Q.$\text{O}_2$ molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms

The key idea here is that fission neutrons are born with energies around 1–2 MeV (million electron volts). The problem gives answers in joules, so we need to recall the conversion factor: $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$. A quick mental check: 1 MeV = $1.6 \times 10^{-13}$ J, so 2 MeV = $3.2 \times 10^{-13}$ J. Now we just match that to the options.

Let’s walk through it step by step:

1. Recall the typical energy of a fission neutron.

   When $^{235}\mathrm{U}$ fissions, it releases about 200 MeV total energy, most of which appears as kinetic energy of the fission fragments. The neutrons emitted (typically 2–3 per fission) have an average kinetic energy around 2 MeV. This is a standard fact from nuclear physics — the neutron spectrum peaks near 0.7 MeV but averages about 2 MeV.

2. Convert MeV to joules.

   The conversion factor is:

Therefore:

For 2 MeV:

3. Match with the given options. The options are:
   • (A) $160 \times 10^{-13}$ J = $1.6 \times 10^{-11}$ J (too large — that’s 100 MeV)
   • (B) $320 \times 10^{-15}$ J = $3.2 \times 10^{-13}$ J (exactly our value) …

Concept & Intuition

Radioactive decay is exponential: in each half‑life, exactly half of the remaining atoms decay. So if you know what fraction remains after a given time, you can count how many half‑lives have elapsed. Here, 96.875% has decayed, meaning only $100\% - 96.875\% = 3.125\%$ remains. That tiny fraction is a power of $\frac12$: $3.125\% = \frac{1}{32} = \left(\frac12\right)^5$. Five half‑lives have passed in 10 days, so one half‑life is $10 \div 5 = 2$ days.

Step‑by‑Step Reasoning

1. Write the decay law The amount remaining after time $t$ is

where $\lambda$ is the decay constant and $N_0$ is the initial amount.

2. Find the fraction remaining If 96.875% has decayed, then the fraction remaining is

So

3. Relate fraction to half‑life The half‑life $T_{1/2}$ satisfies $N(T_{1/2}) = \frac12 N_0$. After $n$ half‑lives, the fraction is $\left(\frac12\right)^n$. Here,

4. Find the half‑life Those 5 half‑lives occur in 10 days, so …

The concept here is exponential decay. When a radioactive element decays, the amount remaining after time $t$ is given by $N = N_0 e^{-\lambda t}$, where $\lambda$ is the decay constant. The half-life $T_{1/2}$ is the time for half the sample to decay, so $N = N_0/2$ when $t = T_{1/2}$. This gives $\lambda = \frac{\ln 2}{T_{1/2}}$.

The problem asks for the time taken for the element to decay from 20% to 80%. Wait — that phrasing is tricky. "Decay from 20% to 80%" means the amount of the element remaining changes from 20% of the original to 80% of the original? That would be an increase, which is impossible in decay. The intended meaning is: the element decays such that its remaining amount goes from 80% of the original down to 20% of the original. So we want the time interval between when 80% is left and when 20% is left.

Let’s work it step by step.

1. Set up the decay equation.

   Let $N_0$ be the initial amount. At any time $t$, the fraction remaining is $\frac{N}{N_0} = e^{-\lambda t}$.

   The half-life is $T_{1/2} = 10$ minutes, so $\lambda = \frac{\ln 2}{10}$.

2. Find the time when 80% remains.

   We want $\frac{N}{N_0} = 0.8$. So:

Taking natural logs:

Since $\ln(0.8) = \ln(8/10) = \ln 8 - \ln 10 = 3\ln 2 - \ln 10$, but we can keep it as is for now.

3. Find the time when 20% remains. Similarly, for $\frac{N}{N_0} = 0.2$:

4. The time interval is the difference. The time taken to go from 80% to 20% is: $\Delta t = t_2 - t_1 = -\frac{\ln(0.2)}{\lambda} + \frac{\ln(0.8)}{\lambda} = \frac{\ln(0.8) - \ln(0.2)}{\lambda}$ …

Concept & Intuition

Power (in watts) is simply energy per second (1 W = 1 J/s). If each fission event releases a fixed amount of energy, then the number of fissions per second needed to produce a given power is just that power divided by the energy released per fission. The only catch is that the energy per fission is given in electronvolts (MeV), while power is in joules per second. So we must convert MeV to joules using the fundamental conversion $1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}$.

Step-by-step solution

1. Identify the given quantities

   • Energy released per fission: $E_{\text{fission}} = 200\ \text{MeV}$.
   • Desired power output: $P = 128\ \text{W} = 128\ \text{J/s}$.
   • We need the number of fissions per second, call it $n$.

2. Convert MeV to joules

Therefore,

3. Relate power to number of fissions per second Power is energy per second:

So

4. Calculate $n$ $n = \frac{128}{3.204} \times 10^{11} \approx 39.95 \times 10^{11} = 3.995 \times 10^{12}.$ …

Concept and Intuition

Nuclear reactors are devices that initiate and control a sustained nuclear chain reaction. The core idea behind a reactor that produces fissile material is to convert naturally abundant, non-fissile (but "fertile") isotopes into fissile ones.

• Fissile materials are isotopes that can undergo fission when struck by a neutron, releasing energy and more neutrons, thus sustaining a chain reaction. Examples include Uranium-235 ($^{235}\text{U}$) and Plutonium-239 ($^{239}\text{Pu}$).
• Fertile materials are isotopes that are not fissile themselves but can be converted into fissile materials by absorbing a neutron and then undergoing radioactive decay. Examples include Uranium-238 ($^{238}\text{U}$) and Thorium-232 ($^{232}\text{Th}$).

The goal of producing fissile material is to extend the supply of nuclear fuel, as naturally occurring fissile Uranium-235 is relatively scarce, while fertile Uranium-238 and Thorium-232 are much more abundant. A reactor designed for this purpose effectively "breeds" new fuel.

Step-by-step Explanation

1. Understanding the Requirement: The question asks for the type of reactor used to produce fissile material. This implies a process where a non-fissile material is transformed into a fissile one.

2. Analyzing Option (A) Batch Reactor:

   • A batch reactor is a type of chemical reactor where reactants are loaded into the reactor, allowed to react for a specific time, and then the products are removed. The process is discontinuous.
   • This term is primarily used in chemical engineering for processes like polymerization, fermentation, or synthesis of fine chemicals. It is not associated with the production of nuclear fissile material.

3. Analyzing Option (B) Breeder Reactor:

   • A breeder reactor is a nuclear reactor that generates more fissile material than it consumes. It achieves this by converting fertile material into fissile material.
   • The most common breeding process involves Uranium-238 ($^{238}\text{U}$) absorbing a neutron to become Uranium-239 ($^{239}\text{U}$), which then undergoes two beta decays to become Plutonium-239 ($^{239}\text{Pu}$). Plutonium-239 is a fissile material.

*   Another breeding cycle involves Thorium-232 ($^{232}\text{Th}$) absorbing a neutron to become Thorium-233 ($^{233}\text{Th}$), which then decays to Uranium-233 ($^{233}\text{U}$), another fissile material. …

Concept & Intuition

Radioactive decay follows an exponential law: after $n$ half-lives, the number of remaining nuclei is the initial number divided by $2^n$. The trick here is that the half-lives of A and B are different, so “three half-lives of A” is a different time interval than “three half-lives of B.” We must first find how many half-lives of B have passed in that same time, then set the remaining numbers equal.

1. Define the half-lives

   Let the half-life of B be $T$. Then the half-life of A is $2T$ (given: “half-life of A is two times the half-life of B”).

2. Time elapsed

   After three half-lives of A, the time elapsed is

3. Number of half-lives of B in that time In time $t = 6T$, the number of half-lives of B is

4. Remaining nuclei after decay For A: after 3 half-lives,

For B: after 6 half-lives,

5. Set them equal The problem states that after three half-lives of A, the numbers are equal:

6. Solve for the ratio …

The key idea here is the empirical nuclear radius formula: for a nucleus with mass number $A$, the radius is given by $R = R_0 A^{1/3}$, where $R_0$ is a constant (roughly $1.2 \times 10^{-15}$ m). This means radius depends on the cube root of the mass number, not linearly. So when $A$ changes by a small percentage, $R$ changes by one-third of that percentage — a direct consequence of the power law.

Let’s work it through step by step.

1. Write the relation for each nucleus:

2. The problem says $A_2$ is larger than $A_1$ by $2\%$. That means:

3. Now find $R_2$ in terms of $R_1$:

4. For small percentage changes, we use the approximation $(1+x)^n \approx 1 + nx$ when $x \ll 1$. Here $x = 0.02$ and $n = \frac{1}{3}$: $(1.02)^{1/3} \approx 1 + \frac{1}{3}(0.02) = 1 + \frac{0.02}{3} = 1 + \frac{2}{300} = 1 + \frac{2}{3}\%$ …

The key to this question is knowing the specific frequency ranges allocated to different types of radio broadcasts. AM (Amplitude Modulation) and FM (Frequency Modulation) are two distinct modulation techniques, and each is assigned a different part of the electromagnetic spectrum by international agreement. The question tests your familiarity with these standard allocations — a factual but essential piece of knowledge for communication systems.

Let’s go through the options one by one.

1. Option (A): $540 - 1600$ KHz

   This is the Medium Wave (MW) band, historically used for AM broadcasting. The lower end is around 530–540 kHz, and the upper end is about 1600–1700 kHz (depending on the region). In India and most of the world, standard AM radio stations operate in this range. This matches the given range exactly.

2. Option (B): $88 - 108$ MHz

   This is the FM broadcast band. FM radio uses VHF (Very High Frequency) waves, and the standard range is 88 MHz to 108 MHz. This is not AM.

3. Option (C): $800 - 900$ MHz

   This range falls in the UHF (Ultra High Frequency) band. It is used for cellular communication (like GSM 900) and some TV broadcasts, but not for standard AM radio.

4. Option (D): $3.7 - 4.2$ GHz …

The key idea is that a nucleus is modelled as a sphere of radius $R = R_0 A^{1/3}$, where $R_0$ is a constant. This formula comes from the fact that nuclear density is roughly constant for all nuclei — so volume is proportional to the number of nucleons $A$. Once you have the radius, the volume is just the sphere volume formula. The trick in this problem is the unit conversion: the answer is asked in cubic angstroms ($\text{Å}^3$), not cubic metres. An angstrom is $10^{-10}$ m, so $1 \text{ m}^3 = 10^{30} \text{ Å}^3$.

Let’s walk through it.

1. Nuclear radius formula The radius of a nucleus with mass number $A$ is

Here $R_0 = 1.0 \times 10^{-15}$ m and $A = 27$. So

Since $27 = 3^3$, we have $27^{1/3} = 3$. Therefore

2. Volume in cubic metres The volume of a sphere is $V = \frac{4}{3}\pi R^3$. Using $\pi \approx 3$:

The $\frac{4}{3} \times 3$ simplifies to $4$. So

because $(3.0 \times 10^{-15})^3 = 27 \times 10^{-45}$.

For an estimate, we round this to $1 \times 10^{-43} \text{ m}^3$.

3. Convert to cubic angstroms Recall $1 \text{ Å} = 10^{-10} \text{ m}$, so $1 \text{ m}^3 = (10^{10})^3 \text{ Å}^3 = 10^{30} \text{ Å}^3$. Therefore …

When two thin lenses are placed in contact, the total power of the combination is simply the sum of their individual powers. This is because the deviation of light produced by each lens adds up directly — think of it as two bends in the ray, one after the other, with no gap between them. Power is measured in diopters ($D$), and focal length in metres is the reciprocal of power: $f = 1/P$.

A positive power means a converging lens (convex), and a negative power means a diverging lens (concave). Here, one lens is diverging ($-1.6\,D$) and the other is converging ($+2.1\,D$). The net effect depends on which is stronger.

1. Find the net power. The formula for lenses in contact is:

Substituting the given values:

2. Convert net power to focal length. Focal length $f$ (in metres) is given by:

So:

3. Express the answer in centimetres. Since the options are in cm, convert: …