datavorous/entrance-exam-dataset huggingface dataset:
A particle starts moving rectilinearly at time $\mathrm{t}=0$ such that its velocity $\mathrm{v}$ changes with time $\mathrm{t}$ according to the equation $v=t^{2}-t$ where $t$ is in seconds and $v$ is in $\mathrm{m} / \mathrm{s}$. Find the time interval for which the particle retards.

Options:
A $\frac{1}{2}<\mathrm{t}<1$
B $\frac{1}{2}>t>1$
C $\frac{1}{4}<\mathrm{t}<1$
D $\frac{1}{2}

The correct answer is:
$\frac{1}{2}<\mathrm{t}<1$

Solution:
Acceleration of the particle $a=2 t-1$ The particle retards when acceleration is opposite to velocity.
$$
\Rightarrow \text { a } \cdot v<0 \Rightarrow(2 t-1)\left(t^{2}-t\right)<0 \Rightarrow t(2 t-1)
$$
$(\mathrm{t}-1)<0$
Now t is always positive
$$
\therefore(2 t-1)(t-1)<0
$$
or $2 \mathrm{t}-1<0$ and $\mathrm{t}-1>0 \Rightarrow \mathrm{t}<\frac{1}{2}$ and $\mathrm{t}>1$.
This is not possible or $2 \mathrm{t}-1>0 \& \mathrm{t}-1<0 \Rightarrow 1 / 2<\mathrm{t}<1$

Topic: Physics, Motion In One Dimension, JEE Main
A particle starts SHM from the mean position. Its amplitude is a and total energy E. At one instant its kinetic energy is 3E4. Its displacement at that instant is

Options:
A a2
B a2
C a32
D a3

The correct answer is:
a2

Solution:
K=12 mω2(a2-y2)
34 E=12 mω2(a2-y2)
34 12 mω2a2= 12 mω2(a2-y2)
y2=a2-34 a2=a24
⇒ y=a2

Topic: Physics, Oscillations, JEE Main
A particle starts with an initial velocity of 10.0 ms-1 along x-direction and accelerates uniformly at the rate of 2.0 m s-2. The time taken by the particle to reach the velocity of 60.0 m s-1 is _____.

Options:
A 25 s
B 3 s
C 6 s
D 30 s

The correct answer is:
25 s

Solution:
The given data is

v=60 m s-1u=10 m s-1a=2 m s-2

Using the equation v=u+at, we get

60=10+2t⇒t=60-102⇒t=25 s

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (06 Apr Shift 2)
A particle starts with speed v0 from x=0 along x-axis with retardation proportional to the square of its displacement. Work done by the force acting on the particle is proportional to

Options:
A x52
B x3
C ex
D x2

The correct answer is:
x3

Solution:
Given a=-Kx2

or –vdvdx=x2

integrating both sides
–12v2–v02=x33

Now, using work-energy theorem W=∆KE

W ∝ x3

Topic: Physics, Work Power Energy, JEE Main
A particle undergoes a one-dimensional motion such that its velocity varies according to vx=Î²x-2n, where Î² and n are constants and x is the position of the particle. The acceleration of the particle as a function of x is given by

Options:
A -2nβ2x-2n-1
B -2nβ2x-4n-1
C -2nβ2x-2n+1
D -2nβ2x-4n+1

The correct answer is:
-2nβ2x-4n-1

Solution:
According to the question
a=vdvdx=vddxÎ²xâ-2n
=-2nÎ²x-2n-1 Î²x-2n
=-2nÎ²2x-4n-1

Topic: Physics, Motion In One Dimension, JEE Main
A particle undergoes uniform circular motion. About which point on the plane of the circle will the angular momentum of the particle remains conserved?

Options:
A Centre of the circle
B On the circumference of the circle
C Inside the circle
D Outside the circle

The correct answer is:
Centre of the circle

Solution:
Force does not produce any torque because it passes through the centre (Point or rotation) and we know that if τ=0 the L = constant.

Topic: Physics, Rotational Motion, NEET
A particle which is experiencing a force, given by F→=3i^-12j^, undergoes a displacement of d→=4i^. If the particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement?

Options:
A 9 J.
B 15 J.
C 12 J.
D 10 J.

The correct answer is:
15 J.

Solution:
Work done by the force,F→.s→=3i^+12j^⋅4i^=12 J
By work energy theorem, Wtotal=∆K
12 J=Kf-Ki12+Ki=Kf
∴Kf=12+3=15 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2019 (10 Jan Shift 2)
A particle which is simultaneously subjected to two perpendicular simple harmonic motions represented by; x=a1cos⁡ωt and y=a2cos⁡2ωt traces a curve given by :

Options:
A

B
C
D

The correct answer is:

Solution:
x=a1cos⁡ωty=a2cos⁡2ωt=a22cos2⁡ωt-1=a22 xa12-1∴ya2=2x2a12-1∴a12y=2a2x2-a12a2which is a parabola symmetrical about y-axis opening in upward direction

Topic: Physics, Oscillations, JEE Main, JEE Main 2014 (09 Apr Online)
A particle with charge $q$ moves with a velocity $v$ in a direction perpendicular to the directions of uniform clectric and magnetic ficlds, $E$ and $B$ respectively, which are mutually perpendicular to each other. Which one of the following gives the condition for which the particle moves undeflected in its original trajectory?

Options:
A $v=\frac{E}{B}$
B $v=\frac{B}{E}$
C $v=\sqrt{\frac{E}{B}}$
D $v=q \frac{B}{E}$

The correct answer is:
$v=\frac{E}{B}$

Solution:
According to the question,
Charge on particle is $=q$
Velocity of particle $=v$
Due to uniform electric field,
electric force on particle
$F_{\text {electric }}=q E$
Due to a uniform magnetic field, magnetic force on particle is given by,
$$
F_{\text {magnetic }}=q(v \times B)
$$
When $v$ is perpendicular to $E$ and $B,$ which are mutually perpendicular to each other, $F_{\text {electric }}=F_{\text {magnetic }}$
From Eqs. (i) and (ii)
$$
q E=q v B
$$
$\therefore$
$$
v=\frac{E}{B}
$$
So, $v=\frac{E}{B}$ is the condition for which the particle moves undeflected in its original trajectory.

Topic: Physics, Magnetic Effects of Current, JEE Main
A particle X of mass m and initial velocity u collides with another particle Y of mass 3m4, which is at rest. The collision is head-on and perfectly elastic. The ratio of de-Broglie wavelengths λY and λX after the collision is,

Options:
A 4:3
B 2:32
C 3:4
D 1:6

The correct answer is:
1:6

Solution:
For perfectly elastic collision,

total momentum before collision = total momentum after collision

i.e. mu+3m4×0=mvx+3m4vy

⇒4u=4vx+3vy ...i

Coefficient of restitution,

e=Velocity of separation after collision Velocity of approach before collision = vy-vxux-uy

⇒1=vy-vxu-0 [∵ ux=u and uy=0]

u=vy-vx ...ii

From equation (i) and (ii), we get

vx=u7 and vy=8u7

∴ λyλx=h34mvyhm×vx=43×vxvy=43u78u7=16

Topic: Physics, Dual Nature of Matter, JEE Main
A particular adsorption process has the following characteristics: (i) It arises due to van der Waals forces and (ii) it is reversible. Identify the correct statement that describes the above adsorption process:

Options:
A Adsorption is monolayer
B Adsorption increases with increase in temperature
C Enthalpy of adsorption is greater than 100 kJ mol-1
D Energy of activation is low

The correct answer is:
Energy of activation is low

Solution:
In the case of physical adsorption there is almost nil energy of activation and very low enthalpy change.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2016 (09 Apr Online)
A particular force F applied on a wire increases its length by 2Ã10-3 m. If we were to increase the wire's length by 4Ã10-3 m, then the applied force is nF. What is the value of n?

Options:

The correct answer is:
2

Solution:
Y=FAlL=F×LA×l [l = Change in length, L = Original length] (where Y is Young's modulus of elasticity)

Since Y, L and A remain same.
⇒F∝l
⇒F1F2=l1l2
⇒FF2=2×10-34×10-3

⇒F2=2F⇒F2=nF⇒n=2

Topic: Physics, Mechanical Properties of Solids, JEE Main
A particular hydrogen like ion emits radiation of frequency 2.92×1015 Hz when it makes transition from n=3 to n=1. The frequency in Hz of radiation emitted in transition from n=2 to n=1 will be:

Options:
A 6.57×1015
B 2.46×1015
C 0.44×1015
D 4.38×1015

The correct answer is:
2.46×1015

Solution:
As we know,

ν∝1n12−1n22

2.92×1015ν=11−1911−14

ν=2.92×1015×34×98

ν=2.46×1015 Hz

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2021 (26 Aug Shift 1)
A particular semiconductor in equilibrium has 1Ã1016 cm-3 donor atoms, 1.1 Ã 1017 cm-3 acceptor atoms. If the intrinsic carrier density ni of the semiconductor is 1012 cm-3, then the electron density in it will be

Options:
A 1016 cm-3
B 1012 cm-3
C 1.1×1017 cm-3
D 107 cm-3

The correct answer is:
107 cm-3

Solution:
From mass action law,
nenh=ni2
and nh=1.1×1017-1016 cm-3= 1017 cm-3
∴ ne=101221017=107 cm-3

Topic: Physics, Semiconductors, JEE Main
A particular semiconductor in equilibrium has 1Ã1016 cm-3 donor atoms, 1.1 Ã 1017 cm-3 acceptor atoms. If the intrinsic carrier density ni of the semiconductor is 1012 cm-3 , then the electron density in it will be

Options:
A 1016 cm-3
B 1012 cm-3
C 1.1×1017 cm-3
D 107 cm-3

The correct answer is:
107 cm-3

Solution:
From mass action law,
nenh=ni2
and nh=1.1×1017-1016 cm-3= 1017 cm-3
∴ ne=101221017=107 cm-3

Topic: Physics, Semiconductors, JEE Main
A partition wall has two layers of different materials A and B in contact with each other. They have the same thickness but the thermal conductivity of layer A is twice that of B. At steady state, if the temperature difference across the layer B is 50 K, then the corresponding temperature difference across the layer A is

Options:
A 50 K
B 12.5 K
C 25 K
D 60 K

The correct answer is:
25 K

Solution:
Let T be the junction temperature
Here, KA=2KB, T-TB=50K
At the steady state HA=HB
⇒ K A A T A −T L 2 = K B A T− T B L 2
â 2KB TA-T=KB T-TB
â TA-T=T-TB2=502=25 K

Topic: Physics, Thermal Properties of Matter, JEE Main
A passenger sitting in a train A moving at 90 km h-1 observes another train B moving in the opposite direction for 8 s. If the velocity of the train B is 54 km h-1, then length of train B is:

Options:
A 120 m
B 320 m
C 80 m
D 200 m

The correct answer is:
320 m

Solution:
Let the velocity of A be vA and the velocity of B be vB.

The relative velocity is

v→BA =v→B-v→A⇒vBA= 54-(-90)=144 km h-1

⇒vBA= 40 m s-1

Time=lengthvBA

⇒length= 40 m s-1 × 8 s=320 m

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (13 Apr Shift 2)
A passenger sitting in a train A moving at $90 \mathrm{~km} /$ $h$ observes another train B moving in the opposite direction for $8 \mathrm{~s}$. If the velocity of the train B is $54 \mathrm{~km} / \mathrm{h}$, then length of train B is :

Options:
A $80 \mathrm{~m}$
B $200 \mathrm{~m}$
C $120 \mathrm{~m}$
D $320 \mathrm{~m}$

The correct answer is:
$320 \mathrm{~m}$

Solution:
Velocity of train A
\(\mathrm{V}_{\mathrm{A}}=90 \frac{\mathrm{km}}{\mathrm{hr}}=90 \times \frac{5}{18}=25 \mathrm{~m} / \mathrm{s}\)
Velocity of train B
\(\mathrm{V}_{\mathrm{B}}=54 \frac{\mathrm{km}}{\mathrm{hr}}=54 \times \frac{5}{18}=15 \mathrm{~m} / \mathrm{s}\)
Velocity of train B w.r.t. train
\(\begin{aligned}
A=\vec{V}_B & -\vec{V}_A \\
& =15-(-25) \mathrm{m} / \mathrm{s} \\
& =40 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
Time of crossing \(=\frac{\text { length of train }}{\text { relative velocity }}\)
\((8)=\frac{\ell}{40}\)
\(\ell=8 \times 40=320\) meter.

Topic: Physics, Motion In One Dimension, JEE Main
A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train of length 120 m travels at a speed of 30 km/hr. The ratio of times taken by the passenger train to completely cross the freight train when: (i) they are moving in the same direction, and (ii) in the opposite directions is:

Options:
A 52
B 32
C 115
D 2511

The correct answer is:
115

Solution:
t1=260+12080-30
t2=260+12080+30
t1t2=115

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2019 (12 Jan Shift 1)
A pendulum bob has a speed of 3 m s-1 at its lowest position. The pendulum is 50 cm long. The speed of bob, when the length makes an angle of 60° to the vertical will be g=10 m s-2 ______ m s-1.

Options:

The correct answer is:
2

Solution:
Applying work energy theorem:

wg+wT=ΔK

-mgl1-cos60°=12mv2-12mu2

v2=u2-2gl1-cos60°

v2=9-2×10×0.512

v2=4

v=2 m s-1

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (25 Jul Shift 1)
A pendulum clock (fitted with a small heavy bob that is connected with a metal rod) is 3 seconds fast each day at a temperature of 15 °C and 2 seconds slow at a temperature of 30 °C. Find the temperature (in °C) at which it will show the correct time.

Options:

The correct answer is:
24.00

Solution:
Fractional loss of time per second =12Î±ÎT

Therefore 12Î±T0â15Ã(24hrs)=3

and 12Î±30âT0Ã(24hrs)=2
on solving T0=24Â°C

Topic: Physics, Thermodynamics, JEE Main
A pendulum clock is 5 s fast at a temperature of 15 Â°C and 10 s slow at a temperature of 30 Â°C. At what temperature does it give the correct time? (take time interval = 24 hours)

Options:
A 18 °C

B 22 Â°C

C 20 Â°C

D 25 °C

The correct answer is:
20 Â°C

Solution:
Î t = 1 2 Î± Î T Ã t

â´ 5 = 1 2 Î± T - 1 5 Ã 8 6 4 0 0 .........(i)

And 1 0 = 1 2 Î± 3 0 - T Ã 8 6 4 0 0 ........(ii)

Dividing both equations,

12=T-1530-T

â 30-T=2T-30

â 3T=60â T = 20 Â°C

Topic: Physics, Thermal Properties of Matter, JEE Main
A pendulum clock loses 12 s a day if the temperature is 40o C and gains 4s a day if the temperature is 20o C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion α of the metal of the pendulum shaft are respectively:

Options:
A 30oC;α=1.85×10-3 / oC
B 55oC;α=1.85×10-2 / oC
C 25oC;α=1.85×10-5 / oC
D 60oC;α=1.85×10-4 / oC

The correct answer is:
25oC;α=1.85×10-5 / oC

Solution:
∆TT=12∆ll=12α∆θ
124=40-θθ-20
3θ-60=40-θ
4θ=100
θ=25oC
Also,∆T=12α ∆θ×T
4=12α 5×86400
85×86400=α
α=1.85×10-5 /oC

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2016 (03 Apr)
A pendulum has a time period $\mathrm{T}$ in air. When it is made to oscillate in water its time period is $\sqrt{2} \mathrm{~T}$. Then the relative density of the material of the bob of the pendulum is (neglect damping)

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

The correct answer is:
$2$

Solution:
The effective acceleration of a bob in water
$$
g^{\prime}=g\left(1-\frac{\sigma}{\rho}\right)
$$
Here $\sigma$ and $\rho$ are the density of water and the bob respectively.
The time period in air is
$$
T=2 \pi \sqrt{\frac{I}{g}}
$$
The time period in water is
$$
T=2 \pi \sqrt{\frac{I}{g^{\prime}}}
$$
Divide equation (2) by (1), we have
$$
\begin{aligned}
& \frac{T^{\prime}}{T}=\sqrt{\frac{I}{g^{\prime}}} \times \sqrt{\frac{g}{I}} \\
& \frac{\sqrt{2} T}{T}=\sqrt{\frac{g}{g\left(1-\frac{\sigma}{\rho}\right)}} \\
& \sqrt{2}=\sqrt{\frac{1}{1-\frac{\sigma}{\rho}}}
\end{aligned}
$$
Since $(\because \sigma=1)$
$$
\begin{aligned}
& 2=\frac{1}{1-\frac{1}{\rho}} \Rightarrow 1-\frac{1}{\rho}=\frac{1}{2} \quad \Rightarrow \frac{1}{\rho}=\frac{1}{2} \\
& \therefore \rho=2
\end{aligned}
$$

Topic: Physics, Oscillations, JEE Main
A pendulum has an angular amplitude Î¸. The tension in the string at extreme position is T1 and at the bottom is T2. If it is known that T2 = 2T1, then what is the value of cosÎ¸?

Options:

The correct answer is:
0.75

Solution:
T 1 = mg cos Î¸ ..... (1)
T 2 - mg = mv 2 l = m l 2 gh
= 2 mg l 1 - cos Î¸ l
= 2 mg 1 - cos Î¸ l
or T 2 = mg + 2 mg 1 - cos Î¸ ....... (2)

Given T2 = 2T1
or mg + 2 mg 1 - cos Î¸ = 2mg cos Î¸
or cos Î¸ = 3 4
= 0.75

Topic: Physics, Laws of Motion, JEE Main
A pendulum is executing simple harmonic motion and its maximum kinetic energy is $\mathrm{K}_{1}$. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is $\mathrm{K}_{2}$

Options:
A $K_{2}=2 K_{1}$
B $\mathrm{K}_{2}=\frac{\mathrm{K}_{1}}{2}$
C $\mathrm{K}_{2}=\frac{\mathrm{K}_{1}}{4}$
D $K_{2}=K$

The correct answer is:
$K_{2}=2 K_{1}$

Solution:
$K=\frac{1}{2} m \omega^{2} x^{2}$

$\Rightarrow K_{\max }=\frac{1}{2} m \omega^{2} A^{2}$

A $=L \theta$

$\omega=\sqrt{\frac{g}{L}}$

$\Rightarrow K=\frac{1}{2} m \cdot \frac{g}{L} \cdot L^{2} \theta^{2}$

$=\frac{1}{2} m g L \theta^{2}$

$\therefore \frac{K_{1}}{K_{2}}=\frac{L}{2 L}=\frac{1}{2} \Rightarrow K_{2}=2 K_{1}$

$\therefore \frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}=\frac{\mathrm{L}}{2 \mathrm{~L}}=\frac{1}{2} \Rightarrow \mathrm{K}_{2}=2 \mathrm{~K}_{1}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A pendulum is oscillating with an angular amplitude of 90Â° as shown in the figure. The value of angle Î±, made by the string with the vertical, at which the acceleration is directed horizontally is

Options:
A 0
B 90°
C cos-1â¡13
D sin-1â¡13

The correct answer is:
cos-1⁡13

Solution:
For horizontal acceleration, net force in vertical direction should be zero.
Loss in potential energy = gain in kinetic energy
mgh=12mv2
mglcos⁡α=12mv2
∴ v=2glcos⁡α
∵ The net force towards O is
T-mgcos⁡α=mv2l
=m2glcos⁡α l=2 mgcos⁡α
∴T=3mgcos⁡α
∴Tcos⁡α=mg
Or 3mg cos2 α=mg
∴α=cos-113

Topic: Physics, Rotational Motion, JEE Main
A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at 60° with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be _____ m s-1. (if g=10 m s-2)

Options:

The correct answer is:
5

Solution:
Velocity will be maximum at the bottom point when potential energy of the pendulum will be zero(considering bottom point as reference for potential energy).

12mv2=mgl-lcos60°v=gl=10×2.5=5 m s-1

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2022 (28 Jun Shift 1)
A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to TM . If the Young's modulus of the material of the wire is Y, then 1Y is equal to:
(g=gravitational acceleration)

Options:
A 1-TTM2AMg
B T M T 2 - 1 Aâ¡ Mgâ¡
C T M T 2 - 1 Mg Aâ¡
D 1-TMT2AMg

The correct answer is:
T M T 2 - 1 A⁡ Mg⁡

Solution:
Initial length =l. Time period T=2πlg ...(i). After suspending mass M,. Youngs modulus Y=StressStrain=FA∆ll=Fl∆l A
Change in length of wire Δl=FlAY
Now time period TM=2πl+Δlg ...(ii)
TTM=2πlg2πl+Δlg ∵ i ii
T2TM2=ll+Δl
⇒Tm2T2=1+∆ll
1+FAY=TmT2putting value of ∆l
1Y=TmT2-1AF
F=Mg
1Y=TmT2-1AMg

Topic: Physics, Mechanical Properties of Solids, JEE Main
A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to TM . If the Young's modulus of the material of the wire is Y, then 1Y is equal to:
(g=gravitational acceleration)

Options:
A 1-TTM2AMg
B T M T 2 - 1 Aâ¡ Mgâ¡
C T M T 2 - 1 Mg Aâ¡
D 1-TMT2AMg

The correct answer is:
T M T 2 - 1 A⁡ Mg⁡

Solution:
Initial length =l. Time period T=2πlg ...(i). After suspending mass M,. Youngs modulus Y=StressStrain=FA∆ll=Fl∆l A
Change in length of wire Δl=FlAY
Now time period TM=2πl+Δlg ...(ii)
TTM=2πlg2πl+Δlg ∵ i ii
T2TM2=ll+Δl
⇒Tm2T2=1+∆ll
1+FAY=TmT2putting value of ∆l
1Y=TmT2-1AF
F=Mg
1Y=TmT2-1AMg

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2015 (04 Apr)
A pendulum of length 2 m consists of a wooden bob of mass 50 g. A bullet of mass 75 g is fired towards the stationary bob with a speed v. The bullet emerges out of the bob with a speed v3 and the bob just completes the vertical circle. The value of v is _____ m s-1

(if g=10 m s-2).

Options:

The correct answer is:
10

Solution:
For the bob to completer the vertical circle, vb=5gl.

From the conservation of momentum,

m1v=m2vb+m1v3⇒75v=505gl+75v3

⇒752v3=505×10×2

⇒752v3=50×10

⇒v=500×3150=50050=10 m s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2022 (27 Jun Shift 1)
A pendulum of length L carries a negative charge -q on the bob. A positive charge +q is held at the point of support. Then, the time period of the bob is

Options:
A Greater thanÂ 2ÏLg
B Less than 2ÏLg
C equal to 2πLg
D Equal to 2Ï2Lg

The correct answer is:
equal to 2πLg

Solution:
There is no change in the restoring force as the electrostatic forces are the central forces. Negative and positive charges at the two extremities of the string affect tension T which does not affect the restoring force

Topic: Physics, Magnetic Properties of Matter, JEE Main
A pendulum with the time period of 1 s is losing energy due to damping. At a certain time, its energy is 45 J. If after completing 15 oscillations its energy has become 15 J, then its damping constant (in s-1) will be

Options:
A 12
B 115ln⁡3
C 130ln⁡3
D 2

The correct answer is:
130ln⁡3

Solution:
Energy of a damping oscillator is E=E0e-2γt at time t where γ=damping constant, E' =Eoe-2γt⇒15=45e-2γ15⇒γ=130ln⁡(3)

Topic: Physics, Oscillations, JEE Main, JEE Main 2015 (11 Apr Online)
A peptide synthesized by the reactions of one molecule each of Glycine, Leucine, Aspartic acid and Histidine will have _____ peptide linkages.

Options:

The correct answer is:
3

Solution:
Two amino acids join together and form peptide bond. Glycine - Leucine - Aspartic acid - Histidine

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2021 (01 Sep Shift 2)
A perfect gas at $27^{\circ} \mathrm{C}$ is heated at constant pressure so as to double its volume. The final temperature of the gas will be, close to

Options:
A
$327^{\circ} \mathrm{C}$

B
$200^{\circ} \mathrm{C}$

C
$54^{\circ} \mathrm{C}$

D
$300^{\circ} \mathrm{C}$

The correct answer is:
$327^{\circ} \mathrm{C}$

Solution:
Given, $V_1=V$
$$
\begin{aligned}
& V_2=2 V \\
& T_1=27^{\circ}+273=300 \mathrm{~K} \\
& T_2=?
\end{aligned}
$$
From charle's law
$$
\frac{V_1}{T_1}=\frac{V_2}{T_2}(\because \text { Pressure is constant })
$$
or, $\frac{V}{300}=\frac{2 V}{T_2}$
$$
\therefore \quad T_2=600 \mathrm{~K}=600-273=327^{\circ} \mathrm{C}
$$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2012 (07 May Online)
A perfect gas at $27^{\circ} \mathrm{C}$ is heated at constant pressure so as to double its volume. The increase in temperature of the gas will be

Options:
A $600^{\circ} \mathrm{C}$
B $327^{\circ} \mathrm{C}$
C $54^{\circ} \mathrm{C}$
D $300^{\circ} \mathrm{C}$

The correct answer is:
$300^{\circ} \mathrm{C}$

Solution:
We know for perfect gas
Here, $\quad \frac{V}{V_{2}}=\frac{T_{1}}{T_{2}}$
According to question,
$V_{1}=V$ then $V_{2}=2 V$ and $T_{1}=300 \mathrm{~K}$
$$
\therefore \quad \begin{aligned}
\frac{1}{2} &=\frac{300}{T_{2}} \\
T_{2} &=600 \mathrm{~K} \\
T_{2} &=327^{\circ} \mathrm{C}
\end{aligned}
$$
So, $\Delta t=327-27=300^{\circ} \mathrm{C}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A perfect gas contained in a cylinder is kept in vacuum. If the cylinder suddenly bursts, then the temperature of the gas

Options:
A remains constant
B becomes zero
C increases
D decreases

The correct answer is:
remains constant

Solution:
During free expansion of a perfect gas no work is done and also no heat is supplied from outside. Therefore from first law of thermodynamics, there is no change in the internal energy. Hence, temperature remains constant.

Topic: Physics, Thermodynamics, NEET
A perfect gas goes from state A to state B by absorbing 8Ã105 J of heat and doing 6.5Ã105 J of external work. It is now transferred between the same two states in another process in which it absorbs 105 J of heat. In the second process,

Options:
A Work done on gas is 105 J
B Work done on gas is 0.5Ã105 J
C Work done by gas is 105 J
D Work done by gas 0.5Ã105Â J

The correct answer is:
Work done on gas is 0.5Ã105 J

Solution:
dU=dQ-dW=8Ã105-6.5Ã105=1.5Ã105 J
In the second process, dU remains the same
â´ dW=dQ-dU=105-1.5Ã105=-0.5Ã105 J

Topic: Physics, Thermodynamics, NEET
A perfect gas of a given mass is heated first in a small vessel and then in a large vessel, such that their volume remains unchanged. The P-T curves are

Options:
A Parabolic with same curvature
B Parabolic with different curvature
C Linear with same slope
D Linear with different slope

The correct answer is:
Linear with different slope

Solution:
Hence PV = nRT
First P1V1=nRT1
Second P2V2=nRT2
So P1V1T1=P2V2T2
Thus, P-T curves are linear but with different slopes.

Topic: Chemistry, States of Matter, NEET
A perfect gas of volume 10 liter is compressed isothermally to a volume of 1 liter. The rms sped of the molecules will

Options:
A decrease 5 time
B remain unchanged
C increase 5 times
D increase 10 times

The correct answer is:
remain unchanged

Solution:
Process is isothermal, hence temperature as well as rms speed remain same.

Topic: Physics, Kinetic Theory of Gases, MHT CET, MHT CET 2021 (20 Sep Shift 1)
A period of geostationary satellite is

Options:
A O \( 24 \) h
B \( 12 \mathrm{~h} \)
C \( 30 \mathrm{~h} \)
D O \( 48 \) h

The correct answer is:
O \( 24 \) h

Solution:
Geostationary satellite orbits around Earth above the equator so that it remains stationary as seen from Earth. A period
of geostationary satellite is \( 24 \mathrm{~h} \).

Topic: Physics, Waves and Sound, KCET, KCET 2014