datavorous/entrance-exam-dataset huggingface dataset:
If the image of the point P1,-2, 3 in the plane, 2x+3y-4z+22=0 measured parallel to the line, x1=y4=z5 is Q, then PQ is equal to:

Options:
A 35
B 242
C 42
D 65

The correct answer is:
242

Solution:
Any line parallel to x1=y4=z5 and passing through P(1,-2,3) is

x-11=y+24=z-35=λ (say)

Any point on above line can be written as (λ+1,4λ-2,5λ+3).

∴ Coordinates of R are (λ+1,4λ-2,5λ+3).

Since, point R lies on the above plane. ∴2(λ+1)+3(4λ-2)-4(5λ+3)+22=0

⇒ λ=1

So, point R is (2,2,8)

Now, PR=(2-1)2+(2+2)2+(8-3)2=42

∴ PQ=2PR=242

Since, point R lies on the above plane.

So, point R is (2,2,8)

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2017 (02 Apr)
If the image of the point P1,-2,3 in the plane 2x+3y-4z+22=0 measured parallel to the line 20x=5y=4z is point Q, then the value of |PQ|2 is

Options:

The correct answer is:
168

Solution:
Since, distance PQ is measured parallel to the line
20x=5y=4z i.e. x1=y4=z5 .
â Direction ratios of line PQ are 1,4,5
Equation of PQ
âx-11=y+24=z-35=Î», (say)
â A point on the line is Î»+1,4Î»-2,5Î»+3â¡M , (say)
Point M lies on the plane 2x+3y-4z+22=0
â2Î»+1+34Î»-2-45Î»+3+22=0
â-6Î»+6=0âÎ»=1
âM=2,2,8
M is the midpoint of PQ
âPQ=2PM
âPQ=22-12+2+22+8-32
=21+16+25
=242
â|PQ|2=4Ã42=168

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the impurity in a metal has a greater affinity for oxygen and is more easily oxidised than the metal, then the purification of metal may be carried out by

Options:
A Poling
B Zone refining
C Electrolytic refining
D Cupellation

The correct answer is:
Cupellation

Solution:
Correct Option is 4 Cupellation
When the impurity in metal has a greater affinity for oxygen and is more readily oxidised than the metal, then purification method carried out is eupellation.
This method is used to separate noble metals like $\mathrm{Au}, \mathrm{Ag}$ from impurities like $\mathrm{Pb}, \mathrm{Cu}, \mathrm{Zn}$. present in ore.
In this method, ores are treated under right temperature and controlled operations to separate noble metals.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
If the incentre of an equilateral triangle is 1,1 and the equation of its one side is 3x+4y+3=0, then the equation of the circumcircle of this triangle is:

Options:
A x2+y2-2x-2y-2=0
B x2+y2-2x-2y+2=0
C x2+y2-2x-2y-7=0
D x2+y2-2x-2y-14=0

The correct answer is:
x2+y2-2x-2y-14=0

Solution:
3x+4y+3=0
Since, for equilateral triangle centroid in center and circumcenter is the same therefore length of AB=31+41+332+42
=105=2
R=2AB=4
As centroid divide median in the ratio 2:1
Centre of the circle would be 1,1
∴ Answer is x2+y2-2x-2y-14=0

Topic: Mathematics, Circle, JEE Main, JEE Main 2015 (11 Apr Online)
If the inequality x2+ax+a2+6a<0 is satisfied for all x∈1,2, then the sum of all the integral values of a must be equal to

Options:
A -10
B -15
C -21
D -28

The correct answer is:
-21

Solution:
fx=x2+ax+a2+6a
f1<0, f2<0 and D>0
f1<0⇒a2+7a+1<0⇒a∈-7-352,-7+352
f2<0⇒a2+8a+4<0⇒a∈-4-23,-4+23
and D>0⇒a2-4×1×a2+6a>0
⇒-3a2-24a>0⇒a2+8a<0⇒a∈-8,0
Taking intersection, we get, a∈-7-352,-4+23
Integral values of a=-6,-5,-4,-3,-2,-1
Hence, the sum of values is -21

Topic: Mathematics, Quadratic Equation, JEE Main
If the initial velocity of a projectile be doubled, keeping the angle of projection same, the maximum height reached by it will

Options:
A Remain the same
B Be doubled
C Be quadrupled
D Be halved

The correct answer is:
Be quadrupled

Solution:
$H=\frac{u^2 \sin ^2 \theta}{2 g} \therefore H \propto u^2$
If initial velocity be doubled then maximum height reached by the projectile will quadrupled.

Topic: Physics, Motion In Two Dimensions, JEE Main
If the input $\mathrm{AC}$ voltage is $10 \mathrm{Vrms}$, find the maximum voltage across the diode of a half wave rectifier with capacitor input filter.

Options:
A $14.1 \mathrm{~V}$
B $20 \mathrm{~V}$
C $10 \mathrm{~V}$
D $7.07 \mathrm{~V}$

The correct answer is:
$14.1 \mathrm{~V}$

Solution:
Concept:
Low-voltage rectifiers require a step-down transformer to reduce the strength of AC voltage.
$\frac{V_1}{V_2}=\frac{N_1}{N_2}$
A step-down transformer is necessary for:
$\mathrm{Vm}$ is the peak voltage
Calculation:
Given
$\begin{aligned}
& \mathrm{Vrms}=10 \mathrm{~V} \\
& \mathrm{Vm}=V_{r m s} \times \sqrt{2} \\
& =10 \times \sqrt{2} \\
& =14.1 \mathrm{~V}
\end{aligned}$

Topic: Physics, Semiconductors, JEE Main
If the input voltage $V_{i}$ to the circuit below is given by $V_{i}(t)=A \cos (2 \pi f t)$, the output voltage is given by $\mathrm{V}_{\mathrm{o}}(\mathrm{t})=\mathrm{B} \cos (2 \pi \mathrm{ft}+\phi)$
Which one of the following four graphs best depict the variation of $\phi$ vs $f$ ?

Options:
A
B
C
D

The correct answer is:

Solution:
Resultant of $V_{C}, V_{R} \& B$ give $V_{i}$ and angle between $V_{i} \& B$ is $\phi .$ When $f$ is very high $X_{C} \rightarrow O$ then $\mathrm{V}_{\mathrm{C}} \rightarrow \mathrm{O} \therefore$ Resultant of $\mathrm{V}_{\mathrm{C}}, \mathrm{V}_{\mathrm{R}} \& \mathrm{~B}$ lie between $\mathrm{B}$ and $\mathrm{V}_{\mathrm{R}}$
B lag behind $\phi$
$\therefore$ At higher frequency $\phi$ become - ve.

Topic: Physics, Alternating Current, JEE Main
If the integral ∫02dxsin⁡x+sin⁡(2-x)=A, then the integral β=∫02xdxsinx+sin(2-x) is equal to

Options:
A sin⁡2A
B 2A
C A
D A2

The correct answer is:
A

Solution:
Applying a+b+x property in Î², we get Î²=â«022-xsinâ¡2-x+sinâ¡xdx
Adding both, we get 2Î²=â«022sinâ¡x+sin2-xdx âÎ²=A

Topic: Mathematics, Definite Integration, JEE Main
If the integral ∫ 5 tan x tan x - 2 dx = x + a log sin x - 2 cos x + k , then the value of a is

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

The correct answer is:
2

Solution:
â« 5 tan x tan x - 2 dx = x + a ln sin x - 2 cos x + k
Differentiating both sides, we get
5 tan x tan x - 2 = 1 + a cos x + 2 sin x sin x - 2 cos x
â 5 sin x sin x - 2 cos x = sin x 1 + 2 a + cos x a - 2 sin x - 2 cos x
â a = 2

Topic: Mathematics, Indefinite Integration, JEE Main
If the integral 525∫0π2sin2x cos112x1+cos52x12dx is equal to n2−64, then n is equal to ________

Options:

The correct answer is:
176

Solution:
Let,

I=∫0π2sin2x cos112x1+cos52x12dx

Let cosx=t2⇒sinxdx=-2t dt

∴I=-4∫10t2⋅t111+t5tdt

⇒I=4∫01t141+t5dt

Now, let 1+t5=k2

⇒5t4dt=2k dk

∴I=4⋅∫12k2−12⋅k·2k5dk

⇒I=85∫12k6−2k4+k2 dk

⇒I=85k77−2k55+k3312

⇒I=85827−825+223−17+25−13

⇒I=85222105−8105

⇒525⋅I=1762−64

Hence, on comparing we get, n=176

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2024 (31 Jan Shift 1)
If the integral
$$
\int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x} d x=A \cos 8 x+k
$$
where $k$ is an arbitrary constant, then $\mathrm{A}$ is equal to:

Options:
A
$-\frac{1}{16}$

B
$\frac{1}{16}$

C
$\frac{1}{8}$

D
$-\frac{1}{8}$

The correct answer is:
$-\frac{1}{16}$

Solution:
Let $\mathrm{I}=\int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x} d x$
Now, $\mathrm{D}^r=\cot 2 x-\tan 2 x=\frac{\cos 2 x}{\sin 2 x}-\frac{\sin 2 x}{\cos 2 x}$
$$
=\frac{\cos ^2 2 x-\sin ^2 2 x}{\sin 2 x \cos 2 x}=\frac{2 \cos 4 x}{\sin 4 x}
$$
$\begin{aligned} \therefore \mathrm{I} &=\int \frac{2 \cos ^2 4 x}{\frac{2 \cos 4 x}{\sin 4 x}} d x=\int \frac{2 \cos ^2 4 x \cdot \sin 4 x}{2 \cos 4 x} d x \\ &=\frac{1}{2} \int \sin 8 x d x=-\frac{1}{2} \frac{\cos 8 x}{8}+k \\ &=-\frac{1}{16} \cdot \cos 8 x+k \\ & \text { Now },-\frac{1}{16} \cdot \cos 8 x+k=\mathrm{A} \cos 8 x+k \\ & \Rightarrow \mathrm{A}=-\frac{1}{16} \end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2013 (25 Apr Online)
If the integral Ι=∫dxx10+x=λlnx91+xμ+C, (where, C is the constant of integration) then the value of 1λ+μ is equal to

Options:
A 81
B 829
C 18
D 8

The correct answer is:
18

Solution:
I=∫x-101+x-9dx
Put 1+x-9=t
⇒-9x-10dx=dt
⇒Ι=∫dt-9t
=-19ln⁡t+C
=-19ln1+x-9+C
=19ln⁡x91+x9+C
⇒λ=19,μ=9
∴1λ+μ=9+9=18

Topic: Mathematics, Indefinite Integration, JEE Main
If the integral Ι=∫esin⁡xcos⁡x⋅x2+2xdx=efxgx+C (where, C is the constant of integration), then the number of solution(s) of fx=gx is/are

Options:
A 0
B 2
C 4
D 6

The correct answer is:
2

Solution:
As, Î=â« e f x f â² x g x + g â² x dx= e f x g x +C
Thus, â«esinâ¡xcosâ¡xâx2+2xdx=esinâ¡xâx2+C
i.e. fx=sinâ¡x & gx=x2
Which intersect at â2â points.

Topic: Mathematics, Indefinite Integration, JEE Main
If the integral Ι=∫ex2x3dx=ex2fx+c, where c is the constant of integration and f1=0, then the value of f2 is equal to

Options:
A 4
B 52
C 32
D 3

The correct answer is:
32

Solution:
Let, x2=t⇒2xdx=dt
∴Ι=∫et⋅t2dt
Using integration by parts, we get,
Ι=12tet-∫1⋅etdt
=12tet-et+c
=ex2x2-12+c
∴fx=x2-12⇒f2=32

Topic: Mathematics, Indefinite Integration, JEE Main
If the integral I=â«019Ïdx1+ecos3â¡x has the value kÏ2, then k2 is equal to

Options:

The correct answer is:
9.5

Solution:
Applying a+b-x and adding,
2I=â«019Ï11+ecos3â¡x+11+e-cos3â¡xdx
â2I=â«019Ï1dx=x019Ï
âI=19Ï2
â´k=19
Hence, k2=192=9.5

Topic: Mathematics, Definite Integration, JEE Main
If the integral I=∫0πsec-1⁡sec⁡x1+tan8⁡xdx, ∀x≠π2, then the value of I is equal to (where · is the greatest integer function)

Options:

The correct answer is:
2

Solution:
Applying a+b-x and adding, we get,

⇒2I=π∫0πdx1+tan8⁡x
⇒2I=π∫0π/211+tan8⁡x+11+tan8⁡π-xdx
⇒2I=2π∫0π2dx1+tan8⁡x=2π∫0π2cos8⁡xcos8⁡x+sin8⁡x

⇒I=π∫0π2cos8⁡xcos8⁡x+sin8⁡x
Applying a+b-x and adding,

2I=π∫0π2sin8⁡x+cos8⁡xsin8⁡x+cos8⁡xdx
⇒2I=π∫0π21dx=ππ2
⇒I=π24⇒I=2

Topic: Mathematics, Definite Integration, JEE Main
If the integral I=∫2x24+x2dx= 2x-fx+c, where f2=π, then the minimum value of y=fx ∀x∈-2,2 is (where, c is the constant of integration)

Options:
A 0
B -π
C 2π
D -4π

The correct answer is:
-π

Solution:
I=2∫x2+4-4x2+4dx
=2∫1-4x2+4dx
=2x-42tan-1⁡x2+c
=2x-4tan-1⁡x2+c
∴fx=4tan-1⁡x2
So, min⁡fx=4tan-1⁡-1=-π

Topic: Mathematics, Indefinite Integration, JEE Main
If the integral I=â«e5âlnâ¡xx6+1-1âdx =lnâ¡x6+1+C, (where, C is the constant of integration) then the value of 1Î» is

Options:

The correct answer is:
6

Solution:
The integral
I=∫e5ln⁡xx6+1-1dx
I=∫eln⁡x5x6+1dx
=∫x5x6+1dx
=16∫6x5dxx6+1
Put x6+1=t
6x5dx=dt
⇒I=16∫dtt=16ln⁡t+C
⇒I=16ln⁡x6+1+C
⇒λ=16
⇒1λ=6

Topic: Mathematics, Indefinite Integration, JEE Main
If the integral In=∫0π2sin⁡2n-1xsin⁡xdx, then the value of I203-I193 is

Options:
A 400
B 200
C 361
D 0

The correct answer is:
0

Solution:
I 20 − I 19 = ∫ 0 π 2 sin 39x −sin 37x sinx dx
⇒I20−I19=∫0π22sinxcos38xsinxdx
⇒ I 20 − I 19 = ∫ 0 π 2 2cos 38x dx
=sin38x19π20
=sin⁡38π219
=0
∴I20=I19

Hence, I203-I193=0

Topic: Mathematics, Definite Integration, JEE Main
If the integral I=∫-sin⁡xx-ln⁡xcos⁡xdx=fx+C (where, C is the constant of integration) and fe=-sin⁡e, then the number of natural numbers less than fπ6 is equal to (where, · is the greatest integer function)

Options:

The correct answer is:
0

Solution:
Given Integral is I=-â«dsinâ¡x⋅lnâ¡x
=-sinâ¡x⋅lnâ¡x+C
â´fx=-sinâ¡x⋅lnâ¡x
âfÏ6=-12lnâ¡Ï6=12lnâ¡6Ï
âfÏ6â0,1âfÏ6=0

Topic: Mathematics, Indefinite Integration, JEE Main
If the integral I=â«tanâ¡x5+7tan2â¡xdx=klnâ¡fx+C (where C is the integration constant) and f0=5, then the value of fÏ4 is equal to

Options:

The correct answer is:
6

Solution:
I=∫sin⁡xcos⁡x5cos2⁡x+7sin2⁡xdx
Let sin2⁡x=t
⇒sin⁡xcos⁡xdx=dt2
I=12∫dt51-t+7t
=12∫dt5+2t
=14ln⁡5+2t+C
=14ln⁡5+2sin2⁡x+C
∴fx=5+2sin2⁡x
⇒fπ4=6

Topic: Mathematics, Indefinite Integration, JEE Main
If the integral I=∫x51+x3dx =Kx3+1x3-2+C, (where, C is the constant of integration), then the value of 9K is equal to

Options:
A 4
B 2
C 6
D 10

The correct answer is:
2

Solution:
Let 1+x3=t2
⇒3x2dx=2tdt
⇒I=∫t2-12t dt3t
⇒I=23∫t2-1dt
=2t39-23t+C
=29tt2-3+C
=291+x3x3-2+C
∴K=29⇒9K=2

Topic: Mathematics, Indefinite Integration, JEE Main
If the integral I=∫xx-3x+3x-1x-2x+1dx=fx+C (where, x>0 and C is the constant of integration) and f1=-13, then the value of f9 is equal to

Options:
A 3
B 6
C 9
D 12

The correct answer is:
9

Solution:
Given integral is I=∫x-13x-12dx
=∫x-1dx
=23x32-x+C
∴fx=23x32-x
⇒f9=18-9=9

Topic: Mathematics, Indefinite Integration, JEE Main
If the integral ∫ln⁡xx3dx=fx4x2+C, where fe=-3 and C is the constant of integration, then the value of fe2 is equal to

Options:
A 3
B -4
C -5
D 5

The correct answer is:
-5

Solution:
∫ln⁡x⏟I1x3⏟IIdx=ln⁡x-12x2-∫1x-12x2dx
(Using integration by parts)
=-ln⁡x2x2+12x-2-2+C
=-14x22ln⁡x+1+C
⇒fx=-2ln⁡x-1
∴fe2=-2ln⁡e2-1
=-4-1=-5

Topic: Mathematics, Indefinite Integration, JEE Main
If the integral $\int \frac{5 \tan x}{\tan x-2} d x=x+a \ln |\sin x-2 \cos x|+k$, then $a$ is equal to

Options:
A
$-1$

B
$-2$

C
$1$

D
$2$

The correct answer is:
$2$

Solution:
$\int \frac{5 \tan x}{\tan x-2} d x=\int \frac{5 \sin x}{\sin x-2 \cos x} d x$
$\Rightarrow \int\left[\frac{2(\cos x+2 \sin x)+(\sin x-2 \cos x)}{\sin x-2 \cos x}\right] d x$
$=2 \int\left(\frac{\cos x+2 \sin x}{\sin x-2 \cos x}\right) d x+\int d x+k$
$=2 \log |\sin x-2 \cos x|+x+k \quad \therefore a=2$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2012 (Offline)
If the integral ∫x4+x2+1x2-x+1dx=fx+C, (where C is the constant of integration andx∈R), then the minimum value of f'x is

Options:
A 1
B 14
C 34
D 2

The correct answer is:
34

Solution:
As x4+x2+1=x2+12-x2=x2+1+xx2+1-x
Thus, ∫x4+x2+1x2-x+1dx=∫x2+x+1dx
i.e., f'x=x2+x+1=x+122+34≥34

Topic: Mathematics, Indefinite Integration, JEE Main
If the integral Î=â«xsinxcosxâlnâ¡x+sinxxdx=fxgx+c âx>0 , then the range of y=gx is (where, c is an arbitrary constant)

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

The correct answer is:
-1,1

Solution:
Let xsin⁡x=t; then,

I=∫dt=t+c=xsin⁡x+c

Hence, range of sinx∈[-1,1]

Topic: Mathematics, Indefinite Integration, JEE Main
If the integrals i1=∫0∞11+x8dx and i2=∫011(1-x8)1/8dx, then

Options:
A i1=2i2
B i2=8i1
C i1=i2
D i1=8i2

The correct answer is:
i1=i2

Solution:
In i1
put x8=tan2Î¸
â8x7dx=2tanâ¡Î¸ sec2Î¸dÎ¸
âi1=â«0Ï/22tanâ¡Î¸ sec2Î¸dÎ¸sec2Î¸Â·8Â·tan7/4Î¸
âi1=14â«0Ï/2tan-3/4Î¸dÎ¸
In i2
put x8=sin2Î¸
âi2=â«0Ï/22sin Î¸ cos Î¸dÎ¸(1-sin2Î¸)1/8Â·8sin74Î¸
=14â«0Ï/2tan-3/4Î¸dÎ¸=i1
âi2=i1

Topic: Mathematics, Definite Integration, JEE Main
If the inverse trigonometric functions take principal values, then cos-1310costan-143+25sintan-143 is equal to

Options:
A 0
B π4
C π3
D π6

The correct answer is:
π3

Solution:
Let tan-143=θ⇒tanθ=43

∴cosθ=35 & sinθ=45

So, cos-1310costan-143+25sintan-143

=cos-1310cosθ+25sinθ

=cos-1310×35+25×45

=cos-1950+825

=cos-112=π3

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2022 (26 Jun Shift 2)
If the ionisation energy and electron affinity of an element are 275 and $86 \mathrm{kcal} \mathrm{mol}^{-1}$ respectively, then the electronegativity of the element on the Mulliken scale is

Options:
A $2.8$
B $0.0$
C $4.0$
D $2.6$

The correct answer is:
$2.8$

Solution:
According to Mulliken, electronegativity of an atom is average of ionization energy and electron affinity (in $\mathrm{eV}$ ).
$\mathrm{n}_{\mathrm{m}}=\frac{\mathrm{IE}+\mathrm{EA}}{2}$
If ionization energy and electron affinity are in kcalmol $^{-1}$,
$\mathrm{n}=\frac{\mathrm{IE}+\mathrm{EA}}{125}=\frac{275+86}{125}=288$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
If the $K . E$. of a body is increased by $300 \%$, its momentum will increase by

Options:
A 100%
B 150%
C $\sqrt{300} \%$
D 175%

The correct answer is:
100%

Solution:
Let initial kinetic energy, $E_1=E$ Final kinetic energy, $E_2=E+300 \%$ of $E=4 E$
$\begin{aligned} & \text { As } P \propto \sqrt{E} \Rightarrow \frac{P_2}{P_1}=\sqrt{\frac{E_2}{E_1}}=\sqrt{\frac{4 E}{E}}=2 \\ & \Rightarrow P_2=P_1+100 \% \text { of } P_1\end{aligned} \Rightarrow P_2=2 P_1$
i.e. Momentum will increase by $100 \%$.

Topic: Physics, Work Power Energy, JEE Main
If the kinetic energies of an electron, an alpha particle and a proton having same de-Broglie wavelength are $\varepsilon_1, \varepsilon_2$ and $\varepsilon_3$ respectively, then

Options:
A $\varepsilon_1>\varepsilon_3>\varepsilon_2$
B $\varepsilon_1=\varepsilon_2=\varepsilon_2$
C $\varepsilon_1 < \varepsilon_3 < \varepsilon_2$
D $\varepsilon_1>\varepsilon_2>\varepsilon_3$

The correct answer is:
$\varepsilon_1>\varepsilon_3>\varepsilon_2$

Solution:
$\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}, \sqrt{\mathrm{mk}}=\mathrm{const}$
$\mathrm{k} \propto \frac{1}{\mathrm{~m}}, \varepsilon_1>\varepsilon_3>\varepsilon_2$

Topic: Physics, Dual Nature of Matter, JEE Main
If the kinetic energy of a body changes by $20 \%$ then its momentum would change by -

Options:
A $20 \%$
B $24 \%$
C $40 \%$
D None of the above

The correct answer is:
None of the above

Solution:
$\frac{\frac{p_f^2}{2 m}-\frac{p_i^2}{2 m}}{\frac{p_i^2}{2 m}} \times 100=20$
$\Rightarrow \frac{p_f}{p_i}=\sqrt{1.2}=1.095 \Rightarrow \frac{p_f-p_i}{p_i}=0.095$
Therefore $\%$ increase $=9.5 \%$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
If the kinetic energy of a free electron doubles. Its deBroglie wavelength changes by the factor

Options:
A
$\frac{1}{2}$

B
2

C
$\frac{1}{\sqrt{2}}$

D
$\sqrt{2}$

The correct answer is:
$\frac{1}{\sqrt{2}}$

Solution:
$\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{Km}}}$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2005
If the kinetic energy of a free electron doubles, it's de-Broglie wavelength changes by the factor

Options:
A 2
B $\frac{1}{2}$
C $\sqrt{2}$
D $\frac{1}{\sqrt{2}}$

The correct answer is:
$\frac{1}{\sqrt{2}}$

Solution:
de-Broglie wavelength,
$$
\begin{array}{l}
\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \cdot \mathrm{m} \cdot(\mathrm{K} \cdot \mathrm{E})}} \\
\therefore \lambda \propto \frac{1}{\sqrt{\mathrm{K} \cdot \mathrm{E}}}
\end{array}
$$
If $\mathrm{K} . \mathrm{E}$ is doubled, then de-Broglie wavelength becomes $\frac{\lambda}{\sqrt{2}}$

Topic: Physics, Dual Nature of Matter, JEE Main
If the kinetic energy of a moving body becomes four times its initial kinetic energy, then the percentage change in its momentum will be:

Options:
A 100%
B 200%
C 300%
D 400%

The correct answer is:
100%

Solution:
K2=4 K1

12mv22=4×12mv12

v2=2v1

P=mv

P2=mv2=2mv1

P1=mv1

% change =ΔPP1×100=2mv1-mv1mv1×100=100%

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2021 (20 Jul Shift 2)
If the kinetic energy of an electron is increased four times, the wavelength of the de-Broglie wave associated with it would become

Options:
A
one fourth

B
half

C
four times

D
two times

The correct answer is:
half

Solution:
de - Broglie wavelength is given by :
$$
\lambda=\frac{h}{m v}
$$
$$
\begin{aligned}
& \text { K.E. }=\frac{1}{2} m v^2 \\
& v^2=\frac{2 K E}{m} \\
& v=\sqrt{\frac{2 K E}{m}}
\end{aligned}
$$
Substituting this in equation (i)
$$
\lambda=\frac{h}{m} \sqrt{\frac{m}{2 K E}}
$$
$\lambda=h \sqrt{\frac{1}{2 m(\text { K.E. })}}$
i.e. $\lambda \propto \frac{1}{\sqrt{K E}}$
$\therefore$ when $K E$ become 4 times wavelength become $1 / 2$.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2012 (19 May Online)
If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is:

Options:
A 25
B 75
C 60
D 50

The correct answer is:
75

Solution:
KEfinal=16KEinitial

⇒12mvf2=16×12mvi2(mass remains same because the particle remains same)

⇒vf=4vi
We know that, De-broglie wavelength (λ)=hmv

∴λf=hmvf=h4mvi

So De-Broglie wave will decreased to 75%

Topic: Physics, Dual Nature of Matter, NEET, NEET 2014
If the L. M. V. T. holds for the function $f(x)=x+\frac{1}{x}, x \in[1,3]$, then c=

Options:
A $\sqrt{3}$
B 3
C 2
D $-\sqrt{3}$

The correct answer is:
$\sqrt{3}$

Solution:
Given $f(x)=x+\frac{1}{x}$ and LMVT holds
$f^{\prime}(x)=1-\frac{1}{x^{2}} \Rightarrow f^{\prime}(c)=1-\frac{1}{c^{2}}$
$f(1)=1+1=2 \text { and } f(3)=3+\frac{1}{3}=\frac{10}{3}$
$\therefore \quad f^{\prime}(c)=1-\frac{1}{c^{2}}=\frac{\frac{10}{3}-2}{(3-1)} \Rightarrow 1-\frac{1}{c^{2}}=\frac{4}{3(2)}=\frac{2}{3}$
$\therefore \quad \frac{1}{c^{2}}=1-\frac{2}{3}=\frac{1}{3} \Rightarrow c^{2}=3 \Rightarrow c=\pm \sqrt{3}$

Topic: Mathematics, Application of Derivatives, JEE Main
If the largest interval of x in which the function fx=x3-3x+1 is decreasing is a,b, then the value of a+2b is equal to

Options:
A –1
B 0
C 1
D 2

The correct answer is:
1

Solution:
As f'x=3x2-3=3x-1x+1

i.e. the largest interval in which fx decreases is -1,1⇒a=-1
b=1
â´a+2b=1

Topic: Mathematics, Application of Derivatives, JEE Main
If the lattice constant of this semiconductor is decreased, then which of the following is correct?

Options:
A
All $E_c, E_g, E_v$ decrease

B
All F $F_{-F}$ increase

C
$\mathrm{E}_{\mathrm{c}}$, and $\mathrm{E}_{\mathrm{v}}$ increase but $\mathrm{E}_{\mathrm{g}}$ decreases

D
$E_c$, and $E_v$, decrease $E_g$ increases

The correct answer is:
$E_c$, and $E_v$, decrease $E_g$ increases

Solution:
$=(8 \times 7.06-7 \times 5.60) \mathrm{MeV}=17.28 \mathrm{MeV}$

Topic: Physics, Semiconductors, JEE Main, JEE Main 2006
If the lattice parameter for a crystalline structure is $3.6 \%$, then the atomic radius in fcc crystal is

Options:
A $1.27 \&$
B $1.81 Å$
C $2.10 Å$
D $2.92 \&$

The correct answer is:
$1.27 \&$

Solution:
$$
2 R=\frac{a}{\sqrt{2}}
$$
$$
\therefore R=\frac{a}{2 \sqrt{2}}
$$

Topic: Chemistry, Solid State, NEET, NEET 2008 (Mains)
If the latus rectum of an hyperbola be 8 and eccentricity be $3 / \sqrt{5}$, then the equation of the hyperbola is

Options:
A $4 x^2-5 y^2=100$
B $5 x^2-4 y^2=100$
C $4 x^2+5 y^2=100$
D $5 x^2+4 y^2=100$

The correct answer is:
$4 x^2-5 y^2=100$

Solution:
$\frac{2 b^2}{a}=8$ and $\frac{3}{\sqrt{5}}=\sqrt{1+\frac{b^2}{a^2}}$ or $\frac{4}{5}=\frac{b^2}{a^2}$ $\Rightarrow a=5, b=2 \sqrt{5}$.
Hence the required equation of hyperbola is $\frac{x^2}{25}-\frac{y^2}{20}=1 \Rightarrow 4 x^2-5 y^2=100$.

Topic: Mathematics, Hyperbola, JEE Main
If the latusrectum of an ellipse subtends a right angle at the centre of that ellipse, then the eccentricity of that ellipse is

Options:
A $\frac{\sqrt{5}+1}{4}$
B $\frac{\sqrt{5}-1}{2}$
C $\frac{\sqrt{10-2 \sqrt{5}}}{5}$
D $\frac{\sqrt{10+2 \sqrt{5}}}{5}$

The correct answer is:
$\frac{\sqrt{5}-1}{2}$

Solution:
Let equation of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Let $L L^{\prime}$ is latusrectum, then co-ordinate of
$$
L\left(a e, \frac{b^2}{a}\right)
$$
$L L^{\prime}$ subtend $\pi / 2$ Angle at the centre, so angle
$$
L C S=\pi / 4
$$
$$
\begin{array}{rlrl}
\Rightarrow & \tan \frac{\pi}{4} & =\frac{\frac{b^2}{a}}{a e} \\
\Rightarrow & 1 & =\frac{b^2}{a^2 e} \\
\Rightarrow & a^2 e & =b^2=a^2\left(1-e^2\right) \\
\Rightarrow & e & =1-e^2 \\
\Rightarrow & e^2+e-1 & =0 \\
\Rightarrow & & e & =\frac{-1 \pm \sqrt{5}}{2} \Rightarrow e=\frac{\sqrt{5}-1}{2} .
\end{array}
$$

Topic: Mathematics, Ellipse, JEE Main
If the least and the largest real values of α, for which the equation z+αz-1+2i=0 z∈C and i=-1 has a solution, are p and q respectively; then 4p2+q2 is equal to_______.

Options:

The correct answer is:
10

Solution:
Put z=x+iy

x+iy+αx+iy-1+2i=0

⇒ x+αx-12+y2+iy+2=0+0i

⇒ y+2=0 and x+αx-12+y2=0

⇒ y=-2 and α2=x2x2-2x+5

Now x2x2-2x+5∈0,54

∴ α2∈0,54⇒α∈-52,52

∴ p=-52;q=52

⇒ 4p2+q2=454+54=10

Topic: Mathematics, Complex Number, JEE Main
If the least and the largest real values of α, for which the equation z+αz-1+2i=0 z∈C and i=-1 has a solution, are p and q respectively; then 4p2+q2 is equal to_______.

Options:

The correct answer is:
10

Solution:
Put z=x+iy

x+iy+αx+iy-1+2i=0

⇒ x+αx-12+y2+iy+2=0+0i

⇒ y+2=0 and x+αx-12+y2=0

⇒ y=-2 and α2=x2x2-2x+5

Now x2x2-2x+5∈0,54

∴ α2∈0,54⇒α∈-52,52

∴ p=-52;q=52

⇒ 4p2+q2=454+54=10

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (24 Feb Shift 1)
If the length of a closed organ pipe is $1.5 \mathrm{~m}$ and velocity of sound is $330 \mathrm{~m} / \mathrm{s}$, then the frequency for the second note is

Options:
A 220 Hz
B 165 HZ
C 110 Hz
D 55 Hz

The correct answer is:
165 HZ

Solution:
For closed pipe second note $=3 \mathrm{v} 41=3 \times 3304 \times 4.5=165 \mathrm{~Hz}$.

Topic: Physics, Waves and Sound, JEE Main