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Grade:12

1 Answers

vikas askiitian expert
509 Points
10 years ago

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

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