datavorous/entrance-exam-dataset huggingface dataset:
If $\vec{u}, \vec{v}, \bar{w}$ are non-coplanar vectors and $p, q$ are real numbers, then the equality $\left[\begin{array}{llll}3 \vec{u} & p \vec{v} & p \vec{w}\end{array}\right]-\left[\begin{array}{lll}p \vec{v} & \vec{w} & q \vec{u}\end{array}\right]-\left[\begin{array}{lll}2 \vec{w} & q \vec{v} & q \vec{u}\end{array}\right]=0$ holds for

Options:
A
exactly one value of $(p, q)$

B
exactly two values of $(p, q)$

C
more than two but not all values of $(p, q)$

D
all values of $(p, q)$

The correct answer is:
exactly one value of $(p, q)$

Solution:
$$
\begin{aligned}
& \left(3 p^2-p q+2 q^2\right)\left[\begin{array}{lll}
\vec{u} & \vec{v} & \bar{w}
\end{array}\right]=0 \\
& \text { But }\left[\begin{array}{lll}
\vec{u} & \vec{v} & \bar{w}
\end{array}\right] \neq 0 \\
& 3 p^2-p q+2 q^2=0 \\
& 2 p^2+p^2-p q+\left(\frac{q}{2}\right)^2+\frac{7 q^2}{4}=0 \Rightarrow 2 p^2+\left(p-\frac{q}{2}\right)^2+\frac{7}{4} q^2=0
\end{aligned}
$$
$$
\Rightarrow p=0, q=0, p=\frac{q}{2}
$$
This possible only when $p=0, q=0$ exactly one value of $(p, q)$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2009
If u=x2+y2 and x=s+3t, y=2s-t, then d2uds2 is equal to (where t is a constant)

Options:
A 12
B 32
C 36
D 10

The correct answer is:
10

Solution:
Given, u=x2+y2,x=s+3t, y=2s-t
Now ,dxds=1,dyds=2…(i)
d2xds2=0,d2yds2=0…(ii)
Now, u=x2+y2
duds=2xdxds+2ydyds
d2uds2=2dxds2+2xd2xds2+2dyds2+2yd2yds2

⇒d2uds2=212+2x0+222+2y0 (from (i) and (ii))

=2+8=10

Topic: Mathematics, Differentiation, JEE Main
If $u=x y^2 \tan ^{-1}\left(\frac{y}{x}\right)$, then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}$ is equal to

Options:
A $2 u$
B $u$
C $3 u$
D $\frac{1}{3} u$

The correct answer is:
$3 u$

Solution:
Given that,
$\begin{aligned} & u=x y^2 \tan ^{-1}\left(\frac{y}{x}\right) \\ & \frac{\partial u}{\partial x}=y^2 \tan ^{-1}\left(\frac{y}{x}\right)+x y^2 \cdot \frac{1}{1+\frac{y^2}{x^2}} \cdot\left(-\frac{y}{x^2}\right) \\ &=y^2 \tan ^{-1}\left(\frac{y}{x}\right)-\frac{x y^3}{x^2+y^2} \\ & \frac{\partial u}{\partial y}=2 x y \tan ^{-1}\left(\frac{y}{x}\right)+x y^2 \cdot \frac{1}{1+\frac{y^2}{x^2}} \cdot\left(\frac{1}{x}\right) \\ &=2 x y \tan ^{-1}\left(\frac{y}{x}\right)+\frac{x^2 y^2}{x^2+y^2} \\ & \therefore \quad x \frac{\partial u}{\partial x}+ y \frac{\partial u}{\partial y}=x y^2 \tan ^{-1}\left(\frac{y}{x}\right)-\frac{x^2 y^3}{x^2+y^2} \\ & x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=3 x y^2 \tan ^{-1} \frac{y}{x}=3 u\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
If uncertainty in position and momentum are equal, then uncertainty in velocity is

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

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

Solution:
$\Delta x=\Delta p$
$$
\Delta p^2=\frac{h}{4 \pi}
$$
or $\Delta p=\frac{1}{2} \sqrt{\frac{h}{\pi}}$
or $m \Delta v=\frac{1}{2} \sqrt{\frac{h}{\pi}}$
or $\Delta v=\frac{1}{2 m} \sqrt{\frac{h}{\pi}}$

Topic: Chemistry, Structure of Atom, NEET, NEET 2008 (Mains)
If uncertainty in the position of an electron is zero, the uncertainty in its momentum would be

Options:
A Zero
B $ \lt \frac{h}{2 \lambda}$
C $\gt\frac{h}{2 \lambda}$
D Infinite

The correct answer is:
Infinite

Solution:
$\Delta x . \Delta p \geq \frac{h}{4 \pi}^{\prime}$ if $\Delta x=0$ then $\Delta p=\infty$.

Topic: Chemistry, Structure of Atom, JEE Main
If unit of length, mass and time each be doubled, the new unit of work done is _____ times the old unit of work

Options:
A 4
B 6
C 8
D 2

The correct answer is:
2

Solution:
n2=n1M2M11 L2L12 T2T1-2
Given, M2=2M1, L2=2L1, T2=2T1
â´n2=n12122 2-2=2n1

Topic: Physics, Units and Dimensions, JEE Main
If VA and VB are the input voltages (either 5 V or 0 V) and V0 is the output voltage then the two gates represented in the following circuits A and B are :

Options:
A OR and NOT Gate
B AND and NOT Gate
C AND and OR Gate
D NAND and NOR Gate

The correct answer is:
OR and NOT Gate

Solution:
In the above circuit is we are applying high voltage in any one of the diodes or in both diodes simultaneously then we get a high out due to forward biasing of diode while we are applied low voltage in both the diode then we get low voltage so this is like OR gate.

In the above circuit which is an NPN transistor in a common emitter configuration having a phase difference of 180° between input voltage and output voltage, when we are applied high voltage then we get low voltage vice-versa so, the circuit is the same as NOT gate.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (31 Aug Shift 2)
If v=At+Bt2+Ct3 where v is velocity, t is time and A, B and C are constants, then the dimensional formula of B is

Options:
A [M0LT0]
B [ML0T0]
C [M0L0T ]
D [M0LT-3]

The correct answer is:
[M0LT-3]

Solution:
Dimension of velocity = [M0L1T-1]
[ v ]=[ B t 2 ]
∴[ B ]= [ v ] [ t 2 ] = [ M 0 L 1 T −1 ] [ T 2 ] =[ L 1 T −3 ]

Topic: Physics, Units and Dimensions, JEE Main
If $v_e$ is escape velocity and $v_o$ is orbital velocity of a satellite for orbit close to the Earth's surface, then these are related by

Options:
A $v_o=\sqrt{2} v_e$
B $v_o=v_e$
C $v_e=\sqrt{2 v_o}$
D $v_e=\sqrt{2} v_o$

The correct answer is:
$v_e=\sqrt{2 v_o}$

Solution:
Orbital velocity of satellite
$$
v_o=\sqrt{\frac{G M}{R}}
$$...(i)
and escape velocity of satellite
$$
v_e=\sqrt{\frac{2 G M}{R}}
$$...(ii)
Hence, $\quad v_e=\sqrt{2} v_o$

Topic: Physics, Gravitation, NEET, NEET 2012 (Mains)
If V is the gravitational potential due to sphere of uniform density on its surface, then its value at the centre of sphere will be:

Options:
A 43 V
B V2
C V
D 3V2

The correct answer is:
3V2

Solution:
The magnitude of gravitational potential of a solid sphere at a distance r from the centre is

V=-GM2R3(3R2-r2)

At r=R, V=-GMR.

At r=0, V'=-GM2R3(3R2)=-3GM2R

Clearly, V'=3V2.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (11 Apr Shift 2)
If $\lambda_v, \lambda_x$ and $\lambda_m$ represent the wavelength of visible light, X-rays and microwaves respectively, then:

Options:
A $\lambda_m>\lambda_{\mathrm{x}}>\lambda_v$
B $\lambda_m>\lambda_{\mathrm{v}}>\lambda_x$
C $\lambda_v>\lambda_x>\lambda_m$
D $\lambda_v>\lambda_m>\lambda_x$

The correct answer is:
$\lambda_m>\lambda_{\mathrm{v}}>\lambda_x$

Solution:
$\quad \lambda_{\mathrm{m}}>\lambda_v>\lambda_x$
In spectrum $\mathrm{X}$-rays has minimum wavelength and microwave has maximum wavelength

Topic: Physics, Electromagnetic Waves, JEE Main
If value of azimuthal quantum number $/$ is 2 , then total possible values of magnetic quantum number will be

Options:
A 7
B 5
C 3
D 2

The correct answer is:
5

Solution:
If value of $l$ is 2 then $m=-2,-1,0,+1,+2, m=-l$ to +1 including zero. (5 values of magnetic quantum number)

Topic: Chemistry, Structure of Atom, JEE Main
If \( \vec{a} \) and \( \vec{b} \) are unit vectors, then the greatest value of \( \sqrt{3}|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}| \) is

Options:

The correct answer is:
4

Solution:
Let angle between a→ and b→=α.
So, 3|a→+b→|+|a→-b→|=32+2cosα+2-2cosα
=32×2cos2α2+2×2sin2α2
=23cosα2+sinα2

As α∈0,π⇒α2∈0,π2

So, given expression is

=23cosα2+sinα2
Thus, maximum value is=232+12=2×2=4

Topic: Mathematics, Vector Algebra, JEE Main
If \((\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2=676\) and \(|\vec{b}|=2\) then \(|\vec{a}|\) is equal to

Options:
A 13
B 26
C 39
D None of these

The correct answer is:
13

Solution:
\(\begin{aligned} &(\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2=676 \\ &(|\vec{a}| \cdot|\vec{b}| \sin \theta \hat{n})^2+(|\vec{a}| \cdot|\vec{b}| \cos \theta)^2=676 \\ & \Rightarrow a^2 b^2 \sin ^2 \theta+a^2 b^2 \cos ^2 \theta=676\left[(\hat{n})^2=1\right] \\ & a^2 b^2\left(\sin ^2 \theta+\cos ^2 \theta\right)=676 \Rightarrow a^2=\frac{676}{b^2}=\frac{676}{4} \\ &|\vec{a}|=\sqrt{\frac{676}{4}} \Rightarrow|\vec{a}|=\frac{26}{2} \Rightarrow|\vec{a}|=13\end{aligned}\)

Topic: Mathematics, Vector Algebra, JEE Main
If \( \vec{a}, \vec{b} \), and \( \vec{c} \) are unit vectors such that \( \vec{a}+2 \vec{b}+2 \vec{c}=\overrightarrow{0} \), then \( |\vec{a} \times \vec{c}| \) equals

Options:
A \( \frac{1}{4} \)
B \( \frac{\sqrt{15}}{16} \)
C \( \frac{15}{16} \)
D \( \frac{\sqrt{15}}{4} \)

The correct answer is:
\( \frac{\sqrt{15}}{4} \)

Solution:
We have,

a→=b→=c→=1

And,

a→+2b→+2c→=0→

⇒a→+2c→=-2b→

⇒a→+2c→·a→+2c→=-2b→·-2b→

⇒a→2+4c→2+4a→·c→=4b→2

⇒1+4+4a→·c→=4

⇒a→·c→=-14

⇒a→c→cosθ=-14

⇒cosθ=-14

Then,

a→×c→=a→c→sinθ

⇒a→×c→=sinθ

⇒a→×c→=1-cos2θ

⇒a→×c→=1-116=154

Topic: Mathematics, Vector Algebra, JEE Main
If \( \vec{a}=2 \hat{i}-\hat{j}, \vec{b}=2 \hat{j}-\hat{k}, \vec{c}=2 \hat{k}-\hat{i} \), then in the reciprocal system of vectors of the vectors \( \vec{a}, \vec{b}, \vec{c} \), reciprocal \( \overrightarrow{a^{\prime}} \) of vector \( \vec{a} \) is

Options:
A \( \frac{4 \hat{i}+\hat{j}+2 \hat{k}}{7} \)
B \( \frac{4 \hat{i}-\hat{j}+2 \hat{k}}{3} \)
C \( \frac{4 \hat{i}+\hat{j}-2 \hat{k}}{7} \)
D \( \frac{2 \hat{i}+2 \hat{j}+2 \hat{k}}{3} \)

The correct answer is:
\( \frac{4 \hat{i}+\hat{j}+2 \hat{k}}{7} \)

Solution:
Here, a'→=b→×c→a→ b→ c→

a→ b→ c→=2-1002-1-102=8-1+0=7

b→×c→=i^j^k^02-1-102=4i^+j^+2k^

∴ a'→=4i^+j^+2k^7

Topic: Mathematics, Vector Algebra, JEE Main
If \( |\vec{a}|=3,|\vec{b}|=4,|\vec{c}|=5 \) each one of \( \vec{a}, \vec{b} \& \vec{c} \) is perpendicular to the sum of the
remaining then \( |\vec{a}+\vec{b}+\vec{c}| \) is equal to

Options:
A \( \frac{5}{\sqrt{2}} \)
B \( \frac{2}{\sqrt{5}} \)
C \( 5 \sqrt{2} \)
D \( \sqrt{5} \)

The correct answer is:
\( 5 \sqrt{2} \)

Solution:
Given that, each one of $\vec{a}, \vec{b}$ and $\vec{c}$ is perpendicular to the sum of the remaining. So,
$\vec{a} \cdot(\vec{b}+\vec{c})=0 \rightarrow(1)$
$\vec{b} \cdot|\vec{c}+\vec{a}|=0 \rightarrow(2)$
$\vec{c} \cdot \mid \vec{a}+\vec{b})=0 \rightarrow(3)$
Now, adding Eqs. (1), (2) and (3), we get
$2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$
Now,
$|a+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$
$\Rightarrow|a+\vec{b}+\vec{c}|^{2}=9+16+25$
$\Rightarrow|a+\vec{b}+\vec{c}|^{2}=50 \Rightarrow|a+\vec{b}+\vec{c}|^{2}=5 \sqrt{2}$

Topic: Mathematics, Vector Algebra, KCET, KCET 2016
If \( |\vec{a}|=7,|\vec{b}|=11,|\vec{a}+\vec{b}|=10 \sqrt{3} \), then \( |\vec{a}-\vec{b}| \) equals

Options:
A \( 10 \)
B \( \sqrt{10} \)
C \( 2 \sqrt{10} \)
D \( 20 \)

The correct answer is:
\( 2 \sqrt{10} \)

Solution:
Given:

a→=7

b→=11

a→+b→=103

Now, we have,

a→+b→2+a→-b→2=a→2+b→2+2a→·b→+a→2+b→2-2a→·b→

⇒a→+b→2+a→-b→2=2a→2+b→2

⇒ 300+a→-b→2=249+121

⇒a→-b→=210

Topic: Mathematics, Vector Algebra, JEE Main
If \( \vec{a}=\frac{1}{\sqrt{10}}(3 i+k) \) and \( \vec{b}=\frac{1}{7}(2 i+3 j-6 k) \), then the value of \( (2 \overrightarrow{\mathrm{a}}-\vec{b}) \cdot[(\vec{a} \times \vec{b}) \times(\vec{a}+2 \vec{b})] \) is

Options:
A \( 5 \)
B \( 3 \)
C \( -5 \)
D \( -3 \)

The correct answer is:
\( -5 \)

Solution:
2 a → - b → · a → × b → × a → + 2 b →
= 2 a → - b → · a → × b → × a → + 2 a → × b → × b →
= 2 a → - b → · a → · a → b → - a → · b → a → + 2 a → · b → b → - 2 b → · b → a → a → · b → = 0 , a → = b → = 1
= 2 a → - b → · b → - 2 a → = - 4 a → · a → - b → · b → = - 5

Topic: Mathematics, Vector Algebra, JEE Main
If \( \vec{a}+\vec{b}=\vec{c} \) and \( a+b=c \), then angle between \( \vec{a} \) and \( \vec{b} \) is

Options:
A \( 90^{\circ} \)
B \( 30^{\circ} \)
C \( 60^{\circ} \)
D \( 0^{0} \)

The correct answer is:
\( 0^{0} \)

Solution:
Given that, a + b = c ,

a→+b→=c→ ,θ = angle between a→ and b→
take magnitude on both sides
a→+b→=c→
a2+b2+2 abcosθ=c2

a+b2 = a2 + b2 + 2ab ⇒ a2 + b2 = a+b2 - 2ab
a+b2-2ab+2abcosθ=c2 ⇒ c2+ 2abcosθ-1 = c2
⇒2 ab cosθ-1=0
cos⁡θ=1 , cos0∘ = 1
then angle between a→ and b→ is,
θ=0∘

Topic: Mathematics, Vector Algebra, JEE Main
If \( \vec{r}=\hat{i}+\hat{j}+t(2 \hat{i}-\hat{j}+\hat{k}) \) and \( \vec{r}=2 \hat{i}-\hat{j}-\hat{k}+t(3 \hat{i}-5 \hat{j}+2 \hat{k}) \) are the vector equations of two lines \( L_{1} \) and \( L_{2} \) then the shortest distance between them is

Options:
A \( \frac{9}{\sqrt[3]{\omega n}} \)
B \( \frac{10}{\sqrt{59}} \)
C \( \frac{11}{\sqrt{59}} \)
D \( 0 \)

The correct answer is:
\( \frac{10}{\sqrt{59}} \)

Solution:
Here, a→1=i^+j^ & b→1=2i^-j^+k^ and a→2=2i^+j^-k^ & b→2=3i^-5j^+2k^.

⇒a→2-a→2=i^-k^ & b→1×b→2=i^j^k^2-113-52=3i^-j^-7k^

The shortest distance between lines L1 and L2 =a→2-a→2·b→1×b→2 b→1×b→2

=i^-k^·3i^-j^-7k^3i^-j^-7k^

=1059

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If vectors $2 \mathrm{i}-\mathrm{j}+\mathrm{k}, \mathrm{i}+2 \mathrm{j}-3 \mathrm{k}$ and $3 \mathrm{i}+\mathrm{aj}+5 \mathrm{k}$ are

coplanar, then the value of a is

Options:
A 2
B -2
C -1
D -4

The correct answer is:
-4

Solution:
If given vectors are coplanar, then there exists two scalar quantities $\mathrm{x}$ and $\mathrm{y}$ such that $2 \hat{i}-\hat{j}+\hat{k}=x(\hat{i}+2 \hat{j}-3 \hat{k})+y(3 \hat{i}+a \hat{j}+5 \hat{k})$

Comparing coefficient of $\hat{i}, \hat{j}$ and $\hat{k}$ on both sides of (1) we get $x+3 y=2,2 x+a y=-1,-3 x+5 y=1 \ldots(2)$

Solving first and third equations, we get $\mathrm{x}=1 / 2, \mathrm{y}=1 / 2$

Since the vectors are coplanar, therefore these values of $x$ and $y$ will satisfy the equation $2 x+a y=-1$

$\therefore 2(1 / 2)+a(1 / 2)=-1 \Rightarrow a=-4$

Topic: Mathematics, Vector Algebra, JEE Main
If vectors 4pâ+qâ, 2pâ-3qâ and 5pâ-3qâ, 5pâ+3qâ are a pair of mutually perpendicular vectors and if the angle between pâ and qâ is Î¸, then the value of sin2Î¸ is equal to

Options:
A 35
B 925
C 12500
D 24992500

The correct answer is:
24992500

Solution:
4p→+q→⋅2p→-3q→=0
8p→2-3q→2=10p→⋅q→ … 1
5p→-3q→⋅5p→+3q→=0
⇒25p→2=9q→2 … 2
cos⁡θ=p→⋅q→p→q→=8p→2-3q→210p→q→=8925-310×35
⇒cos⁡θ=-3256=-150
⇒sin2θ=24992500 ∵θ∈0,π

Topic: Mathematics, Vector Algebra, JEE Main
If vectors a1→=xi^-j^+k^ and a2→=i^+yj^+zk^ are collinear, then a possible unit vector parallel to the vector xi^+yj^+zk^ is:

Options:
A 12-j^+k^
B 13i^+j^-k^
C 12i^-j^
D 13i^-j^+k^

The correct answer is:
13i^-j^+k^

Solution:
a→1 and a→2 are collinear

So, x1=-1y=1z

Unit vector in direction of

xi^+yj^+zk^=±13i^-j^+k^

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (26 Feb Shift 2)
If vectors $\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \bar{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\bar{c}=-3 \hat{i}+\hat{j}+2 \hat{k}$ are such that, $\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$ is perpendicular to $\overline{\mathrm{c}}$, then $\lambda=$

Options:
A -14
B 14
C 2
D -2

The correct answer is:
-14

Solution:
From given data, we write
$$
\bar{a}+\lambda \bar{b}=(2-\lambda)
\hat{i}+(2+2 \lambda)
\hat{j}+(3+\lambda)
\hat{k}
$$
Since (1) is $\perp$ er to $\bar{c}$, we write
$$
\begin{aligned}
& (2-\lambda)(3)+(2+2 \lambda)(1)+(3+\lambda)(2)=0 \\
& \therefore 6-3 \lambda+2+2 \lambda+6+2 \lambda=0 \Rightarrow \lambda=-14
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main
If velocity, force and time are taken as the fundamental quantities, then using dimensional analysis choose the correct dimensional formula for mass among the following. [ K is a dimensionless constant ]

Options:
A Q=K vâ1 F T
B Q=K v3 F T2
C Q=2 K vâ1 F T
D Q=3 K v2 F T

The correct answer is:
Q=K vâ1 F T

Solution:
Let the quantity be Q, then,

Q=f(v,F,T)
Assuming that the function is the product of power functions of v,F and T,
Q = Kvx Fy Tz ...(i)
where K is a dimensionless constant of proportionality. The above equation dimensionally becomes
[Q] = [LT-1]x[MLT-2]y[T]z
i.e., [Q] = [MyL(x+y)T(-x-2y+z)], âNow
Q = mass i.e., [Q]=[M]
So Equation (ii) becomes
[M] = [MyL(x+y)T(-x-2y+z)]
its dimensional correctness requires
y=1, x+y=0 and - x - 2y + z = 0
which on solving yields
x=-1, y=1 and z=1
Substituting it in Equation (i), we get
Q=K v-1 F T

Topic: Physics, Units and Dimensions, JEE Main
If velocity, force and time are taken to be fundamental quantities, then the dimensional formula for mass is

Options:
A Q = Kv-1 FT
B Q = Kv3 FT2
C Q = 2Kv-2 FT
D Q = -3Kv2 FT

The correct answer is:
Q = Kv-1 FT

Solution:
Let the quantity be Q, then,
Q = f (v, F, T)
Assuming that the function is the product of power functions of v, F and T,
Q = Kvx Fy Tz ...(i)
where K is a dimensionless constant of proportionality. The above equation dimensionally becomes
[Q] = [LT-1]x[MLT-2]y[T]z
i.e., [Q] = [My][Lx+yT-x-2y+z] ...(ii)
âNow
Q = mass i.e., [Q] = [M]
So Equation (ii) becomes
[M] = [MyLx+yT-x-2y+z]
its dimensional correctness requires
y = 1, x + y = 0 and - x - 2y + z = 0
which on solving yields
x = -1, y = 1 and z = 1
Substituting it in Equation (i), we get
Q = Kv-1 FT

Topic: Physics, Units and Dimensions, JEE Main
If velocity V time T and force F are chosen as the base quantities, the dimensions of the mass will be :

Options:
A FVT-1
B FT-1V-1
C FT2 V
D FTV-1

The correct answer is:
FTV-1

Solution:
[M]1=[V]a[T]b[F]c
[M]1=LT-1a[T]bMLT-2c
[M]1=[M]c[L]a+c[T]-a+b-2c
Thus,

c=1,a+c=0 andb-a-2c=0⇒a=-1⇒b=-1+2=1
Thus, [M]=[V]-1[ T]1[ F]1

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2021 (31 Aug Shift 2)
If vertex of parabola is 2,-1 and equation of its directrix is 4x-3y=21, then the length of latus rectum is

Options:
A 2
B 8
C 12
D 16

The correct answer is:
8

Solution:
Plotting the graph with given data we have,

We know that, in any parabola, the length of the latus rectum =4a and perpendicular distance from vertex to the directrix =a

So, a=8+3-2142+-32

⇒a=-105=2

Therefore, the length of the latus rectum 4a=4×2=8

Topic: Mathematics, Parabola, JEE Main, JEE Main 2022 (28 Jun Shift 2)
If \(V_H, V_N\) and \(V_O\) denote the root mean square velocities of molecules of hydrogen, nitrogen and oxygen respectively at a given temperature, then:

Options:
A \(V_N > V_O > V_H\)
B \(V_H > V_N > V_O\)
C \(V_O=V_N=V_H\)
D \(V_O > V_H > V_N\)

The correct answer is:
\(V_H > V_N > V_O\)

Solution:
\(\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{KT}}{\mathrm{m}}}\)
\(\mathrm{m}=\) Molar Mass of molecule
Now, \(\mathrm{m}_{\mathrm{H}} < \mathrm{m}_{\mathrm{N}} < \mathrm{m}_{\mathrm{O}}\)
\(\therefore \mathrm{V}_{\mathrm{H}} > \mathrm{V}_{\mathrm{N}} > \mathrm{V}_{\mathrm{O}}\)

Topic: Physics, Kinetic Theory of Gases, JEE Main
If vibrations of a sting are to be increased by a factor of two,then tension in the string must be made

Options:
A Half
B Twice
C Four times
D Eight times

The correct answer is:
Four times

Solution:
\(\begin{aligned} & n \propto \sqrt{T} \\ & \frac{n_1}{T_1}=\frac{n_2}{T_2} \\ & \sqrt{T_2}=\frac{n_2 \sqrt{T_1}}{n_1} \Rightarrow \sqrt{T_2}=2 \sqrt{T_1} \\ & T_2=4 T_1\end{aligned}\)

Topic: Physics, Waves and Sound, JEE Main
If \( V_{r} \) denotes the sum of the first \( r \) terms of an arithmetic progression (AP), whose first term is \( r \) and the common difference is \( (2 r-1) \). Then, the sum of \( V_{1}+V_{2}+\ldots+V_{n} \) is

Options:
A \( \frac{1}{12} n(n+1)\left(3 n^{2}-n+1\right) \)
B \( \frac{1}{12} n(n+1)\left(3 n^{2}+n+2\right) \)
C \( \frac{1}{2} n\left(2 n^{2}-n+1\right) \)
D \( \frac{1}{3}\left(2 n^{3}-2 n+3\right) \)

The correct answer is:
\( \frac{1}{12} n(n+1)\left(3 n^{2}+n+2\right) \)

Solution:
Vr= Sum of first r terms of an AP whose first term is r and common difference is 2r-1

=r22r+r-12r-1

=r22r+2r2-3r+1

=r22r2-r+1

=122r3-r2+r

Now, ∑r=1nVr=122∑r=1nr3-∑r=1nr2+∑r=1nr

=122·nn+122-nn+12n+16+nn+12

=12n2n+122-nn+12n+16+nn+12

=nn+14nn+1-2n+13+1

=nn+143n2+3n-2n-1+33

=nn+1123n2+n+2

Topic: Mathematics, Sequences and Series, JEE Main
If W1,W2 and W3 represent the work done in moving a particle from A to B along three different paths 1,2 and 3 (as shown in figure) in the gravitational field of the point mass ‘ m ’. Find the correct relation between ‘ W1 ’, ‘ W2 ’ and ‘ W3 ’

Options:
A W1<W3<W2
B W1<W2<W3
C W1=W2=W3
D W1>W3>W2

The correct answer is:
W1=W2=W3

Solution:
∵  Gravitation field in conservative in nature.
∴  Work done will not depend on path
∴ W1  = W2 = W3

Topic: Physics, Gravitation, JEE Main
If W1, W2 and W3 represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively (as shown) in the gravitational field of a point mass m. Find the correct relation between W1, W2 and W3.

Options:
A W1>W2>W3
B W1=W2=W3
C W1<W2<W3
D W2>W1>W3

The correct answer is:
W1=W2=W3

Solution:
Gravitational field is a conservative force field. In a conservative force field work done is path independent

Topic: Physics, Work Power Energy, JEE Main
If wattless current flows in the AC circuit, then the circuit is :

Options:
A Purely Resistive circuit
B Purely Inductive circuit
C LCR series circuit
D RC series circuit only

The correct answer is:
Purely Inductive circuit

Solution:
The current in an AC circuit containing only either a capacitor or an inductor is said to be wattless as the average power is zero. As wattless current has phase difference of ϕ=90° with voltage, therefore power factor cosϕ will be zero.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (25 Jun Shift 1)
If wavelength of the first line of the Paschen series of hydrogen atom is 720 nm, then the wavelength of the second line of this series is _______ nm. (Nearest integer)

Options:

The correct answer is:
492

Solution:
Paschen series are spectral lines coming to n=3

1st line of Paschen series makes a transition from, 4→3

So, 2nd line of Paschen series makes a transition from, 5→3

We know that,

1λ=R1n12-1n22Z2

For 1st line, 1λ=R1(3)2-1(4)2Z2

and for 2nd line,

1λ'=R1(3)2-1(5)2Z2

or 1λ1λ'=R79×16Z2R169×25Z2

Given, λ=720 nm

or 17201λ'=175256

or λ'720=175256

or λ' ≈ 492 nm

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (24 Jan Shift 1)
If we consider that $\frac{1}{6}$, in place of $\frac{1}{12}$; mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Options:
A
Decrease twice

B
Increase two fold

C
Remain unchanged

D
Be a function of the molecular mass of the substance

The correct answer is:
Remain unchanged

Solution:
No Solution Available

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2005
If we need a magnification of 375 from a compound microscope of tube length 150mm and an objective of focal length 5mm, the focal length of the eye-piece, should be close to:

Options:
A 22mm
B 2mm
C 4mm
D 33mm

The correct answer is:
22mm

Solution:
M.P.=Lf01+Dfe;375=15051+25fe
37530=1+25fe
34530=25fe
fe=750345=2.17cm;fe≈22mm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (07 Jan Shift 1)
If we remove half of the forest cover of earth, the crisis that will occur is

Options:
A Many new species would be formed
B Population, pollution and ecological imbalance will rise
C Energy crisis will commence
D The remaining forest will correct the imbalance

The correct answer is:
Population, pollution and ecological imbalance will rise

Solution:
Forests are very important to us. They cover about 23.68% of our earth and help in population control. They also help us by providing useful food and thus play an important role in ecological balance.

Topic: Biology, Environmental Issues, NEET
If \( w=\frac{z}{z-\frac{1}{3} i} \) and \( |w|=1 \), then \( z \) lies on

Options:
A A parabola
B A straight line
C A circle
D An ellipse

The correct answer is:
A straight line

Solution:
Since, zz-i/3 = 1
⇒ 3z=3z-i
⇒ 3x+iy=3x+iy-i [put z=x+iy]
⇒ 3x2+y2=3x2+3y-12
⇒ 9x2+9y2=9x2+9y2+1-6y
⇒ y=16
Which shows that z lies on a straight line.

Topic: Mathematics, Complex Number, JEE Main
If x∈0,1, then the value of 2tan-1⁡1-x22x+2cos-1⁡1-x21+x2 is equal to

Options:
A -π2
B 0
C π2
D π

The correct answer is:
π

Solution:
∴ Let y=2tan-1⁡1-x22x+2cos-1⁡1-x21+x2
Putting x=tan⁡θ

As, x∈0,1, θ∈0,π4 and π2-2θ∈0,π2
∴y=2tan-1⁡cot⁡2θ+2cos-1⁡cos⁡2θ

=2tan-1⁡tan⁡π2-2θ+2cos-1⁡cos⁡2θ=2π2-2θ+4θ=π

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main