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  • A small steel ball falls through a syrup at a constant speed of <img alt="1 \mathrm{~m} / \mathrm{s}" src="https://entrancecorner.oncodecogs.com/gif.latex?1%20%5Cmathrm%7B%7Em%7D%20/%20%5Cmathrm%7B

A small steel ball falls through a syrup at a constant speed of 1 \mathrm{~m} / \mathrm{s}. If the steel ball is pulled upward with a force equal to twice its effective weight, how fast will it move upward?
 

Option: 1

2 \mathrm{~m} / \mathrm{s}


Option: 2

1 \mathrm{~m} / \mathrm{s}


Option: 3

4 \mathrm{~m} / \mathrm{s}


Option: 4

3 \mathrm{~m} / \mathrm{s}


Answers (1)

best_answer

Viscous force, F_v=k v, and effective weight =W_e,

Then Initally,
\begin{aligned} W_e =k v~~~~~...(i) \\ \end{aligned}

& Finally

\begin{aligned} 2 W_e & =W_e+k v^{\prime} \\ W_e & =k v^{\prime}~~~~~~~...(ii) \\ \end{aligned}

Put the value W_e in equation (i)
\begin{aligned} k v^{\prime} & =k v \\ v^{\prime} & =v=1 \mathrm{~m} / \mathrm{s} \end{aligned}

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chirag

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