Find the length of TS if PQ = 6 cm, QR = 10 cm and PS = 8 cm.

Given,
PS = 8 cm, PQ = 6 cm, QR = 10 cm
Now from the figure,
$\mathrm{PR} = \mathrm{QR} + \mathrm{PQ}$
$\mathrm{PR} = 10 + 6 = 16 \mathrm{cm}$
Now, we know that when a chord of a circle intersects externally, then the product of segments of chords the same in the measure.
So, 

$\Rightarrow \mathrm{PS} \times \mathrm{PT} = \mathrm{PQ} \times \mathrm{PR}$
$\Rightarrow 8 \times \mathrm{PT}= 6 \times 16$
$\Rightarrow \mathbf{PT}\mathbf{=}\mathbf{12}\mathbf{ }\mathbf{cm}$
Here, $\mathrm{PT}= \mathrm{PS} + \mathrm{TS}$
$\mathrm{TS} = \mathrm{PT}- \mathrm{PS}$
$\mathrm{TS} = 12 - 8$
$\mathbf{TS}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{4}\mathbf{ }\mathbf{cm}$

Therefore, the correct answer is 4 cm.