CBSE Class 11-science Answered

Following are the solutions to your queries:

1.

Let angle of string with vertical be θ.

T is the tension in the string.

Resolving T along Y and X axis and as per Newton’s 2^nd law we get :

T cos θ = mg

hence T = mg / cos θ

Along horizontal axis :

T sin θ = ma            where acceleration a = g / √3

Substituting :

mg tan θ = mg / √3

tan θ = 1/ √3

hence θ = Π/6

2.

Let the initial velocity tangent to the surface of the sphere be u . Object is at point A at the top of sphere. O is the center of the sphere.

Object detaches from the surface at point P on the surface which is vertically at a distance d below A.

Radius of the sphere is = r

Angle between OA and OP = θ

Velocity when the object detaches from the surface of sphere = v.

Force along the direction of the normal to the sphere surface = F.

There is no friction.

d = r - r cos θ = r ( 1 - cos θ )

Resultant force towards the center of the sphere (normal to the sphere surface at P ) is equal to centripetal force.

Hence

mg cos θ - F = mv² / r

When the object detaches from the surface of sphere, F = 0.

Hence

mg cos θ = mv² / r                         

r cos θ = v²/g                                           (I)

Applying conservation of energy :

1/2 mv² - 1/2 mu² = difference in potential energy at 2 points.  

1/2 mv² - 1/2 mu² = mgr - mgr cos θ = mg r( 1 - cos θ )

Hence

r  -  r cos θ  = 1/2g * (v² - u²)                (II)     

r - v²/g = 1/2g * (v² - u²)                       using (I) and (II)

r + u²/2g = 3/2g * v²

v² = 2g/3 * ( r + u²/2g)                           (III)

Now

d = r - r cos θ = r - v²/g                      . . . as per (I)

Substituting v² as per (III) :

d = r - 1/g * [ 2g/3 * ( r + u²/2g ) ]

    = r - 2/3 (r + u²/2g)  =  r (1/3 - u²/3gr)

Hence

d = r/3 (1 - u²/gr)

If initial velocity u =0,

d = r/3.

Height = h = 2r - d.

cos θ = (r-d) / r

We hope that clarifies your query.
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