Well let the time period of pendulum at A be T1 and at B be T2.
wkt time lost by a pendulum in "t" secs is given by \(\frac{T2-T1}{T2}\times t\)

Your question states that time lost is 150 s in a day i.e. in 86400 secs.

Hence 150= \(\frac{T2-T1}{T2}\times 86400\)
Using this you can find T2 in terms of T1.
You would get \(\frac{T1}{T2} =\frac{86250}{86400}\)
now you know that T1 is inversly propotional to \(\sqrt{gA}\) and T2 is inversly proportional to
\(\sqrt{gB}\) therefore using it.
\(\frac{\sqrt{gB}}{\sqrt{gA}}=\frac{86250}{86400}\)
hence gB=0.9965308 * gA
HENCE gB = 9.7660 m/s^2.

@Saurabh Dubey
–
i dnt knw whether the ans is right bcoz this Q came in our phy exam 2 days ago and no one could do it.............our teacher didnt teach us that time period formula also..............anyway thanks for the ans......................r u trying for IIT??????

@Rajath Krishna R
–
IITJEE!! thats too tough..
I am weak at chemistry I'am trying though.
btw i hope you got the other graph question that you askd.
This answer i'am 90 percent sure.
Let me know the answer once you get to know it.

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TopNewestIs the answer 9.766 m/s^2 ?

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but how to do it??????????

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Well let the time period of pendulum at A be T1 and at B be T2. wkt time lost by a pendulum in "t" secs is given by

\(\frac{T2-T1}{T2}\times t\)Your question states that time lost is 150 s in a day i.e. in 86400 secs.

Hence 150=

\(\frac{T2-T1}{T2}\times 86400\)Using this you can find T2 in terms of T1.

You would get \(\frac{T1}{T2} =\frac{86250}{86400}\)

now you know that T1 is inversly propotional to \(\sqrt{gA}\) and T2 is inversly proportional to \(\sqrt{gB}\) therefore using it.

\(\frac{\sqrt{gB}}{\sqrt{gA}}=\frac{86250}{86400}\) hence gB=0.9965308 * gA HENCE gB = 9.7660 m/s^2.

Let me know if the answer is corerect.

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is it 158.11

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i think there will only be a small change............................points A and B are two places on earth....

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Solution provided by saurabh is correct.

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