B.

30⁰

Let AB be the tower and BC be its shadow. Let $\mathrm{\theta }$ be the angle of elevation of the sun.

According to the given information,

$$\begin{gathered} \mathrm{BC} = \sqrt{3} \mathrm{AB} ..........\left(1\right) \\ \mathrm{In} ∆\mathrm{ABC}, \\ \mathrm{tan\theta } = \frac{\mathrm{AB}}{\mathrm{BC}} = \frac{\mathrm{AB}}{\sqrt{3} \mathrm{AB}} = \frac{1}{\sqrt{3}} [\mathrm{Using} (1\left)\right] \\ \mathrm{We} \mathrm{know} \mathrm{that} \tan {30}^0 = \frac{1}{\sqrt{3}} \\ \therefore \mathrm{\theta } = {30}^0 \end{gathered}$$

Hence, the angle of elevation of the sun is 30⁰.

B.

26

Diameter of two circles are given as 10 cm and 24 cm.

Radius of one circle = r₁ = 5 cm,  Radius of other circle = r₂ = 12 cm.

According to the given information,

$$\begin{gathered} Area of the larger circle = \mathrm{\pi }({\mathrm{r}}_1{)}^2 + \mathrm{\pi }({\mathrm{r}}_2{)}^2 \\ = \mathrm{\pi }(5{)}^2 + \mathrm{\pi }(12{)}^2 \\ = \mathrm{\pi }(25 + 144) \\ = 169\mathrm{\pi } \\ = \mathrm{\pi }(13{)}^2 \end{gathered}$$

$\therefore$ Radius of the larger circle = 13 cm

Hence, the diameter of larger circle = 26 cm.

C.

1 : 4

Let the original radius and the height of the cylinder be r and h respectively.

$$\begin{gathered} Volume of the original cylinder = {\mathrm{\pi r}}^2\mathrm{h} \\ \mathrm{Radius} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{cylinder} = \frac{\mathrm{r}}{2} \\ \mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{cylinder} = \mathrm{h} \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{cylinder} = \mathrm{\pi }{\left(\frac{\mathrm{r}}{2}\right)}^2\mathrm{h} = \frac{{\mathrm{\pi r}}^2\mathrm{h}}{4} \\ \mathrm{Required} \mathrm{ratio} = \frac{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{cylinder}}{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{original} \mathrm{cylinder}} = \frac{ \frac{{\mathrm{\pi r}}^2\mathrm{h}}{4}}{ {\mathrm{\pi r}}^2\mathrm{h}} = \frac{1}{4} = 1 : 4 \end{gathered}$$

C.

$\frac{1}{6}$

When two dice are thrown together, the total number of outcomes is 36.

favourable outcomes = { (1,1),  (2,2),  (3,3),  (4,4),  (5,5),  (6,6) }

$\therefore \mathrm{Required} \mathrm{probability} = \frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}} = \frac{6}{36} = \frac{1}{6}$

B.

( 3, 5 )

It is given that the point P divides AB in the ratio  2 : 1.

Using the section formula, the coordinates of the point P are

$$\begin{gathered} \left( \frac{\left(1\mathrm{x}1\right) + \left(2\mathrm{x}4\right)}{2 + 1} , \frac{\left(1\mathrm{x}3\right) + \left(2\mathrm{x}6\right)}{2 + 1}\right) \\ = \left(\frac{1 + 8}{3}, \frac{3 + 12}{3}\right) = (3,5) \end{gathered}$$

Hence, the coordinates of the point P are (3,5).