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Naba Ansari Grade: 11
        the hour of a clock is 6cm long.the magnitude of the displacement of the tip of hour hand between 1:00 pm and 5:00 pm is :  

options are
A. 6cm
B. 6sqrt3
C.12 cm
D.3 sqrt3cm
please explain me
7 years ago

Answers : (2)

AskiitiansExpert Abhinav Batra
25 Points
										

Dear Naba


The angle traced by hour hand in 12 hours =360 (as the hour hand is back to its original position after 12 hours)


thus Angle traced in one hor =360/12=30


Angle traced by hour hand in 4 hours(between 1 and 5 pm) =4* 30=120


So the vector of length 6cm is displaced by 120° in 4 hrs


Since for any two vectors u and v with angle θ between them


|v-u|2 = |v|2 + |u|-2uvcosθ


Here v and u have mod = 6 cm and angle =120°


so |v-u|2=36+36-72*-1/2=36*3


so magnitude of displacement = |v-u| =6√3


1896_14504_Untitled.jpg

7 years ago
Naba Ansari
18 Points
										

thank you very much...............................but i had a query why did we used (v-u)2 why not (v+u)2 formula........

7 years ago
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