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Solution :

Let at any instant, the boby is at angular position `theta` with respect to the vertical line drow from the center of mirror , if `phi` is the angular displacement of the ball about its senter, then <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V06_C04_S01_059_S01.png" width="80%"> <br> Restoring tarque acting on the ball `tau = - mg sin theta xxr` <br> For small `theta, sin theta` `:. tau = - mg sin theta xxr` <br> or `tau = - mg [(r)/(R - r)] phi r = mg (r^(2)/(R - r)) (- phi)` <br> Angular acceleration `alpha = (mg)/(l) ((r^(2)))/(R - r)) (- phi)` <br> Now comparing above equation with atandard eqation of SHM, `alpha = - omega ^(2) phi, we get omega = sqrt((mg)/(l) ((r^(2))/(R - r))` <br> Here `l` is moment of inertia of the rolling ball about the point of constant which is `l = 7//5 m r^(2)` <br> `sqrt((mg)/((7)/(5) m r^(2) (r^(2)/(R - r))) = sqrt((5)/(7) (g)/((R - r))))` <br> and `T = (2 pi)/(omega) = 2 pi sqrt((7(R - r))/(5 g))` <br> Alternate method: We can that when the `CM` of the boby is shifted slightly by an angle `theta` relative to the center of curvature of the surface, the magnitude of acceleration of the `CM` of the boby can be given as <br> `a = (g sin theta)/(1 + (k^(2))/(r^(2))` <br> (because the body is accelerating down the inclined plane of instancous angle `theta`) <br> Subastituting `sin theta = theta = (x)/(R - r)` , we have <br> `a = (- gx)/((R - r) (1 + (k^(2))/(r^(2)))` <br> (negative sign is given because `bara` and `barx` are oppositely directed) <br> Comparing the above equation with `a = - omega^(2) x`, we have <br> `omega = sqrt(( g)/((R - r) (1 + (k^(2))/(r^(2)))` <br> For solid sphere `k^(2) = 2//5 r^(2)`. <br> Hence `omega = sqrt((5)/(7) (g)/((R - r)))` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V06_C04_S01_059_S02.png" width="80%">