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rajan jha Grade: 12
        

2097_23993_Photo0146.jpg

6 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

dear rajat , this is a case of conical pendulam ....


let T is tension in thread & @ is angle of inclination of string with verticle then


Tsin@ = mw2r          ..........1                                      (r is radius of circle)


Tcos@ = mg     ...............2


length of string is L then r = Lsin@


now eq 1 becomes


Tsin@ = mw2Lsin@


T = mw2L                ................3


divide 2 by 3


cos@ = mg/mw2L = g/w2L  ...........4


now w = 2pi*(number of revolution per sec)=2pi(2/pi) = 4rad/sec                                               (number of revolution per sec is 2/pi (given))


now cos@ = 10/16 = 5/8


       @ = cos-1(5/8)


option d) is correct

6 years ago
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