JEE Main 25 July 2022 Shift 2: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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Maths · Complex Numbers and Quadratic Equations · Single correct
For $z \in \mathbb{C}$ if the minimum value of $$\left( |z - 3\sqrt{2}| + |z - p\sqrt{2}i| \right)$$ is $5\sqrt{2}$, then a value of $p$ is
Answer: (c)
4
Maths · Determinants · Single correct
The number of real values $\lambda$, such that the system of linear equations $$2x - 3y + 5z = 9$$ $$x + 3y - z = -18$$ $$3x - y + (\lambda^2 - |\lambda|)z = 16$$ has no solution, is :-
Answer: (c)
2
Maths · Permutations and Combinations · Single correct
The number of bijective functions $f : \{1, 3, 5, 7, \ldots, 99\} \to \{2, 4, 6, 8, \ldots, 100\}$, such that $f(3) \geq f(9) \geq f(15) \geq f(21) \geq \ldots \geq f(99)$, is
Answer: (b)
{ "rawText": "", "answer": "b" }
Maths · Binomial Theorem · Single correct
The remainder when $(11)^{1011} + (1011)^{11}$ is divided by 9 is
Answer: (d)
```json { "rawText": "", "answer": "d" } ```
Maths · Sequences and Series · Single correct
The sum $$\sum_{n=1}^{21} \frac{3}{(4n-1)(4n+3)}$$ is equal to
Answer: (b)
{ "rawText": "", "answer": "b" }
Maths · Limits and Derivatives · Single correct
$$\lim_{x \to \frac{\pi}{4}} \frac{8\sqrt{2} - (\cos x + \sin x)^7}{\sqrt{2} - \sqrt{2} \sin 2x}$$ is equal to
Answer is A
Maths · Sequences and Series · Single correct
\[ \lim_{n\to\infty}\frac{1}{n} \left( \frac{1}{\sqrt{1-\frac{1}{2^n}}} + \frac{1}{\sqrt{1-\frac{2}{2^n}}} + \frac{1}{\sqrt{1-\frac{3}{2^n}}} +\cdots+ \frac{1}{\sqrt{1-\frac{2^n-1}{2^n}}} \right) \]
Answer: (c)
2
Maths · Probability · Single correct
If A and B are two events such that $$\mathrm{P}(A) = \frac{1}{3}, \mathrm{P}(B) = \frac{1}{5} and \mathrm{P}(A \cup B) = \frac{1}{2},$$ then $\mathrm{P}(A | B') + \mathrm{P}(B | A')$ is equal to
Answer: (b)
B
Maths · Integrals · Single correct
Let [t] denote the greatest integer less than or equal to t. Then the value of the integral $$\int_{-3}^{101} ( [\sin(\pi x)] + e^{[\cos(2 \pi x)]} ) \, dx$$ is equal to
Answer: (b)
$\frac{52}{e}$
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let the point P $(\alpha, \beta)$ be at a unit distance from each of the two lines $L_1 : 3x - 4y + 12 = 0$, and $L_2 : 8x + 6y + 11 = 0$. If P lies below $L_1$ and above $L_2$, then $100\,(\alpha + \beta)$ is equal to
Answer: (d)
{ "rawText": "", "answer": "d" }
Maths · Differential Equations · Single correct
Let a smooth curve $y = f(x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\($ ( $\frac{-y}{x}$ ) $\)$. If the curve passes through the point $(1, 2)$ and $(8, 1)$, then \[ |\, y\!(\frac{1}{8})| \] is equal to
Answer: (b)
{ "rawText": "", "answer": "b" }
Maths · Conic Sections · Single correct
If the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ meets the line $\frac{x}{7} + \frac{y}{2\sqrt{6}} = 1$ on the x-axis and the line $\frac{x}{7} - \frac{y}{2\sqrt{6}} = 1$ on the y-axis, then the eccentricity of the ellipse is
Answer: (a)
{ "rawText": "", "answer": "a" }
Maths · Conic Sections · Single correct
The tangents at the point A(1, 3) and B(1, -1) on the parabola $y^2 - 2x - 2y = 1$ meet at the point P. Then the area (in unit$^2$) of the triangle PAB is :-
Answer: (d)
D
Maths · Conic Sections · Single correct
Let the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{7} = 1$ and the hyperbola $\frac{x^2}{144} - \frac{y^2}{\alpha} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:-
answer is D
Maths · Three Dimensional Geometry · Single correct
A plane E is perpendicular to the two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$, and passes through the point $P(1, -1, 1)$. If the distance of the plane E from the point $Q(a, a, 2)$ is $3\sqrt{2}$, then $(PQ)^2$ is equal to
Answer: (c)
{ "rawText": "", "answer": "c" }
Maths · Three Dimensional Geometry · Single correct
The shortest distance between the lines $\($ $\frac{x+7}{-6}$ = $\frac{y-6}{7}$ = z $\)$ and $\($ $\frac{7-x}{2}$ = y-2 = z-6 $\)$ is
Answer: (a)
{ "rawText": "", "answer": "a" }
Maths · Vector Algebra · Single correct
Let $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{b}$ be a vector such that $\vec{a} \times \vec{b} = 2\hat{i} - \hat{k}$ and $\vec{a} \cdot \vec{b} = 3$. Then the projection of $\vec{b}$ on the vector $\vec{a} - \vec{b}$ is :-
Answer: (a)
$\frac{2}{\sqrt{21}}$
Maths · Statistics · Single correct
If the mean deviation about median for the number 3, 5, 7, 2k, 12, 16, 21, 24 arranged in the ascending order, is 6 then the median is
Answer: (d)
{ "rawText": "", "answer": "d" }
Maths · Trigonometric Functions · Single correct
2 $\sin(\frac{\pi}{22}) \sin(\frac{3\pi}{22}) \sin(\frac{5\pi}{22}) \sin(\frac{7\pi}{22})$ $\sin(\frac{9\pi}{22})$ is equal to
Answer: (b)
$\frac{1}{16}$
Maths · Mathematical Reasoning · Single correct
Consider the following statements : P : Ramu is intelligent Q : Ramu is rich R : Ramu is not honest The negation of the statement "Ramu is intelligent and honest if and only if Ramu is not rich" can be expressed as :
Answer is D
Maths · Sets · Numerical
Let $A = \{1, 2, 3, 4, 5, 6, 7\}$. Define $B = \{T \subseteq A : \text{either } 1 \notin T \text{ or } 2 \in T\}$ and $C = \{T \subseteq A : \text{the sum of all the elements of } T \text{ is a prime number}\}$. Then the number of elements in the set $B \cup C$ is
Answer: 107
Q1 (107) (107)
Maths · Complex Numbers and Quadratic Equations · Fill in the blank
Let f(x) be a quadratic polynomial with leading coefficient 1 such that f(0) = p, p $\neq$ 0 and f(1) = $\frac{1}{3}$. If the equation f(x) = 0 and fofofof(x) = 0 have a common real root, then f(-3) is equal to....................
Answer: 25
Maths · Matrices · Numerical
Let $A = \begin{bmatrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{bmatrix}, a, b \in \mathbb{R}$. If for some $n \in \mathbb{N}$, $$A^n = \begin{bmatrix} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{bmatrix}$$ then $n + a + b$ is equal to
Answer: 24
Answer is D
Maths · Applications of Derivatives · Numerical
The sum of the maximum and minimum values of the function $f(x) = |5x - 7| + [x^2 + 2x]$ is the interval $\left[ \frac{5}{4}, 2 \right]$, where $[t]$ is the greatest integer $\leq t$ is ____________
Answer: 15
{ "rawText": "", "answer": "" }
Maths · Differential Equations · Numerical
Let y = y(x) be the solution of the differential equation $\frac{dy}{dx} = \frac{4y^3 + 2yx^2}{3xy^2 + x^3}$, $y(1) = 1$. If for some $n \in \mathbb{N}, y(2) \in [n-1, n)$, then n is equal to
Answer: 3
{ "rawText": "", "answer": "3" }
Maths · Integrals · Numerical
Let f be a twice differentiable function on R. If $f'(0) = 4$ and $$f(x) + \int_{0}^{x} (x-t) f'(t) \, dt = (e^{2x} + e^{-2x}) \cos 2x + \frac{2}{a} x,$$ then $(2a + 1)^5 \, a^2$ is equal to
Answer: 8
Maths · Integrals · Numerical
Let $a_n = \int_{-1}^{n} \left( 1 + \frac{x}{2} + \frac{x^2}{3} + \frac{x^3}{4} + \ldots + \frac{x^{n-1}}{n} \right) \, dx$ for $n \in \mathbb{N}$. Then the sum of all the elements of the set $\{ n \in \mathbb{N} : a_n \in (2,30) \}$ is
Answer: 5
Maths · Conic Sections · Numerical
If the circles $x^2 + y^2 + 6x + 8y + 16 = 0$ and $x^2 + y^2 + 2(3 - \sqrt{3})x + x + 2(4 - \sqrt{6})y = k + 6\sqrt{3} + 8\sqrt{6}, k > 0$, touch internally at the point $P(\alpha, \beta)$, then $(\alpha + \sqrt{3})^2 + (\beta + \sqrt{6})^2$ is equal to
Answer: 25
25
Maths · Applications of Derivatives · Numerical
Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve $4x^3 - 3xy^2 + 6x^2 - 5xy - 8y^2 + 9x + 14 = 0$ at the point $(-2, 3)$ be $A$. Then $8A$ is equal to
Answer: 170
(170)
Maths · Inverse Trigonometric Functions · Numerical
Let $x = \sin(2 \tan^{-1} \alpha)$ and $y = \sin\left(\frac{1}{2} \tan^{-1} \frac{4}{3}\right)$. If $S = \{\alpha \in \mathbb{R} : y^2 = 1-x\}$, then $\sum_{\alpha \in S} 16\alpha^3$ is equal to
Answer: 130
answer is 130
Physics · Current Electricity · Single correct
In AM modulation, a signal is modulated on a carrier wave such that maximum and minimum amplitude are found to be 6V and 2V respectively. The modulation index is
Answer: (d)
The modulation index is given by $$\frac{V_{max} - V_{min}}{V_{max} + V_{min}} \times 100\%$$ Substituting the values: $$= \frac{6 - 2}{6 + 2} \times 100\% = 50\%$$
Physics · Moving Charges and Magnetism · Single correct
The electric current in a circular coil of 2 turns produces a magnetic induction $B_1$ at its centre. The coil is unwound and is rewound into a circular coil of 5 turns and the same current produces a magnetic induction $B_2$ at its centre. The ratio of $\frac{B_2}{B_1}$ is :
Answer: (b)
Given $B = \frac{N \mu_0 i}{2R}$. $B_1 = \frac{N_1 \mu_0 i}{2R_1}$. For $N_2 = 5$. Radius of coil $= R_2 = \frac{N_1 \times R_1}{N_2}$. $B_2 = \frac{N_2 \mu_0 i}{R_2}$. $$\frac{B_2}{B_1} = \frac{N_2}{N_1} \frac{R_1}{R_2} = \frac{N_2}{N_1} \times \frac{N_2}{N_1}; \frac{B_2}{B_1} = \frac{25}{4}$$
Physics · Mechanical Properties of Fluids · Single correct
A drop of liquid of density $\rho$ is floating half immersed in a liquid of density $\sigma$ and surface tension $7.5 \times 10^{-4} \, \mathrm{N/cm}^{-1}$. The radius of drop in cm will be : (Take : $g = 10 \, \mathrm{m/s}^2$)
Answer: (a)
Buoyant force plus surface tension equals $mg$. $$\frac{\sigma V}{2} g + 2 \pi R T = \rho V g$$ $$2 \pi R T = \frac{(2 \rho - \sigma)}{2} \frac{4}{3} \pi R^3 g; \left[ V = \frac{4}{3} \pi R^3 \right]$$ $$R^3 = \frac{3 T}{(2 \rho - \sigma) g} \implies R = \sqrt{\frac{3 \times 7.5 \times 10^{-2} \, \mathrm{N} \, \mathrm{m}^{-1}}{(2 \rho - \sigma) \times 10}}$$ $$R = \frac{3}{20 \sqrt{(2 \rho - \sigma)}} m = \frac{15}{\sqrt{2 \rho - \sigma}} \, \mathrm{cm}$$
Physics · Laws of Motion · Single correct
Two billiard balls of mass 0.05 kg each moving in opposite directions with 10 $\mathrm{ms}^{-1}$ collide and rebound with the same speed. If the time duration of contact is t = 0.005 $\mathrm{s}$, then what is the force exerted on the ball due to each other?
Answer: (b)
Change in momentum of any one ball $$|\Delta \vec{P}| = 2 \times 0.05 \times 10$$ $$|\Delta \vec{P}| = 1$$ $$|\vec{F}_{av}| = \frac{|\Delta \vec{P}|}{\Delta t}$$ $$F_{av} = 200 \, \mathrm{N}$$
Physics · Laws of Motion · Single correct
For a free body diagram shown in the figure, the four forces are applied in the 'x' and 'y' directions. What additional force must be applied and at what angle with positive x-axis so that the net acceleration of body is zero?
Answer: (a)
Let addition force required is $\vec{F}$. $$\vec{F} + 5\hat{i} - 6\hat{i} + 7\hat{j} - 8\hat{j} = 0$$ $$\vec{F} = \hat{i} + \hat{j}, \; |\vec{F}| = \sqrt{2}$$ Angle with x-axis: $\tan \theta = \frac{y component}{x component} = \frac{1}{1}$ $\theta = 45^\circ$
Physics · Electrostatic Potential and Capacitance · Single correct
Capacitance of an isolated conducting sphere of radius $R_1$ becomes $n$ times when it is enclosed by a concentric conducting sphere of radius $R_2$ connected to earth. The ratio of their radii $\left( \frac{R_2}{R_1} \right)$ is:
Answer: (a)
Capacitance of isolated conducting sphere is $4\pi \varepsilon_0 R_1$. By enclosing inside another sphere of radius $R_2$, the new capacitance is $$\frac{4\pi \varepsilon_0 R_1 R_2}{(R_2 - R_1)}.$$ Given: $$\frac{4\pi \varepsilon_0 R_1 R_2}{(R_2 - R_1)} = n \times 4\pi \varepsilon_0 R_1$$ which implies $$\frac{R_2}{(R_2 - R_1)} = n \implies \frac{R_2}{R_1} - 1 = n$$ leading to $$\frac{R_2}{R_1} = n + 1.$$ Therefore, $$\frac{R_2}{R_1} = \frac{n}{(n-1)}.$$
Physics · Dual Nature of Radiation and Matter · Single correct
The ratio of wavelengths of proton and deuteron accelerated by potential $V_p$ and $V_d$ is $1 : \sqrt{2}$. Then, the ratio of $V_p$ to $V_d$ will be
Answer: (d)
Kinetic energy gained by a charged particle accelerated by a potential $V$ is $qV$. KE = $qV$ $$\Rightarrow \frac{p^2}{2m} = qV \Rightarrow p = \sqrt{2mqV}$$ $$p = \frac{h}{\lambda}, thus \lambda = \frac{h}{\sqrt{2mqV}}$$ Now $$\frac{\lambda_p}{\lambda_d} = \sqrt{\frac{m_d V_d}{m_p V_p}}$$ $$\Rightarrow \frac{1}{\sqrt{2}} = \sqrt{\frac{2V_d}{V_p}} \Rightarrow \frac{V_p}{V_d} = 4$$
Physics · Ray Optics and Optical Instruments · Single correct
For an object placed at a distance 2.4 m from a lens, a sharp focused image is observed on a screen placed at a distance 12 cm from the lens. A glass plate of refractive index 1.5 and thickness 1 cm is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?
Answer: (b)
Applying lens formula $$\frac{1}{0.12} + \frac{1}{2.4} = \frac{1}{f} \implies \frac{1}{f} = \frac{210}{24}$$ Upon putting the glass slab, shift of image is $$\Delta x = t \left(1 - \frac{1}{\mu}\right) = \frac{1}{3} \, cm$$ Now $v = 12 - \frac{1}{3} = \frac{35}{3} \, cm$ Again apply lens formula $$\frac{1}{0.12} + \frac{1}{u} = \frac{1}{f} = \frac{210}{24}$$ Solving $u = -5.6 \, m$ Thus shift of object is $$5.6 - 2.4 = 3.2 \, m$$
Physics · Electromagnetic Waves · Single correct
Light wave traveling in air along x-direction is given by $E_y = 540 \sin \pi \times 10^4(x - ct) \, \mathrm{Vm}^{-1}$. Then, the peak value of magnetic field of wave will be (Given $c = 3 \times 10^8 \, \mathrm{ms}^{-1}$)
Answer: (a)
Given $E_y = 540 \sin \pi \times 10^4 (x - ct) \, \mathrm{Vm^{-1}}$. $E_0 = 540 \, \mathrm{Vm^{-1}}$. $$B_0 = \frac{E_0}{C} = \frac{540}{3 \times 10^8} = 18 \times 10^{-7} \, \mathrm{T}$$
Physics · Alternating Current · Single correct
When you walk through a metal detector carrying a metal object in your pocket, it raises an alarm. This phenomenon works on
Answer: (b)
Metal detector works on the principle of transmitting an electromagnetic signal and analyses a return signal from the target. So it works on the principle of resonance in AC circuit.
Physics · Magnetism and Matter · Single correct
An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of $1 \times 10^{-4} \, \mathrm{Wbm}^{-2}$. The frequency of revolution of the electron will be (Take mass of electron = $9.0 \times 10^{-31} \, \mathrm{kg}$)
Answer: (c)
The frequency $f$ is given by the equation $$f = \frac{1}{T} = \frac{eB}{2\pi m}$$ where $e$ is the charge, $B$ is the magnetic field, and $m$ is the mass. Substituting the given values, we have $$f = \frac{1.6 \times 10^{-19} \times 10^{-4}}{2\pi \times 9 \times 10^{-31}} = 2.8 \times 10^{6} \, \mathrm{Hz}$$
Physics · Current Electricity · Single correct
A current of 15 $\mathrm{mA}$ flows in the circuit as shown in figure. The value of potential difference between the points A and B will be
Answer: (d)
Given $i = 15 \, \mathrm{mA}$. Calculate $i_1$ using the current division rule: $$i_1 = \frac{5}{10 + 5} \times 15 \, \mathrm{mA} = 5 \, \mathrm{mA}$$ Using Kirchhoff's voltage law: $$V_A - 5i - 10i_1 - 10i = V_B$$ Substitute the values: $$V_A - V_B = 75 + 50 + 150 = 275 \, \mathrm{V}$$
Physics · Oscillations · Single correct
The length of a seconds pendulum at a height $h = 2R$ from earth surface will be: (Given: $R$ = Radius of earth and acceleration due to gravity at the surface of earth $g = \pi^2 \, \mathrm{m/s^2}$)
Answer: (d)
Given $T = 2\pi \sqrt{\frac{L}{g}}$, $g' = \frac{GM}{9R^2} = \frac{g}{9} = \frac{\pi^2}{9}$. Then, $$2 = 2\pi \sqrt{\frac{L}{\pi^2}} \times 9$$ This implies $$1 = \pi \sqrt{L} \times \frac{3}{\pi}$$ Therefore, $L = \frac{1}{9} \, \mathrm{m}$.
Physics · Kinetic Theory · Single correct
Sound travels in a mixture of two moles of helium and $n$ moles of hydrogen. If rms speed of gas molecules in the mixture is $\sqrt{2}$ times the speed of sound, then the value of $n$ will be
Answer: (b)
Given $v_s = \sqrt{\frac{\gamma RT}{M}}$ and $v_{rms} = \sqrt{\frac{3RT}{M}}$. The ratio $\frac{v_s}{v_{rms}} = \frac{\sqrt{\frac{\gamma}{3}}}{\sqrt{1}} = \frac{1}{\sqrt{2}} \implies \frac{\gamma}{3} = \frac{1}{2} \implies \gamma = \frac{3}{2}$. We have $\gamma = 1 + \frac{2}{f_{mix}}$. For $f_{mix}$, $f_{mix} = \frac{2 \times 3 + n \times 5}{n + 2} = \frac{6 + n \times 5}{n + 2}$. Thus, $\gamma = 1 + \frac{2(n + 2)}{6 + n \times 5} = \frac{6 + 5n + 2n + 4}{6 + 5n}$. This simplifies to $\gamma = \frac{7n + 10}{6 + 5n} = \frac{3}{2}$. Solving $14n + 20 = 18 + 15n$ gives $n = 2$.
Physics · Thermodynamics · Single correct
Let $\eta_1$ is the efficiency of an engine at $T_1 = 447^\circ C$ and $T_2 = 147^\circ C$ while $\eta_2$ is the efficiency at $T_1 = 947^\circ C$ and $T_2 = 47^\circ C$. The ratio $\frac{\eta_1}{\eta_2}$ will be:
Answer: (b)
Efficiency $\eta = 1 - \frac{T_L}{T_H}$ $$\eta_1 = 1 - \frac{147 + 273}{447 + 273} = 1 - \frac{420}{720}$$ $$\eta_1 = \frac{300}{720}$$ $$\eta_2 = 1 - \frac{47 + 273}{947 + 273} = 1 - \frac{320}{1220}$$ $$\eta_2 = \frac{900}{1220}$$ $$\frac{\eta_1}{\eta_2} = \frac{300}{720} \times \frac{1220}{900} = \frac{122}{72 \times 3}$$ $$\frac{\eta_1}{\eta_2} = 0.56$$
Physics · Gravitation · Single correct
An object is taken to a height above the surface of earth at a distance $\frac{5}{4} R$ from the centre of the earth. Where radius of earth, $R = 6400 \, \mathrm{km}$. The percentage decrease in the weight of the object will be
Answer: (a)
Given $g_{eff} = \frac{g}{\left(1 + \frac{h}{R}\right)^2}$; $g_{eff} = \frac{g}{\left(1 + \frac{1}{4}\right)^2} = \frac{16g}{25}$. The change is $\frac{g_{eff} - g}{g} \times 100 = \frac{16}{25} - 1 \times 100$. $$= \frac{-9}{25} \times 100 = -36\%$$ Hence % decrease in the weight = 36$\%$.
Physics · Work, Energy and Power · Single correct
A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of 200 g travelling with speed 10 $\mathrm{ms}^{-1}$ gets embedded in it, then loss of kinetic energy will be
Answer: (b)
Given $P_i = P_f$ (no any external force) $$0.2 \times 10 = 10 \times v$$ $$v = 0.2 \, \mathrm{m/s}$$ Loss in K.E. $$= \frac{1}{2} \times (0.2) \times 10^2 - \frac{1}{2} \times 10 (0.2)^2$$ $$= \frac{1}{2} \times 10 \times (0.2) [10 - 0.2]$$ $$= 9.8 \, \mathrm{J}$$
Physics · Motion in a Plane · Single correct
A ball is projected from the ground with a speed 15 $\mathrm{ms^{-1}}$ at an angle $\theta$ with horizontal so that its range and maximum height are equal, then 'tan $\theta$' will be equal to
Answer: (d)
Given $R = H$. $$\frac{2v_x \times v_y}{g} = \frac{v_y^2}{2g}$$ $v_x = \frac{v_y}{4}$; $u \cos \theta = \frac{u \sin \theta}{4}$ $\tan \theta = 4$
Physics · Mathematics in Physics · Single correct
The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are 1$\%$, 2$\%$ and 3$\%$ respectively. The maximum percentage error in the detection of the dissipated heat will be:
Answer: (d)
Given $E_H = I^2 R \times t$. $$\frac{\Delta E}{E} \times 100 = \frac{2 \Delta I}{I} \times 100 + \frac{\Delta R}{R} \times 100 + \frac{\Delta T}{T} \times 100$$ $$= 2 \times 2 + 1 + 3 = 8$$
Physics · Atoms · Single correct
Hydrogen atom from excited state comes to the ground by emitting a photon of wavelength $\lambda$. The value of principal quantum number 'n' of the excited state will be : (R : Rydberg constant)
Answer: (b)
The energy of the nth level is given by $$E_n = \frac{-Rch}{n^2} (1).$$ The energy of the first level is $$E_1 = \frac{-Rch}{(1)^2} (1).$$ The energy of the photon is $$E_{photon} = E_n - E_1.$$ Substituting the expressions for $E_n$ and $E_1$, we have $$\frac{-Rch}{(n)^2} + \frac{Rch}{1} = \frac{hc}{\lambda}.$$ Simplifying, we get $$\frac{-R}{n^2} + R = \frac{1}{\lambda}.$$ Rearranging terms gives $$R - \frac{1}{\lambda} = \frac{R}{n^2}.$$ Further simplification leads to $$\frac{\lambda R - 1}{\lambda} = \frac{R}{n^2}.$$ Solving for $n^2$, we find $$n^2 = \frac{\lambda R}{\lambda R - 1} \implies n = \sqrt{\frac{\lambda R}{\lambda R - 1}}.$$
Physics · Motion in a Straight Line · Numerical
A particle is moving in a straight line such that its velocity is increasing at $5 \, \mathrm{ms^{-1}}$ per meter. The acceleration of the particle is $\mathrm{ms^{-2}}$ at a point where its velocity is $20 \, \mathrm{ms^{-1}}$.
Answer: 100
Given $\($ $\frac{dv}{ds}$ = 5 $\)$. \[ a = v \frac{dv}{ds} = 20 \times 5 = 100 \mathrm{m/sec^2} \]
Physics · System of Particles and Rotational Motion · Numerical
Three identical spheres each of mass $M$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to $3 \, \mathrm{m}$ each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be $\sqrt{x} \, \mathrm{m}$. The value of $x$ is
Answer: 2
The position vectors of the masses are given as $(0, 3)$, $(0, 0)$, and $(3, 0)$. The center of mass $\vec{r}_{com}$ is calculated as: $$\vec{r}_{com} = \frac{M(0\hat{i} + 0\hat{j}) + M(3\hat{i}) + M(3\hat{j})}{3M}$$ Simplifying, we get: $$\vec{r}_{com} = \hat{i} + \hat{j}$$ The magnitude of $\vec{r}_{com}$ is: $$|\vec{r}_{com}| = \sqrt{2} = \sqrt{x}$$ Solving for $x$, we find: $$x = 2$$
Physics · Thermal Properties of Matter · Numerical
A block of ice of mass $120\,\mathrm{g}$ at temperature $0^\circ\mathrm{C}$ is put in $300\,\mathrm{g}$ of water at $25^\circ\mathrm{C}$. The $x\,\mathrm{g}$ of ice melts as the temperature of the water reaches $0^\circ\mathrm{C}$. The value of $x$ is [Use: Specific heat capacity of water $= 4200\,\mathrm{J\,kg^{-1}\,K^{-1}}$, Latent heat of ice $= 3.5 \times 10^5\,\mathrm{J\,kg^{-1}}$]
Answer: 90
Energy released by water $$= 0.3 \times 25 \times 4200 = 31500 \, \mathrm{J}$$ Let $m$ kg ice melts $$m \times 3.5 \times 10^5 = 31500$$ $$m = \frac{31500 \times 10^{-5}}{3.5} = 9000 \times 10^{-5}$$ $$m = 0.09 \, \mathrm{kg} = 90 \, \mathrm{gm}$$ $$x = 90$$
Physics · Atoms · Numerical
$\frac{x}{x+4}$ is the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) third permitted energy level to the second level and (ii) the highest permitted energy level to the second permitted level. The value of x will be
Answer: 5
Given the equation: $$\frac{13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)}{13.6 \left( \frac{1}{2^2} - 0 \right)} = \frac{x}{x+4}$$ Simplifying, we have: $$\frac{\frac{1}{4} - \frac{1}{9}}{\frac{1}{4}} = \frac{x}{x+4}$$ This simplifies to: $$\frac{5}{9} = \frac{x}{x+4}$$ Solving for $x$, we get: $$5x + 20 = 9x$$ $$4x = 20$$ $$x = 5$$
Physics · Current Electricity · Numerical
In a potentiometer arrangement, a cell of emf 1.20 $\mathrm{V}$ gives a balance point at 36 $\mathrm{cm}$ length of wire. This cell is now replaced by another cell of emf 1.80 $\mathrm{V}$. The difference in balancing length of potentiometer wire in above conditions will be $\mathrm{cm}$.
Answer: 18
Given: $$1.2 = (Potential Gradient) \times 36$$ $$1.8 = (Potential Gradient) \times x$$ On dividing, we get $$\frac{2}{3} = \frac{36}{x}$$ Solving for $x$: $$x = 18 \times 3 = 54 \, cm$$ Hence difference $= 54 - 36 = 18 \, cm$
Physics · Current Electricity · Numerical
Two ideal diodes are connected in the network as shown in figure. The equivalent resistance between A and B is _____ $\Omega$.
Answer: 25
The forward biased diode will conduct while the reverse biased will not. Therefore, equivalent resistance is $10 \, \Omega + 15 \, \Omega = 25 \, \Omega$.
Physics · Oscillations · Numerical
Two waves executing simple harmonic motion travelling in the same direction with same amplitude and frequency are superimposed. The resultant amplitude is equal to the $\sqrt{3}$ times of amplitude of individual motions. The phase difference between the two motions is ______ (degree)
Answer: 60
The resultant amplitude is given by $$A_{resultant} = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$$ Simplifying, we have $$\sqrt{3}A = \sqrt{A^2 + A^2 + 2A^2 \cos \phi}$$ This leads to $$3A^2 = 2A^2 + 2A^2 \cos \phi$$ Solving for $\cos \phi$, we get $$\cos \phi = \frac{1}{2}$$ Therefore, $$\phi = 60^\circ$$ Thus, the phase difference is 60 degrees.
Physics · Electrostatic Potential and Capacitance · Numerical
Two parallel plate capacitors of capacity $C$ and $3C$ are connected in parallel combination and charged to a potential difference $18 \, \mathrm{V}$. The battery is then disconnected and the space between the plates of the capacitor of capacity $C$ is completely filled with a material of dielectric constant $9$. The final potential difference across the combination of capacitors will be _______ V
Answer: 6
Initial charge on $C = 18 \, \mathrm{CV}$. Initial charge on $3C = 54 \, \mathrm{CV}$. Let final common potential difference $= V'$. $$9CV' + 3CV' = 18 \, \mathrm{CV} + 54 \, \mathrm{CV}$$ $$\Rightarrow 12CV' = 72 \, \mathrm{CV} \Rightarrow V' = 6 \, \mathrm{V}$$
Physics · Ray Optics and Optical Instruments · Numerical
A convex lens of focal length 20 cm is placed in front of convex mirror with principal axis coinciding each other. The distance between the lens and mirror is 10 cm. A point object is placed on principal axis at a distance of 60 cm from the convex lens. The image formed by combination coincides the object itself. The focal length of the convex mirror is _____ cm.
Answer: 10
For lens $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$ $$\Rightarrow \frac{1}{v} - \frac{1}{(-60)} = \frac{1}{20} \Rightarrow \frac{1}{v} + \frac{1}{60} = \frac{1}{20}$$ $$v = 30 \, cm$$ For final image to be formed on the object itself, after refraction from lens the ray should meet the mirror perpendicularly and the image by lens should be on the centre of curvature of mirror. $$R = 30 - 10 = 20 \, cm$$ Focal length of mirror = $R/2 = 10 \, cm$
Physics · Electromagnetic Induction · Numerical
Magnetic flux (in weber) in a closed circuit of resistance $20 \, \Omega$ varies with time $t(\mathrm{s})$ as $\phi = 8t^2 - 9t + 5$. The magnitude of the induced current at $t = 0.25 \, \mathrm{s}$ will be
Answer: 250
Given $\phi = 8t^2 - 9t + 5$. The emf is given by $$emf = -\frac{d\phi}{dt} = -(16t - 9).$$ At $t = 0.25 \, s$, $$Emf = -[(16 \times 0.25) - 9] = 5 \, V.$$ The current is given by $$Current = \frac{Emf}{Resistance} = \frac{5 \, V}{20 \, \Omega}.$$ This simplifies to $$= \frac{1}{4} \, A = \frac{1000}{4} \, mA = 250 \, mA.$$
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|}{(molecule)} & \multicolumn{2}{c|}{(hybridization; shape)} \\ \hline A. & XeO$_3$ & I. & $sp^3d$ ; linear \\ \hline B. & XeF$_2$ & II. & $sp^3$ ; pyramidal \\ \hline C. & XeOF$_4$ & III. & $sp^3d^3$ ; distorted octahedral \\ \hline D. & XeF$_5$ & IV. & $sp^3d^2$ ; square pyramidal \\ \hline \end{tabular} Choose the correct answer from the options given below:
Answer: (a)
Chemistry · Solutions · Single correct
Two solutions A and B are prepared by dissolving 1 g of non-volatile solutes X and Y, respectively in 1 kg of water. The ratio of depression in freezing points for A and B is found to be 1 : 4. The ratio of molar masses of X and Y is:
Answer: (b)
The ratio of $\Delta T_{fx}$ to $\Delta T_{fy}$ is given by: $$\frac{\Delta T_{fx}}{\Delta T_{fy}} = \frac{k_f \cdot m_x}{k_f \cdot m_y} = \frac{\frac{1}{M_x}}{\frac{1}{M_y}}$$ This simplifies to: $$\frac{1}{4} = \frac{M_y}{M_x}$$ Therefore, the ratio $M_x : M_y$ is $1 : 0.25$.
Chemistry · Equilibrium · Single correct
$K_{a_1}$, $K_{a_2}$ and $K_{a_3}$ are the respective ionization constants for the following reactions (a), (b), and $(c)$. (a) $\mathrm{H_2C_2O_4} \rightleftharpoons \mathrm{H^+} + \mathrm{HC_2O_4^-}$ (b) $\mathrm{HC_2O_4^-} \rightleftharpoons \mathrm{H^+} + \mathrm{C_2O_4^{2-}}$ $(c)$ $\mathrm{H_2C_2O_4} \rightleftharpoons 2\mathrm{H^+} + \mathrm{C_2O_4^{2-}}$ The relationship between $K_{a_1}$, $K_{a_2}$ and $K_{a_3}$ is given as
Answer: (d)
The dissociation of $\mathrm{H_2C_2O_4}$ is shown in the following steps: $$\mathrm{H_2C_2O_4 \rightleftharpoons H^+ + HC_2O_4^-} K_{a_1}$$ $$\mathrm{HC_2O_4^- \rightleftharpoons H^+ + C_2O_4^{2-}} K_{a_2}$$ Overall dissociation: $$\mathrm{H_2C_2O_4 \rightleftharpoons 2H^+ + C_2O_4^{2-}} K_{a_3} = K_{a_1} \times K_{a_2}$$
Chemistry · Electrochemistry · Single correct
The molar conductivity of a conductivity cell filled with 10 moles of 20 mL NaCl solution is $\Lambda_{m1}$ and that of 20 moles another identical cell heaving 80 mL NaCl solution is $\Lambda_{m2}$, The conductivities exhibited by these two cells are same. The relationship between $\Lambda_{m2}$ and $\Lambda_{m1}$ is
Answer: (a)
Given $\Lambda_m = \kappa \times \frac{1000}{M}$. Therefore, $\Lambda_m \propto \frac{1}{M}$. $$\frac{\Lambda_{m_1}}{\Lambda_{m_2}} = \frac{M_2}{M_1} = \frac{\frac{20}{80}}{\frac{10}{20}} = \frac{1}{4} \times \frac{2}{1} = \frac{1}{2}$$ Thus, $\Lambda_{m_2} = 2 \Lambda_{m_1}$.
Chemistry · Surface Chemistry · Single correct
For micelle formation, which of the following statements are correct? (A) Micelle formation is an exothermic process. (B) Micelle formation is an endothermic process. $(C)$ The entropy change is positive. (D) The entropy change is negative.
Answer: (c)
For micelle formation, $\Delta S > 0$ (hydrophobic effect). This is possible because the decrease in entropy due to clustering is offset by increase in entropy due to desolvation of the surfactant. Also $\Delta H > 0$.
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The first ionization enthalpies of Be, B, N and O follow the order
Answer: (d)
The first ionization energy (1st I.E.) order is given as follows: $$N \,(2p^3) > O \,(2p^4) > Be \,(2s^2) > B \,(2p^1).$$
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Given below are two statements. Statement I: Pig iron is obtained by heating cast iron with scrap iron. Statement II: Pig iron has a relatively lower carbon content than that of cast iron. In the light of the above statements, choose the correct answer from the options given below.
Answer: (b)
Statement–I is incorrect because cast iron is obtained by heating pig iron with scrap iron. Statement–II is also incorrect because pig iron has more carbon content (approximately 4$\%$) than cast iron (approximately 3$\%$).
Chemistry · Hydrogen · Single correct
High purity (>99.95$\%$) dihydrogen is obtained by
Answer: (c)
High purity (>99.95%) dihydrogen is obtained by electrolysis of warm aqueous $\mathrm{Ba(OH)_2}$ solution between Ni-electrodes.
Chemistry · The s-Block Elements · Single correct
The correct order of density is
Answer: (c)
In IIA group density decreases down the group till Ca and after that it increases. Correct order of density is Sr > Be > Mg > Ca
Chemistry · Chemical Bonding and Molecular Structure · Single correct
The total number of acidic oxides from the following list is: $NO, N_2O, B_2O_3, N_2O_5, CO, SO_3, P_4O_{10}$
Answer: (b)
Neutral Oxides — $\mathrm{N_2O}$, $\mathrm{NO}$, $\mathrm{CO}$ Acidic Oxides — $\mathrm{B_2O_3}$, $\mathrm{N_2O_5}$, $\mathrm{SO_3}$, $\mathrm{P_4O_{10}}$
Chemistry · Co-ordination Compounds · Single correct
The correct order of energy of absorption for the following metal complexes is A: $[\mathrm{Ni(en)}_3]^{2+}$, B: $[\mathrm{Ni(NH}_3)_6]^{2+}$, C: $[\mathrm{Ni(H}_2\mathrm{O})_6]^{2+}$
Answer: (a)
Stronger the ligand, larger the splitting and higher the energy of absorption. $$[\mathrm{Ni(en)_3}]^{+2} > [\mathrm{Ni(NH_3)_6}]^{+2} > [\mathrm{Ni(H_2O)_6}]^{+2}$$
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Match List I with List II. Choose the correct answer from the options given below:
Answer: (c)
A-Sulphate – III (Laxative effect) B-Fluoride – II (Bending of bones) C-Nictoine – I (pesticides) D-Sodium Arsinite – IV (herbicide)
Chemistry · Alcohols, Phenols and Ethers · Single correct
Major product of the following reaction is
Answer: (d)
The reaction involves the addition of HBr to the given compound. The first step is the addition of HBr across the double bond, resulting in the formation of a bromo ketone. In the second step, another molecule of HBr adds to the remaining double bond, leading to the final product with two bromine atoms added.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
What is the major product of the following reaction?
Answer: (b)
The reaction begins with the deprotonation of the aldehyde by $\mathrm{OH}^-$, forming an enolate ion. This enolate ion is in equilibrium with its keto form. The enolate ion then attacks another molecule of the aldehyde, leading to the formation of a new carbon-carbon bond. After protonation, the aldol product is formed. Aldol formation takes place.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Arrange the following in decreasing acidic strength.
Answer: (a)
The correct order of acid strength is
Chemistry · Amines · Single correct
$CH_3CH_2CN \xrightarrow[\mathrm{Ether}]{CH_3MgBr} A \xrightarrow{H_3O^+} B \xrightarrow[\mathrm{HCl}]{Zn-Hg} C$ The correct structure of C is
Answer: (a)
Chemistry · Polymers · 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|}{Polymer} & \multicolumn{2}{c|}{used for items} \\ \hline A. & Nylon 6,6 & I. & Buckets \\ \hline B. & Low density polythene & II. & Non-stick utensils \\ \hline C. & High density polythene & III. & Bristles of brushes \\ \hline D. & Teflon & IV. & Toys \\ \hline \end{tabular} Choose the correct answer from the options given below:
Answer: (b)
LDPE $\rightarrow$ Toys HDPE $\rightarrow$ Buckets (As per NCERT)
Chemistry · Biomolecules · Single correct
Glycosidic linkage between $C_1$ of $\alpha$-glucose and $C_2$ of $\beta$-fructose is found in
Answer: (b)
Theoretical
Chemistry · Chemistry in Everyday Life · Single correct
Some drugs bind to a site other than, the active site of an enzyme. This site is known as
Answer: (b)
Theoretical
Chemistry · Redox Reactions · Single correct
In base vs. Acid titration, at the end point methyl orange is present as
Answer: (a)
The reaction involves the conversion of the azo compound to its quinonoid form upon the addition of $\mathrm{H}^+$. The structure changes from $\mathrm{Me_2N{-}C_6H_4{-}N{=}N{-}C_6H_4{-}SO_3^-Na^+}$ to $\mathrm{Me_2N^+{-}C_6H_4{-}N{=}NH{-}C_6H_4{-}SO_3^-Na^+}$, indicating the formation of the quinonoid form.
Chemistry · Some Basic Concepts of Chemistry · Numerical
56.0 $\mathrm{L}$ of nitrogen gas is mixed with excess of hydrogen gas and it is found that 20 $\mathrm{L}$ of ammonia gas is produced. The volume of unused nitrogen gas is found to be _____ $\mathrm{L}$.
Answer: 46
Given the reaction: $\mathrm{N_2} + 3\mathrm{H_2} \rightarrow 2\mathrm{NH_3}$. Initially, there are $56 \, \mathrm{L}$ of $\mathrm{N_2}$ and an excess of $\mathrm{H_2}$. During the reaction, $10 \, \mathrm{L}$ of $\mathrm{N_2}$ is consumed and $30 \, \mathrm{L}$ of $\mathrm{H_2}$ is consumed. As a result, $20 \, \mathrm{L}$ of $\mathrm{NH_3}$ is produced. The remaining $\mathrm{N_2}$ is $46 \, \mathrm{L}$ and the remaining $\mathrm{NH_3}$ is $20 \, \mathrm{L}$.
Chemistry · States of Matter · Numerical
A sealed flask with a capacity of 2 $\mathrm{dm}^3$ contains 11 $\mathrm{g}$ of propane gas. The flask is so weak that it will burst if the pressure becomes 2 $\mathrm{MPa}$. The minimum temperature at which the flask will burst is _______ $\degree \mathrm{C}$. [Nearest integer] (Given: R = 8.3 $\mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}$. Atomic masses of C and H are 12u and 1u respectively.) (Assume that propane behaves as an ideal gas.)
Answer: 1655
Moles of $\mathrm{C_3H_8} = \frac{11}{44} = 0.25$ moles $PV = nRT$ $\Rightarrow \ 2 \times 10^6 \times 2 \times 10^{-3} = 0.25 \times 8.3 \times T$ $\Rightarrow \ T = 1927.710 \, \mathrm{K} = 1654.56 \, ^\circ \mathrm{C}$
Chemistry · Structure of Atom · Numerical
When the excited electron of a H atom from $n = 5$ drops to the ground state, the maximum number of emission lines observed are
Answer: 4
Since only a single H atom is present, maximum number of spectral lines = 4
Chemistry · Thermodynamics · Numerical
While performing a thermodynamics experiment, a student made the following observations, $$\mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O} \Delta H = -57.3 \, \mathrm{kJ \, mol^{-1}}$$ $$\mathrm{CH_3COOH} + \mathrm{NaOH} \rightarrow \mathrm{CH_3COONa} + \mathrm{H_2O} \Delta H = -55.3 \, \mathrm{kJ \, mol^{-1}}$$ The enthalpy of ionization of $\mathrm{CH_3COOH}$ as calculated by the student is ______ kJ mol$^{-1}$. (nearest integer)
Answer: 2
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
For the decomposition of azomethane. $$\mathrm{CH_3N_2CH_3(g) \rightarrow CH_3CH_3(g) + N_2(g)}$$ a first order reaction, the variation in partial pressure with time at 600 K is given as The half life of the reaction is _____ $\times 10^{-5}$ s. [Nearest integer]
Answer: 2
For first order reaction $$k = \frac{1}{t} \ln \left( \frac{P_0}{P} \right)$$ $$\ln \left( \frac{P_0}{P} \right) = kt$$ $$t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{3.465 \times 10^4} = 2 \times 10^{-5}$$
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Fill in the blank
The sum of number of lone pairs of electrons present on the central atoms of ${XeO3}$, ${XeOF4}$ and ${XeF6}$ is ____________
Answer: 3
Chemistry · The d-and f-Block Elements · Numerical
The spin-only magnetic moment value of $\mathrm{M}^{n+}$ ion (in gaseous state) from the pairs $\mathrm{Cr}^{3+}/\mathrm{Cr}^{2+}$, $\mathrm{Mn}^{3+}/\mathrm{Mn}^{2+}$, $\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}$ and $\mathrm{Co}^{3+}/\mathrm{Co}^{2+}$ that has negative standard electrode potential, is B.M. [Nearest integer]
Answer: 4
The standard electrode potential for the reaction is given by: $$E^0_{\mathrm{Cr^{+3}} | \mathrm{Cr^{+2}}} = -0.41 \, \mathrm{V}$$ The electronic configuration of $\mathrm{Cr^{+3}}$ is: $$[\mathrm{Cr^{+3}}] = 4s^0 \, 3d^3$$ The magnetic moment $\mu$ is calculated as: $$\mu = \sqrt{n(n+2)} \, \mathrm{B.M}$$ Substituting $n = 3$: $$= \sqrt{15} \, \mathrm{B.M} \approx 4 \, \mathrm{B.M}$$
Chemistry · States of Matter · Numerical
A sample of 4.5 mg of an unknown monohydric alcohol, R–OH was added to methylmagnesium iodide. A gas is evolved and is collected and its volume measured to be 3.1 mL. The molecular weight of the unknown alcohol is ____ g/mol. [Nearest integer]
Answer: 33
ROH + $\mathrm{CH_3MgI}$ $\rightarrow$ $\mathrm{ROMgI}$ + $\mathrm{CH_4(g)}$ Moles of $\mathrm{CH_4}$ = moles of ROH $$\Rightarrow \frac{V}{22400} = \frac{m}{\mathrm{M.M}} (Assuming NTP Condition)$$ $$\Rightarrow \frac{3.1}{22400} = \frac{4.5 \times 10^{-3}}{\mathrm{M.M}}$$ $$\Rightarrow \mathrm{M.M} = 32.51$$ Nearest Integer = 33
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Numerical
The separation of two coloured substances was done by paper chromatography. The distances travelled by solvent front, substance A and substance B from the base line are 3.25 cm, 2.08 cm and 1.05 cm respectively. The ratio of $R_f$ values of A to B is
Answer: 2
The ratio of $R_{FA}$ to $R_{FB}$ is calculated as follows: $$\frac{R_{FA}}{R_{FB}} = \frac{\frac{2.08}{3.25}}{\frac{1.05}{3.25}} = \frac{2.08}{1.05} \approx 2$$
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Numerical
The total number of monobromo derivatives formed by the alkanes with molecular formula $C_5H_{12}$ is (excluding stereo isomers)_____
Answer: 8
The Alkanes and their monobromoderivative are: 1. Alkane: Hexane $\rightarrow$ 1-Bromohexane, 2-Bromohexane, 3-Bromohexane 2. Alkane: Isobutane $\rightarrow$ 1-Bromo-2-methylpropane, 2-Bromo-2-methylpropane, 1-Bromo-3-methylbutane 3. Alkane: Neopentane $\rightarrow$ 1-Bromo-2,2-dimethylpropane