JEE Physics 2023 Question with Solution

Given: The height of the TV tower is increased by $21\%$.

Find: The percentage change in the transmission range.

For line-of-sight communication,

$$d = \sqrt{2Rh}$$

So,

$$d \propto \sqrt{h}$$

If the new height is

$$h' = 1.21h$$

then the new range becomes

$$d' \propto \sqrt{1.21h} = \sqrt{1.21}\sqrt{h} = 1.1\sqrt{h}$$

Hence,

$$d' = 1.1d$$

Therefore, the percentage change in range is

$$\frac{d' - d}{d} \times 100 = \frac{1.1d - d}{d} \times 100 = 10\%$$

Therefore, the transmission range increases by $10\%$. The correct option is D.

Given: Original tower height is $h$ and new height is $1.21h$.

Find: By what percentage the transmission range changes.

The range in line-of-sight communication varies as the square root of height. Using the relation given in the solution:

$$R = \sqrt{2 \times h \times d}$$

When height becomes $1.21h$, the new range is

$$R' = \sqrt{2 \times 1.21h \times d}$$ $$R' = \sqrt{1.21}\, \sqrt{2hd}$$ $$R' = 1.1R$$

Now calculate the percentage increase:

$$\frac{R' - R}{R} \times 100 = \frac{1.1R - R}{R} \times 100$$ $$= 0.1 \times 100 = 10\%$$

Therefore, the transmission range is affected by $10\%$, that is, it increases by $10\%$.