Question-59380

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

Answered by tanmay last updated on 09/May/19

N_((cup andball)) cosθ=(mg)_(ball) N_((cup and ball)) sinθ=N_(ball and ball) (N_(cup and ball) /N_(ball and ball) )=(1/(sinθ)) cos2θ=(((3r)^2 +(3r)^2 −(2r)^2 )/(2×3r×3r))=((14r^2 )/(18r^2 ))=(7/9) 1−2sin^2 θ=(7/9) (2/9)=2sin^2 θ→sinθ=(1/3) So answer=(1/(sinθ))=3

$${N}_{\left({cup}\:{andball}\right)} {cos}\theta=\left({mg}\right)_{{ball}} \\ $$$${N}_{\left({cup}\:{and}\:{ball}\right)} {sin}\theta={N}_{{ball}\:{and}\:{ball}} \\ $$$$\frac{{N}_{{cup}\:{and}\:{ball}} }{{N}_{{ball}\:{and}\:{ball}} }=\frac{\mathrm{1}}{{sin}\theta} \\ $$$${cos}\mathrm{2}\theta=\frac{\left(\mathrm{3}{r}\right)^{\mathrm{2}} +\left(\mathrm{3}{r}\right)^{\mathrm{2}} −\left(\mathrm{2}{r}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{3}{r}×\mathrm{3}{r}}=\frac{\mathrm{14}{r}^{\mathrm{2}} }{\mathrm{18}{r}^{\mathrm{2}} }=\frac{\mathrm{7}}{\mathrm{9}} \\ $$$$\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \theta=\frac{\mathrm{7}}{\mathrm{9}} \\ $$$$\frac{\mathrm{2}}{\mathrm{9}}=\mathrm{2}{sin}^{\mathrm{2}} \theta\rightarrow{sin}\theta=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\boldsymbol{{S}}{o}\:{answer}=\frac{\mathrm{1}}{{sin}\theta}=\mathrm{3} \\ $$

Commented by tanmay last updated on 09/May/19

$$\mathrm{2}\theta={angle}\:{between}\:\mathrm{3}{r}\:{and}\:\mathrm{3}{r} \\ $$$$ \\ $$

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

please check sir: sin θ=(1/(3−1))=(1/2) ⇒answer (b)

$${please}\:{check}\:{sir}: \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{3}−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{answer}\:\left({b}\right) \\ $$

Commented by tanmay last updated on 09/May/19