JEE Main 24 January 2023 Shift 1 question paper with solutions
JEE Main 24 January 2023 Shift 1: all 90 questions in paper order (Physics, Chemistry, Maths) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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The maximum vertical height to which a man can throw a ball is $136 \, \mathrm{m}$. The maximum horizontal distance upto which he can throw the same ball is
$192 \, \mathrm{m}$
$136 \, \mathrm{m}$
$272 \, \mathrm{m}$
$68 \, \mathrm{m}$
Answer: (c)
Solution
Given the maximum height $H_{max} = \frac{v^2}{2g} = 136 \, \mathrm{m}$. The maximum range $R_{max} = \frac{v^2}{g} = 2H_{max}$. Therefore, $R_{max} = 2(136) = 272 \, \mathrm{m}$.
Question 3
Physics · Laws of Motion · Single correct
As per given figure, a weightless pulley P is attached on a double inclined frictionless surface. The tension in the string (massless) will be (if g = 10 m s$^{-2}$)
(4$\sqrt{3}$ + 1) N
4$\sqrt{3}$ N
4($\sqrt{3}$ - 1) N
(4$\sqrt{3}$ - 1) N
Answer: (b)
Solution
Given the equations: $$4g \sin 60^\circ - T = 4a ...(1)$$ $$T - g \sin 30^\circ = a ...(2)$$ Solving (1) and (2) we get: $$20\sqrt{3} - T = 4T - 20$$ $$T = 4(\sqrt{3} + 1) \mathrm{N}$$
Question 4
Physics · Laws of Motion · Single correct
Given below are two statements : Statement-I : An elevator can go up or down with uniform speed when its weight is balanced with the tension of its cable. Statement-II : Force exerted by the floor of an elevator on the foot of a person standing on it is more than his/her weight when the elevator goes down with increasing speed. In the light of the above statements, choose the correct answer from the options given below :
Both statement I and statement II are false
Statement I is true but Statement II is false
Both Statement I and Statement II are true
Statement I is false but Statement II is true
Answer: (b)
Solution
Statement-1 When elevator is moving with uniform speed $T = F_g$. Statement-2 When elevator is going down with increasing speed, its acceleration is downward. Hence $$W - N = \frac{W}{g} \times a$$ $$N = W \left( 1 - \frac{a}{g} \right)$$ i.e. less than weight.
Question 5
Physics · Gravitation · Single correct
The weight of a body at the surface of earth is 18 N. The weight of the body at an altitude of 3200 km above the earth’s surface is (given, radius of earth $R_e = 6400 \, \mathrm{km}$)
9.8 N
4.9 N
19.6 N
8 N
Answer: (d)
Solution
Acceleration due to gravity at height $h$ $$g' = \frac{g}{\left[1 + \frac{h}{R}\right]^2}$$ So weight at given height $$mg' = \frac{mg}{\left[1 + \frac{h}{R}\right]^2} = \frac{18}{\left[1 + \frac{1}{2}\right]^2} = 8 \, \mathrm{N}$$
Question 6
Physics · Mechanical Properties of Solids · Single correct
A 100 m long wire having cross-sectional area $6.25 \times 10^{-4} \, \mathrm{m}^2$ and Young's modulus is $10^{10} \, \mathrm{Nm}^{-2}$ is subjected to a load of $250 \, \mathrm{N}$, then the elongation in the wire will be:
1g of a liquid is converted to vapour at $3 \times 10^5$ Pa pressure. If 10\% of the heat supplied is used for increasing the volume by 1600 cm$^3$ during this phase change, then the increase in internal energy in the process will be :
4320 $\mathrm{J}$
432000 $\mathrm{J}$
4800 $\mathrm{J}$
\[ 4.32\times10^{8}\ \text{J} \]
Answer: (a)
Solution
Work done = $P \Delta V$ $$= 3 \times 10^5 \times 1600 \times 10^{-6}$$ $$= 480 \, \mathrm{J}$$ Only 10$\%$ of heat is used in work done. Hence $\Delta Q = 4800 \, \mathrm{J}$. The rest goes in internal energy, which is 90$\%$ of heat. Change in internal energy = $0.9 \times 4800 = 4320 \, \mathrm{J}$.
Question 8
Physics · Kinetic Theory · Single correct
Given below are two statements : Statements I : The temperature of a gas is $-73^\circ \mathrm{C}$. When the gas is heated to $527^\circ \mathrm{C}$, the root mean square speed of the molecules is doubled. Statement II : The product of pressure and volume of an ideal gas will be equal to translational kinetic energy of the molecules. In the light of the above statements, choose the correct answer from the options given below :
A travelling wave is described by the equation $y(x, t) = [0.05 \sin (8x - 4t)] \, \mathrm{m}$ The velocity of the wave is : [all the quantities are in SI unit]
$4\,\mathrm{ms}^{-1}$
$2\,\mathrm{m\,s}^{-1}$
$0.5\,\mathrm{m\,s}^{-1}$
$8\,\mathrm{m\,s}^{-1}$
Answer: (c)
Solution
From the given equation $k = 8 \, \mathrm{m^{-1}}$ and $\omega = 4 \, \mathrm{rad/s}$. Velocity of wave $= \frac{\omega}{k}$ $$v = \frac{4}{8} = 0.5 \, \mathrm{m/s}$$
Question 10
Physics · Electric Charges and Fields · Single correct
If two charges $q_1$ and $q_2$ are separated with distance $d$ and placed in a medium of dielectric constant $K$. What will be the equivalent distance between charges in air for the same electrostatic force?
$d\sqrt{K}$
$K\sqrt{d}$
$1.5d\sqrt{K}$
$2d\sqrt{K}$
Answer: (a)
Solution
The force $F$ in a medium is given by $$F = \frac{1}{(4\pi \varepsilon_0) \cdot k \cdot d^2} q_1 q_2 (in medium)$$ The force in air $F_{Air}$ is given by $$F_{Air} = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{d'^2}$$ Equating the forces, we have $$F = F_{Air}$$ $$\frac{q_1 q_2}{4\pi \varepsilon_0 k d^2} = \frac{q_1 q_2}{4\pi \varepsilon_0 d'^2}$$ Solving for $d'$, we get $$d' = d \sqrt{k}$$
Question 11
Physics · Current Electricity · Single correct
As shown in the figure, a network of resistors is connected to a battery of 24 V with an internal resistance of 3Ω. The currents through the resistors $R_4$ and $R_5$ are $I_4$ and $I_5$ respectively. The values of $I_4$ and $I_5$ are:
Physics · Moving Charges and Magnetism · Single correct
Two long straight wires P and Q carrying equal current 10A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is $F_1$. If distance between wires is halved and currents on them are doubled, force $F_2$ on 10 cm length of wire P will be:
8 $F_1$
10 $F_1$
$F_1/8$
$F_1/10$
Answer: (a)
Solution
Force per unit length between two parallel straight wires is given by $$\frac{\mu_0 i_1 i_2}{2 \pi d}$$. The ratio of forces is given by $$\frac{F_1}{F_2} = \frac{\mu_0 (10)^2}{2 \pi (5 \mathrm{cm})} \div \frac{\mu_0 (20)^2}{2 \pi \left( \frac{5 \mathrm{cm}}{2} \right)} = \frac{1}{8}$$ Therefore, $$F_2 = 8F_1$$.
Question 13
Physics · Moving Charges and Magnetism · Single correct
A circular loop of radius $r$ is carrying current $I \, \mathrm{A}$. The ratio of magnetic field at the centre of circular loop and at a distance $r$ from the center of the loop on its axis is:
$1 : 3\sqrt{2}$
$3\sqrt{2} : 2$
$2\sqrt{2} : 1$
$1 : \sqrt{2}$
Answer: (c)
Solution
Magnetic field due to current carrying circular loop on its axis is given as $$\frac{\mu_0 i r^2}{2(r^2 + x^2)^{3/2}}$$ At centre, $x = 0$, $B_1 = \frac{\mu_0 i}{2r}$ At $x = r$, $B_2 = \frac{\mu_0 i}{2 \times 2 \sqrt{2} r}$ $$\frac{B_1}{B_2} = 2\sqrt{2}$$
Question 14
Physics · Electromagnetic Induction · Single correct
A conducting loop of radius $\frac{10}{\sqrt{\pi}} \, \mathrm{cm}$ is placed perpendicular to a uniform magnetic field of $0.5 \, \mathrm{T}$. The magnetic field is decreased to zero in $0.5 \, \mathrm{s}$ at a steady rate. The induced emf in the circular loop at $0.25 \, \mathrm{s}$ is:
emf = 1 mV
emf = 10 mV
emf = 100 mV
emf = 5 mV
Answer: (b)
Solution
EMF is given by $$EMF = \frac{d\phi}{dt} = \frac{BA - 0}{t}$$. The area is $$A = \pi r^2 = \pi \left( \frac{0.1^2}{\pi} \right) = 0.01$$. The magnetic field is $$B = 0.5$$. Therefore, $$EMF = \frac{(0.5)(0.01)}{0.5} = 0.01 \, V = 10 \, mV$$.
Question 15
Physics · Electromagnetic Waves · Single correct
If $\vec{E}$ and $\vec{K}$ represent electric field and propagation vectors of the EM waves in vacuum, then magnetic field vector is given by : ($\omega$ - angular frequency) :
$\frac{1}{\omega} (\vec{K} \times \vec{E})$
$\omega (\vec{E} \times \vec{K})$
$\omega (\vec{K} \times \vec{E})$
$\vec{K} \times \vec{E}$
Answer: (a)
Solution
Magnetic field vector will be in the direction of $\hat{K} \times \hat{E}$. The magnitude of $\mathbf{B} = \frac{\mathbf{E}}{C} = \frac{K}{\omega} \mathbf{E}$. Or $\mathbf{B} = \frac{1}{\omega} (\vec{K} \times \vec{E})$.
Question 16
Physics · Wave Optics · Single correct
Given below are two statements : Statement I : If the Brewster’s angle for the light propagating from air to glass is $\theta_B$, then Brewster’s angle for the light propagating from glass to air is $\frac{\pi}{2} - \theta_B$. Statement II : The Brewster’s angle for the light propagating from glass to air is $\tan^{-1}(\mu_g)$ where $\mu_g$ is the refractive index of glass. In the light of the above statements, choose the correct answer from the options given below :
Both Statements I and Statement II are true.
Statement I is true but Statement II is false.
Both Statement I and Statement II are false.
Statement I is false but Statement II is true.
Answer: (b)
Solution
Given $\mu_a \sin i_1 = \mu_g \sin(90 - i_1)$. $$\tan i_1 = \frac{\mu_g}{\mu_a}$$ When going from glass to air $$\tan i_2 = \frac{\mu_a}{\mu_g} = \cot i_1$$ Hence $$i_2 = \frac{\pi}{2} - i_1$$
Question 17
Physics · Dual Nature of Radiation and Matter · Single correct
From the photoelectric effect experiment, following observations are made. Identify which of these are correct A. The stopping potential depends only on the work function of the metal. B. The saturation current increases as the intensity of incident light increases. C. The maximum kinetic energy of a photo electron depends on the intensity of the incident light. D. Photoelectric effect can be explained using wave theory of light. Choose the correct answer from the options given below:
B, C only
A, C, D only
B only
A, B, D only
Answer: (c)
Solution
(A) Stopping potential depends on both frequency of light and work function. (B) Saturation current is proportional to intensity of light. (C) Maximum kinetic energy depends on frequency. (D) Photoelectric effect is explained using particle theory.
Question 18
Physics · Nuclei · Single correct
Consider the following radioactive decay process $$^{218}_{84}A \xrightarrow{\alpha} A_1 \xrightarrow{\beta^-} A_2 \xrightarrow{\gamma} A_3 \xrightarrow{\alpha} A_4 \xrightarrow{\beta^+} A_5 \xrightarrow{\gamma} A_6$$ The mass number and the atomic number $A_6$ are given by:
210 and 82
210 and 84
210 and 80
211 and 80
Answer: (c)
Solution
The decay process is as follows: For the first sequence: $$^{218}_{84}\mathrm{A} \xrightarrow{\alpha} \, ^{214}_{82}\mathrm{A}_1 \xrightarrow{\beta^-} \, ^{214}_{83}\mathrm{A}_2 \xrightarrow{\gamma} \, ^{214}_{83}\mathrm{A}_3$$ For the second sequence: $$^{214}_{83}\mathrm{A}_3 \xrightarrow{\alpha} \, ^{210}_{81}\mathrm{A}_4 \xrightarrow{\beta^+} \, ^{210}_{80}\mathrm{A}_5 \xrightarrow{\gamma} \, ^{210}_{80}\mathrm{A}_6$$
Question 19
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: Photodiodes are preferably operated in reverse bias condition for light intensity measurement. Reason R: The current in the forward bias is more than the current in the reverse bias for a p-n junction diode. In the light of the above statement, choose the correct answer from the options given below:
A is false but R is true
Both A and R are true but R is NOT the correct explanation of A
A is true but R is false
Both A and R are true and R is the correct explanation of A
Answer: (b)
Solution
Photodiodes are operated in reverse bias as fractional change in current due to light is more easy to detect in reverse bias.
Question 20
Physics · Communication Systems · Single correct
A modulating signal is a square wave, as shown in the figure. If the carrier wave is given as $c(t) = 2 \sin (8\pi t)$ volts, the modulation index is :
1/4
1
1/3
1/2
Answer: (d)
Solution
Modulation index is given by the ratio of the amplitude of the modulating signal to the amplitude of the carrier wave. $$Modulation\ index = \frac{Amplitude\ of\ modulating\ signal}{Amplitude\ of\ carrier\ wave}$$ Given that the modulation index $\mu$ is $\frac{1}{2}$, we have: $$\mu = \frac{1}{2}$$
Question 21
Physics · Mathematics in Physics · Numerical
Vectors $ai + bj + \hat{k}$ and $2\hat{i} - 3\hat{j} + 4\hat{k}$ are perpendicular to each other when $3a + 2b = 7$, the ratio of $a$ to $b$ is $\frac{x}{2}$. The value of $x$ is _______.
Answer: 1
Solution
For two perpendicular vectors $$(ai + bj + k) \cdot (2i - 3j + 4k) = 0$$ $$2a - 3b + 4 = 0$$ On solving, $2a - 3b = -4$. Also given $$3a + 2b = 7$$ We get $a = 1$, $b = 2$. $$\frac{a}{b} = \frac{x}{2} \Rightarrow x = \frac{2a}{b} = \frac{2 \times 1}{2}$$ $$\Rightarrow x = 1$$
Question 22
Physics · Work, Energy and Power · Numerical
A spherical body of mass $2 \, \mathrm{kg}$ starting from rest acquires a kinetic energy of $10000 \, \mathrm{J}$ at the end of $5^{th}$ second. The force acted on the body is _____ N.
Answer: 40
Solution
Given $\($ $\frac{1}{2}$ $\times$ 2 $\times$ v^2 = 10000 $\)$. $\($ $\Rightarrow$ v^2 = 10000 $\)$ $\($ $\Rightarrow$ v = 100 $\,$ $\mathrm{m/s}$ $\)$ $\($ $\Rightarrow$ v = at = a $\times$ 5 = 100 $\)$ $\($ $\Rightarrow$ a = 20 $\,$ $\mathrm{m/s^2}$ $\)$ $\($ F = ma = 2 $\times$ 20 = 40 $\,$ $\mathrm{N}$ $\)$
Question 23
Physics · System of Particles and Rotational Motion · Numerical
Solid sphere A is rotating about an axis PQ. If the radius of the sphere is 5 cm then its radius of gyration about PQ will be $\sqrt{x}$ cm. The value of $x$ is _____.
Physics · Thermal Properties of Matter · Numerical
A hole is drilled in a metal sheet. At 27°C, the diameter of hole is 5 cm. When the sheet is heated to 177°C, the change in the diameter of hole is $d \times 10^{-3}$ cm. The value of $d$ will be ______ if coefficient of linear expansion of the metal is $1.6 \times 10^{-5}/°\mathrm{C}$.
Answer: 12
Solution
Given $d_0$ at $27^\circ \mathrm{C}$ and $d_1$ at $177^\circ \mathrm{C}$. $d_1 = d_0 (1 + \alpha \Delta T)$ $d_1 - d_0 = 5 \times 1.6 \times 10^{-5} \times 150 \, \mathrm{cm}$ $= 12 \times 10^{-3} \, \mathrm{cm}$
Question 25
Physics · Oscillations · Numerical
A block of mass 2 kg is attached with two identical springs of spring constant 20 N/m each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is $\frac{\pi}{\sqrt{x}}$ in SI unit. The value of $x$ is ________.
Answer: 5
Solution
Given $F = -2kx$, $a = -\frac{2kx}{m}$, $\omega = \sqrt{\frac{2k}{m}} = \sqrt{\frac{2 \times 20}{2}}$. This simplifies to $\sqrt{20} \, \mathrm{rad/s}$. The period $T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{20}} = \frac{\pi}{\sqrt{5}}$. Given $x = 5$.
Question 26
Physics · Electric Charges and Fields · Numerical
A stream of a positively charged particles having $\frac{q}{m} = 2 \times 10^{11} \, \mathrm{C/kg}$ and velocity $\vec{v}_0 = 3 \times 10^7 \, \hat{i} \, \mathrm{m/s}$ is deflected by an electric field $1.8 \hat{j} \, \mathrm{kV/m}$. The electric field exists in a region of $10 \, \mathrm{cm}$ along $x$ direction. Due to the electric field, the deflection of the charge particles in the $y$ direction is mm.
Answer: 2
Solution
Given $V_0 = 3 \times 10^7 \, \mathrm{m/s}$, $E = 1.8 \times 10^3 \, \mathrm{N/m}$, and $\ell = 10 \, \mathrm{cm}$. The acceleration $a$ is given by $a = \frac{F}{m} = \frac{qE}{m} = (2 \times 10^{11})(1.8 \times 10^3) = 3.6 \times 10^{14} \, \mathrm{m/s^2}$. Time to cross plates is $\frac{d}{v}$. $$t = \frac{0.10}{3 \times 10^7}$$ The displacement $y$ is given by $y = \frac{1}{2} a t^2 = \frac{1}{2} (3.6 \times 10^{14}) \left( \frac{0.01}{9 \times 10^{14}} \right)$. $$= 0.2 \times 0.01$$ $$= 0.002 \, \mathrm{m}$$ $$= 2 \, \mathrm{mm}$$
Question 27
Physics · Current Electricity · Numerical
A hollow cylindrical conductor has length of 3.14 $\,$ $\mathrm{m}$, while its inner and outer diameters are 4 $\,$ $\mathrm{mm}$ and 8 $\,$ $\mathrm{mm}$ respectively. The resistance of the conductor is $n \times 10^{-3} \, \Omega$. If the resistivity of the material is $2.4 \times 10^{-8} \, \Omega \mathrm{m}$. The value of $n$ is
Answer: 2
Solution
Given $R = \rho \frac{\ell}{A}$, the cross-sectional area is $\pi(b^2 - a^2)$. $$R = \rho \frac{\ell}{\pi(b^2 - a^2)} = \frac{2.4 \times 10^{-8} \times 3.14}{3.14 \times (4^2 - 2^2) \times 10^{-6}}$$ $$= 2 \times 10^{-3} \, \Omega$$ Therefore, $n = 2$.
Question 28
Physics · Alternating Current · Numerical
In the circuit shown in the figure, the ratio of the quality factor and the band width is ______ s.
Physics · Ray Optics and Optical Instruments · Numerical
As shown in the figure, a combination of a thin plano concave lens and a thin plano convex lens is used to image an object placed at infinity. The radius of curvature of both the lenses is $30\,\mathrm{cm}$ and refraction index of the material for both the lenses is $1.75$. Both the lenses are placed at distance of $40\,\mathrm{cm}$ from each other. Due to the combination, the image of the object is formed at distance $x=$ $\underline{\hspace{2cm}}$ $\mathrm{cm}$, from concave lens.
Answer: 120
Solution
Given $\($ $\frac{1}{f_1}$ = (1.75 - 1) $\left$( -$\frac{1}{30}$ $\right$) $\)$ which implies $\($ f_1 = -40 $\,$ cm $\)$. For $\($ $\frac{1}{f_2}$ = (1.75 - 1) $\left$( $\frac{1}{30}$ $\right$) $\)$ which implies $\($ f_2 = 40 $\,$ cm $\)$. Image from $\($ L_1 $\)$ will be virtual and on the left of $\($ L_1 $\)$ at focal length 40 cm. So the object for $\($ L_2 $\)$ will be 80 cm from $\($ L_2 $\)$, which is $\($ 2f $\)$. Final image is formed at 80 cm from $\($ L_2 $\)$ on the right. So $\($ x = 120 $\)$.
Question 30
Physics · Atoms · Numerical
Assume that protons and neutrons have equal masses. Mass of a nucleon is $1.6 \times 10^{-27} \, \mathrm{kg}$ and radius of nucleus is $1.5 \times 10^{-15} \, A^{1/3} \, \mathrm{m}$. The approximate ratio of the nuclear density and water density is $n \times 10^{13}$. The value of $n$ is _______.
Answer: 11
Solution
The density of nuclei is given by the mass of nuclei divided by the volume of nuclei. $$\rho = \frac{1.6 \times 10^{-27} \, A}{\frac{4}{3} \pi (1.5 \times 10^{-15})^3 \, A}$$ This simplifies to: $$\rho = \frac{1.6 \times 10^{-27}}{14.14 \times 10^{-45}} = 0.113 \times 10^{18}$$ Given $\rho_w = 10^3$, we have: $$\frac{\rho}{\rho_w} = 11.31 \times 10^{13}$$ Hence
Chemistry
Question 31
Chemistry · Structure of Atom · Single correct
It is observed that characteristic X-ray spectra of elements show regularity. When frequency to the power 'n' i.e. $\nu^n$ of X-rays emitted is plotted against atomic number 'Z', following graph is obtained. The value of 'n' is
1
2
$\frac{1}{2}$
3
Answer: (c)
Solution
According to Henry Moseley $\sqrt{\nu} \alpha z - b$. So $n = \frac{1}{2}$.
Question 32
Chemistry · Hydrogen · Single correct
Decreasing order of the hydrogen bonding in following forms of water is correctly represented by A. Liquid water B. Ice C. Impure water
A = B > C
B > A > C
C > B > A
A > B > C
Answer: (b)
Solution
Ice $>$ Liquid water $>$ Impure water Due to impurity extent of H-Bonding decreases.
Question 33
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Order of Covalent bond; A. KF > KI; LiF > KF B. KF KF C. SnCl4 > SnCl2; CuCl > NaCl D. LiF > KF; CuCl NaCl
C, E only
B, C only
B, C, E only
A, B only
Answer: (c)
Solution
According to Fajan's Rule, A. $\mathrm{KF > KI}$ – False; $\mathrm{LiF > KF}$ – True B. $\mathrm{KF KF}$ – True C. $\mathrm{SnCl_4 > SnCl_2}$ – True; $\mathrm{CuCl > NaCl}$ – True D. $\mathrm{LiF > KF}$ – True; $\mathrm{CuCl NaCl}$ – True
Question 34
Chemistry · Co-ordination Compounds · Single correct
Match List I with List II. Choose the correct answer from the options given below :
A – III, B – I, C – II, D – IV
A – II, B – I, C – III, D – IV
A – III, B – IV, C – I, D – II
A – II, B – III, C – IV, D – I
Answer: (a)
Solution
Chlorophyll: $\mathrm{Mg^{+2}}$ complex Soda ash: $\mathrm{Na_2CO_3}$ Dentistry, Ornamental work: $\mathrm{CaSO_4}$ Used in white washing: $\mathrm{Ca(OH)_2}$
Question 35
Chemistry · The s-Block Elements · Single correct
Reaction of BeO with ammonia and hydrogen fluoride gives 'A' which on thermal decomposition gives $\mathrm{BeF_2}$ and $\mathrm{NH_4F}$. What is 'A'?
(NH_4)_2BeF_4
H_3NBeF_3
(NH_4)BeF_3
(NH_4)Be_2F_5
Answer: (a)
Solution
The reaction is given as follows: $$\mathrm{BeO} + 2\mathrm{NH_3} + 4\mathrm{HF} \rightarrow (\mathrm{NH_4})_2\mathrm{BeF_4} + \mathrm{H_2O}$$ Upon heating, the compound decomposes: $$(\mathrm{NH_4})_2\mathrm{BeF_4} \xrightarrow{\Delta} \mathrm{BeF_2} + \mathrm{NH_4F}$$
Question 36
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Increasing order of stability of the resonance structure is:
C, D, B, A
C, D, A, B
D, C, A, B
D, C, B, A
Answer: (b)
Solution
No option is matching the correct answer. Order should be: C < A < B < D
Question 37
Chemistry · Environmental Chemistry · Single correct
Which of the following is true about freons?
These are chlorofluorocarbon compounds
These are chemicals causing skin cancer
These are radicals of chlorine and chlorine monoxide
All radicals are called freons
Answer: (a)
Solution
Fact
Question 38
Chemistry · Solutions · Multiple correct
In the depression of freezing point experiment A. Vapour pressure of the solution is less than that of pure solvent B. Vapour pressure of the solution is more than that of pure solvent C. Only solute molecules solidify at the freezing point D. Only solvent molecules solidify at the freezing point
A and D only
B and C only
A and C only
A only
Answer: (a)
Solution
Vapour pressure (V.P.) of solvent is greater than vapour pressure (V.P.) of solution. Only solvent freezes.
Question 39
Chemistry · Surface Chemistry · Single correct
Statement I : For colloidal particles, the values of colligative properties are of small order as compared to values shown by true solutions at same concentration. Statement II : For colloidal particles, the potential difference between the fixed layer and the diffused layer of same charges is called the electrokinetic potential or zeta potential. In the light of the above statements, choose the correct answer from the options given below.
Statement I is true but Statement II is false
Statement I is false but Statement II is true
Both Statement I and Statement II are true
Both Statement I and Statement II are false
Answer: (c)
Solution
Statement I: For colloidal particles, the values of colligative properties are of small order as compared to values shown by true solutions at same concentration. True Statement II: For colloidal particles, the potential difference between the fixed layer and the diffused layer of same charges is called the electrokinetic potential or zeta potential. True
Question 40
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Match List I with List II.
A – IV, B – II, C – I, D – III
A – I, B – IV, C – II, D – III
A – I, B – III, C – II, D – IV
A – III, B – IV, C – I, D – II
Answer: (a)
Solution
Reverberatory furnace: Used for roasting of Copper. Electrolytic cell: For reactive metal: Al Blast furnace: Hematite to Pig Iron Zone Refining furnace: For semiconductors: Si
Question 41
Chemistry · Analytical Chemistry · Single correct
An ammoniacal metal salt solution gives a brilliant red precipitate on addition of dimethylglyoxime. The metal ion is:
$\mathrm{Cu^{2+}}$
$\mathrm{Co^{2+}}$
$\mathrm{Fe^{2+}}$
$\mathrm{Ni^{2+}}$
Answer: (d)
Solution
The reaction is given by: $$\mathrm{Ni^{2+} + 2DMG \rightarrow [Ni(DMG)_2]}$$ This forms a Rosy Red complex.
Question 42
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Which of the Phosphorus oxoacid can create silver mirror from $\mathrm{AgNO_3}$ solution?
$(\mathrm{HPO_3})_n$
$\mathrm{H_4P_2O_5}$
$\mathrm{H_4P_2O_6}$
$\mathrm{H_4P_2O_7}$
Answer: (b)
Solution
Oxyacid having P–H bond can reduce $\mathrm{AgNO_3}$ to Ag.
Question 43
Chemistry · The d-and f-Block Elements · Single correct
The magnetic moment of a transition metal compound has been calculated to be 3.87 $\mathrm{B.M.}$ The metal ion is
Chemistry · Co-ordination Compounds · Single correct
The primary and secondary valencies of cobalt respectively in $[Co(NH_3)_5Cl]Cl_2$ are:
3 and 5
2 and 6
2 and 8
3 and 6
Answer: (d)
Solution
[$\mathrm{Co(NH_3)_5Cl}$]$\mathrm{Cl}$_2 Oxidation number of Co is $+3$. So primary valency is $3$. It is an octahedral complex so secondary valency $6$ or Co-ordination number $6$.
Question 45
Chemistry · Haloalkanes and Haloarenes · Single correct
Assertion A: Hydrolysis of an alkyl chloride is a slow reaction but in the presence of NaI, the rate of the hydrolysis increases. Reason R: $\mathrm{I}^-$ is a good nucleophile as well as a good leaving group. In the light of the above statements, choose the correct answer from the options given below.
A is false but R is true
A is true but R is false
Both A and R are true and R is the correct explanation of A
Both A and R are true but R is NOT the correct explanation of A
Answer: (c)
Solution
The rate of hydrolysis of alkyl chloride improves because of better nucleophilicity of $\mathrm{I}^-$.
Question 46
Chemistry · Hydrocarbons · Single correct
In the following given reaction 'A' is
Answer: (d)
Solution
The reaction involves the addition of HBr to the alkene. The alkene is 1-methylcyclobutene. Upon reaction with HBr, the more stable carbocation is formed by the addition of H to the less substituted carbon. This leads to the formation of a secondary carbocation. The bromide ion then attacks the carbocation, resulting in the formation of 1-bromo-1-methylcyclopentane.
Question 47
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
'A' and 'B' formed in the following set of reactions are:
Answer: (d)
Solution
Question 48
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
'R' formed in the following sequence of reaction is:
Answer: (b)
Solution
The given reaction sequence involves the conversion of a chlorobenzaldehyde derivative to different products. First, the chlorobenzaldehyde reacts with $\mathrm{NaCN}$ and $\mathrm{HOAc}$ to form product (P), which is a cyanohydrin derivative. Next, the cyanohydrin derivative undergoes esterification with $\mathrm{EtOH}$ in the presence of an acid catalyst $\mathrm{H^+}$ to form product (Q), an ethyl ester. Finally, the ethyl ester reacts with two equivalents of methylmagnesium bromide $\mathrm{(2MeMgBr)}$ followed by hydrolysis with $\mathrm{H_3O^+}$ to form product (R), a tertiary alcohol.
Question 49
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Compound (X) undergoes following sequence of reactions to give the Lactone (Y).
Answer: (a)
Solution
The reaction begins with an aldol condensation. The starting materials are acetaldehyde and acetone. The aldol product is formed by the reaction of these two compounds. The aldol product then undergoes a cyanohydrin formation reaction. This involves the addition of hydrogen cyanide to the carbonyl group of the aldol product, resulting in the formation of a cyanohydrin.
Question 50
Chemistry · Chemistry in Everyday Life · Single correct
Given below are two statements : Statement I : Noradrenaline is a neurotransmitter. Statement II : Low level of noradrenaline is not the cause of depression in human. In the light of the above statements, choose the correct answer from the options given below
Statement I is correct but Statement II is incorrect
Statement I is incorrect but Statement II is correct
Both Statement I and Statement II are correct
Both Statement I and Statement II are incorrect
Answer: (a)
Solution
Fact
Question 51
Chemistry · Structure of Atom · Numerical
If wavelength of the first line of the Paschen series of hydrogen atom is $720 \, \mathrm{nm}$, then the wavelength of the second line of this series is _______ nm. (Nearest integer)
For independent process at 300 K. The number of non-spontaneous process from the following is
Answer: 2
Solution
Given $\Delta G = \Delta H - T \Delta S$. A: $\Delta G \, (\mathrm{J \, mol^{-1}}) = -25 \times 10^3 + 80 \times 300 : -\mathrm{ve}$ B: $\Delta G \, (\mathrm{J \, mol^{-1}}) = -22 \times 10^3 - 40 \times 300 : -\mathrm{ve}$ C: $\Delta G \, (\mathrm{J \, mol^{-1}}) = 25 \times 10^3 + 300 \times 50 : +\mathrm{ve}$ D: $\Delta G \, (\mathrm{J \, mol^{-1}}) = 22 \times 10^3 - 20 \times 300 : +\mathrm{ve}$ Processes C and D are non-spontaneous.
Question 53
Chemistry · Equilibrium · Numerical
The dissociation constant of acetic acid is $x\times10^{-5}$. When $25\,\mathrm{mL}$ of $0.2\ \mathrm{M}\ \mathrm{CH_3COONa}$ solution is mixed with $25\,\mathrm{mL}$ of $0.02\ \mathrm{M}\ \mathrm{CH_3COOH}$ solution, the pH of the resultant solution is found to be equal to $5$. The value of $x$ is ______.
Answer: 10
Solution
Buffer of HOAc and NaOAc. pH = pKa + $\log$ $\frac{0.1}{0.01}$ 5 = pKa + 1 pKa = 4 $K_a$ = $10^{-4}$ x = 10
Question 54
Chemistry · Biomolecules · Numerical
Uracil is base present in RNA with the following structure. % of N in uracil is _______. Given: Molar mass of $\mathrm{N} = 14\,\mathrm{g\,mol^{-1}}$; $\mathrm{O} = 16\,\mathrm{g\,mol^{-1}}$; $\mathrm{C} = 12\,\mathrm{g\,mol^{-1}}$; $\mathrm{H} = 1\,\mathrm{g\,mol^{-1}}$.
When $\mathrm{Fe}_{0.93}\mathrm{O}$ is heated in presence of oxygen, it converts to $\mathrm{Fe}_2\mathrm{O}_3$. The number of correct statement/s from the following is ________.
The equivalent weight of $\mathrm{Fe}_{0.93}\mathrm{O}$ is $\($ $\frac{Molecular weight}{0.79}$ $\)$.
The number of moles of $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$ in 1 mole of $\mathrm{Fe}_{0.93}\mathrm{O}$ is 0.79 and 0.14 respectively.
$\mathrm{Fe}_{0.93}\mathrm{O}$ is metal deficient with lattice comprising of cubic closed packed arrangement of $\mathrm{O}^{2-}$ ions.
The $\%$ composition of $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$ in $\mathrm{Fe}_{0.93}\mathrm{O}$ is 85$\%$ and 15$\%$ respectively.
5 $\mathrm{g}$ of NaOH was dissolved in deionized water to prepare a 450 $\mathrm{mL}$ stock solution. What volume (in mL) of this solution would be required to prepare 500 $\mathrm{mL}$ of 0.1 $\mathrm{M}$ solution? Given: Molar Mass of Na, O and H is 23, 16 and 1 $\mathrm{g \, mol^{-1}}$ respectively
At 298 K, a 1 litre solution containing 10 $\mathrm{mmol}$ of $\mathrm{Cr_2O_7^{2-}}$ and 100 $\mathrm{mmol}$ of $\mathrm{Cr^{3+}}$ shows a pH of 3.0. Given : $\mathrm{Cr_2O_7^{2-} \rightarrow Cr^{3+}}$; $E^0$ = 1.330 $\mathrm{V}$ and $\frac{2.303 \ RT}{F}$ = 0.059 $\mathrm{V}$ The potential for the half cell reaction is $x \times 10^{-3}$ V. The value of $x$ is
Answer: 917
Solution
The reaction is given by: $$\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O}$$ The equation for the potential is: $$E = 1.33 - \frac{0.059}{6} \log \frac{(0.1)^2}{(10^{-2})(10^{-3})^{14}}$$ Simplifying the expression: $$E = 1.33 - \frac{0.059}{6} \times 42 = 0.917$$ Thus, the potential is: $$E = 917 \times 10^{-3}$$ Finally, we have: $$x = 917$$
Question 58
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
The number of correct statement/s from the following is ________.
Larger the activation energy, smaller is the value of the rate constant.
The higher is the activation energy, higher is the value of the temperature coefficient.
At lower temperatures, increase in temperature causes more change in the value of k than at higher temperature.
A plot of $\ln k$ vs $\frac{1}{T}$ is a straight line with slope equal to $\frac{-E_a}{R}$
Answer: c
Solution
$(A)$: $k = Ae^{-E_a/RT}$. As $E_a$ increases, $k$ decreases. $(B)$: Temperature coefficient $= \frac{k_{(T+10)}}{k_T}$. $(C)$: Option $(C)$ is wrong. $\Delta k$ may be greater or lesser depending on temperature. $(D)$: $\ln k = \ln A - \frac{E_a}{RT}$.
Question 59
Chemistry · Co-ordination Compounds · Fill in the blank
The d-electronic configuration of $[\mathrm{CoCl}_4]^{2-}$ in tetrahedral crystal field is $e^m t_2^n$. Sum of 'm' and 'number of unpaired electrons is _______.
Answer: 7
Solution
$\mathrm{Co}^{2+}: 3d^7\,4s^0,\qquad \mathrm{Cl^-}:\ \text{Weak Field Ligand}$ $\begin{array}{c c c} \underline{\uparrow} & \underline{\uparrow} & \underline{\uparrow} \\ t_2 & & \\ \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} & \\ e & & \end{array}$ Configuration: $e^4\,t_2^3$ Number of unpaired electrons $=3$ $m=4$ Therefore, the answer is $7$.
Question 60
Chemistry · Haloalkanes and Haloarenes · Numerical
Number of moles of AgCl formed in the following reaction is _______.
Answer: 2
Solution
Benzylic and tertiary carbocations are stable.
Maths
Question 61
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $p, q \in \mathbb{R}$ and $\left(1 - \sqrt{3}i\right)^{200} = 2^{199} \left(p + iq\right)$, $i = \sqrt{-1}$ Then $p + q + q^2$ and $p - q + q^2$ are roots of the equation.
For three positive integers $p$, $q$, $r$, $x^{pq} = y^{qr} = z^{p^r}$ and $r = pq + 1$ such that $3$, $3 \log_y x$, $3 \log_z y$, $7 \log_x z$ are in A.P. with common difference $\frac{1}{2}$. Then $r - p - q$ is equal to
2
6
12
-6
Answer: (a)
Solution
Given the equations: $$pq^2 = \log_x \lambda$$ $$qr = \log_y \lambda$$ $$p^2 r = \log_z \lambda$$ We have: $$\log_y x = \frac{qr}{pq^2} = \frac{r}{pq} .......(1)$$ $$\log_x z = \frac{pq^2}{p^2 r} = \frac{q^2}{pr} .......(2)$$ $$\log_z y = \frac{p^2 r}{qr} = \frac{p^2}{q} .......(3)$$ The sequence $3, \frac{3r}{pq}, \frac{3p^2}{q}, \frac{7q^2}{pr}$ is in arithmetic progression (A.P). Solving: $$\frac{3r}{pq} - 3 = \frac{1}{2}$$ From this, we find: $$r = \frac{7}{6} pq .......(4)$$ Finally, we have: $$r = pq + 1$$
Question 63
Maths · Binomial Theorem · Single correct
The value $\sum_{r=0}^{22} {}^{22}C_{r}\;{}^{23}C_{r}$ is
Let a tangent to the curve $y^2 = 24x$ meet the curve $xy = 2$ at the points A and B. Then the mid points of such line segments AB lie on a parabola with the
directrix $4x = 3$
directrix $4x = -3$
Length of latus rectum $\frac{3}{2}$
Length of latus rectum 2
Answer: (a)
Solution
Given $y^2 = 24x$, $a = 6$, $xy = 2$. $AB \equiv ty = x + 6t^2 \ldots (1)$ $AB \equiv T = S_1$ $kx + hy = 2hk \ldots (2)$ From (1) and (2) $$\frac{k}{1} = \frac{h}{-t} = \frac{2hk}{-6t^2}$$ Then locus is $y^2 = -3x$. Therefore directrix is $4x = 3$.
Question 65
Maths · Limits and Derivatives · Single correct
$\lim_{t\to0} \Bigg( \frac{1}{1^{\sin^2 t}} +\frac{1}{2^{\sin^2 t}} +\cdots+ \frac{1}{n^{\sin^2 t}} \Bigg)^{{\sin^2 t}}$ is equal to
$n^2 + n$
n
$\frac{n(n+1)}{2}$
$n^2$
Answer: (b)
Solution
The limit is given by $$\lim_{t \to 0} \left( 1^{\csc^2 t} + 2^{\csc^2 t} + \ldots + n^{\csc^2 t} \right) \sin^2 t$$ which simplifies to $$\lim_{t \to 0} n \left( \left( \frac{1}{n} \right)^{\csc^2 t} + \left( \frac{2}{n} \right)^{\csc^2 t} + \ldots + 1 \right)^{\sin^2 t}$$ and further simplifies to $$n.$$
Question 66
Maths · Mathematical Reasoning · Single correct
The compound statement $\left(\sim (P \land Q)\right) \lor \left((\sim P) \land Q\right) \Rightarrow \left((\sim P) \land (\sim Q)\right)$ is equivalent to
Let $r = (\sim (P \land Q)) \lor ((\sim P) \land Q)$; $s = ((\sim P) \land (\sim Q))$. Option (A): $((\sim P) \lor Q) \land ((\sim Q) \lor P)$ is equivalent to (not of only P) $\land$ (not of only Q) $=$ (Both P, Q) and (neither P nor Q).
Question 67
Maths · Relations and Functions · Single correct
The relation $R = \{(a, b) : \gcd(a, b) = 1, 2a \neq b, a, b \in \mathbb{Z}\}$ is: ___
transitive but not reflexive
symmetric but not transitive
reflexive but not symmetric
neither symmetric nor transitive
Answer: (d)
Solution
Reflexive: $(a, a) \Rightarrow \gcd(a, a) = 1$ which is not true for every $a \in \mathbb{Z}$. Symmetric: Take $a = 2$, $b = 1 \Rightarrow \gcd(2, 1) = 1$. Also $2a = 4 \neq b$. Now when $a = 1$, $b = 2 \Rightarrow \gcd(1, 2) = 1$. Also now $2a = 2 = b$. Hence $a = 2b$. $\Rightarrow R$ is not Symmetric. Transitive: Let $a = 14$, $b = 19$, $c = 21$. $$\gcd(a, b) = 1$$ $$\gcd(b, c) = 1$$ $$\gcd(a, c) = 7$$ Hence not transitive. $\Rightarrow R$ is neither symmetric nor transitive.
Question 68
Maths · Matrices · Single correct
If $A$ and $B$ are two non-zero $n \times n$ matrices such that $A^2 + B = A^2 B$, then
$AB = I$
$A^2 B = I$
$A^2 = I$ or $B = I$
$A^2 B = B A^2$
Answer: (d)
Solution
Given $A^2 + B = A^2 B$. $$\left( A^2 - I \right) \left( B - I \right) = I \ldots (1)$$ $A^2 + B = A^2 B$ implies $$A^2 \left( B - I \right) = B$$ Thus, $$A^2 = B \left( B - I \right)^{-1}$$ Also, $$A^2 = B \left( A^2 - I \right)$$ This gives $$A^2 = B A^2 - B$$ Rearranging, $$A^2 + B = B A^2$$ Therefore, $$A^2 B = B A^2$$
Question 69
Maths · Probability · Single correct
Let N denote the number that turns up when a fair die is rolled. If the probability that the system of equations $$x + y + z = 1$$ $$2x + Ny + 2z = 2$$ $$3x + 3y + Nz = 3$$ has unique solution is $\frac{k}{6}$, then the sum of value of k and all possible values of N is
18
19
20
21
Answer: (c)
Solution
Given the system of equations: $$x + y + z = 1$$ $$2x + Ny + 2z = 2$$ $$3x + 3y + Nz = 3$$ The determinant is: $$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & N & 2 \\ 3 & 3 & N \end{vmatrix}$$ This simplifies to: $$(N - 2)(N - 3)$$ For a unique solution, $\Delta \neq 0$. So $N \neq 2, 3$. Therefore, $P$ (system has unique solution) $= \frac{4}{6}$. So $k = 4$. Therefore, the sum $= 4 + 1 + 4 + 5 + 6 = 20$.
Question 70
Maths · Determinants · Single correct
Let $\alpha$ be a root of the equation $(a-c)x^2 + (b-a)x + (c-b) = 0$ where $a, b, c$ are distinct real numbers such that the matrix $$\begin{vmatrix} \alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c \end{vmatrix}$$ is singular. Then the value of $$\frac{(a-c)^2}{(b-a)(c-b)} + \frac{(b-a)^2}{(a-c)(c-b)} + \frac{(c-b)^2}{(a-c)(b-a)}$$ is
6
3
9
12
Answer: (b)
Solution
Given $\Delta = 0 = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \end{vmatrix}$. This implies $\alpha^2 (c-b) - \alpha (c-a) + (b-a) = 0$. It is singular when $\alpha = 1$. $$\frac{(a-c)^2}{(b-a)(c-b)} + \frac{(b-a)^2}{(a-c)(c-b)} + \frac{(c-b)^2}{(a-c)(b-a)}$$ $$= \frac{(a-b)^3 + (b-c)^3 + (c-a)^3}{(a-b)(b-c)(c-a)}$$ $$= \frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)} = 3$$
Question 71
Maths · Inverse Trigonometric Functions · Single correct
$\tan^{-1}\!\left(\dfrac{1+\sqrt3}{3+\sqrt3}\right) +\sec^{-1}\!\left( \sqrt{\dfrac{8+4\sqrt3}{6+3\sqrt3}} \right)$ is equal to
$\frac{\pi}{4}$
$\frac{\pi}{2}$
$\frac{\pi}{3}$
$\frac{\pi}{6}$
Answer: (c)
Solution
Given $$\tan^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right) + \sec^{-1}\left(\frac{\sqrt{8+4\sqrt{3}}}{\sqrt{6+3\sqrt{3}}}\right)$$ This simplifies to $$= \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) + \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{3}$$
Question 72
Maths · Relations and Functions · Single correct
The equation $x^2 - 4x + [x] + 3 = x[x]$, where $[x]$ denotes the greatest integer function, has:
exactly two solutions in $(-\infty, \infty)$
no solution
a unique solution in $(-\infty, 1]$
a unique solution in $(-\infty, \infty)$
Answer: (d)
Solution
Given $x^2 - 4x + [x] + 3 = x[x]$. This implies $x^2 - 4x + 3 = x[x] - [x]$. Therefore, $(x-1)(x-3) = [x](x-1)$. This gives $x = 1$ or $x - 3 = [x]$. Thus, $x - [x] = 3$. This implies $\{x\} = 3$ (Not Possible). Only one solution $x = 1$ in $(-\infty, \infty)$.
Question 73
Maths · Continuity and Differentiability · Single correct
Let $f(x) = \begin{cases} x^2 \sin \left( \frac{1}{x} \right), & x \neq 0 \\ 0, & x = 0 \end{cases};$ Then at $x = 0$
Maths · Applications of Integrals · Single correct
The area enclosed by the curves $y^2 + 4x = 4$ and $y - 2x = 2$ is :
$\frac{25}{3}$
$\frac{22}{3}$
9
$\frac{23}{3}$
Answer: (c)
Solution
The equations of the curves are given by $y^2 + 4x = 4$ and $y^2 = -4(x-1)$. The area $A$ is calculated as follows: $$A = \int_{-4}^{2} \left( \frac{4-y^2}{4} - \frac{y-2}{2} \right) \, dy = 9.$$
Question 75
Maths · Differential Equations · Single correct
Let $y = y(x)$ be the solution of the differential equation $x^3 \, dy + (xy - 1) \, dx = 0$, $x > 0$, $y\left(\frac{1}{2}\right) = 3 - e$. Then $y(1)$ is equal to
Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that $\frac{QA}{AR} = \frac{RB}{BP} = \frac{PC}{CQ} = \frac{1}{2}$. Then $\frac{Area(\triangle PQR)}{Area(\triangle ABC)}$ is equal to
4
3
2
$\frac{5}{2}$
Answer: (b)
Solution
Let $P$ is $\vec{0}$, $Q$ is $\vec{q}$ and $R$ is $\vec{r}$. A is $\frac{2\vec{q} + \vec{r}}{3}$, B is $\frac{2\vec{r}}{3}$ and C is $\frac{\vec{q}}{3}$. Area of $\triangle PQR$ is $\frac{1}{2} \left| \vec{q} \times \vec{r} \right|$. Area of $\triangle ABC$ is $\frac{1}{2} \left| \overrightarrow{AB} \times \overrightarrow{AC} \right|$. $\overrightarrow{AB} = \frac{\vec{r} - 2\vec{q}}{3}$, $\overrightarrow{AC} = \frac{\vec{r} - \vec{q}}{3}$. Area of $\triangle ABC = \frac{1}{6} \left| \vec{q} \times \vec{r} \right|$. $$\frac{Area(\triangle PQR)}{Area(\triangle ABC)} = 3$$
Question 78
Maths · Three Dimensional Geometry · Single correct
The distance of the point $(7, -3, -4)$ from the plane passing through the points $(2, -3, 1)$, $(-1, 1, -2)$ and $(3, -4, 2)$ is:
4
5
5$\sqrt{2}$
4$\sqrt{2}$
Answer: (c)
Solution
Equation of Plane is $$\begin{vmatrix} x-2 & y+3 & z-1 \\ -3 & 4 & -3 \\ 4 & -5 & 4 \end{vmatrix} = 0$$ $$x - z - 1 = 0$$ Distance of P (7, -3, -4) from Plane is $$d = \frac{|7 + 4 - 1|}{\sqrt{2}} = 5\sqrt{2}$$
Question 79
Maths · Three Dimensional Geometry · Single correct
The distance of the point (-1, 9, -16) from the plane $2x + 3y - z = 5$ measured parallel to the line $$\frac{x+4}{3} = \frac{2-y}{4} = \frac{z-3}{12}$$ is
13$\sqrt{2}$
31
26
20$\sqrt{2}$
Answer: (c)
Solution
Equation of line $$\frac{x+1}{3} = \frac{y-9}{-4} = \frac{z+16}{12}$$ G.P on line $$(3\lambda - 1, -4\lambda + 9, 12\lambda - 16)$$ Point of intersection of line and plane $$6\lambda - 2 - 12\lambda + 27 - 12\lambda + 16 = 5$$ $$\lambda = 2$$ Point $$(5, 1, 8)$$ Distance $$= \sqrt{36 + 64 + 576} = 26$$
Question 80
Maths · Probability · Single correct
Let $\Omega$ be the sample space and A $\subseteq$ $\Omega$ be an event. Given below are two statements: (S1): If P(A) = 0, then A = $\phi$ (S2): If P(A) = 1, then A = $\Omega$ Then
only (S1) is true
only (S2) is true
both (S1) and (S2) are true
both (S1) and (S2) are false
Answer: (d)
Solution
Let $\Omega$ be the sample space and $A$ be an event. If $P(A) = 0$, then $A = \emptyset$. If $P(A) = 1$, then $A = \Omega$. Therefore, both statements are true.
Question 81
Maths · Complex Numbers and Quadratic Equations · Numerical
Let $\lambda\in\mathbb{R}$ and let the equation E be $$|x|^2-2|x|+\lambda-3=0$$. Then the largest element in the set $S=\{x+\lambda:x\text{ is an integer solution of }E\}$ is
Answer: 5
Solution
$|x|^2-2|x|+|\lambda-3|=0$ $|x|^2-2|x|+|\lambda-3|-1=0$ $(|x|-1)^2+|\lambda-3|=1$ At $\lambda=3$, $x=0$ and $2$. At $\lambda=4$ or $2$, then $x=1$ or $-1$. So, the maximum value of $x+\lambda$ =5
Question 82
Maths · Permutations and Combinations · Numerical
A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is
Answer: 546
Solution
For at most two language courses $$= \binom{5}{2} \times \binom{7}{3} + \binom{5}{1} \times \binom{7}{4} + \binom{7}{5} = 546$$
Question 83
Maths · Permutations and Combinations · Numerical
The number of 9 digit numbers, that can be formed using all the digits of the number 123412341 so that the even digits occupy only even places, is
Answer: 60
Solution
Even digits occupy at even places $$\frac{4!}{2!2!} \times \frac{5!}{2!3!} = \frac{24 \times 120}{4 \times 12} = 60$$
Question 84
Maths · Sequences and Series · Numerical
The $4^{th}$ term of GP is 500 and its common ratio is $\frac{1}{m}$, m $\in$ $\mathbb{N}$. Let $S_n$ denote the sum of the first n terms of this GP. If $S_6 > S_5 + 1$ and $S_7 < S_6 +$ $\frac{1}{2}$, then the number of possible values of m is ______
Answer: 12
Solution
Given $T_4 = 500$ where $a =$ first term, $r =$ common ratio $= \frac{1}{m}$, $m \in \mathbb{N}$. $$ar^3 = 500$$ $$\frac{a}{m^3} = 500$$ $$S_n - S_{n-1} = ar^{n-1}$$ $$S_6 > S_5 + 1 and S_7 - S_6 1$$ $$\frac{a}{m^6} 1$$ $$m^3 > 10^3$$ $$\frac{500}{m^2} > 1 m > 10 \ldots (2)$$ $$m^2 < 500 \ldots (1)$$ From (1) and (2) $m = 11, 12, 13, \ldots, 22$ So number of possible values of $m$ is 12
Question 85
Maths · Binomial Theorem · Numerical
Suppose $\sum_{r=0}^{2023} r^2\;{}^{2023}C_{r}=2023\times\alpha\times2^{2022}$ Then the value of $\alpha$ is
Answer: 1012
Solution
Using the result $\sum_{r=0}^{n} r^2\;{}^{n}C_{r}=n(n+1)\cdot2^{\,n-2}$ Then $\sum_{r=0}^{2023} r^2\;{}^{2023}C_{r}=2023\times2024\times2^{2021}$ $=2023\times1012\times2^{2022}$ So, $\Rightarrow \alpha=1012$
Question 86
Maths · Conic Sections · Numerical
Let a tangent to the Curve $9x^2 + 16y^2 = 144$ intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is
Answer: 7
Solution
Equation of tangent at point $P(4 \cos \theta, 3 \sin \theta)$ is $$\frac{x \cos \theta}{4} + \frac{y \sin \theta}{3} = 1$$ So $A$ is $(4 \sec \theta, 0)$ and point $B$ is $(0, 3 \csc \theta)$. Length $AB = \sqrt{16 \sec^2 \theta + 9 \csc^2 \theta}$ $$= \sqrt{25 + 16 \tan^2 \theta + 9 \cot^2 \theta} \geq 7$$
Question 87
Maths · Conic Sections · Numerical
Let C be the largest circle centred at (2, 0) and inscribed in the ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$. If (1, $\alpha$) lies on C, then $10 \alpha^2$ is equal to _______
Answer: 118
Solution
Equation of normal of ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$ at any point $P \,(6 \cos \theta, 4 \sin \theta)$ is $3 \sec \theta x - 2 \csc \theta y = 10$. This normal is also the normal of the circle passing through the point $(2, 0)$. So, $6 \sec \theta = 10$ or $\sin \theta = 0$ (Not possible) $\cos \theta = \frac{3}{5}$ and $\sin \theta = \frac{4}{5}$ so point $P = \left( \frac{18}{5}, \frac{16}{5} \right)$ So the largest radius of circle $$r = \frac{\sqrt{320}}{5}$$ So the equation of circle $(x-2)^2 + y^2 = \frac{64}{5}$ Passing it through $(1, \alpha)$ Then $\alpha^2 = \frac{59}{5}$ $10 \alpha^2 = 118$
Question 88
Maths · Integrals · Numerical
The value of $$\frac{8}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} \, dx$$ is _______.
The value of $12 \int_{0}^{3} \left| x^2 - 3x + 2 \right| \, dx$ is ______
Answer: 22
Solution
Given $$12 \int_0^3 \left| x^2 - 3x + 2 \right| \, dx$$ This is equal to $$12 \int_0^3 \left( \left( x - \frac{3}{2} \right)^2 - \frac{1}{4} \right) \, dx$$ If $$x - \frac{3}{2} = t$$ then $$dx = dt$$ Thus, $$= 24 \int_0^{3/2} \left| t^2 - \frac{1}{4} \right| \, dt$$ This can be split as $$= 24 \left[ -\int_0^{1/2} \left( t^2 - \frac{1}{4} \right) \, dt + \int_{1/2}^{3/2} \left( t^2 - \frac{1}{4} \right) \, dt \right] = 22$$
Question 90
Maths · Three Dimensional Geometry · Numerical
The shortest distance between the lines $\frac{x-2}{3} = \frac{y+1}{2} = \frac{z-6}{2}$ and $\frac{x-6}{3} = \frac{1-y}{2} = \frac{z+8}{0}$ is equal to
Answer: 14
Solution
The shortest distance between the lines is given by the determinant: $ \begin{vmatrix} 4 & 2 & -14 \\ 3 & 2 & 2 \\ 3 & -2 & 0 \end{vmatrix} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 3 & -2 & 0 \end{vmatrix} $ Calculating the determinant, we have: $$ = \frac{16 + 12 + 168}{\sqrt{(-4)^2 + 6^2 + (-12)^2}} = \frac{196}{14} = 14 $$