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3-2.Motion in Plane
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The second's hand of a watch has length $6\,\, cm$. Speed of end point and magnitude of difference of velocities at two perpendicular positions will be
A$6.28$ and $0 \,\,mm/s$
B$8.88$ and $4.44 \,\,mm/s$
C$8.88$ and $6.28 \,\,mm/s$
D$6.28 $ and $8.88 \,\,mm/s$
Solution
(d) $v = r\omega = \frac{{r \times 2\pi }}{T} = \frac{{0.06 \times 2\pi }}{{60}} = 6.28\,mm/s$
Magnitude of change in velocity = $|\overrightarrow {{v_2}} – \overrightarrow {{v_1}} |$
$ = \sqrt {v_1^2 + v_2^2} = 8.88\,mm/s$ $\left( {{\rm{As\,}}{v_1} = {v_2} = 6.28\,mm/s} \right)$
Magnitude of change in velocity = $|\overrightarrow {{v_2}} – \overrightarrow {{v_1}} |$
$ = \sqrt {v_1^2 + v_2^2} = 8.88\,mm/s$ $\left( {{\rm{As\,}}{v_1} = {v_2} = 6.28\,mm/s} \right)$
Standard 11
Physics
Similar Questions
A particle is rotating in a circle of radius $1\,m$ with constant speed $4\,m / s$. In time $1\,s$, match the following (in $SI$ units) columns.
| Colum $I$ | Colum $II$ |
| $(A)$ Displacement | $(p)$ $8 \sin 2$ |
| $(B)$ Distance | $(q)$ $4$ |
| $(C)$ Average velocity | $(r)$ $2 \sin 2$ |
| $(D)$ Average acceleration | $(s)$ $4 \sin 2$ |
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