A boy ties a stone of mass 100 g to the end of 2 m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 8 N. If the maximum speed with which the stone can revolve is $\frac{K}{π}$ rev. /min. The value of K is :

(Assume the string is massless and unstretchable)

Option 1 - 400

Option 2 - 300

Option 3 - 600

Option 4 - 800

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Vishal Baghel | Contributor-Level 10

6 months ago

Correct Option - 3

Detailed Solution:

By N L M : T sin θ = $^{}$ $m ω^{2} R$ - (1)

& R = $l$ sin θ - (2)

From (1) & (2)

$\Rightarrow T s i n θ = m ω^{2} l s i n θ$

$\Rightarrow T = m ω^{2} l$

$\Rightarrow 80 = (\frac{100}{1000}) ω^{2} * 2$

$\Rightarrow \frac{800}{2} = ω^{2}$

$\Rightarrow ω^{2} = 400$

$ω = 20 r a d / s e c$

And, $ω = \frac{2 π N}{2 π}$ $\frac{}{}$

$N = \frac{20 * 60}{2 π}$

$N = \frac{600}{π} r p m = \frac{k}{π}$

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