datavorous/entrance-exam-dataset huggingface dataset:
The potential energy of a particle varies as

Ux=E0 for 0â¤xâ¤1

=0 for x>1

For 0â¤xâ¤1 , the de-Broglie wavelength is Î»1 and for x>1 the de-Broglie wavelength is Î»2. The total energy of the particle is 2E0. Find Î» 1 Î» 2

Options:
A 3
B 7
C 2
D 5

The correct answer is:
2

Solution:
For 0≤x≤1 , U=E0
∴ Kinetic energy, K1 = Total energy - U
=2E0-E0=E0
∴λ1=h2mE0
For x>1, U=0
∴ Kinetic energy K2 = Total energy -U

=2E0
∴ λ 2 = h⁡ 2 m⁡ ( 2E⁡ 0 ) ...(ii)
From Eqs. (i) and (ii), we have
λ 1 λ 2 = 2

Topic: Physics, Dual Nature of Matter, JEE Main
The potential energy of a particle varies with distance x from a fixed origin as U=Ax x 2 +B where A and B are constants. The dimensions of A are -

Options:
A ML52T-2
B M1L2T-2
C M32 L52T-2
D M1L72T-2

The correct answer is:
M1L72T-2

Solution:
U=Axx2+B .....(i)

Where U = Potential energy

x= distance

Dimensional formula of potential energy, U=M1 L2 T-2

Dimensional formula of distance, x=M0 L1 T0

From equation (i), we can write

[x2]=[B]

[B]= M0 L1 T02

[B]=M0 L2 T0

Dimensional formula of B=M0 L2 T0

We can write equation (i) as,

U=Axx2

U=Ax32

A=Ux32

A=M1 L2 T-2×M0 L1 T032

A=M1 L72 T-2

Topic: Physics, Units and Dimensions, JEE Main
The potential energy of a particle with displacement $X$ is $U(X)$. The motion is simple harmonic, when ( $K$ is a positive constant)

Options:
A $U=\frac{k x^2}{2}$
B $U=K X^2$
C $U=K$
D $U=K X$

The correct answer is:
$U=\frac{k x^2}{2}$

Solution:
$\begin{aligned} & F=-k x \Rightarrow d W=F d x=-k x d x \\ & \text { So } \int_0^W d W=\int_0^x-k x d x \Rightarrow W=U=-\frac{1}{2} k x^2\end{aligned}$

Topic: Physics, Oscillations, JEE Main
The potential energy of a simple harmonic oscillator of mass $2 \mathrm{~kg}$ at its mean position is $5 \mathrm{~J}$. If its total energy is $9 \mathrm{~J}$ and amplitude is $1 \mathrm{~cm}$, then its time period is

Options:
A $\frac{\pi}{100} \mathrm{~s}$
B $\frac{\pi}{50} \mathrm{~s}$
C $\frac{\pi}{20} \mathrm{~s}$
D $\frac{\pi}{10} \mathrm{~s}$

The correct answer is:
$\frac{\pi}{100} \mathrm{~s}$

Solution:
Given, total energy $=9 \mathrm{~J}$
$\mathrm{PE}$ at mean position $=5 \mathrm{~J}$
So, maximum $\mathrm{KE}=9 \mathrm{~J}-5 \mathrm{~J}=4 \mathrm{~J}$
Now, in SHM
Maximum (at mean) $\mathrm{KE}=$ Maximum PE (at extremes)
$$
\begin{aligned}
& \therefore \quad \frac{1}{2} k a^2=4 \mathrm{~J} \\
& \Rightarrow \quad k=\frac{8}{a^2}=\frac{8}{10^{-4}}=8 \times 10^4 \mathrm{~J} / \mathrm{m}^2 \\
&
\end{aligned}
$$
Now, time period
$$
T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \times \sqrt{\frac{2}{8 \times 10^4}} \Rightarrow T=\frac{\pi}{100} \mathrm{~s}
$$

Topic: Physics, Oscillations, JEE Main
The potential energy of a system increase if work is done

Options:
A by the system against a conservative force
B by the system against a nonconservative force
C upon the system by a conservative force
D upon the system by a nonconservative force

The correct answer is:
by the system against a conservative force

Solution:
The potential energy of a system increases if work is done by the system against a conservative force.
$$
-\Delta U=W_{\text {conservative face }}
$$

Topic: Physics, Work Power Energy, JEE Main
The potential energy of charged parallel plate capacitor is $v_0$. If a slab of dielectric constant $\mathrm{K}$ is inserted between the plates, then the new potential energy will be

Options:
A $\frac{v_0}{K}$
B ${v_0}{K^2}$
C $\frac{v_0}{K^2}$
D ${v_0}^2$

The correct answer is:
$\frac{v_0}{K}$

Solution:
We know, $v_0=\frac{Q^2}{2 C}$
On inserting the slab of dielectric constant $\mathrm{k}$, the new capacitance $\mathrm{C}^{\prime}=\mathrm{KC}$
$\therefore \quad$ New potential energy $v_0^{\prime}=\frac{\mathrm{Q}^2}{2 \mathrm{C}^{\prime}}$
$v_0^1=\frac{Q^2}{2 \mathrm{KC}}=\frac{v_0}{\mathrm{~K}}$

Topic: Physics, Capacitance, JEE Main
The potential gradient along the length of a uniform wire is 10 volt/metre $B$ and $C$ are the two points at $30 \mathrm{~cm}$ and $60 \mathrm{~cm}$ point on a meter scale fitted along the wire. The potential difference between $B$ and $C$ will be

Options:
A 3 volt
B 0.4 volt
C 7 volt
D 4 volt

The correct answer is:
3 volt

Solution:
Potential gradient $=$ Change in voltage per unit length
$\therefore 10=\frac{V_2-V_1}{30 / 100} \Rightarrow V_2-V_1=3$ volt

Topic: Physics, Current Electricity, JEE Main
The potential (in volts) of a charge distribution is given by
Vz=30-5z2 for z≤1 m
Vz=35-10 z for z≥1 m .
Vz does not depend on x and y. If this potential is generated by a constant charge per unit volume ρ0 (in units of ϵ0 ) which is spread over a certain region, then choose the correct statement.

Options:
A Ï0=20Â Ïµ0 in the entire region
B Ï0=10 Ïµ0 for zâ¤1 m and Ï0=0 else where
C Ï0=20Â Ïµ0 for zâ¤1Â m and Ï0=0 else where
D Ï0=40Â Ïµ0 in the entire region

The correct answer is:
Ï0=10 Ïµ0 for zâ¤1 m and Ï0=0 else where

Solution:
Electric field due to given potentials is

E1=-dVdz=10 z for ∣z∣≤1 m. Here E1 is proportional to z, so charge distribution is continuous for z≤1 m
E2=-dVdz=10 for z≥1 m. Here, E2 is constant.

So, using the Poisson’s equation ∇2V=-ρε0, we get ρ=0.
∴ The source is an infinity large non-conducting thick plate.

Hence, consider a cylindrical Gaussian surface:

Applying Gauss's law for ∣z∣≤1 m

E×2A=ρ×2z×Aε0

⇒10z2A=ρ(2Az)ε0
⇒ρ0=10ε0 for z≤1 m.

Topic: Physics, Electrostatics, JEE Main
The potential (in volts) of a charge distribution is given by
Vz=30-5z2 for z≤1 m
Vz=35-10 z for z≥1 m .
Vz does not depend on x and y. If this potential is generated by a constant charge per unit volume ρ0 (in units of ϵ0 ) which is spread over a certain region, then choose the correct statement.

Options:
A Ï0=20Â Ïµ0 in the entire region
B Ï0=10 Ïµ0 for zâ¤1 m and Ï0=0 else where
C Ï0=20Â Ïµ0 for zâ¤1Â m and Ï0=0 else where
D Ï0=40Â Ïµ0 in the entire region

The correct answer is:
Ï0=10 Ïµ0 for zâ¤1 m and Ï0=0 else where

Solution:
Electric field due to given potentials is

E1=-dVdz=10 z for ∣z∣≤1 m. Here E1 is proportional to z, so charge distribution is continuous for z≤1 m
E2=-dVdz=10 for z≥1 m. Here, E2 is constant.

So, using the Poisson’s equation ∇2V=-ρε0, we get ρ=0.
∴ The source is an infinity large non-conducting thick plate.

Hence, consider a cylindrical Gaussian surface:

Applying Gauss's law for ∣z∣≤1 m

E×2A=ρ×2z×Aε0

⇒10z2A=ρ(2Az)ε0
⇒ρ0=10ε0 for z≤1 m.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2016 (09 Apr Online)
The potential is varying with distance $(\mathrm{x})$ as $\mathrm{V}=\frac{1}{2}$ $\left(\mathrm{y}^2-4 \mathrm{x}\right)$ volt. The electric field at $\mathrm{x}=1 \mathrm{~m}$ and $\mathrm{y}=1 \mathrm{~m}$ is

Options:
A $2 \hat{\mathrm{i}}+\hat{\mathrm{j}} \mathrm{Vm}^{-1}$
B $-2 \hat{i}+\hat{j} V m^{-1}$
C $2 \hat{\mathrm{i}}-\hat{\mathrm{j}} \mathrm{Vm}^{-1}$
D $-2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}} \mathrm{Vm}^{-1}$

The correct answer is:
$2 \hat{\mathrm{i}}-\hat{\mathrm{j}} \mathrm{Vm}^{-1}$

Solution:
$\begin{aligned} & V=\frac{1}{2}\left(y^2-4 x\right) \\ & E=-\vec{\nabla} V \\ & =-\left[-\frac{4}{2} \hat{i}+\frac{2 y}{2} \hat{j}\right] \\ & =-[-2 \hat{i}+y \hat{j}] \text { at } x=1, y=1 \\ & =2 \hat{i}-\hat{j} V / m\end{aligned}$

Topic: Physics, Electrostatics, JEE Main
The potential of the electric field produced by point charge at any point (x,y,z) is given by V=3x2+5, where x,y are in metres and V is in volts. The intensity of the electric field at (-2 m, 1 m, 0 m) is

Options:
A +17Â Vm-1
B -17Â Vm-1
C +12 Vm-1
D -12 Vm-1

The correct answer is:
+12 Vm-1

Solution:
Potential (V)=3x2+5
Intensity of the electric field=
-dVdx=6x
∴E at x=-2
=-6(-2)=+12Vm-1

Topic: Physics, Magnetic Properties of Matter, JEE Main
The power applied to a particle \( P=\left[3 t^{2}-2 t+1\right] \mathrm{W} \), where \( t \) is in seconds. Find the change in kinetic energy of particle
from \( t=2 \mathrm{~s} \) to \( 4 \mathrm{~s} \).

Options:
A \( 46 \mathrm{~J} \)
B \( 52 \mathrm{~J} \)
C \( 92 \mathrm{~J} \)
D \( 104 \mathrm{~J} \)

The correct answer is:
\( 46 \mathrm{~J} \)

Solution:
ΔK=∫24Pdt=t3 – t2 + t24=46J

Topic: Physics, Work Power Energy, JEE Main
The power dissipated in the adjacent circuit is

Options:
A \( \frac{V_{0}^{2}}{R} \)
B \( \frac{V_{0}^{2}}{5 R} \)
C \( \frac{V_{0}^{2}}{\sqrt{5} R} \)
D \( \frac{V_{0}^{2}}{10 R} \)

The correct answer is:
\( \frac{V_{0}^{2}}{10 R} \)

Solution:
From given, we can write the given data as,

Resistance=R

Inductive reactance is, XL=3R

Capacitive reactance is, XC=R

Emf of AC source is, V=V0sinωt

V0 is the peak value of AC emf.

Here, XL>XC

The impedance of the circuit is,

Z=R2+XL-XC2

=R2+3R-R2

Z=R2+4R2⇒Z=5R

RMS value of emf of AC source is,

Vrms=V02

Power factor of given AC circuit is,

cosϕ=RZ=R5R

cosϕ=15

The power dissipated in the given circuit can be written as,

P=Vrms2Zcosϕ

Substitute Vrms, Z and cosϕ values in above equation,

Then we get,

P=V0225R15=V02215R

P=V0210R

Topic: Physics, Alternating Current, JEE Main
The power dissipated in the circuit shown in the figure is 30 Watt. The value of $R$ is

Options:
A $20 \Omega$
B $15 \Omega$
C $10 \Omega$
D $30 \Omega$

The correct answer is:
$10 \Omega$

Solution:
Here, $R_1=R=$ ?
$$
R_2=5 \Omega, V=10 \mathrm{~V}
$$
and $\quad P=30 \mathrm{~W}$
Hence
$$
\begin{aligned}
P & =\frac{V^2}{R_1}+\frac{V^2}{R_2} \\
\frac{10^2}{R} & =30-\frac{10^2}{5} \\
\frac{100}{R} & =30-20 \\
R & =10 \Omega
\end{aligned}
$$

Topic: Physics, Current Electricity, NEET, NEET 2012 (Mains)
The power dissipated (in watt) in 3 Î© resistance in the following circuit is

Options:

The correct answer is:
0.75

Solution:
From circuit, 1R=13+16=2+16=36 Î©
âR1=2 Î©
R2=(R2+R2)â¥R4=2 Î©
The internal resistance of battery =1 Î©
So, the equivalent resistance of circuit=2 Î©+1 Î©=3 Î©
The current in the battery,
I=VR=4.53=1.5 A
Current in 3 Î© is, 1.52Ã23 A=12 A
The power dissipated in 3 Î© resistance,
P=I2ÃR
=122Ã3
=14Ã3
=0.75 W

Topic: Physics, Current Electricity, JEE Main
The power factor of a good choke coil is

Options:
A Nearly zero
B Exactly zero
C Nearly one
D Exactly one

The correct answer is:
Nearly zero

Solution:
$\cos \phi=\frac{R}{Z}$. In choke coil $\phi \approx 90^{\circ}$ so $\cos \phi \approx 0$

Topic: Physics, Alternating Current, JEE Main
The power factor of an $A C$ circuit having resistance $(R)$ and inductance $(L)$ connected in series and an angular velocity $\omega$ is

Options:
A
$\mathrm{R} / \omega \mathrm{L}$

B
$\mathrm{R} /\left(\mathrm{R}^2+\omega^2 L^2\right)^{1 / 2}$

C
$\omega \mathrm{L} / \mathrm{R}$

D
$R /\left(R^2-\omega^2 L^2\right)^{1 / 2}$

The correct answer is:
$\mathrm{R} /\left(\mathrm{R}^2+\omega^2 L^2\right)^{1 / 2}$

Solution:
No Solution Available

Topic: Physics, Alternating Current, JEE Main, JEE Main 2002
The power factor of an ac circuit having resistance $(R)$ and inductance $(L)$ connected in series and an angular velocity $\omega$ is

Options:
A $R / \omega L$
B $R /\left(R^2+\omega^2 L^2\right)^{1 / 2}$
C $\omega L / R$
D $R /\left(R^2-\omega^2 L^2\right)^{1 / 2}$

The correct answer is:
$R /\left(R^2+\omega^2 L^2\right)^{1 / 2}$

Solution:
Reactance due to inductor \(=\omega \mathrm{L}\)
\(\mathrm{Z}=\sqrt{\mathrm{R}^2+\omega \mathrm{L}^2}\)
Power factor: \(\cos \varphi=\frac{\mathrm{R}}{\mathrm{Z}}\)
\(\cos \varphi=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\omega \mathrm{L}^2}}\)

Topic: Physics, Alternating Current, JEE Main
The power factor of an L-R series circuit is 0.5 and that of a C-R series circuit is 0.2. If the elements (L, C and R) of the two circuits are joined in series and connected to the same ac source, the power factor of this circuit is found to be 1. The ratio of the resistance in the L-R circuit to the resistance in the C-R circuit is

Options:
A 2
B 2
C 22
D 4

The correct answer is:
22

Solution:
R1XL2+R12=0.5
2R12=R12+XL2
3R12=XL2 .....(i)
R2XC2+R22=0.2
25R22=XC2+R22
24R22=XC2=XL2 (Since the LCR circuit is in resonance, the capacitive reactance and the inductive reactance are equal)
3R12=24RL2
R1R2=22

Topic: Physics, Alternating Current, JEE Main
The power factor of $L C R$ circuit at resonance is

Options:
A 0.707
B 1
C Zero
D 0.5

The correct answer is:
1

Solution:
At resonance, $L C R$ circuit behaves as purely resistive circuit, for purely resistive circuit power factor $=1$

Topic: Physics, Alternating Current, JEE Main
The power obtained in a reactor using U235 disintegration is 1000 kW. The mass decay of U235 per hour is

Options:
A 10 μg

B 20 Î¼g
C 40 Î¼g
D 1 μg

The correct answer is:
40 Î¼g

Solution:
According to Einstein's mass energy relation
E⁡ = mc⁡ 2 or m⁡ = E⁡ c 2
Mass decay per second
=Δm⁡Δt=1c2ΔE⁡Δt=P⁡c2=1000×103W⁡3×108m/s2=1069×1016 kg s-1
Mass decay per hour
=Δm⁡Δt×60×60=1069×1016 kg s-13600 s⁡
= 4 × 1 0 - 8 kg = 4 0 × 1 0 - 6 g = 4 0 μ g

Topic: Physics, Nuclear Physics, JEE Main
The power of a lens (biconvex) is 1.25 m-1 in particular medium. Refractive index of the lens is 1.5 and radii of curvature are 20 cm and 40 cm respectively. The refractive index of surrounding medium:

Options:
A 1.0
B 97
C 32
D 43

The correct answer is:
43

Solution:
Using Len's maker formula,

P=1f=μ1-μ21R1-1R2

Where, μ1 is refractive index of lens and μ2 is of surrounding medium.

Putting the values,

⇒1.25=1.5-μ210.2+10.4

⇒1.25×0.080.6=1.5-μ2

The refractive index of surrounding medium is μ2=43.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (28 Jul Shift 2)
The power radiated by a black-body is P0 and it radiates maximum energy around the wavelength Î»0. If the temperature of the black-body is now changed so that it radiates maximum energy around a wavelength 3Î»04, then the power radiated by it will be

Options:
A 43P0
B 169P0
C 6427P0
D 25681P0

The correct answer is:
25681P0

Solution:
Let T0 be the initial temperature of the black body
∴ λ0T0=b (Wien's law)
Power radiated, P0=CT04, where, C is constant.
If T is new temperature of black body, then
3λ04T=b=λ0T0 or T=43 T0
Power radiated, P=CT4=CT04434
P=25681P0

Topic: Physics, Thermal Properties of Matter, JEE Main
The power radiated from a linear antenna of length l is proportional to
(Given, λ= Wavelength of wave):

Options:
A lλ
B lλ2
C lλ2
D l2λ

The correct answer is:
lλ2

Solution:
Linear antennas are considered to be those antennas that imply the use of electrically thin conductors having diameter much less than the wavelength of the electromagnetic wave passing through them.

The formula to calculate the effective power radiated by a linear antenna is given by

Peff=kl2λ2

where, k is a constant.

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The power supplied by a force acting on a particle moving in a straight line is constant. The velocity of the particle varies with the displacement x as

Options:
A x⁡ 1 / 2
B x⁡
C x⁡ 2
D x⁡ 1 / 3

The correct answer is:
x⁡ 1 / 3

Solution:
Pâ¡ = F vâ¡ = ma vâ¡ = mâ¡ dâ¡ 2 xâ¡ dâ¡ tâ¡ 2 dâ¡ xâ¡ dâ¡ tâ¡
Since, power is constant
dâ¡ 2 xâ¡ dâ¡ tâ¡ 2 dâ¡ xâ¡ dâ¡ tâ¡ = kâ¡
or dâ¡ dâ¡ tâ¡ dâ¡ xâ¡ dâ¡ tâ¡ 2 = kâ¡
dâ¡ xâ¡ dâ¡ tâ¡ 2 = kâ¡ 1 tâ¡
dâ¡ xâ¡ dâ¡ tâ¡ = kâ¡ 1 tâ¡
dâ¡ xâ¡ dâ¡ tâ¡ = kâ¡ 2 tâ¡ 1 / 2 âµ kâ¡ 1 1 / 2 = kâ¡ 2
xâ¡ = kâ¡ 3 tâ¡ 3 / 2 âµ kâ¡ 3 = 2 3 kâ¡ 2
Hence dâ¡ xâ¡ dâ¡ tâ¡ ∝ tâ¡ 1 / 2 ∝ xâ¡ 1 / 3

Topic: Physics, Work Power Energy, JEE Main
The ppm level of $\mathrm{F}^{-}$in a $500 \mathrm{~g}$ sample of a tooth paste containing $0.2 \mathrm{~g} \mathrm{~F}^{-}$is

Options:
A
400

B
1000

C
250

D
200

The correct answer is:
400

Solution:
$\begin{aligned} \text { ppm } & =\frac{\text { wt. of solute }}{\text { wt. of solvent }} \times 10^6 \\ & =\frac{0.2}{500} \times 10^6=400\end{aligned}$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2012 (12 May Online)
The precipitation occurs if ionic concentration is

Options:
A Less than solubility product
B More than solubility product
C Equal to solubility product
D None of these

The correct answer is:
More than solubility product

Solution:
Precipitation occurs when ionic product > solubility.

Topic: Chemistry, Ionic Equilibrium, JEE Main
The precursor of tissue macrophages are

Options:
A Lymphocytes
B Eosinophils
C Monocytes
D None of these

The correct answer is:
Monocytes

Solution:
Monocyte is a type of leucocyte of the agranulocyte group that is produced from stem cells in the bone marrow. Monocytes remain in the blood for a short time and then migrate to other tissues as macrophages, moving particularly to those areas invaded by bacteria and other foreign materials, where they ingest large particles by phagocytosis.

Topic: Biology, Structural Organisation in Animals, NEET
The predominant form of histamine present in human blood is pKa, Histidine=60

Options:
A
B
C
D

The correct answer is:

Solution:
Structure of histamine
Blood is slightly basic in nature (pH=7.35) approx. At this pH terminal NH2 will get protonated due to more basic nature.
∴ Predominant structure of histamine is

Topic: Chemistry, Biomolecules, JEE Main
The predominant form of histamine present in human blood is pKa, Histidine=60

Options:
A
B
C
D

The correct answer is:

Solution:
Structure of histamine
Blood is slightly basic in nature (pH=7.35) approx. At this pH terminal NH2 will get protonated due to more basic nature.
∴ Predominant structure of histamine is

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2018 (08 Apr)
The preparation of ethers from alcohols by using sulphuric acid is called as

Options:
A Williamson's ether synthesis
B Continuous etherification process
C Diazomethane method
D Wurtz synthesis

The correct answer is:
Continuous etherification process

Solution:
Ether can be prepared with the help of alcohol by using sulphuric acid at 140C°. This process is called the continuous etherification process. In this process, the alcohol acts as a nucleophile.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The presence of \( \mathrm{F} \)-centres in a crystal makes it

Options:
A conducting.
B colourless.
C non-conducting.
D coloured.

The correct answer is:
coloured.

Solution:
The presence of F-centres in a crystal is responsible for colours in the crystal. They are basically unpaired electrons which occupy the anionic sites in the crystal. The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals.

Topic: Chemistry, Solid State, JEE Main
The presence of $\mathrm{NH}_4 \mathrm{Cl}$ in the test solution while precipitating group III-A hydroxides (in qualitative inorganic analysis) helps in

Options:
A Lowering $\left\lfloor\mathrm{OH}^{-}\right\rfloor$
B Lowering $\left[\mathrm{NH}_4 \mathrm{OH}\right]$
C Increasing $\left\lfloor\mathrm{OH}^{-}\right\rfloor$
D Increasing $\left[\mathrm{NH}_4 \mathrm{OH}\right]$

The correct answer is:
Lowering $\left\lfloor\mathrm{OH}^{-}\right\rfloor$

Solution:
$\begin{aligned} & \mathrm{NH}_4 \mathrm{OH} \rightarrow \mathrm{NH}_4^{+}+\mathrm{OH}^{-} \\ & \mathrm{NH}_4 \mathrm{Cl} \rightarrow \mathrm{NH}_4^{+}+\mathrm{Cl}^{-}\end{aligned}$
$\mathrm{NH}_4 \mathrm{Cl}$ decreases the dissociation of $\mathrm{NH}_4 \mathrm{OH}$ by common ion effect

Topic: Chemistry, Practical Chemistry, JEE Main
The presence of ozone in troposphere:

Options:
A protects us from the UV radiation
B protects us from greenhouse effect
C protects us from the X-ray radiation
D generates photochemical smog

The correct answer is:
generates photochemical smog

Solution:
The presence of ozone in troposphere generates photochemical smog.

Ozone acts as an important infrared absorber (greenhouse gas), particularly in the upper troposphere, and also is an absorber of solar ultraviolet (UV) radiation.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The presence of soluble fluoride ion upto 1 ppm concentration in drinking water, is :

Options:
A harmful for teeth
B harmful to skin
C harmful to bones
D safe for teeth

The correct answer is:
safe for teeth

Solution:
Soluble fluoride is often added to drinking water to make concentration upto 1 ppm because F– ions make the enamel on teeth much harder. But above 2 ppm cause brown mottling of teeth. Cause harmful effect on bones and teeth.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2020 (06 Sep Shift 1)
The presence or absence of hydroxyl group on which carbon atom of sugar differentiates RNA and DNA ?

Options:
A
$2^{\text {nd }}$

B
$3^{\text {rd }}$

C
$4^{\text {th }}$

D
$1^{\mathrm{st}}$

The correct answer is:
$2^{\text {nd }}$

Solution:
In RNA, the sugar is $\beta-D$-Ribose, where as in DNA the Sugar is $\beta$-D-2-deoxy Ribose

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2011
The pressure acting on a submarine is 3×105 Pa at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be: (Assume that atmospheric pressure is 1×105 Pa density of water is 103 kg m-3, g=10 ms-2)

Options:
A 2003%
B 2005%
C 5200%
D 3200%

The correct answer is:
2003%

Solution:
P1=ρgd+P0=3×105 Pa

∴ ρgd=2×105 Pa

P2=2ρgd+P0

=4×105+105=5×105 Pa

% increase =P2-P1P1×100

=5×105-3×1053×105×100=2003%

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2021 (16 Mar Shift 1)
The pressure and density of a diatomic gas $\left(\gamma=\frac{7}{5}\right)$ changes adiabatically from $(P, d)$ to $\left(P^{\prime}, d^{\prime}\right)$. If $\frac{d^{\prime}}{d}=32$. Then, $\frac{P^{\prime}}{P}$ is

Options:
A $\frac{1}{128}$
B $32$
C $128$
D $256$

The correct answer is:
$128$

Solution:
$\gamma=\frac{7}{5}$
For adiabatic process $P V^\gamma=$ constant where,
$V=$ volume
$\therefore \quad P d^{-7}=$ constant
$\left(\because v \propto \frac{1}{d}\right)$
where $d=$ density
$\begin{aligned} P_1 d_1^{-7} & =P_2 d_2^{-\gamma} \\ \frac{P_1}{P_2} & =\left(\frac{d_2}{d_1}\right)^{-\gamma} \Rightarrow \frac{P_1}{P_2}=\left(\frac{d_1}{d_2}\right)^\gamma \\ \frac{P}{P^{\prime}} & =\left(\frac{d}{d^{\prime}}\right)^{7 / 5} \Rightarrow \frac{P^{\prime}}{P}=\left(\frac{d^{\prime}}{d}\right)^{7 / 5} \\ & =(32)^{7 / 5} \\ & =\left(2^5\right)^{7 / 5}=2^7=128\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
The pressure and density of a diatomic gas $(\gamma=7 / 5)$ change adiabatically from $(P, d)$ to $\left(P^{\prime}, d^{\prime}\right)$. If $\frac{d^{\prime}}{d}=32$, then $\frac{P^{\prime}}{P}$ should be

Options:
A $1 / 128$
B $32$
C $128$
D None of the above

The correct answer is:
$128$

Solution:
$V=\frac{m}{d}$ and using $P V^\gamma=$ constant
We get $\frac{P^{\prime}}{P}=\left(\frac{V}{V^{\prime}}\right)^\gamma=\left(\frac{d^{\prime}}{d}\right)^\gamma=(32)^{7 / 5}=128$

Topic: Physics, Thermodynamics, JEE Main
The pressure and density of a given mass of a diatomic gas $\left(\gamma=\frac{7}{5}\right)$ change adiabatically from $(P, d)$ to $\left(P^{\prime}, d^{\prime}\right)$. If $\frac{d^{\prime}}{d}=32$, then $\frac{P^{\prime}}{P}$ is ( $\gamma=$ ratio of specific heats $)$

Options:
A $\frac{1}{128}$
B $\frac{1}{64}$
C $64$
D $128$

The correct answer is:
$128$

Solution:
$\begin{aligned} & p V^\gamma=\frac{p}{\rho^\gamma}=\text { constant } \\ & \therefore \quad \frac{p^{\prime}}{p}=\left(\frac{\rho^{\prime}}{\rho}\right)=(32)^{7 / 5} \\ & =32 \times 4=128 \\ & \end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
The pressure and temperature of $4 \mathrm{dm}^{3}$ of carbon dioxide gas are double(d) Then, the volume of carbon dioxide gas would be

Options:
A $4 \mathrm{dm}^{3}$
B $8 \mathrm{dm}^{3}$
C $2 \mathrm{dm}^{3}$
D $3 \mathrm{dm}^{3}$

The correct answer is:
$4 \mathrm{dm}^{3}$

Solution:
According to gas equation,
$\begin{aligned}
& p V &=n R T \\
\therefore & V &=\frac{n R T}{p}
\end{aligned}$
As both $p$ and $T$ are double, $V$ remains constant i.e., $4 \mathrm{dm}^{3}$.

Topic: Chemistry, States of Matter, COMEDK, COMEDK 2016
The pressure and volume of an ideal gas are related as PV32=K (Constant). The work done when the gas is taken from state AP1, V1, T1 to state BP2, V2, T2 is :

Options:
A 2(P1V1-P2V2)
B 2(P2V2-P1V1)
C 2P1V1−P2V2
D 2P2V2−P1V1

The correct answer is:
2(P1V1-P2V2)

Solution:
For PVx= constant, work done by the gas is given by,

W=nRΔT1−x=P2V2-P1V11−x.

Here x=32, therefore we can write

∴W=P2V2−P1V1−12

=2P1V1−P2V2

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2024 (01 Feb Shift 1)