $\mathrm{P}(E_2) = \frac{2 - p}{8}$ and $\mathrm{P}(E_3) = \frac{1 - p}{2}$. If the maximum and minimum values of $p$ are $p_1$ and $p_2$, then $(p_1 + p_2)$ is equal to:
$\frac{2}{3}$
$\frac{5}{3}$
$\frac{5}{4}$
1
Answer: (d)
Solution
Given $0 \leq \mathrm{P}(E_i) \leq 1$ for $i = 1, 2, 3$. This implies $-2/3 \leq p \leq 1$. $E_1$, $E_2$, and $E_3$ are mutually exclusive. Therefore, $\mathrm{P}(E_1) + \mathrm{P}(E_2) + \mathrm{P}(E_3) \leq 1$. This implies $2/3 \leq p \leq 1$. Given $p_1 = 1$, $p_2 = 2/3$. Therefore, $p_1 + p_2 = 5/3$.
Question 18
Maths · Trigonometric Functions · Single correct
Let $$S = \{ \theta \in [0, 2\pi] : 8^{2\sin^2 \theta} + 8^{2\cos^2 \theta} = 16 \}.$$ Then $$n(S) + \sum_{\theta \in S} \left( \sec \left( \frac{\pi}{4} + 2\theta \right) \csc \left( \frac{\pi}{4} + 2\theta \right) \right)$$ is equal to:
The expression $\sim (p \iff \sim q) \land q \equiv (p \iff q) \land q$. This simplifies to $(p \iff q) \land q \equiv p \land q$. The Venn diagrams illustrate the logical equivalence.
Question 21
Maths · Complex Numbers and Quadratic Equations · Numerical
If for some $p, q, r \in \mathbb{R}$, not all have same sign, one of the roots of the equation $(p^2 + q^2)x^2 - 2q(p + r)x + q^2 + r^2 = 0$ is also a root of the equation $x^2 + 2x - 8 = 0$, then $\frac{q^2 + r^2}{p^2}$ is equal to-
The number of 5-digit natural numbers, such that the product of their digits is 36, is
Answer: 180
Solution
The expression is given by $$3 \times \frac{5!}{2!2!} + \frac{5!}{3! \times 2!} + \frac{5!}{2!} + \frac{5!}{3!} = 180.$$
Question 23
Maths · Sequences and Series · Numerical
The series of positive multiples of 3 is divided into sets: $\{3\}$, $\{6, 9, 12\}$, $\{15, 18, 21, 24, 27\}$, $\ldots$ Then the sum of the elements in the $11^{\text{th}}$ set is equal to:
Answer: 6993
Solution
Given $S_{11} = 3[101 + 102 + \ldots + 121]$. This simplifies to $$= \frac{3}{2}(222) \times 21 = 6993.$$
Question 24
Maths · Complex Numbers and Quadratic Equations · Numerical
The number of distinct real roots of the equation $$x^5(x^3 - x^2 - x + 1) + x(3x^3 - 4x^2 - 2x + 4) - 1 = 0$$ is
Answer: 3
Solution
Given the equation $$x^5 (x^3 - x^2 - x + 1) + x (3x^3 - 4x^2 - 2x + 4) - 1 = 0$$ we have: $$\Rightarrow (x - 1)^2 (x + 1) \left(x^5 + 3x - 1\right) = 0$$ Let $f(x) = x^5 + 3x - 1$. The derivative $f'(x) > 0 \; \forall \; x \in \mathbb{R}$. Hence, there are 3 real distinct roots.
Question 25
Maths · Binomial Theorem · Numerical
If the coefficients of $x$ and $x^2$ in the expansion of $(1 + x)^p (1 - x)^q$, $p, q \leq 15$, are $-3$ and $-5$ respectively, then the coefficient of $x^3$ is equal to
Answer: 23
Solution
Since coefficient of $x$ is $-3$ $\[$ $\Rightarrow$ $\binom{p}{1}$ - $\binom{q}{1}$ = -3 $\]$ $\[$ $\Rightarrow$ p - q = -3 $\]$ Comparing coefficients of $x^2$ $\[$ -$\binom{p}{1}$$\binom{q}{1}$ + $\binom{p}{2}$ + $\binom{q}{2}$ = -5 $\]$ $\[$ -pq + $\frac{p(p-1)}{2}$ + $\frac{q(q-1)}{2}$ = -5 $\]$ Solving (1) and (2) $\[$ p = 8, $\ $q = 11 $\]$ Coefficient of $x^3$ is $\[$ -$\binom{q}{3}$ + $\binom{p}{3}$ + $\binom{p}{1}$$\binom{q}{2}$ - $\binom{p}{2}$$\binom{q}{1}$ $\]$ $\[$ = -$\binom{11}{3}$ + $\binom{8}{3}$ + $\binom{8}{1}$$\binom{11}{2}$ - $\binom{8}{2}$$\binom{11}{1}$ $\]$ $\[$ = 23 $\]$
Question 26
Maths · Integrals · Fill in the blank
If $$n(2n+1) \int_0^1 (1-x^n)^{2n} \, dx = 1177 \int_0^1 (1-x^n)^{2n+1} \, dx,$$ then $$n \in \mathbb{N}$$ is equal to ______
Let a curve $y = y(x)$ pass through the point $(3, 3)$ and the area of the region under this curve, above the x-axis and between the abscissae 3 and $x(>3)$ be $\left(\frac{y}{x}\right)^3$. If this curve also passes through the point $\left(\alpha, 6\sqrt{10}\right)$ in the first quadrant, then $\alpha$ is equal to _______
Maths · Straight Lines and Pair of Straight Lines · Numerical
The equations of the sides $AB$, $BC$ and $CA$ of a triangle $ABC$ are $2x + y = 0$, $x + py = 15a$ and $x - y = 3$ respectively. If its orthocentre is $(2, a)$, $\frac{1}{2} < a < 2$, then $p$ is equal to
Answer: 3
Solution
Coordinates of $A(1, -2)$, $B\left( \frac{15a}{1-2p}, \frac{-30a}{1-2p} \right)$ and orthocentre $H(2, a)$. Slope of $AH = p$. $a + 2 = p$ $\($1) Slope of $BH = -1$. $31a - 2ab = 15a + 4p - 2$ $\($2) From (1) and (2), $a = 1$ and $p = 3$.
Question 29
Maths · Conic Sections · Numerical
Let the function $f(x) = 2x^2 - \log_e x$, $x > 0$, be decreasing in $(0, a)$ and increasing in $(a, 4)$. A tangent to the parabola $y^2 = 4ax$ at a point $P$ on it passes through the point $(8a, 8a - 1)$ but does not pass through the point $\left(-\frac{1}{a}, 0\right)$. If the equation of the normal at $P$ is $\frac{x}{\alpha} + \frac{y}{\beta} = 1$, then $\alpha + \beta$ is equal to-
Answer: 45
Solution
Given $f'(x) = 4x - \frac{1}{x}$. $a = \frac{1}{2}$. Let $P(x_1, y_1)$ be any point on $y^2 = 4ax$. $$\frac{1}{y_1} = \frac{3 - y_1}{4 - x_1} \implies y_1^2 - 6y_1 + 8 = 0$$ $y_1 = 2, 4$. $\Rightarrow P(8, 4)$ as $P(2, 2)$ rejected. Equation of normal at $P$. $y - 4 = -4(x - 8)$. $$\frac{x}{9} + \frac{y}{36} = 1$$ $\alpha = 9$, $\beta = 36$. $\alpha + \beta = 45$
Question 30
Maths · Three Dimensional Geometry · Fill in the blank
Let Q and R be two points on the line $\frac{x+1}{2} = \frac{y+2}{3} = \frac{z-1}{2}$ at a distance $\sqrt{26}$ from the point P(4, 2, 7). Then the square of the area of the triangle PQR is __________.
Answer: 153
Solution
Let $(2\lambda - 1, 3\lambda - 2, 2\lambda + 1)$ be any point on the line $$(2\lambda - 5)^2 + (3\lambda - 4)^2 + (2\lambda - 6)^2 = 26$$ $\lambda = 1, 3$ $Q (1, 1, 3); R (5, 7, 7); P (4, 2, 7)$ Area of triangle $PQR = \frac{1}{2} \left| \overrightarrow{PQ} \times \overrightarrow{PR} \right|$ $$= \sqrt{153}$$
Physics
Question 31
Physics · Laws of Motion · Single correct
Three masses $M = 100 \, \mathrm{kg}$, $m_1 = 10 \, \mathrm{kg}$ and $m_2 = 20 \, \mathrm{kg}$ are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force $F$ is applied on the system so that the mass $m_2$ moves upward with an acceleration of $2 \, \mathrm{ms^{-2}}$. The value of $F$ is: (Take $g = 10 \, \mathrm{ms^{-2}}$)
3360 N
3380 N
3120 N
3240 N
Answer: (a)
Solution
Let acceleration of 100 kg block be $a_1$. FBD of 100 kg block with respect to ground: $$F - T - N_1 = 100 \, a_1 (i)$$ FBD of 20 kg block with respect to 100 kg: $$T - 20g = 20(2)$$ $$T = 240 (ii)$$ $$N_1 = 20a_1 (iii)$$ FBD of 10 kg block with respect to 100 kg: $$10a_1 - 240 = 10(2)$$ $$a_1 = 26 \, \mathrm{m/s^2}$$ $$F - 240 - 20(26) = 100 \times 26$$ $$\Rightarrow F = 3360 \, \mathrm{N}$$
Question 32
Physics · Communication Systems · Single correct
A radio can tune to any station in 6 $\mathrm{MHz}$ to 10 $\mathrm{MHz}$ band. The value of corresponding wavelength bandwidth will be:
The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minute. 10 minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is : (Take $\log_{10} 1.88 = 0.274$)
0.02 $\mathrm{min}$^{-1}
2.7 $\mathrm{min}$^{-1}
0.063 $\mathrm{min}$^{-1}
6.3 $\mathrm{min}$^{-1}
Answer: (c)
Solution
At $t=0$, the disintegration rate is $4250 \, dpm$. At $t=10$, the disintegration rate is $2250 \, dpm$. The equation is given by $$A = A_o e^{-\lambda t}$$ Substituting the values, $$2250 = 4250 e^{-\lambda (10)}$$ Solving for $\lambda$, $$\Rightarrow \lambda (10) = \ln \left( \frac{4250}{2250} \right)$$ $$\Rightarrow \lambda = 0.063 \, min^{-1}$$
Question 34
Physics · Dual Nature of Radiation and Matter · Single correct
A parallel beam of light of wavelength 900 nm and intensity 100 Wm^{-2} is incident on a surface perpendicular to the beam. Tire number of photons crossing 1 cm^2 area perpendicular to the beam in one second is:
3 $\times$ 10^{16}
4.5 $\times$ 10^{16}
4.5 $\times$ 10^{17}
4.5 $\times$ 10^{20}
Answer: (b)
Solution
Wavelength of incident beam $\lambda = 900 \times 10^{-9} \, \mathrm{m}$. Intensity of incident beam $I = 100 \, \mathrm{W/m^2}$. Number of photons crossing per unit second $$n = \frac{E_{net}}{E_{single photon}} = \frac{IA\lambda}{hc}$$ $$= \frac{(100)(1 \times 10^{-4})(900 \times 10^{-9})}{6.62 \times 10^{-34} \times 3 \times 10^8} = 4.5 \times 10^{16}$$
Question 35
Physics · Wave Optics · Single correct
In young’s double slit experiment, the fringe width is $12 \, \mathrm{mm}$. If the entire arrangement is placed in water of refractive index $\frac{4}{3}$, then the fringe width becomes (in mm)
16
9
48
12
Answer: (b)
Solution
For a given light wavelength corresponding a medium of refractive index $\mu$ $$\lambda_{med} = \frac{\lambda_{vacuum}}{\mu}$$ and we know that fringe width $\beta = \frac{\lambda D}{d}$ Therefore, $$\beta_{med} = \frac{\beta_{vacuum}}{\mu} = \frac{12}{\frac{4}{3}} = 9 \, \mathrm{mm}$$
Question 36
Physics · Electromagnetic Waves · Single correct
The magnetic field of a plane electromagnetic wave is given by $$\mathbf{B} = 2 \times 10^{-8} \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t) \hat{\mathbf{j}} \, \mathrm{T}$$ The amplitude of the electric field would be
$6 \, \mathrm{Vm}^{-1}$ along x-axis
$3 \, \mathrm{Vm}^{-1}$ along z-axis
$6 \, \mathrm{Vm}^{-1}$ along z-axis
$2 \times 10^{-8} \, \mathrm{Vm}^{-1}$ along z-axis
Answer: (c)
Solution
Given $c=\frac{E_0}{B_0}$, it follows that $E_0=cB_0$. $$ E_0=\left(3\times10^8\right)\left(2\times10^{-8}\right) $$ $$ E_0=6\,\mathrm{V\,m^{-1}} $$ As $\vec{B}$ is along the y-axis and $\vec{v}$ is along the negative x-axis, $\vec{E}_0$ is along the z-axis.
Question 37
Physics · Alternating Current · Single correct
In a series LR circuit $X_L = R$ and power factor of the circuit is $P_1$. When capacitor with capacitance $C$ such that $X_L = X_C$ is put in series, the power factor becomes $P_2$. The ratio $$\frac{P_1}{P_2}$$ is
$\frac{1}{2}$
$\frac{1}{\sqrt{2}}$
$\frac{\sqrt{3}}{\sqrt{2}}$
2 : 1
Answer: (b)
Solution
In case of L-R circuit, $$Z = \sqrt{X_L^2 + R^2}$$ and power factor $$P_1 = \cos \phi = \frac{R}{Z}$$ As $X_L = R$, $$\Rightarrow Z = \sqrt{2R}$$ $$\Rightarrow P_1 = \frac{R}{\sqrt{2R}} \Rightarrow P_1 = \frac{1}{\sqrt{2}}$$ In case of L-C-R circuit, $$Z = \sqrt{R^2 + (X_L - X_C)^2}$$ As $X_L = X_C$, $$\Rightarrow Z = R$$ $$\Rightarrow P_2 = \cos \phi = \frac{R}{R} = 1$$ $$\Rightarrow \frac{P_1}{P_2} = \frac{1}{\sqrt{2}}$$
Question 38
Physics · Moving Charges and Magnetism · Single correct
A charge particle is moving in a uniform magnetic field $\left(2\hat{i} + 3\hat{j}\right)\, \mathrm{T}$. If it has an acceleration of $\left(\alpha \hat{i} - 4\hat{j}\right)\, \mathrm{m/s^2}$, then the value of $\alpha$ will be
3
6
12
2
Answer: (b)
Solution
As $\vec{F} = q(\vec{v} \times \vec{B})$ $$\vec{a} = \frac{q}{m}(\vec{v} \times \vec{B})$$ So, $\vec{a}$ and $\vec{B}$ are $\perp$ to each other. Hence, $\vec{a} \cdot \vec{B} = 0$ $$\left( \alpha \hat{i} - 4 \hat{j} \right) \cdot \left( 2 \hat{i} + 3 \hat{j} \right) = 0$$ $$\alpha (2) + (-4)(3) = 0$$ $$\alpha = \frac{12}{2} \implies \alpha = 6$$
Question 39
Physics · Moving Charges and Magnetism · Single correct
$B_X$ and $B_Y$ are the magnetic field at the centre of two coils of two coils X and Y respectively, each carrying equal current. If coil X has 200 turns and 20 cm radius and coil Y has 400 turns and 20 cm radius, the ratio of $B_X$ and $B_Y$ is
Given circuit is balanced wheatstone bridge. Hence $2\, \Omega$ can be neglected. $$R_{net} = 4\, \Omega$$ $$I = \frac{40}{4}$$ $$I = 10\, A$$
Question 41
Physics · Electrostatic Potential and Capacitance · Single correct
The total charge on the system of capacitance $C_1 = 1 \, \mu \mathrm{F}$, $C_2 = 2 \, \mu \mathrm{F}$, $C_3 = 4 \, \mu \mathrm{F}$ and $C_4 = 3 \, \mu \mathrm{F}$ connected in parallel is (Assume a battery of $20 \, \mathrm{V}$ is connected to the combination)
When a particle executes simple Harmonic motion, the nature of graph of velocity as function of displacement will be :
Circular
Elliptical
Sinusoidal
Straight line
Answer: (b)
Solution
For a particle in SHM, its speed depends on position as $$v = \omega \sqrt{A^2 - x^2}$$ Where $\omega$ is angular frequency and $A$ is amplitude. Now $v^2 = \omega^2 A^2 - \omega^2 x^2$ So, $$\frac{v^2}{(\omega A)^2} + \frac{x^2}{(A)^2} = 1$$ So graph between $v$ and $x$ is elliptical.
Question 43
Physics · Thermodynamics · Single correct
7 mole of certain monoatomic ideal gas undergoes a temperature increase of 40K at constant pressure. The increase in the internal energy of the gas in this process is (Given R = $8.3 \, \mathrm{JK^{-1}}$ $\mathrm{mol^{-1}}$)
5810 J
3486 J
11620 J
6972 J
Answer: (b)
Solution
Question 44
Physics · Thermodynamics · Single correct
A monoatomic gas at pressure $P$ and volume $V$ is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be:
$P$
$8P$
$32P$
$64P$
Answer: (c)
Solution
Question 45
Physics · Mechanical Properties of Fluids · Single correct
A water drop of radius 1 cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/cm, then the gain in surface energy upto first decimal place will be: [Given $\pi = 3.14$]
$8.5 \times 10^{-4} \mathrm{J}$
$8.2 \times 10^{-4} \mathrm{J}$
$7.5 \times 10^{-4} \mathrm{J}$
$5.3 \times 10^{-4} \mathrm{J}$
Answer: (c)
Solution
Initial surface energy = TA Where T is surface tension and A is surface area $$U_i = \left( \frac{75 \times 10^{-5} \, \mathrm{N}}{10^{-2} \, \mathrm{m}} \right) \times \left[ 4 \pi \left( 1 \times 10^{-2} \right)^2 \right]$$ $$= 75 \times 10^{-3} \times 4 \pi \times 10^{-4} = 942 \times 10^{-7} \, \mathrm{J}$$ To get final radius of drops by volume conservation $$\frac{4}{3} \pi R^3 = 729 \left( \frac{4}{3} \pi r^3 \right)$$ R = Initial radius r = final radius $$r = \frac{R}{(729)^{1/3}} = \frac{R}{9} = \frac{1}{9} \, \mathrm{cm}$$ Final surface energy $$U_f = 729 \left[ TA \right]$$ $$= 729 \left[ \frac{75 \times 10^{-5} \, \mathrm{N}}{10^{-2} \, \mathrm{m}} \times \left[ 4 \pi \left( \frac{1}{9} \times 10^{-2} \right)^2 \right] \right]$$ $$= 729 \left[ 75 \times 10^{-3} \times \frac{4 \pi \times 10^{-4}}{81} \right]$$ $$= 9 \left[ 942 \times 10^{-7} \, \mathrm{J} \right]$$ Gain in surface energy $$\Delta U = 9 \times 942 \times 10^{-7} - 942 \times 10^{-7}$$ $$= 8 \times 942 \times 10^{-7} = 7536 \times 10^{-7} \, \mathrm{J}$$
Question 46
Physics · Gravitation · Single correct
The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be : (Radius of earth = 6400 km)
1 %
3 %
4 %
0.5 %
Answer: (a)
Solution
Acceleration due to gravity at a height $h \ll R$ is $$g' = g \left( 1 - \frac{2h}{R} \right)$$ Therefore, $$\frac{\Delta g}{g} = \frac{2h}{R}$$ Thus, $$\frac{\Delta g}{g} \times 100 = \frac{2h}{R} \times 100$$ $$= 2 \times \frac{32}{6400} \times 100 = 1\%$$
Question 47
Physics · Oscillations · Single correct
As per the given figure, two blocks each of mass 250 g are connected to a spring of spring constant 2 $\mathrm{Nm}^{-1}$. If both are given velocity v in opposite directions, then maximum elongation of the spring is :
$\frac{v}{2\sqrt{2}}$
$\frac{v}{2}$
$\frac{v}{4}$
$\frac{v}{\sqrt{2}}$
Answer: (b)
Solution
Using energy conservation, $\($ $\frac{1}{2}$ mv^2 $\times$ 2 = $\frac{1}{2}$ kx^2 $\)$. Therefore, $\($ $\frac{1}{4}$ v^2 = $\frac{1}{2}$ $\times$ 2 $\times$ x^2 $\)$. Thus, $\($ x = $\frac{v}{2}$ $\)$.
Question 48
Physics · Laws of Motion · Single correct
A monkey of mass 50 kg climbs on a rope which can withstand the tension (T) of 350 N. If monkey initially climbs down with an acceleration of 4 m/s^2 and then climbs up with an acceleration of 5 m/s^2. Choose the correct option (g = 10 m/s^2)
T = 700 N while climbing upward
T = 350 N while going downward
Rope will break while climbing upward
Rope will break while going downward
Answer: (c)
Solution
F.B.D of monkey while moving downward. Using Newton's second law, $mg - T = ma_1$. Therefore, $500 - T = 50 \times 4 \implies T = 300 \, \mathrm{N}$. F.B.D of monkey while moving up. Using Newton's second law of motion, $T - mg = ma_2$. Therefore, $T - 500 = 50 \times 5$. Thus, $T = 750 \, \mathrm{N}$. Breaking strength of string = $350 \, \mathrm{N}$. String will break while monkey is moving upward.
Question 49
Physics · Motion in a Plane · Single correct
Two projectile thrown at $30^0$ and $45^0$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is
1 : $\sqrt{2}$
2 : 1
$\sqrt{2}$ : 1
1 : 2
Answer: (c)
Solution
Time taken to reach maximum height $$t = \frac{u \sin \theta}{g}$$ Therefore, $$\frac{u_1 \sin \theta_1}{g} = \frac{u_2 \sin \theta_2}{g}$$ $$\Rightarrow u_1 \sin 30 = u_2 \sin 45$$ $$\Rightarrow \frac{u_1}{u_2} = \frac{1/\sqrt{2}}{1/2} = \frac{\sqrt{2}}{1}$$
Question 50
Physics · Experimental Physics · Single correct
A screw gauge of pitch 0.5 mm is used to measure the diameter of uniform wire of length 6.8 cm, the main scale reading is 1.5 mm and circular scale reading is 7. The calculated curved surface area of wire to appropriate significant figures is: [Screw gauge has 50 divisions on the circular scale]
$6.8\,\mathrm{cm^2}$
$3.4\,\mathrm{cm^2}$
$3.9\,\mathrm{cm^2}$
$2.4\,\mathrm{cm^2}$
Answer: (b)
Solution
L.C. $=\dfrac{P}{N}=\dfrac{0.5\,\mathrm{mm}}{50}=0.01\,\mathrm{mm}$ Length of wire $=6.8\,\mathrm{cm}$ Diameter of wire $=1.5\,\mathrm{mm}+7\times\mathrm{L.C.}$ $=1.5\,\mathrm{mm}+7\times0.01\,\mathrm{mm}$ $=1.57\,\mathrm{mm}$ Curved surface area $=\pi D\ell$ $=3.14\times1.57\times10^{-1}\times6.8\,\mathrm{cm^2}$ $=3.352\,\mathrm{cm^2}$ $\approx 3.4\,\mathrm{cm^2}$
Question 51
Physics · Motion in a Plane · Numerical
If the initial velocity in horizontal direction of a projectile is unit vector $\hat{i}$ and the equation of trajectory is $y = 5x(1-x)$. The y component vector of the initial velocity is _____ $\hat{j}$ (Take $g = 10 \, \mathrm{m/s^2}$)
Answer: 5
Solution
Given $u_x = 1$ and $y = 5x(1-x)$. The derivative is $$\frac{dy}{dt} = 5 \frac{dx}{dt} - 10x \frac{dx}{dt}.$$ For the initial $y$-component of velocity, $$u_y = \left( \frac{dy}{dt} \right)_{x=0} \Rightarrow 5(1) = 5.$$ Therefore, $$\bar{u}_y = 5 \hat{j}.$$
Question 52
Physics · System of Particles and Rotational Motion · Numerical
A disc of mass 1 kg and radius R is free of rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be $$4 \sqrt{\frac{x}{3R}} rad s^{-1}$$ where $x = $ $$\left( g = 10 ms^{-2} \right)$$
Physics · Physical World, Units and Measurements · Numerical
In an experiment of determine the Young’s modulus of wire of a length exactly 1m, the extension in the length of the wire is measured as 0.4mm with an uncertainty of $\pm 0.02 \, \mathrm{mm}$ when a load of 1kg is applied. The diameter of the wire is measured as 0.4mm with an uncertainty of $\pm 0.01 \, \mathrm{mm}$. The error in the measurement of Young’s modulus ($\Delta Y$) is found to be $x \times 10^{10} \, \mathrm{Nm}^{-2}$. The value of $x$ is [Take $g = 10 \, \mathrm{m/s}^2$]
Answer: 2
Solution
Given $L = 1 \, \mathrm{m}$, $\Delta L = 0.4 \times 10^{-3} \, \mathrm{m}$, $m = 1 \, \mathrm{kg}$, $d = 0.4 \times 10^{-3} \, \mathrm{m}$. The formula is $\frac{F}{A} = Y \frac{\Delta L}{L}$. Therefore, $$Y = \frac{FL}{A \Delta L} = \frac{(mg) \cdot (1)}{\left( \frac{\pi d^2}{4} \right) 0.4 \times 10^{-3}}$$ This simplifies to $$\Rightarrow \frac{10 \times 4}{\pi (0.4 \times 10^{-3})^2 \times 0.4 \times 10^{-3}}$$ Hints and Solutions: $$Y = \frac{40 \times 7}{22 \times 64 \times 10^{-3} \times 10^{-9}}$$ $$Y = 0.199 \times 10^{-12} \, \mathrm{N/m^2}$$ The change in $Y$ is given by $$\frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta L}{L} + \frac{\Delta A}{A} + \frac{\Delta (\Delta L)}{(\Delta L)}$$ This becomes $$= \frac{0.02}{0.4} + 2 \frac{\Delta d}{d} = \frac{0.2}{4} + 2 \times \frac{0.01}{0.4}$$ $$= \frac{0.1}{2} + \frac{0.1}{2} = 0.1$$ Thus, $$\Rightarrow \Delta Y = 0.1 \times Y$$ Finally, $$= 0.199 \times 10^{11} = 1.99 \times 10^{10}$$
Question 54
Physics · Waves · Numerical
When a car is approaching the observer, the frequency of horn is 100 Hz. After passing the observer, it is 50 Hz. If the observer moves with the car, the frequency will be $\frac{x}{3}$ Hz where $x =$
Answer: 200
Solution
Given $f_1 = 100 = f_0 \left( \frac{C}{C - V_s} \right)$. $C$ is the speed of sound. $V_s$ is the speed of the source. Given $f_2 = 50 = f_0 \left( \frac{C}{C + V_s} \right)$. From the equations: $$C - V_s = \frac{C}{2}$$ Solving for $V_s$: $$2C - 2V_s = C + V_s$$ $$3V_s = C$$ $$V_s = \frac{C}{3}$$ Substitute $V_s$ back: $$100 = f_0 \frac{C}{2C/3} = \frac{3}{2} f_0$$ Solving for $f_0$: $$f_0 = \frac{200}{3}$$
Question 55
Physics · Electrostatic Potential and Capacitance · Fill in the blank
A composite parallel plate capacitor is made up of two different dielectric materials with different thickness ($t_1$ and $t_2$) as shown in figure. The two different dielectric material are separated by a conducting foil F. The voltage of the conducting foil is ____ V.
Answer: 60
Solution
Capacitance of each capacitor $$C_1 = \frac{A 3 \varepsilon_0}{\frac{1}{2}} = 6 A \varepsilon_0$$ $$C_2 = A 4 \varepsilon_0 = 4 A \varepsilon_0$$ Equivalent capacitance $$C_{eq} = \frac{C_1 C_2}{C_1 + C_2} \Rightarrow \frac{24}{10} A \varepsilon_0$$ $$q_{net} = C_{eq} (\Delta V) \Rightarrow 240 A \varepsilon_0$$ $$\Delta V_2 = \frac{240 A \varepsilon_0}{4 A \varepsilon_0} = 60 V$$ $$(\Delta V_2 = Potential drop across C_2)$$ $$V_{foil} = 60 V$$
Question 56
Physics · Current Electricity · Numerical
Resistance are connected in a meter bridge circuit as shown in the figure. The balancing length $l_1$ is $40\,\mathrm{cm}$. Now an unknown resistance $x$ is connected in series with $P$ and new balancing length is found to be $80\,\mathrm{cm}$ measured from the same end. Then the value of $x$ will be $\Omega$
The effective current $I$ in the given circuit at very high frequencies will be ____ A
Answer: 44
Solution
At very high frequencies, $X_C = \frac{1}{\omega C} \approx 0$. Also $X_L = \omega L \approx \infty$. Thus, the equivalent circuit can be redrawn as shown in the diagram. The total impedance $Z = 1 + 2 + 2 = 5 \, \Omega$. The current $I = \frac{220 \, \mathrm{V}}{5 \, \Omega} = 44 \, \mathrm{A}$.
Question 58
Physics · Ray Optics and Optical Instruments · Numerical
The graph between $\frac{1}{u}$ and $\frac{1}{v}$ for a thin convex lens in order to determine its focal length is plotted as shown in the figure. The refractive index of length is 1.5 and its both the surfaces have same radius of curvatures R. The value of R will be ____ cm. (Where $u =$ object distance , $v =$ image distance)
In a hydrogen spectrum, $\lambda$ be the wavelength of first transition line of Lyman series. The wavelength difference will be "$a\lambda$" between the wavelength of $3^{\text{rd}}$ transition line of Paschen series and that of $2^{\text{nd}}$ transition line of Balmer series where $a =$
Answer: 5
Solution
For the first line of Lyman $$\frac{1}{\lambda} = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right)$$ Thus, $$\lambda = \frac{4}{3R} \ldots (1)$$ For the 3rd line (Paschen) $$\frac{1}{\lambda_3} = R \left( \frac{1}{3^2} - \frac{1}{6^2} \right) = \frac{R}{9} \times \frac{3}{4}$$ For the 2nd line (Balmer) $$\frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = \frac{R}{4} \times \frac{3}{4}$$ Thus, $a \lambda = \lambda_3 - \lambda_2 = \frac{12}{R} - \frac{16}{3R} = \frac{20}{3R}$ putting (1) $$a \left( \frac{4}{3R} \right) = \frac{20}{3R} \implies a = 5$$
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Match List - I with List - II. \begin{tabular}{|c|c|c|c|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{(Compound)} & \multicolumn{2}{c|}{(Shape)} \\ \hline (A) & BrF$_5$ & (I) & bent \\ \hline (B) & [CrF$_6$]$^{3-}$ & (II) & square pyramidal \\ \hline (C) & O$_3$ & (III) & trigonal bipyramidal \\ \hline (D) & PCl$_5$ & (IV) & octahedral \\ \hline \end{tabular} Choose the correct answer from the options given below:
(A)- (I), (B) - (II), $(C)$ - (III), (D) - (IV)
(A)- (IV), (B) - (III), $(C)$ - (II), (D) - (I)
(A)- (II), (B) - (IV), $(C)$ - (I), (D) - (III)
(A)- (III), (B) - (IV), $(C)$ - (II), (D) - (I)
Answer: (c)
Solution
Question 62
Chemistry · Surface Chemistry · Single correct
Match List - I with List - II. \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{(Processes/Reactions)} & \multicolumn{2}{c|}{(Catalyst)} \\ \hline (A) & $2SO_2(g)+O_2(g)\rightarrow2SO_3(g)$ & (I) & Fe(s) \\ \hline (B) & $4NH_3(g)+5O_2(g)\rightarrow4NO(g)+6H_2O(g)$ & (II) & Pt(s)-Rh(s) \\ \hline (C) & $N_2(g)+3H_2(g)\rightarrow2NH_3(g)$ & (III) & V$_2$O$_5$ \\ \hline (D) & Vegetable oil$(l)+H_2\rightarrow$ Vegetable ghee$(s)$ & (IV) & Ni(s) \\ \hline \end{tabular} Choose the correct answer from the options given below:
(A)- (III), (B) - (I), $(C)$ - (II), (D) - (IV)
(A)- (III), (B) - (II), $(C)$ - (I), (D) - (IV)
(A)- (IV), (B) - (III), $(C)$ - (I), (D) - (II)
(A)- (IV), (B) - (II), $(C)$ - (III), (D) - (I)
Answer: (b)
Solution
The reaction $2\mathrm{SO_2} (g) + \mathrm{O_2} (g) \xrightarrow{\mathrm{V_2O_5}} 2\mathrm{SO_3} (g)$ is the contact process. The reaction $4\mathrm{NH_3} (g) + 5\mathrm{O_2} (g) \xrightarrow{\mathrm{Pt(s) - Rh(s)}} 4\mathrm{NO} (g) + 6\mathrm{H_2O} (g)$ is Ostwald's process. The reaction $\mathrm{N_2} (g) + 3\mathrm{H_2} (g) \xrightarrow{\mathrm{Fe(s)}} 2\mathrm{NH_3} (g)$ is Haber’s process. The reaction $Vegetable oil (l) + \mathrm{H_2} (g) \xrightarrow{\mathrm{Ni(s)}} vegetable ghee$ is Hydrogenation.
Question 63
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Given two statements below : Statement I : In $\mathrm{Cl}_2$ molecule the covalent radius is double of the atomic radius of chlorine. Statement II : Radius of anionic species is always greater than their parent atomic radius. Choose the most appropriate answer from options given below :
Both Statement I and Statement II are correct.
Both Statement I and Statement II are incorrect.
Statement I is correct but Statement II is incorrect.
Statement I is incorrect but Statement II is correct.
Answer: (d)
Solution
In $\mathrm{Cl_2}$ molecule, the covalent radius is half of the internuclear distance, so statement (I) is false. For the same element, anion has lower effective nuclear charge than atom $\Rightarrow$ so anion is larger than atom. $\Rightarrow$ statement (II) is correct.
Question 64
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Refining using liquation method is the most suitable for metals with:
Low melting point
High boiling point
High electrical conductivity
Less tendency to be soluble in melts than impurities
Answer: (a)
Solution
Liquation is used to purify metals having lower melting point than impurities present in them.
Question 65
Chemistry · The s-Block Elements · Single correct
Which of the following can be used to prevent the decomposition of $\mathrm{H_2O_2}$?
Urea
Formaldehyde
Formic acid
Ethanol
Answer: (a)
Solution
Urea acts as stabiliser for $\mathrm{H_2O_2}$.
Question 66
Chemistry · The s-Block Elements · Single correct
Reaction of BeCl$_2$ with LiAlH$_4$ gives : (A) AlCl$_3$ (B) BeH$_2$ $(C)$ LiH (D) LiCl (E) BeAlH$_4$ Choose the correct answer from the options given below :
, (D) and (E)
, (B) and (D)
and (E)
, $(C)$ and (D)
Answer: (b)
Solution
The reaction is given by: $$2\mathrm{BeCl_2} + \mathrm{LiAlH_4} \rightarrow 2\mathrm{BeH_2} + \mathrm{LiCl} + \mathrm{AlCl_3}$$
Question 67
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
Borazine, also known as inorganic benzene, can be prepared by the reaction of 3-equivalents of "X" with 6-equivalents of "Y". "X" and "Y", respectively are :
$B(OH)_3$ and $NH_3$
$B_2H_6$ and $NH_3$
$B_2H_6$ and $HN_3$
$NH_3$ and $B_2O_3$
Answer: (b)
Solution
The reaction is given by: $$3\mathrm{B_2H_6} + 6\mathrm{NH_3} \xrightarrow{\Delta} 2\mathrm{B_3N_3H_6} + 12\mathrm{H_2}$$
Question 68
Chemistry · Redox Reactions · Single correct
Which of the given reactions is not an example of disproportionation reaction?
The reaction $2\mathrm{H_2O_2} \rightarrow 2\mathrm{H_2O} + \mathrm{O_2}$ is a disproportionation reaction. The reaction $2\mathrm{NO_2} + \mathrm{H_2O} \rightarrow \mathrm{HNO_3} + \mathrm{HNO_2}$ is a disproportionation reaction. The reaction $\mathrm{MnO_4^-} + 4\mathrm{H^+} + 3\mathrm{e^-} \rightarrow \mathrm{MnO_2} + 2\mathrm{H_2O}$ is a reduction reaction. The reaction $3\mathrm{MnO_4^{2-}} + 4\mathrm{H^+} \rightarrow 2\mathrm{MnO_4^-} + \mathrm{MnO_2} + 2\mathrm{H_2O}$ is a disproportionation reaction.
Question 69
Chemistry · Redox Reactions · Single correct
The dark purple colour of $\mathrm{KMnO_4}$ disappears in the titration with oxalic acid in acidic medium. The overall change in the oxidation number of manganese in the reaction is:
5
1
7
2
Answer: (a)
Solution
In acidic medium, $$\mathrm{MnO_4^-} \rightarrow \mathrm{Mn^{+2}}$$ change in ox. no. = 5
Question 70
Chemistry · Haloalkanes and Haloarenes · Single correct
\[ \overset{\bullet}{\mathrm{Cl}} + \mathrm{CH}_4 \rightarrow \mathrm{A} + \mathrm{B} \] A and B in the above atmospheric reaction step are
$\mathrm{C_2H_6}$ and $\mathrm{Cl_2}$
$\overset{\bullet}{\mathrm{CHCl_2}}$ and $\mathrm{H_2}$
$\overset{\bullet}{\mathrm{CH_3}}$ and $\mathrm{HCl}$
$\mathrm{C_2H_6}$ and $\mathrm{HCl}$
Answer: (d)
Solution
The reaction is between $\overset{\bullet}\mathrm{Cl} + \mathrm{CH}_4$ to form $\overset{\bullet} \mathrm{CH}_3 + \mathrm{HCl}$.
Question 71
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Which technique among the following, is most appropriate in separation of a mixture of 100 mg of p-nitrophenol and picric acid?
Steam distillation
2-5 ft long column of silica gel
Sublimation
Preparative TLC (Thin Layer Chromatography)
Answer: (d)
Solution
Solvent polarity has been related to $R_f$ value of nitrocompounds. 100 mg p-nitrophenol and picric acid have different $R_f$ value on silica gel plate. Therefore, preparative TLC is best to separate 100 mg of para nitrophenol and picric acid.
Question 72
Chemistry · Alcohols, Phenols and Ethers · Single correct
The difference in the reaction of phenol with bromine in chloroform and bromine in water medium is due to:
Hyperconjugation in substrate
Polarity of solvent
Free radical formation
Electromeric effect of the substrate
Answer: (b)
Solution
Difference in reactions is observed due to solvent polarity, which (i) Ionizes phenol to make more reactive phenoxide ion (ii) Increases electrophilicity of bromine.
Question 73
Chemistry · Hydrocarbons · Single correct
Which of the following compounds is not aromatic?
Answer: (c)
Solution
Annulene, although follow $(4n + 2)\pi$ electron rule, but it is non-aromatic due to its non planar nature. It is nonplanar due to repulsion of C-H bonds present inside the ring.
Question 74
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The products formed in the following reaction, A and B are
Answer: (c)
Solution
$NaBH_4$ does not reduce carboxylic acid.
Question 75
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Which reactant will give the following alcohol on reaction with one mole of phenyl magnesium bromide (PhMgBr) followed by acidic hydrolysis?
$CH_3 - C \equiv$ N
$Ph - C \equiv$ N
Answer: (d)
Solution
The reaction involves the addition of phenylmagnesium bromide (PhMgBr) to acetophenone (Ph-CO-CH3) followed by acid workup with H+. This results in the formation of a tertiary alcohol with the structure Ph-C(OH)(CH3)-Ph.
Question 76
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The major product of the following reaction is
Answer: (a)
Solution
Given reaction is an example of birch reduction.
Question 77
Chemistry · Amines · Single correct
The correct stability order of the following diazonium salt is
(A) > (B) > ($C$) > (D)
(A) > ($C$) > (D) > (B)
($C$) > (A) > (D) > (B)
($C$) > (D) > (B) > (A)
Answer: (b)
Solution
Since diazonium ion is a cation hence it is stabilized by electron donating groups and destabilized by electron withdrawing group. Hence Stability order should be A > C > D > B.
Question 78
Chemistry · Chemistry in Everyday Life · Single correct
Stearic acid and polyethylene glycol react to form which one of the following soap/s detergents?
Cationic detergent
Soap
Anionic detergent
Non-ionic detergent
Answer: (d)
Solution
Stearic acid $\mathrm{CH_3(CH_2)_{16}COOH}$ reacts with polyethylene glycol $\mathrm{OH(CH_2CH_2O)_nCH_2CH_2OH}$ to form a non-ionic detergent. The reaction involves the removal of water $-\mathrm{H_2O}$ and results in the formation of the ester
Question 79
Chemistry · Biomolecules · Single correct
Which of the following is reducing sugar?
Answer: (a)
Solution
If any sugar is having free $-OH$ group at anomeric carbon then it will be a reducing sugar.
Question 80
Chemistry · Amines · Single correct
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason $(R)$. Assertion (A) : Experimental reaction of $\mathrm{CH_3Cl}$ with aniline and anhydrous $\mathrm{AlCl_3}$ does not give o and p-methylaniline. Reason $(R)$ : The $\mathrm{-NH_2}$ group of aniline becomes deactivating because of salt formation with anhydrous $\mathrm{AlCl_3}$ and hence yields m-methyl aniline as the product. In the light of the above statements, choose the most appropriate answer from the options given below :
Both (A) and $(R)$ are true and $(R)$ is the correct explanation of (A).
Both (A) and $(R)$ are true but $(R)$ is not the correct explanation of (A).
is true, but $(R)$ is false.
is false, but $(R)$ is true.
Answer: (c)
Solution
Friedel Craft Alkylation does not occur on this deactivated ring.
Question 81
Chemistry · Some Basic Concepts of Chemistry · Numerical
Chlorophyll extracted from the crushed green leaves was dissolved in water to make 2 $\mathrm{L}$ solution of Mg of concentration 48 $\mathrm{ppm}$. The number of atoms of Mg in this solution is $x \times 10^{20}$ atoms. The value of $x$ is ________. (Nearest Integer) (Given : Atomic mass of Mg is $24 \, \mathrm{g \, mol^{-1}}$, $N_A = 6.02 \times 10^{23} \, \mathrm{mol^{-1}}$)
Answer: 24
Solution
Given $\left(\mathrm{ppm}=\frac{W_{Mg}}{V_{soln}}\times10^6=48\right)$ $\Rightarrow W_{Mg}=\frac{48\times2\times1000}{10^6}$ $=48\times2\times10^{-3}\,\mathrm{g}$ $n_{Mg}=\frac{W_{Mg}}{24}=\frac{48\times2\times10^{-3}}{24}$ $=4\times10^{-3}$ Number of Mg atoms $\left(=4\times10^{-3}\times6.02\times10^{23}\right)$ $=4\times6.02\times10^{20}$ $=24.08\times10^{20}$ $\therefore x=24.08$
Question 82
Chemistry · States of Matter · Numerical
A mixture of hydrogen and oxygen contains 40$\%$ hydrogen by mass when the pressure is $2.2 \, \mathrm{bar}$. The partial pressure of hydrogen is $\mathrm{bar}$. (Nearest Integer)
The wavelength of an electron and a neutron will become equal when the velocity of the electron is x times the velocity of neutron. The value of x is__________. (Nearest Integer) (Mass of electron is $9.1 \times 10^{-31} \, \mathrm{kg}$ and mass of neutron is $1.6 \times 10^{-27} \, \mathrm{kg}$)
Answer: 1758
Solution
Given $\nu_e = x \nu_N$ and $\lambda_e = \lambda_N$. Therefore, $$\frac{h}{m_e \nu_e} = \frac{h}{m_N \nu_N}$$ This implies $$\nu_e = \frac{m_N}{m_e} \cdot \nu_N$$ Substituting the values, $$\nu_e = \frac{1.6 \times 10^{-27}}{9.1 \times 10^{-31}} \nu_N$$ Thus, $$\nu_e = 1758.24 \times \nu_N$$ Therefore, $x = 1758.24$.
Question 84
Chemistry · Thermodynamics · Numerical
2.4 $\mathrm{g}$ coal is burnt in a bomb calorimeter in excess of oxygen at 298 $\mathrm{K}$ and 1 $\mathrm{atm}$ pressure. The temperature of the calorimeter rises from 298 $\mathrm{K}$ to 300 $\mathrm{K}$. The enthalpy change during the combustion of coal is -x $\mathrm{kJ}$ $\mathrm{mol}^{-1}$. The value of x is . (Nearest Integer) (Given: Heat capacity of bomb calorimeter 20.0 $\mathrm{kJ}$ $\mathrm{K}^{-1}$. Assume coal to be pure carbon)
Answer: 200
Solution
The reaction is given by $\($ $\mathrm{C (s) + O_2 (g) \rightarrow CO_2 (g)}$ $\)$ with $\($ $\Delta$ H = -x $\)$ kJ/mole. The heat $\($ Q $\)$ is calculated as $\($ Q = C $\Delta$ T = 20 $\mathrm{kJ}$ $\times$ 2 $\)$. Therefore, 40 kJ of heat is released for 2.4 g of C. For 1 mole of C, $\($ Q = $\frac{40}{2.4}$ $\times$ 12 $\)$. This simplifies to: $$ \frac{400}{24} \times 12 = 200 \, \mathrm{kJ/mole} $$ Thus, $\($ Q = $\Delta$ E = $\Delta$ H = 200 $\mathrm{kJ}$ $\)$ (since $\($ $\Delta$ n_g = 0 $\)$). Therefore, $\($ x = 200 $\)$.
Question 85
Chemistry · Some Basic Concepts of Chemistry · Numerical
When 800 $\mathrm{mL}$ of 0.5 $\mathrm{M}$ nitric acid is heated in a beaker, its volume is reduced to half and 11.5 $\mathrm{g}$ of nitric acid is evaporated. The molarity of the remaining nitric acid solution is x $\times$ $10^{-2}$ $\mathrm{M}$. (Nearest Integer) (Molar mass of nitric acid is 63 $\mathrm{g \, mol^{-1})}$
At 298 $\mathrm{K}$, the equilibrium constant is $2 \times 10^{15}$ for the reaction: $Cu(s)+2Ag^+(aq)\rightleftharpoons Cu^{2+}(aq)+2Ag(s)$ The equilibrium constant for the reaction $\dfrac12Cu^{2+}(aq)+Ag(s)\rightleftharpoons\dfrac12Cu(s)+Ag^+(aq)$ is $x \times 10^{-8}$. The value of $x$ is _______. (Nearest Integer)
Answer: 2
Solution
Given $$K'_{eq} = \frac{1}{\sqrt{K_{eq}}} = \frac{1}{\sqrt{2 \times 10^{15}}} = x \times 10^{-8}$$ This implies $$\frac{1}{\sqrt{20}} \times \frac{1}{10^7} = x \times 10^{-8}$$ Therefore, $$\frac{1}{\sqrt{20}} \times 10^{-7} = x \times 10^{-8}$$ Simplifying, $$\frac{10}{\sqrt{20}} = x$$ Thus, $$x = \frac{\sqrt{10}}{\sqrt{2}} = \sqrt{5} = 2.236$$ Approximately, $$\approx 2.24$$
Question 87
Chemistry · Electrochemistry · Numerical
The amount of charge in F (Faraday) required to obtain one mole of iron from $\mathrm{Fe_3O_4}$ is _____. (Nearest Integer)
Answer: 3
Solution
The reaction is given as $\mathrm{Fe_3O_4} + 8e^- \rightarrow 3\mathrm{Fe}$. Charge for 1 mole of $\mathrm{Fe}$ is $\frac{8}{3} \, F$. Therefore, the total charge is $2.67 \, F$.
Question 88
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
For a reaction A $\rightarrow$ 2B + C the half lives are 100 $\mathrm{s}$ and 50 $\mathrm{s}$ when the concentration of reactant A is 0.5 and 1.0 $\mathrm{mol \, L^{-1}}$ respectively. The order of the reaction is . (Nearest Integer)
Answer: 2
Solution
The half-life $t_{\frac{1}{2}}$ is proportional to $\frac{1}{[A_0]^{n-1}}$. For $[100]$, it is proportional to $\frac{1}{(0.5)^{n-1}}$ and for $[50]$, it is proportional to $\frac{1}{(1)^{n-1}}$. Thus, $[2]^1 = \left[ \frac{1}{0.5} \right]^{n-1}$. This simplifies to $[2]^1 = [2]^{n-1}$. Therefore, $n - 1 = 1$, which gives $n = 2$. Thus, the order is 2.
Question 89
Chemistry · Co-ordination Compounds · Numerical
The difference between spin only magnetic moment values of $\mathrm{[Co(H_2O)_6]Cl_2}$ and $\mathrm{[Cr(H_2O)_6]Cl_3}$ is
Answer: 0
Solution
For $[\mathrm{Co(H_2O)_6}]^{2+}$, the electron configuration of $\mathrm{Co}^{2+}$ is shown with 3 unpaired electrons. The magnetic moment $\mu$ is given by $\mu = \sqrt{15} \mathrm{BM}$. For $[\mathrm{Cr(H_2O)_6}]^{3+}$, the electron configuration of $\mathrm{Cr}^{3+}$ also shows 3 unpaired electrons. The magnetic moment $\mu$ is $\mu = \sqrt{15} \mathrm{BM}$. The difference in spin only magnetic moment is 0.
Question 90
Chemistry · Hydrocarbons · Numerical
In the presence of sunlight, benzene reacts with $\mathrm{Cl}_2$ to give product, X. The number of hydrogens in X is _________.
Answer: 6
Solution
The reaction shown is the chlorination of benzene in the presence of sunlight. This leads to the formation of hexachlorocyclohexane, where all hydrogen atoms in benzene are replaced by chlorine atoms.