Pendulum clock taken underground If a pendulum clock is taken 1 km below the Earth's surface (into a mine), it will _____ time compared with at the surface.

Introduction / Context:
Pendulum clocks use gravitational acceleration g to set their period. Moving the clock to locations where g changes will affect its timekeeping. Underground, g is slightly smaller than at the surface, and this influences the oscillation period.

Given Data / Assumptions:

• Pendulum length L is constant.
• Small-angle approximation holds: simple pendulum model.
• At a depth of 1 km, effective g is reduced compared with surface g.

Concept / Approach:
The period of a simple pendulum is T = 2π * sqrt(L / g_eff). When g_eff decreases, the period increases (oscillations take longer). A clock that counts oscillations will then register fewer cycles per true second, meaning it loses time relative to a standard.

Step-by-Step Solution:

Verification / Alternative check:
Numerically, even a small percentage decrease in g produces a proportional increase in period because T ∝ g^(−1/2). Precision clocks correct for local gravity; pendulum regulators are adjusted when moved geographically or in elevation.

Why Other Options Are Wrong:

• Gain: Would require larger g (e.g., at the poles vs equator, not underground).
• Same rate: Contradicts T dependence on g.
• Stop / alternate faster-slower: No physical basis under steady conditions.

Common Pitfalls:
Thinking gravity is stronger underground. Inside a uniform Earth model, gravitational pull decreases with depth due to the shell theorem.

Final Answer:
lose