datavorous/entrance-exam-dataset huggingface dataset:
The time for half life period of a certain reaction $A \rightarrow$ products is 1 hour. When the initial concentration of the reactant ' $A$ ', is $2.0 \mathrm{~mol} \mathrm{~L}^{-1}$, how much time does it take for its concentration to come from $0.50$ to $0.25 \mathrm{~mol} \mathrm{~L}^{-1}$ if it is a zero order reaction?

Options:
A
$4 \mathrm{~h}$

B
$0.5 \mathrm{~h}$

C
$0.25 \mathrm{~h}$

D
$1 \mathrm{~h}$

The correct answer is:
$0.25 \mathrm{~h}$

Solution:
For a zero order reaction $k=\frac{x}{t}$
Where $x=$ amount decomposed
$\mathrm{k}=$ zero order rate constant
for a zero order reaction
$$
k=\frac{[\mathrm{A}]_0}{2 \mathrm{t}_{\frac{1}{2}}}
$$
Since $\left[A_0\right]=2 M, t_{1 / 2}=1 \mathrm{hr} ; \mathrm{k}=1$
$\therefore$ from equation (1)
$\mathrm{t}=\frac{0.25}{1}=0.25 \mathrm{hr}$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2010
The time in which a force of $2 \mathrm{~N}$ produces a change of momentum of $0.4 \mathrm{~kg}-\mathrm{ms}^{-1}$ in the body is

Options:
A $0.2 \mathrm{~s}$
B $0.02 \mathrm{~s}$
C $0.5 \mathrm{~s}$
D $0.05 \mathrm{~s}$

The correct answer is:
$0.2 \mathrm{~s}$

Solution:
$\vec{F}=\frac{\Delta \vec{p}}{\Delta t} \Rightarrow \Delta t=\frac{|\Delta \vec{p}|}{|\vec{F}|}=\frac{0.4}{2}=0.2 \mathrm{~s}$

Topic: Physics, Laws of Motion, JEE Main
The time of half change of a first order reaction in ....... initial concentration

Options:
A Proportional to
B Inversely proportional to
C Independent of
D Equal to

The correct answer is:
Independent of

Solution:
For $\mathrm{I}^{\text {st }}$ order reaction half life is independent of concentration.

Topic: Chemistry, Chemical Kinetics, JEE Main
The time period of a bob performing simple harmonic motion in water is 2 s. If density of bob is 43×103 kg m-3, then time period of bob performing simple harmonic motion in air will be

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

The correct answer is:
1 s

Solution:
Given, density of bob, ρ=43×103 kg m-3
Density of water, σ=103 kg m-3

If g' be gravitational acceleration in water then,

g'=g 1-σρ=g 1-10343×103=g4

As, Tair =2πIg,

Similarly, Twater =2πIg' Twater =2 s g'=g4

2=2π Ig/4=2π Ig ·2

2=2.Tair

⇒ Tair =1 s

Topic: Physics, Oscillations, JEE Main
The time period of a charged particle undergoing a circular motion in a uniform magnetic field is
independent of its

Options:
A
speed

B
mass

C
charge

D
magnetic induction

The correct answer is:
speed

Solution:
$\mathrm{T}=2 \pi \mathrm{m} / \mathrm{Bq}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2002
The time period of a geostationary satellite is

Options:
A 24 hours
B 12 hours
C 365 days
D One month

The correct answer is:
24 hours

Solution:
24 hours
A geostationary satellite is an earthorbiting satellite, placed at an altitude of approximately 35,800 kilometers directly over the equator, that revolves in the same direction the earth rotates (west to east). At this altitude, one orbit takes 24 hours, the same length of time as the earth requires to rotate once on its axis.
The term geostationary comes from the fact that such a satellite appears nearly stationary in the sky as seen by a groundbased observer.

Topic: Physics, Gravitation, JEE Main
The time period of a physical pendulum about some pivot point is T. When we take another pivot point, opposite of the first one such that the centre of mass of the physical pendulum lies on the line joining these two pivot points, we obtain the same time period. If the two points are separated by a distance L, then the time period T is

Options:
A 2πLg
B πLg
C 2πL2g
D π2Lg

The correct answer is:
2πLg

Solution:
T=2πIC+mx2mg x=2πIC+m(L-x)2mg (L-x)
ICx+mx=ICL-x+m(L-x)
ICL-2xx(L-x)=m(L-2x) IC=mx (L-x)
T=2πmxL-mx2+mx2mgx=2πLg

Topic: Physics, Oscillations, JEE Main
The time period of a satellite in a circular orbit of the radius R is T. The period of another satellite in a circular orbit of the radius 9R is:

Options:
A 9T
B 27T
C 12T
D 3T

The correct answer is:
27T

Solution:
T2∝R3.
T'T2=9RR3
T2=T2×93
T'=T×33
T'=27 T

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become

Options:
A
10 hours

B
80 hours

C
40 hours

D
20 hours

The correct answer is:
40 hours

Solution:
$\mathrm{T}^2 \propto \mathrm{R}^3 \Rightarrow\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^2=\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^3$
$\Rightarrow\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)=\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^{3 / 2}=\left(\frac{1}{4}\right)^{3 / 2} \quad \Rightarrow \frac{\mathrm{T}_2}{\mathrm{~T}_1}=(4)^{3 / 2}=8$
$\Rightarrow \mathrm{T}_2=8 \times \mathrm{T}_1=8 \times 5=40=40$ hours

Topic: Physics, Gravitation, JEE Main, JEE Main 2003
The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.

Options:
A 4 hours
B 6 hours
C 12 hours
D 3 hours

The correct answer is:
3 hours

Solution:
From Kepler's law of periods, we know

T2∝R3

⇒T12T22=R13R23⇒T1T22=RR43

⇒T12T22=64

∴T2=248=3 hours

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The time period of a satellite revolving around earth in a given orbit is 7 hours. If the radius of orbit is increased to three times its previous value, then approximate new time period of the satellite will be

Options:
A 36 hours
B 40 hours
C 30 hours
D 25 hours

The correct answer is:
36 hours

Solution:
From Kepler's Third Law(The Law of Periods), we know

T2∝R3 ⇒T22T12=R2R13=3RR3

Therefore, T2=27T1

=5.19×7≈36 hours

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (29 Jun Shift 2)
The time period of a simple pendulum is given by T=2πlg. The measured value of the length of the pendulum is 10 cm known to a 1 mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1s resolution. The percentage accuracy in the determination of g using this pendulum is x. The value of x to the nearest integer is:-

Options:
A 2%
B 3%
C 5%
D 4%

The correct answer is:
3%

Solution:
g=4π2lT2
Δgg=Δll+2ΔTT=0.110+212000.5
Δgg=1100+150
Δgg×100=3%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The time period of a simple pendulum is T1 when its point of suspension is at rest with respect to the ground. When the point of suspension moves vertically upwards according to the law y=kt2 where k=1 ms-2, its new time period is T2, then the value of T12T22 is [Take g=10 ms-2]

Options:
A 45
B 65
C 56
D 1

The correct answer is:
65

Solution:
Acceleration of the point of suspension
a=d2ydt2=2k=2m/s2
T=2ÏLgeff âT1=2ÏLg
T2=2ÏLg+2
T12T22=g+2g=10+210=65

Topic: Physics, Oscillations, JEE Main
The time period of a simple pendulum is $T$ remaining at rest inside a lift. Find the time period of pendulum when lift starts to move up with an acceleration of $g / 4$.

Options:
A $T$
B $T / 2$
C $2 T / 5$
D $2 T / \sqrt{5}$.

The correct answer is:
$2 T / \sqrt{5}$.

Solution:
When the lift is at rest, the time period of the simple pendulum is $T=2 \pi \sqrt{\frac{l}{g}}$
Now, if the lift is moving up with an acceleration $\frac{g}{4}$ then the effective acceleration due to gravity
$$
\begin{aligned}
& g^{\prime}=g+\frac{g}{4}=\frac{5 g}{4} \\
& \therefore \text { Time period } T^{\prime}=2 \pi \sqrt{\frac{l}{g^{\prime}}} \\
&=2 \pi \sqrt{\frac{l}{5 g / 4}} \\
&=\frac{2}{\sqrt{5}} \times 2 \pi \sqrt{\frac{l}{g}}=\frac{2}{\sqrt{5}} T
\end{aligned}
$$

Topic: Physics, Oscillations, AIIMS, AIIMS 2001
The time period of a spring-mass system is T. If this spring is cut into two parts, whose lengths are in the ratio 1:2, and the same mass is attached to the longer part, the new time period will be

Options:
A 13 T
B 23T
C 3T2
D 3T

The correct answer is:
23T

Solution:
The spring-mass system oscillates in SHM. Its time period is given by

T=2πmk

When spring is cut in ratio 1:2, the new spring constant is k'=3k​/2 (for the longer part)

∴ TT'=3k2k

⇒ T'=23T

Topic: Physics, Oscillations, JEE Main
The time period of an earth satellite in circular orbit is independent of

Options:
A
the mass of the satellite

B
radius of its orbit

C
both the mass and radius of the orbit

D
neither the mass of the satellite nor the radius of its orbit.

The correct answer is:
the mass of the satellite

Solution:
The time period of satellite is given by
$$
T=2 \pi \sqrt{\frac{(\mathrm{R}+\mathrm{h})^3}{\mathrm{GM}}}
$$
where, $\mathrm{R}+\mathrm{h}=$ radius of orbit satellite, $\mathrm{M}=$ mass of earth.

Topic: Physics, Gravitation, JEE Main, JEE Main 2004
The time period of mass suspended from a spring is $T$. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be:

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

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

Solution:
Let $K=$ force constant of spring and $K^{\prime}=$ force constant of each part then
$$
\begin{gathered}
\frac{1}{K}=\frac{4}{K^{\prime}} \\
\Rightarrow \quad K^{\prime}=4 K \\
\therefore \text { Time period }=2 \pi \sqrt{\frac{m}{4 K}} \\
=\frac{1}{2} \times 2 \pi \sqrt{\frac{m}{4 K}}=\frac{T}{2}
\end{gathered}
$$

Topic: Physics, Oscillations, JEE Main
The time period of oscillation of a simple pendulum is given by T=2πlg . The length of the pendulum is measured as l=10±0.01 cm and the time period as T=0.5±0.02 s . The percentage error in the value of g is

Options:
A 5%
B 8%
C 7%
D None of these

The correct answer is:
8%

Solution:
ÎggÃ100=ÎllÃ100+2ÎTTÃ100
Îl=0.01 cm
ÎT=0.02s
l=10 cm
T=0.5 s
Îl l Ã100= 0.01 10 Ã100=0.1
ÎT T Ã100= 0.02 0.5 Ã100=4
Δgg×100=0.1+2×4≈8%

Topic: Physics, Units and Dimensions, JEE Main
The time period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination α, is given by :

Options:
A 2πLgcosα
B 2πLgsinα
C 2πLg
D 2πLgtanα

The correct answer is:
2πLgcosα

Solution:
Acceleration of vehicle down the incline will be a=gsinα
In the frame of the vehicle, we have to apply pseudo force. The Force diagram is shown in figure.
Component of mg along the inclined (i.e. mgsinα) will balance the pseudo force. Therefore, geff=gcosα

Hence,

T=2πLgeff=2πLgcosα.

Topic: Physics, Oscillations, JEE Main, JEE Main 2022 (29 Jul Shift 1)
The time period of oscillations of a block attached to a spring is t1. When the spring is replaced by another spring, the time period of the block is t2. If both the springs are connected in series and the block is made to oscillate using the combination, then the time period of the block is

Options:
A T=t1+t2
B T2=t12+t22
C T-1=t1-1+t2-1
D T-2=t1-2+t2-2

The correct answer is:
T2=t12+t22

Solution:
When springs are in series, k=k1k2k1+k2
For first spring, t1=2Ïmk1
For Second spring, t2=2Ïmk2
â´ t12+t22=4Ï2mk1+4Ï2mk2=4Ï2mk1+k2k1k2
or t12+t22=2Ïmk1+k2k1k22
or t12+t22=T2 .

Topic: Physics, Oscillations, JEE Main
The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6×10-16s. The frequency of revolution of the electron in its first excited state (in s-1 ) is:

Options:
A 1.6×1014
B 7.8×1014
C 6.2×1015
D 5.6×1012

The correct answer is:
7.8×1014

Solution:
T∝rv∝n2z×nz∝n3z2
T1T2=n13n23=18
T2=8T1

now,
f2=1T2
f2=112.8×10-16≈7.8×1014

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2020 (07 Jan Shift 1)
The time period of simple harmonic motion of mass M in the given figure is παM5K, where the value of α is _______.

Options:

The correct answer is:
12

Solution:
For series combination of the spring, equivalent spring constant is given by, 1keq=1k1+1k2.

Therefore, for combined system net spring constant keq=2k×k2k+k+k=5k3

Angular frequency of oscillation ω=keqm

⇒ω=5k3m

Period of oscillation τ=2πω=2π3m5k

=π12m5k

Therefore, α=12.

Topic: Physics, Oscillations, JEE Main, JEE Main 2024 (31 Jan Shift 2)
The time period (T) of small oscillations of the surface of a liquid drop depends on its surface tension (s), the density (Ï) of the liquid and it's mean radius (r) as T = c s x Ï y r z. If in the measurement of the mean radius of the drop, the error is 2%, the error in the measurement of surface tension and density both are of 1%. Find the percentage error in measurement of the time period.

Options:

The correct answer is:
4

Solution:
T=csxρyrz

[T]=[s]x[ρ]y[r]z=[MT−2]x[ML−3]y[L]z⇒[T]=Mx+yL−3y+zT−2xx+y=0

−2x=1⇒  x=−12,  y=12−3y+z=0 , z=32

T=cρr3s⇒ΔTT=12Δρρ+32Δrr+12Δss=12+3+12=4

Topic: Physics, Units and Dimensions, JEE Main
The time required for $100 \%$ completion of a zero order reaction is

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

The correct answer is:
$\frac{a}{k}$

Solution:
Half-life of a zero order reaction $=\frac{a}{2 k}$
Time taken for completion $=2 t_{1 / 2}=2 \times \frac{a}{2 k}$
$$
=\frac{a}{k}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main
The time required to pass the light through a glass slab of $2 \mathrm{~mm}$ thick is $\left(\mu_{\text {glass }}=1.5\right.$ )

Options:
A $10^{-5} \mathrm{~s}$
B $10^{-11} \mathrm{~s}$
C $10^{-9} \mathrm{~s}$
D $10^{-13} \mathrm{~s}$

The correct answer is:
$10^{-11} \mathrm{~s}$

Solution:
Correct option is 2. $10^{-11} s$
As we know,
$t=\frac{\mu x}{c}=\frac{1.5 \times 2 \times 10^{-3}}{3 \times 10^8}=10^{-11} \mathrm{sec}$

Topic: Physics, Ray Optics, JEE Main
The time taken for $10 \%$ completion of a first order reactin is 20 mins. Then, for $19 \%$ completion, the reaction will take

Options:
A $40 \mathrm{mins}$
B 60 mins
C $30 \mathrm{mins}$
D 50 mins

The correct answer is:
$40 \mathrm{mins}$

Solution:
$t=\frac{2.303}{\lambda} \log \frac{\mathrm{N}_0}{\mathrm{~N}}$
$\begin{aligned} & 20=\frac{2.303}{\lambda} \log \frac{100}{90} \\ & t=\frac{2.303}{\lambda} \log \frac{100}{81}\end{aligned}$
equation (i) / (ii)
$\therefore t=40 \mathrm{~min}$.

Topic: Chemistry, Chemical Kinetics, JEE Main
The time taken for an electron to complete one revolution in Bohr orbit of hydrogen atom is

Options:
A $\frac{4 m^{2} \pi r^{2}}{n^{2} h^{2}}$
B $\frac{n^{2} h^{2}}{4 m r^{2}}$
C $\frac{4 \pi^{2} m r^{2}}{n h}$
D $\frac{n h}{4 \pi^{2} m r^{2}}$

The correct answer is:
$\frac{4 \pi^{2} m r^{2}}{n h}$

Solution:
Let the distance travelled in $T$ time $=2 \pi r$ (Circumference of orbit)
$\therefore \quad$ Velocity $(v)=\frac{2 \pi r}{T}$
Also, $v=\frac{n \cdot h}{2 \pi m r}$
$\therefore \quad \frac{1}{T \text { (Time period) }}=$ frequency $v=\text{v} \times \frac{1}{2 \pi r}$
From Eq.
$$
\begin{array}{l}
\mathrm{v}=\frac{1}{T}=\frac{n h}{2 \pi \times 2 \pi m r} \\
v=\frac{n h}{4 \pi^{2} r^{2} m}
\end{array}
$$
$T$ (time taken in one revolution)
$$
=\frac{4 \pi^{2} r^{2} m}{n h}
$$

Topic: Chemistry, Structure of Atom, JEE Main
The time taken for the magnetic energy to reach 25% of its maximum value, when a solenoid of resistance R, inductance L is connected to a battery, is :

Options:
A LRln5
B infinite
C LRln2
D LRln10

The correct answer is:
LRln2

Solution:
Magnetic energy =12Li2=25%

ME⇒25%⇒i=i02

i=i01-R-Rt/L for charging

t=LRln2

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (18 Mar Shift 2)
The time-period of a simple pendulum of length 5 m suspended in a car moving with uniform acceleration of 5 m s-2 in a horizontal straight road is g=10 m s-2

Options:
A 2Ï5 s
B Ï5Â s
C 5ÏÂ s
D 4ÏÂ s

The correct answer is:
2Ï5 s

Solution:
Given, Length of simple pendulum, l=5 m
Acceleration of car a=5 m s-2
Now geffective=a2+g2
=52+102
=125=55
T=2Ïlgeffective
T=2Ï555
T=2Ï5 s

Topic: Physics, Oscillations, JEE Main
The top of a water tank is open to air and its water level is maintained. It is giving out 0.74 m3 water per minute through a circular opening of 2 cm radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to:

Options:
A 2.9 m
B 4.8 m
C 6.0 m
D 9.6 m

The correct answer is:
4.8 m

Solution:
Volume of water flowing out=Av, where A is area of opening and v is speed of water from the opening.
Volume=πr2v
0.74 m3min=3.14×2×10-22×2×10×H m3s×60 smin
⇒H=0.7460×3.14×4×10-42×120
⇒H=4.8 m

Hence, the height of water in the tank was 4.8 m from the bottom of the tank.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The tops of two poles of height 40 m and 25 m are connected by a wire of length 3023-1 m. If the wire makes an angle Î± with the horizontal, then the value of 2sinÎ± is equal to (take, 3=1.7)

Options:

The correct answer is:
0.35

Solution:
Let, AB=40 and CD=25 be the two poles.
AC=l is the length of the wire.
CE is the horizontal through C and ∠ACE=α
AEl=sin⁡α⇒l=15sin⁡α=3023-1⇒sin⁡α=3-122

Hence, 2sinα=0.35

Topic: Mathematics, Heights and Distances, JEE Main
The tops of two poles of heights 40m and 25m are connected by a wire. If the wire makes an angle 30o with the horizontal, then the length of the wire is

Options:
A 30m
B 20m
C 15m
D 25m

The correct answer is:
30m

Solution:
Let AB=40 and CD=25 be the two poles.
AC=l is the length of the wire.
CE, the horizontal through C, ∠ACE=30o
AEl=sin⁡30o=12
∴l=2AE=30m

Topic: Mathematics, Heights and Distances, JEE Main
The torque of a force 5i^+3j^-7k^ about the origin is τ. If the force acts on a particle whose position vector is 2i^+2j^+k^, then the value of τ will be

Options:
A 11i^+19j^-4k^
B -11i^+9j^-16k^
C -17i^+19j^-4k^
D 17i^+9j^+16k^

The correct answer is:
-17i^+19j^-4k^

Solution:
Using the determinant rule to find the torque

τ→=r→×F→=i^j^k^22153-7

=i^-14-3-j^-14-5+k^6-10

Thus, the value of τ→=-17i^+19j^-4k^

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2022 (29 Jul Shift 2)
The torque required to hold a small circular coil of cross-sectional area \( 1 \mathrm{~mm}^{2} \), carrying a current of \( \frac{21}{44} \mathrm{~A} \) and having \( 10 \) turns in the middle of a long solenoid of \( 10^{3} \) turns / \( \mathrm{m} \) carrying current of \( 2.5 \mathrm{~A} \) with its axis perpendicular to the axis of the solenoid is Solenoid Axis

Options:
A \( 1.5 \times 10^{-6} \mathrm{Nm} \)
B \( 1.5 \times 10^{-8} \mathrm{Nm} \)
C \( 1.5 \times 10^{6} \mathrm{Nm} \)
D \( 1.5 \times 10^{8} \mathrm{Nm} \)

The correct answer is:
\( 1.5 \times 10^{-8} \mathrm{Nm} \)

Solution:
Here, For small circular coil,
Number of turns, N=10, Area, A=1 mm2=1×10-6 m2
Current, I1=2144 A
For a long solenoid,
Number of turns per metre, n=103 per m
Current I2=2.5 A
Magnetic field due to a long solenoid on its axis is B=μ0nI2 …(i)
Magnetic moment of a circular coil is M=NAI1 …(ii)
Torque, τ→=m→×B→=MBsin⁡θ
L τ=MBsin⁡θ=MB ∵θ=90o Given
τ=NAI1μ0nI2 (Using (i) and (ii) )
τ=10×1×10-6×2144×4×227×10-7×103×2.5=1.5×10-8 N m

Topic: Physics, Magnetic Effects of Current, JEE Main
The total change in the oxidation state of manganese involved in the reaction of KMnO4 and potassium iodide in the acidic medium is ________.

Options:

The correct answer is:
5

Solution:
In the reaction of KMnO4 and potassium iodide (KI) in an acidic medium, the KMnO4 acts as an oxidizing agent, and the iodide ions (I-) in KI act as reducing agents.

KMnO4+KI+H+⟶Mn+2+I2+H2O

In KMnO4, the oxidation state of manganese is +7, which is the highest possible oxidation state for manganese in this compound.

In the final product, MnSO4, the oxidation state of manganese is +2.

Therefore, the total change in the oxidation state of manganese is:

+7 (initial) - +2 (final) = 5

Hence the change in O.S. of Mn is 5.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The total charge enclosed in an incremental volume of 2×10-9 m3 located at the origin is nC, if electric flux density of its field is found asD=e-xsinyi^-e-xcosyj^+2zk^ C m-2

Options:

The correct answer is:
4

Solution:
Electric flux density,

D→= charge Area ×r^=Q4πr2r^=ϵ0Q4πϵ0r2r^

⇒E→=D→ϵ0=e-xsinyi^-e-xcos(j^+2zk)^ϵ0

Also, by Gauss's law,

ρϵ0=∂∂xi^+∂∂yj^+∂∂zk^·E→=∂∂xi^+∂∂yj^+∂∂zk^·D→ϵ0

⇒ρ=∂∂xe-xsiny+∂∂y-e-xcosy+∂∂z(2z)

ρ=-e-xsiny+e-xsiny+2
At origin ρ=-e-0sin0+e-0sin0+2
ρ=2 C m-3

Charge =ρ× volume =2×2×10-9=4×10-9=
4 nC

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (22 Jul Shift 1)
The total charge on the system of capacitance C1=1μF,C2=2μF,C3=4μF and C4=3μF connected in parallel is
(Assume a battery of 20 V is connected to the combination)

Options:
A 200μC
B 200C
C 10μC
D 10C

The correct answer is:
200μC

Solution:
As the given capacitors are in parallel combination, therefore,Ceq=1+2+3+4 μF=10 μF

Hence, charge flown through the switch∴ Q=20 V×10 μF=200 μC

Topic: Physics, Capacitance, JEE Main, JEE Main 2022 (26 Jul Shift 1)
The total current supplied to the circuit by the battery is

Options:
A
$1 \mathrm{~A}$

B
$2 \mathrm{~A}$

C
$4 A$

D
$6 \mathrm{~A}$

The correct answer is:
$4 A$

Solution:
The given circuit can be written as
$$
\mathrm{I}=\frac{6 \mathrm{~V}}{1.5 \Omega}=4 \mathrm{~A}
$$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2004
The total current supplied to the circuit by the battery is

Options:
A \( 1 \mathrm{~A} \)
B \( 2 \mathrm{~A} \)
C \( 4 \mathrm{~A} \)
D \( 6 \mathrm{~A} \)

The correct answer is:
\( 4 \mathrm{~A} \)

Solution:
Let us assume that the battery of 6 V is connected across points A and B. Let the centre point be C.

Using the point-potential method, the points having the same potential should have the same name.

After redrawing the circuit, the figure looks as shown below,

The current supplied by the battery is

i=VRAB

V=6 V.

Now, to find the equivalent resistance between A and B,

RAB=3∥RAC+RCB

RAC=6×26+2

RAC=1.5 Ω

RBC=1.5 Ω

RAB=3∥1.5+1.5

RAB=3∥3

RAB=3×33+3

RAB=1.5 Ω.

Thus, the net resistance between A and B is 1.5 Ω.

Now, the current flowing through the battery is

i=61.5 A.

Hence, the current flowing through the battery is 4 A.

Topic: Physics, Current Electricity, JEE Main
The total duration of the cardiac cycle is 0.8 seconds. If a person is suffering from fever and their heartbeat rate increases, then the duration of their cardiac cycle will

Options:
A remain the same
B increase
C decrease
D become irregular

The correct answer is:
decrease

Solution:
When heart rate increases, the duration of each cardiac cycle decreases, including the contraction and relaxation phases. The duration of the action potential and the period of contraction (systole) also decrease, but not by as great a percentage as does the relaxation phase (diastole). At a normal heart rate of 72 beats/min, systole comprises about 0.4 of the entire cardiac cycle. At three times the normal heart rate, systole is about 0.65 of the entire cardiac cycle. This means that the heart beating at a very fast rate does not remain relaxed long enough to allow complete filling of the cardiac chambers before the next contraction.

Topic: Biology, Body Fluids and Circulation, NEET
The total electrostatic energy stored in both the capacitors (in μJ) is

Options:

The correct answer is:
9

Solution:
According to circuit diagram, capacitors are connected in series combination.
So, equivalent capacitance Ceq=3Ã63+6
=189=2 Î¼F
We know that,
Total electrostatic energy U=12CeqV2
=12Ã2Ã-32
=9 Î¼J

Topic: Physics, Electrostatics, JEE Main
The total energy of an electron in the first excited state of the hydrogen is about $-3.4 \mathrm{eV}$.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed?

Options:

None

Solution:
We know, Kinetic energy of electron
$$
=\frac{1}{4 \pi \varepsilon_0} \frac{7 e^2}{2 \mathrm{r}}
$$
Potential energy $=-2($ K. E. $)$
Total energy $=$ P. E. $+\mathrm{K}$. E. $=-\mathrm{K}$. E. where electric potential would be zero at infinity.
(a) Total energy $=-3.4 \mathrm{eV}$
$$
\Rightarrow \mathrm{K} . \mathrm{E}=-(-3.4)=+3.4 \mathrm{eV} \text {. }
$$
(b) P. E. $=-2 \mathrm{~K}, \mathrm{E}=-6.8 \mathrm{eV}$
(c) If zero of potential is changed then K. E. which is independent of this choice remains same whereas P. E. and hence total energy of the state would change.

Topic: Physics, Atomic Physics