datavorous/entrance-exam-dataset huggingface dataset:
The supporting cells of nervous tissue is

Options:
A neurons.
B neuroglia.
C myelinated fibres.
D both A & C

The correct answer is:
neuroglia.

Solution:
Neurons and glial cells make the nervous system. Neurons are specialized cells that are responsible for transmission of signals. glial cells support neurons. Neurons are main conducting cells but occupy less that one of the volume while supporting cells occupy more than half of the volume. These cells do no show conduction but can regenerate. Oligodendrocytes, Schwann cells, microglial cells are its examples. These cells protect neurons by phagocytosis.

Topic: Biology, Structural Organisation in Animals, NEET
The supportive skeletal structures in the human external ears and in the nose tip are examples of

Options:
A ligament
B areolar tissue
C bone
D cartilage

The correct answer is:
cartilage

Solution:
Yellow elastic cartilage is found in external ear or pinna, external auditory canal, eustachian tube, tip of nose, epiglottis and some laryngeal components. It brings about the quick recovery of the organ as soon as the stress is released.

Topic: Biology, Structural Organisation in Animals, NEET, NEET 2012 (Mains)
The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is

Options:
A 9
B 7
C 5
D 3

The correct answer is:
9

Solution:
Let surface area of the spherical balloon A=4πr2

dAdt=8πrdrdt=k (let) ...1

On integrating on both sides w.r.t t , we get

4πr2=kt+C .

Given that, at t=0, r=3.

⇒36π=C

Also given that, at t=5, r=7

⇒4π×49=5k+36π

⇒5k=4π49-9 ⇒5k=4π×40

⇒k=32π

On substituting k value in equation 1we get, 4πr2=32πt+36π

⇒r2=8t+9

Given t=9.

⇒r2=81⇒r=9.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (24 Jun Shift 1)
The surface charge density of a thin charged disc of radius $\mathrm{R}$ is $\sigma$. The value of the electric field at the centre of the disc is $\frac{\sigma}{2 \in_0}$. With respect to the field at the centre, the electric field along the axis at a distance $\mathrm{R}$ from the centre of the disc:

Options:
A
reduces by $70.7 \%$

B
reduces by $29.3 \%$

C
reduces by $9.7 \%$

D
reduces by $14.6 \%$

The correct answer is:
reduces by $70.7 \%$

Solution:
Electric field intensity at the centre of the disc.
$$
\mathrm{E}=\frac{\sigma}{2 \epsilon_0} \quad \text { (given) }
$$
Electric field along the axis at any distance $x$ from the centre of the disc
$$
\mathrm{E}^{\prime}=\frac{\sigma}{2 \epsilon_0}\left(1-\frac{\mathrm{x}}{\sqrt{\mathrm{x}^2+\mathrm{R}^2}}\right)
$$
From question, $\mathrm{x}=\mathrm{R}$ (radius of disc)
$$
\begin{aligned}
\therefore \mathrm{E}^{\prime} &=\frac{\sigma}{2 \epsilon_0}\left(1-\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{R}^2}}\right) \\
&=\frac{\sigma}{2 \epsilon_0}\left(\frac{\sqrt{2} \mathrm{R}-\mathrm{R}}{\sqrt{2} \mathrm{R}}\right)
\end{aligned}
$$
$$
=\frac{4}{14} \mathrm{E}
$$
$\therefore \%$ reduction in the value of electric field
$$
=\frac{\left(E-\frac{4}{14} E\right) \times 100}{E}=\frac{1000}{14} \%=70.7 \%
$$

Topic: Physics, Electrostatics, JEE Main
The surface charge density of a thin charged disc of radius $\mathrm{R}$ is $\sigma$. The value of the electric field at the centre of the disc is $\frac{\sigma}{2 \in_0}$. With respect to the field at the centre, the electric field along the axis at a distance $\mathrm{R}$ from the centre of the disc:

Options:
A
reduces by $70.7 \%$

B
reduces by $29.3 \%$

C
reduces by $9.7 \%$

D
reduces by $14.6 \%$

The correct answer is:
reduces by $70.7 \%$

Solution:
Electric field intensity at the centre of the disc.
$$
\mathrm{E}=\frac{\sigma}{2 \epsilon_0} \quad \text { (given) }
$$
Electric field along the axis at any distance $x$ from the centre of the disc
$$
\mathrm{E}^{\prime}=\frac{\sigma}{2 \epsilon_0}\left(1-\frac{\mathrm{x}}{\sqrt{\mathrm{x}^2+\mathrm{R}^2}}\right)
$$
From question, $\mathrm{x}=\mathrm{R}$ (radius of disc)
$$
\begin{aligned}
\therefore \mathrm{E}^{\prime} &=\frac{\sigma}{2 \epsilon_0}\left(1-\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{R}^2}}\right) \\
&=\frac{\sigma}{2 \epsilon_0}\left(\frac{\sqrt{2} \mathrm{R}-\mathrm{R}}{\sqrt{2} \mathrm{R}}\right)
\end{aligned}
$$
$$
=\frac{4}{14} \mathrm{E}
$$
$\therefore \%$ reduction in the value of electric field
$$
=\frac{\left(E-\frac{4}{14} E\right) \times 100}{E}=\frac{1000}{14} \%=70.7 \%
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2013 (25 Apr Online)
The surface of a metal is illuminated alternately with photons of energies E1=4 eV and E2=2.5 eV respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal in (eV) is..........

Options:

The correct answer is:
2

Solution:
Given,

Energy of two photons are E1=4 eV and E2=2.5 eV

The ratio of maximum speeds of the photoelectrons emitted in the two cases is v1v2=2

Using Einstein equation of photoelectric effect,

KEmax=12mv2=E-ϕ ...(1)

Where, ϕ is the work function of metal and E is the energy of photon

Now using equation for both the cases we get,

12mv12=4-ϕ …(2)

12mv22=2.5-ϕ ...(3)

Dividing equation (2) and (3) and substitute given values, we get,

v12v22=4-ϕ2.5-ϕ=22

⇒3ϕ=6

⇒ϕ=2 eV

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2020 (05 Sep Shift 2)
The surface of a metal is illuminated with the light of $400 \mathrm{~nm}$. The kinetic energy of the ejected photoelectrons was found to be $1.68 \mathrm{eV}$. The work function of the metal is $(\mathrm{hc}=1240 \mathrm{eV} \mathrm{nm})$

Options:
A
$3.09 \mathrm{eV}$

B
$1.41 \mathrm{eV}$

C
$151 \mathrm{eV}$

D
$1.68 \mathrm{eV}$

The correct answer is:
$1.41 \mathrm{eV}$

Solution:
$$
\begin{aligned}
& \frac{1}{2} \mathrm{mv}^2=\mathrm{eV}_{\mathrm{o}}=1.68 \mathrm{eV} \Rightarrow \mathrm{hv}=\frac{\mathrm{hc}}{\lambda}=\frac{1240 \mathrm{evnm}}{400 \mathrm{~nm}}=3.1 \mathrm{eV} \Rightarrow 3.1 \mathrm{eV}=\mathrm{W}_0+1.6 \mathrm{eV} \\
& \therefore \quad \mathrm{W}_0=1.42 \mathrm{eV}
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2009
The surface of certain metal is first illuminated with light of wavelength Î»1=350 nm and then, by a light of wavelength Î»2=540 nm. It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to
(Energy of photon =1240Î»in nm eV )

Options:
A 2.5
B 1.8
C 5.6
D 1.4

The correct answer is:
1.8

Solution:
λ1=350 nm,  λ2=540 nm, V1V2=2⇒V1=2V and V2=V
hcλ1=ϕ+12m2V2
hcλ2=ϕ+12mV2
hcλ1-4hcλ2=ϕ-4ϕ
hc4540-1340=3ϕ
ϕ=1.8 eV

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The surface of certain metal is first illuminated with light of wavelength Î»1=350 nm and then, by a light of wavelength Î»2=540 nm. It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to
(Energy of photon =1240Î»in nm eV )

Options:
A 2.5
B 1.8
C 5.6
D 1.4

The correct answer is:
1.8

Solution:
Î»1=350 nm,ââÎ»2=540ânm, V1V2=2âV1=2V and V2=V
hcÎ»1=Ï+12m2V2
hcÎ»2=Ï+12mV2
hcÎ»1-4hcÎ»2=Ï-4Ï
hc4540-1340=3Ï
Ï=1.8 eV

Topic: Physics, Dual Nature of Matter, JEE Main
The surface of copper gets tarnished by the formation of copper oxide. N2 gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the N2 gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below:

2Cus+ H2Ogâ Cu2O(s) + H2(g)

PH2 Is the minimum partial pressure of H2 (in bar) needed to prevent the oxidation at 1250 K. The value of lnâ¡pH2 is ____.

(Given: total pressure = 1 bar, R (universal gas constant) = 8 J K-1 mol-1,lnâ¡10 = 2.3. Cu(s) and Cu2O(s) are mutually immiscible.

At 1250 K: 2Cus+12 O2gâ Cu2O(s); âGâ = - 78,000 J mol-1

H2g+12O2gâ H2O(g); ââ = - 1,78,000 J mol-1; G is the Gibbs energy)

Options:

The correct answer is:
-14.6

Solution:
2Cus+14O2gâ1 Cu2Os âG0=-78 kJ
H2g+12O2âH2Og âGo=-178 kJÃ-1
Hence, 2Cus+H2OgâCu2O+H2g âGo=+100kJ
âG=âGo+RTlnâ¡Q
0=+100+81000Ã1250lnâ¡PH2PH2O
-100Ã10008=1250lnâ¡PH21100Ã1
lnâ¡PH2=-14.6

Topic: Chemistry, Thermodynamics (C), JEE Main
The surface of the metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV, the work function of the metal is

Options:
A 1.51 eV

B 1.42 eV

C 3.0 eV

D 1.68 eV

The correct answer is:
1.42 eV

Solution:
As we know
ϕ0=hcλ-Ek (Einstein's equation)
= ϕ0=1240 eV nm400 nm-1.68 eV
= 3.1-1.68=1.42 eV

Topic: Physics, Dual Nature of Matter, JEE Main
The surface of water in a water tank of cross section area 750 cm2 on the top of a house is h m. above the tap level. The speed of water coming out through the tap of cross section area 500 mm2 is 30 cm s-1. At that instant, dhdt is x×10-3 m s-1. The value of x will be ______.

Options:

The correct answer is:
2

Solution:
Using continuity equation

A1v1=A2v2

We have,

750×10-4v1=500×10-6×0.3

v1=500×3×10-3750

=2×10-3 m s-1

Thus, dhdt=-2×10-3 m s-1

Hence, the value of x=2.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2023 (01 Feb Shift 2)
The surface tension and vapour pressure of water at 20°C are 7.28×10-2 Nm-1 and 2.33×103 Pa respectively. At this temperature, the radius of the smallest possible drop, which can form without evaporating is

Options:
A 1.25Ã10-5 m
B 6.25×10-5 m
C 4.3×10-5 m
D 3.4×10-5 m

The correct answer is:
6.25×10-5 m

Solution:
Here, S=7.28×10-2Nm-1,P=2.33×103Pa
The drop will evaporate if the water pressure is greater than the vapour pressure P
As P=2SR where R is the radius of drop
∴R=2SP=2×7.28×10-22.33×103=6.25×10-5

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The surface tension of liquid is $0.5 \mathrm{~N} / \mathrm{m}$. If a film is held on a ring of area $0.02 \mathrm{~m}^2$, its surface energy is

Options:
A $5 \times 10^{-2}$ joule
B $2.0 \times 10^{-2}$ joule
C $4 \times 10^{-4}$ joule
D $0.8 \times 10^{-1}$ joule

The correct answer is:
$2.0 \times 10^{-2}$ joule

Solution:
Surface energy $=T \times \Delta A=0.5 \times 2 \times(0.02)=2 \times 10^{-2} J$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The surface tension of soap solution is 3.5×10-2 N m-1. The amount of work done required to increase the radius of soap bubble from 10 cm to 20 cm is _____×10-4 J . (take π=227)

Options:

The correct answer is:
264

Solution:
The amount of work done is equal to the change in the surface energy to increase the radius of the bubble.

The formula to calculate the change in surface energy is given by

ΔU=2(T×ΔA) ...1

Substitute the values of the known parameters into equation (1) to calculate the required work done.

∆U=3.5×10-2×4π(0.2)2-(0.1)2×2 J=3.5×10-2×4×227×0.03×2=2×1.32×10-2=264×10-4 J

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2023 (11 Apr Shift 2)
The susceptibility of a magnetism at $300 \mathrm{~K}$ is $1.2 \times 10^{-5}$. The temperature at which the susceptibility increases to $1.8 \times 10^{-5}$ is

Options:
A $150 \mathrm{~K}$
B $200 \mathrm{~K}$
C $250 \mathrm{~K}$
D $20 \mathrm{~K}$

The correct answer is:
$200 \mathrm{~K}$

Solution:
$\begin{aligned} & \chi=\frac{\mathrm{C}}{\mathrm{T}} \\ & \Rightarrow \frac{\chi_{1}}{\chi_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} \Rightarrow \frac{1.2 \times 10^{-5}}{1.8 \times 10^{-5}}=\frac{\mathrm{T}_{2}}{300} \\ & \Rightarrow \mathrm{T}_{2} \Rightarrow \frac{12}{18} \times 300=200 \mathrm{~K} \end{aligned}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
The susceptibility of a paramagnetic material is 99. The permeability of the material in Wb/A-m, is [Permeability of free space μ0=4π×10-7 Wb/A-m]

Options:
A 4π×10-5
B 4π×10-6
C 4π×10-7
D 4π×10-4

The correct answer is:
4π×10-5

Solution:
Using the relation of magnetic susceptibility χ and relative permeabilityμr.

μr=1+χ=1+99=100

Magnetic permeability, μ=μ0μr

=4π×10-7×100

μ=4π×10-5

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2022 (27 Jun Shift 1)
The symbolic representation of four logic gates are given below :
(i)
(ii)
(iii)
(iv)
The logic symbols for OR, NOT and NAND gates are respectively :

Options:
A (i), (iii), (iv)
B (iii), (iv), (ii)
C (iv), (i), (iii)
D (iv), (ii), (i)

The correct answer is:
(iv), (ii), (i)

Solution:

Topic: Physics, Semiconductors, JEE Main
The symbolic representation of four logic gates are given below :
(i)
(ii)
(iii)
(iv)
The logic symbols for OR, NOT and NAND gates are respectively :

Options:
A (i), (iii), (iv)
B (iii), (iv), (ii)
C (iv), (i), (iii)
D (iv), (ii), (i)

The correct answer is:
(iv), (ii), (i)

Solution:

Topic: Physics, Semiconductors, NEET, NEET 2009 (Mains)
The symmetric equation of lines $3 x+2 y+z-5=0$ and $x+y-2 z-3=0$, is

Options:
A $\frac{x-1}{5}=\frac{y-4}{7}=\frac{z-0}{1}$
B $\frac{x+1}{5}=\frac{y+4}{7}=\frac{z-0}{1}$
C $\frac{x+1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}$
D $\frac{x-1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}$

The correct answer is:
$\frac{x+1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}$

Solution:
Let $a, b, c$ be the direction ratios of required line. $\therefore 3 a+2 b+c=0$ and $a+b-2 c=0$
$\Rightarrow \quad \frac{a}{-4-1}=\frac{b}{1+6}=\frac{c}{3-2}$
$\Rightarrow$
$\frac{a}{-5}=\frac{b}{7}=\frac{c}{1}$
In order to find a point on the required line we put $z=0$ in the two given equations to obtain, $3 x+2 y=5$ and $x+y=3$. Solving these two equations, we get $x=-1, y=4$. $\therefore$ Coordinates of point on required line are $(-1,4,0)$
Hence, required line is $\frac{x+1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The synonym for water gas when used in the production of methanol is:

Options:
A syn gas
B laughing gas
C natural gas
D fuel gas

The correct answer is:
syn gas

Solution:
Water gas is produced by the reaction of hot coke with blast of steam.

C+H2O→CO+H2 (Water gas or Syn gas);

Water gas is also called syn gas because it is used for synthesis of methanol.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2019 (10 Apr Shift 1)
The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are

Options:
A
$\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and $$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}
$$

B
$\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and
$$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}
$$

C
$\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and $$
\mathrm{CH}_3 \mathrm{C} \equiv \equiv \mathrm{CH}
$$

D
$\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and
$$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}
$$a)

The correct answer is:
$\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and
$$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}
$$a)

Solution:
$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_2$ $-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$ There are two (marked) possibility of halides
Thus, in the given context
$$
\begin{array}{r}
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}+\mathrm{NaNH}_2 \longrightarrow \\
\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{C} \equiv \overline{\mathrm{CNa}^{+}}
\end{array}
$$
$$
\begin{aligned}
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \overline{\mathrm{CNa}} \\
\longrightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \\
\text { 3-octyne } \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3
\end{aligned}
$$
Hydrocarbon
Conceptual, reaction understanding II

Topic: Chemistry, Hydrocarbons, JEE Main
The synthesis of alkyl fluorides is best accomplished by:

Options:
A Swarts reaction
B Free radical fluorination
C Sandmeyer's reaction
D Finkelstein reaction

The correct answer is:
Swarts reaction

Solution:
Alkyl fluorides is best accomplished by swarts reaction i.e. heating an alkyl chloride/bromide in the presence of metallic fluoride such as AgF, Hg2F2, CoF2, SbF3.
CH 3 Br + AgF → CH 3F + AgBr

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2015 (04 Apr)
The synthesis of non-essential amino acids like glutamic acid, aspartic acid, etc. takes place during

Options:
A catabolism.
B anabolism.
C both (a) and (b)
D none of these

The correct answer is:
anabolism.

Solution:
The formation or synthesis of molecules is called as anabolic reactions. Glutamic acid is the main amino acid from which the transfer of NH2, the amino group takes place and other amino acids are formed through transamination. The enzyme transaminase catalyzes all such reactions. For example, the two most important amides â asparagine and glutamine â found in plants are a structural part of proteins. They are formed from two amino acids, namely aspartic acid and glutamic acid, respectively, by addition of another amino group to each.

Topic: Biology, Mineral Nutrition, NEET
The system is released form rest with both the springs in unstretched position. Mass of each block is 1 kg and force constant of each spring is 10 N/m. Assume pulleys and strings are massless and all contacts are smooth.Then which of the following is incorrect. (g = 10 m/s2)

Options:
A Extension of horizontal spring in equilibrium is 2/5 m
B Extension of vertical spring in equilibrium is 1/5 m
C Maximum speed of the block A is 8 5 m/s
D Maximum speed of the block A is 1 4 m/s

The correct answer is:
Maximum speed of the block A is 1 4 m/s

Solution:
The system is released...

T = 2 kx ⇒ 5 kx = mg
x = mg 5 k = 1 × 1 0 5 × 1 0 = 1 5
(ii) 1 g x = 1 2 1 v 2 + 1 2 1 2 v 2 x + 1 2 1 0 0.2 2 + 1 2 1 0 0.4 2
2 = V 2 2 1 + 4 + 1 2 0.4 + 1.6 ⇒ v = 2 5
v A = 2 v = 2 5 2 = 8 5
So only wrong option is (iv)

Topic: Physics, Work Power Energy, JEE Main
The system of equations $2 x+y-5=0$, $x-2 y+1=9,2 x-14 y-a=0$, is consistent. Then, $a$ is equal to

Options:
A 1
B 2
C 5
D None of these

The correct answer is:
None of these

Solution:
Given system of equations are
$$
\begin{array}{l}
2 x+y-5=0 ...(i)\\
x-2 y+1=0...(ii)
\end{array}
$$
and $2 x-14 y-a=0...(ii)$
This system is consistent
$$
\begin{array}{l}
\therefore\left|\begin{array}{ccc}
2 & 1 & -5 \\
1 & -2 & 1 \\
2 & -14 & -\mathrm{a}
\end{array}\right|=0 \\
\Rightarrow 2(2 \mathrm{a}+14)-1(-\mathrm{a}-2)-5(-14+4)=0 \\
\Rightarrow 4 \mathrm{a}+28+\mathrm{a}+2+50=0 \\
\Rightarrow 5 \mathrm{a}=-80 \Rightarrow \mathrm{a}=-16
\end{array}
$$

Topic: Mathematics, Determinants, JEE Main
The system of equations $3 x+2 y+z=6$, $3 x+4 y+3 z=14$ and $6 x+10 y+8 z=a$, has infinite number of solutions, if $a$ is equal to

Options:
A $8$
B $12$
C $24$
D $36$

The correct answer is:
$36$

Solution:
Given system of equation is
$$
\begin{gathered}
3 x+2 y+z=6 \\
3 x+4 y+3 z=14 \\
6 x+10 y+8 z=a \\
A=\left[\begin{array}{rrr}
3 & 2 & 1 \\
3 & 4 & 3 \\
6 & 10 & 8
\end{array}\right], B=\left[\begin{array}{r}
6 \\
14 \\
a
\end{array}\right]
\end{gathered}
$$
Here, $\quad A=\left[\begin{array}{rrr}3 & 2 & 1 \\ 3 & 4 & 3 \\ 6 & 10 & 8\end{array}\right], B=\left[\begin{array}{r}6 \\ 14 \\ a\end{array}\right]$
$$
\begin{aligned}
& C_{11}=(32-30)=2, C_{12}=-(24-18)=-6, \\
& C_{13}=(30-24)=6 \\
& C_{21}=-(16-10)=-6, C_{22}=(24-6)=18, \\
& C_{23}=-(30-12)=-18 \\
& C_{31}=(6-4)=2, C_{32}=-(9-3)=-6, \\
& C_{33}=(12-6)=6
\end{aligned}
$$
$$
\begin{aligned}
\operatorname{adj} A & =\left[\begin{array}{lll}
C_{11} & C_{12} & C_{13} \\
C_{21} & C_{22} & C_{23} \\
C_{31} & C_{32} & C_{33}
\end{array}\right]^{\prime}=\left[\begin{array}{rrr}
2 & -6 & 6 \\
-6 & 18 & -18 \\
2 & -6 & 6
\end{array}\right]^{\prime} \\
& =\left[\begin{array}{rrr}
2 & -6 & 2 \\
-6 & 18 & -6 \\
6 & -18 & 6
\end{array}\right]
\end{aligned}
$$
So,
$$
\begin{aligned}
& \text { A) } B=\left[\begin{array}{rrr}
2 & -6 & 2 \\
-6 & 18 & -6 \\
6 & -18 & 6
\end{array}\right]\left[\begin{array}{r}
6 \\
14 \\
a
\end{array}\right] \\
& =\left[\begin{array}{c}
12-84+2 a \\
-36+252-6 a \\
36-252+6 a
\end{array}\right]=\left[\begin{array}{c}
-72+2 a \\
216-6 a \\
-216+6 a
\end{array}\right] \\
& |A|=\left|\begin{array}{rrr}
3 & 2 & 1 \\
3 & 4 & 3 \\
6 & 10 & 8
\end{array}\right| \\
& =3(32-30)-2(24-18)+1(30-24) \\
& =3(2)-2(6)+6=6-12+6=0 \\
&
\end{aligned}
$$
and
We know that, if $|A|=0$ and $(\operatorname{adj} A) \cdot B=0$, then the system of equations is consistent and has an infinite number of solutions.
$$
\begin{gathered}
(\operatorname{adj} A) \cdot B=0 \\
\Rightarrow \quad\left[\begin{array}{c}
-72+2 a \\
216+6 a \\
-216+6 a
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]
\end{gathered}
$$
On comparing, we get
$$
\begin{aligned}
2 a-72 & =0 \\
a & =36
\end{aligned}
$$
$$
\therefore \quad a=36
$$

Topic: Mathematics, Determinants, JEE Main
The system of equations
-kx+3y-14z=25

-15x+4y-kz=3

-4x+y+3z=4

Question: is consistent for all k in the set

Options:
A R
B R--11,13
C R--13
D R--11,11

The correct answer is:
R--11,11

Solution:
For consistent solutions ∆≠0

⇒-k3-14-154-k-413≠0

⇒ -k12+k-3-45-4k-14-15+16≠0

⇒ -12k-k2+135+12k-14≠0

⇒ k2≠121

⇒ k≠11,-11

Topic: Mathematics, Determinants, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The system of equations kx+k+2y+k-2z=0, k+2x+ky+k+4z=0 and k-2x+k+4y+kz=0 has a non-trivial solution for

Options:
A exactly one real value of k
B exactly two real values of k
C infinite real values of k
D no real value of k

The correct answer is:
exactly one real value of k

Solution:
For a non-trivial solution,
∆=0
kk+2k-2k+2kk+4k-2k+4k=0
⇒kk2-k+42-k+2k2+2k-k2+2k-8+k-2k2+6k+8-k2+2k=0
⇒-8k2-16k-8k-16+8k2-8k-16=0
⇒-32k-32=0⇒k=-1

Topic: Mathematics, Determinants, JEE Main
The system of equations kx+y+z=1, x+ky+z=k and x+y+zk=k2 has no solution if k is equal to:

Options:
A 0
B 1
C -1
D -2

The correct answer is:
-2

Solution:
kx+y+z=1

x+ky+z=k

x+y+zk=k2

Δ=K111K111K=KK2-1-1K-1+11-K

=K3-K-K+1+1-K

=K3-3 K+2

=(K-1)2 K+2

For K=1

Δ=Δ1=Δ2=Δ3=0

But for K=-2, at least one out of Δ1,Δ2,Δ3 are not zero

Hence for no solution, K=-2

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The system of equations \( k x+y+z=1, x+k y+z=k \) and \( x+y+z k=k^{2} \) has no solution if \( k \) is equal to:

Options:
A \( 0 \)
B \( 1 \)
C \( -1 \)
D \( -2 \)

The correct answer is:
\( -2 \)

Solution:
kx+y+z=1

x+ky+z=k

x+y+zk=k2

Δ=K111K111K=KK2-1-1K-1+11-K

=K3-K-K+1+1-K

=K3-3 K+2

=(K-1)2 K+2

For K=1

Δ=Δ1=Δ2=Δ3=0

But for K=-2, at least one out of Δ1,Δ2,Δ3 are not zero

Hence for no solution, K=-2

Topic: Mathematics, Determinants, JEE Main
The system of equations αx-1+y+z=-1, x+αy-1+z=-1 and x+y+αz-1=-1 has no solution, if α is equal to

Options:
A -2 or 1
B -2
C 1
D -1

The correct answer is:
-2

Solution:
Rearranging the given equations, we get,
αx+y+z=α-1
x+αy+z=α-1
x+y+αz=α-1
For no solution, Δ=0 and atleast one of Δ1,Δ2,Δ3 must be non zero.
Δ=α111α111α=αα2-1-1α-1+11-α
=α-1α2+α-1-1=0
⇒α+2α-12=0
⇒α=-2,1
Δ1=α-111α-1α1α-11α=α-1α2-1-1α-12-1α-12
=α-12α+1-2=α-13
Δ2=α-13=Δ3 for no solution α=-2

Topic: Mathematics, Determinants, JEE Main
The system of equations x+py=0, y+pz=0 and z+px=0 has infinitely many solutions for

Options:
A p=1
B p=0
C p=-1
D no real value of p

The correct answer is:
p=-1

Solution:
For infinite solutions, 1p001pp01=1+p3=0
Hence, p=-1

Topic: Mathematics, Determinants, JEE Main