A couple when applied to a body it produce 

Consider the point at which resultant of the forces will act, is 'x' m from the point A.

Thus the point is (1.5 - x) m far from the point B.

Thus, using the formula Torque = Distance x Force acting

$20x = 30(1.5 - x)$

or,$ 2x = 4.5 - 3x$

or,$ 5x = 4.5$

or, $x = 0.9m or 90cm$

Thus the resultant point will be 0.9 m or 90 cm far from point A

A couple consists of two parallel forces that are equal in magnitude, opposite in sense and do not share a line of action. It does not produce any translation, but only rotational motion.

A couple consists of two equal and opposite forces whose lines of action are parallel and laterally separated by the some distance. Therefore, net force (or resultant) of a couple is null vector, hence there is no translational acceleration, only rotational motion will be there.

True , because a couple donot  produces any linear acceleration it only produces only rotation but single force will produce linear acceleration too.

A couple consists of two equal and opposite forces acting at a separation, so that net force becomes zero. When a couple acts on a body it rotates the body but does not produce any translatory motion. Hence, only rotational motion is produced.

A couple has to be applied to the tap in order to open it. A couple is the combination of two equal and opposite parallel forces acting at different axes. Thus option D is correct.

$\left[ MOI \right] =\left[ M{ R }^{ 2 } \right] =\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ 0 } \right] \ \left[ Pressure \right] =\left[ N/{ M }^{ 2 } \right] =\left[ { M }^{ 1 }{ L }^{ -1 }{ T }^{ -2 } \right] \ \left[ Couple \right] =\left[ N.{ M } \right] =\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 } \right] $

$Power\quad P=100kW\quad =100000W\ w=1800\times \cfrac { 2\pi  }{ 60 } \quad rad/s\ \quad =60\pi \quad rad/s\ P=torque\times w\ torque=530.5\quad Nm$

If principle of moments hold good, then the net torque about a given point is zero (usually CM or the pivoted point is zero). Hence the object does not rotate and is said to be in equilibrium

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.

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

$K.E=\cfrac{1}{2}mv^2\rightarrow(1)\v=r\omega$