Solve the following :
Question:

The radius of gyration of a uniform disc about a line perpendicular to the disc equals its radius. Find the distance of the line from the centre.

Solution:

$I_{a x i s}=I_{C O M}+m d^{2}$

$m k^{2}=\frac{m r^{2}}{2}+m d^{2}$

$\Rightarrow m r^{2}=\frac{m r^{2}}{2}+m d^{2}$

$\Rightarrow d=\sqrt{2}$

Solve the following
Question:

(i) Is 68 a term of the A.P. 7, 10, 13, …?

(ii) Is 302 a term of the A.P. 3, 8, 13, …?

Solution:

(i) 7, 10, 13…

Here, we have:

a = 7

$d=(10-7)=3$

Let $a_{n}=68$

$\Rightarrow a+(n-1) d=68$

$\Rightarrow 7+(n-1)(3)=68$

$\Rightarrow(n-1)(3)=61$

$\Rightarrow(n-1)=\frac{61}{3}$

$\Rightarrow n=\frac{61}{3}+1=\frac{64}{3}$

Since n is not a natural number.So, 68 is not a term of the given A.P.

(ii) 3, 8, 13…

Here, we have:

a  = 3

$d=(8-3)=5$

Let $a_{n}=302$

$\Rightarrow a+(n-1) d=302$

$\Rightarrow 3+(n-1) 5=302$

$\Rightarrow(n-1) 5=299$

$\Rightarrow(n-1)=\frac{299}{5}$

$\Rightarrow n=\frac{299}{5}+1=\frac{304}{5}$

Since n is not a natural number.So, 302 is not a term of the given A.P.