JEE MAIN - Physics (2007 - No. 23)
A sound absorber attenuates the sound level by $$20$$ $$dB$$. The intensity decreases by a factor of
$$100$$
$$1000$$
$$10000$$
$$10$$
Explanation
We have, $${L_1} = 10\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_0}}}} \right);$$
$${L_2} = 10\,\log \left( {{{{{\rm I}_2}} \over {{{\rm I}_0}}}} \right)$$
$$\therefore$$ $$\,\,{L_1} - {L_2} = 10\,\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_0}}}} \right) - 10\,\log \left( {{{{{\rm I}_2}} \over {{{\rm I}_0}}}} \right)$$
or, $$\Delta L = 10\,\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_0}}} \times {{{{\rm I}_0}} \over {{{\rm I}_2}}}} \right)$$
or, $$\Delta L = 10\,\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_2}}}} \right)$$
or, $$20 = 10\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_2}}}} \right)$$
or, $$2 = \log \left( {{{{{\rm I}_1}} \over {{{\rm I}_2}}}} \right)$$
or, $${{{{\rm I}_1}} \over {{{\rm I}_2}}} = {10^2}$$
or, $${{\rm I}_2} = {{{{\rm I}_1}} \over {100}}.$$
$$ \Rightarrow $$ Intensity decreases by a factor $$100.$$
$${L_2} = 10\,\log \left( {{{{{\rm I}_2}} \over {{{\rm I}_0}}}} \right)$$
$$\therefore$$ $$\,\,{L_1} - {L_2} = 10\,\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_0}}}} \right) - 10\,\log \left( {{{{{\rm I}_2}} \over {{{\rm I}_0}}}} \right)$$
or, $$\Delta L = 10\,\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_0}}} \times {{{{\rm I}_0}} \over {{{\rm I}_2}}}} \right)$$
or, $$\Delta L = 10\,\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_2}}}} \right)$$
or, $$20 = 10\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_2}}}} \right)$$
or, $$2 = \log \left( {{{{{\rm I}_1}} \over {{{\rm I}_2}}}} \right)$$
or, $${{{{\rm I}_1}} \over {{{\rm I}_2}}} = {10^2}$$
or, $${{\rm I}_2} = {{{{\rm I}_1}} \over {100}}.$$
$$ \Rightarrow $$ Intensity decreases by a factor $$100.$$
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