How many times more intense is a 90 dB sound than a 40 dB sound?

Question

How many times more intense is a 90 dB sound than a 40 dB sound? (A) 2.5 (B) 5 (C) 50 (D) 105

Tardigrade Admin · Accepted Answer

90=10 log 10(( I 1/ I 0)) ⇒ ( I 1/ I 0)=109 ldots (i) Again, 40=10 log 10(( I 2/ I 0)) ⇒ ( I 2/ I 0)=104 ldots (ii) From Eqs.(i) and (ii), we get ( I 1/ I 2)=105

Q. How many times more intense is a $90 d B$ sound than a $40 d B$ sound?

$2.5$

$5$

$50$

$1 0^{5}$

$90 = 10 lo g_{10} (\frac{I _{1}}{I _{0}})$
$\Rightarrow \frac{I _{1}}{I _{0}} = 1 0^{9} \dots$ (i)
Again, $40 = 10 lo g_{10} (\frac{I _{2}}{I _{0}})$
$\Rightarrow \frac{I _{2}}{I _{0}} = 1 0^{4} \dots$ (ii)
From Eqs.(i) and (ii), we get
$\frac{I _{1}}{I _{2}} = 1 0^{5}$

Q. How many times more intense is a 90dB sound than a 40dB sound?

2.5

5

50

105

90=10log10​(I0​I1​​) ⇒I0​I1​​=109… (i) Again, 40=10log10​(I0​I2​​) ⇒I0​I2​​=104… (ii) From Eqs.(i) and (ii), we get I2​I1​​=105

Q. How many times more intense is a $90 d B$ sound than a $40 d B$ sound?

$2.5$

$5$

$50$

$1 0^{5}$

$90 = 10 lo g_{10} (\frac{I _{1}}{I _{0}})$
$\Rightarrow \frac{I _{1}}{I _{0}} = 1 0^{9} \dots$ (i)
Again, $40 = 10 lo g_{10} (\frac{I _{2}}{I _{0}})$
$\Rightarrow \frac{I _{2}}{I _{0}} = 1 0^{4} \dots$ (ii)
From Eqs.(i) and (ii), we get
$\frac{I _{1}}{I _{2}} = 1 0^{5}$