172841
The intensity level of sound $A$ is $30 \mathrm{~dB}$ greater than of $B$. How many times more intense is the sound A than B?
1 10
2 100
3 1000
4 2
Explanation:
C Since, we know that- The intensity level in sound is given by $\mathrm{L}=10 \log _{10} \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{o}}}$ Where $\mathrm{I}_{\mathrm{o}}=$ initial intensity But given $\quad \mathrm{L}_{\mathrm{A}}=30+\mathrm{L}_{\mathrm{B}}$ Therefore, $10 \log _{10} \frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{o}}}=30+10 \log _{10} \frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{o}}}$ $10 \log _{10} \frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=30 \quad\left(\because \log _{10} \mathrm{a}=\mathrm{b} \Rightarrow \mathrm{a}=10^{\mathrm{b}}\right)$ $\log _{10} \frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=3$ $\frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=(10)^{3}$ $\frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=1000$ $\mathrm{I}_{\mathrm{A}}=1000 \mathrm{I}_{\mathrm{B}}$ Thus, it is clear that sound $\mathrm{A}$ is 1000 times more intense than sound $\mathrm{B}$.
J and K CET- 2005
WAVES
172845
A tuning fork of frequency $480 \mathrm{~Hz}$ produces 10 beats/sec, when sounded with a sonometer string.A slight increase in tension in the sonometer string produces fewer beats/sec than before. What was the frequency of sonometer string?
1 $470 \mathrm{~Hz}$
2 $490 \mathrm{~Hz}$
3 $480 \mathrm{~Hz}$
4 $460 \mathrm{~Hz}$
Explanation:
A For sonometer, $\mathrm{f}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}$ Beats per second $=$ difference between frequency $\mathrm{f}_{1}=480$ $10=\left|\mathrm{f}_{1}-\mathrm{f}_{2}\right|$ $\mathrm{f}_{2}=470 \mathrm{~Hz} \text { or } 490 \mathrm{~Hz}$ - $x=f_{1}-f_{2}$, if we increase $f_{2}$ then $x$ will decrease. - $x=f_{2}-f_{1}$, if we increase tension $f_{2}$ will increase so beat frequency will increase so not possible, $\mathrm{f}_{2}=470 \mathrm{~Hz}$
J and K CET- 2003
WAVES
172849
Two waves of wavelength $50 \mathrm{~cm}$ and $51 \mathrm{~cm}$ produce $12 \mathrm{beat} / \mathrm{s}$. The speed of sound is
172850
A tuning fork produces 4 beats/sec with another fork of frequency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats/sec. The unknown frequency is
1 $286 \mathrm{cps}$
2 $292 \mathrm{cps}$
3 $284 \mathrm{cps}$
4 $290 \mathrm{cps}$
Explanation:
B Let, the frequency of the unknown fork be $n$ but according to question, it gives 4 beats with $288 \mathrm{cps}$. $\therefore \quad|\mathrm{n}-288|=4$ $n-288= \pm 4$ $n=292 \text { or } 284$ Since, the waxing of fork get decreased its frequency therefore, the beat frequency also drops. Thus, the unknown frequency will be $\mathrm{n}=292 \mathrm{cps}$ Hence, (b) is the correct.
172841
The intensity level of sound $A$ is $30 \mathrm{~dB}$ greater than of $B$. How many times more intense is the sound A than B?
1 10
2 100
3 1000
4 2
Explanation:
C Since, we know that- The intensity level in sound is given by $\mathrm{L}=10 \log _{10} \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{o}}}$ Where $\mathrm{I}_{\mathrm{o}}=$ initial intensity But given $\quad \mathrm{L}_{\mathrm{A}}=30+\mathrm{L}_{\mathrm{B}}$ Therefore, $10 \log _{10} \frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{o}}}=30+10 \log _{10} \frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{o}}}$ $10 \log _{10} \frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=30 \quad\left(\because \log _{10} \mathrm{a}=\mathrm{b} \Rightarrow \mathrm{a}=10^{\mathrm{b}}\right)$ $\log _{10} \frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=3$ $\frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=(10)^{3}$ $\frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=1000$ $\mathrm{I}_{\mathrm{A}}=1000 \mathrm{I}_{\mathrm{B}}$ Thus, it is clear that sound $\mathrm{A}$ is 1000 times more intense than sound $\mathrm{B}$.
J and K CET- 2005
WAVES
172845
A tuning fork of frequency $480 \mathrm{~Hz}$ produces 10 beats/sec, when sounded with a sonometer string.A slight increase in tension in the sonometer string produces fewer beats/sec than before. What was the frequency of sonometer string?
1 $470 \mathrm{~Hz}$
2 $490 \mathrm{~Hz}$
3 $480 \mathrm{~Hz}$
4 $460 \mathrm{~Hz}$
Explanation:
A For sonometer, $\mathrm{f}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}$ Beats per second $=$ difference between frequency $\mathrm{f}_{1}=480$ $10=\left|\mathrm{f}_{1}-\mathrm{f}_{2}\right|$ $\mathrm{f}_{2}=470 \mathrm{~Hz} \text { or } 490 \mathrm{~Hz}$ - $x=f_{1}-f_{2}$, if we increase $f_{2}$ then $x$ will decrease. - $x=f_{2}-f_{1}$, if we increase tension $f_{2}$ will increase so beat frequency will increase so not possible, $\mathrm{f}_{2}=470 \mathrm{~Hz}$
J and K CET- 2003
WAVES
172849
Two waves of wavelength $50 \mathrm{~cm}$ and $51 \mathrm{~cm}$ produce $12 \mathrm{beat} / \mathrm{s}$. The speed of sound is
172850
A tuning fork produces 4 beats/sec with another fork of frequency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats/sec. The unknown frequency is
1 $286 \mathrm{cps}$
2 $292 \mathrm{cps}$
3 $284 \mathrm{cps}$
4 $290 \mathrm{cps}$
Explanation:
B Let, the frequency of the unknown fork be $n$ but according to question, it gives 4 beats with $288 \mathrm{cps}$. $\therefore \quad|\mathrm{n}-288|=4$ $n-288= \pm 4$ $n=292 \text { or } 284$ Since, the waxing of fork get decreased its frequency therefore, the beat frequency also drops. Thus, the unknown frequency will be $\mathrm{n}=292 \mathrm{cps}$ Hence, (b) is the correct.
172841
The intensity level of sound $A$ is $30 \mathrm{~dB}$ greater than of $B$. How many times more intense is the sound A than B?
1 10
2 100
3 1000
4 2
Explanation:
C Since, we know that- The intensity level in sound is given by $\mathrm{L}=10 \log _{10} \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{o}}}$ Where $\mathrm{I}_{\mathrm{o}}=$ initial intensity But given $\quad \mathrm{L}_{\mathrm{A}}=30+\mathrm{L}_{\mathrm{B}}$ Therefore, $10 \log _{10} \frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{o}}}=30+10 \log _{10} \frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{o}}}$ $10 \log _{10} \frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=30 \quad\left(\because \log _{10} \mathrm{a}=\mathrm{b} \Rightarrow \mathrm{a}=10^{\mathrm{b}}\right)$ $\log _{10} \frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=3$ $\frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=(10)^{3}$ $\frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=1000$ $\mathrm{I}_{\mathrm{A}}=1000 \mathrm{I}_{\mathrm{B}}$ Thus, it is clear that sound $\mathrm{A}$ is 1000 times more intense than sound $\mathrm{B}$.
J and K CET- 2005
WAVES
172845
A tuning fork of frequency $480 \mathrm{~Hz}$ produces 10 beats/sec, when sounded with a sonometer string.A slight increase in tension in the sonometer string produces fewer beats/sec than before. What was the frequency of sonometer string?
1 $470 \mathrm{~Hz}$
2 $490 \mathrm{~Hz}$
3 $480 \mathrm{~Hz}$
4 $460 \mathrm{~Hz}$
Explanation:
A For sonometer, $\mathrm{f}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}$ Beats per second $=$ difference between frequency $\mathrm{f}_{1}=480$ $10=\left|\mathrm{f}_{1}-\mathrm{f}_{2}\right|$ $\mathrm{f}_{2}=470 \mathrm{~Hz} \text { or } 490 \mathrm{~Hz}$ - $x=f_{1}-f_{2}$, if we increase $f_{2}$ then $x$ will decrease. - $x=f_{2}-f_{1}$, if we increase tension $f_{2}$ will increase so beat frequency will increase so not possible, $\mathrm{f}_{2}=470 \mathrm{~Hz}$
J and K CET- 2003
WAVES
172849
Two waves of wavelength $50 \mathrm{~cm}$ and $51 \mathrm{~cm}$ produce $12 \mathrm{beat} / \mathrm{s}$. The speed of sound is
172850
A tuning fork produces 4 beats/sec with another fork of frequency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats/sec. The unknown frequency is
1 $286 \mathrm{cps}$
2 $292 \mathrm{cps}$
3 $284 \mathrm{cps}$
4 $290 \mathrm{cps}$
Explanation:
B Let, the frequency of the unknown fork be $n$ but according to question, it gives 4 beats with $288 \mathrm{cps}$. $\therefore \quad|\mathrm{n}-288|=4$ $n-288= \pm 4$ $n=292 \text { or } 284$ Since, the waxing of fork get decreased its frequency therefore, the beat frequency also drops. Thus, the unknown frequency will be $\mathrm{n}=292 \mathrm{cps}$ Hence, (b) is the correct.
172841
The intensity level of sound $A$ is $30 \mathrm{~dB}$ greater than of $B$. How many times more intense is the sound A than B?
1 10
2 100
3 1000
4 2
Explanation:
C Since, we know that- The intensity level in sound is given by $\mathrm{L}=10 \log _{10} \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{o}}}$ Where $\mathrm{I}_{\mathrm{o}}=$ initial intensity But given $\quad \mathrm{L}_{\mathrm{A}}=30+\mathrm{L}_{\mathrm{B}}$ Therefore, $10 \log _{10} \frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{o}}}=30+10 \log _{10} \frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{o}}}$ $10 \log _{10} \frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=30 \quad\left(\because \log _{10} \mathrm{a}=\mathrm{b} \Rightarrow \mathrm{a}=10^{\mathrm{b}}\right)$ $\log _{10} \frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=3$ $\frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=(10)^{3}$ $\frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=1000$ $\mathrm{I}_{\mathrm{A}}=1000 \mathrm{I}_{\mathrm{B}}$ Thus, it is clear that sound $\mathrm{A}$ is 1000 times more intense than sound $\mathrm{B}$.
J and K CET- 2005
WAVES
172845
A tuning fork of frequency $480 \mathrm{~Hz}$ produces 10 beats/sec, when sounded with a sonometer string.A slight increase in tension in the sonometer string produces fewer beats/sec than before. What was the frequency of sonometer string?
1 $470 \mathrm{~Hz}$
2 $490 \mathrm{~Hz}$
3 $480 \mathrm{~Hz}$
4 $460 \mathrm{~Hz}$
Explanation:
A For sonometer, $\mathrm{f}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}$ Beats per second $=$ difference between frequency $\mathrm{f}_{1}=480$ $10=\left|\mathrm{f}_{1}-\mathrm{f}_{2}\right|$ $\mathrm{f}_{2}=470 \mathrm{~Hz} \text { or } 490 \mathrm{~Hz}$ - $x=f_{1}-f_{2}$, if we increase $f_{2}$ then $x$ will decrease. - $x=f_{2}-f_{1}$, if we increase tension $f_{2}$ will increase so beat frequency will increase so not possible, $\mathrm{f}_{2}=470 \mathrm{~Hz}$
J and K CET- 2003
WAVES
172849
Two waves of wavelength $50 \mathrm{~cm}$ and $51 \mathrm{~cm}$ produce $12 \mathrm{beat} / \mathrm{s}$. The speed of sound is
172850
A tuning fork produces 4 beats/sec with another fork of frequency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats/sec. The unknown frequency is
1 $286 \mathrm{cps}$
2 $292 \mathrm{cps}$
3 $284 \mathrm{cps}$
4 $290 \mathrm{cps}$
Explanation:
B Let, the frequency of the unknown fork be $n$ but according to question, it gives 4 beats with $288 \mathrm{cps}$. $\therefore \quad|\mathrm{n}-288|=4$ $n-288= \pm 4$ $n=292 \text { or } 284$ Since, the waxing of fork get decreased its frequency therefore, the beat frequency also drops. Thus, the unknown frequency will be $\mathrm{n}=292 \mathrm{cps}$ Hence, (b) is the correct.
