Two equal heavy spheres, each of radius r , are in equilibrium within a smooth cup of radius 3 r . r

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4-1.Newton's Laws of Motion

hard

Two equal heavy spheres, each of radius $r$, are in equilibrium within a smooth cup of radius $3 r$. The ratio of reaction between the cup and one sphere and that between the two sphere is

A

$1$

B

$2$

C

$3$

D

$4$

Solution

(b)

$\sin \theta=\frac{1}{2}$

Thus, $N _1 \sin \theta= N _2$

$\therefore \frac{N_1}{N_2}=\frac{1}{\sin \theta}=2$

Standard 11

Physics

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