Q.Choose the wrong option.

The key concept here is that the time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$, where $g$ is the acceleration due to gravity at the pendulum’s location. For a pendulum of fixed length $L$, the period changes only if $g$ changes. When we move away from Earth’s surface, gravity decreases according to the inverse-square law: $g \propto \frac{1}{r^2}$, where $r$ is the distance from Earth’s center. So the ratio of periods at two different heights is simply the square root of the inverse ratio of the corresponding $g$ values.

Let’s work it through step by step.

1. Find the distances from Earth’s center.

   The radius of Earth is $R_E$. At a height $h$ above the surface, the distance from the center is $r = R_E + h$.

   • For height $h_1 = 2R_E$: $r_1 = R_E + 2R_E = 3R_E$
   • For height $h_2 = 3R_E$: $r_2 = R_E + 3R_E = 4R_E$

2. Relate gravity to distance.

   Gravity at a distance $r$ from Earth’s center (outside Earth) is $g(r) = \frac{GM}{r^2}$, so $g \propto \frac{1}{r^2}$.

   Therefore,

3. Write the period ratio. …

The key concept here is fundamental forces and their properties. In physics, there are four known fundamental forces: gravitational, electromagnetic, strong nuclear, and weak nuclear. Each has a specific range, strength, and type of particle it affects. The question specifies "any two objects by virtue of their masses"—this immediately rules out forces that depend on electric charge or act only inside atomic nuclei.

1. Gravitational force is the attraction between any two objects that have mass. It is universal—every mass attracts every other mass—and its strength is proportional to the product of the masses (Newton’s law: $F = G \frac{m_1 m_2}{r^2}$). This matches the description exactly.

2. Electromagnetic force acts between charged particles. While massive objects often contain charge, the force is not due to mass itself but to electric charge. Two neutral objects (like two uncharged bricks) experience negligible electromagnetic force, so this is not the correct answer.

3. Strong nuclear force binds protons and neutrons inside atomic nuclei. It operates only over extremely short distances (about $10^{-15}$ m) and does not act between everyday objects or at macroscopic scales. It is not “any two objects.” …

The question asks about the force that acts between any two objects purely because they have mass. This is a fundamental concept in physics: mass attracts mass, and the force responsible is gravity. The other forces listed — electromagnetic, weak nuclear, and strong nuclear — have very different ranges and properties. Electromagnetic force acts between charged particles, not all masses. Weak and strong nuclear forces operate only inside atomic nuclei, over distances far smaller than the size of an atom. Gravity, by contrast, is universal and acts between all objects with mass, no matter how far apart they are.

1. Identify the force that depends on mass. Newton’s law of universal gravitation states that every particle of matter in the universe attracts every other particle with a force proportional to the product of their masses and inversely proportional to the square of the distance between them:

Here, $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses, and $r$ is the separation. This force is always attractive and acts between any two objects with mass.

2. Eliminate the other options.
   • (A) Electromagnetic force: Acts only between charged particles (like protons and electrons). It can be attractive or repulsive, and it does not depend on mass — a neutral object experiences no net electromagnetic force from another neutral object. …

The universe is governed by four fundamental forces: the strong nuclear force, the electromagnetic force, the weak nuclear force, and the gravitational force. These forces dictate how particles interact and how structures form, from atomic nuclei to galaxies. Their "relative strengths" are typically compared by assigning the strongest force (the strong nuclear force) a value of 1, and then expressing the others as fractions of this value. This allows us to understand their comparative dominance in different physical phenomena.

1. Understanding the Fundamental Forces and Their Strengths:

   The four fundamental forces have vastly different strengths and ranges. Here's a quick overview of their approximate relative strengths, taking the strong nuclear force as 1:

   • Strong Nuclear Force: This is the strongest force, responsible for holding atomic nuclei together. Its relative strength is $1$.
   • Electromagnetic Force: This force acts between charged particles and is responsible for all chemical interactions, light, and electricity. Its relative strength is about $10^{-2}$.
   • Weak Nuclear Force: This force is responsible for certain types of radioactive decay (like beta decay). Its relative strength is about $10^{-13}$.
   • Gravitational Force: This is the weakest force, acting between all particles with mass or energy. It is responsible for large-scale structures in the universe, like planets, stars, and galaxies. Its relative strength is about $10^{-39}$.
   Important

   The relative strengths of the fundamental forces are:

   • Strong Nuclear Force: $1$
   • Electromagnetic Force: $10^{-2}$
   • Weak Nuclear Force: $10^{-13}$
   • Gravitational Force: $10^{-39}$

2. Identifying the Strengths for the Ratio: …

The key idea is that gravity varies with distance from Earth’s center, not height above the surface. Many students mistakenly plug heights directly into the formula, forgetting to add Earth’s radius. Here, we compute the distances from the center, then apply the inverse-square law.

1. Understand the formula for gravitational acceleration The acceleration due to gravity at a distance $r$ from Earth’s center is

where $G$ is the gravitational constant and $M$ is Earth’s mass. This shows $g$ is inversely proportional to the square of the distance.

2. Convert heights to distances from Earth’s center Earth’s radius $R = 6400$ km.
   • At height $h_1 = 1280$ km:

• At height $h_2 = 3200$ km:

3. Set up the ratio of accelerations Since $g \propto 1/r^2$, the ratio is

Substitute the distances:

4. Simplify the ratio Cancel common factors. Divide numerator and denominator by 100 to get:

But it’s easier to simplify the fraction before squaring:

Concept and Intuition

This question tests your understanding of the four fundamental forces in nature: gravitational, electromagnetic, strong nuclear, and weak nuclear. The trick is to recall the sign (attractive or repulsive) each force can have.

• Gravitational force arises from mass; masses always attract each other (there’s no “negative mass” to cause repulsion).
• Electromagnetic force arises from electric charge; like charges repel, opposite charges attract — so it can be either.
• Strong nuclear force holds protons and neutrons together in the nucleus; it is attractive at short distances (about 1 femtometer) but becomes repulsive at even shorter distances (to prevent collapse). However, in everyday nuclear physics, it’s primarily attractive.
• Weak nuclear force is involved in radioactive decay; it’s not typically described as simply attractive or repulsive.

Thus, only one option is universally true.

Step-by-Step Reasoning

1. Evaluate option (A): “Electromagnetic forces are always attractive.”

   This is false. Electromagnetic forces between like charges (e.g., two electrons) are repulsive. Only opposite charges attract. So (A) is incorrect.

2. Evaluate option (B): “Electromagnetic forces are always repulsive.”

   This is also false. Opposite charges attract, so the force can be attractive. Hence (B) is incorrect.

3. Evaluate option (C): “Gravitational forces are always attractive.”

   This is true. Gravity depends on mass, and all masses attract each other. There is no known “negative mass” that would cause repulsion. This matches our everyday experience (objects fall down, planets orbit the Sun). So (C) is correct. …

The key concept is the range of a fundamental force. In physics, the range of a force is determined by the mass of its mediating particle. For gravity, the mediator is the graviton (still hypothetical), which is massless. A massless force carrier leads to a potential that falls off as $1/r^2$ (or $1/r$ for the potential) without any exponential suppression factor like $e^{-r/\lambda}$. That means the force never truly vanishes; it just gets weaker with distance but extends to infinity.

Let’s walk through the reasoning:

1. Recall the form of a force with finite range. For a force mediated by a particle of mass $m$, the potential is the Yukawa potential:

where $\lambda = \frac{\hbar}{mc}$ is the range. If $m > 0$, the exponential kills the force beyond distances $\sim \lambda$. For example, the weak nuclear force has a range of about $10^{-18}$ m because its mediator (the W/Z boson) is massive.

2. Apply this to gravity. The graviton is massless (special relativity and general relativity require it). So $m = 0$ implies $\lambda = \infty$. The exponential factor becomes $e^{-r/\infty} = 1$, and the potential reduces to the familiar Newtonian form:

The force is $F \propto 1/r^2$, which never goes to zero for any finite $r$.

3. Compare with the given options.
   • (A) $10^{-15}$ m is the typical range of the strong nuclear force.
   • (B) $10^{-39}$ m is a number that appears in some quantum gravity contexts (Planck length is $10^{-35}$ m), but it is not the range of gravity. …

Concept and Intuition

Satellites are broadly classified based on their orbits, which dictate their applications. The two main types relevant here are geostationary satellites and polar satellites. Understanding their orbital characteristics is key to knowing their uses.

Geostationary Satellites:

These satellites orbit the Earth at a very specific altitude (approximately $35,786 \text{ km}$ above the equator) and in the same direction as the Earth's rotation. Their orbital period is exactly $24$ hours, matching the Earth's rotational period. This means they appear stationary from a point on the Earth's surface.

• Intuition for use: Because they stay over a fixed point, they are excellent for continuous monitoring of a specific region and for communication. A ground antenna can be pointed at them permanently.
• Primary uses: Telecommunication (TV broadcasting, phone calls), weather monitoring for a specific region, navigation system augmentation.

Polar Satellites:

These satellites orbit the Earth in a north-south direction, passing over the Earth's poles. They typically orbit at much lower altitudes (around $500 \text{ km}$ to $1000 \text{ km}$) compared to geostationary satellites. Their orbital period is much shorter, usually around $90$ to $100$ minutes. As the Earth rotates beneath them, they cover different strips of the Earth's surface with each orbit, eventually covering the entire globe over several orbits.

• Intuition for use: Their low altitude allows for high-resolution imaging, and their polar orbit ensures they can scan the entire Earth's surface.
• Primary uses: Remote sensing (detailed imaging of land, oceans, atmosphere), environmental monitoring (deforestation, ice caps, pollution), weather forecasting (global), mapping, military reconnaissance.

Step-by-step Evaluation

1. Analyze Statement A: "Geostationary satellites are used for remote sensing of earth"
   • Geostationary satellites are positioned high above the equator and appear stationary relative to a point on Earth. This provides a continuous view of a large, fixed area.
   • While they can perform remote sensing of this specific, fixed region (e.g., for continuous weather monitoring of a continent), they cannot provide global coverage or detailed, high-resolution images of the entire Earth's surface due to their high altitude and limited field of view. …

The key idea is that the gravitational field due to a ring is not uniform along its axis — it starts at zero at the center, rises to a maximum, then falls off to zero at infinity. We find that maximum by treating the field as a function of distance and using calculus.

1. Set up the gravitational field expression Consider a point on the axis at distance $x$ from the ring’s center. Every mass element $dm$ on the ring is at distance $\sqrt{R^2 + x^2}$ from the point. The gravitational field due to $dm$ has magnitude $G\,dm/(R^2+x^2)$, but only the component along the axis adds (the perpendicular components cancel by symmetry). The axial component is $\cos\theta$, where $\theta$ is the angle between the line from the point to $dm$ and the axis. Since $\cos\theta = x/\sqrt{R^2+x^2}$, the field from $dm$ along the axis is

2. Sum over the whole ring Integrating over all $dm$ gives the total field magnitude:

This is the function we need to maximize.

3. Find the maximum by differentiation Treat $E(x)$ as a function of $x$ (with $G, M, R$ constant). Differentiate using the quotient or product rule:

Set $E'(x)=0$ (the constant factor $GM$ can be dropped):

4. Simplify the equation Multiply both sides by $(R^2+x^2)^{5/2}$: $R^2 + x^2 = 3x^2.$ …

The key idea is that in quantum field theory, every fundamental force is transmitted by a "force carrier" particle (a boson). The range of the force is inversely proportional to the mass of that carrier. If the carrier has zero mass, the force can reach arbitrarily far — it has infinite range. If the carrier has mass, the force dies off exponentially beyond a distance related to the Compton wavelength of that particle.

Let’s apply this to each force:

1. Gravitational force — The exchange particle is the graviton, which is massless (and travels at the speed of light). Therefore, gravity has infinite range. It falls off as $1/r^2$ but never truly vanishes.

2. Weak Nuclear force — The carriers are the $W^+$, $W^-$, and $Z^0$ bosons, which are very massive (about 80–90 GeV/$c^2$). Because of this mass, the weak force has a very short range — roughly $10^{-18}$ m. So it is not infinite.

3. Electromagnetic force — The carrier is the photon, which is massless. Hence electromagnetism also has infinite range (the familiar $1/r^2$ Coulomb force). …