Problem:
A pendulum clock which keeps correct time at sea level loses 15 seconds per day when taken to the top of a mountain. If the radius of the earth is 6400 km, the height of the mountain is:
a) 1.1km
b) 2.2km
c) 3.3km
d) 4.4km

Solution:
Let's assume that the clock loses time due to a decrease in the acceleration due to gravity at the top of the mountain. We can use the following formula to calculate the acceleration due to gravity at the top of the mountain:

g' = g(1 - 2h/R)

where g is the acceleration due to gravity at sea level, h is the height of the mountain, and R is the radius of the earth.

We can use the following formula to calculate the time period of a pendulum:

T = 2π√(l/g)

where l is the length of the pendulum and g is the acceleration due to gravity.

Since the length of the pendulum remains constant, we can calculate the change in time period due to the change in acceleration due to gravity:

ΔT = T(1 - √(g'/g))

We know that the clock loses 15 seconds per day, which is equivalent to a loss of 15/86400 = 0.0001736 days per day. Therefore, we can calculate the change in time period in one day:

ΔT/day = ΔT(86400)

We can substitute the values of g, R, and ΔT/day into the above equations to find the height of the mountain:

g = 9.8 m/s²
R = 6400 km = 6400000 m
ΔT/day = 0.0001736 days/day = 15 seconds/day = 15/86400 days/day

Calculations:
g' = g(1 - 2h/R) = 9.8(1 - 2h/6400000)
ΔT = T(1 - √(g'/g)) = T(1 - √((1 - 2h/6400000)))
ΔT/day = ΔT(86400) = T(86400)(1 - √((1 - 2h/6400000)))

Substituting the given values, we get:

9.8(1 - 2h/6400000) = 9.8 - 0.0001736
1 - √((1 - 2h/6400000)) = 0.999998009
ΔT/day = T(86400)(0.000001991)

Solving the first equation, we get:

h = 2200 m

Therefore, the height of the mountain is 2.2 km.

Answer:
The height of the mountain is 2.2 km (b).

T/T'=√g'/√gWKT g'=g/(1+h/r)2put the value T'-T as15sec T as60×60×24=86400 &on solving u get h=1.1