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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
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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