Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Two cars moving in opposite directions approach each other with speed of $22 \text{ m/s}$ and $16.5 \text{ m/s}$ respectively. The driver of the first car blows a horn having a frequency $400 \text{ Hz}$ . The frequency heard by the driver of the second car is [velocity of sound $340 \text{ m/s}$ ]

$350 \text{ Hz}$

$361 \text{ Hz}$

$411 \text{ Hz}$

$448 \text{ Hz}$

Step-by-Step Solution

Identify the Doppler Effect Formula: The apparent frequency ( $f'$ ) heard by an observer when both the source and observer are in relative motion along the line joining them is given by $f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right)$ , where $v$ is the velocity of sound in the medium, $v_o$ is the velocity of the observer, and $v_s$ is the velocity of the source .
Determine Signs for Approaching Bodies: Since both cars are moving towards each other, they both contribute to an increase in the apparent frequency.

The observer approaching the source means the numerator is $(v + v_o)$ .
The source approaching the observer means the denominator is $(v - v_s)$ .

Substitute the Given Values: Given $v = 340 \text{ m/s}$ , $v_o = 16.5 \text{ m/s}$ (second car), $v_s = 22 \text{ m/s}$ (first car), and $f = 400 \text{ Hz}$ . $f' = 400 \left( \frac{340 + 16.5}{340 - 22} \right)$ $f' = 400 \left( \frac{356.5}{318} \right)$ $f' = 400 \times 1.12107... \approx 448.4 \text{ Hz}$ Rounding off to the nearest integer, the frequency heard by the driver of the second car is $448 \text{ Hz}$ .

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