Angle of intersection of two circle having distance between their centres `d` is given by : (A) `cos theta = (r_1^2 + r_2^2 - d)/(2r_1^2 + r_2^2)` (B) `sec theta = (r_1^2 + r_2^2 + d^2)/(2r_1 r_2)` (C) `sec theta = (2r_1 r_2)/(r_1^2 + r_2^2 - d^2` (D) none of these

The distance between the centres of the two circles of radii r_1 and r_2 is d. They will thouch each other internally if

The total surface area of frustum shaped glass tumbler is (r_1 gt r_2) (a) pi r_1 l + pi r_2 l (b) pi l (r_1 + r_2) + pi (r_2)^2 (c) 1/3 pi h ( (r_1)^2 + (r_2)^2 + r_1r_2 ) (d) sqrt (h^2 + (r_1 - r_2)^2)

If the sum of the areas of two circles with radii r_1 and r_2 is equal to the area of a circle of radius r , then r_1^2 +r_2^2 (a) > r^2 (b) =r^2 (c)

In the given loop the magnetic field at the centre O is 1) (mu_0I)/(4) ( (r_1 + r_2)/(r_1 r_2) ) out of the page 2) (mu_0I)/(4) ( (r_1 + r_2)/(r_1 r_2)) into the page 3) (mu_0I)/(4) ( (r_1 - r_2)/(r_1 r_2)) out of the page 4) (mu_0 I)/(4) ( (r_1 - r_2)/(r_1r_2) ) into the page

If (r)/(r_(1))=(r_(2))/(r_(3)) ,then A=90^(@)(b)B=90^(@)C=90^(@) (d) none of these

Show that 16R^(2)r r_(1) r _(2) r_(3)=a^(2) b ^(2) c ^(2)