datavorous/entrance-exam-dataset huggingface dataset:
The population p(t) at a time t of a certain mouse species satisfies the differential equation dptdt=0.5p(t)-450. If p(0)=850, then the time at which the population becomes zero is

Options:
A 12ln18

B ln18

C 2ln18

D ln9

The correct answer is:
2ln18

Solution:
Let, p=p(t)
dp dt = 0.5 p - 450
= p - 9 0 0 2
⇒∫850p2p-900dp=∫0tdt
⇒2logp-900850p=t
⇒2logp-900-log-50=t
⇒ 2 log p - 900 - 5 0 = t ...(1)
⇒ | p - 900 - 5 0 |= e t/2
⇒ p = 9 0 0 - 5 0 e t/2
Let, when t=T, p=0 (using (1))
⇒T=2ln⁡18

Topic: Mathematics, Differential Equations, JEE Main
The population $\mathrm{p}(\mathrm{t})$ at time $t$ of a certain mouse species satisfies the differential equation $\frac{\mathrm{dp}(\mathrm{t})}{\mathrm{dt}}=0.5 \mathrm{~p}(\mathrm{t})$ $-450$. If $p(0)=850$, then the time at which the population becomes zero is

Options:
A
$2 \ln 18$

B
$\ln 9$

C
$\frac{1}{2} \ln 18$

D
$\ln 18$

The correct answer is:
$2 \ln 18$

Solution:
$\frac{d(p(t))}{d t}=\frac{1}{2} p(t)-450$
$\frac{d(p(t))}{d t}=\frac{p(t)-900}{2}$
$2 \int \frac{d(p(t))}{p(t)-900}=\int d t$
$2 \ln |p(t)-900|=t+c$
$t=0 \quad \Rightarrow 2 \ln 50=0+c \quad \Rightarrow c=2 \ln 50$
$\therefore 2 \ln |\mathrm{p}(\mathrm{t})-900|=\mathrm{t}+2 \ln 50$
$P(\mathrm{t})=0 \quad \Rightarrow 2 \ln 900=\mathrm{t}+2 \ln 50$
$t=2(\ln 900-\ln 50)=2 \ln \left(\frac{900}{50}\right)=2 \ln 18$.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2012 (Offline)
The population \( p(t) \) at a time \( t \) of a certain mouse species satisfies the differential equation \( \frac{d p(t)}{d t}=0.5 p(t)-450 \). If \( p(0)=850 \), then the time at which the population becomes zero is

Options:
A \( \frac{1}{2} \ln 18 \)
B \( \ln 18 \)
C \( 2 \ln 18 \)
D \( \ln 9 \)

The correct answer is:
\( 2 \ln 18 \)

Solution:
Let, p=pt

dpdt=0.5p-450

= p-9002

⇒ ∫850p2p-900dp=∫0tdt

⇒ 2logp−900p850=t

⇒ 2logp−900−log−50=t

⇒ 2logp-900-50=t ...1

⇒ p-900-50=et/2

⇒ p=900-50et/2

Let, when t=T, p=0 (using 1)

⇒ T=2 ln 18

Topic: Mathematics, Differential Equations, JEE Main
The portion of the line 4x+5y=20 in the first quadrant is trisected by the lines L1 and L2 passing through the origin. The tangent of an angle between the lines L1 and L2 is :

Options:
A 85
B 2541
C 25
D 3041

The correct answer is:
3041

Solution:
Using section formula

Co-ordinates of A=1×5+2×01+2,1×0+2×41+2=53,83
Co-ordinates of B=2×5+1×01+2,2×0+1×41+2=103,43
Slope of OA=m1=83-053-0=85
Slope of OB=m2=43-0103-0=25

We know that, tanθ=m1-m21+m1m2

⇒tanθ=651+1625

⇒tanθ=3041

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (27 Jan Shift 1)
The position and velocity of a particle executing simple harmonic motion at t=0 are given by 3 cm and 8 cm s-1 respectively. If the angular frequency of the particle is 2 rad s-1, then the amplitude of oscillation (in cm) is

Options:
A 3
B 4
C 5
D 6

The correct answer is:
5

Solution:
Given, the position and velocity of the particle executing SHM .
y=3 cm
v=8 cm s-1
Angular frequency, ω=2 rad s-1
The velocity,
v=ωa2-y2
8=2a2-32
4=a2-32
16=a2-9
a2=25
a=5 cm

Topic: Physics, Oscillations, JEE Main
The position co-ordinates of a particle moving in a 3D coordinate system is given by
x=acos⁡ωt
y=asin⁡ωt
and z=aωt
The speed of the particle is:

Options:
A 2 aÏ
B aÏ
C 2aÏ
D 3Â aÏ

The correct answer is:
2 aÏ

Solution:
x=acosâ¡Ït
y=asinâ¡Ït
z=a Ït
vx=-aÏsinâ¡Ït
vy=aÏcosâ¡Ït
v2=aÏ
v=vx2+vy2+vz2
=-aÏsinâ¡Ït2+aÏcosâ¡Ït2+aÏ2
=aÏ2

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The position of a moving car at time t is given by f(t)=at2+bt+c, t>0, where a, b and c are real numbers greater than 1. Then the average speed of the car over the time interval t1,t2 is attained at the point:

Options:
A t2-t12
B at2-t1+b
C t1+t22
D 2at1+t2+b

The correct answer is:
t1+t22

Solution:
Given: position of the moving car at time t is =ft=at2+bt+c

So,

vavg=ft2-ft1t2-t1

⇒vavg=at22-t12+bt2-t1t2-t1

⇒vavg=at1+t2+b

The instantaneous speed is given by:

f't=2at+b

So, to get the point where average speed is equal to car's actual speed,

f't=vavg

⇒at1+t2+b=at+b

⇒t=t1+t22

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (06 Sep Shift 1)
The position of a particle as a function of time t, is given by xt=at+bt2-ct3 where a, b and c are constants. When the particles zero acceleration, then its velocity will be:

Options:
A a+b23c
B a+b22c
C a+b2c
D a+b24c

The correct answer is:
a+b23c

Solution:
x(t)=at+bt2-ct3
velocity v(t)=ddt(x)=ddtat+bt2-ct3
=a+2bt-3ct2
Acceleration =ddtv(t)=2b-6tc
acceleration=0⇒2b-6tc=0
t=b3c
∴ velocity when t=b3c,
vt=b3c=a+2bb3c-3b3c2
=a+2b23c-b23c
=a+b23c

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The position of a particle at time $t$ is given by the equation $x(t)=\frac{v_0}{A}\left(1-e^{A t}\right) v_o=$ constant and $A>0$. Dimensions of $v_o$ and A respectively are

Options:
A $\left[\mathrm{M}^0 \mathrm{LT}^0\right]$ and $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]$
B $\left[\mathrm{M}^0 \mathrm{LT}^{-1}\right]$ and $\left[\mathrm{M}^0 \mathrm{LT}^{-2}\right]$
C $\left[\mathrm{M}^0 \mathrm{LT}^{-1}\right]$ and $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}\right]$
D $\left[\mathrm{M}^0 \mathrm{LT}^{-1}\right]$ and $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]$

The correct answer is:
$\left[\mathrm{M}^0 \mathrm{LT}^{-1}\right]$ and $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]$

Solution:
The dimensions of $x=$ dimensions of $\frac{v_0}{A}$
Therefore, out of the given options $v_0$ has dimensions equal to $\left[\mathrm{M}^{\circ} \mathrm{LT}^{-1}\right]$ and $A$ has dimensions equal to $\left[\mathrm{M}^{\circ} \mathrm{L}^{\circ} \mathrm{T}^{-1}\right]$
So, that
$\begin{aligned}
\frac{\left[v_0\right]}{[A]} & =\frac{\left[\mathrm{M}^{\circ} \mathrm{LT}^{-1}\right]}{\left[\mathrm{M}^{\circ} \mathrm{L}^{\circ} \mathrm{T}^{-1}\right]}=[\mathrm{L}] \\
& =\text { dimension of } x .
\end{aligned}$

Topic: Physics, Units and Dimensions, JEE Main
The position of a particle is given by \( x=2\left(t-t^{2}\right) \), where \( t \) is expressed in seconds and \( x \) is in meter. The total distance travelled by the particle between \( \mathrm{t}=0 \) to \( \mathrm{t}=1 \mathrm{~s} \) is:

Options:
A \( 0 \mathrm{~m} \)
B \( 1 \mathrm{~m} \)
C \( 2 \mathrm{~m} \)
D \( \frac{1}{2} \mathrm{~m} \)

The correct answer is:
\( 1 \mathrm{~m} \)

Solution:
Let the total distance be s between for 0<t<1

We have

x=2(t-t2)

For 0<t<12

the displacement is s1=12

for 12<t<1

the displacement is s2=-1 2

distance

s=s1+s2s=12+12=1 m

Topic: Physics, Motion In One Dimension, JEE Main
The position of a particle related to time is given by x=(5t2-4t+5) m. The magnitude of velocity of the particle at t=2 s will be :

Options:
A 06 m s-1
B 14 m s-1
C 10 m s-1
D 16 m s-1

The correct answer is:
16 m s-1

Solution:
It is given that

x=5t2-4t+5

The velocity is given by

v=dxdt=d(5t2-4t+5)dt=10t -4

At t=2 s,

v=10(2)-4=16 m s-1

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The position of a projectile launched from the origin at t = 0 is given by r → = 4 0 i⁡ ^ + 5 0 j⁡ ^ m at t = 2s. If the projectile was launched at an angle θ from the horizontal, then θ is (take g = 10 ms-2).

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

The correct answer is:
tan - 1 7 4

Solution:
From question, Horizontal velocity (initial),
ux=402=20 m/s
Vertical velocity (initial), 50=uyt+12gt2

⇒50= uy×2+12(-10)×4
50=2uy-20
uy=702=35 m/s
∴ tanθ=uyux=3520=74
⇒ θ=tan-174

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2014 (09 Apr Online)
The position of some metals in the electrochemical series in decreasing electropositive character is given as $M g \gt A l \gt Z n \gt C u \gt A g$. What will happen, if a copper spoon is used to stir a solution of aluminium nitrate

Options:
A The spoon will get coated with $A I$
B An alloy of and is formed
C The solution becomes blue
D There is no reaction

The correct answer is:
There is no reaction

Solution:
The metal placed below in electrochemical series does not react with that metal salt solution which metal is placed above in series.

Topic: Chemistry, Electrochemistry, JEE Main
The position of the centre of mass of a uniform semi-circular wire of radius R placed in x-y plane with its centre at the origin and the line joining its ends as x-axis is given by, 0, xRπ. Then, the value of |x| is .

Options:

The correct answer is:
2

Solution:
COM of semi-circular ring is at 2Rπ

Distance from centre ⇒x=2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2021 (22 Jul Shift 1)
The position of the point $(1,3)$ with respect to the ellipse $4 x^2+9 y^2-16 x-54 y+61=0$

Options:
A Outside the ellipse
B On the ellipse
C On the major axis
D On the minor axis

The correct answer is:
On the major axis

Solution:
$E \equiv 4+9(3)^2-16(1)-54(3)+61 \lt 0$
Therefore, the point is inside the ellipse.
$\frac{4(x-2)^2}{36}+\frac{9(y-3)^2}{36}=1$
Equation of major axis is $y-3=0$ and point $(1,3)$ lies on it.

Topic: Mathematics, Ellipse, JEE Main
The position of the point $(4,-3)$ with respect to the ellipse $2 x^2+5 y^2=20$ is

Options:
A Outside the ellipse
B On the ellipse
C On the major axis
D None of these

The correct answer is:
Outside the ellipse

Solution:
Since $S_1 \gt 0$. Hence the point is outside the ellipse.
\(\begin{aligned} & \text {as, } 2(4)^2+5(-3)^2-20=S_1 \\ & \Rightarrow S_1=32+45-20 > 0 \text { (Outside) } \end{aligned}\)

Topic: Mathematics, Ellipse, JEE Main
The position of the point $(8,-9)$ with respect to the lines $2 x+3 y-4=0$ and $6 x+9 y+8=0$ is

Options:
A Point lies on the same side of the lines
B Point lies on the different sides of the line
C Point lies on one of the line
D None of these

The correct answer is:
Point lies on the same side of the lines

Solution:
$\begin{aligned}
& L_1(8,-9)=2(8)+3(-9)-4=-15 \\
& L_2(8,-9)=6(8)+9(-9)+8=-25
\end{aligned}$
Hence point lies on same side of the lines.

Topic: Mathematics, Straight Lines, JEE Main
The position of the points $(3,4)$ and $(2,-6)$ with respect to the line $3 x-4 y=8$ are

Options:
A On the same side of the line
B On different side of the line
C One point on the line and the other outside the line
D Both point on the line

The correct answer is:
On different side of the line

Solution:
$\begin{aligned}
& L \equiv 3 x-4 y-8=0 \\
& L_{(3,4)}=9-16-8 \lt 0 \text { and } L_{(2,-6)}=6+24-8\gt0
\end{aligned}$
Hence, the points lie on different side of the line.

Topic: Mathematics, Straight Lines, JEE Main
The position vector of 1 kg object is r→=3i^-j^ m and its velocity v→=3j^+k^ m s-1. The magnitude of its angular momentum is x N m s, where x is

Options:

The correct answer is:
91

Solution:
Angular momentum of a particle is given by,

L→=r→×P→=r→×mv→ also m=1 kg.

⇒L→=3i^-j^×3j^+k^=9k^-3j^-i^ N m s

The magnitude of the angular momentum will be,

⇒L→=92+32+12=91 N m s

Therefore, the value of x=91.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2022 (28 Jun Shift 1)
The position vector of a particle changes with time according to the relation r→t=15t2i^+4-20t2j^. What is the magnitude of the acceleration at t=1?

Options:
A 40
B 25
C 100
D 50

The correct answer is:
50

Solution:
r→=15t2i^+4-20t2j^
v→=30 ti^-40t j^
a→=30 i^-40j^
At t=1 ,
a→=30 i^-40j^
a→=50

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The position vector of a particle is determined by the expression r→=2t2i^+3tj^+9k^. The magnitude of its displacement (in m) in first two seconds is

Options:

The correct answer is:
10

Solution:
r→=2t2i^+3tj^+9k^
At t=0,r1→=9k^
At t=2s,r2→=8i^+6j^+9k^
∴ Δr→=r2→-r1→=8i^+6j^
∴Δr→=r2→-r1→=82+62=10m

Topic: Physics, Motion In Two Dimensions, JEE Main
The position vector of a particle related to time t is given by r→=10ti^+15t2j^+7k^m. The direction of net force experienced by the particle is :

Options:
A Positive x-axis
B In x-y plane
C Positive y-axis
D Positive z-axis

The correct answer is:
Positive y-axis

Solution:
Given, the position vector of the particle is

r→=10ti^+15t2j^+7k^

The velocity of the particle can be calculated as follows:

v→=dr→dt=ddt10ti^+15t2j^+7k^= 10i^+30tj^

And, the acceleration of the particle is given by

a→=dv→dt=ddt10i^+30tj^=30j^

Hence, the force on the particle can be written as

F→=ma→=30mj^

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The position vector of the particle is rt=acos⁡ωti^+asin⁡ωtj^, where a and ω are real constants of suitable dimensions. The acceleration is

Options:
A Perpendicular to the velocity
B Parallel to the velocity
C Directed away from the origin
D Perpendicular to the position vector

The correct answer is:
Perpendicular to the velocity

Solution:
Given that, rt=acosâ¡Ïti^+a sinâ¡Ïtj^
âµvâ=drtdt=-aÏsinâ¡Ït i^+aÏcosâ¡Ït i^j
aâ=dvâdt=-aÏ2cosâ¡Ïti^-aÏ2cosâ¡Ïtj^
aâ.vâ=a2Ï3sinâ¡Ïtcosâ¡Ït-a2Ï3sinâ¡Ïtcosâ¡Ït
â aâ.vâ=0
Above result implies that acceleration is perpendicular to velocity.

Topic: Physics, Motion In Two Dimensions, JEE Main
The position vector of the point P is Ïâ=xi^+yj^ +zk^, where x,y,zâN. If aâ=i^+j^+k^ and Ïâ.aâ=10, then the number of possible positions of point P is equal to

Options:
A 36
B 72
C 66
D 54

The correct answer is:
36

Solution:
π→.a→=10
⇒x+y+z=10
The required number of positions
= The number of positive integral solutions of x+y+z=10
=C3-110-1
=9×82=36

Topic: Mathematics, Permutation Combination, JEE Main
The position vector of the points which divides internally in the ratio $2: 3$ the join of the points $2 \mathbf{a}-3 \mathbf{b}$ and $3 \mathbf{a}-\mathbf{2 b}$, is

Options:
A $\frac{12}{5} \mathbf{a}+\frac{13}{5} \mathbf{b}$
B $\frac{12}{5} \mathbf{a}-\frac{13}{5} \mathbf{b}$
C $\frac{3}{5} \mathbf{a}-\frac{2}{5} \mathbf{b}$
D None of these

The correct answer is:
$\frac{12}{5} \mathbf{a}-\frac{13}{5} \mathbf{b}$

Solution:
Position vectors of the points which divides internally is,
$\frac{3(2 a-3 b)+2(3 a-2 b)}{5}=\frac{12 a-13 b}{5} \text {. }$

Topic: Mathematics, Vector Algebra, JEE Main
The position vectors of $A$ and $B$ are $2 \mathbf{i}-9 \mathbf{j}-4 \mathbf{k}$ and $6 \mathbf{i}-3 \mathbf{j}+8 \mathbf{k}$ respectively, then the magnitude of $\overrightarrow{A B}$ is

Options:
A 11
B 12
C 13
D 14

The correct answer is:
14

Solution:
$\begin{aligned}
& \overrightarrow{A B}=(6-2) \mathbf{i}+(-3+9) \mathbf{j}+(8+4) \mathbf{k}=4 \mathbf{i}+6 \mathbf{j}+12 \mathbf{k} \\
& |\overrightarrow{A B}|=\sqrt{16+36+144}=14
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
The position vectors of $A$ and $B$ are $\mathbf{i}-\mathbf{j}+2 \mathbf{k}$ and $3 \mathbf{i}-\mathbf{j}+3 \mathbf{k}$. The position vector of the middle point of the line $A B$ is

Options:
A $\frac{1}{2} \mathbf{i}-\frac{1}{2} \mathbf{j}+\mathbf{k}$
B $2 \mathbf{i}-\mathbf{j}+\frac{5}{2} \mathbf{k}$
C $\frac{3}{2} \mathbf{i}-\frac{1}{2} \mathbf{j}+\frac{3}{2} \mathbf{k}$
D None of these

The correct answer is:
$2 \mathbf{i}-\mathbf{j}+\frac{5}{2} \mathbf{k}$

Solution:
$\frac{\mathbf{a}+\mathbf{b}}{2}=2 \mathbf{i}-\mathbf{j}+\frac{5}{2} \mathbf{k}$.

Topic: Mathematics, Vector Algebra, JEE Main
The position vectors of four points $P, Q, R, S$ are $2 \mathbf{a}+4 \mathbf{c}, \quad 5 \mathbf{a}+3 \sqrt{3} \mathbf{b}+4 \mathbf{c},-2 \sqrt{3} \mathbf{b}+\mathbf{c}$ and $2 \mathbf{a}+\mathbf{c}$ respectively, then

Options:
A $\overrightarrow{P Q}$ is parallel to $\overrightarrow{R S}$
B $\overrightarrow{P Q}$ is not parallel to $\overrightarrow{R S}$
C $\overrightarrow{P Q}$ is equal to $\overrightarrow{R S}$
D $\overrightarrow{P Q}$ is parallel and equal to $\overrightarrow{R S}$

The correct answer is:
$\overrightarrow{P Q}$ is parallel to $\overrightarrow{R S}$

Solution:
$\overrightarrow{P Q}=3 \mathbf{a}+3 \sqrt{3} \mathbf{b}$ and $\overrightarrow{R S}=2 \mathbf{a}+2 \sqrt{3} \mathbf{b}$
Hence $\overrightarrow{P Q }~||~ \overrightarrow{R S}$

Topic: Mathematics, Vector Algebra, JEE Main
The position vectors of $P$ and $Q$ are $5 \mathbf{i}+4 \mathbf{j}+a \mathbf{k}$ and $-\mathbf{i}+2 \mathbf{j}-2 \mathbf{k}$ respectively. If the distance between them is 7 , then the value of $a$ will be

Options:
A $-5,1$
B 5,1
C 0,5
D 1,0

The correct answer is:
$-5,1$

Solution:
$\begin{aligned} & 7=\sqrt{(5+1)^2+(4-2)^2+(a+2)^2} \Rightarrow a+2= \pm 3 \\ & \text { or } a=-5,1 .\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
The position vectors of $P$ and $Q$ are respectively $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$. If $\mathrm{R}$ is a point on $\overrightarrow{\mathrm{PQ}}$ such that $\overrightarrow{\mathrm{PR}}=5 \overrightarrow{\mathrm{PQ}}$, then the position vector of $R$ is

Options:
A $5 \overrightarrow{\mathbf{b}}-4 \overrightarrow{\mathbf{a}}$
B $5 \overrightarrow{\mathbf{b}}+4 \overrightarrow{\mathbf{a}}$
C $4 \overrightarrow{\mathbf{b}}-5 \overrightarrow{\mathbf{a}}$
D $4 \overrightarrow{\mathbf{b}}+5 \overrightarrow{\mathbf{a}}$

The correct answer is:
$5 \overrightarrow{\mathbf{b}}-4 \overrightarrow{\mathbf{a}}$

Solution:
Given that,
It mean $R$ divides $P Q$ externally in the ratio $5: 4$.
$\begin{aligned}
\therefore \text { Position vector of } R & =\frac{5 \overrightarrow{\mathrm{b}}-4 \overrightarrow{\mathbf{a}}}{5-4} \\
& =5 \overrightarrow{\mathrm{b}}-4 \overrightarrow{\mathbf{a}}
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
The position vectors of radius are $2 \hat{i}+\hat{j}+\hat{k}$ and $2 \hat{i}-3 \hat{j}+\hat{k}$ while those of linear momentum are $2 \hat{i}+3 \hat{j}-\hat{k}$. Then the angular momentum is

Options:
A $2 \hat{i}-4\hat{j}+2\hat{k}$
B $2 \hat{i}-4\hat{j}$
C $4 \hat{i}-8\hat{k}$
D \(4 \hat{i}+8 \hat{k}\)

The correct answer is:
$4 \hat{i}-8\hat{k}$

Solution:
Correct option is 2. $4 \hat{i}-8 \hat{k}$
Radius vector $r=r_2-r_1=(2 \hat{i}-3 \hat{j}+\hat{k})-(2 \hat{i}+\hat{j}+\hat{k})$
$\therefore r-4 j$
Linear momentum $\vec{p}=2 \hat{i}+3 \hat{j}-\hat{k}$
$\begin{aligned} & L=r \times p=(-4 j) \times(2 \hat{i}+3 \hat{j}-\hat{k}) \\ & =\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & -4 & 0 \\ 2 & 3 & -1\end{array}\right|=4 \hat{i}-8 \hat{k}\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main
The position vectors of the vertices A, B and C of a triangle are 2i^-3j^+3k^, 2i^+2j^+3k^ and -i^+j^+3k^ respectively. Let l denotes the length of the angle bisector AD of ∠BAC where D is on the line segment BC, then 2l2 equals :

Options:
A 49
B 42
C 50
D 45

The correct answer is:
45

Solution:
Given: A2,-3,3, B2,2,3 and C-1,1,3 are vertices of a △ABC.

AB=2-22+2+32+3-32

⇒AB=5

⇒AC=2+12+-3-12+3-32

⇒AC=9+16

⇒AC=5

Also, AD is the angle bisector of ∠BAC.

⇒ABAC=BDCD

⇒BD=CD

∴ D is midpoint of BC

So, using mid-point formula,

⇒D≡2-12,2+12,3+32

⇒D≡12,32,3

⇒AD=l=2-122+-3-322+(3-3)2

⇒l=322+-922

⇒l=94+814

⇒l=452

⇒l2=452

⇒2l2=45

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The position vectors of vertices of a $\Delta A B C$ are $\begin{array}{lll}4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}, & \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \quad \text { and } & -\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}}\end{array}$
respectively, then $\angle A B C$ is equal to

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{4}$
C $\frac{\pi}{3}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{\pi}{2}$

Solution:
Here $, \overrightarrow{\mathbf{A B}}=-3 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, \overrightarrow{\mathbf{B C}}=-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
and $\overrightarrow{\mathbf{A B}} \cdot \overrightarrow{\mathbf{B C}}=6+6-12=0$
$$
\angle A B C=\frac{\pi}{2}
$$

Topic: Mathematics, Vector Algebra, JEE Main
The position $\mathrm{x}$ of a particle varies with time $(\mathrm{t})$ as $\mathrm{x}=\mathrm{A} \mathrm{t}^{2}-\mathrm{B} \mathrm{t}^{3}$. The acceleration at time $\mathrm{t}$ of the particle will be equal to zero. What is the value of $t ?$

Options:
A $\frac{2 \mathrm{~A}}{3 \mathrm{~B}}$
B $\frac{\mathrm{A}}{\mathrm{B}}$
C $\frac{A}{3 B}$
D zero

The correct answer is:
$\frac{A}{3 B}$

Solution:
Given that $x=A t^{2}-B t^{3}$
$$
\therefore \text { velocity }=\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{~A} \mathrm{t}-3 \mathrm{~B} \mathrm{t}^{2}
$$
and acceleration
$$
=\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)=2 \mathrm{~A}-6 \mathrm{Bt}
$$
For acceleration to be zero $2 \mathrm{~A}-6 \mathrm{Bt}=0$.
$$
\therefore \quad \mathrm{t}=\frac{2 \mathrm{~A}}{6 \mathrm{~B}}=\frac{\mathrm{A}}{3 \mathrm{~B}}
$$

Topic: Physics, Motion In One Dimension, JEE Main
The position x of a particle with respect to time t along the x-axis is given by xâ=â9 t 2 â t 3 where x is in metres and t in second. What will be the position of this particle (in m) when it achieves maximum speed along the positive x-direction?

Options:

The correct answer is:
54

Solution:
Let us first find speed.

Speed, vâ=âdxdtâ=âddt(9t2ât3)â=â18tâ3t2

For maximum speed, using dvdt=0

dvdtâ=â0âââ18â6tâ=â0âââtâ=â3 ,

Now finding displacement at that time, we have :

ââxmaxâ=â81â27â=â54 m

Topic: Physics, Motion In One Dimension, JEE Main
The positive charge of carbocation can be delocalised over how many oxygen atoms in the resonating structures?

Options:

The correct answer is:
2

Solution:
In the given structure, carbocation is at the benzylic position. Hence, it can be delocalized at ortho- and para- positions of the ring. Therefore, the oxygen atoms which are directly connected to the ortho- and para- positions of the ring, will delocalise and stabilized the carbocation.

Topic: Chemistry, General Organic Chemistry, JEE Main
The positive difference between the local maximum value and the local minimum value of the function fx=x3-3x-1, âxâ-2,3 is

Options:
A 20
B 4
C 14
D 22

The correct answer is:
4

Solution:
As, f â² x =3 x 2 â1
=3x-1x+1
â´ Local maximum =-1+3-1=1
And local minimum =-3
â´ The positive difference =4

Topic: Mathematics, Application of Derivatives, JEE Main
The positive feedback is required by an amplifier to act as an oscillator. The feedback here means

Options:
A External input is necessary to sustain ac signal in output.
B A portion of the output power is returned back to the input.
C Feedback can be achieved by LR network.
D The base-collector junction must be forward biased.

The correct answer is:
A portion of the output power is returned back to the input.

Solution:
When the amplifier connects with positive feedback, it acts as the oscillator. The feedback employed here is positive feedback, this means some amount of the output voltage is given to the input. Hence, the circuit continues its process without any input to the oscillator. The positive feedback is also known as regenerative feedback because the system sustains or regenerates itself without any input source. Hence, the oscillator is nothing but an amplifier with the positive feedback.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2022 (26 Jun Shift 2)
The positive integer just greater than $(1+0.0001)^{10000}$ is

Options:
A
4

B
5

C
2

D
3

The correct answer is:
2

Solution:
No Solution Available

Topic: Mathematics, Binomial Theorem, JEE Main
The positive integer just greater than $(1+0.0001)^{10000}$ is

Options:
A
4

B
5

C
2

D
3

The correct answer is:
2

Solution:
No Solution Available

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2002