Here, W = 40 kg = 40\(\times\)9.8 N = 392 N, AB = AC = 1.6 m, BD
= \(\frac { 1 }{ 2 } \)\(\times\)1.6 m = 0.8 m
BF = 1.2 m and DE = 0.5 m
In \(\triangle\)ADE and \(\triangle\)ABC are similar triangles ,hence
BC = \(DE\times \frac { AB }{ AD } =\frac { 0.5\times 1.6 }{ 0.8 } \)=1.0 m
Now, considering equilibrium at point B, Σፒ = 0
∴ W\(\times\)(MB) = N_c\(\times\)(CB)                 ...(i)
But MB= \(\frac { KB\times BF }{ BA } =\frac { 0.5\times 1.2 }{ 1.6 } \) = 0.375 m
Substituting this value in (i), we get
∴  N_c=\(\frac { W\times (MB) }{ (CB) } =\frac { 392\times 0.375 }{ 1 } \) =147N
Again considering equilibrium at point C in similar manner, we have
∴ W\(\times\)(MB) = N_B\(\times\)(CB)
N_B=\(\frac { W\times (MB) }{ (BC) } \frac { W\times (BC-BM) }{ (BC) } \)
= \(\frac { 392\times (1-0.375) }{ 1 } \)= 245 N
Now, it can be easily shown that tension in the string T = N_B - N_C = 245- 147 = 98 N.

Work done by torque in turing rotor by angle d \(\theta \) is
= \(\tau \)  d \(\theta \)
So, power delivered by engine
 P = \(\frac { Work\ done }{ Time\ taken } =\tau \frac { d\theta }{ dt } \)
[dt= time for turing by angle dθ]     
or   P = \(\tau \omega \)
So, power required = 180 x 200 = 36000 W
= 36 k W     [ 1 k W = 1000 W]

Here, mass of the disc, M = 5 kg, Radius of the disc R = 50 cm = 1/2 m.
Linear velocity of the disc, v = 10 ms^-1.
As \(v=R \omega \quad \therefore 10=\frac{1}{2} \omega\)
or ω = 10 x 2 = 20 radian/sec.
Also, moment of inertia of disc about or axis through its centre.
\(\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}\)
K.E. of the disc
\(=\frac{1}{2} \mathrm{I} \omega^{2}+\frac{1}{2} \mathrm{M} v^{2}\)
\(=\frac{1}{2} \frac{\mathrm{MR}^{2}}{2} \omega^{2}+\frac{1}{2} \mathrm{M} v^{2}\)
\(=\frac{1}{4} \times 5 \times\left(\frac{1}{2}\right)^{2} \times(20)^{2}+\frac{1}{2} \times 5 \times(10)^{2}\)
= 375 J.

Consider a couple as shown in Fig acting on a rigid body. The forces F and -F act respectively at points B and A. These points have position vectors r₁ and r₂ with respect to origin O. Let us take the moments of the forces about the origin.
The moment of the couple = sum of the moments of the two forces making the couple
= r₁ x (–F) + r₂ x F
= r₂ x F – r₁ x F
= (r₂ –r₁ ) x F
But r₁ + AB = r₂ , and hence AB = r₂ – r₁ .
The moment of the couple, therefore, is AB x F
Clearly this is independent of the origin, the point about which we took the moments of the forces.

Angular displacement is
\(\Delta \theta ={ \theta }_{ f }-{ \theta }_{ i }=6-(-2)=8rad\)

Let M and R be the mass and radius of the sphere and hollow cylinder.
Let torque \(\tau \) of equal magnitude be applied on hollow cylinder and solid sphere. The angular accelerations produced in it are\({ \alpha }_{ 1 }\ and\ { \alpha }_{ 2 }\) , respectively.         
 \(\therefore \ \tau ={ I }_{ 1 }{ \alpha }_{ 1 }\ and\ \tau ={ I }_{ 2 }{ \alpha }_{ 2 }\)  
Therfore,  \({ I }_{ 1 }{ \alpha }_{ 1 } ={ I }_{ 2 }{ \alpha }_{ 2 }\) 
or    \(\frac { { \alpha }_{ 1 } }{ { \alpha }_{ 2 } } =\frac { { I }_{ 1 } }{ { I }_{ 2 } } =\frac { \frac { 2 }{ 5 } { MR }^{ 2 } }{ { MR }^{ 2 } } =\frac { 2 }{ 5 } \)
or     \({ \alpha }_{ 2 }=\frac { 5 }{ 2 } { \alpha }_{ 1 }=2.5{ \alpha }_{ 1 }\) 
Let after time t,\({ \omega }_{ 1 }{ and\omega }_{ 2 }\) be the angular speeds of the hollow cylinder and solid sphere, respectively.
 \(\therefore { \omega }_{ 1 }={ \omega }_{ 0 }+{ \alpha }_{ 1 }t\)      
\(\\ and\ \ { \omega }_{ 2 }={ \omega }_{ 0 }+{ \alpha }_{ 2 }t\)
From  Eqs. (ii)  and (iii), we get \({ \omega }_{ 2 }{ >\omega }_{ 1 }\) 
Therefore, solid sphere will acquire a greater angular speed after a given time.

When the planet moves along the line joining the sun and the planet it sweeps some area given by 
\(\mathrm{A}=\frac{1}{2} r^{2} \theta\) where θ is the angular displacement.
\(\therefore \quad \frac{d \mathrm{~A}}{d t}=\frac{1}{2} r^{2} \frac{d \theta}{d t}=\frac{1}{2} r^{2} \omega\)
\(\frac{d \mathrm{~A}}{d t}=\frac{1}{2 m} m r^{2} \omega=\frac{\mathrm{L}}{2 m}\)
Since no torque acts, angular momentum Lis a constant, so \(\frac{d \mathrm{~A}}{d t}\) is a constant, i.e., the line Joining t e sun and the planet sweeps equal areas in equal intervals of time.

To distinguish between a hard boiled egg and a raw egg, we spin each on a table top. The egg which spins at a slower rate shall be a raw egg. This is because in a raw egg, fiquid matter inside tries to get away from the axis of rotation. Therefore, its moment of inertia I increases. As τ = Iα = constant, therefore, a decreases, i.e., raw egg will spin with smaller angular acceleration. The reverse is true for a hard boiled egg which will rotate more or less like a rigid body.

The units of moment of inertia are kg m² and its dimensional formula is [M¹L²T⁰].No, it is not a vector quantity.

531 N-m.

(i) Moment of inertia of a body about a given axis is a scalar quantity.
(ii) Moment of inertia of a body depends on
(i) Mass of the body
(ii) Size and shape of the body
(iii) axis of rotation of the body.
(iii) \(I_{d i s c}=\frac{1}{2} M R^{2}, I_{r i n g}=M R^{2}\) 
(iv) \(\mathrm{I}=\frac{1}{2} M R^{2}=\frac{1}{2} \times 40\left(\frac{1}{2}\right)^{2}=5 \mathrm{~kg} \mathrm{~m}^{2}\) 
(v) K.E of rotation = \(\frac{1}{2} I \omega^{2}=\frac{1}{2} I(2 \pi n)^{2}\) 
\(\frac{1}{2} \times 5\left(2 \pi \times \frac{120}{60}\right)^{2}=394.8 \mathrm{~J}\) 
(vi) \(I=\frac{M L^{2}}{2}=M K^{2} \text { or } K=\frac{L}{2 \sqrt{3}}\)  
(vii) \(I_{1}=M K_{1}^{2}=\frac{5}{4} M R^{2} \Rightarrow K_{1}=\sqrt{\frac{5}{4}} R\) 
M.I of circular ring of same radius about a tangential axis in the plane of the ring.
\(I_{1}=M K_{2}^{2}=\frac{3}{2} M R^{2}\) 
\(K_{2}=\sqrt{\frac{3}{2} R}\) 
\(\therefore \quad \frac{K_{1}}{K_{2}}=\sqrt{\frac{5}{4} \times \frac{2}{3}}=\sqrt{\frac{5}{6}}\) 
 

(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.