JEE Main Part Test - 6 - JEE MCQ

Since, P_r​=P_b​ 
r for red and b for blue.
P_r​=P_b​
or, n_r​× (hc/λr)  ​=nb​× (hc​/λ_b)​
or,  ​(n_r/n_b)​​= λ_r​​/λ_b ​
Since, the wavelength of red bulb is greater than the wavelength of blue bulb.
or,  n_r​>n_b

Current is 0.16×10⁻⁶ Amp it means 0.16×10−6 Coulomb charge is flowing per second
So, n=0.16×10−6C /1.6×10⁻¹⁹C ​=10¹² electrons are generated per second
Now we notice that one photon has energy E, E=hc/λ=​=(6.62×10⁻³⁴Js×3×10⁸ms^-1)/(5000×10−10m) ​=3.972×10⁻¹⁹Joule
So, number of photon in 10−3W will be N=10⁻³/3.972×10⁻¹⁹ ​=0.25×10¹⁶ this is number of photons incident per second
So required percentage is (n/N)×100=10¹⁴/(0.25×10¹⁶) ​=0.04%

The maximum kinetic energy for the photoelectrons is 
E_max​=hν−ϕ
where, ν is the frequency of incident light and ϕ is photoelectric work function of metal.
eV_0​=hν−ϕ ...................(1)
where, V₀​ is the stopping potential and e is the electronic charge.
When, the frequency of light in a photoelectric experiment is doubled, 
eV₀′​=2hν−ϕ 
eV₀′​=2[hν−(ϕ/2​)].........................(2)
From the above two equations we can say that the K.E. in (2) is more than double of K.E in (1). Hence, when the frequency of light in a photoelectric experiment is doubled, the stopping potential becomes more than double.
So, the answer is option (C).
 

As we know that the stopping potential of the photoelectron is equal to the maximum kinetic energy of the photoelectron,
KEmax​=10.4V
Now, in photoelectric effect,
Energy of incident radiation (E_in​) = work function + K.Emax​
⇒ E_in​=1.7+10.4
⇒ E_in​=12.1eV
Now, for 0 hydrogen atom,
Energy of first energy level, E₁​=−13.6eV
Energy of second energy level, E₂​=−3.4eV
Energy of third energy level, E₃​=−1.5eV
Hence, a transition from third to first energy level will result in emission of radiation of energy = E₃​−E₁​=12.1eV which is same as the energy of incident radiation of above photoelectric effect.
Thus, correct answer is n=3 to 1
 

When photon strikes with the electron it completely transfers it’s energy to the electron as during photoelectric experiment the threshold frequency required is used by the electron to eject from the atom which is also called the work function and the remaining energy in electron is kinetic energy which can be measured. Now (before collision) since photon is a particle it must have mass and thus it has energy equivalent to E=mc2 .
After collision when it completely transfers it’s energy to electron thus E=0
And therefore 0=mc2 thus i guess photon vanishes.
 

When the source is moved away from the emitter, intensity of the incident radiation decreases but frequency remains the same so there will be no change in the stopping potential. Thus, it remains constant.

Saturation current is the maximum current possible and it will be directly proportional to the number of number of electrons falling on collector plate per second which depend on number of photons
 incident on the cathode as one photon contribute in one electron and the number of photons is actually
 proportional to intensity which varies 
As intensity I∝1/r2​, where r is the distance 
So the correct graph will be decreasing with power2 of distance and it will be rapidly decreasing with a higher value of r.
 

Given, the maximum kinetic energy: K_max​=4eV
If V₀​ be the stopping potential, then K_max​=eV_0​
⇒eV_0​=4eV 
⇒V₀​=4V

As,
Maximum K.E= incident photon energy − work function
(½)​mV₁²​=2W−W=W⟶(1)
and
V_e​mV₂²​=5W−W=4W⟶(2)
from (1) and (2)
V₁/V₂​ ​​=1/2​
V_1​:V₂​=1:2

We know that,
eV=(hc/λ)-w
V=(hc/eλ)-(w/e)
Arithmetic progression =>V₂=(V₁+V₂)/2
Now,
(hc/eλ₂)-w/e=1/2[(hc/eλ₁)-(w/e) +(hc/eλ₃) -(w/e)]
=>1/ λ₂=1/2[(1/ λ₁)+(1/λ₃)]
=>2/ λ₂=1/ λ₁ + 1/λ₃
Hence the correct answer is harmonic Progression.

The energy released by the fission of one uranium atom, E = 200 MeV
E = 200 × 10⁶ × 1.6 × 10^-19 J
The number of fissions required per second, n/t = P/E.

E = mc² = (0.5/1000) × (3 × 10⁸)² = 4.5 × 10¹³ Ws
1000 W = 1 kW
3600 seconds = 1 hour
So, by unit conversion, E = 1.25 × 10⁷ kWh

Neutrons are not emitted by a radioactive substance.

Negative β decay is expressed by the following equation:
stands for antineutrino.
Hence, the correct choice is (c).

Since 15 minutes = 5 x 3 minutes = 5 half-lives, the number of nuclei left after 15 minutes = 1/2⁵ = 1/32 of the original number

Therefore, the mass of 16 g sample left after 15 minutes = 16/32 = 0.5 g

Hence, the correct choice is (d).

We know that,
BE(B) = [ZM_p + NM_n - M(N, Z)]c²
Where Z is number of protons
N is number of neutrons
M_p is mass of a proton
and M_n is mass of the nucleus

In the given circuit only diode D₁ will allow the current to pass through as it is forward-biased.
Hence, the current through the 100 ohm resistance is (∵ resistance of D₁ = 50 Ω)

= 0.02 A

Current, I = V/R
If the radius becomes twice, the area becomes 4 times and resistance becomes R/4, as resistance is inversely proportional to the area of wire.
New current = 4V/R = 4I

To obtain a p-type semiconductor the impurity must be trivalent, i.e. it must be doped with foreign atoms whose valency is 3. In p-type semiconductors, holes are the majority carriers and electrons are the minority carriers. P-type semiconductors are created by doping an intrinsic semiconductor with acceptor impurities. A common p-type dopant for silicon is boron.

When a monochromatic light is incident on hydrogen atoms in the ground state (n = 1), the hydrogen atoms can absorb energy and transition to higher energy levels. When the atoms return to lower energy levels, they emit radiation of different wavelengths corresponding to the energy differences between the energy levels.

The energy levels of the hydrogen atom are given by the formula:

where E_n is the energy of the nth level and n is the principal quantum number.

Since the hydrogen atoms emit radiation of six different wavelengths, there must be six different transitions from the excited states back to lower energy levels.

The six transitions correspond to the following energy level changes:

From n = 2 to n = 1
From n = 3 to n = 1
From n = 3 to n = 2
From n = 4 to n = 1
From n = 4 to n = 2
From n = 4 to n = 3

The highest energy level involved is n = 4. Therefore, the incident light must have a frequency high enough to excite the hydrogen atoms from the ground state (n = 1) to n = 4.

The energy difference between these levels is:

The frequency of the incident light is related to the energy difference by the equation:
ΔE = hv
where h is the Planck's constant and v is the frequency.
Now, we can solve for the frequency:

So, the value of x is 3.

Let the masses of the two nuclear parts be m₁ and m₂ and their respective speeds be v₁ and v₂. 
According to the problem, the ratio of their nuclear sizes is 1 : 2^1/3. Since the nuclear size is proportional to the cube root of the mass, we can write: 

Now, according to the conservation of linear momentum, the momentum before disintegration is equal to the momentum after disintegration:

From the problem statement, the ratio of their respective speeds is n : 1, so we can write:

Substitute the expression for v₁ into the momentum conservation equation:

We know the mass ratio, so substitute that into the equation:

Divide both sides by 

Now, solve for n: 
n = 2
Thus, the value of n is 2.

The total binding energy of a nucleus is the binding energy per nucleon multiplied by the number of nucleons (protons and neutrons), which is the mass number.

The initial total binding energy of the nucleus is 242 × 7.6 MeV.

After the break, each fragment has a total binding energy of 121 × 8.1 MeV.

Since there are two such fragments, the final total binding energy is 2 × 121 × 8.1MeV.

The gain in binding energy is the final total binding energy minus the initial total binding energy. Therefore, the gain in binding energy is:

2 × 121 × 8.1 MeV – 242 × 7.6 MeV = 1960.2 MeV – 1839.2 MeV = 121 MeV.

• Higher aldehydes are soluble in Benzene. Therefore, statement (B) is correct.
• Lower Aldehydes are pungent but higher Aldehydes are fragrant. Therefore, statement (A) is incorrect.
• Acetone is more polar than n-Butane and weight is comparable, so acetone has higher boiling point. Therefore, statement (C) is incorrect.
• Methanal is soluble in water. Therefore, statement (D) is incorrect.

Presence of one vinyl (or ethenyl) group gives formaldehyde as one of the product in ozonolysis.

CH₂=CH- on ozonolysis will give HCHO.

• Ozonolysis is an organic reaction where the unsaturated bonds of alkenes, alkynes, or azo compounds are cleaved with ozone. 
• Alkenes and alkynes form organic compounds in which the multiple carbon–carbon bond has been replaced by a carbonyl group while azo compounds form nitrosamine.

The correct order of increasing acidic strength is Ethanol < Phenol < Acetic acid < Chloroacetic acid.

• Phenol is more acidic than ethanol because in phenol, the phenoxide ion obtained on deprotonation is stabilized by resonance which is not possible in case of ethanol.
• Also carboxylic acids are more acidic than alcohols and phenols as the carboxylate ion is stabilized by resonance. Chloroacetic acid is more acidic than acetic acid due to inductive effect of chlorine atom which stabilizes the carboxylate anion.

Baeyer's reagent (alk. KMnO₄) which is pink in colour decolourises due to the presence of unsaturation. Thus, it shows the presence of unsaturation in an organic compound.

• For aldol condensation to occur, α -hydrogen atom is required for carbonyl compound.
• For formaldehyde, no α -carbon and no α -hydrogen is present. Therefore, H C H O will not undergo aldol condensation.
• An alpha (α) hydrogen is the hydrogen which is bonded to alpha carbon. Alpha carbon is the first carbon which is bonded to the functional group.