Two equal heavy sphers, 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

Question

Two equal heavy sphers, 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

Tardigrade Admin · Accepted Answer

sin θ=(1/2) Thus, N1 sin θ=N2 ∴ (N1/N2)=(1/ sin θ)=2.

Q. Two equal heavy sphers, 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

1

2

3

4

$sin θ = \frac{1}{2}$
Thus, $N_{1} sin θ = N_{2}$
$∴ \frac{N _{1}}{N _{2}} = \frac{1}{s i n θ} = 2$ .

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

1

2

3

4

sinθ=21​ Thus, N1​sinθ=N2​ ∴N2​N1​​=sinθ1​=2.

Q. Two equal heavy sphers, 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

$sin θ = \frac{1}{2}$
Thus, $N_{1} sin θ = N_{2}$
$∴ \frac{N _{1}}{N _{2}} = \frac{1}{s i n θ} = 2$ .