datavorous/entrance-exam-dataset huggingface dataset:
If the vectors $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}=\lambda \hat{\mathrm{i}}+\hat{\mathrm{j}}+\mu \hat{\mathrm{k}}$ are mutually orthogonal, then $(\lambda, \mu)=$

Options:
A
$(2,-3)$

B
$(-2,3)$

C
$(3,-2)$

D
$(-3,2)$

The correct answer is:
$(-3,2)$

Solution:
$$
\begin{aligned}
& \vec{a} \cdot \vec{b}=0, \quad \vec{b} \cdot \vec{c}=0, \quad \vec{c} \cdot \vec{a}=0 \\
& \Rightarrow 2 \lambda+4+\mu=0 \quad \lambda-1+2 \mu=0 \\
&
\end{aligned}
$$
Solving we get: $\lambda=-3, \mu=2$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2010
If the vectors a→=λi^+μj^+4k^,b→=-2i^+4j^-2k^ and c→=2i^+3j^+k^ are coplanar and the projection of a→ on the vector b→ is 54 units, then the sum of all possible values of λ+μ is equal to

Options:
A 0
B 6
C 24
D 18

The correct answer is:
24

Solution:
Given,

a→=λi^+μj^+4k^,b→=-2i^+4j^-2k^ and c→=2i^+3j^+k^ are coplanar

So, λμ4-24-2231=0

⇒λ(10)-μ(2)+4(-14)=0

⇒10λ-2μ=56

⇒5λ-μ=28 .....1

And given projection of a→ on b→ is 54

⇒a→·b→|b→|=54

⇒-2λ+4μ-824=54

⇒-2λ+4μ-8=54×24 ....2

By solving equation 1 & 2 we get,

⇒λ+μ=24

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (29 Jan Shift 1)
If the vectors AB→=3i^+4k^ and AC→=5i^-2j^+4k^ are the sides of a triangle ABC, then the length of the median through A is:

Options:
A 3 3
B 4 5
C 1 8
D 7 2

The correct answer is:
3 3

Solution:
Length of median =AB→+AC→2
=8i^-2j^+8k^2
= 3 3

Topic: Mathematics, Vector Algebra, JEE Main
If the vectors AB→=3i^+4k^ and AC→=5i^-2j^+4k^ are the sides of a triangle ABC, then the length of the median through A is:

Options:
A 3 3
B 4 5
C 1 8
D 7 2

The correct answer is:
3 3

Solution:
Length of median =AB→+AC→2
=8i^-2j^+8k^2
= 3 3

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2013 (07 Apr)
If the vectors $\mathbf{i}+3 \mathbf{j}-2 \mathbf{k}, 2 \mathbf{i}-\mathbf{j}+4 \mathbf{k}$ and $3 \mathbf{i}+2 \mathbf{j}+x \mathbf{k}$ are coplanar, then the value of $x$ is

Options:
A -2
B 2
C 1
D 3

The correct answer is:
2

Solution:
Vectors $\mathbf{i}+3 \mathbf{j}-\mathbf{2} ; \mathbf{k} \mathbf{i} \mathbf{i} \mathbf{j}+\mathbf{4} \mathbf{k}$ and $3 \mathbf{i}+2 \mathbf{j}+\mathbf{x} \mathbf{k}$. We know that as the vectors are coplanar, therefore
$\begin{aligned}
& \left|\begin{array}{rrr}
1 & 3 & -2 \\
2 & -1 & 4 \\
3 & 2 & x
\end{array}\right|=0 \Rightarrow 1(-x-8)-3(2 x-12)-2(4+3)=0 \\
& \Rightarrow-x-8-6 x+36-14=0 \Rightarrow 7 x=14 \Rightarrow x=2
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
If the vectors $\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}},-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$ represents
the diagonals of a parallelogram, then its area will be

Options:
A $\sqrt{21}$
B $\frac{\sqrt{21}}{2}$
C $2 \sqrt{21}$
D $\frac{\sqrt{21}}{4}$

The correct answer is:
$\frac{\sqrt{21}}{2}$

Solution:
Let
$\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$
Now, $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\left|\begin{array}{rrr}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \mathbf{k} \\ 1 & -3 & 2 \\ -1 & 2 & 0\end{array}\right|$
$=-4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
Hence, required area $=\frac{1}{2}|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|$
$=\frac{1}{2} \sqrt{16+4+1}=\frac{\sqrt{21}}{2}$

Topic: Mathematics, Vector Algebra, JEE Main
If the vectors $\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \lambda \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ are orthogonal to each other, then $\lambda$ is equal to

Options:
A $5$
B $-5$
C $8$
D $-8$

The correct answer is:
$8$

Solution:
Let $\mathbf{a}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and
$\mathbf{b}=\lambda \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+\hat{\mathbf{k}}$
Since, $\mathbf{a}$ and $\mathbf{b}$ are orthogonal to each other
$\begin{array}{lc}
\therefore & \mathbf{a} \cdot \mathbf{b}=0 \\
\Rightarrow & (\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(\lambda \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\hat{\mathbf{k}})=0 \\
\Rightarrow & \lambda-12+4=0 \Rightarrow \lambda=8
\end{array}$

Topic: Mathematics, Vector Algebra, JEE Main
If the vectors, p→=a+1i^+aj^+ak^,q→=ai^+a+1j^+ak^ and r→=ai^+aj^+a+1k^a∈R are coplanar and 3p→.q→2-λr→×q→2=0 , then the value of λ is ________

Options:

The correct answer is:
1

Solution:
a+1aaaa+1aaaa+1=0⇒a+1+a+a=0⇒a=-13
Pâ=23i^-13j^-k^
Qâ=13-i^+2j^-k^
Râ=13-i^-j^+2k^
Pâ.Qâ=19-2-2+1=-13
RâÃQâ=19ijk-12-1-1-12=19i4-1-j-2-1+k1+2=193i+3j+3k=i+j+k3
RâÃQâ=133⇒RâÃQâ2=13 3Pâ.Qâ2-Î»RâÃQâ2=0
3.19-Î».13=0⇒Î»=1

Topic: Mathematics, Vector Algebra, JEE Main
If the vectors, p→=a+1i^+aj^+ak^,q→=ai^+a+1j^+ak^ and r→=ai^+aj^+a+1k^a∈R are coplanar and 3p→.q→2-λr→×q→2=0 , then the value of λ is ________

Options:

The correct answer is:
1

Solution:
a+1aaaa+1aaaa+1=0⇒a+1+a+a=0⇒a=-13
Pâ=23i^-13j^-k^
Qâ=13-i^+2j^-k^
Râ=13-i^-j^+2k^
Pâ.Qâ=19-2-2+1=-13
RâÃQâ=19ijk-12-1-1-12=19i4-1-j-2-1+k1+2=193i+3j+3k=i+j+k3
RâÃQâ=133⇒RâÃQâ2=13 3Pâ.Qâ2-Î»RâÃQâ2=0
3.19-Î».13=0⇒Î»=1

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2020 (09 Jan Shift 1)
If the vectors
( x 2 â1 ) i ^ +2( x 2 â1 ) j ^ â3( x 2 â1 ) k ^ ,
( 2 x 2 â1 ) i ^ +( 2 x 2 +1 ) j ^ + x 2 k ^ ââ and
( 3 x 2 +2 ) i ^ +( x 2 +4 ) j ^ +( x 2 +1 ) k ^
are non-coplanar, then the number of real values x cannot take is

Options:
A 1
B 2
C 4
D 6

The correct answer is:
2

Solution:
For non-coplanar vectors

x2-12x2-1-3x2-12x2-12x2+1x23x2+2x2+4x2+1≠0
x2-112-32x2-12x2+1x23x2+2x2+4x2+1≠0
x2-112x4+3x2+1-x4-4x2-22x4+x2-1-3x4-2x2-32x4+7x2-4-6x4-7x2-2≠0
x2-1x4-x2+1+2x4+2x2+2+12x4+18≠0
x2-115x4+x2+21≠0
⇒x≠+1,-1

Topic: Mathematics, Vector Algebra, JEE Main
If the velocity of a body related to displacement x is given by v=5000+24x m s-1, then the acceleration of the body is _________ m s-2.

Options:

The correct answer is:
12

Solution:
We know that the rate of change of linear velocity is called linear acceleration, a=dvdt=dv×dxdt×dx=vdvdx

⇒a=5000+24x×125000+24x×24=12 m s-2.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (27 Aug Shift 1)
If the velocity of light $C$, the gravitational constant $G$ and Planck's constant $h$ are chosen as the fundamental units, the dimension of density in the new system is

Options:
A $C^3 G^{-2} h^1$
B $C^5 G^{-2} h^{-1}$
C $C^{-3 / 2} G^{-1 / 2} h^{1 / 2}$
D $C^{9 / 2} G^{-1 / 2} h^{-1 / 2}$

The correct answer is:
$C^5 G^{-2} h^{-1}$

Solution:
Let $\rho=\mathrm{kC}^{\mathrm{a}} \mathrm{G}^{\mathrm{b}} \mathrm{h}^{\mathrm{c}}$
$\begin{aligned}
& \Rightarrow \quad[\rho]=[\mathrm{C}]^{\mathrm{a}}[\mathrm{G}]^{\mathrm{b}}[\mathrm{h}]^{\mathrm{c}} \\
& \Rightarrow \quad \mathrm{ML}^{-3}=\mathrm{L}^{\mathrm{a}} \mathrm{T}^{-\mathrm{a}} \mathrm{M}^{-\mathrm{b}} \mathrm{L}^{3 \mathrm{~b}} \mathrm{~T}^{-2 \mathrm{~b}} \mathrm{M}^{\mathrm{c}} \mathrm{L}^{2 \mathrm{c}} \mathrm{T}^{-\mathrm{c}} \\
& \Rightarrow \quad 1=-\mathrm{b}+\mathrm{c},-3=\mathrm{a}+3 \mathrm{~b}+2 \mathrm{c}, 0=-\mathrm{a}-2 \mathrm{~b}-\mathrm{c} \\
& \Rightarrow \quad \mathrm{a}=5, \mathrm{~b}=-2 \text { and } \mathrm{c}=-1
\end{aligned}$

Topic: Physics, Units and Dimensions, JEE Main
If the velocity of sound in air is $350 \mathrm{~m} / \mathrm{s}$. Then the fundamental frequency of an open organ pipe of length $50 \mathrm{~cm}$, will be

Options:
A 350 Hz
B 175 Hz
C 900 Hz
D 750 Hz

The correct answer is:
350 Hz

Solution:
Given $; l=50 \mathrm{~cm}, v=350 \mathrm{~m} / \mathrm{s}$
Fundamental frequency of open pipe
$\nu=\frac{v}{2 l}=\frac{350}{2 \times 0.5}=350 \mathrm{~Hz}$

Topic: Physics, Waves and Sound, JEE Main
If the velocity-time graph has the shape AMB, what would be the shape of the corresponding acceleration-time graph?

Options:
A
B
C
D

The correct answer is:

Solution:
Slope of v-t graph gives acceleration

⇒Acceleration will be

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (24 Feb Shift 1)
If the vertex of a parabola be at origin and directrix be $x+5=0$, then its latus rectum is

Options:
A $5$
B $10$
C $20$
D $40$

The correct answer is:
$20$

Solution:
$S \equiv(5,0)$. Therefore, latus rectum $=4 a=20$.

Topic: Mathematics, Parabola, JEE Main
If the vertex opposite to the hypotenuse of a right-angled triangle lies on the straight line 2x+y-10=0 and the two other vertices are 2,-3 and 4,1, then the area of the triangle (in sq. units) is equal to

Options:
A 10
B 3
C 335
D 11

The correct answer is:
3

Solution:
m1m2=-1â9-2aa-4Ã13-2aa-2=-1
â5a2-50a+125=0

So, a=5 and B is 5,0
Hence, required area =12ABÃAC=122Ã32=3 sq. units

Topic: Mathematics, Straight Lines, JEE Main
If the vertices of a hyperbola be at -2, 0 and 2, 0 and one of its foci be at -3, 0, then which one of the following points does not lie on this hyperbola ?

Options:
A 6, 52
B -6, 210
C 26, 5
D 4, 15

The correct answer is:
6, 52

Solution:
Since, the vertex of the hyperbola are A′-2, 0 and A2, 0.

Hence, the transverse axis of the hyperbola is x- axis and centre is O0,0.

Equation of hyperbola can be taken as

x2a2-y2b2=1

⇒a=2 and ae=3

⇒e=32

Since, b2=a2e2-1⇒b2=a2e2-a2=9-4=5

Hence, equation of hyperbola is x24-y25=1

Clearly, point 6, 52 doesn't lies on the hyperbola.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2019 (12 Jan Shift 1)
If the vertices of a triangle are $(-2,6),(3,-6)$ and $(1,5)$, then the area of the triangle is

Options:
A 40 sq. units
B 15.5 sq. units
C 30 sq. units
D 35 sq. units

The correct answer is:
15.5 sq. units

Solution:
Given vertices of triangle are $(-2,6),(3,-6)$ and $(1,5)$.
$$
\begin{aligned}
& \text { Area of triangle }=\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right| \\
&= \frac{1}{2}\left|\begin{array}{ccc}
-2 & 6 & 1 \\
3 & -6 & 1 \\
1 & 5 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}
-5 & 12 & 0 \\
2 & -11 & 0 \\
1 & 5 & 1
\end{array}\right| \\
& {\left[R_1 \leftrightarrow R_1-R_2, R_2 \leftrightarrow R_2-R_3\right] } \\
&= \frac{1}{2}[1(55-24)]=\frac{31}{2}=15.5 \text { sq. units }
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Main
If the volume of a gas is doubled while keeping the temperature constant, how does this affect the molar volume (Vm)?

Options:
A Vm decreases.
B Vm increases.
C Vm remains the same.
D Vm becomes zero.

The correct answer is:
Vm remains the same.

Solution:
Molar volume (Vm) is defined as the volume per mole of substance and remains constant when the volume is changed while keeping temperature constant.

Topic: Chemistry, Thermodynamics (C), NEET
If the volume of a parallelopiped, whose coterminous edges are given by the vectors a→=i^+j^+nk^ , b→=2i^+ 4j^- nk^ and,c→=i^+nj^+3k^ (n≥0) is 158 cubic units, then :

Options:
A a→·c→=17
B b→·c→=10
C n=7
D n=9

The correct answer is:
b→·c→=10

Solution:
The volume of the parallelepiped is, v=a→ b→ c→

⇒11n24-n1n3=±158

⇒112+n2-1(6+n)+n(2n-4)=±158

⇒3n2-5n-152=0 or 3n2-5n+164=0

(∵ D<0, no real roots of 3n2-5n+164=0)

n=8, -193 ⇒ n=8, as n≥0

then, b→·c→=2+4n-3n=10

a→·c→=1+n+3n=33

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2020 (05 Sep Shift 1)
If the volume of a spherical ball is increasing at the rate of 4π cc/sec then the rate of increase of its radius in cm/sec, when the volume is 288π cc is

Options:
A 1 9
B 1 6
C 1 2 4
D 1 3 6

The correct answer is:
1 3 6

Solution:
Volume of sphere of radius r is V=43πr3

Given V=288π cc

⇒288π=43πr3

⇒r3=288×34=216

⇒r=6 cm.

Again, V=43πr3

Differentiate w.r.t. 't'

⇒dVdt=4πr2drdt

Put the given values, to get

⇒4π=4π×62×drdt

⇒drdt=136 cm/sec.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2014 (19 Apr Online)
If the volume of parallelepiped formed by the vectors i^+λj^+k^,j^+λk^ and λi^+k^ is minimum, then λ is equal to:

Options:
A -13
B -3
C 3
D 13

The correct answer is:
13

Solution:
Volume of parallelopiped =A→B→C→
Let f(λ)=1λ101λλ01 =λ3-λ+1
f'(λ)=3λ2-1=3λ+13λ-1
sign conversion for f'λ
at λ=13,f'λ changes its sign from - ve to + ve
∴ λ=13, is the point of minima.
⇒ At λ=13 volume of parallelopiped is minimum.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2019 (12 Apr Shift 1)
If the volume of parallelopiped with conterminus edges $4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}},-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}+p \hat{\mathbf{k}}$ is 34 cubic units, then $p$ is equal to :

Options:
A 4
B -13
C 13
D 6

The correct answer is:
-13

Solution:
Coterminus edges of a parallelopiped are $4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}},-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}+p \hat{\mathbf{k}}$
Volume of parallelopiped $=34$
$\Rightarrow \quad\left|\begin{array}{rrr}4 & 5 & 1 \\ 0 & -1 & 1 \\ 3 & 9 & p\end{array}\right|=34$
$\Rightarrow \quad 4\left|\begin{array}{cc}-1 & 1 \\ 9 & p\end{array}\right|-5\left|\begin{array}{cc}0 & 1 \\ 3 & p\end{array}\right|+1\left|\begin{array}{cc}0 & -1 \\ 3 & 9\end{array}\right|=34$
$\Rightarrow \quad 4(-p-9)-5(-3)+1(3)=34$
$\Rightarrow \quad-4 p-36+15+3=34$
$\Rightarrow \quad 4 p=-36+18-34$
$\Rightarrow \quad p=-\frac{52}{4}=-13$

Topic: Mathematics, Vector Algebra, JEE Main
If the volume of the given mass of a gas is increased four times, the temperature is raised from $27^{\circ} \mathrm{C}$ to $127^{\circ} \mathrm{C}$. The elasticity will become

Options:
A 4 times
B $1 / 4$ times
C 3 times
D $1 / 3$ times

The correct answer is:
$1 / 3$ times

Solution:
From the ideal gas equation
$\frac{E_2}{E_1}=\frac{P_2}{P_1}=\frac{V_1}{V_2} \times \frac{T_2}{T_1}=\left(\frac{1}{4}\right) \times\left(\frac{400}{300}\right)=\frac{1}{3}=E_2=\frac{E_1}{3}$
i.e. elasticity will become $\frac{1}{3}$ times.

Topic: Physics, Mechanical Properties of Solids, JEE Main
If the volume of the parallelepiped formed by the vectors aâÃbâ, bâÃcâ and câÃaâ is 36 cubic units, then the volume (in cubic units) of the tetrahedron formed by the vectors aâ+bâ, bâ+câ and câ+aâ is equal to

Options:
A 12
B 6
C 2
D 1

The correct answer is:
2

Solution:
Given, a→×b→ b→×c→ c→×a→=36
⇒a→b→c→2=36
⇒a→b→c→=±6
Volume of the tetrahedron =16a→+b→ b→+c→ c→+a→

=13a→b→c→=2 cubic units

Topic: Mathematics, Vector Algebra, JEE Main
If the volume of the parallelopiped formed by three non-coplanar vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ is 4 cu units, then $[\mathbf{a} \times \mathbf{b} \mathbf{b} \times \mathbf{c} \mathbf{c} \times \mathbf{a}]$ is equal to

Options:
A 64
B 16
C 4
D 8

The correct answer is:
16

Solution:
In scalar triple product, the position of dot and cross can be changed provided the cyclic order is maintained i.e.,
$$
[\mathbf{a} \mathbf{b} \mathbf{c}]=(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}=\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})
$$
Put $\mathbf{c} \times \mathbf{a}=\mathbf{n}$
$\therefore[\mathbf{a} \times \mathbf{b} \mathbf{b} \times \mathbf{c} \mathbf{c} \times \mathbf{a}]=(\mathbf{a} \times \mathbf{b}) \cdot\{(\mathbf{b} \times \mathbf{c}) \times \mathbf{n}\}$
$=(\mathbf{a} \times \mathbf{b}) \cdot\{(\mathbf{n} \cdot \mathbf{b}) \mathbf{c}-(\mathbf{n} \cdot \mathbf{c}) \mathbf{b}\}$
$=(\mathbf{a} \times \mathbf{b}) \cdot[\{(\mathbf{c} \times \mathbf{a}) \cdot \mathbf{b}\} \mathbf{c}-\{(\mathbf{c} \times \mathbf{a}) \cdot \mathbf{c}\} \mathbf{b}]$
$=(\mathbf{a} \times \mathbf{b}) \cdot\{[\mathbf{c} \mathbf{a} \mathbf{b}] \mathbf{c}-[\mathbf{c} \mathbf{a} \mathbf{c}] \mathbf{b}\}$
$=[(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}][\mathbf{c} \mathbf{a} \mathbf{b}]-0$
$=[\mathbf{a b c}]$ [abc]
$=[\mathbf{a b c}]^{2}=(4)^{2}=16$
Alternate Method
By properties of Scalar triple product,
$[\mathbf{a} \times \mathbf{b} \mathbf{b} \times \mathbf{c} \mathbf{c} \times \mathbf{a}]=\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}^{2}\end{array}\right.$
$$
=(4)^{2}=16
$$

Topic: Mathematics, Vector Algebra, KCET, KCET 2012
If the wave equation of a particle is represented by, \( y=0.8 \sin \left(20 \pi t-5 \pi x+\frac{\pi}{4}\right) \), where \( x \) and \( y \) are measured in metre and \( t \) is measured in seconds, then the wave equation represents a wave travelling in

Options:
A the positive \( x \)-direction with velocity \( 4 \mathrm{~m} \mathrm{~s}^{-1} \) and wavelength \( 0.1 \mathrm{~m} \).
B the positive \( x \)-direction with velocity \( 4 \mathrm{~m} \mathrm{~s}^{-1} \) and wavelength \( 0.4 \mathrm{~m} \).
C the negative \( x \)-direction with velocity \( 1 \mathrm{~m} \mathrm{~s}^{-1} \) and wavelength \( 0.4 \mathrm{~m} \).
D the negative \( x \)-direction with velocity \( 4 \mathrm{~m} \mathrm{~s}^{-1} \) and wavelength \( 0.1 \mathrm{~m} \).

The correct answer is:
the positive \( x \)-direction with velocity \( 4 \mathrm{~m} \mathrm{~s}^{-1} \) and wavelength \( 0.4 \mathrm{~m} \).

Solution:
Equation of a wave is given by,

y=Asinwt-kx+ϕ

After comparing from given equation, we got,

w=20π, k=5π,v=wkv=20π5π=5 m s-1

λ=2πkλ=2π5π=25=0.4 m

So that the question represents a wave in the positive x-direction with velocity 4 m s-1 and wavelength 0.4 m

Topic: Physics, Waves and Sound, JEE Main
If the wavelength is brought down from $6000 Å$ to $4000 Å$ in a photoelectric experiment then what will happen?

Options:
A The work function of the metal will increase
B The threshold frequency will decrease
C No change will take place
D Cut off voltage will increase

The correct answer is:
Cut off voltage will increase

Solution:
When wavelength decreases, frequency increases. Also, we know that cut-off voltage (or stopping potential) increases when frequency increases.
Hence, option (d) is correct. Note that work function and threshold frequency are constant for a given metal.

Topic: Physics, Dual Nature of Matter, JEE Main
If the wavelength of the first line of the Balmer series of hydrogen is \( 6561 \mathrm{~A} \), the wavelength of the second line of the series
should be?

Options:
A \( 13122 Å \)
B \( 3280 Å \)
C \( 4860 \mathrm{~A} \)
D \( 2187 Å \)

The correct answer is:
\( 4860 \mathrm{~A} \)

Solution:
Balmer series wavelength is given by,

1λ=R122-1n2

For the first line, transition from n=3, or

1λ1=R122-132=5R36 ...i

For the second line, transition from n=4, then

1λ2=R122-142=3R16 ...ii

Dividing equation i by ii, we get,

λ2λ1=2027⇒λ2=2027×6561=4860 A∘

Topic: Physics, Atomic Physics, JEE Main
If the wavelength of the first member of Lyman series of hydrogen is λ. The wavelength of the second member will be

Options:
A 2732λ
B 3227λ
C 275λ
D 527λ

The correct answer is:
2732λ

Solution:
For Lyman series:

For first member, 1λ=13.6Z2hc112-122 ….. (i)

For second member, 1λ'=13.6Z2hc112-132 ….. (ii)

On dividing (i) and (ii), we get

λ'=2732λ

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (31 Jan Shift 1)
If the wavelength of the $n^{\text {th }}$ line of Lyman series is equal to the de-Broglie wavelength of electron in initial orbit of a hydrogen like element $(Z=11)$. Find the value of $n$.

Options:

The correct answer is:
24

Solution:
$n^{\text {th }}$ line of Lyman series means transition from $(n+1)$ th state to first state.
$$
\begin{aligned}
& \frac{1}{\lambda}=R Z^2\left[1-\frac{1}{(n+1)^2}\right] \\
& \text { de-Broglie wavelength, } \quad \lambda=\frac{h}{m v}=\frac{h r}{m v r}=\frac{(2 \pi)(h r)}{(n+1) h}=\frac{2 \pi r}{(n+1)} \\
& \frac{1}{\lambda}=\frac{(n+1)}{2 \pi r} \\
&
\end{aligned}
$$
or
Equating (i) and (ii), we get
$$
\left(\frac{n+1}{2 \pi r}\right)=R Z^2\left[\frac{n(n+2)}{(n+1)^2}\right]
$$
Now, as
$$
\begin{aligned}
& r \propto \frac{n^2}{Z} \\
& r=\frac{(n+1)^2}{11} r_0
\end{aligned}
$$
$$
\therefore \quad r=\frac{(n+1)^2}{11} r_0
$$
Substituting in equations (iii), we get
$$
\frac{11}{2 \pi r_0}=\frac{R(11)^2(n)(n+2)}{(n+1)}
$$
or
$$
(n+1)=\left(1.09 \times 10^7\right)(11)(2 \pi) \times\left(0.529 \times 10^{-10}\right)\left(n^2+2 n\right)
$$
Solving this equation we get, $n=24$

Topic: Physics, Atomic Physics, JEE Main
If the wavelenth of Kα radiation emitted by an atom of atomic number Z=41 is λ, then the atomic number for an atom that emits Kα radiation with wavelength 4λ, is

Options:
A 21
B 28
C 11
D 36

The correct answer is:
21

Solution:
1λ∝Z-12
∴λ1λ2=Z2-12Z1-12
14=Z2-141-12

12=Z2-140

Solving this, we get, Z2=21

Topic: Physics, Atomic Physics, JEE Main
If the work done in blowing a soap bubble of volume V is W, then the work done in blowing a soap bubble of volume 2V under the same conditions will be

Options:
A W
B 2W
C 2 W
D 41/3W

The correct answer is:
41/3W

Solution:
The work done will be against surface tension and we know that surface tension is related to work as W=TÃâA
Since, the surface area of sphere is 4ÏR2 and there are two free surfaces, we have

W=TÃ8ÏR2 ...(i)
and volume of sphere =43ÏR3

i.e. V=43ÏR3
âR=3V4Ï1/3 ...(ii)
From Eqs. (i) and (ii), we get
W=TÃ8ÏÃ3V4Ï2/3
â WâV2/3
â´ W1âV12/3
and, W2âV22/3
W2W1=2V1V12/3
âW2=22/3W1=41/3W

Topic: Physics, Mechanical Properties of Fluids, JEE Main
If the work function for a certain metal is $3.2 \times 10^{-19}$ joule and it is illuminated with light of frequency $8 \times 10^{14} \mathrm{~Hz}$. The maximum kinetic energy of the photoelectrons would be $\left(h=6.63 \times 10^{-34} \mathrm{Js}\right)$.

Options:
A $2.1 \times 10^{-19} \mathrm{~J}$
B $8.5 \times 10^{-19} \mathrm{~J}$
C $5.3 \times 10^{-19} \mathrm{~J}$
D $3.2 \times 10^{-19} \mathrm{~J}$

The correct answer is:
$2.1 \times 10^{-19} \mathrm{~J}$

Solution:
$\begin{aligned} & \text { K. E.max }=\text { incident energy - work function } \\ & =\mathrm{h} \gamma-3.2 \times 10^{-19} \\ & =6.63 \times 10^{-34} \times 8 \times 10^{14}-3.2 \times 10^{-19} \\ & =5.3 \times 10^{-19}-3.2 \times 10^{-19} \\ & =2.1 \times 10^{-19} \mathrm{~J}\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main
If the work function of a metal is 6.63×10-19 J, the maximum wavelength of the photon required to remove a photoelectron from the metal is____nm. Nearest integer)
[Given : h=6.63×10-34 J s, and c=3×108 m s-1]

Options:

The correct answer is:
300

Solution:
Eincident=W0+K.E.

As K.E is zero

Eincident=W0

6.63×10-19=hcλ=6.63×10-34×3×108λ

λ=3×10-7 m=300 nm.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2022 (28 Jun Shift 1)
If the work functions of three photosensitive materials are 1 eV, 2 eV and 3 eV respectively, then the ratio of the respective frequencies of light that produce photoelectrons of the maximum kinetic energy of 1 eV from each of them is

Options:
A 1Â :2Â :3
B 2 :3 :4
C 1Â :1Â :1
D 3Â :2Â :1

The correct answer is:
2 :3 :4

Solution:
From Einstein photo electric equation
hv=KE+Ï
Where, Ï= work function =hv0
KE=1 eV
hv1=1+1=2 eV
hv2=1+2=3 eV
hv3=1+3=4 eV
â v1 :v2 :v3=2 :3 :4

Topic: Physics, Dual Nature of Matter, JEE Main
If the x-intercept of a focal chord of the parabola y2=8x+4y+4 is 3, then the length of this chord is equal to _____ .

Options:

The correct answer is:
16

Solution:
Given,

The x-intercept of a focal chord of the parabola y2=8x+4y+4 is 3,

Now simplifying equation of the given parabola we get,

y2=8x+4y+4

⇒y-22=8x+1

Now on comparing with standard parabola y2=4ax we get,

a=2, X=x+1, Y=y-2

Hence, the vertex is -1,2 and focus is 1,2

So, equation of line passing through focus is given by y-2=mx-1

Now, putting 3,0 {as given x-intercept of focal chord is 3} in the above line we get, m=-1

So, equation of chord will be y=3-x,

Now putting the value of y=3-x in parabola y-22=8x+1 we get,

⇒1-x2=8x+1

⇒x2-10x-7=0

⇒x1-x2=102+4×7=128

⇒y1-y2=3-x1-3-x2=x2-x1=128

Length of focal chord x2-x12+y2-y12=256=16.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2023 (01 Feb Shift 2)