Choose the wrong option. (a) Inertial mass is a measure of difficulty of accelerating a body by an external fo

NCERT Exemplar · Q7

Q. Choose the wrong option.

(a) Inertial mass is a measure of difficulty of accelerating a body by an external force whereas the gravitational mass is relevant in determining the gravitational force on it by an external mass.

(b) That the gravitational mass and inertial mass are equal is an experimental result.

(c) That the acceleration due to gravity on earth is the same for all bodies is due to the equality of gravitational mass and inertial mass.

(d) Gravitational mass of a particle like proton can depend on the presence of neighouring heavy objects but the inertial mass cannot.

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Concept understanding — Equivalence Principle

The Equivalence Principle: Why Gravity Feels Like Acceleration

Imagine you're in a windowless elevator. The cable snaps, and you're in free fall. You let go of a pen — it floats beside you. Your coffee cup doesn't fall to the floor; it just hangs in the air. You feel weightless.

Now imagine a different elevator, far out in deep space, far from any star or planet. This elevator is being pulled upward by a rope with a constant force, accelerating at $9.8\ \text{m/s}^2$ . You stand on the floor. You drop a pen — it falls to the floor exactly as it would on Earth. Your coffee cup sits on the table, pressing down just like at home.

Here's the key question: Can you do any experiment inside that elevator — dropping objects, measuring forces, swinging a pendulum — that would tell you which situation you're actually in?

The answer, according to Einstein, is no. No local experiment can distinguish between being in a uniform gravitational field and being in an accelerating reference frame. That's the heart of the Equivalence Principle.

The Precise Statement

Einstein's Equivalence Principle (strong form): In a sufficiently small region of spacetime, the effects of a uniform gravitational field are indistinguishable from the effects of a constant acceleration. Conversely, free fall in a gravitational field is locally equivalent to being in an inertial (non-accelerating) frame with no gravity.

There are two key layers here:

Weak Equivalence Principle (already known to Galileo): All objects fall with the same acceleration in a gravitational field, regardless of their mass or composition. A feather and a hammer fall at the same rate in vacuum.
Einstein's Equivalence Principle (the leap): Not just falling objects, but all laws of physics — electromagnetism, nuclear forces, everything — behave identically in a uniformly accelerating rocket and in a uniform gravitational field.

Why This Matters

This single idea forced Einstein to completely rethink gravity. If gravity is locally indistinguishable from acceleration, then gravity isn't a "force" in the Newtonian sense at all. Instead, gravity is a manifestation of the curvature of spacetime itself.

Note

The "sufficiently small region" is crucial. Over large distances, real gravitational fields aren't uniform — they vary (e.g., Earth's gravity is weaker at the top of a mountain than at sea level). The equivalence principle holds only locally, in a patch small enough that the field appears uniform.

A Common Misunderstanding …

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The key idea is that gravitational mass and inertial mass are fundamentally distinct concepts, but experiments show they are equal — and option (D) falsely claims gravitational mass can change with nearby objects, which violates the equivalence principle.

Let’s start with the core physics. Inertial mass ( $m_i$ ) appears in Newton’s second law: $F = m_i a$ . It measures how much a body resists acceleration — the bigger the inertial mass, the harder it is to push. Gravitational mass ( $m_g$ ), on the other hand, appears in the law of gravitation: $F = G M m_g / r^2$ . It determines how strongly a body feels or produces a gravitational field. These are two different properties, like a person’s weight and their stubbornness — not obviously related.

The deep insight is that all objects fall with the same acceleration in a given gravitational field. For a body near Earth, $F = G M_E m_g / R_E^2 = m_i a$ . Cancel the masses: $a = (G M_E / R_E^2) \cdot (m_g / m_i)$ . If $a$ is the same for all bodies, then $m_g / m_i$ must be the same constant for everything. By choosing units, we set that constant to 1, so $m_g = m_i$ . This equality is not a logical necessity — it’s an experimental fact, and it’s the heart of the equivalence principle that Einstein built general relativity on.

Now let’s examine each option.

Option (A) says inertial mass measures difficulty of acceleration, gravitational mass determines gravitational force. That’s exactly right — they are defined differently. This is correct.
Option (B) states that their equality is an experimental result. Yes — from Galileo’s leaning tower experiments to modern torsion-balance tests, no difference has ever been found. This is correct.
Option (C) says that $g$ being the same for all bodies is due to the equality of gravitational and inertial mass. That’s the logical consequence we just derived: if $m_g = m_i$ , then $a = g$ for any object. This is correct. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2020Set A-11 markMCQ
$x$
›Reveal solutionSolution
For small heights and depths, gravity decreases linearly with height and depth, but with different slopes. The height of 10 km corresponds to a depth of 20 km for the same change in $g$ .

The key idea is that near the Earth's surface, the variation of $g$ with height and with depth follows two different linear laws. For a height $h$ (small compared to Earth's radius $R$ ), gravity decreases because you are farther from the centre. For a depth $d$ , gravity decreases because the mass above you no longer contributes — only the mass of the inner sphere of radius $R-d$ matters. These two effects are not symmetric: the rate of decrease with depth is twice as fast as the rate of decrease with height. So the same fractional change in $g$ occurs at a depth that is twice the corresponding height.

Let’s work it out precisely.

Gravity at a height $h$ (small $h$ ) The acceleration due to gravity at a distance $R+h$ from the centre is

$g_h = \frac{GM}{(R+h)^2} = \frac{GM}{R^2}\left(1+\frac{h}{R}\right)^{-2}.$
For $h \ll R$ , use the binomial approximation $(1+x)^{-2} \approx 1 - 2x$ :
$g_h \approx g_0\left(1 - \frac{2h}{R}\right),$
where $g_0 = GM/R^2$ is the surface value.

So the fractional decrease is $\Delta g / g_0 = -2h/R$ .

Gravity at a depth $d$ Inside a uniform sphere, only the mass inside the radius $R-d$ contributes. That mass is

$M' = \frac{4}{3}\pi (R-d)^3 \rho = M\left(1 - \frac{d}{R}\right)^3.$
The distance from the centre is $R-d$ , so

$g_d = \frac{G M'}{(R-d)^2} = \frac{G M}{(R-d)^2}\left(1 - \frac{d}{R}\right)^3 = g_0 \left(1 - \frac{d}{R}\right).$ …
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Q.Choose the wrong option.

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