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Question
A particle is kept fixed on a turntable rotating uniformly. As seen from the ground the particle goes in a circle, its speed is 20 cm/s and acceleration is 20 cm/s2. The particle is now shifted to a new position to make the radius half of the original value. The new value of the speed and acceleration will be
Options
10 cm/s, 10 cm/s2
10 cm/s, 80 cm/s2
40 cm/s, 10 cm/s2
40 cm/s, 40 cm/s2
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Solution
(a) 10 cm/s, 10 cm/s2
It is given that the turntable is rotating with uniform angular velocity. Let the velocity be \[\omega\] .
We have:
\[\text{v = r}\omega\]
\[ \Rightarrow \text{v} \propto \text{r} (\text{P for constant }\omega)\]
\[\frac{v}{v'} = \frac{r}{r'}\]
\[ \Rightarrow \text{v}' = \frac{\text{v}}{2} = 10 \text{ cm/s}\]
Similarly, we have: \[a' = \frac{a}{2} = 10\text{ cm/ s}^2\]
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