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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|2  -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

```
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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