Question

if a pendulum clock keeping at sea level is taken at depth h=1km below sea level then the clock approximately gives time is a.gain 13.5 s per day b.loses 13.5s per day c.loses 7 s per day d.gains 7 s per day

Answer 1

T = 2 pi sqrt(L/g)
I could compute T for the two values of g but the change in g is small so better to use calculus

for sport let's make L = 1 meter

T = 2 pi sqrt L (g^-.5)
=2 pi sqrt(1/9.81)
= 2 Seconds for L = 1 and g = 9.81

dT/dg = 2 pi sqrt L(-.5)(g^-1.5)
so dT = - pi dg /g^1.5

Now what is dg ?
g = 9.81 (Re^2/r^2) where Re is R earth
dg/dr = 9.81 Re^2 (-2 r)/r^4
so
dg = 9.81 (Re^2/r^2)(-2/r) dr
for this small change in r, r is about Re and dr = -1000 meters
so
dg = 9.81 (-2/6.4*10^6) (-10^3)
dg = 3.06*10^-3

now use that dg to find the change in our period
Remember we got:
dT = - pi dg /g^1.5
so
dT = - pi (3.06*10^-3) /(9.81)^1.5
dT= - .313 * 10^-3
the negative sign means the pendulum swings faster , shorter period
our fractional change in period is
-.313*10^-3 seconds/ 2 seconds
=-.156*10^-3 seconds per second
multiply that by the number of seconds in a day to find out how much it gets ahead in a day
-.156*10^-3 * 3600 * 24 = 13.5 seconds per day fast

Answer 2

Answers : (1)
At sea level,
gravitational constant that is
g = GM/R
and Time period,
T = 2pi*root(l/g)
at below 1 km,
g’ = GM/(R-1000)
T’ = 2pi*root(l/g’)
T/T’ = root(g’/g) = > root(R/(R-1000))
T’ = T/ root(R/(R-1)) => T/root(6400/6399) = > T/root(1.000156274) => T/1.000078134
or
T’ = 0.999921872*T
putting value of T for appropriaate T we get
T’ approx 13.6 seconds.