JEE MainPhysics

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Electromagnetic Waves: JEE Main Physics Question with Solution

Q: Why can't we just use ?

The differential approximation is only accurate for very small percentage changes (typically ). For , exact calculation with gives the precise integer .

By what percentage will the transmission range of a TV tower be affected when the height of the tower is increased by

21\%

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

14\%

12\%

10\%

15\%

Step-by-step solutionView

Correct answer

The transmission range is proportional to the square root of the tower's height, so a $21\%$ increase in height produces a $10\%$ increase in range.

Option analysis

Why each option works or fails

A ·

14\%

Believing that the percentage increase in range equals approximately two-thirds of the height increase or making an arithmetic error when approximating $\sqrt{1.21}$ . Express the new height as $1.21h$ and evaluate $\sqrt{1.21} = 1.10$ , which represents exactly a $10\%$ increase.

B ·

12\%

Miscalculating $\sqrt{1.21} - 1$ as $0.12$ instead of $0.10$ . Recall that $1.1^2 = 1.21$ , so $\sqrt{1.21} = 1.10$ , corresponding to a $10\%$ change.

C ·

10\%

None. This is the correct option. Since $d = \sqrt{2Rh}$ , $d \propto \sqrt{h}$ . With $h' = 1.21h$ , $d' = \sqrt{1.21}d = 1.1d$ , yielding a $10\%$ increase.

D ·

15\%

Assuming a linear or incorrect fractional power proportionality between height and transmission range. Use the standard radio horizon formula $d = \sqrt{2Rh}$ , which establishes a square-root dependence on height.

Reviewed route

Solution

StepWorking

01given

Initial tower height $h_1$ , final height $h_2 = 1.21 h_1$ (since height is increased by $21\%$ ).

02find

Percentage change in transmission range: $\frac{d_2 - d_1}{d_1} \times 100\%$ .

03strategise

Transmission range from a TV tower of height $h$ on Earth of radius $R_E$ is given by $d = \sqrt{2 R_E h}$ . Thus, $d \propto \sqrt{h}$ . Find $d_2 / d_1 = \sqrt{h_2 / h_1}$ .

04execute

Calculate the ratio and percentage increase: $\frac{d_2}{d_1} = \sqrt{1.21} = 1.1$ . Percentage increase is $(1.1 - 1) \times 100 = 10\%$ .

✓verify

Since $d \propto h^{1/2}$ , small-change approximation would give $\approx \frac{1}{2}(21\%) = 10.5\%$ . The exact value must be slightly less than $10.5\%$ , and $10\%$ matches exactly since $1.1^2 = 1.21$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why can't we just use

\Delta d / d = \frac{1}{2} \Delta h / h = 10.5\%

The differential approximation $\Delta d/d \approx \frac{1}{2}\Delta h/h$ is only accurate for very small percentage changes (typically $< 5\%$ ). For $21\%$ , exact calculation with $\sqrt{1.21} = 1.1$ gives the precise integer $10\%$ .

Answer

The transmission range is proportional to the square root of the tower's height, so a $21\%$ increase in height produces a $10\%$ increase in range.

Why each option works or fails

A: $14\%$ - Believing that the percentage increase in range equals approximately two-thirds of the height increase or making an arithmetic error when approximating $\sqrt{1.21}$ . Express the new height as $1.21h$ and evaluate $\sqrt{1.21} = 1.10$ , which represents exactly a $10\%$ increase.
B: $12\%$ - Miscalculating $\sqrt{1.21} - 1$ as $0.12$ instead of $0.10$ . Recall that $1.1^2 = 1.21$ , so $\sqrt{1.21} = 1.10$ , corresponding to a $10\%$ change.
C · correct: $10\%$ - None. This is the correct option. Since $d = \sqrt{2Rh}$ , $d \propto \sqrt{h}$ . With $h' = 1.21h$ , $d' = \sqrt{1.21}d = 1.1d$ , yielding a $10\%$ increase.
D: $15\%$ - Assuming a linear or incorrect fractional power proportionality between height and transmission range. Use the standard radio horizon formula $d = \sqrt{2Rh}$ , which establishes a square-root dependence on height.

Step-by-step solution

given: Initial tower height $h_1$ , final height $h_2 = 1.21 h_1$ (since height is increased by $21\%$ ).
find: Percentage change in transmission range: $\frac{d_2 - d_1}{d_1} \times 100\%$ .
strategise: Transmission range from a TV tower of height $h$ on Earth of radius $R_E$ is given by $d = \sqrt{2 R_E h}$ . Thus, $h_2 = 1.21 h_1$ . Find $d_2 / d_1 = \sqrt{h_2 / h_1}$ .
execute: Calculate the ratio and percentage increase: $\frac{d_2}{d_1} = \sqrt{1.21} = 1.1$ . Percentage increase is $(1.1 - 1) \times 100 = 10\%$ .
verify: Since $d \propto h^{1/2}$ , small-change approximation would give $\approx \frac{1}{2}(21\%) = 10.5\%$ . The exact value must be slightly less than $10.5\%$ , and $10\%$ matches exactly since $\frac{d_2}{d_1} = \sqrt{1.21} = 1.1$ .

Shortcut:

When to use it: When the percentage increase is a perfect square decimal like $21\%$ (factor $1.21$ ) or $44\%$ (factor $1.44$ ).

quick_kill: Transmission range $d \propto \sqrt{h}$ . Height becomes $1.21h$ , so range becomes $\sqrt{1.21}d = 1.1d$ , an increase of $(1.1 - 1) \times 100\% = 10\%$ .

Electromagnetic Waves: JEE Main Physics Question with Solution

What feels right?

The transmission range is proportional to the square root of the tower's height, so a $21\%$ increase in height produces a $10\%$ increase in range.

Why each option works or fails

Solution

01given

02find

03strategise

04execute

✓verify

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