A simple pendulum of length 1m is taken to height R (radius of the earth) from the earth’s surface. The time period of small oscillation of the pendulum at that point is 

Complete Solution:

we calculate the time period of the pendulum at a height equal to the Earth's radius R above the surface.

Formula for the Time Period:

T = 2π √(L / g_effective)

T = Time period

L = Length of the pendulum (L = 1 m given)

g_effective = Effective acceleration due to gravity at height h

Step 1: Effective Gravity at Height R

At a height R (equal to Earth's radius), gravity decreases according to the formula:

g_effective = g × (R / (R + h))²

Here, h = R, so:

g_effective = g × (R / 2R)² = g × (1/2)² = g / 4

Step 2: Substitute g_effective into the Formula:

T = 2π √(L / g_effective)

Substitute g_effective = g / 4:

T = 2π √(L / (g / 4)) = 2π √(4L / g)

Given L = 1 m:

T = 2π √(4 × 1 / g)

Step 3: Simplify:

At Earth's surface, g ≈ 9.8 m/s². So:

T = 2π √(4 / 9.8) ≈ 4.01 seconds

Final Answer:

The closest value is 4 seconds. Thus, the correct option is 4 s