datavorous/entrance-exam-dataset huggingface dataset:
The direction cosines of the normal to the plane $2 x+3 y-6 z=5$ are

Options:
A $2,3,-6$
B $\frac{2}{7}, \frac{3}{7},-\frac{6}{7}$
C $\frac{2}{5}, \frac{3}{5},-\frac{6}{5}$
D None of these

The correct answer is:
$\frac{2}{7}, \frac{3}{7},-\frac{6}{7}$

Solution:
Direction ratios of normal to the plane are $(2,3,-6)$
$\therefore$ Direction cosines are, $\left(\frac{2}{\sqrt{49}}, \frac{3}{\sqrt{49}},-\frac{6}{\sqrt{49}}\right)$ or $\left(\frac{2}{7}, \frac{3}{7}, \frac{-6}{7}\right)$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The direction cosines of the normal to the plane $2 x+3 y-6 z=5$ are

Options:
A $2,3,-6$
B $\frac{2}{7}, \frac{3}{7},-\frac{6}{7}$
C $\frac{2}{5}, \frac{3}{5},-\frac{6}{5}$
D None of these

The correct answer is:
$\frac{2}{7}, \frac{3}{7},-\frac{6}{7}$

Solution:
Direction ratios of normal to the plane are $(2,3,-6)$
$\therefore$ Direction cosines are,
$\left(\frac{2}{\sqrt{49}}, \frac{3}{\sqrt{49}},-\frac{6}{\sqrt{49}}\right) \text { or }\left(\frac{2}{7}, \frac{3}{7}, \frac{-6}{7}\right)$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The direction cosines of two lines satisfy 2l+2m-n=0 and lm+mn+nl=0. The angle between these lines is

Options:
A π4
B π2
C π6
D π3

The correct answer is:
π2

Solution:
Substituting n=2l+m in lm+mn+nl=0, we get
lm+2l+m2=0⇒2l2+5lm+2m2=0
⇒2l+ml+2m=0
Hence m=-2l⇒n=-2l or m=-l2⇒n=l
So D.C’s of the lines are (l,-2l,-2l) or l,-l2,l
Hence, the angle between them is 90o

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The direction of induced current is such that it opposes the very cause that has produced it. This is the law of

Options:
A Lenz
B Faraday
C Kirchhoff
D Fleming

The correct answer is:
Lenz

Solution:
Lenz's Law: If an induced current flows, its direction is always such that it will oppose the change which produced it.

Topic: Physics, Electromagnetic Induction, JEE Main
The direction of induced e.m.f. during electromagnetic induction is given by

Options:
A Faraday's law
B Lenz's law
C Maxwell's law
D Ampere's law

The correct answer is:
Lenz's law

Solution:
Lenz's Law:
According to this law, the emf will be induced in a coil due to a changing magnetic flux in such a way that it will oppose the cause which has produced it
This law states that the induced emf in a conductor dueto a changing magnetic flux is such that the magnetic field created by the induced emf opposes the change in a magnetic field.
$\Rightarrow e m f=-N\left(\frac{d \phi}{d t}\right)$
where $\mathrm{N}=$ number of loops and $\mathrm{d} \phi=$ Change in magnetic flux
The above equation is given by Faraday's law, but the negative sign is a result of Lenz's law.
EXPLANATION:
The Lenz law states that the induced emf in a conductor dueto a changing magnetic flux is such that the magnetic field created by the induced emf opposes the change in a magnetic field.
So the direction of induced emf during electromagnetic induction is given by Lenz's law. Hence, option 2 is correct.

Topic: Physics, Electromagnetic Induction, JEE Main
The direction ratios of normal to the plane through the

points (0,-1,0) and (0,0,1) and making an angle $\frac{\pi}{4}$

with the plane $y-z+5=0$ are;

2,-1,1

$2, \sqrt{2}-\sqrt{2}$

$\sqrt{2}, 1,-1$

$2 \sqrt{3}, 1,-1$

Options:
A option 1 and 2
B option 2 and 3
C option 3 and 4
D all the options

The correct answer is:
option 2 and 3

Solution:
Let the d.r's of the normal be $\langle a, b, c\rangle$ Equation of the plane is $a(x-0)+b(y+1)+c(z-0)=0$

$\because$ It passes through (0,0,1)

$\therefore \quad b+c=0$

Also $\frac{0 \cdot a+b-c}{\sqrt{a^{2}+b^{2}+c^{2} \cdot \sqrt{2}}}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$

$\Rightarrow \quad b-c=\sqrt{a^{2}+b^{2}+c^{2}}$

And $b+c=0$

$\Rightarrow \quad b=\pm \frac{1}{\sqrt{2}} a$

$\therefore \quad$ The d.r's are $\sqrt{2}, 1,-1$ or $2, \sqrt{2},-\sqrt{2}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The direction ratios of the line joining the points $(4,3,-5)$ and $(-2,1,-8)$ are

Options:
A $\frac{6}{7}, \frac{2}{7}, \frac{3}{7}$
B $6,2,3$
C $2,4,-13$
D None of these

The correct answer is:
$6,2,3$

Solution:
Direction ratios are, $l=4-(-2)=6, m=3-1=2$ and $n=-5+8=3$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The direction ratios of the line perpendicular to the lines having direction ratios $2,3,1$
and $1,2,1$ are

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

The correct answer is:
$1,-1,1$

Solution:
(D)
Let $\bar{a}$ and $\bar{b}$ be the vectors along the lines whose direction ratios are $2,3,1$ and $1,2,1$ respectively.
$\therefore \overline{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}} \quad \text { and } \quad \overline{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
A vector perpendicular to both $\bar{a}$ and $\bar{b}$ is given by,
$\bar{a} \times \bar{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 1 \\
1 & 2 & 1
\end{array}\right|=\hat{i}(3-2)-\hat{j}(2-1)+\hat{k}(4-3)=\hat{i}-\hat{j}+\hat{k}$
Hence d.r.s are $1,-1,1$

Topic: Mathematics, Vector Algebra, JEE Main
The direction ratios of the normal to the plane passing through the points 1, -2, 3, -1, 2, -1 and parallel to the line x-22=y+13=z4 are proportional to

Options:
A 2, 3, 4
B 4, 0, 7
C -2, 0, -1
D 2, 0, -1

The correct answer is:
2, 0, -1

Solution:
Let, a,b,c be the DRs.

The equation of a plane passing through 1, -2, 3 is
ax-1+by+2+cz-3=0 .... (i)
Since, it passes through -1, 2,-1 and is parallel to the given line
∴ a-2+b4+c-4=0
Also, 2a+3b+4c=0
⇒ a28=b0=c-14
⇒ a2=b0=c-1
Hence, a :b :c=2 :0 : -1

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The directrix of the hyperbola is $\frac{x^2}{9}-\frac{y^2}{4}=1$

Options:
A $x=9 / \sqrt{13}$
B $y=9 / \sqrt{13}$
C $x=6 / \sqrt{13}$
D $y=6 / \sqrt{13}$

The correct answer is:
$x=9 / \sqrt{13}$

Solution:
Directrix of hyperbola $\quad x=\frac{a}{e}$,
where $e=\sqrt{\frac{b^2+a^2}{a^2}}=\frac{\sqrt{b^2+a^2}}{a}$
Directrix is, $\quad x=\frac{a^2}{\sqrt{a^2+b^2}}=\frac{9}{\sqrt{9+4}} \Rightarrow x=\frac{9}{\sqrt{13}}$

Topic: Mathematics, Hyperbola, JEE Main
The disc of mass M with uniform surface mass density σ is shown in the figure. The center of mass of the quarter disc (the shaded area) is at the position xa3π, xa3π where x is _______ . (Round off to the Nearest Integer) [a is an area as shown in the figure]

Options:

The correct answer is:
4

Solution:
C.O.M of quarter disc is at 4a3π,4a3π

=4

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2021 (17 Mar Shift 2)
The discriminant of the quadratic 2x+12+3x+22+4x+32+… n terms =0, ∀n>3, x∈R is

Options:
A positive
B zero
C negative
D depends on n

The correct answer is:
negative

Solution:
2x+12+3x+22+4x+32+…n terms =0,x∈R
This is possible only when individually every perfect square is zero, but this can’t happen here.
So, the above equation doesn’t have any real roots.
⇒D<0

Topic: Mathematics, Quadratic Equation, JEE Main
The disintegration energy $Q$ for the nuclear fission of ${ }^{235} \mathrm{U} \rightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+n$ is ______ $\mathrm{MeV}$.
Given atomic masses of ${ }^{235} \mathrm{U}: 235.0439 u ;{ }^{140} \mathrm{Ce}: 139.9054 u$,
${ }^{94} \mathrm{Zr}: 93.9063 u ; n: 1.0086 u \text {, }$
Value of $c^2=931 \mathrm{MeV} / \mathrm{u}$

Options:

The correct answer is:
208

Solution:
${ }^{235} \mathrm{U} \rightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+\mathrm{n}$
Disintegration energy
$\begin{aligned}
\mathrm{Q} & =\left(\mathrm{m}_{\mathrm{R}}-\mathrm{m}_{\mathrm{p}}\right) \mathrm{c}^2 \\
\mathrm{~m}_{\mathrm{R}} & =235.0439 \mathrm{u} \\
\mathrm{m}_{\mathrm{p}} & =139.9054 \mathrm{u}+93.9063 \mathrm{u}+1.0086 \mathrm{u} \\
& =234.8203 \mathrm{u} \\
\mathrm{Q} & =(235.0439 \mathrm{u}-234.8203 \mathrm{u}) \mathrm{c}^2 \\
& =0.2236 \mathrm{c}^2 \\
& =0.2236 \times 931 \\
\mathrm{Q} & =208.1716
\end{aligned}$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2024 (04 Apr Shift 2)
The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minute. 10 minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is

(Take loge1.88=0.63)

Options:
A 0.02 min-1
B 2.7min-1
C 0.063 min-1
D 6.3 min-1

The correct answer is:
0.063 min-1

Solution:
At t=0 disintegration rate A0=4250 dpm

At t=10 min disintegration rate At=2250dpm

The activity of a sample at any time is given by,

At=A0e-λt

⇒2250=4250e-λ10

⇒10λ=ln42502250=ln1.88

⇒λ=0.063 min-1

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2022 (26 Jul Shift 1)
The disperse phase in colloidal iron (III) hydroxide and colloidal gold is positively and negatively charged, respectively, which of the following statements is NOT correct?

Options:
A
magnesium chloride solution coagulates, the gold sol more readily than the iron (III) hydroxide sol.

B
sodium sulphate solution causes coagulation in both sols

C
mixing the sols has no effect

D
coagulation in both sols can be brought about by electrophoresis

The correct answer is:
mixing the sols has no effect

Solution:
No Solution Available

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2005
The disperse phase in colloidal iron (III) hydroxide and colloidal gold is positively and negatively charged, respectively. Which of the following statements is NOT correct?

Options:
A Magnesium chloride solution coagulates, the gold solution more readily than the iron (III) hydroxide solution
B Sodium sulphate solution causes coagulation in both the solutions
C Mixing of the solutions has no effect
D Coagulation in both solutions can be brought about by electrophoresis

The correct answer is:
Mixing of the solutions has no effect

Solution:
Opposite charges attract each other. Hence on mixing, coagulation of two solutions may take place.

Topic: Chemistry, Surface Chemistry, JEE Main
The displacement and the increase in the velocity of a moving particle in the time interval of t to (t+1) s are 125 m and 50 m s-1, respectively. The distance travelled by the particle in (t+2)th s is ___________ m.

Options:

The correct answer is:
175

Solution:
Considering acceleration is constant, we can write

v=u+at

As increase in velocity in 1 s is given as 50 m s-1, therefore

⇒a=50 m s-2

Now, displacement covered between given time interval is,

125=ut+12at2

⇒125=u+a2

⇒u=100 m s-1

Velocity after 1 s will be, v=100+50=150 m s-1

Therefore, displacement covered in t+1 s to t+2 s will

∴Snth=vt'+12at'2, where t'=1 s

=175 m

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (30 Jan Shift 1)
The displacement current of 4.425 μA is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 106 V s-1. The area of each plate of the capacitor is 40 cm2. The distance between each plate of the capacitor is x×10-3 m. The value of x is ,
(Permittivity of free space, ε0=8.85×10-12 C2 N-1 m-2) _______

Options:

The correct answer is:
8

Solution:
Displacement current is given by id=ε0dϕEdt

Or id=ε0ddtEA, where E=qAε0

Or id=ε0ddtqAAε0=dqdt=ddtCV

Or id=CdVdt=ε0AddVdt

Putting the values, we have

4.425×10-6=8.85×10-12×40×10-4×106 d

d=2×10-6×10-4×106×40

d=80×10-4=8×10-3 m

Hence, value of x=8.

Topic: Physics, Capacitance, JEE Main, JEE Main 2022 (29 Jun Shift 2)
The displacement current was first postulated by,

Options:
A Maxwell
B Marconi
C Ampere
D Hertz

The correct answer is:
Maxwell

Solution:
The concept of displacement current was postulated by Maxwell because if there was no displacement current, the Ampere circuital law would not hold true in a capacitor.

Displacement current is given by,

Id=ε0dϕedt

where, ε0 is the permittivity of the free space.

dϕedt is the rate of change of electric flux with respect to time.

Topic: Physics, Electromagnetic Waves, JEE Main
The displacement equations of two interfering waves are given by y1=10 sin ωt+π3 cm, y2=5sin ωt+3 cos ωt cm respectively. The amplitude of the resultant wave is _____ cm.

Options:

The correct answer is:
20

Solution:
Displacement equation of second wave can be written as,y2=5sinωt+3cosωt=10sinωt×12+32×cosωt=10sinωt×cosπ3+sinπ3×cosωt

=10sinωt+π3

As we can see, the phase difference between the waves is 0. Therefore, resultant amplitude will be sum of the amplitudes of the waves.

So A=A1+A2=10 cm+10 cm=20 cm

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2023 (31 Jan Shift 2)
The displacement is given by $x=2 t^2+t+5$, the acceleration at $t=2 s$ is

Options:
A $4 \mathrm{~m} / \mathrm{s}^2$
B $8 \mathrm{~m} / \mathrm{s}^2$
C $10 \mathrm{~m} / \mathrm{s}^2$
D $15 \mathrm{~m} / \mathrm{s}^2$

The correct answer is:
$4 \mathrm{~m} / \mathrm{s}^2$

Solution:
Displacement $x=2 t^2+t+5$
Velocity $=\frac{d x}{d t}=4 t+1$
Acceleration $=\frac{d^2 x}{d t^2}=4$ i.e. independent of time
Hence acceleration $=4 \mathrm{~m} / \mathrm{s}^2$

Topic: Physics, Motion In One Dimension, JEE Main
The displacement of a damped harmonic oscillator is given by xt=e-0.1t cos10πt+φ, where t is the time in seconds. The time taken for the amplitude of the oscillator to drop to half of its initial value is close to

Options:
A 13 s
B 7 s
C 4 s
D 27 s

The correct answer is:
7 s

Solution:
A02=A0e-0.1t⇒e-0.1t=2⇒0.1t=ln2

t=ℓn20.1=10en2=6.93=7s

Topic: Physics, Oscillations, JEE Main
The displacement of a damped harmonic oscillator is given by xt=e-0.1tcos⁡10πt+φ. Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to:

Options:
A 27 s
B 4 s
C 13 s
D 7 s

The correct answer is:
7 s

Solution:
Amplitude A=A0e-kt
From given condition, A02=A0e-0.1×t
⇒ln2=0.1×t
t=ln20.1=0.6930.10=6.935≈7 s

Topic: Physics, Oscillations, JEE Main, JEE Main 2019 (10 Apr Shift 1)
The displacement of a particle executing SHM is given by $x=10 \sin \left(w t+\frac{\pi}{3}\right) m$. The time period of motion is $3.14 \mathrm{~s}$. The velocity of the particle at $t=0$ is ______$\mathrm{m} / \mathrm{s}$.

Options:

The correct answer is:
10

Solution:
Given,
$\begin{aligned}
& \mathrm{T}=3.14=\frac{2 \pi}{\omega} \\
& \omega=2 \mathrm{rad} / \mathrm{s} \\
& \mathrm{x}=10 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right) \\
& \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=10 \omega \cos \left(\omega \mathrm{t}+\frac{\pi}{3}\right) \\
& \text { at } \mathrm{t}=0 \\
& \mathrm{v}=10 \omega \cos \left(\frac{\pi}{3}\right)=10 \times 2 \times \frac{1}{2}[\mathrm{using} \omega=2 \mathrm{rad} / \mathrm{s}] \\
& \mathrm{v}=10 \mathrm{~m} / \mathrm{s}
\end{aligned}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2024 (04 Apr Shift 2)
The displacement of a particle executing SHM is given by
$y=5 \sin \left(4 t+\frac{\pi}{3}\right) .$
If $T$ is the time period and the mass of the particle is $2 \mathrm{~g}$, the kinetic energy of the particle when $t=\frac{T}{4}$ is given by

Options:
A 0.4 J
B 0.5 J
C 3 J
D 0.3 J

The correct answer is:
0.4 J

Solution:
The displacement of particle, executing SHM
Velocity of particle
$\begin{aligned}
\left(\frac{d y}{d t}\right) & =\frac{5 d}{d t} \sin \left(4 t+\frac{\pi}{3}\right) \\
& =5 \cos \left(4 t+\frac{\pi}{3}\right) \cdot 4 \\
& =20 \cos \left(4 t+\frac{\pi}{3}\right)
\end{aligned}$
Velocity at $t=\left(\frac{T}{4}\right)$
$\left(\frac{d y}{d t}\right)_{t=\frac{T}{4}}=20 \cos \left(4 \times \frac{T}{4}+\frac{\pi}{3}\right)$
Comparing the given equation with standard equation of SHM, given by
We get
$\begin{aligned}
& y=a \sin (\omega t+\phi) \\
& \omega=4
\end{aligned}$
As
$\omega=\frac{2 \pi}{T}$
$\begin{aligned} & \Rightarrow \quad T=\frac{2 \pi}{\omega} \\ & \text { or } & T=\frac{2 \pi}{4} \\ & \text { or } & T=\left(\frac{\pi}{2}\right) \\ & \end{aligned}$
Now, putting value of $T$ in Eq. (ii), we get
$\begin{aligned}
u & =20 \cos \left(\frac{\pi}{2}+\frac{\pi}{3}\right) \\
& =-20 \sin \frac{\pi}{3} \\
& =-20 \times \frac{\sqrt{3}}{2} \\
& =-10 \times \sqrt{3}
\end{aligned}$
The kinetic energy of particle,
$\begin{aligned}
& \mathrm{KE}=\frac{1}{2} m u^2 \\
& \because \quad m=2 \mathrm{~g}=2 \times 10^{-3} \mathrm{~kg} \\
& =\frac{1}{2} \times 2 \times 10^{-3} \times(-10 \sqrt{3})^2 \\
& =10^{-3} \times 100 \times 3 \\
& =3 \times 10^{-1} \\
& \mathrm{KE}=0.3 \mathrm{~J} \\
&
\end{aligned}$

Topic: Physics, Oscillations, JEE Main
The displacement of a particle is given at time t, by:

$x=A \sin (-2 \omega t)+B \sin ^{2} \omega t \quad$ Then,

Options:
A the motion of the particle is SHM with an amplitude of $\sqrt{A^{2}+\frac{B^{2}}{4}}$
B the motion of the particle is not SHM, but oscillatory with a time period of $T=\pi / \omega$
C the motion of the particle is oscillatory with a time period of $T=\pi / 2 \omega$
D the motion of the particle is a periodic.

The correct answer is:
the motion of the particle is SHM with an amplitude of $\sqrt{A^{2}+\frac{B^{2}}{4}}$

Solution:
The displacement of the particle is given by:

$$

\begin{aligned}

x &=A \sin (-2 \omega t)+B \sin ^{2} \omega t \\

&=-A \sin 2 \omega t+\frac{B}{2}(1-\cos 2 \omega t) \\

&=-\left(A \sin 2 \omega t+\frac{B}{2} \cos 2 \omega t\right)+\frac{B}{2}

\end{aligned}

$$

This motion represents SHM with an amplitude: $\sqrt{A^{2}+\frac{B^{2}}{4}},$ and mean position

$$

\frac{B}{2}

$$

Topic: Physics, Oscillations, JEE Main
The displacement of a particle is proportional to the cube of time elapsed. How does the acceleration of the particle depends on time obtained

Options:
A $a \propto t^2$
B $a \propto 2 t$
C $a \propto t^3$
D $a \propto t$

The correct answer is:
$a \propto t$

Solution:
$\begin{aligned} & x \propto t^3 \quad \therefore x=K t^3 \\ & \Rightarrow v=\frac{d x}{d t}=3 K t^2 \text { and } a=\frac{d v}{d t}=6 K t \\ & \text { i.e } a \propto t\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main
The displacement of a particle, moving in a straight line, is given by $s=2 t^2+2 t+4$ where s in metres and ${ }^t$ in seconds. The acceleration of the particle is

Options:
A $2 \mathrm{~m} / \mathrm{s}^2$
B $4 \mathrm{~m} / \mathrm{s}^2$
C $6 \mathrm{~m} / \mathrm{s}^2$
D $8 \mathrm{~m} / \mathrm{s}^2$

The correct answer is:
$4 \mathrm{~m} / \mathrm{s}^2$

Solution:
$s=2 t^2+2 t+4, a=\frac{d^2 s}{d t^2}=4 \mathrm{~m} / \mathrm{s}^2$

Topic: Physics, Motion In One Dimension, JEE Main
The displacement of a particle moving in a straight line is given by the expression $x=A t^3+B t^2+C t+D$ in metres, where $t$ is in seconds and $A, B, C$ and $D$ are constants. The ratio between the initial acceleration and initial velocity is

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

The correct answer is:
$\frac{2 B}{C}$

Solution:
Displacement of a particle moving in a straight line $(x)=A t^3+B t^2+C t+D$
$\therefore$ Velocity,$\quad v=\frac{d x}{d t}=3 A t^2+2 B t+C$
$$
v_{\text {initial }}=C
$$
and
$$
\begin{aligned}
a & =\frac{d^2 x}{d t^2}=6 A t+2 B \\
a_{\text {initial }} & =2 B
\end{aligned}
$$
$\therefore$ Ratio between initial acceleration and initial velocity is $\frac{2 B}{C}$.

Topic: Physics, Motion In One Dimension, JEE Main
The displacement of a particle of mass 3 g executing simple harmonic motion is given by Y=3 sinâ¡(0.2t) in SI units. The KE of the particle at a point which is at a distance equal to 1/3 of its amplitude from its mean position is

Options:
A 12Ã10-3Â J
B 25Ã10-3Â J
C 0.48Ã10-3 J
D 0.24Ã10-3Â J

The correct answer is:
0.48Ã10-3 J

Solution:
Displacement of particle in the case of SHM
y=Asinâ¡(Ït) ...(i)
y=3sinâ¡(0.2 t) ...(ii)(given)
Comparing Eqs. (i) and (ii), we get
A=3, Ï=0.2
Now, particle distance x=A3=1
Kinetic energy in SHM
= 12mÏ2(A2-x2)
= 12Ã3Ã10-30.22[32-12]
=0.48Ã10-3 J

Topic: Physics, Oscillations, JEE Main
The displacement of a particle varies according to the relation $x=4(\cos \pi t+\sin \pi t)$. The amplitude of the particle is

Options:
A
$-4$

B
$4$

C
$4 \sqrt{2}$

D
$8$

The correct answer is:
$4 \sqrt{2}$

Solution:
$\left.x=4(\cos \pi t+\sin \pi t)=4\left[\sin \left(\frac{\pi}{2}-\pi t\right)\right]+\sin \pi t\right]=4\left[2 \times \sin \left(\frac{\pi t-\frac{\pi}{2}-\pi t}{2}\right) \cos \left(\frac{\pi \mathrm{t}-\frac{\pi}{2}+\pi \mathrm{t}}{2}\right)\right]$
$=8\left[\sin \frac{\pi}{4} \cdot \cos \left(-\frac{\pi}{4}+\pi t\right)\right]$
$=\frac{8}{\sqrt{2}} \cdot \cos \left[\pi t-\frac{\pi}{4}\right]=4 \sqrt{2} \cos \left[\pi t-\frac{\pi}{4}\right]$
Comparing it with standard equation
$X=A \cos (w t-K x) \quad \Rightarrow A=4 \sqrt{2}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2003
The displacement of an object attached to a spring and executing simple harmonic motion is given by $x=2 \times 10^{-2} \cos \pi t$ metres. The time at which the maximum speed first occurs is

Options:
A
$0.5 \mathrm{~s}$

B
$0.75 \mathrm{~s}$

C
$0.125 \mathrm{~s}$

D
$0.25 \mathrm{~s}$

The correct answer is:
$0.5 \mathrm{~s}$

Solution:
$\begin{aligned}
& \mathrm{x}=2 \times 10^{-2} \cos \pi t \\
& \mathrm{v}=-0.02 \pi \sin \pi t
\end{aligned}$
$\mathrm{v}$ is maximum at $\mathrm{t}=\frac{1}{2}=0.5 \mathrm{~sec}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2007
The displacement of an object of mass 3 kg is given by the relation S=13t2 , where t is time in seconds. If the work done by the net force on the object in 2 s is pq joule, where p and q are smallest integer values, then what is the value of p+q?

Options:

The correct answer is:
11

Solution:
S=t23
v=dSdt
dSdt=23t=v;
W=âK=Kt=2-Kt=0
W=12mvt=22-vt=02=12â3169-0= 8/3 J
âÂ Â p+q=11

Topic: Physics, Work Power Energy, JEE Main
The displacement of simple harmonic oscillator after 3 seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is

Options:
A 12 s
B 8 s
C 36 s
D 6 s

The correct answer is:
36 s

Solution:
Displacement equation is given by x=Asinωt

Now, when amplitude is halved, A2=Asinωt

Or ωt=π6

At, t=3 s, ω3=π6⇒ω=π18

Time period, T=2πω=2ππ18=36 s

Topic: Physics, Oscillations, JEE Main, JEE Main 2022 (27 Jun Shift 1)
The displacement (x) is plotted with time (t) for a particle executing SHM as shown below. The correct equation of SHM is

Options:
A x=4sin⁡πt6+π6
B x=4sin⁡πt6-π6
C x=2sin⁡2π6+π6
D x=2sin⁡πt6-π6

The correct answer is:
x=4sin⁡πt6-π6

Solution:
x=4sinâ¡(Ït +Ï)
Ï=2ÏT=2Ï12=Ï6
x=4sinâ¡Ït6+Ï
-2=4sinâ¡Ï
-12=sinâ¡Ï
Ï=-Ï6
x=4sinâ¡Ït6-Ï6

Topic: Physics, Oscillations, JEE Main
The displacement $x$ of a particle varies with time $t$ as $x=a e^{-a t}+B^{\beta x}$, where $a, b, \alpha$ and $\beta$ are positive constants. The velocity of the particle will:

Options:
A be independent of $\beta$
B drop to zero when $\alpha=\beta$
C go on decreasing with time
D go on increasing with time

The correct answer is:
go on increasing with time

Solution:
From question
$$
\begin{aligned}
& x=a e^{-\alpha \mathrm{t}}+b e^{\beta t} \\
& V=\frac{d x}{d t}=-a \alpha e^{-x t}+b \beta_e^{\mathrm{B}+} \\
& \Rightarrow \mathrm{V}=-a \alpha e^{-t}+b \beta \varepsilon^{\beta t}
\end{aligned}
$$
$\therefore$ Velocity will decrease for certain value of $t$.

Topic: Physics, Motion In One Dimension, JEE Main
The displacement y of a particle, if given by y=4cos2â¡t2sinâ¡1000 t. This expression may be considered to be a result of the superposition of how many simple harmonic motions?

Options:
A 4
B 3
C 2
D 5

The correct answer is:
3

Solution:
âµ Equation of displacement of particle is
y=4cos2â¡t2sinâ¡1000t .....(i)
Or y=41+cost2sin1000t
âµ cos2â¡x=1+cosâ¡2x2
Or y=2sinâ¡1000t+2costsin1000t
Or y=2 sin1000t+sin1001t+sint 999t
2sinâ¡AcosB=sinâ¡A+B+sinâ¡A-B
Hence, equation (i) is the result of the superposition of three simple harmonic motion.

Topic: Physics, Oscillations, JEE Main
The displacement y of a particle in a medium can be expressed as, $y=10^{-6} \sin \left(100 t+20 x+\frac{\pi}{4}\right) m$ where $t$ is in second and $x$ in meter. The speed of the wave is

Options:
A $20 \mathrm{~m} / \mathrm{s}$
B $5 \mathrm{~m} / \mathrm{s}$
C $2000 \mathrm{~m} / \mathrm{s}$
D $5 \pi \mathrm{m} / \mathrm{s}$

The correct answer is:
$5 \mathrm{~m} / \mathrm{s}$

Solution:
From equation given,

$\begin{array}{l}

\omega=100 \\

\text { and } \mathrm{k}=20 \\

\mathrm{v}=\frac{\omega}{\mathrm{k}}=\frac{100}{20}=5 \mathrm{~m} / \mathrm{s}

\end{array}$

Topic: Physics, Waves and Sound, JEE Main