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Question Number 59380 by rahul 19 last updated on 09/May/19

Answered by tanmay last updated on 09/May/19

$$N_{\left(cupandball\right)}cos\theta ={\left(mg\right)}_{ball}$$$$N_{\left(cupandball\right)}sin\theta =N_{ballandball}$$$$\frac{N_{cupandball}}{N_{ballandball}}=\frac{1}{sin\theta }$$$$cos2\theta =\frac{{\left(3r\right)}^2+{\left(3r\right)}^2-{\left(2r\right)}^2}{2\times 3r\times 3r}=\frac{14r^2}{18r^2}=\frac{7}{9}$$$$1-2{sin}^2\theta =\frac{7}{9}$$$$\frac{2}{9}=2{sin}^2\theta \to sin\theta =\frac{1}{3}$$$$\mathit{S}oanswer=\frac{1}{sin\theta }=3$$

Commented by tanmay last updated on 09/May/19

$$2\theta =anglebetween3rand3r$$$$$$

Commented by mr W last updated on 09/May/19

$$pleasechecksir:$$$$\sin \theta =\frac{1}{3-1}=\frac{1}{2}$$$$\Rightarrow answer\left(b\right)$$

Commented by tanmay last updated on 09/May/19

$$oksirishalldrawandcheck...plssuggestapp$$$$fordrawing..$$

Commented by mr W last updated on 09/May/19

$$usuallyiuse‘‘photo{editor}^{″}tomodify$$$$images.$$

Commented by tanmay last updated on 09/May/19

$$oksir...$$

Commented by mr W last updated on 09/May/19

Commented by rahul 19 last updated on 10/May/19

$$thankUbothsirs!$$

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