Acceleration a = g*sin(theta) / (1 + I/MR^2). The object with the smallest I/MR^2 has the largest acceleration and reaches the bottom first. I/MR^2 values: SS (2/5), SC (1/2), HC (1). Order: SS, SC, HC.

Toppling occurs when the torque due to pseudo-force (ma) about the edge exceeds the torque due to gravity. The condition is m*a*h/2 = m*g*r. With a = d^2x/dt^2 = 2bt, we get m*(2bt)*h/2 = m*g*r, so t = rg/bh.

The velocity of a point on the rim is v_p = v_cm + omega x r. For vertical velocity, the horizontal components must cancel. The height y = R - v/omega is a standard result for the point where horizontal velocity is zero.

Acceleration a = g*sin(theta) / (1 + I/MR^2). The sphere has I = (2/5)MR^2, so a = g*sin(theta) / 1.4. The disc has I = (1/2)MR^2, so a = g*sin(theta) / 1.5. The sphere has higher acceleration and arrives first.

By conservation of angular momentum, I1*omega1 = I2*omega2. Given I2 = (2/5)I1, then omega2 = (5/2)omega1 = 2.5 * 40 = 100 rev/min. KE2 = (1/2)I2*omega2^2 = (1/2)(2/5)I1(2.5*omega1)^2 = 2.5 * KE1.

The whole system does not continue to rotate about P because the moment is balanced. Thus option B is wrong. And the center of mass of the system lies at pivot point P. Thus option C and D are wrong. As the force applied at the two points of suspension is different $\tau$ is different.

$\begin{array}{l} You\, \, have\, \, to\, \, use\, \, the\, \, equation, \ \; { { { \omega  } }^{ { 2 } } }\; ={ { { \omega  } } _{ { 0 } } }^{ { 2 } }\; +{ { 2\alpha \theta  } }\; \, \, for\, \, finding\, \, the\, \, angular\, \, acceleration\; \, \alpha \, \, and \ hence\, \, the\, \, number\, \, of\, \, further\, \, rotations. \ Note\, \, that\, \, this\, \, equation\, \, is\, \, the\, \, rotational\, \, analogue\, \, of\, \, the\, equation \ { v^{ 2 } }\; ={ v _{ 0 } }^{ 2 }+2as{ {  } }(or,\; { v^{ 2 } }\; ={ u^{ 2 } }\; +2as)\, \, in\, \, linear\, \, motion. \ Since\, \, the\, \, angular\, \, velocity\, \, has\, \, reduce\, \, to\, \, half\, \, of\, \, the\, \, initial\, \, value\, \, { \omega _{ 0 } }\, \, after\, \, 36\, \, rotations,\, \, we\, \, have \ { \left( { { \omega _{ 0\;  } }/2 } \right) _{ \;  } }^{ 2 }={ \omega _{ 0 } }^{ 2 }+2\alpha \times 36\, \, from\, \, which\, \; \alpha =--\; { \omega _{ 0 } }^{ 2 }/96 \ \left[ { We\, \, have\, \, expressed\, \, the\, \, angular\, \, displacement\, \, \theta \, \, in\, \, rotations\, \, itself\, \, for\, \, convenience } \right]  \ If\, \, the\, \, additional\, \, number\, \, of\, \, rotations\, \, is\, \, x,\, \, we\, \, have \ 0={ \left( { { \omega _{ 0\;  } }/2 } \right) _{ \;  } }^{ 2 }\; +\; 2\alpha x\; =\; { \left( { { \omega _{ 0\;  } }/2 } \right) _{ \;  } }^{ 2 }\; +\; 2\times (--\; { \omega _{ 0 } }^{ 2 }/96)x \ This\, \, gives, \ x\; =12 \end{array}$

For a disc rolling down an incline, the final linear velocity v = sqrt(2gh / (1 + k^2/R^2)). For a uniform disc, k^2/R^2 = 0.5. Since the velocity depends only on height h and the shape (moment of inertia factor), the mass and radius do not affect the final velocity. Thus, the ratio is 1:1.

According to the law of conservation of energy 
Total energy of the system (KE $+$ PE) will be equal at both  , topmost and bottom most point.
Therefore, $\bigtriangleup PE =\bigtriangleup KE$
 $KE _{b}-KE _{t} = PE _{t} - PE _{b}$
$PE _{t} - PE _{b} = m\times g\times (2\times r)$
$\bigtriangleup KE = 0.4 \ mg$ $joule$

Newton's colour disc, as the name suggests, was invented by Sir Issac Newton and is a disc with seven segments in rainbow colours.
When the disc is rotated, the colours fade to white. In this way, Sir Isaac Newton demonstrated that white light is a combination of the seven different colours found in a rainbow.

Hence, the correct answer is OPTION A.