physics

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

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  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

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    posted by Damon
  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.

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