The square root of $$\frac{\left(3\frac{ 1}4\right)^4-\left(4\frac{ 1}3\right)^4}{\left(3 \frac{1}4\right)^2-\left(4\frac{ 1}3\right)^2 }$$​ is:

Correct option is B

Given:​

​$$\frac{\left(3\frac{ 1}4\right)^4-\left(4\frac{ 1}3\right)^4}{\left(3 \frac{1}4\right)^2-\left(4\frac{ 1}3\right)^2 }$$ 

Formula Used: 

​$$a^4 -b^4 = (a^2-b^2)(a^2+b^2)$$ 

Solution: 

​$$\frac{\left(3\frac{ 1}4\right)^4-\left(4\frac{ 1}3\right)^4}{\left(3 \frac{1}4\right)^2-\left(4\frac{ 1}3\right)^2 } \\ \ \\ = \frac{\left[\left(3\frac{ 1}4\right)^2-\left(4\frac{ 1}3\right)^2\right]\left[\left(3\frac{ 1}4\right)^2+\left(4\frac{ 1}3\right)^2\right]}{\left(3 \frac{1}4\right)^2-\left(4\frac{ 1}3\right)^2 } \\ \ \\ = \left[\left(3\frac{ 1}4\right)^2+\left(4\frac{ 1}3\right)^2\right] \\ \ \\ =\left(\frac{ 13}4\right)^2+\left(\frac{ 13}3\right)^2\\ \ \\ =\frac{169}{16}+\frac{169}{9} \\ \ \\ = 169\left(\frac{9 + 16}{144}\right) \\ \ \\ = \frac{169 \times 25}{144}$$ 

Now , square root ;

​$$=\sqrt{\frac{169 \times 25}{144}} \\ \ \\ = \frac{65}{12} \\ \ \\ = 5\frac{5}{12}$$​​