A circular wire of radius 15 cm is cut and bent so as to lie along the circumference of a loop of radius 120 cm.

Circumference of the circle of radius 15 cm:

$2 \pi \mathrm{r}$

$=2 \times 3.14 \times 15 \mathrm{~cm}$

$=94.2 \mathrm{~cm}$

Now, 94.2 cm will be the length of arcl">(l)l for the circle with radius 120 cm.

We know:

$l=r \theta$

Here, $\theta$ is measured in radians.

$\therefore 94.2=120 \times \theta$

$\Rightarrow \theta=0.785$ radians

$45^{\circ}=\frac{\pi}{4}=\frac{22}{7 \times 4}=0.785$ radians

Therefore, the angle subtended by it at the centre of the loop is $45^{\circ}$.