JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 18)
The pressure wave, P = 0.01 sin [1000t – 3x] Nm–2,
corresponds to the sound produced by a vibrating blade
on a day when atmospheric temperature is 0°C. On
some other day, when temperature is T, the speed of
sound produced by the same blade and at the same
frequency is found to be 336 ms–1 . Approximate value
of T is
12°C
15°C
4°C
11°C
Explanation
Speed of wave from wave equation
$$v = - {{\left( {coeffecient{\rm{ }}of{\rm{ }}t} \right)} \over {\left( {coeffecient{\rm{ }}of{\rm{ }}x} \right){\rm{ }}}}$$
$$v = - {{1000} \over {( - 3)}} = {{1000} \over 3}$$
Since speed of wave $$ \propto \sqrt T $$
So $$ = {{1000} \over {{3 \over {336}}}} = \sqrt {{{273} \over T}} $$
$$ \Rightarrow $$ T = 277.41 K
T = 4.41°C
$$v = - {{\left( {coeffecient{\rm{ }}of{\rm{ }}t} \right)} \over {\left( {coeffecient{\rm{ }}of{\rm{ }}x} \right){\rm{ }}}}$$
$$v = - {{1000} \over {( - 3)}} = {{1000} \over 3}$$
Since speed of wave $$ \propto \sqrt T $$
So $$ = {{1000} \over {{3 \over {336}}}} = \sqrt {{{273} \over T}} $$
$$ \Rightarrow $$ T = 277.41 K
T = 4.41°C
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