datavorous/entrance-exam-dataset huggingface dataset:
When the battery across the plates of a charged condenser is disconnected and a dielectric slab is introduced between its plates then the energy stored

Options:
A becomes infinity
B does not change
C increases
D decreases

The correct answer is:
decreases

Solution:
Since the battery is disconnected, the charge on the capacitor remains constant. When the dielectric slab is introduced, the capacitance increases $\mathrm{k}$ times.
Energy stored, $\mathrm{U}_{\mathrm{i}}=\frac{\mathrm{q}^2}{2 \mathrm{C}}$
$$
\mathrm{U}_{\mathrm{f}}=\frac{\mathrm{q}^2}{2 \mathrm{kC}} < \mathrm{U}_{\mathrm{i}}
$$

Topic: Physics, Capacitance, MHT CET, MHT CET 2021 (22 Sep Shift 1)
When the concentration of alkyl halide is tripled and concentration of nucleophile is reduced to half, the rate of SN2 reaction is increased by

Options:
A 3 times
B 1.5 times
C 2 times
D 6 times

The correct answer is:
1.5 times

Solution:
Rate of SN2 reaction is as follows:
Rate = [alkyl halide] Ã [nucleophile]
Rate (initially) =C1C2
At new concentration
Ratefinally=3C1C22=1.5ÃC1C2
Rate final  = 1.5 × Rate initial

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
When the concentration of O2 is doubled while keeping the concentration of NO constant, how does the rate change?

Options:
A It decreases by a factor of two
B It increases by a factor of two
C It remains constant
D It becomes unpredictable

The correct answer is:
It increases by a factor of two

Solution:
When the concentration of O2 is doubled while the concentration of NO is kept constant, the rate also gets doubled.

Topic: Chemistry, Chemical Kinetics, NEET
When the current changes from $+2 \mathrm{~A}$ to $-2 \mathrm{~A}$ in $0.05$ second, an e.m.f. of $8 \mathrm{~V}$ is induced in a coil. The coefficient of self-induction of the coil is

Options:
A
$0.2 \mathrm{H}$

B
$0.4 \mathrm{H}$

C
$0.8 \mathrm{H}$

D
$0.1 \mathrm{H}$

The correct answer is:
$0.1 \mathrm{H}$

Solution:
$e=-\mathrm{L} \frac{\mathrm{dl}}{\mathrm{dt}} \Rightarrow 8=(\mathrm{L}) \frac{2-(-2)}{0.05} \Rightarrow \mathrm{L}=0.1 \mathrm{H}$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2003
When the current in a coil changes from 5 A to 2 A in 0.1 s, an average voltage of 50 V is produced. The self-inductance of the coil is

Options:
A 1.67 H
B 6 H
C 3 H
D 0.67 H

The correct answer is:
1.67 H

Solution:
Magnetic flux
Ï=LI â âÏât=L âIât
â´ Îµindaverage= âÏât=L âIât
â 50=L Ã 5-20.1
â 53=L
â L=1.67 H

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2015 (10 Apr Online)
When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is x8, where x=_________.

Options:

The correct answer is:
9

Solution:
The total energy of the harmonic oscillator is given by the formula

E=12kA2 ...1

where, k is the force constant and A is the amplitude.

The potential energy of the oscillator at the given displacement can be written as

U=12kA32=kA22×9=E9

Hence, its kinetic energy is given by

KE=E-E9=8E9

Thus, the required ratio is

EKE=E8E9=98

Hence, x=9.

Topic: Physics, Oscillations, JEE Main, JEE Main 2024 (29 Jan Shift 1)
When the distance between the charged particles is halved, the force between them becomes

Options:
A One-fourth
B Half
C Double
D Four times

The correct answer is:
Four times

Solution:
Step 1: Assumption
As given the distance between them which is cut in half is \(\frac{\mathrm{r}}{2}\)
Let \(F^\prime\) be the force between the charges when the distance is cut in half
Step 2: Formula used
The force between the two charged particles can be written as:
\(F=\frac{k q_1 q_2}{r^2}\)
where, \(q_1\) and \(q_2\) are the charges, \(r\) is the distance, \(k\) is constant of proportionality(Coulomb's constant), \(F\) is the force between the charges.
Step 3: Calculation of the force when the distance between the charges is cut in half
When the distance is halved the force becomes
\(\begin{aligned}
& F=\frac{k q_1 q_2}{r^2} \\
& F^{\prime}=\frac{k q_1 q_2}{\left(\frac{r}{2}\right)^2} \\
& F^{\prime}=\frac{4 k q_1 q_2}{r^2} \\
& F^{\prime}=4 F^{\prime}
\end{aligned}\)
So, when the distance between the two charges is halved, the force is increased four times.

Topic: Physics, Electrostatics, JEE Main
When the distance separations between the slit and screen is doubled, the angular separation between the fringes in slit diffraction experiment.

Options:
A First increase and then decreases
B Remains same
C Increases
D Decreases

The correct answer is:
Remains same

Solution:
Angular separation is $\theta=\frac{\beta}{D}=\frac{\left(\frac{D \lambda}{d}\right)}{D}=\frac{\lambda}{d}$
Since $\theta$ is independent of $D$, angular separation would remain the same.

Topic: Physics, Wave Optics, MHT CET, MHT CET 2022 (07 Aug Shift 1)
When the electron orbiting in hydrogen atom in its ground state moves to third excited state, the de-Broglie wavelength associated with it

Options:
A will decrease.
B remain same.
C will increase.
D will be zero.

The correct answer is:
will increase.

Solution:
Will increase.

Topic: Physics, Atomic Physics, MHT CET, MHT CET 2020 (20 Oct Shift 2)
When the elevation of the sun changes from 45o to 30o , the shadow of a tower increases by 60 units, then the height of the tower is

Options:
A 303 units
B 302 units
C 303+1 units
D 302+1 units

The correct answer is:
303+1 units

Solution:
tan⁡45o=ha⇒h=a
tan⁡30o=ha+60⇒h=a+603
⇒3 h=h+60
⇒h=603-1=603+12=303+1 units

Topic: Mathematics, Heights and Distances, JEE Main
When the excited electron of a H atom from n=5 drops to the ground state, the maximum number of emission lines observed are

Options:

The correct answer is:
10

Solution:
Maximum number of emission lines

=ΔnΔn+12Δn=n2-n1n2=5n1=1

Maximum number of emission lines=44+12=202=10

These lines correspond to the lyman series of the hydrogen spectrum

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2022 (25 Jul Shift 2)
When the following aldohexose exists in its D-configuration, the total number of its open chain stereoisomers are

Options:

The correct answer is:
4.00

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
When the following aldohexose exists in Pyranose form, the total number of stereoisomers in its pyranose form is

Options:

The correct answer is:
16

Solution:
Pyranose form is cyclic structure containing four chiral atoms.
Hence number of stereoisomers =2n
=24=16

Topic: Chemistry, General Organic Chemistry, JEE Main
When the following reaction was carried out in a bomb calorimeter, ÎU is found to be ââ740.0âkJ/mol of NH 2 CN(s) (s) at 300âK .
NH2CN(s)+ 32O2gâN2g+ CO2g+ H2O(l)
Calculate ÎH 300âK for the reaction.

Options:
A -738.75 kJ
B +738.75 kJ
C -824.75 kJ
D -919.57 kJ

The correct answer is:
-738.75 kJ

Solution:
ΔH = ΔU +  Δn g RT
ΔH=−740000+0.5×8.3×300=−738755 J

Topic: Chemistry, Thermodynamics (C), JEE Main
When the forward bias voltage of a diode is changed from 0.6 V to 0.7 V, the current changes from 5 mA to 15 mA. Then its forward bias resistance is

Options:

The correct answer is:
10

Solution:
Forward biased resistance = ∆V∆I=0.7-0.615-5×10-3
=0.110×10-3=10 Ω

Topic: Physics, Semiconductors, JEE Main
When the gap between two identical concave thin lenses Î¼=32,f=10 cm placed in contact is filled with certain liquid, the image of an object placed at 15 cm from lens combination shifts away from the lens by 54 cm. If the refractive index of the liquid is Î¼, then write 3Î¼ as your answer.

Options:

The correct answer is:
4

Solution:
In case (i) (without liquid)
1v-1-15=2f
⇒ 1v=-15-115=-415
⇒ v=-3.75 cm
When the gap is filled with liquid
v'=-3.75-1.25=-5 cm
⇒ -15-1-15=-15+(μl-1)(2R)
Here, 110=32-12R ⇒R=10 cm
Hence, -215=-15+μl-15
⇒ μl=43

Topic: Physics, Ray Optics, JEE Main
When the hydrogen ion concentration H+changes by a factor of 1000 , the value of pH of the solution

Options:
A increases by 1000 units
B decreases by 3 units
C decreases by 2 units
D increases by 2 units

The correct answer is:
decreases by 3 units

Solution:
Let the initial concentration of H+ be 1

∴ H+i = 1 ⇒ pH = 0

It changes by 1000 units

∴ H+f = 103 ⇒ pH = -3

So, pH decreases by 3 units.

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2023 (25 Jan Shift 2)
When the ideal gas is compressed isothermally, then its pressure

Options:
A increases
B decreases
C remains the same
D first increases and then decreases

The correct answer is:
increases

Solution:
According to isothermal condition,
$p V=K$ (constant)
i.e. $\quad p \propto \frac{1}{V} \quad\left[\because p=\frac{K}{V}\right]$
So, in isothermal compression, when volume decreases, pressure increases.

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2021 (24 Aug Shift 2)
When the length and area of cross-section both are doubled, then its resistance

Options:
A Will become half
B Will be doubled
C Will remain the same
D Will become four times

The correct answer is:
Will remain the same

Solution:
$\begin{aligned} & R_1 \propto \frac{1}{A} \Rightarrow R_2 \propto \frac{21}{2 A} \text { i.e. } R_2 \propto \frac{1}{A} \\ & \therefore R_1=R_2\end{aligned}$

Topic: Physics, Current Electricity, JEE Main
When the mass and speed of the body are doubled, the kinetic energy of the body

Options:
A becomes double
B becomes four times
C becomes eight times
D remains unchanged

The correct answer is:
becomes eight times

Solution:
Given that, the mass and speed of body are doubled.
Then, final kinetic energy,
$$
\mathrm{KE}_f=\frac{1}{2}(2 m)(2 v)^2=\left(\frac{1}{2} m v^2\right) 8
$$
where, $m$ and $v$ are initial mass and speed.
Here, initial kinetic energy, $\mathrm{KE}_i=\frac{1}{2} m v^2$
$$
\therefore \quad \mathrm{KE}_f=8 \mathrm{KE}_i
$$

Topic: Physics, Work Power Energy, AP EAMCET, AP EAMCET 2021 (24 Aug Shift 1)
When the momentum of a photon is chagned by an amount p' then the corresponding change in the de-Broglie wavelength is found to be 0.2%. Then, the original momentum of the photon was

Options:
A 300pâ²
B 500p′
C 400p′
D 600p′

The correct answer is:
500p′

Solution:
âµ Î»=hp
â´ Î»â1p
â Îpp= -ÎÎ»Î»
â´ Îpp=ÎÎ»Î»
â pâ²p=0.20100=1500
â p=500pâ²

Topic: Physics, Atomic Physics, JEE Main
When the number of chromosomes is already reduced to half in the first reduction division of meiosis. What is the necessity of the second mitotic division?

Options:
A The division is required for the formation of four gametes.
B The division ensures equal distribution of genes on the chromosomes.
C The division ensures equal distribution of haploid chromosomes.
D The division is required for segregation of replicated chromosomes.

The correct answer is:
The division is required for segregation of replicated chromosomes.

Solution:
There are two stages in the meiotic division, meiosis I and meiosis II. These stages mainly involve the reductional and equational division of the cells. In the first meiotic division, the chromosome number is reduced to half (haploid condition) and the chromatids move to the opposite poles. In the second meiotic division which resembles normal mitosis, there is the formation of four gametes and the chromosomes are distributed to each gamete, therefore the second meiotic division is required for the segregation of the replicated chromosome.

Topic: Biology, Cell Cycle and Cell Division, NEET
When the object is self-luminous, the resolving power of a microscope is given by the expression

Options:
A $\frac{2 \mu \sin \theta}{1.22 \lambda}$
B $\frac{\mu \sin \theta}{\lambda}$
C $\frac{2 \mu \cos \theta}{1.22 \lambda}$
D $\frac{2 \mu}{\lambda}$

The correct answer is:
$\frac{2 \mu \sin \theta}{1.22 \lambda}$

Solution:
For a normal object, resolving power of microscope
is given by $\Rightarrow R \cdot P \cdot=\frac{2 \mu \sin \theta}{\lambda}$
Self-luminous objects are those which can emit their own light. egg. $\rightarrow$ torch etc.
Resolving power of self luminous object will be $\Rightarrow$
$=\frac{(R \cdot P .)_{\text {normal }}}{1.22}=\frac{2 \mu \sin \theta}{1.22 \lambda}$
Resolving power for self-luminous= $\frac{2 \mu \sin \theta}{1.22 \lambda}$
Option (1) is the correct answer.

Topic: Physics, Ray Optics, JEE Main
When the radioactive isotope ${ }_{88} R a^{226}$ decays in a series by emission of three alpha $(\alpha)$ and a beta ( $\beta$ ) particle, the isotope $X$ which remains undecay is

Options:
A ${ }_{83} X^{214}$
B ${ }_{84} X^{218}$
C ${ }_{84} X^{220}$
D ${ }_{87} X^{223}$

The correct answer is:
${ }_{83} X^{214}$

Solution:
Due to emission of $3 \alpha$ particles, the mass number is reduced by 12 while atomic number is decreased by 6 . Due to emission of $\beta$ particle, the atomic number is increased by one while there is no change in mass number. Now, resulting mass will be decreased by 12 and atomic number is decreased by 5. So, isotope $X$ is ${ }_{83} X^{214}$.

Topic: Physics, Nuclear Physics, JIPMER, JIPMER 2017
When the rms voltages V L ,  V C and V R are measured respectively across the inductor L, the capacitor C and the resistor R in a series LCR circuit connected to an AC source, it is found that the ratio V L : V C : V R =1:2:3. If the rms voltage of the AC source is 100 V, then V R is close to :

Options:
A 50 V
B 70 V
C 100 V
D 90 V

The correct answer is:
90 V

Solution:
V L : V C : V R =1:2:3
V rms =100 V then V R =?
Let V L =x, then V C =2x and V R =3x
using V= V R 2 + ( V c − V L ) 2
1 0 0 = 3 x 2 + 2 x - x 2
1 0 0 = 1 0 x 2
x = 1 0 0 1 0 =10 10
∴ V R = 3 x = 30 10 ≈90 V

Topic: Physics, Alternating Current, JEE Main
When the rms voltages V L ,  V C and V R are measured respectively across the inductor L, the capacitor C and the resistor R in a series LCR circuit connected to an AC source, it is found that the ratio V L : V C : V R =1:2:3. If the rms voltage of the AC source is 100 V, then V R is close to :

Options:
A 50 V
B 70 V
C 100 V
D 90 V

The correct answer is:
90 V

Solution:
V L : V C : V R =1:2:3
V rms =100 V then V R =?
Let V L =x, then V C =2x and V R =3x
using V= V R 2 + ( V c − V L ) 2
1 0 0 = 3 x 2 + 2 x - x 2
1 0 0 = 1 0 x 2
x = 1 0 0 1 0 =10 10
∴ V R = 3 x = 30 10 ≈90 V

Topic: Physics, Alternating Current, JEE Main, JEE Main 2014 (09 Apr Online)
When the sample of copper with zinc impurity is to be purified by electrolysis, the appropriate electrodes are

Options:
A
cathode - pure zinc anode - pure copper

B
cathode - impure sample anode - pure copper

C
cathode - impure zinc anode - impure sample

D
cathode - pure copper anode - impure sample

The correct answer is:
cathode - pure copper anode - impure sample

Solution:
Pure metal always deposits at cathode

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2002
When the source of sound moves towards a stationary observer, the wavelength of sound received by him

Options:
A decreases while frequency increases.
B remains the same whereas frequency increases.
C increases and frequency also increases.
D decreases while frequency remains the same.

The correct answer is:
decreases while frequency increases.

Solution:
n a =n[ ÏÂ± Ï 0 Ïâ Ï s ] .
Ï 0 =0 .
n a =n[ Ï Ïâ Ï s ]
So frequency increase, wavelength decreases.

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2018
When the source of sound moves towards a stationary observer, the wavelength of sound received by him

Options:
A decreases while frequency increases.
B remains the same whereas frequency increases.
C increases and frequency also increases.
D decreases while frequency remains the same.

The correct answer is:
decreases while frequency increases.

Solution:
n a =n[ ÏÂ± Ï 0 Ïâ Ï s ] .
Ï 0 =0 .
n a =n[ Ï Ïâ Ï s ]
So frequency increase, wavelength decreases.

Topic: Physics, Waves and Sound, NEET
When the temperature is increased, surface tension of water

Options:
A increases
B decreases
C Remain constant
D Shows irregular behaviour

The correct answer is:
decreases

Solution:
An increase in temperature of water (or any liquid) leads to an increase in kinetic energy of molecules. Thus, intermolecular forces among molecules are lowered and a decrease in surface tension is noticed.

Topic: Chemistry, States of Matter, NEET
When the temperature is increased the angle of contact of a liquid

Options:
A Increases
B Decreases
C Remains the same
D First increases and then decreases

The correct answer is:
Decreases

Solution:
As the temperature increases, the surface tension and cohesive force among liquid molecules weakens, so the angle of contact with solid surface decreases.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
When the temperature of 2 moles of an ideal gas is increased by 20°C at constant pressure, find the work involved in the process.

Options:
A 5R
B 40R
C 15R
D 20R

The correct answer is:
40R

Solution:
We know that for ideal gas

PV=nRT

Now taking small changes

∆(PV)=nR∆T

as pressure remains constant. So

P∆V=nR∆T

Work done in increasing the kinetic energy of gas is given by

W=P∆V

From the above expression, work can be modified as

W=nRΔTW=2×R×20W=40 R

Topic: Chemistry, Thermodynamics (C), AP EAMCET, AP EAMCET 2021 (19 Aug Shift 2)
When the temperature of a black body increases, it is observed that the wavelength corresponding of maximum energy changes from 0.26 Î¼m to 0.13 Î¼m. The ratio of emissive powers of the body at the respective temperature is

Options:
A 161
B 41
C 14
D 116

The correct answer is:
116

Solution:
The body would appear red â´ Î»r>Î»y
Î»y1Î»y2=T2T1=0.260.13=2
â´ E1E2=T1T24=124=116

Topic: Physics, Thermal Properties of Matter, JEE Main
When the temperature of a gas is increased from 30°C to 930°C, the root mean square speed of the gas would

Options:
A get doubled
B remain constant
C gets halved
D increase by 5.5 times

The correct answer is:
get doubled

Solution:
Root-mean-square is directly proportional to the square root of temperature.

urms=3RTM

At the different temperature of same gas, the mass will be same.

Temperature T1=30+273=303K

Temperature T2=930+273=1203K

So,

u1u2=T1T2=3031203=17.413.68u2=2u1

Hence, the root mean square speed of the gas would get doubled.

Topic: Chemistry, States of Matter, AP EAMCET, AP EAMCET 2021 (20 Aug Shift 2)
When the temperature of a metal wire is increased from 0ºC to 10ºC, its length increases by 0.02%.The percentage change in its mass density will be closed to:

Options:
A 0.06
B 2.3
C 0.008
D 0.8

The correct answer is:
0.06

Solution:
Δℓ=ℓαΔt
α=ΔℓℓΔT=0.02100×10; α=2×10-5
γ=3α=6×10-5
ΔVV=γ×ΔT; ΔVV×100=6×10-5×10×100=6×10-2
Volume increase by 0.06% therefore density decrease by 0.06%

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2020 (02 Sep Shift 2)
When the temperature of reactions will increase then the effect on $\mathrm{pH}$ value will

Options:
A increase
B decrease
C first increases then decreases
D remains same.

The correct answer is:
decrease

Solution:
As the temperature increases, the dissociation of water in solution increases. Hence the $\mathrm{H}^{+}$ion conc. increases and $\mathrm{pH}$ of the solution decreases with increase in temperature.

Topic: Chemistry, Ionic Equilibrium, AIIMS, AIIMS 2001
When the torque acting upon a system is zero, the parameter that remains constant is

Options:
A force
B linear momentum
C angular momentum
D linear impulse

The correct answer is:
angular momentum

Solution:
Torque acting upon a rotating body is given as
$$
\tau=\frac{d \mathbf{L}}{d t}
$$
where, $L=$ angular momentum.
When $\tau=0$, then
$$
\begin{aligned}
& \frac{d \mathbf{L}}{d t}=0 \\
& \mathbf{L}=\text { constant }
\end{aligned}
$$
$$
\Rightarrow \quad \mathbf{L}=\text { constant }
$$
Hence, when the torque acting upon a system is zero, then its angular momentum remains constant.

Topic: Physics, Rotational Motion, AP EAMCET, AP EAMCET 2020 (22 Sep Shift 2)
When the value of $R$ in the balanced Nheatstone bridge, shown in the figure, is increased from $5 \Omega$ to $7 \Omega$, the value of shas to increased by $3 \Omega$ in order to maintain the palance. What is the initial values of $S$ ?

Options:
A $2.5 \Omega$
B $3 \Omega$
C $5 \Omega$
D $7.5 \Omega$

The correct answer is:
$7.5 \Omega$

Solution:
According to the balanced condition of Wheatstone bridge.
In the first case,
$$
\begin{array}{l}
\frac{P}{Q}=\frac{R}{S} \\
\frac{P}{Q}=\frac{5}{S}....(i)
\end{array}
$$
In the second case
$$
\frac{P}{Q}=\frac{7}{S+3}....(ii)
$$
From Eqs. (i) and (ii), we get
$\begin{aligned} \frac{5}{S} &=\frac{7}{S+3} \\ 5(S+3) &=7 S \\ 5 S+15 &=7 S \\ 2 S &=15 \\ S &=7.5 \Omega \end{aligned}$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2019
When the vapours of tertiary butyl alcohol are passed through heated copper at \( 573 \mathrm{~K} \), the product formed is

Options:
A But-2-ene
B 2-Butanone
C 2-Methyl propene
D Butanal

The correct answer is:
2-Methyl propene

Solution:
When the tertiary butyl alcohol vapours pass through heated copper, 2-methyl propene is formed with the removal of one water molecule.

According to the Saytzeff rule, more substituted alkene will be more stable.

Topic: Chemistry, Hydrocarbons, JEE Main
When the volume of a sealed container with a liquid-vapour equilibrium is suddenly increased, what happens to the rates of evaporation and condensation initially?

Options:
A The rate of evaporation increases, and the rate of condensation decreases.
B The rate of evaporation decreases, and the rate of condensation increases.
C The rate of evaporation exceeds the rate of condensation.
D The rates of evaporation and condensation remain the same.

The correct answer is:
The rate of evaporation exceeds the rate of condensation.

Solution:
Correct Option is : (C)
The rate of evaporation exceeds the rate of condensation.

Topic: Chemistry, Ionic Equilibrium, NEET
When the wavelength of radiation falling on a metal is changed from 500nm to 200nm, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to:

Options:
A 0.81eV
B 1.02eV
C 0.52eV
D 0.61eV

The correct answer is:
0.61eV

Solution:
KEmax=hcλ-ϕ=hc500-ϕ .....(i)

Now,

3KEmax=hc200-ϕ

.....(ii)

From equation (i) and (ii)

iii=31=hc200-ϕhc500-ϕ

Put the value of hc=1237.5 and solving ϕ=0.61eV

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2020 (03 Sep Shift 1)
When the work function of a metal increases, maximum kinetic energy of emitted
photoelectrons

Options:
A first decreases and then increases.
B increases.
C remains same.
D decreases.

The correct answer is:
decreases.

Solution:
(K.E.) $_{\max }=h v-w_{0}$
Hence if $w_{0}$ increase (K.E.) $_{\max }$
Decreases

Topic: Physics, Dual Nature of Matter, MHT CET, MHT CET 2020 (13 Oct Shift 2)
When there is an electric current through a conducting wire along its length, then an electric field must exist

Options:
A Outside the wire but normal to it
B Outside the wire but parallel to it
C Inside the wire but parallel to it
D Inside the wire but normal to it

The correct answer is:
Inside the wire but parallel to it

Solution:
As current is flowing through it, it means charges are flowing along its length therefore heir must be some electric field parallel to the length of the wire.
Hence correct option is C

Topic: Physics, Current Electricity, JEE Main
When three identical bulbs of 60 watt, 200 volt rating are connected in series to a 200 volt supply, the power drawn by them will be:

Options:
A 60 watt
B 180 watt
C 10 watt
D 20 watt

The correct answer is:
20 watt

Solution:
The resistance of each bulb is $\frac{V^2}{P}$ $=\frac{(200)^2}{60} \Omega$ when three bulbs are connected in series then equivalent resistance is $\frac{3 \bullet(200)^2}{60}$
$\therefore \quad$ Power $P=\frac{V^2}{R_{e q}}=20 \mathrm{~W}$

Topic: Physics, Current Electricity, JEE Main
When tin is treated with concentrated nitric acid

Options:
A It is converted into stannous nitrate
B It is converted into stannic nitrate
C It is converted into metastannic acid
D It becomes passive

The correct answer is:
It is converted into metastannic acid

Solution:
Tin is oxidised to meta stannic acid when it is treated with nitric acid.
$\mathrm{S} n+4 \mathrm{HNO}_3 \rightarrow \mathrm{H}_2 \mathrm{SnO}_3+4 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main