6.83m

Join OT. Let it intersect PQ at the point R. Then \(\Delta\) TPQ is isosceles and TO is the angle bisector of \(\angle\) PTQ. So, OT \(\perp\) PQ and therefore, OT bisects PQ which gives PR = RQ = 4 cm.
Also, \(\mathrm{OR}=\sqrt{\mathrm{OP}^{2}-\mathrm{PR}^{2}}=\sqrt{5^{2}-4^{2}} \mathrm{~cm}=3 \mathrm{~cm}\) 
Now, \(\angle \mathrm{TPR}+\angle \mathrm{RPO}=90^{\circ}=\angle \mathrm{TPR}+\angle \mathrm{PTR}\) 
So, \(\angle \mathrm{RPO}=\angle \mathrm{PTR}\) 
Therefore, right triangle TRP is similar to the right triangle PRO by AA similarity
This gives \(\frac{\mathrm{TP}}{\mathrm{PO}}=\frac{\mathrm{RP}}{\mathrm{RO}}, \text { i.e., } \frac{\mathrm{TP}}{5}=\frac{4}{3} \text { or } \mathrm{TP}=\frac{20}{3} \mathrm{~cm}\) 
Note : TP can also be found by using the Pythagoras Theorem, as follows:
Let TP = x and TR = y. Then
x² = y² + 16 (Taking right \(\Delta PRT\) )              (1)
x² + 5² = (y + 3)²(Taking right \(\Delta \mathrm{OPT}\) )        (2)
Subtracting (1) from (2), we get
\(25=6 y-7 \quad \text { or } \quad y=\frac{32}{6}=\frac{16}{3}\) 
Therefore, \(x^{2}=\left(\frac{16}{3}\right)^{2}+16=\frac{16}{9}(16+9)=\frac{16 \times 25}{9}\) [From (1)] 
or \(x=\frac{20}{3}\) 

Here the possible outcomes are all the numbers between 0 and 2. This is the portion of the number line from 0 to 2.

Let E be the event that ‘the music is stopped within the first half-minute’.
The outcomes favourable to E are points on the number line from 0 to \(\frac{1}{2}\).
The distance from 0 to 2 is 2, while the distance from 0 to \(\frac{1}{2}\) is \(\frac{1}{2}\).
Since all the outcomes are equally likely, we can argue that, of the total distance of 2, the distance favourable to the event E is \(\frac{1}{2}\).
So, \(\mathrm{P}(\mathrm{E})=\frac{\text { Distance favourable to the event } \mathrm{E}}{\text { Total distance in which outcomes can lie }}=\frac{\frac{1}{2}}{2}=\frac{1}{4}\)