As the temperature is increased, the time period of a pendulum (a) increases as its effective length increases

NCERT Exemplar · Q5

Q. As the temperature is increased, the time period of a pendulum

(a) increases as its effective length increases even though its centre of mass still remains at the centre of the bob.

(b) decreases as its effective length increases even though its centre of mass still remains at the centre of the bob.

(d) decreases as its effective length remains same but the centre of mass shifts above the centre of the bob.

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Concept understanding — Thermal Expansion

Thermal Expansion: Why Things Grow When Heated

You already know that matter is made of atoms or molecules. In a solid, these particles are locked into a lattice, but they are not still — they vibrate around fixed positions. When you heat a substance, you give its particles more energy. They vibrate faster and with larger amplitude. That larger vibration pushes the particles slightly farther apart from each other. The result? The whole object gets bigger. That is thermal expansion.

The key intuition: heat increases atomic vibration, which increases the average distance between atoms, which makes the object expand in every direction.

The Precise Physics

For small temperature changes, the expansion is directly proportional to the temperature change. This is an excellent approximation for most engineering and exam problems.

Linear expansion (change in one dimension — length):

\Delta L = \alpha L_0 \Delta T

where $\alpha$ is the coefficient of linear expansion, $L_0$ is the original length, and $\Delta T$ is the temperature change.

Area expansion (change in two dimensions):

\Delta A = \beta A_0 \Delta T

where $\beta$ is the coefficient of area expansion.

Volume expansion (change in three dimensions):

\Delta V = \gamma V_0 \Delta T

where $\gamma$ is the coefficient of volume expansion.

\alpha : \beta : \gamma = 1 : 2 : 3

This ratio is not a coincidence — it follows directly from geometry. Imagine a cube of side $L$ . Its area is $L^2$ and its volume is $L^3$ . If each side expands by a factor $(1 + \alpha \Delta T)$ , then:

New area = $L^2 (1 + \alpha \Delta T)^2 \approx L^2 (1 + 2\alpha \Delta T)$ → so $\beta = 2\alpha$
New volume = $L^3 (1 + \alpha \Delta T)^3 \approx L^3 (1 + 3\alpha \Delta T)$ → so $\gamma = 3\alpha$

The approximations hold because $\alpha \Delta T$ is tiny (typically $10^{-5}$ per °C), so squares and cubes of it are negligible.

Watch out

This ratio only holds for isotropic materials — solids that expand equally in all directions. For anisotropic crystals (like wood along vs. across the grain), the coefficients differ in different directions. In Indian exams, assume isotropy unless told otherwise.

Why Liquids Are Different

Liquids have no fixed shape, so they have no "linear" or "area" expansion — only volume expansion matters. A liquid expands to fill its container. The coefficient of volume expansion for a liquid is typically much larger than for a solid (about 10 times larger for water compared to steel).

A classic exam point: when a liquid is heated in a container, both the liquid and the container expand. The apparent expansion you see (liquid rising in a graduated cylinder) is the difference between the liquid's true expansion and the container's expansion.

Tip

For exam problems: if a liquid overflows on heating, the volume spilled = $\gamma_{\text{liquid}} V_0 \Delta T - \gamma_{\text{container}} V_0 \Delta T$ . The container's expansion "makes room" for some of the liquid's expansion.

Real-World Examples You Already Know …

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When temperature rises, the pendulum rod expands thermally, increasing the effective length from pivot to center of mass; since $T \propto \sqrt{L}$ , the time period increases. The answer is (A).

Why thermal expansion affects the pendulum

A simple pendulum's time period depends on its effective length—the distance from the pivot to the center of mass of the bob. The formula is

T = 2\pi\sqrt{\frac{L}{g}}

where $L$ is this effective length. When you heat the pendulum, the metal rod undergoes linear thermal expansion. Every part of the rod stretches proportionally, so the bob moves farther from the pivot. The center of mass of the bob itself (assuming it's a uniform sphere or similar) remains at its geometric center; what changes is where that center sits relative to the pivot.

The key insight: thermal expansion increases $L$ , and since the period grows with the square root of length, $T$ must increase.

Step-by-step reasoning

Identify what expands. The pendulum rod (or string, if it's a metal wire) has length $L_0$ at temperature $T_0$ . When temperature rises by $\Delta T$ , the new length is

L = L_0(1 + \alpha \Delta T)

where $\alpha$ is the coefficient of linear expansion. For typical metals, $\alpha \sim 10^{-5}\,\text{K}^{-1}$ , so the change is small but measurable.

Recognize that the bob's center of mass doesn't shift within the bob.

The bob itself may also expand slightly, but its center of mass remains at its geometric center. The bob is not deforming asymmetrically. What matters is that the distance from the pivot to this center increases because the rod is longer.
Apply the period formula.

The effective length $L$ has increased. Substituting into $T = 2\pi\sqrt{L/g}$ , we see

T_{\text{new}} = 2\pi\sqrt{\frac{L_0(1 + \alpha \Delta T)}{g}} = T_0\sqrt{1 + \alpha \Delta T} \approx T_0\left(1 + \frac{\alpha \Delta T}{2}\right)

for small $\alpha \Delta T$ . The period increases.

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Q.As the temperature is increased, the time period of a pendulum

Why thermal expansion affects the pendulum

Step-by-step reasoning

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