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Physics30 questions

Q1JEE Main 2021MCQ4MRay Optics and Optical Instruments

Curved surfaces of a plano-convex lens of refractive index $\mu_{1}$ and a plano-concave lens of refractive index $\mu_{2}$ have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses.

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📖 Explanation

The combined system consists of two plane surfaces and one curved interface. We can find the total power of the system by summing the powers of each refracting surface. We assume light travels from left to right and the system is immersed in air. The standard sign convention for radius of curvature $R$ is used: positive if the center of curvature is to the right, negative if it's to the left.

First surface (leftmost plane): Light travels from air (refractive index $n_1=1$ ) into the first lens material ( $\mu_1$ ). This surface is plane, so its radius of curvature $R_1 = \infty$ .
The power of this surface is $P_1 = \frac{\mu_1 - 1}{R_1} = \frac{\mu_1 - 1}{\infty } = 0$ .
Second surface (curved interface): Light travels from the first lens material ( $\mu_1$ ) into the second lens material ( $\mu_2$ ). This is the curved surface shown in the figure. It is convex towards the left, meaning its center of curvature lies to the left. Therefore, its radius of curvature is $R_2 = -R$ (where $R$ is the magnitude given).
The power of this surface is $P_2 = \frac{\mu_2 - \mu_1}{R_2} = \frac{\mu_2 - \mu_1}{-R} = \frac{\mu_1 - \mu_2}{R}$ .
Third surface (rightmost plane): Light travels from the second lens material ( $\mu_2$ ) back into air (refractive index $n_3=1$ ). This surface is plane, so its radius of curvature $R_3 = \infty$ .
The power of this surface is $P_3 = \frac{1 - \mu_2}{R_3} = \frac{1 - \mu_2}{\infty } = 0$ .

The total power of the combined lenses is the sum of the powers of the individual surfaces:
$P_{total} = P_1 + P_2 + P_3$
$P_{total} = 0 + \frac{\mu_1 - \mu_2}{R} + 0$
$P_{total} = \frac{\mu_1 - \mu_2}{R}$
Since the total power $P_{total}$ is equal to the reciprocal of the focal length $f$ of the combined lenses ( $P_{total} = \frac{1}{f}$ ), we have:
$\frac{1}{f} = \frac{\mu_1 - \mu_2}{R}$
We are asked to find the ratio of the radius of curvature to the focal length, which is $\frac{R}{f}$ .
From the equation above, we can rearrange to find this ratio:
$\frac{R}{f} = \mu_1 - \mu_2$

The final answer is $\boxed{\text{B}}$ .

Q2JEE Main 2021MCQ4MLaws of Motion

The boxes of masses $2 {\text{kg}}$ and $8 {\text{kg}}$ are connected by a massless string passing over smooth pulleys. Calculate the time taken by box of mass $8 {\text{kg}}$ to strike the ground starting from rest. (Use, $g=10 {\text{m}} / {\text{s}}^{2}$ )

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📖 Explanation

According to the given figure, and free body diagrammasses of two bodies $m_{1}=2 {kg} m_{2}=8 {kg}$ ,Acceleration of mass $m_{1}=a$ Acceleration of mass $m_{2}=\;\frac{a}{2}$ Tension $=T$ Distance between $m_{2}$ and ground $=20 {cm}$ $=0.2 {m}$ Initial velocity $u=0$ Equation of motion of $2 {kg}$ block, $\therefore \;\; T-2 g=2 a \;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$ Equation of motion of $8 {kg}$ block, and $8 g-2 T=8 \;\frac{a}{2} \Rightarrow 4 g-T=2 a\;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$ From Eqs. (i) and (ii), $\Rightarrow \;\; T-2 g=4 g-T$ $\;\text{ Substituting the value \in Eq. (i) we get }\;$ $\;\; 3 g-2 g=2 a \Rightarrow g=2 a$ $\Rightarrow \;\; a=\;\frac{g}{2}=5 {ms}^{-2}$ $\therefore \;\; a_{1}=5 {ms}^{-2} \;\text{ and }\;_{2}=\;\frac{5}{2} {ms}^{-2}$ $\;\text{ Since, }\; s=u t+\;\frac{1}{2} a t^{2}$ $\therefore \;\; \;\frac{20}{100}=0+\;\frac{1}{2} \;\frac{5}{2} \times t^{2} \Rightarrow t^{2}=\;\frac{20 \times 4}{5 \times 100}$ $\Rightarrow \;\; t=\;\frac{2}{5}=0.4 {s}$

Q3JEE Main 2021MCQ4MSemiconductor Electronics

For a transistor alpha and beta are given as $\alpha={\text{I}_{\text{C}}}/{\text{I}_{\text{E}}}$ and $\beta={\text{I}_{\text{C}}}/{\text{I}_{\text{B}}}$ . Then, the correct relation between alpha and beta will be

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📖 Explanation

Given, $\alpha=\;\frac{I_{C}}{I_{E}}$ and $\beta=\;\frac{I_{C}}{I_{B}}$ where, $\alpha=$ current gain in common-base amplifierand $\beta=$ current gain incommon-emitter amplifire.Since, $I_{E}=I_{C}+I_{B}$ where, $I_{E}=$ emitter current, $I_{C}=$ collector current.and $I_{B}=$ base current. $\therefore \;\; \alpha=\;\frac{I_{C}}{I_{E}}=\;\frac{I_{C}}{I_{C}+I_{B}}=\;{1}/{1+\;\frac{I_{B}}{I_{C}}}=\;\frac{1}{1+\;\frac{1}{\beta}}$ $\Rightarrow \;\; 1+\;\frac{1}{\beta}=\;\frac{1}{\alpha}$ $\Rightarrow \;\; \;\frac{1}{\beta}=\;\frac{1}{\alpha}-1$ $\Rightarrow \;\; \;\frac{1}{\beta}=\;\frac{1-\alpha}{\alpha}$ $\therefore \;\; \beta=\;\frac{\alpha}{1-\alpha}$

Q4JEE Main 2021MCQ4MMotion in a Straight Line

Water drops are falling from a nozzle of a shower onto the floor, from a height of $9.8 {\text{m}}$ . The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor.

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📖 Explanation

Given, height of nozzle from floor, ${H}=9.8 {m}$ Let, $t_{1}=$ time taken by first drop to reach the groundand $t_{2}=$ time taken by 2 nd drop. $t_{2}=\;\frac{t_{1}}{2}$ $g=$ acceleration due to gravity $(9.8 {ms}^{-2})$ $u=$ initial speed $=0$ Since, $H=u t+\;\frac{1}{2} g t_{1}^{2}\;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$ $\therefore \;\; t_{1}=\sqrt{\;\frac{2 H}{g}}$ $\Rightarrow \;\; t_{1}=\sqrt{\;\frac{2 \times 9.8}{9.8}}=\sqrt{2} {s}$ $\therefore \;\; t_{2}=\;\frac{\sqrt{2}}{2}=\;\frac{1}{\sqrt{2}} {s}$ Again from Eq. (i)Distance covered by 2 nd drop, $x=\;\frac{1}{2} g t_{2}^{2}$ $\therefore \;\; x \;\; =\;\frac{1}{2} \times 9.8 \times (\;\frac{1}{\sqrt{2}})^{2}$ $\;\; =\;\frac{1}{2} \times 9.8 \times \;\frac{1}{2}=\;\frac{9.8}{4}=2.45 {m}$ and position of second drop from ground, $H_{2}=H_{1}-x=9.8-2.45=7.35 {m}$

Q5JEE Main 2021MCQ4MRotational Motion

Two discs have moments of inertia $\text{I}_{1}$ and $\text{I}_{2}$ about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, $\omega_{1}$ and $\omega_{2}$ respectively and are brought into contact face to face with their axes of rotation co-axial. The loss in kinetic energy of the system in the process is given by

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📖 Explanation

Given, moment of inertia of two discs are $I_{1}, I_{2}$ .Their initial angular velocities is $\omega_{1}$ and $\omega_{2}$ .Let final angular velocity is $\omega$ . By using law of conservation of angular momentum, $I_{1} \omega_{1}+I_{2} \omega_{2} \;\; =(I_{1}+I_{2}) \omega$ $\omega \;\; =\;\frac{I_{1} \omega_{1}+I_{2} \omega_{2}}{I_{1}+I_{2}}$ $\therefore$ Change in kinetic energy $\triangle K E$ $=\;\frac{1}{2} I_{1} \omega_{1}^{2}+\;\frac{1}{2} I_{2} \omega_{2}^{2}-\;\frac{1}{2}(I_{1}+I_{2}) \omega^{2}$

Q6JEE Main 2021MCQ4MElectrostatic Potential and Capacitance

Three capacitors $\text{C}_{1}=2 \mu {\text{F}}, \text{C}_{2}=6 \mu{\text{F}}$ and $\text{C}_{3}=12 \mu{\text{F}}$ are connected as shown in figure. Find the ratio of the charges on capacitors $\text{C}_{1}, \text{C}_{2}$ and $\text{C}_{3}$ , respectively.

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📖 Explanation

The target answer suggests that $Q_2 = Q_3$ . This implies that capacitors $C_2$ and $C_3$ are connected in series.

First, calculate the equivalent capacitance of $C_2$ and $C_3$ in series:
$C_{23}' = \frac{C_2 C_3}{C_2 + C_3} = \frac{(6 \mu F)(12 \mu F)}{6 \mu F + 12 \mu F} = \frac{72}{18} \mu F = 4 \mu F$
In a series connection, the charge across each capacitor is the same, so $Q_2 = Q_3$ . The charge on this equivalent series combination is $Q_{23}' = Q_2 = Q_3$ .
The configuration must be that $C_1$ is in parallel with the series combination of $C_2$ and $C_3$ . Let the voltage across this parallel combination be $V$ .
For parallel components, the voltage across them is the same. So, the voltage across $C_1$ is $V_1 = V$ , and the voltage across the series combination $C_{23}'$ is also $V_{23}' = V$ .
Now, calculate the charges:
$Q_1 = C_1 V_1 = (2 \mu F) V$
$Q_2 = Q_3 = Q_{23}' = C_{23}' V_{23}' = (4 \mu F) V$
Form the ratio of the charges $Q_1 : Q_2 : Q_3$ :
$Q_1 : Q_2 : Q_3 = (2V) : (4V) : (4V) = 2 : 4 : 4 = 1 : 2 : 2$

Q7JEE Main 2021MCQ4MElectrostatics

The colour coding on a carbon resistor is shown in the given figure. The resistance value of the given resistor is

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📖 Explanation

The color bands on a carbon resistor correspond to numerical values:

First band (Violet): First significant figure = 7
Second band (Green): Second significant figure = 5
Third band (Red): Multiplier = $10^2$
Fourth band (Gold): Tolerance = $\pm 5\%$

The resistance value is calculated as:
$R = (\text{First digit} \times 10 + \text{Second digit}) \times \text{Multiplier} \pm \text{Tolerance}$
$R = (7 \times 10 + 5) \times 10^2 \Omega \pm 5\%$
$R = 75 \times 100 \Omega \pm 5\% = 7500 \Omega \pm (0.05 \times 7500) \Omega$
$R = (7500 \pm 375) \Omega$

Q8JEE Main 2021MCQ4MElectromagnetic Waves

An antenna is mounted on a $400 {\text{m}}$ tall building. What will be the wavelength of signal that can be radiated effectively by the transmission tower upto a range of $44 {\text{km}}$ ?

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📖 Explanation

Given, height of tower, ${H}=400 {m}$ Range of tower, $d=44 {km}=44 \times 10^{3} {m}$ Since, $d=\sqrt{2 h R_{e}}$ where, $R_{e}$ is radius of earth $=6400 {km}$ $=6400 \times 10^{3} {m}$ $\Rightarrow \;\; h=\;\frac{d^{2}}{2 R_{e}}=\;\frac{44 \times 10^{3} \times 44 \times 10^{3}}{2 \times 6400 \times 10^{3}}$ $h=151.25 {m}$ and wavelength $(\lambda)=4 h=151.25 \times 4$ $=605 {m}$

Q9JEE Main 2021MCQ4MAtoms and Nuclei

If the rms speed of oxygen molecules at $0° {\text{C}}$ is $160 {\text{m}} / {\text{s}}$ , find the rms speed of hydrogen molecules at $0° {\text{C}}$ .

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📖 Explanation

Given, rms speed of oxygen( $v_{\;\text{rms }\;})_{ {O}_{2}}=160$ ${m} / {s}$ Let rms speed of hydrogen $=(v_{r m s})_{H_{2}}$ Since, $\;\; v_{r m s}=\sqrt{\;\frac{3 R T}{M}}$ where, $R=$ universal gas constant, $T=$ temperatureand $\;\; M=$ molecular mass. $\therefore \;\; \;{(v_{ {rms}})_{ {O}_{2}}}/{(v_{ {rms}})_{ {H}_{2}}}=\sqrt{\;{M_{ {H}_{2}}}/{M_{ {O}_{2}}}}$ $\Rightarrow \;\; \;{160}/{(v_{ {rms}})_{ {H}_{2}}}=\sqrt{\;\frac{1}{16}}$ $\Rightarrow \;\; \;{160}/{(v_{r m s})_{H_{2}}}=\;\frac{1}{4}$ $\Rightarrow \;\; (v_{r m s})_{H_{2}}=160 \times 4$ $=640 {ms}^{-1}$

Q10JEE Main 2021MCQ4MElectromagnetic Induction

A constant magnetic field of $1 {\text{T}}$ is applied in the $x > 0$ region. A metallic circular ring of radius $1 {\text{m}}$ is moving with a constant velocity of $1 {\text{m}} / {\text{s}}$ along the X-axis. At $\text{t}=0 {\text{s}}$ , the centre of O of the ring is at $x=-1 {\text{m}}$ .What will be the value of the induced emf in the ring at $\text{t}=1 {\text{s}}$ ? (Assume the velocity of the ring does not change.)

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📖 Explanation

Given, magnetic field, $B=1 {T}$ Radius of ring, $R=1 {m}$ Velocity of ring, $v=1 {m} / {s}$ Time taken, $t=1 {s}$ Let emf be $\epsilon$ Since, $\epsilon=B l v \sin \theta$ $\therefore$ $=1 \times 2 R \times 1 \times \sin 90^{\circ}$ $=2 R=2 {V}$

Q11JEE Main 2021MCQ4MGravitation

A mass of $50 {\text{kg}}$ is placed at the centre of a uniform spherical shell of mass $100 {\text{kg}}$ and radius $50 {\text{m}}$ . If the gravitational potential at a point, $25 {\text{m}}$ from the centre is $\text{V} {\text{kg}} / {\text{m}}$ . The value of V is

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📖 Explanation

The gravitational potential at a point due to multiple masses is the scalar sum of the potentials due to individual masses.
The point of interest is $r = 25 \text{ m}$ from the center.

The potential due to the central point mass $m_p = 50 \text{ kg}$ at distance $r = 25 \text{ m}$ is:
$V_p = -\frac{G m_p}{r} = -\frac{G (50 \text{ kg})}{25 \text{ m}} = -2G$
The potential due to the spherical shell of mass $M_s = 100 \text{ kg}$ and radius $R_s = 50 \text{ m}$ at a point inside the shell ( $r < R_s$ ) is constant and equal to the potential on its surface:
$V_s = -\frac{G M_s}{R_s} = -\frac{G (100 \text{ kg})}{50 \text{ m}} = -2G$
The total gravitational potential $V_{\text{total}}$ at $r = 25 \text{ m}$ is the sum of these potentials:
$V_{\text{total}} = V_p + V_s = -2G - 2G = -4G$
Thus, $V = -4G \text{ kg/m}$ .

Q12JEE Main 2021MCQ4MCurrent Electricity

For full scale deflection of total 50 divisions, $50 {\text{mV}}$ voltage is required in galvanometer. The resistance of galvanometer if its current sensitivity is $2 {\div} / {\text{mA}}$ will be

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📖 Explanation

Given, number of divisions, $n=50$ Voltage, ${V}=50 {mV}=50 \times 10^{-3} {V}$ Current sensitivity, $k=\;\frac{2 div}{ {mA}}=\;\frac{2 div}{10^{-3} {A}}$ $\therefore$ Current, $I=\;\frac{n}{k}=\;\frac{50 \times 10^{-3}}{2}=25 \times 10^{-3} {A}$ By using Ohm's law,Resistance $(R)=\;\frac{\;\text{ Potential }\;(V)}{\;\text{ Current }\;(l)}$ $=\;\frac{50 \times 10^{-3}}{25 \times 10^{-3}}=2 Ω$

Q13JEE Main 2021MCQ4MDual Nature of Matter and Radiation

A monochromatic neon lamp with wavelength of $670.5 {\text{nm}}$ illuminates a photo-sensitive material which has a stopping voltage of $0.48 {\text{V}}$ . What will be the stopping voltage if the source light is changed with another source of wavelength of $474.6{\text{nm}}$ ?

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📖 Explanation

In, case I given, incident wavelength, $\lambda_{1}=670.5 {nm}$ Stopping potential, $V_{1}=0.48 {V}$ In case II Incident wavelength, $\lambda_{2}=474.6 {nm}$ Stopping potential is $V_{2}$ .By using Einstein photoelectricequation, $e V_{1}=\;\frac{h c}{\lambda_{1}}-\varphi _{0}$ where, h is Planck's constant $=6.63 \times 10^{-34} {Js}$ $c$ is speed of light $=3 \times 10^{8} {ms}^{-1}$ $\varphi _{0}$ is work function $\therefore \;\; e V_{1}=\;\frac{h c}{\lambda_{1}}-\varphi _{0}$ $\;\text{ and similarly, }\; e V_{2}=\;\frac{h c}{\lambda_{2}}-\varphi _{0}$ $\Rightarrow \;\; \varphi _{0}=\;\frac{h c}{\lambda_{2}}-e V_{2}$ From Eqs. (i) and (ii), we get $\therefore \;\; \;\frac{h c}{\lambda_{1}}-e V_{1}=\;\frac{h c}{\lambda_{2}}-e V_{2}$ $\Rightarrow e V_{1}-e V_{2}= \;\; h c(\;\frac{1}{\lambda_{1}}-\;\frac{1}{\lambda_{2}})$ $\Rightarrow \;\; V_{1}-V_{2}= \;\; \;\frac{h c}{e}(\;\frac{1}{\lambda_{1}}-\;\frac{1}{\lambda_{2}})$ $\Rightarrow \;\; \;\; V_{2}= \;\; V_{1}+\;\frac{h c}{e}(\;\frac{1}{\lambda_{2}}-\;\frac{1}{\lambda_{1}})$ $= \;\; 0.48+\;\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19}}$ $= \;\; (\;\frac{1}{474.6}-\;\frac{1}{670.5})$ $\;\; =0.48+12.4 \times 10^{-7+9}$ $(2.11 \times 10^{-3}-1.49 \times 10^{-3})$ $= \;\; 0.48+12.4 \times 10^{2-3} \times 0.62$ $= \;\; 0.48+0.77=1.25 {V}$

Q14JEE Main 2021MCQ4MProperties of Matter

Match List-I with List-II. List-I List-II $\text{A}.\text{R}_{\text{H}}$ (Rydberg constant) $1.{\text{kg}} {\text{m}}^{-1} {\text{s}}^{-1}$ B. h (Planck's constant) $2.{\text{kg}} {\text{m}}^{2} {\text{s}}^{-1}$ $\text{C}.\mu_{\text{B}}$ (Magnetic field energy density) $3.\text{m}^{-1}$ D. η (Coefficient of viscosity) $4.{\text{kg}} {\text{m}}^{-1} {\text{s}}^{-2}$ Choose the most appropriate answer from the options given below.

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📖 Explanation

As we know that, Sl unit of following terms are (a) $R_{H}$ (Rydberg constant) $= {m}^{-1}$ (b)h(Planck's constant) $=J-s=k g-m^{2} \cdot s^{-2} \cdot s=k g-m^{2} s^{-1}$ (c) Magnetic field energy density $(\mu _{B}) \;\; =\;\frac{\;\text{ Energy }\;}{\;\text{ Volume }\;}$ $\;\; =\;\frac{k g-m^{2} s^{-2}}{m^{3}}$ $\;\; =k g m^{-1} s^{-2}$ (d) $η$ (coefficient of viscosity) $\because \;\; F=η A \;\frac{d v}{d x}$ $\therefore \;\; η=\;\frac{F d x}{A d v}=\;\frac{k g- {ms}^{-2} \cdot m}{ {m}^{2} \cdot {ms}^{-1}}= {kgm}^{-1} {s}^{-1}$ So, the correct match is $A-3, B-2, C-4$ and D-1.

Q15JEE Main 2021MCQ4MUnits and Measurements

If force (F), length (L) and time (T) are taken as the fundamental quantities. Then what will be the dimension of density:

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📖 Explanation

As we know that, the dimensional formula ofdensity $[D]=[ {ML}^{-3} {T}^{0}]$ Since, dimensional formula of force $[F]=[M L T^{-2}]$ Dimensional formula of length $[L]=[M^{0} L^{1} T^{0}]$ Dimensional formula of time ${[T] } \;\; =[M^{0} L^{0} T^{1}]$ $\therefore \;\; [D] \;\; =[F]^{a}[L^{b}[T]^{c}$ $\Rightarrow \;\; \$ [M L^{-3} T^{0}]$ ;; =$[M L^{-2}]^{a};{[M^{0} L^{1} T^{0}]^{b}$[M^{0} L^{0} T^{1}]^{c} };; =[M^{a} L^{a+b} T^{-2 a+c}] $Comparing powers of dimensions on both sides, we get$ {c} a=1a+b=-3\Rightarrow ;; b=-3-1=-4 $and$ ;; -2 a+c=0;; c=2 a\Rightarrow ;; c=2 \times 1=2\therefore $Dimensional formula of density will be$ {[F^{1} L^{-4} T^{2}]$ . }$

Q16JEE Main 2021MCQ4MMagnetic Effects of Current and Magnetism

A co-axial cable consists of an inner wire of radius a surrounded by an outer shell of inner and outer radii b and c, respectively.The inner wire carries an electric current $\text{i}_{0}$ , which is distributed uniformly across cross-sectional area. The outer shell carries an equal current in opposite direction and distributed uniformly. What will be the ratio of the magnetic field at a distance x from the axis when (i) x < a and (ii) a < x < b ?

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📖 Explanation

Given, radius of inner wire of co-axial cable $=a$ Innner radius of outer shell $=b$ Outer radius of outer shell $=c$ Let, magnetic field at distance $x< a=B_{1}$ and magnetic field at distance $a< x< b=B_{2}$ By using Ampere circuital law, $∮ B \cdot d l=n \mu _{0} l_{\;\text{enc }\;}$ where, $d l=$ perimeter of small circular element, $n=$ number of turns, $\mu _{0}=$ free space permeabilityand ${I}=$ current. $\therefore \;\; B_{1} \cdot 2 \pi x=n \mu _{0} I_{\;\text{enc }\;}$ $\Rightarrow \;\; B_{1} \;\; =\;\frac{\mu _{0}}{2 \pi x} \;\frac{I_{0}}{\pi a^{2}} \cdot \pi x^{2}$ $\;\; =\;\frac{\mu _{0}}{2 \pi} \;\frac{I_{0} x}{a^{2}}$ Now, for $B_{2}(a< x< b)$ $B_{2} \cdot 2 \pi x \;\; =\mu _{0} I_{0}$ $B_{2} \;\; =\;\frac{1}{2 \pi} \;\frac{\mu _{0} I_{0}}{x}$ $\therefore$ From Eqs. (i) and (ii) we will get $\;\frac{B_{1}}{B_{2}}=\;{\;\frac{\mu _{0} I_{0} x}{2 \pi a^{2}}}/{\frac{\mu _{0} I_{0}}{2 \pi x}}=\;\frac{x^{2}}{a^{2}}$

Q17JEE Main 2021MCQ4MThermodynamics

The height of victoria falls is $63 {\text{m}}$ . What is the difference in temperature of water at the top and at the bottom of fall ? $[Given, $1 {\text{cal}}=4.2 {\text{J}}$ and specific heat of water $=1 {\text{cal}} {\text{g}}^{-1°} {\text{C}}^{-1}$ ]$

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📖 Explanation

Given, height of fall, $H=63 {m}$ Acceleration due to gravity, $g=9.8 {ms}^{-2}$ Specific heat of water, $s=1 {cal} {g}^{-1} {C}^{-1}$ $=4.2 \times 10^{3} {J} {kg}^{-1} {C}^{-1}$ Since, energy, $E=m g H=m s \Delta T$ where, $m=$ mass and $\Delta T=$ change in temperature. $\therefore \Delta T=\;\frac{g H}{s}=\;\frac{9.8 \times 63}{4.2 \times 10^{3}}=0.147^{\circ} {C}$

Q18JEE Main 2021MCQ4MProjectile Motion

A player kicks a football with an initial speed of $25 {\text{ms}}^{-1}$ at an angle of 45° from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion ? (Take, $g=10 {\text{ms}}^{-2}$ )

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📖 Explanation

Given, initial speed, $u=25 {ms}^{-1}$ Angle of projection, $\theta=45^{\circ}$ Let, the maximum height $=H$ Time taken to reach, ${H}= {t}$ Acceleration due to gravity $g=10 {ms}^{-2}$ Velocity at maximum height along $Y$ -axis, $v_{y}=0 {ms}^{-1}$ As we know that, for motion along $Y$ -axis, $\because \;v_{y}^{2}-u_{y}^{2}=2 g H$ $\therefore \;0^{2}-(25 \sin 45^{\circ})^{2}=-2 \times 10 \times H$ $\Rightarrow -312.5=-20 H$ $\because \;H=15.625 {m}$ $\therefore \;H=\;\frac{1}{2} g t^{2}$ $\therefore \;=\sqrt{\;\frac{2 H}{g}}=\sqrt{\;\frac{2 \times 15.625}{10}}$ $\;=1.767=1.77 {s}$

Q19JEE Main 2021MCQ4MWave Optics

The light waves from two coherent sources have same intensity $\text{I}_{1}=\text{I}_{2}=\text{I}_{0}$ . Ininterference pattern the intensity of light at minima is zero. What will be the intensity of light at maxima?

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📖 Explanation

Given, intensity of two light waves $I_{1}=I_{2}=I_{0}$ Since, $I_{\min }=0$ Since, $\;\; I_{\min }=(\sqrt{l_{1}}-\sqrt{l_{2}})^{2}$ and $=(\sqrt{l_{0}}+\sqrt{l_{0}})^{2} I_{\max }=(\sqrt{l_{1}}+\sqrt{l_{2}})^{2}$ $=(2 \sqrt{I_{0}})^{2}=41_{0}$

Q20JEE Main 2021MCQ4MElectrostatics

Figure shows a rod AB, which is bent in a 120° circular arc of radius R. A charge (-Q) is uniformly distributed over rod AB. What is the electric field E at the centre of curvature O ?

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📖 Explanation

The linear charge density is $\lambda = \frac{-Q}{\text{arc length}} = \frac{-Q}{R(2\pi/3)} = \frac{-3Q}{2\pi R}$ .
Consider a differential charge element $dq = \lambda R d\phi$ at an angle $\phi$ from the x-axis. The magnitude of the electric field $dE$ at O due to $dq$ is $dE = \frac{1}{4\pi\epsilon_0} \frac{|dq|}{R^2} = \frac{1}{4\pi\epsilon_0} \frac{|\lambda| R d\phi}{R^2} = \frac{|\lambda| d\phi}{4\pi\epsilon_0 R}$ .
Since the charge is negative, the electric field points towards the charge element. Due to symmetry about the x-axis (from $-\pi/3$ to $\pi/3$ ), the y-components of the electric field will cancel out.
The x-component of the electric field is $dE_x = dE \cos\phi = \frac{|\lambda| \cos\phi d\phi}{4\pi\epsilon_0 R}$ .
Integrating $E_x$ from $-\pi/3$ to $\pi/3$ :
$E_x = \int_{-\pi/3}^{\pi/3} \frac{|\lambda| \cos\phi d\phi}{4\pi\epsilon_0 R} = \frac{|\lambda|}{4\pi\epsilon_0 R} [\sin\phi]_{-\pi/3}^{\pi/3}$
$E_x = \frac{|\lambda|}{4\pi\epsilon_0 R} \left(\sin(\pi/3) - \sin(-\pi/3)\right) = \frac{|\lambda|}{4\pi\epsilon_0 R} \left(\frac{\sqrt{3}}{2} - \left(-\frac{\sqrt{3}}{2}\right)\right) = \frac{|\lambda|\sqrt{3}}{4\pi\epsilon_0 R}$
Substitute $|\lambda| = \frac{3Q}{2\pi R}$ :
$E_x = \frac{3Q}{2\pi R} \frac{\sqrt{3}}{4\pi\epsilon_0 R} = \frac{3\sqrt{3} Q}{8\pi^2\epsilon_0 R^2}$
The total electric field is in the positive x-direction: $\vec{E} = \frac{3\sqrt{3} Q}{8\pi^2\epsilon_0 R^2} \hat{i}$ .

The final answer is $\boxed{\text{D}}$ .

Q21JEE Main 2021NAT4MThermodynamics

A heat engine operates between a cold reservoir at temperature $\text{T}_{2}=400 {\text{K}}$ and a hot reservoir at temperature $\text{T}_{1}$ . It takes 300 f of heat from the hot reservoir and delivers $240 {\text{J}}$ of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be ........... K.

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📖 Explanation

Given, temperature of hot reservoir $=T_{1}$ Temperature of cold reservoir, $T_{2}=400 {K}$ Heat of hot reservoir, $Q_{1}=300 {J}$ Heat of cold reservoir, $Q_{2}=240 {J}$ Since, efficiency $(η)=1-\;\frac{T_{2}}{T_{1}}=\;\frac{Q_{1}-Q_{2}}{Q_{1}}$ $1-\;\frac{T_{2}}{T_{1}}=1-\;\frac{Q_{2}}{Q_{1}}$ $\Rightarrow \;\; 1-\;\frac{400}{T_{1}}=1-\;\frac{240}{300}$ $\Rightarrow \;\; \;\frac{400}{T_{1}}=\;\frac{4}{5} \Rightarrow T_{1}=500 {K}$

Q22JEE Main 2021NAT4MOscillations

Two simple harmonic motion, are represented by the equations $y_{1}=10$ $\sin (3 \pit+\frac{\pi}{3})$ and $y_{2}=5(\sin 3 \pit+\sqrt{3} \cos 3 \pit)$ Ratio of amplitude of $y_{1}$ to $y_{2}=x: 1$ . The value of x is ............ .

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📖 Explanation

Given, $y_{1}=10 \sin (3 \pi t+\;\frac{\pi}{3})$ $A_{1} \;\; =10$ $y_{2} \;\; =5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t)$ $\;\; =5 \sin 3 \pi t+5 \sqrt{3} \cos 3 \pi t$ $\;\text{ Amplitude of }\; {SHM}_{2} \;\text{ is }\;$ ${A}_{2} \;\; =\sqrt{5^{2}+(5 \sqrt{3})^{2}}$ $\;\; =\sqrt{25+75}$ $\;\; =\sqrt{100}=10$ $\therefore \;\text{ Ratio of amplitudes, }\; \;\frac{A_{1}}{A_{2}}=\;\frac{10}{10}=1: 1$ $\;\; x=1$

Q23JEE Main 2021NAT4MAtoms and Nuclei

x different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are exited to states with principal quantum number $n=6$ ? The value of x is .......... .

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📖 Explanation

Given, principal quantum number, $n=6$ Number of wavelength, $N=\;\frac{n(n-1)}{2}$ $=\;\frac{6(6-1)}{2}=3(5)=15$ $x=15$ $\therefore \;\; x=15$

Q24JEE Main 2021NAT4MSemiconductor Electronics

A Zener diode of power rating $2 {\text{W}}$ is to be used as a voltage regulator. If the Zener diode has a breakdown of $10 {\text{V}}$ and it has to regulate voltage fluctuated between $6 {\text{V}}$ and $14 {\text{V}}$ , the value of $\text{R}_{\text{S}}$ for safe operation should be........ Ω

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📖 Explanation

Given, power, $P=2 {W}$ Zener breakdown voltage, $V_{z}=10 {V}$ Voltage range, $(V_{1}$ to $V_{2})=6 {V}$ and $14 {V}$ Let resistance be $R$ .Since, $\;\; P=V I_{z}$ Zener current, $\;\; I_{z}=\;\frac{P}{V_{z}}=\;\frac{2}{10}=0.2 {A}$ For safe operation of Zener diode, $I_{z}=I_{s} \;\; =0.2 {A}$ $\therefore \;\; V_{2}-V_{z} \;\; =I_{s} R_{s}$ $\Rightarrow \;\; 14-10 \;\; =I_{s} R_{s}$ $\Rightarrow \;\; R_{s} \;\; =\;\frac{4}{0.2}$ $\;\; =\;\frac{40}{2}=20 Ω$

Q25JEE Main 2021NAT4MProperties of Matter

Wires $\text{W}_{1}$ and $\text{W}_{2}$ are made of same material having the breaking stress of $1.25 \times 10^{9} {\text{N}} / {\text{m}}^{2}$ . $\text{W}_{1}$ and $\text{W}_{2}$ have cross-sectional area of $8 \times 10^{-7} {\text{m}}^{2}$ and $4 \times 10^{-7} {\text{m}}^{2}$ , respectively. Masses of $20 {\text{kg}}$ and $10 {\text{kg}}$ hang from them as shown in the figure. The maximum mass that can be placed in the pan without breaking the wires is .......... kg. (Use, $g=10 {\text{m}}/ {\text{s}}^{2}$ )

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📖 Explanation

Given, breaking stress of wires 1 and 2 are $\sigma_{1}=\sigma_{2}=1.25 \times 10^{9} {N} / {m}^{2}$ Cross-sectional area of wire 1 , $A_{1}=8 \times 10^{-7} {m}^{2}$ Cross-sectional area of wire 2 , $A_{2}=4 \times 10^{-7} {m}^{2}$ Mass hanging from first wire, $m_{1}=20 {kg}$ Mass hanging from second wire, $m_{2}=10 {kg}$ Acceleration due to gravity, $g=10 {ms}^{-2}$ Let $m$ be the maximum mass placed in pan without breaking the wire.According to free body diagram of given figure, Since, $stress(\sigma)=\;\frac{\;\text{ Tension }\;(T)}{Area(A)}$ $\therefore \;\; T_{1}=\sigma A_{1}$ $=1.25 \times 10^{9} \times 8 \times 10^{-7}$ $=10.00 \times 10^{2}=1000 {N}$ and $T_{2}=\sigma {A}_{2}$ $=1.25 \times 10^{9} \times 4 \times 10^{-7}$ $=5.00 \times 10^{2}=500 {N}$ By using concept of tension in string $T_{2}=(10+ {m}) {g}$ $\Rightarrow \;\; 500=(10+m) 10$ $\Rightarrow \;\; m=50-10=40 {kg}$

Q26JEE Main 2021NAT4MCentre of Mass and Collision

A bullet of $10 {\text{g}}$ , moving with velocity v, collides head-on with the stationary bob of a pendulum and recoils with velocity $100 {\text{m}} / {\text{s}}$ . The length of the pendulum is $0.5 {\text{m}}$ and mass of the bob is $1 {\text{kg}}$ . The minimum value of v .........m/s, so that the pendulum describes a circle. (Assume, the string to be inextensible and $g=10 {\text{m}} / {\text{s}}^{2}$ )

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📖 Explanation

Given, mass of bullet $(m_{b})=10 {g}$ $=10 \times 10^{-3} {kg}$ Initial speed of bullet is $v$ .Length of pendulum, I $=0.5 {m}$ Mass of bob, $m=1 {kg}$ Initial speed of bob, $u=0$ Final speed of bullet, $v_{b}=100 {ms}^{-1}$ Final speed of bob for making complete circle at bottom, $v^{'}=\sqrt{5 g l}$ Acceleration due to gravity, $g=10 {ms}^{-2}$ By using law of conservation of momentum $\;m_{b} u_{b}+m u \;\; =-m_{b} v_{b}+m v^{'}$ $\Rightarrow \;\;\frac{10}{1000} v+0 \;\; =\;\frac{-10}{1000} \times 100+1 \sqrt{5 g l}$ $\Rightarrow \;\;\frac{v}{100} \;\; =-1+\sqrt{5 \times 10 \times \;\frac{5}{10}}$ $\Rightarrow \;\;\frac{v}{100} \;\; =-1+5$ $\Rightarrow \;v \;\; =4 \times 100=400 {ms}^{-1}$

Q27JEE Main 2021NAT4MAlternating Current

An AC circuit has an inductor and a resistor of resistance R in series, such that $\text{X}_{\text{L}}=3 \text{R}$ . Now, a capacitor is added in series such that $\text{X}_{\text{C}}=2 \text{R}$ . The ratio of new power factor with the old power factor of the circuit is $\sqrt{5}: x$ The valueof x is.

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📖 Explanation

Given, resistance of resistor $=R$ Inductance, $X_{L}=3 R$ Capacitance $X_{C}=2 R$ As we know that,Power factor, $\cos \varphi =\;\frac{R}{Z}$ where, $Z$ is impedance, $Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}$ For case I, an inductance connected in series with resistance $\cos \varphi _{1} \;\; =\;\frac{R}{Z_{1}}=\;{R}/{\sqrt{R^{2}+(X_{L}-0)^{2}}}$ $\;\; =\;{R}/{\sqrt{R^{2}+(3 R)^{2}}}=\;\frac{R}{R \sqrt{10}}=\;\frac{1}{\sqrt{10}} \;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$ For case II, A capacitor is also connected in series with resistance $\therefore \;\; \cos \varphi _{2} \;\; =\;\frac{R}{Z_{2}}$ $\Rightarrow \;\; \cos \varphi _{2} \;\; =\;{R}/{\sqrt{R^{2}+(3 R-2 R)^{2}}}$ $\;\; =\;{R}/{\sqrt{R^{2}+R^{2}}}=\;\frac{R}{R \sqrt{2}}=\;\frac{1}{\sqrt{2}} . . . \;\text{ (ii) }\;$ Now, ratio of Eqs. (ii) and (i), we get $\;\; \;\frac{\cos \varphi _{2}}{\cos \varphi _{1}} \;\; =\;{\;\frac{1}{\sqrt{2}}}/{\frac{1}{\sqrt{10}}}=\sqrt{\;\frac{10}{2}}=\sqrt{5}: 1$ $\therefore \;\; x \;\; =1$

Q28JEE Main 2021NAT4MCurrent Electricity

The ratio of the equivalent resistance of the network (shown in figure) between the points a and b when switch is open and switch is closed is $x: 8$ . The value of x is

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📖 Explanation

According to given circuit diagram When switch is open, then combination $R_{1}$ and $R_{2}$ will be in series and also combination $R_{3}$ and $R_{4}$ will be in series and these branches will be in parallel. $\therefore$ Equivalent resistance $(R_{\;\text{eq }\;})=\;\frac{3 R \cdot 3 R}{3 R+3 R}=\;\frac{9 R^{2}}{6 R}=\;\frac{3}{2} R$ When switch is closed, then combination $R_{1}$ and $R_{3}$ will be in parallel and also, combination $R_{2}$ and $R_{4}$ will be in parallel. After that, both will be in series. $\therefore$ Equivalent resistance, $({R^{'}}_{\text{eq}}) \;\; =\;\frac{R \cdot 2 R}{R+2 R}+\;\frac{R \cdot 2 R}{R+2 R}=\;\frac{2 R^{2}}{3 R}+\;\frac{2 R^{2}}{3 R}$ $\;\; =\;\frac{2 R}{3}+\;\frac{2 R}{3}=\;\frac{4 R}{3}$ Now, dividing Eq. (i) by Eq. (ii), we get ${c} \;{R_{\;\text{eq }\;}}/{R_{\;\text{eq }\;}{ }^{'}}=\;\frac{3 \frac{R}{2}}{4 \;\frac{R}{3}}=\;\frac{9}{8}$ $x=9$

Q29JEE Main 2021NAT4MElectromagnetic Waves

A plane electromagnetic wave with frequency of $30 {\text{MHz}}$ travels in free space. At particular point in space and time, electric field is $6 {\text{V}} / {\text{m}}$ . The magnetic field at this point will be $x \times 10^{-8} {\text{T}}$ . The value of x is.

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📖 Explanation

Given, frequency of electromagnetic wave $(f)$ $\;\; =30 {MHz}$ $\;\; =30 \times 10^{6} {Hz}$ Electric field, $E=6 {V} / {m}$ Magnetic field, ${B}=x \times 10^{-8} {T}$ Speed of light in air, $c=3 \times 10^{8} {ms}^{-1}$ We know that, $B=\;\frac{E}{C}$ $\Rightarrow B=\;\frac{6}{3 \times 10^{8}}$ $\Rightarrow B=2 \times 10^{-8} {T}$ $\therefore \;x=2$

Q30JEE Main 2021NAT4MWaves

A tuning fork is vibrating at $250 {\text{Hz}}$ . The length of the shortest closed organ pipe that will resonate with the tuning fork will be ........ cm. (Take, speed of sound in air as $340 {\text{ms}}^{-1}$ )

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📖 Explanation

For a closed organ pipe, the shortest length ( $L$ ) corresponds to the fundamental frequency, where the length is one-quarter of the wavelength ( $\lambda$ ).
Given frequency $f = 250 {\text{ Hz}}$ and speed of sound $v = 340 {\text{ m/s}}$ .
The relationship between speed, frequency, and wavelength is $v = f\lambda$ .
For the fundamental mode in a closed organ pipe:
$L = \frac{\lambda}{4}$
Substitute $\lambda = \frac{v}{f}$ into the equation for $L$ :
$L = \frac{v}{4f}$
Plugging in the values:
$L = \frac{340 {\text{ m/s}}}{4 \times 250 {\text{ Hz}}} = \frac{340}{1000} {\text{ m}} = 0.34 {\text{ m}}$
To convert meters to centimeters:
$L = 0.34 \times 100 {\text{ cm}} = 34 {\text{ cm}}$
[CORRECT_OPTION: N/A]

Chemistry30 questions

Q31JEE Main 2021MCQ4Mp Block Elements

Choose the correct statement from the following.

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📖 Explanation

The $\Delta\text{H}_{\text{f}}$ of alkali metal halides is shown through this graph. This graph shows that formation of metal halide is negative, i.e. energy is released during the formation of metal halides. (A) ${\text{table} \text{M}-\text{F};\text{M}-\text{Cl};\text{M}-\text{Br};\text{M}-\text{I} }↓$ On moving down the group, the size of halogens increase which results in the lengthening of bond, making the bond weaker. Therefore, $\Delta\text{H}_{\text{f}}$ is less negative on moving down the group as the stability of metal halide decreases down the group. (B) ${\text{table} \text{Li}-\text{X};\text{Na}-\text{X};\text{K}-\text{X};\text{Rb}-\text{X}}↓$ On moving down the group, the electropositivity of metal increases making the attraction between the oppositely charged ions even stronger, thus increasing the strength of bond. Increase in strength is more than the weakness produced due to lengthening of bond because of increase in size of metal ions down the group. This makes $\Delta\text{H}_{\text{f}}$ more negative on moving down the group. Option (a) is incorrect, due to reason produced in (B). Option (b) is incorrect. As CsI has $\text{Cs}^{⊕}$ and $\text{I}^{⊖}$ an constituent ions and both the ions are larger in size. The lattice enthalpy is not so high as shown in graph. But it has low solubility due to less hydration energy released when hydration of larger ions takes place. Option (c) is correct. As LiF has highest lattice enthalpy as shown in graph. This is due to the size of ions that is smallest in their respective groups. Smaller is the size, shorter is the bond length and stronger is the bond. The energy required is break the bond and release the constituents ions is more than the energy released in the hydration of ions. So, LiF is least soluble in water. Option (d) is incorrect. LiF has most negative enthalpy in all metal fluorides as shown in graph.

Q32JEE Main 2021MCQ4MCoordination Compounds

The addition of dilute ${\text{NaOH}}$ to ${\text{Cr}}^{3+}$ salt solution will give

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📖 Explanation

The addition of dil. ${\text{NaOH}}$ to ${\text{Cr}}^{3+}$ solution will produce ppt. of ${\text{Cr}}_{2} {\text{O}}_{3}({\text{H}}_{2} {\text{O}})_{n}$ . ${\text{Cr}}^{3+}+{\text{NaOH}} \to {{\text{Cr}}_{2} {\text{O}}_{3}({\text{H}}_{2} {\text{O}})_{n}}_{(\text{Precipitate})}$ Therefore, the option (b) is correct.

Q33JEE Main 2021MCQ4MAlcohols, Phenols and Ethers

Given below are two statements Statement I Ethyl pent-4-yn-oate on reaction with ${\text{CH}}_{3} {\text{MgBr}}$ gives a 3° alcohol. Statement II In this reaction, one mole of ethyl pent-4-yn-oate utilizes two moles of ${\text{CH}}_{3} {\text{MgBr}}$ . In the light of the above statements, choose the most appropriate answer from the options given below.

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📖 Explanation

Ethyl pent-4-yn-oate on reaction with ${\text{CH}}_{3} {\text{MgBr}}$ gives 3° alcohol. ∴ Statement I is true. Mechanism 3 molecules of ${\text{CH}}_{3} {\text{MgBr}}$ is used for this conversion. ∴ Statement Il is false.

Q34JEE Main 2021MCQ4MSome Basic Concepts of Chemistry

In stratosphere most of the ozone formation is assisted by

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📖 Explanation

In the stratosphere, the UV-light causes the splitting of ${\text{O}}_{2}$ molecule into oxygen atoms. These atoms react with oxygen present in the air to form ozone. $\text{O}_2(g){\to }^{\text{UV} \text{radiation}} \text{O}(g) {\to }^{{\text{O}}_{2}(g)+\text{UV} \text{radiation}}{\text{O}_3(g)}_{\text{Ozone}}$ Therefore, the option (c) is correct.

Q35JEE Main 2021MCQ4MAlcohols, Phenols and Ethers

The compound/s which will show significant intermolecular H-bonding is/are

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Q36JEE Main 2021MCQ4MAmines

Which one of the following chemicals is responsible for the production of HCl in the stomach leading to irritation and pain?

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Q37JEE Main 2021MCQ4MRedox Reactions

The oxide that gives ${\text{H}}_{2} {\text{O}}_{2}$ most readily on treatment with ${\text{H}}_{2} {\text{O}}$ is

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📖 Explanation

When sodium peroxide reacts with water, it is hydrolysed and forms sodium hydroxide and hydrogen peroxide. $\text{Na}_2\text{O}_2+2\text{H}_2\text{O}\to 2\text{NaOH}+\text{H}_2\text{O}_2$ The reaction is highly exothermic. Therefore, the option (b) is correct.

Q38JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Which one of the following reactions will not yield propionic acid?

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📖 Explanation

Option (d) is correct as shown in mechanism, it does not produce propanoic acid. When bromopropane react with ${\text{Mg}}, {\text{CO}}_{2}$ , dry ether and ${\text{H}}_{3} {\text{O}}^{+}$ , product formed is butanoic acid. $\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}{\to }^{\text{Mg},\text{CO}_2,\text{dry} \text{ether}}_{\text{H}_3\text{O}^{⊕}}{\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}}_{\text{Butanoic} \text{acid}}$ Mechanism

Q39JEE Main 2021MCQ4MIonic Equilibrium

The correct order of ionic radii for the ions, ${\text{P}}^{3-}, {\text{S}}^{2-}, {\text{Ca}}^{2+}, {\text{K}}^{+}, {\text{Cl}}^{-}$ is

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📖 Explanation

The electronic configuration of given ions are as follows: ${\text{K}}_{19}=2,8,8,1$ ${\text{K}}^{+}=2,8,8$ ${\text{Ca}}_{20}=2,8,8,2$ ${\text{Ca}}^{2+}=2,8,8$ $\text{P}_{15}=2,8,5$ $\text{P}^{3-}=2,8,8$ ${\text{S}}_{16}=2,8,6$ ${\text{S}}^{2-}=2,8,8$ ${\text{Cl}}_{17}=2,8,7$ ${\text{Cl}}^{-}=2,8,8$ As the effective nuclear charge increases from P to Ca, the atomic size decreases in the same order. Order of ionic radii ${\text{P}}^{3-}_{15} > {\text{S}}^{2-}_{16} > {\text{Cl}}^{-}_{17} > {\text{K}}^{+}_{19} > {\text{Ca}}^{2+}_{20}$ Therefore, option (a) is correct.

Q40JEE Main 2021MCQ4MOrganic Chemistry - Some Basic Principles

Which one of the following is the major product of the given reaction?

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📖 Explanation

The reaction of keto and cyanide group with Grignard reagent and further hydrolysis gives alcohol and keto group respectively. On further heating with ${\text{H}}_{2} {\text{SO}}_{4}$ at $-{\text{OH}}$ group is eliminated resulting in formation of alkene. Therefore, option(a) is correct.

Q41JEE Main 2021MCQ4MHaloalkanes and Haloarenes

The major product (A) formed in the reaction given below is

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📖 Explanation

On reaction of haloalkane with methoxide ion, alkene is formed. Mechanism ● ${\text{O}}^{⊖}{\text{CH}}_{3}$ will act as a base due to its small size and high electron density and therefore, abstracts proton to form double bond which is in conjugation with aromatic ring. ● The ${\text{O}}^{⊖}{\text{CH}}_{3}$ when acts as nucleophile undergoes nucleophilic substitution and replaces ${\text{Br}}^{⊖}$ to form ether, which is a minor product. So, option (b) is correct.

Q42JEE Main 2021MCQ4MCoordination Compounds

Which one of the following is used to remove most of plutonium from spent nuclear fuel?

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📖 Explanation

Dioxygen difluoride is used to remove the plutonium from spent nuclear fuel. $3\text{F}-\text{O}-\text{O}-\text{F}+{\text{Pu}}_{(\text{Fromi} \text{nuclear} \text{fuel})}\to \text{Pu}\text{F}_6$ $+3\text{O}_2↑$ Therefore, the option (b) is correct.

Q43JEE Main 2021MCQ4MSolutions

Lyophilic sols are more stable than lyophobic sols because

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Q44JEE Main 2021MCQ4MHaloalkanes and Haloarenes

The major product of the following reaction, if it occurs by ${\text{S}}_{{\text{N}}} 2$ mechanism is

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📖 Explanation

In ${\text{S}}_{{\text{N}}} 2$ reaction, no formation of carbocation takes place. The nucleophile here which is phenoxide attacks from the back side to the carbon bearing the leaving group (bromide ion). In ${\text{S}}_{{\text{N}}} 2$ mechanism Here, in the question So, here option (d) is correct.

Q45JEE Main 2021MCQ4MRedox Reactions

Potassium permanganate on heating at $513 {\text{K}}$ gives a product which is

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📖 Explanation

When ${\text{KMnO}}_{4}$ is heated at $513 {\text{K}}$ , it forms ${\text{K}}_{2} {\text{MnO}}_{4}$ and ${\text{MnO}}_{2}$ and ${\text{O}}_{2}$ . ${2\text{KMnO}_4}^{+7} {\to }^{\Delta}_{513 \text{K}}{\text{K}_2\text{MnO}_4}^{+6}_{(\text{Green})}+{\text{MnO}_2}^{+4}_{(\text{Black} \text{residue})}+\text{O}_2$ ${\text{Mn}}_{25}=1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6}, 4 s^{2}, 3 d^{5}$ ${\text{Mn}}^{+6}=1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6}, 4 s^{1}$ ; 1 unpaired electron, so paramagnetic. So, option (d) is correct.

Q46JEE Main 2021MCQ4MPractical Organic Chemistry

Which one of the following tests used for the identification of functional groups in organic compounds does not use copper reagent?

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📖 Explanation

Barfoed's test is used to detect monosaccharides. In this test, the aldehyde group of monosaccharides reduces ${\text{Cu}}({\text{II}})$ acetate to ${\text{Cu}}({\text{I}})$ oxide which result in the formation of brick red ppt. $\text{RCHO}+2\text{Cu}^{2+}+\text{H}_2\text{O} \to \text{RCOOH}+{\text{Cu}\text{O}}_{(\text{ppt}.)}↓+4\text{H}^{+}$ Benedict's test is used to test the presence of reducing sugar. Copper (II) sulphate is used in Barfoed's test. Biuret test is used to test the presence of peptide bond. The biuret reagent contains hydrated copper sulphate, sodium hydroxide and Rochelle salt (sodium-potassium tartarate). When the aqueous, solution of protein is treated with this reagent, the solution turns purple confirming the presence of amide bond. Chelated complex of ${\text{Cu}}^{2+}$ ions formed gives purple colour to the solution. So, correct answer is option (b) Seliwanoff's test which is used to distinguish between ketoses and aldoses. The Seliwanoff's reagent contains resorcinol and ${\text{HCl}}$ .

Q47JEE Main 2021MCQ4MBiomolecules

Hydrolysis of sucrose gives

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📖 Explanation

Hydrolysis of sucrose gives, Sucrose $+{\text{H}}_{2} {\text{O}} ⇌ \alpha-{\text{D}}-(+)-$ glucose $+\beta-{\text{D}}-(-)-$ fructose Therefore, option (d) is correct answer. 'D' represents the orientation of hydroxyl group at the chiral carbon that is farthest from the highest oxidised carbon (aldehyde group in case of glucose and keto group in fructose) when the - OH group is on the right side, 'L' is used when - ${\text{OH}}$ group is on the left side. $(+)$ and $(-)$ represents the direction of rotation of plane polarised light $(+)$ is clockwise rotation and $(-)$ is anticlockwise rotation.alpha-is used when - ${\text{OH}}$ group at anomeric carbon is below the ring in Haworth projection, beta is used when - ${\text{OH}}$ group at anomeric carbon is above the rings in Haworth projection.

Q48JEE Main 2021MCQ4MSalt Analysis

Match List-I with List-II. List-I (Name ofore/mineral) List-II (Chemical formula) A. Calamine 1. ZnS B. Malachite $2.{\text{FeCO}}_{3}$ C.Siderite $3.{\text{ZnCO}}_{3}$ D. Sphalerite $4.{\text{CuCO}}_{3} . {\text{Cu}}({\text{OH}})_{2}$ Choose the most appropriate answer from the options given below

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📖 Explanation

Ore and their chemical formula are given below (A) Calamine (3) ${\text{ZnCO}}_{3}$ (B) Malachite (4) ${\text{CuCO}}_{3} .{\text{Cu}}({\text{OH}})_{2}$ (C) Siderite (2) ${\text{FeCO}}_{3}$ (D) Sphalerite (1)ZnS Therefore, the correct match is A → 3, B → 4, C → 2, D → 1.

Q49JEE Main 2021MCQ4MSolid State

Which one of the following is formed (mainly) when red phosphorus is heated in a sealed tube at $803 {\text{K}}$ ?

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📖 Explanation

When red phosphorus is heated in a sealed tube at $803 {\text{K}}$ , alpha -black phosphorus is formed. Therefore, option (d) is correct.

Q50JEE Main 2021MCQ4MAmines

The correct structures of A and B formed in the following reactions are

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📖 Explanation

Since, only one equivalent of acetic anhydride is present and $-{\text{NH}}_{2}$ group is more nucleophilic than - ${\text{OH}}$ group due to lower electronegativity of nitrogen. So, it attacks at the carbonyl center of acetic anhydride forming the product B. Mechanism Therefore, the option (d) is correct.

Q51JEE Main 2021NAT4MChemical Kinetics

The first order rate constant for the decomposition of ${\text{CaCO}}_{3}$ at $700 {\text{K}}$ is $6.36 \times 10^{-3} {\text{s}}^{-1}$ and activation energy is $209 {\text{kJ}} {\text{mol}}^{-1}$ . Its rate constant (in ${\text{s}}^{-1}$ ) at $600 {\text{K}}$ is $x \times 10^{-6}$ . The value of x is ......... (Nearest integer) $[Given, $\text{R}=8.31 \text{JK}^{-1}\text{mol}^{-1},\log6.36× 10^{-3}$ $=-2.19,10^{-4.79}=1.62× 10^{-5}$ ]$

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📖 Explanation

According to Arrhenius theory, $k=\text{A} {e}^{-\text{E}_{a} / \text{RT}}$ (k is the rate constant, A is Arrhenius factor, $\text{E}_{a}=$ Activation energy, T is absolute temperature and R is gas constant.) $\ln k=\ln \text{A}-\frac{\text{E}_{a}}{\text{RT}}$ At temperature $\text{T}_{1}$ , $\ln \text{k}_{1}=\ln \text{A}-{\text{E}_{{a}}}/{\text{RT}_{1}}$ ...(i) At temperature $\text{T}_{2}$ , $\ln \text{k}_{2}=\ln \text{A}-\frac{\text{E}_{a}}{\text{RT}_{2}}$ ...(ii) Subtracting Eqs. (ii) from (i), $\ln \text{k}_{1}-\ln \text{k}_{2}=\frac{\text{E}_{a}}{\text{R}}(\frac{1}{\text{T}_{2}}-\frac{1}{\text{T}_{1}})$ $\ln \frac{\text{k}_{1}}{\text{k}_{2}}=\frac{\text{E}_{a}}{\text{R}}(\frac{1}{\text{T}_{2}}-\frac{1}{\text{T}_{1}})$ $2.303 \log \frac{\text{k}_{1}}{\text{k}_{2}}=\frac{\text{E}_{a}}{\text{R}}(\frac{1}{\text{T}_{2}}-\frac{1}{\text{T}_{1}})$ $\log \frac{\text{k}_{1}}{\text{k}_{2}}=\frac{\text{E}_{a}}{2.303 \text{R}}(\frac{1}{\text{T}_{2}}-\frac{1}{\text{T}_{1}})$ At, $\text{T}_{1}=700 {\text{K}}, \text{k}_{1}=6.36 \times 10^{-3} {\text{s}}^{-1}, \text{E}_{a}$ $=209 {\text{kJmol}}^{-1}$ , $\text{R}=8.314 {\text{JK}}^{-1} {\text{mol}}^{-1}$ At, $\text{T}_{2}=600 {\text{K}}, \text{k}_{2}=x \times 10^{-6} {\text{s}}^{-1}$ $\log \frac{6.36 \times 10^{-3}}{x \times 10^{-6}}=\frac{209}{2.303 \times 8.314}(\frac{1}{600}-\frac{1}{700})$ $=\frac{209}{2.303 \times 8.314}(\frac{100}{420000})$ x = 16.

Q52JEE Main 2021NAT4MCoordination Compounds

The number of optical isomers possible for $[{\text{Cr}}({\text{C}}_{2} {\text{O}}_{4})_{3}]^{3-}$ is ...

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📖 Explanation

The structure of $[{\text{Cr}}({\text{C}}_{2} {\text{O}}_{4})_{3}]^{3-}$ is Oxalate is bidentate symmetrical ligand. The complex does not possess any element of symmetry and are optically active. The two optical isomers are shown above. ∴ Answer is two (2).

Q53JEE Main 2021NAT4MGas Laws and Ideal Gases

Two flasks I and II shown below are connected by a valve of negligible volume. When the valve is opened, the final pressure of the system in bar is $x \times 10^{-2}$ . The value of x is ........... .(Integer answer) $[Assume, Ideal gas, 1 bar $=10^{5} {\text{Pa}}$ , molar mass of ${\text{N}}_{2}=28.0 {\text{g}} {\text{mol}}^{-1} ; \text{R}=8.31 {\text{J}} {\text{mol}}^{-1} {\text{K}}^{-1}$ ]$

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On mixing the temperature of gas becomes T. Internal energy lost by gas A is the internal energy gained by gas B. $\Delta\text{U}_{\text{A}}=\Delta\text{U}_{\text{B}}$ $\frac{\text{f}}{2} n_{\text{A}} \text{R} \Delta\text{T}_{\text{A}}=\frac{\text{f}}{2} n_{\text{B}} \text{R} \Delta\text{T}_{\text{B}}$ (As both gases are diatomic, f is same on both sides) $n_{\text{A}} \Delta\text{T}_{\text{A}}=n_{\text{B}} \Delta\text{T}_{\text{B}}$ $\frac{2.8}{28} \times (\text{T}_{1}-\text{T})=\frac{0.2}{28}(\text{T}-\text{T}_{2})$ $\frac{1}{10}(300-\text{T})=\frac{1}{140}(\text{T}-60)$ $\text{T}=284 {\text{K}}$ On mixing the temperature becomes $284 {\text{K}}$ and volume becomes $3 {\text{L}}$ and total number of moles becomes $(n_{\text{A}}+n_{\text{B}})=0.10$ moles. $\text{pV}=(n_{\text{A}}+n_{\text{B}}) \text{RT}$ $=\frac{0.10 \times 8.3 \times 284}{3}$ $=84.18 {\text{atm}}$ or $84.18 {\bar}$ This is final pressure of gas.

Q54JEE Main 2021NAT4MHydrocarbons

$100 {\text{g}}$ of propane is completely reacted with $1000 {\text{g}}$ of oxygen. The mole fraction of carbon dioxide in the resulting mixture is $x \times 10^{-2}$ . The value of x is ......... (Nearest integer) [Atomic weight : ${\text{H}}=1.008, {\text{C}}=12.00, {\text{O}}=16.00$ ]

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For the reaction, ${\text{CH}_3\text{CH}_2\text{CH}_3(g)}_{1}{+}_{:}{5\text{O}_2(g)}_{5}\to {3\text{CO}_2(g)}_{3}{+}_{:}{4\text{H}_2\text{O}(g)}_{4}$ 1 mole of propane reacts completely with 5 moles of oxygen to form 3 moles of carbon dioxide and 4 moles of steam. 44 g of propane = 1 mole of propane $100 {\text{g}}$ of propane $=\frac{1}{44} \times 100=2.27 {\text{mol}}$ $32 {\text{g}}$ of ${\text{O}}_{2}=1$ mole of ${\text{O}}_{2}$ $1000 {\text{g}}$ of ${\text{O}}_{2}=\frac{1}{32} \times 1000=31.25$ moles ∴ 2.27 moles of propane requires $5 \times 2.27=11.35$ moles of ${\text{O}}_{2}$ moles of ${\text{CO}}_{2}$ formed $=3 \times 2.7=6.681$ mol of ${\text{CO}}_{2}$ $\text{table} _{}_{\text{at} \text{t}=0}_{\text{at}=\text{t}}{\text{CH}_3\text{CH}_2\text{CH}_3}_{1}_{2.27 \text{mol}}_{0}\;\;{+}_{:}{5\text{O}_2}_{5}_{31.25 \text{mol}}_{31.25−(5\times 2.27)}_{19.90 \text{mol}}\to {3\text{CO}_2}_{3}_{0}_{3\times 2.27}_{6.81 \text{mol}}{+}_{:}{4\text{H}_2\text{O}}_{4}_{0}_{4\times 2.27}_{9.08 \text{mol}}$ When reaction is completed 19.90 moles of ${\text{O}}_{2}$ , 6.81 moles of ${\text{CO}}_{2}$ and 9.08 moles of steam are left in the flask. Mole fraction of ${\text{CO}}_{2}={{ \text{Moles} \text{of} } {\text{CO}}_{2}({\text{g}})}/{{ \text{Total} \text{number} \text{of} \text{moles} }}$ $=\frac{6.81}{19.90+6.81+9.08}=0.19$ $x\times 10^{-2}=0.19$ $x=19$

Q55JEE Main 2021NAT4MSolutions

$40 {\text{g}}$ of glucose (Molar mass $=180$ ) is mixed with $200 {\text{mL}}$ of water. The freezing point of solution is ........... K. (Nearest integer) $[Given, $\text{K}_{\text{f}}=1.86 {\text{K}} {\text{kg}} {\text{mol}}^{-1}$ , density of water $= 1.00 {\text{g}} {\text{cm}}^{-3}$ , freezing point of water $=273.15 {\text{K}}$ ]$

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40 g of glucose mixed with 200 mL of water 180 g of glucose = 1 moles of glucose 40 g of glucose mol = 0.22 mol $1 {\text{mL}}$ of water $=1 {\text{g}}$ of water $[{d}=1 {\text{g}} / {\text{cm}}^{2}, 1 {\text{mL}}=1 {\text{cm}}^{3}]$ $200 {\text{mL}}$ of water $=200 {\text{g}}$ of water $\Delta\text{T}_{\text{f}}=\text{K}_{\text{f}}\ \text{m}$ where, $\Delta\text{T}_{\text{f}}=$ depression in freezing point, $\text{K}_{\text{f}}=$ molal elevation constant $=1.86 {\text{Kkg}} {\text{mol}}^{-1}$ and m = molality of solution $\Delta\text{T}_{\text{f}}=\frac{1.86 \times 0.22}{200} \times 1000$ , $(\because 1000 {\text{g}}=1 {\text{kg}})$ $\Delta\text{T}_{\text{f}}=2 {\text{K}}$ ∴ $(\text{T}_{\text{f}}-\text{T}_{\text{f}}')=2$ $\begin{bmatrix}\text{T}_\text{f}=\text{freezing} \text{point} \text{of} \text{water} (273 {\text{K}}) \\ \text{T}_\text{f}'=\text{freezing} \text{point} \text{of} \text{solution}\end{bmatrix}$ $273 {\text{K}}-\text{T}_{\text{f}}'=2 {\text{K}}$ $\text{T}_{\text{f}}'=(273-2) {\text{K}}=271 {\text{K}}$

Q56JEE Main 2021NAT4MElectrochemistry

The resistance of a conductivity cell with cell constant $1.14 {\text{cm}}^{-1}$ , containing $0.001 \;{\text{M}} {\text{KCl}}$ at $298 {\text{K}}$ is $1500 \Omega$ . The molar conductivity of $0.001 \;{\text{M}} {\text{KCl}}$ solution at $298 {\text{K}}$ in ${\text{S}} {\text{cm}}^{2} {\text{mol}}^{-1}$ is ......... .(Integer answer)

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Given, $\text{C}{*}=1.14 {\text{cm}}^{-1}=1.14 \times 10^{3} {\text{m}}^{-1}$ R = 1500 Ω C = 0.001 M ${\text{G}}{*}=\text{R} \times \κ$ [where, R = resistance $\text{G}{*}=$ cell constant $\κ=$ conductivity $\text{C}=$ concentration] $\therefore \κ=\frac{\text{G}{*}}{\text{R}}={1.14 \times 10^{3} {\text{m}}^{-1}}/{1500}$ $\Lambda{{\text{m}}}=\frac{\κ}{\text{C}}$ [where , $\Lambda_m=$ molar conductivity] $=\frac{1.14 \times 10^{3}}{1500 \times 0.001}=760 {\text{S}} {\text{cm}}^{2} {\text{mol}}^{-1}$ .

Q57JEE Main 2021NAT4MSome Basic Concepts of Chemistry

The number of photons emitted by a monochromatic (single frequency) infrared range finder of power $1 {\text{mW}}$ and wavelength of $1000 {\text{nm}}$ , in $0.1$ second is $x \times 10^{13}$ . The value of x is ......... (Nearest integer) $(h=6.63 \times 10^{-34} {\text{Js}}, c=3.00 \times 10^{8} {\text{ms}}^{-1})$

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Power of the source $=1 {\text{mW}}=10^{-3} {\text{W}}$ Energy of one photon, $\text{E}=h v=\frac{h c}{\lambda}$ where, $h=6.6 \times 10^{-34} {\text{Js}}$ $\text{c}=3 \times 10^{8} {\text{m}} / {\text{s}}$ and lambda = 1000 nm $=1000 \times 10^{-9} {\text{m}}$ $(1 {\text{nm}}=10^{-9} {\text{m}})$ $=10^{-6} {\text{m}}$ $\text{E}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{10^{-6}} ≅ 20 \times 10^{-20} {\text{J}}$ Power = number of photons emitted in one second × energy of 1 photon. $10^{-3}=n \times 20 \times 10^{-20}$ $\Rightarrow n=\frac{10^{-3}}{20\times 10^{-20}}=0.5 \times 10^{16}$ Number of photons emitted in $1 {\text{s}}=0.5 \times 10^{16}$ Number of photons emitted in $0.1 {\text{s}}=0.5 \times 10^{16} \times 0.1=50 \times 10^{13}$ $x \times 10^{13}=50 \times 10^{13}$ x = 50

Q58JEE Main 2021NAT4MChemical Equilibrium

When $5.1 {\text{g}}$ of solid ${\text{NH}}_{4} {\text{HS}}$ is introduced into a two litre evacuated flask at $27° {\text{C}}, 20 \%$ of the solid decomposes into gaseous ammonia and hydrogen sulphide. The $\text{K}_{\text{p}}$ for the reaction at $27° {\text{C}}$ is $x \times 10^{-2}$ . The value of x is ........... (Integer answer) $[Given, $\text{R}=0.082 {\text{L}} {\text{atm}} {\text{K}}^{-1} {\text{mol}}^{-1}$ ]$

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$51 {\text{g}}$ of ${\text{NH}}_{4} {\text{HS}}=1 {\text{mol}}$ $5.1 {\text{g}}$ of ${\text{NH}}_{4} {\text{HS}}=\frac{1}{51} \times 5.1=0.1 {\text{mol}}$ ${\text{NH}}_{4} {\text{HS}}(s) ⇌ {\text{NH}}_{3}(g)+{\text{H}}_{2} {\text{S}}(g)$ $\text{table} \text{At} \text{t} =0{,},,,0.1;\text{At} \text{t}=\text{t}{,},,,0.1(1-\alpha),,,0.1\alpha,,,0.1\alpha$ It is given that dissociation is $20 \%$ from 100 moles. ∴ 20 moles get dissociated. $20 \%$ dissociation from $1 {\text{mol}} 20 / 100=0.2$ moles get dissociated. alpha = 0.2 \because $\text{K}_{\text{C}}={[{\text{NH}}_{3}][{\text{H}}_{2} {\text{S}}]}/{[{\text{NH}}_{4} {\text{HS}}]}={[{\text{NH}}_{3}][{\text{H}}_{2} {\text{S}}]}/{1}$ (Concentration of solid is assumed as 1) $[{\text{NH}}_{3}]={{ \text{Number} \text{of} \text{moles} \text{of} } {\text{NH}}_{3}(g)}/{{ \text{Volume} }({ \in } {\text{L}})}$ $=\frac{0.1 \alpha}{2}$ $[{\text{H}}_{2} {\text{S}}]={{ \text{Number} \text{of} \text{moles} \text{of} } {\text{H}}_{2} {\text{S}}}/{{ \text{Volume} (\in \text{L}) }}$ $=\frac{0.1 \alpha}{2}$ $\text{K}_{\text{C}}={0.1 \alpha}{2} \times \frac{0.1 \alpha}{2}=\frac{0.1 \times 0.2}{2} \times \frac{0.1 \times 0.2}{2}$ $=10^{-4}$ $\therefore \text{K}_{\text{p}}=\text{K}_{\text{C}}(\text{RT})^{\Delta\text{n}}$ [Δn =change in the number of gaseous moles = 2] $\text{K}_{\text{p}}=10^{-4} \times (0.08 \times 300)^{2}$ $\text{K}_{\text{p}}=10^{-4} \times (24)^{2}=0.06$ $x \times 10^{-2}=0.06$ $\Rightarrow x=\frac{0.06}{10^{-2}}=0.06 \times 10^{2}=6$ $x=6$ is the answer.

Q59JEE Main 2021NAT4MChemical Bonding and Molecular Structure

The number of species having non-pyramidal shape among the following is (i) ${\text{SO}}_{3}$ (ii) ${\text{NO}}_{3}^{-}$ (iii) ${\text{PCl}}_{3}$ (iv) ${\text{CO}}_{3}^{2-}$

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Q60JEE Main 2021NAT4MChemical Thermodynamics

Data given for the following reaction is as follows. ${\text{FeO}}(s)+{\text{C}}_{{(\text{graphite}) }} \to {\text{Fe}}(s)+{\text{CO}}(g)$ Substance $\Delta{\text{H}}°({\text{kJ}} { \text{mol}}^{-1})$ $\Delta{\text{S}}°({\text{J}} \;{ \text{mol}}^{-1}\text{K}^{-1})$ FeO(s) -266.3 57.49 $\text{C}_(\text{graphite})$ 0 5.74 Fe(s) 0 27.28 CO(g) -110.5 197.6 The minimum temperature in K at which the reaction becomes spontaneous is......... (Integer answer)

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${\text{FeO}}(s)+{\text{C}}_{{(\text{graphite}) }} \to {\text{Fe}}(s)+{\text{CO}}(g)$ $\Delta\text{H}°_{{\text{reaction} }=\Delta\text{H}°_{\text{f} { (\text{product}) }}-\Delta\text{H}°_{\text{f} { (\text{reactants}) }}$ $=[\Delta\text{H}°_{\text{f}({\text{Fe}})}+\Delta\text{H}°_{\text{f}({\text{CO}})}]-[\Delta\text{H}°_{\text{f}({\text{FeO}})}-\Delta\text{H}°_{\text{f}({\text{C}})}]$ $=[0+(-110.5)]-[-266.3-0]$ $=156 {\text{kJ}}^{-1} {\text{mol}}^{-1}$ $\Delta\text{S}°_{{\text{reaction} }}=\Delta\text{S}°_{{\text{product} }}-\Delta\text{S}°_{{\text{reactant} }}$ $=[\Delta\text{S}°_{({\text{Fe}})}+\Delta\text{S}°_{({\text{CO}})}]-[\Delta\text{S}°_{({\text{FeO}})}-\Delta\text{S}°_{({\text{O}})}]$ $=[27.28+197.6]-[57.49+5.79]$ $=161 {\text{JK}}^{-1} {\text{mol}}^{-1}$ According to Gibb's equation, $\Delta\text{G}°=\Delta\text{H}°-\text{T} \Delta\text{S}°$ The reaction becomes spontaneous when $\Delta\text{G}°$ is atleast zero or negative. $0=\Delta\text{H}°-\text{T} \Delta\text{S}°$ ${\text{T}} \Delta{\text{S}}°=\Delta\text{H}°$ $\Rightarrow \text{T}=\frac{\Delta\text{H}°}{\Delta\text{S}°}={156 {\text{kJ}} {\text{mol}}^{-1}}/{161 {\text{JK}}^{-1} {\text{mol}}^{-1}}$ $={156000 {\text{mol}}^{-1}}/{161 {\text{JK}}^{-1} {\text{mol}}^{-1}}=964 {\text{K}}$ The temperature at which reaction becomes spontaneous is 964 K.

Mathematics30 questions

Q61JEE Main 2021MCQ4MStraight Lines and Pair of Straight Lines

The angle between the straight lines, whose direction cosines are given by the equations $2 \text{I}+2 m-n=0$ and $m n+n \text{I}+\text{Im}=0$ , is

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Given, 2l + 2m − n = 0 ....(i) mn + nl + lm = 0 ...(ii) From Eq. (i), we get n = 2l + 2m ...(iii) Substituting, n = 2l + 2m in Eq. (ii), we have m(2l + 2m) + l(2l + 2m) + lm = 0 $\Rightarrow 2 \text{Im}+2m^{2}+2 \text{I}^{2}+2 \text{Im}+\text{Im}=0$ $\Rightarrow 2 \text{I}^{2}+4 \text{Im}+\text{Im}+2 m^{2}=0$ ⇒ 2l(l + 2m) + m(l + 2m) = 0 ⇒ (2l + m) (l + 2m) = 0 When 2l = − m From Eq (iii), n = m $\Rightarrow \frac{2 \text{I}}{-1}=\frac{m}{1}=\frac{n}{1}$ $\Rightarrow \frac{\text{I}}{-1 / 2}=\frac{m}{1}=\frac{n}{1}$ or $\frac{\text{I}}{1}=\frac{m}{-2}=\frac{n}{-2}$ \Rightarrow (l, m, n) = (1, − 2, − 2) When l = − 2m From Eq. (iii), n = − 2m ⇒ l = − 2m = n $\Rightarrow \frac{\text{I}}{-2}=\frac{m}{1}=\frac{n}{-2}$ \Rightarrow (l, m, n) = (− 2,1, − 2) ∴ Angles between straight lines $\cos \theta=\frac{(\hat{i}-2 \hat{j}-2 \hat{k})(-2 \hat{i}+\hat{j}-2 \hat{k})}{|\hat{i}-2 \hat{j}-2 \hat{k}|-2 \hat{i}+\hat{j}-2 \hat{k} |}$ $\cos \theta=\frac{-2-2+4}{9}=0$ $\Rightarrow \theta=\frac{\pi}{2}$

Q62JEE Main 2021MCQ4MMatrices and Determinants

Let $\text{A}=(\text{table} [x+1],[x+2],[x+3];[x],[x+3],[x+3];[x],[x+2],[x+4])$ , where [t] denotes the greatest integer less than or equal to t. If det(A) = 192, then the set of values of $x$ is the interval

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Given, $\text{A}=(\text{table} [x+1],[x+2],[x+3];[x],[x+3],[x+3];[x],[x+2],[x+4])$ $\text{A}=(\text{table} [x]+1,[x]+2,[x]+3;[x],[x]+3,[x]+3;[x],[x]+2,[x]+4)$ (\because [x + n] = n + [x], n in I) Applying $\text{R}_1 \to \text{R}_1 − \text{R}_3, \text{R}_2 \to \text{R}_2 − \text{R}_3$ $\text{A}=\begin{bmatrix}1 & 0 & -1 \\ 0 & 1 & -1 \\ [x\end{bmatrix},[x]+2,[x]+4]$ det(A) = 1([x] + 4 + [x] + 2) − 1(−[x]) = 3[x] + 6 ∵ det(A) = 192 ⇒ 3[x] + 6 = 192 ⇒ [x] = 62 ⇒ 62 ≤ x < 6 ⇒ x in [62,63)

Q63JEE Main 2021MCQ4MInverse Trigonometric Functions

Let M and m respectively be the maximum and minimum values of the function $\text{f}(x)=\tan ^{-1}(\sin x+\cos x)$ in $[0, \frac{\pi}{2}]$ , then the value of $\tan (\text{M}-\text{m})$ is equal to

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We have, $\text{f}(x)=\tan ^{-1}(\sin x+\cos x)$ \because $x \in[0, \frac{\pi}{2}]$ $\Rightarrow 1 \le \sin x+\cos x \le \sqrt{2}$ $[\because -\sqrt{\text{A}^{2}+\text{B}^{2}} \le \text{A} \sin x+\text{B} \cos x \le \sqrt{\text{A}^{2}+\text{B}^{2}}]$ $\Rightarrow \tan ^{-1}(1) \le \tan ^{-1}(\sin x+\cos x) \le \tan ^{-1}(\sqrt{2})$ $\therefore m=\tan ^{-1}(1)$ and $\text{M}=\tan ^{-1}(\sqrt{2})$ $\therefore \text{M}-m=\tan ^{-1} \sqrt{2}-\tan ^{-1}(\sqrt{1})$ $=\tan ^{-1}(\frac{\sqrt{2}-1}{1+\sqrt{2}})=\tan ^{-1}({3}-2 \sqrt{2})$ $\Rightarrow \tan (\text{M}-\text{m})=3-2 \sqrt{2}$

Q64JEE Main 2021MCQ4MProbability

Each of the persons A and B independently tosses three fair coins. The probability that both of them get the same number of heads is

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Let X = Number of heads $\text{P}(\text{X}=r)={ }^{n} \text{C}_{r} p^{r} q^{n-r}$ $\Rightarrow { }^{3} \text{C}_{x}(\frac{1}{2})(\frac{1}{2})$ X = x 0 1 2 3 P(x) $1/8$ $3/8$ $3/8$ $1/8$ $\text{Required} \text{probability} =\text{P}_{\text{A}}(0) \times \text{P}_{\text{B}}(0)+\text{P}_{\text{A}}(1) \times \text{P}_{\text{B}}(1)+\text{P}_{\text{A}}(2) \times \text{P}_{\text{B}}(2)+\text{P}_{\text{A}}(3) \times \text{P}_{\text{B}}(3)$ $=\frac{1}{8} \times \frac{1}{8}+\frac{3}{8} \times \frac{3}{8}+\frac{3}{8} \times \frac{3}{8}+\frac{1}{8} \times \frac{1}{8}=\frac{5}{16}$

Q65JEE Main 2021MCQ4MDifferential Equations

A differential equation representing the family of parabolas with axis parallel to Y-axis and whose length of latus rectum is the distance of the point $(2,-3)$ from the line $3 x+4 y=5$ , is given by

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Let (h, k) be the vertex of parabola. Then, equation of parabola parallel to Y-axis is $(x-h)^{2}=4 a(y-k)$ ...(i) Also, Length of latusrectum = Distance of point (2, − 3) from the line 3x + 4y = 5 $\Rightarrow 4 a={|6-12-5|}/{\sqrt{3^{2}+4^{2}}}$ $\Rightarrow 4 a=\frac{11}{5}$ ∴ From Eq. (i), $(x-h)^{2}=\frac{11}{5}(y-k)$ Differentiating w.r.t. x, we get $2(x-h)=\frac{11}{5} \frac{d y}{d x}$ Again, differentiating w.r.t. x $2=\frac{11}{5} \frac{d^{2} y}{d x^{2}}$ $\Rightarrow 11 \frac{d^{2} y}{d x^{2}}=10$

Q66JEE Main 2021MCQ4MParabola

If two tangents drawn from a point P to the parabola $y^{2}=16(x-3)$ are at right angles, then the locus of point P is

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Q67JEE Main 2021MCQ4MThree Dimensional Geometry

The equation of the plane passing through the line of intersection of the planes ${r} .(\hat{i}+\hat{j}+\hat{k})=1$ and ${r} .(2 \hat{i}+3 \hat{j}-\hat{k})+4=0$ and parallel to the X-axis is

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Given, equation of planes ${r} .(\hat{i}+\hat{j}+\hat{k})=1$ ...(i) ${r} .(2 \hat{i}+3 \hat{j}-\hat{k})+4=0$ ...(ii) Equation of plane passing through the intersection of the planes Eqs. (i) and (ii) is given by (x + y + z − 1) + lambda(2x + 3y - z + 4) = 0 or (1 + 2lambda)x + (1 + 3lambda)y + (1 − lambda)z + (− 1 + 4lambda) = 0 ...(iii) Plane (iii) in parallel to X-axis 1 + 2lambda = 0 [Coefficient of x = 0] $\Rightarrow \lambda=\frac{-1}{2}$ ∴ From Eq. (iii) becomes y − 3z + 6 = 0 or ${r} .(\hat{j}-3 \hat{k})+6=0$

Q68JEE Main 2021MCQ4MDifferential Equations

If the solution curve of the differential equation $(2 x-10 y^{3}) d y+y d x=0$ , passes through the points $(0,1)$ and $(2, \beta)$ , then beta is a root of the equation

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Given, differential equation $(2 x-10 y^{3}) d y+y d x=0$ $\Rightarrow \frac{d x}{d y}+\frac{2 x}{y}=10 y^{2}$ ...(i) This is Linear differential equation Integrating factor ${\text{IF}}={e}^{2 \int\frac{1}{y} d y}={y}^{2}$ Solution of differential Eq. (i), $x .y^{2}=\int10 y^{2} .y^{2} d y+\text{C}$ $\Rightarrow x y^{2}=2 y^{5}+\text{C}$ ...(ii) Solution Eq. (ii) passes through (0, 1) $\Rightarrow 0 .1^{2}=2 1^{5}+\text{C}$ ⇒ C = − 2 ∴ Solution of Eq. (i) is $x y^{2}=2 y^{5}-2$ Now, this equation passes through (2, beta). $\therefore 2 .\beta^{2}=2 \beta^{5}-2$ $\Rightarrow \beta^{5}-\beta^{2}-1=0$ ⇒ beta is root of the equation $y^{5}-y^{2}-1=0$

Q69JEE Main 2021MCQ4MArea Under The Curves

Let $\text{A}(a, 0), \text{B}(b, 2 b+1)$ and $\text{C}(0, b), b \ne 0,|b| \ne 1$ , be points such that the area of $\Delta\text{ABC}$ is $1 {\text{sq}}$ . unit, then the sum of all possible values of a is

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A(a, 0), B(b, 2b + 1), C(0, b) Area of $\Delta\text{ABC}=\frac{1}{2}{\begin{vmatrix}a & 0 & 1 \\ b & 2b+1 & 1 \\ 0 & b & 1\end{vmatrix}}=\pm 1$ $\Rightarrow \frac{1}{2}\$ [a(b+1)+b^{2}]$=\pm 1\Rightarrow a=\frac{2-b^{2}}{b+1} $or$ \frac{-2-b^{2}}{b+1} $Sum of all possible values of$ a=\frac{-2 b^{2}}{b+1}$

Q70JEE Main 2021MCQ4MSets and Relations

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Given, system of equations x + y + z = 4 3x + 2y + 5z = 3 9x + 4y + (28 + |lambda|)z = |lambda| $\Delta = \begin{vmatrix}1 & 1 & 1 \\ 3 & 2 & 5 \\ 9 & 4 & 28+{\end{vmatrix}\lambda|}|$ = 1(56 + 2|lambda| − 20) − 1(84 + 3|lambda| − 45) + 1(− 6) = − (|lambda| + 9) If \Delta ≠ 0 i.e |lambda| + 9 ≠ 0, then system of equation has unique solution. If $|\lambda|+9=0$ , then $\Delta_{1}=\Delta_{2}=\Delta_{3}=0$ , the system of equation has infinite solution. ⇒ lambda in R

Q71JEE Main 2021MCQ4MSets and Relations

The set of all values of $k > -1$ , for which the equation $(3 x^{2}+4 x+3)^{2}-(k+1)(3 x^{2}+4 x+3)$ $(3 x^{2}+4 x+2)+k(3 x^{2}+4 x+2)^{2}=0$ has real roots, is

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Given, $(3 x^{2}+4 x+3)^{2}-(k+1)(3 x^{2}+4 x+3)$ $(3 x^{2}+4 x+2)+k(3 x^{2}+4 x+2)^{2}=0$ Let $y=3 x^{2}+4 x+2$ Then, given equation becomes $(y+1)^{2}-(k+1) y(y+1)+k y^{2}=0$ $\Rightarrow y^{2}+2 y+1-k y^{2}-k y-y^{2}-y+k y^{2}=0$ $\Rightarrow y+1-k y=0$ $\Rightarrow y (1 − k) = − 1$ $\Rightarrow y=\frac{1}{k-1}$ $\Rightarrow 3 x^{2}+4 x+2-\frac{1}{k-1}=0$ For real roots, D ≥ 0 $\Rightarrow 16-4 .3 .(2-\frac{1}{k-1}) \ge 0$ $\Rightarrow -8+\frac{12}{k-1} \ge 0$ $\Rightarrow \frac{3}{k-1} \ge 2$ $\Rightarrow \frac{3-2 k+2}{k-1} \ge 0$ $\Rightarrow \frac{2 k-5}{k-1} \le 0$ $\Rightarrow k \in(1, \frac{5}{2}]$ [∵ k ≠ 1]

Q72JEE Main 2021MCQ4MApplication of Derivatives

A box open from top is made from a rectangular sheet of dimension a × b by cutting squares each of side x from each of the four corners and folding up the flaps. If the volume of the box is maximum, then x is equal to

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Length of box = a − 2x Breadth of box = b − 2x Height of box = x Volume of box, V = (a − 2x) (b − 2x)x $\Rightarrow \text{V}=4 x^{3}-20 x^{2}-2 b x^{2}+a b x$ Differentiating V w.r.t. x, $\text{V}'_{(x)}=12 x^{2}-4(a+b) x+a b$ Critical Point, $\text{V}'_{(x)} = 0$ $\Rightarrow 12 x^{2}-4(a+b) x+a b=0$ $\Rightarrow x={4(a+b) \\pm \sqrt{16(a+b)^{2}-4 .12 .a b}}/{2(12)}$ $x={(a+b) \\pm \sqrt{a^{2}+b^{2}-a b}}/{6}$ $\text{V}′ ′(x) =24x − 4(a + b)$ For $x={(a+b)-\sqrt{a^{2}+b^{2}-a b}}/{6}$ , V′ ′(x) < 0 Hence, for maximum volume $x={a+b-\sqrt{a^{2}+b^{2}-a b}}/{6}$

Q73JEE Main 2021MCQ4MSets and Relations

The Boolean expression $(p \\wedge q) \Rightarrow((r \\wedge q) \\wedge p)$ is equivalent to

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Q74JEE Main 2021MCQ4MSets and Relations

Let Z be the set of all integers, $\text{A}=\{(x, y) \in\text{Z} \times \text{Z}:(x-2)^{2}+y^{2} \\le 4\}$ , $\text{B}=\{(x, y) \in\text{Z} \times \text{Z}: x^{2}+y^{2} \\le 4\}$ and $\text{C}=\{(x, y) \in\text{Z} \times \text{Z}:(x-2)^{2}+(y-2)^{2} \\le 4\}$ If the total number of relation from $\text{A} \⋂\text{B}$ to $\text{A}\⋂\text{B}$ is $2^{p}$ , then the value of p is

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📖 Explanation

$\text{A}=\{(x, y) \in\text{Z} \times \text{Z}:(x-2)^{2}+y^{2} \\le 4\}$ , $\text{B}=\{(x, y) \in\text{Z} \times \text{Z}: x^{2}+y^{2} \\le 4\}$ $\text{C}=\{(x, y) \in\text{Z} \times \text{Z}:(x-2)^{2}+(y-2)^{2} \\le 4\}$ ∴ A cap B = {(0, 0), (1, 0), (2, 0), (1,1), (1, −1)} n(A cap B) = 5 Similarly, n(A cap C) = 5 Number of relation from (A cap B) to (A C) $=2^{5\times 5}=2^\text{p}$ ⇒ p = 25

Q75JEE Main 2021MCQ4MArea Under The Curves

The area of the region bounded by the parabola $(y-2)^{2}=(x-1)$ , the tangent to it at the point whose ordinate is 3 and the X-axis is

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📖 Explanation

Given parabola $(y-2)^{2}=(x-1)$ Since, Ordinate = y = 3 Then, x = 2 Point on parabola (2, 3) Differentiating Eq. (i) w.r.t. x, we get $2(y-2) \frac{d y}{d x}=1$ $\frac{d y}{d x}=\frac{1}{2(y-2)}$ At (2, 3) $\frac{d y}{d x}=\frac{1}{2}$ Equation of tangent at (2, 3) $y-3=\frac{1}{2}(x-2)$ or x − 2y + 4 = 0 Intersection point of parabola on X-axis is y = 0, x = 5 i.e (5, 0) Intersection point of tangent and X-axis y = 0, x = − 4 i.e (− 4, 0) Area of shaded region $=\int_{0}^{3}\$ [(y-2)^{2}+1-(2 y-4)]$ d y=\int_{0}^{3}(y^{2}-6 y+9) d y=(\frac{y^{3}}{3}-3 y^{2}+9 y)_{0}^{3}=9$ sq. unit.

Q76JEE Main 2021MCQ4MInverse Trigonometric Functions

If $y(x)=\cot ^{-1}(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}})$ , $x \in(\frac{\pi}{2}, \pi)$ , then $\frac{d y}{d x}$ at $x=\frac{5 \pi}{6}$ is

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📖 Explanation

$y(x)=\cot ^{-1}(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}})$ , $x \in(\frac{\pi}{2}, \pi)$ $=\cot ^{-1} \frac{|\cos \frac{x}{2}+\sin \frac{x}{2}|+|\cos {x}{2}-\sin \frac{x}{2}|}{|\cos \frac{x}{2}+\sin \frac{x}{2}|-|\cos \frac{x}{2}-\sin \frac{x}{2}|}$ $=\cot ^{-1}(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}-\sin \frac{x}{2}+\cos \frac{x}{2}})$ $[\because \frac{x}{2} \in(\frac{\pi}{4}, \frac{\pi}{2})]$ $=\cot ^{-1}(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}})=\cot ^{-1}(\tan \frac{x}{2})=\cot ^{-1}(\cot (\frac{\pi}{2}-\frac{x}{2}))$ $\Rightarrow y(x)=\frac{\pi}{2}-\frac{x}{2}$ $\therefore y'(x)=-1/2$

Q77JEE Main 2021MCQ4MStraight Lines and Pair of Straight Lines

Two poles, AB of length a m and CD of length $a+b(b \ne a)$ metres are erected at the same horizontal level with bases at B and D. If $\text{BD}=x$ and $\tan \angle \text{ACB} !=\frac{1}{2}$ , then

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📖 Explanation

In \Delta ACE $\tan (\theta+\alpha)=\frac{x}{b}$ In \Delta ACD $\tan \alpha=\frac{x}{a+b}$ and $\tan \theta=\frac{1}{2}$ Then, $\tan (\theta+\alpha)=\frac{\tan \theta+\tan \alpha}{1-\tan \theta.\tan \alpha}$ $\Rightarrow \frac{\frac{1}{2}+\frac{x}{a+b}}{1-\frac{1}{2} .\frac{x}{a+b}}=\frac{x}{b}$ $\Rightarrow \frac{a+b+2 x}{2(a+b)-x}=\frac{x}{b}$ $\Rightarrow a b+b^{2}+2 b x=2 a x+2 b x-x^{2}$ $\Rightarrow ^{2}-2 a x+a b+b^{2}=0$ $\Rightarrow x^{2}-2 a x+b(a+b)=0$

Q78JEE Main 2021MCQ4MSequences and Series

If $0 < x< 1$ and $y=\frac{1}{2} x^{2}+\frac{2}{3} x^{3}+\frac{3}{4} x^{4+...}$ then the value of $e^{1+y}$ at $x=\frac{1}{2}$ is

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📖 Explanation

$y=\frac{1}{2} x^{2}+\frac{2}{3} x^{3}+\frac{3}{4} x^{4}+...$ $\Rightarrow y=(1-\frac{1}{2}) x^{2}+(1-\frac{1}{3}) x^{3}+(1-\frac{1}{4}) x^{4}+...$ $=((x^{2}+x^{3}+x^{4}+...)-(\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4}+...)$ $=\frac{x^{2}}{1-x}+x-(x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4} ...)$ $\therefore y=\frac{x}{1-x}+\ln (1-x)$ Put $x=\frac{1}{2}$ , we get y = 1 − ln 2 Then, $e^{1+y}=e^{1+1-\ln 2}=e^{2-\ln 2}$ $=e^{2} .e^{\ln 2^{-1}}$ $=\frac{1}{2} e^{2}$

Q79JEE Main 2021MCQ4MDefinite Integrals

The value of the integral $\int ^{1}_{0} \frac{\sqrt{x} d x}{(1+x)(1+3 x)(3+x)}$ is

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📖 Explanation

$\int ^{1}_{0} \frac{\sqrt{x} d x}{(1+x)(1+3 x)(3+x)}$ Put $\sqrt{x}=t$ $\Rightarrow x=t^{2}$ or dx = 2t dt $\therefore \text{I}=\int_{0}^{1} \frac{2 t^{2} d t}{(t^{2}+1)(3 t^{2}+1)(t^{2}+3)}$ $=\int_{0}^{1} \frac{(3 t^{2}+1)-(t^{2}+1)}{(t^{2}+1)(3 t^{2}+1)(t^{2}+3)} d t$ $=\int_{0}^{1}\$ [\frac{1}{(t^{2}+3)(t^{2}+1)}-\frac{1}{(t^{2}+3)(3 t^{2}+1)}]$ d t=\int_{0}^{1}$[\frac{1}{2(t^{2}+1)}-\frac{1}{2(t^{2}+3)}-\frac{3}{8(3 t^{2}+1)}+\frac{1}{8(t^{2}+3)}]$ d t=\int_{0}^{1} \frac{d t}{2(t^{2}+1)}=\int_{0}^{1} \frac{3}{8} \frac{d t}{(3 t^{2}+1)}-\int_{0}^{1} \frac{3}{8} \frac{d t}{(t^{2}+3)}=(\frac{1}{2} \tan ^{-1} t){0}^{1}-(\frac{3}{8} \frac{\sqrt{3}}{3} \tan ^{-1} \sqrt{3} t){0}^{1}-(\frac{3}{8 \sqrt{3}} \tan ^{-1} \frac{t}{\sqrt{3}})_{0}^{1}=\frac{\pi}{8}-\frac{\sqrt{3}}{8} \frac{\pi}{3}-\frac{\sqrt{3}}{8} \frac{\pi}{6}=\frac{\pi}{8}-\frac{\sqrt{3} \pi}{16}=\frac{\pi}{8}(1-\frac{\sqrt{3}}{2})$

Q80JEE Main 2021MCQ4MLimits, Continuity and Differentiability

If ${\lim}_{x\to \infty }(\sqrt{x^{2}-x+1}-a x)=b$ , then the ordered pair (a, b) is

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📖 Explanation

Given, ${\lim}_{x\to \infty }(\sqrt{x^{2}-x+1}-a x)=b$ $\Rightarrow {\lim}_{x\to \infty }(\sqrt{x^{2}-x+1}-a x) {(\sqrt{x^{2}-x+1})+a x}/{(\sqrt{x^{2}-x+1}+a x)}$ $=b$ $\Rightarrow {\lim}_{x\to \infty } {x^{2}-x+1-a^{2} x^{2}}/{\sqrt{x^{2}-x+1}+a x}=b$ Limit exists only if $a^{2}=1$ ∴ a = \pm 1 $\Rightarrow {\lim}_{x\to \infty } {-x+1}/{\sqrt{x^{2}-x+1}+a x}=b$ $\Rightarrow {\lim}_{x\to \infty } {-1+\frac{1}{x}}/{\sqrt{1-\frac{1}{x}+\frac{1}{x^{2}}}+a}=b$ $\Rightarrow \frac{-1}{1+a}=b$ But a ≠ −1 ⇒ a = 1 $\therefore b=-\frac{1}{2}$ $(a, b)=(1,-\frac{1}{2})$

Q81JEE Main 2021NAT4MTrigonometric Ratios and Identities

Section B : Numerical Type Questions Let S be the sum of all solutions (in radians) of the equation $\sin ^{4} \theta+\cos ^{4} \theta-\sin \theta\cos \theta=0$ in $[0,4 \pi]$ . Then, $\frac{85}{\pi}$ is equal to

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📖 Explanation

$\sin ^{4} \theta+\cos ^{4} \theta-\sin \theta\cos \theta=0$ in $[0,4 \pi]$ $\Rightarrow 1-2 \sin ^{2} \theta \cos ^{2} \theta-\sin \theta\cos \theta=0$ $\Rightarrow 2-\sin ^{2} 2 \theta-\sin 2 \theta=0$ $\Rightarrow \sin ^{2} 2 \theta+\sin 2 \theta-2=0$ $\Rightarrow (\sin2\theta + 2) (\sin2\theta − 1) = 0$ $\Rightarrow \sin2\theta = 1, 2\theta \in[0, 8 \pi]$ $\theta=\frac{\pi}{4}, \frac{5 \pi}{4}, \frac{9 \pi}{4}, \frac{13 \pi}{4}$ Sum of solutions $\text{S}=\frac{28 \pi}{4}$ Then, $\frac{8 \text{S}}{\pi}=\frac{8}{\pi} \times \frac{28 \pi}{4}=56$

Q82JEE Main 2021NAT4MThree Dimensional Geometry

Let S be the mirror image of the point Q $(1,3,4)$ with respect to the plane $2 x-y+z+3=0$ and let $\text{R}(3,5, \gamma)$ be a point of this plane. Then the square of the length of the line segment SR is

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📖 Explanation

Let point S(a, b, c) Then, $\frac{a-1}{2}=\frac{b-3}{-1}=\frac{c-4}{1}=\frac{-2(2-3+4+3)}{4+1+1}$ $=-2$ \Rightarrow a = − 3, b = 5, c = 2 ∴ S (− 3, 5, 2) and point R(3, 5, gamma) lies on the plane 2x − y + z + 3 = 0 ⇒ 6 − 5 + gamma + 3 = 0 ⇒ gamma = − 4 ∴ R (3, 5, − 4) Now, $\text{SR}^{2}=6^{2}+0^{2}+(6)^{2}$ $= 36 + 0 + 36 = 72$

Q83JEE Main 2021NAT4MProbability

The probability distribution of random variable X is given by X 1 2 3 4 5 P(X) K 2K 2K 3K K Let $p=\text{P}(1<\text{X}<4 |\text{X}<3)$ . If $5 p=\lambda\text{K}$ then, lambdaequal to

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📖 Explanation

$\Σ\text{P}_(x) = 1$ \Rightarrow K + 2K + 2K + 3K + K = 1 $\Rightarrow \text{K}=\frac{1}{9}$ $\text{p}=\text{P}(1<\text{X}<4 |\text{X}<3)=\frac{\text{P}((1<\text{X}<4) \cap(\text{X}<3))}{\text{p}(\text{X}<3)}$ $=\frac{\text{P}(\text{X}=2)}{\text{P}(\text{X}=1)+\text{P}(\text{X}=2)}$ $=\frac{2 \text{K}}{\text{K}+2 \text{K}}=\frac{2}{3}$ \because 5p = lambdaK $5 \times \frac{2}{3}=\lambda\times \frac{1}{9}$ ⇒ lambda = 30

Q84JEE Main 2021NAT4MComplex Numbers

Let $\text{Z}_{1}$ and $\text{Z}_{2}$ be two complex numbers such that $\arg (\text{Z}_{1}-\text{Z}_{2})=\frac{\pi}{4}$ and $\text{Z}_{1}, \text{Z}_{2}$ satisfy the equation $|\text{Z}-3|={\text{Re}}(\text{Z})$ . Then, the imaginary part of $\text{Z}_{1}+\text{Z}_{2}$ is equal to

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📖 Explanation

Let $\text{Z}_{1}=a_{1}+i b_{1}, \text{Z}_{2}=a_{2}+i b_{2}$ $\text{Z}_{1}-\text{Z}_{2}=(a_{1}-a_{2})+i(b_{1}-b_{2})$ $\arg (\text{Z}_{1}-\text{Z}_{2})=\frac{\pi}{4}$ $\Rightarrow \tan ^{-1}(\frac{b_{1}-b_{2}}{a_{1}-a_{2}})=\frac{\pi}{4}$ ⇒ $b_{1}-b_{2}=a_{1}-a_{2}$ ...(i) Also, $|\text{Z}_{1}-3|={\text{Re}}(\text{Z}_{1})$ $\Rightarrow (a_{1}-3)^{2}+b_{1}^{2}=a_{1}^{2}$ and $|\text{Z}_{2}-3|={\text{Re}}(\text{Z}_{2})$ $\Rightarrow (a_{2}-3)^{2}+b_{2}^{2}=a_{2}^{2}$ $\Rightarrow (a_{1}-3)^{2}-(a_{2}-3)^{2}+b_{1}^{2}-b_{2}^{2}=a_{1}^{2}- a_{2}^{2}$ $\Rightarrow (a_{1}-a_{2})(a_{1}+a_{2}-6)+(b_{1}-b_{2})(b_{1}+b_{2}) =(a_{1}-a_{2})(a_{1}+a_{2})$ $\Rightarrow a_{1}+a_{2}-6+b_{1}+b_{2}=a_{1}+a_{2}$ [using Eq. (i), $\Rightarrow b_{1}+b_{2}=6$ $\Rightarrow {\text{Im}}(\text{Z}_{1}+\text{Z}_{2})=6$

Q85JEE Main 2021NAT4MSets and Relations

Let $\text{S}=\{1,2,3,4,5,6,9\}$ . Then, the number of elements in the set $\text{T}=\{\text{A} \subseteq \text{S}: \text{A} \\ne \phi$ and the sum of all the elements of A is not a multiple of 3} is

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📖 Explanation

S = {1, 2, 3, 4, 5, 6, 9} 3n Type numbers 3, 6, 9 3n −1 Type numbers 2, 5 3n − 2 Type numbers 1, 4 Let $\text{N}_\text{P} =$ Number of Subset of S containing p element which are not divisible by 3. For P = 1 ${ }^{2} \text{C}_{1}+{ }^{2} \text{C}_{1}=4$ For P = 2 ${ }^{3} \text{C}_{1}{ }^{2} \text{C}_{1}+{ }^{3} \text{C}_{1}{ }^{2} \text{C}_{1}+{ }^{2} \text{C}_{2}+{ }^{2} \text{C}_{2}=14$ For P = 3 ${ }^{3} \text{C}_{1}({ }^{2} \text{C}_{2}+{ }^{2} \text{C}_{2})+{ }^{3} \text{C}_{2}({ }^{2} \text{C}_{1}+{ }^{2} \text{C}_{1})+{ }^{2} \text{C}_{2}{ }^{2} \text{C}_{1}+{ }^{2} \text{C}_{1}{ }^{2} \text{C}_{2}=22$ For P = 4 ${ }^{3} \text{C}_{1}\$ [{ }^{2}\text{C}2 { }^{2}\text{C}1 + { }^{2}\text{C}1 { }^{2}\text{C}2]$+{ }^{3} \text{C}{2}({ }^{2} \text{C}{2}+{ }^{2} \text{C}{2})+{ }^{3} \text{C}{3}({ }^{2} \text{C}{1}+{ }^{2} \text{C}{1})=22 $For P = 5$ { }^{3} \text{C}{2}({ }^{2} \text{C}{2}{ }^{2} \text{C}{1}+{ }^{2} \text{C}{1}{ }^{2} \text{C}{2})+{ }^{3} \text{C}{3}({ }^{2} \text{C}{2}+{ }^{2} \text{C}{2})=14 $For P = 6$ { }^{3} \text{C}{3}({ }^{2} \text{C}{2}{ }^{2} \text{C}{1}+{ }^{2} \text{C}{1}{ }^{2} \text{C}_{2})=4 $Total Subsets$ = 4 + 14 + 22 + 22 + 14 + 4 = 80$

Q86JEE Main 2021NAT4MHyperbola

Let $\text{A}(\sec \theta, 2 \tan \theta)$ and $\text{B}(\sec \phi, 2 \tan \phi)$ ,where $\theta+\phi=\pi/ 2$ , be two points on the hyperbola $2 x^{2}-y^{2}=2$ . If $(\alpha, \beta)$ is the point of the intersection of the normals to the hyperbola at A and B, then $(2 \beta)^{2}$ is equal to

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📖 Explanation

General form of hyperbola is $x^2/a^{2-}y^2/b^2=1$ ...(i) Given equation of hyperbola is $2x^{2-}y^2=2$ $\Rightarrow x^{2-}y^2/2=1$ ...(ii) comparing equation (i) with (ii), we get $a^2=1,b^2=2$ General equation of normal, ${a^2x}/x_1+{b^2y}/y_1=a^{2+}b^2$ ...(iii) Equation of normal at point $\text{A}(\sec \theta, 2 \tan \theta)$ $\frac{1\times x}{\sec\theta}+\frac{2\times y}{2\tan\theta}=1+2$ $\Rightarrow \frac{ x}{\sec\theta}+\frac{y}{\tan\theta}=3$ $\Rightarrow \frac{x}{\sec \theta}+\frac{y}{\frac{\sin \theta}{\cos\theta}}=3$ $\Rightarrow \frac{x}{\sec\theta}+{y \cos \theta} \times {\text{cosec}} \theta=3$ $\Rightarrow \frac{x}{\sec \theta}+\frac{y}{\sec \theta} {\text{cosec}} \theta=3$ $\Rightarrow {x}+{y} {\text{cosec}} \theta=3{\sec \theta}$ ...(iv) Similarly, equation of normal at point $\text{B}(\sec \phi, 2 \tan \phi)$ $x+y\text{cosec}\phi=3\sec\phi$ $\Rightarrow x+y\text{cosec}(\pi/2-\theta)=3\sec(\pi/2-\theta)$ $[ \because \theta+\phi=\pi/2]$ $\Rightarrow x+y\sec\theta=3\text{cosec}\theta$ ...(v) subtract equation (v) from (iv), we get $\Rightarrow y({\text{cosec}} \theta-\sec \theta)=3({\sec \theta-{\text{cosec}} \theta})$ $\Rightarrow y({\text{cosec}} \theta-\sec \theta)=-3({\text{cosec}} \theta-\sec \theta)$ $\Rightarrow y=-3$ Now, from equation (ii), we get $x^{2}-\frac{y^{2}}{2}=1$ $x^{2}-\frac{9}{2}=1$ $[\because y=-3]$ $x^{2}=1+\frac{9}{2}$ $x=\sqrt{\frac{11}{2}}$ $\therefore (x,y)=(\alpha,\beta)=(\sqrt{11/2},-3)$ $\therefore (2\beta)^2=(2\times (-3))^2=36$

Q87JEE Main 2021NAT4MCircles

Two circles each of radius 5 units touch each other at the point $(1,2)$ . If the equation of their common tangent is $4 x+3 y=10$ and $\text{C}_{1}(\alpha, \beta)$ and $\text{C}_{2}(\gamma, \delta), \text{C}_{1} \ne \text{C}_{2}$ are their centres, then $|(\alpha+\beta)(\gamma+\delta)|$ is equal to

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📖 Explanation

Equation of circles $(x-1)^{2}+(y-2)^{2}+\lambda(4 x+3 y-10)=0$ $x^{2}+y^{2}+(2 \lambda-1) 2 x+(\frac{3}{2} \lambda-2) 2 y+5-10 \lambda$ $=0$ $r=\sqrt{(2 \lambda-1)^{2}+(\frac{3}{2} \lambda-2)^{2}-5+10 \lambda}=5$ $\Rightarrow 4 \lambda^{2}+1-4 \lambda+\frac{9}{4} \lambda^{2}+4-6 \lambda-5+10 \lambda=25$ $\Rightarrow \frac{25}{4} \lambda^{2}-25=0$ ⇒ lambda = \pm 2 For lambda = 2 $x^{2}+y^{2}+6 x+2 y-15=0, \text{C}_{1}(-3,-1)$ For lambda = −2 $x^{2}+y^{2}-10 x-10 y+25=0, \text{C}_{2}(5,5)$ $|(\alpha+\beta)(\gamma+\delta)|=|-4 \times 10|=40$

Q88JEE Main 2021NAT4MProbability

$3 \times 7^{22}+2 \times 10^{22}-44$ when divided by 18 leaves the remainder

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📖 Explanation

$3 \times 7^{22}+2 \times 10^{22}-44$ $=3 \times (6+1)^{22}+2 \times (9+1)^{22}-44$ Now, $(1+6)^{22}=1+{ }^{22} \text{C}_{1} 6+{ }^{22} \text{C}_{2} .6^{2}+...+{ }^{22} \text{C}_{22} 6^{22}$ = (1+ 6lambda) $(1+9)^{22}=1+{ }^{22} \text{C}_{1} 9+{ }^{22} \text{C}_{2} .9^{2}+...+{ }^{22} \text{C}_{22} 6^{22}$ = (1+ 9µ) $\therefore = 3(1+ 6\lambda) + 2(1+ 9\mu) - 44$ $= 18\lambda + 3 + 18\mu + 2 − 44$ $=18 \delta-39=18 \alpha+15$ $3 \times 7^{22}+2 \times 10^{22}-44$ , when divided by 18 leaves remainder 15.

Q89JEE Main 2021NAT4MStatistics

An online exam is attempted by 50 candidates out of which 20 are boys. The average marks obtained by boys is 12 with a variance 2. The variance of marks obtained by 30 girls is also 2 . The average marks of all 50 candidates is 15. If µ is the average marks of girls and $\sigma^{2}$ is the variance of marks of 50 candidates, then $\mu+\sigma^{2}$ is equal to

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📖 Explanation

Number of boys = 20 Number of girls = 30 $\sigma_{\text{B}}^{2}=2, \ov {\text{X}}_{\text{B}}=12, \sigma_{\text{G}}^{2}=2$ $\ov {\text{X}}_{\text{G}}=\frac{50 \times 15-20 \times 12}{30}=\frac{510}{30}=17=\mu$ Variance of 50 candidates $\sigma^{2}={20 \sigma_{\text{B}}^{2}+30 \sigma_{\text{C}}^{2}}/{50}+\frac{20 .30}{(20+30)^{2}}(\ov {\text{X}}_{\text{B}}-\ov {\text{X}}_{\text{C}})^{2}$ $=\frac{20 \times 2+30 \times 2}{50}+\frac{600}{2500} \times 25=8$ $\therefore \mu+\sigma^{2}=17+8=25$

Q90JEE Main 2021NAT4MIndefinite Integrals

$\text{If}\int\frac{2 e^{x}+3 e^{-x}}{4 e^{x}+7 e^{-x}} d x=\frac{1}{14}(u x+v \log _{e}$ $(4 e^{x}+7 e^{-x}))+\text{C},$ where C is a constant of integration, then u + v is equal to

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📖 Explanation

$\text{I}=\int\frac{2 e^{x}+3 e^{-x}}{4 e^{x}+7 e^{-x}} d x=\int\frac{2 e^{2 x}+3}{4 e^{2 x}+7} d x$ Let $2 e^{2 x}+3=\text{A} \frac{d}{d x}(4 e^{2 x}+7)+\text{B}(4 e^{2 x}+7)$ $\Rightarrow 2 e^{2 x}+3=(8 \text{A}+4 \text{B}) e^{2 x}+7 \text{B}$ Comparing both sides $\text{B}=\frac{3}{7}$ and $\text{A}=\frac{1}{28}$ $\therefore \text{I}=\int\frac{\frac{1}{28}(8 {e}^{2 x})+\frac{3}{7}(4 {e}^{2 x}+7)}{4 {e}^{2 x}+7} {dx}$ $=\frac{1}{28} \ln |4 e^{2 x}+7|+\frac{3}{7} x+\text{C}$ $=\frac{1}{28} \ln |e^{x}(4 e^{x}+7 e^{-x})|+\frac{3}{7} x+\text{C}$ $=\frac{1}{28} x+\frac{1}{28} \ln |4 e^{x}+7 e^{-x}|+\frac{3}{7} x+\text{C}$ $=\frac{1}{14}(\frac{13}{2} x+\frac{1}{2} \ln |4 e^{x}+7 e^{-x}|)+\text{C}$ $\Rightarrow u=\frac{13}{2}$ and $v=\frac{1}{2}$ $\therefore u+v=\frac{13}{2}+\frac{1}{2}=7$

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