MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 11
        A simple pendulum has a time period T1 when on the earths surface and T2 when taken to height R above the earths surface where R is the radius of the earth the value of T1 /T2 is
2 years ago

Answers : (2)

Arun
23374 Points
							
Dear Student
 
 
T=2pi*sqrt(l/g)
g is inversly proportional to (R+h)2 …..........h is height above the earth’s  surface
therefore T is proportional to R+h
T2/T1= (R+h)/R =2............ putting h=R
Hence 
T1 / T2  = 1/2 RegardsArun (askIITians forum expert)
2 years ago
bharat
29 Points
							
time period of a pendulum on earth is 
T=2\prod*sqrt(r/g)
g is the acceleration due to gravity
as now the distance is r+h 
since g is inversely proportional to (r+h)^2  
where h is the height above the earth’s surface
hence t is proportinal to r+h
now since h=r
t2/t1=(r+h)/r=2 
hence 
t1/t2=1/2
2 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details