Class 11th

For no toppling

$F\left(\frac{a}{2}+b\right)\le mg\frac{a}{2}$

$\mu \frac{\mathrm{a}}{2}+\mu \mathrm{b}\le \frac{\mathrm{a}}{2}$

$0.2a+0.4b\le 0.5a$

$0.4b\le 0.3a$

$\mathrm{b}\le \frac{3\mathrm{a}}{4}$

$\mathrm{b}\le 0.75\mathrm{a}\mathrm{}$ (in limiting case)

But is not possible as maximum value of $b$ can be equal to $0.5a$ only.

$\therefore {\left(100\frac{\mathrm{b}}{\mathrm{a}}\right)}_{\mathrm{m}\mathrm{a}\mathrm{x}}=50.00$

Let C be the centre and M be the mid point of AB

$\Delta APC:sin\theta =\frac{13/2}{pc}=\frac{5}{13}\Rightarrow pC=\frac{169}{10}$                

$\Delta AMC:cos\theta =\frac{6}{13/2}=\frac{12}{13}$              

PC = $\frac{169}{10},MC=\frac{13}{2}sin\theta =\frac{13}{2}\cdot \frac{5}{13}$  

PM = PC – MC = $\frac{169}{10}-\frac{5}{2}=\frac{144}{10}$  

5PM = 72

The direction of torque and angular momentum defines how and in which orientation an object will rotate or sustain its spin. This is important to understand in machines, athletic movements, and even natural phenomena, such as planetary motion. 

Both torque and angular momentum have direction. We determine their direction based on the right-hand rule. That makes them vector quantities in rotational motion. 

Yes, an object can have angular momentum without torque. The physics behind it is that if no external force act on an object, its angular momentum is constant. That is based on the law of conservation of momentum. 

Torque is the measure of rotational force on an object (rigid or extended body). It changes the object's angular momentum. To calculate in physics, torque equals the rate of change of angular momentum. The formula for that is

$?=\frac{dL}{dt}$

x² – x – 4 = 0

  $P_n={\alpha }^n-{\beta }^n$                    

$?=\frac{P_{15}{P}_{16}-P_{14}{P}_{16}-P_{15}^2+P_{14}{P}_{15}}{P_{13}{P}_{14}}$

$=\frac{\left(P_{15}-P_{14}\right)\left(P_{16}-P_{15}\right)}{P_{13}{P}_{14}}$

$=\frac{4P_{13}\cdot 4P_{14}}{P_{13}\cdot P_{14}}=16$

P_n = aⁿ - bⁿ   

$={\alpha }^{n-1}\cdot \alpha -{\beta }^{n-1}\cdot \beta$

$={\alpha }^{n-1}\left({\alpha }^2-4\right)-{\beta }^{n-1}\left(\beta -4\right)$

$P_n={\alpha }^{n+1}-{\beta }^{n+1}-4\left({\alpha }^{n-1}-{\beta }^{n-1}\right)$

$P_n=P_{n+1}-4P_{n-1}\Rightarrow P_{n+1}-P_n=4P_{n-1}$