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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 T2T2−T1×t
Your question states that time lost is 150 s in a day i.e. in 86400 secs.
Hence 150= T2T2−T1×86400
Using this you can find T2 in terms of T1.
You would get T2T1=8640086250
now you know that T1 is inversly propotional to gA and T2 is inversly proportional to
gB therefore using it. gAgB=8640086250
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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Comments
Is 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
T2T2−T1×t
Your question states that time lost is 150 s in a day i.e. in 86400 secs.
Hence 150= T2T2−T1×86400
Using this you can find T2 in terms of T1.
You would get T2T1=8640086250
now you know that T1 is inversly propotional to gA and T2 is inversly proportional to gB therefore using it.
gAgB=8640086250 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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