172409
A uniform metal wire has length ' $L$ ', mass ' $M$ ' density ' $\rho$ '. It is under tension ' $T$ ' and ' $v$ ' is the speed of transverse wave along the wire. The area of cross-section $A$ of the wire is
1 $T^{2} \rho v$
2 $\frac{v^{2} \rho}{T}$
3 $\frac{T}{v^{2} \rho}$
4 $T v^{2} \rho$
Explanation:
C Given, length of wire $=\mathrm{L}$ Mass of wire $=M$ Density of wire $=\rho$, Tension of wire $=\mathrm{T}$ Velocity, $v=\sqrt{\frac{\mathrm{T}}{\mu}} \quad\left[\because \mu=\frac{\mathrm{M}}{\mathrm{L}}\right]$ Where, $\mu=$ mass per unit length $\mathrm{v}=\sqrt{\frac{\mathrm{TL}}{\mathrm{M}}} \quad[\because \mathrm{M}=\rho \times \text { volume }=\rho \mathrm{AL}]$ $\mathrm{v}=\sqrt{\frac{\mathrm{TL}}{\rho \mathrm{AL}}}$ On squaring both sides $\mathrm{v}^{2}=\frac{\mathrm{T}}{\mathrm{A} \rho}$ Cross-section, $A=\frac{T}{v^{2} \rho}$
MHT-CET 2020
WAVES
172410
A piano wire with a diameter of $0.90 \mathrm{~mm}$ is replaced by another wire of diameter $0.93 \mathrm{~mm}$ of the same material. If tension of wire is kept the same, then the percentage change in frequency of fundamental tone is
172519
Standing waves are produced in $10 \mathrm{~m}$ long stretched string. If the string vibrates in 5 segments and wave velocity of $20 \mathrm{~m} / \mathrm{s}$, then its frequency will be :
1 $5 \mathrm{~Hz}$
2 $2 \mathrm{~Hz}$
3 $10 \mathrm{~Hz}$
4 $2 \mathrm{~Hz}$
Explanation:
A Given that, $l=10 \mathrm{~m}$ No. of segments $=5$ $\therefore \quad$ Length of one loop $=\frac{10}{5}=2 \mathrm{~m}$ $\mathrm{v}=20 \mathrm{~m} / \mathrm{s} \text {. }$ Since, one complete wave in a standing wave pattern consists of two loops. $\therefore \quad \lambda=2 \times 2 \mathrm{~m}= 4 \mathrm{~m}$ $\text { Now, frequency, } \mathrm{f} =\frac{\mathrm{v}}{\lambda}$ $=\frac{20 \mathrm{~m} / \mathrm{s}}{4 \mathrm{~m}}$ $\mathrm{f} =5 \mathrm{~Hz}$
AIIMS-1998
WAVES
172477
A string has a length of $5 \mathrm{~m}$ between fixed points and has fundamental frequency of $\mathbf{2 0}$ Hz. What is the frequency of the second overtone?
1 $40 \mathrm{~Hz}$
2 $50 \mathrm{~Hz}$
3 $60 \mathrm{~Hz}$
4 $30 \mathrm{~Hz}$
Explanation:
C We know that, Length of string between fixed ends $=5 \mathrm{~m}$ Fundamental frequency $(\mathrm{f})=20 \mathrm{~Hz}$ For second overtone $(\mathrm{n})=3$ $\therefore \quad \mathrm{f}^{\prime} =3 \mathrm{f}$ $\mathrm{f}^{\prime} =3 \times 20$ $\mathrm{f}^{\prime} =60 \mathrm{~Hz}$
172409
A uniform metal wire has length ' $L$ ', mass ' $M$ ' density ' $\rho$ '. It is under tension ' $T$ ' and ' $v$ ' is the speed of transverse wave along the wire. The area of cross-section $A$ of the wire is
1 $T^{2} \rho v$
2 $\frac{v^{2} \rho}{T}$
3 $\frac{T}{v^{2} \rho}$
4 $T v^{2} \rho$
Explanation:
C Given, length of wire $=\mathrm{L}$ Mass of wire $=M$ Density of wire $=\rho$, Tension of wire $=\mathrm{T}$ Velocity, $v=\sqrt{\frac{\mathrm{T}}{\mu}} \quad\left[\because \mu=\frac{\mathrm{M}}{\mathrm{L}}\right]$ Where, $\mu=$ mass per unit length $\mathrm{v}=\sqrt{\frac{\mathrm{TL}}{\mathrm{M}}} \quad[\because \mathrm{M}=\rho \times \text { volume }=\rho \mathrm{AL}]$ $\mathrm{v}=\sqrt{\frac{\mathrm{TL}}{\rho \mathrm{AL}}}$ On squaring both sides $\mathrm{v}^{2}=\frac{\mathrm{T}}{\mathrm{A} \rho}$ Cross-section, $A=\frac{T}{v^{2} \rho}$
MHT-CET 2020
WAVES
172410
A piano wire with a diameter of $0.90 \mathrm{~mm}$ is replaced by another wire of diameter $0.93 \mathrm{~mm}$ of the same material. If tension of wire is kept the same, then the percentage change in frequency of fundamental tone is
172519
Standing waves are produced in $10 \mathrm{~m}$ long stretched string. If the string vibrates in 5 segments and wave velocity of $20 \mathrm{~m} / \mathrm{s}$, then its frequency will be :
1 $5 \mathrm{~Hz}$
2 $2 \mathrm{~Hz}$
3 $10 \mathrm{~Hz}$
4 $2 \mathrm{~Hz}$
Explanation:
A Given that, $l=10 \mathrm{~m}$ No. of segments $=5$ $\therefore \quad$ Length of one loop $=\frac{10}{5}=2 \mathrm{~m}$ $\mathrm{v}=20 \mathrm{~m} / \mathrm{s} \text {. }$ Since, one complete wave in a standing wave pattern consists of two loops. $\therefore \quad \lambda=2 \times 2 \mathrm{~m}= 4 \mathrm{~m}$ $\text { Now, frequency, } \mathrm{f} =\frac{\mathrm{v}}{\lambda}$ $=\frac{20 \mathrm{~m} / \mathrm{s}}{4 \mathrm{~m}}$ $\mathrm{f} =5 \mathrm{~Hz}$
AIIMS-1998
WAVES
172477
A string has a length of $5 \mathrm{~m}$ between fixed points and has fundamental frequency of $\mathbf{2 0}$ Hz. What is the frequency of the second overtone?
1 $40 \mathrm{~Hz}$
2 $50 \mathrm{~Hz}$
3 $60 \mathrm{~Hz}$
4 $30 \mathrm{~Hz}$
Explanation:
C We know that, Length of string between fixed ends $=5 \mathrm{~m}$ Fundamental frequency $(\mathrm{f})=20 \mathrm{~Hz}$ For second overtone $(\mathrm{n})=3$ $\therefore \quad \mathrm{f}^{\prime} =3 \mathrm{f}$ $\mathrm{f}^{\prime} =3 \times 20$ $\mathrm{f}^{\prime} =60 \mathrm{~Hz}$
172409
A uniform metal wire has length ' $L$ ', mass ' $M$ ' density ' $\rho$ '. It is under tension ' $T$ ' and ' $v$ ' is the speed of transverse wave along the wire. The area of cross-section $A$ of the wire is
1 $T^{2} \rho v$
2 $\frac{v^{2} \rho}{T}$
3 $\frac{T}{v^{2} \rho}$
4 $T v^{2} \rho$
Explanation:
C Given, length of wire $=\mathrm{L}$ Mass of wire $=M$ Density of wire $=\rho$, Tension of wire $=\mathrm{T}$ Velocity, $v=\sqrt{\frac{\mathrm{T}}{\mu}} \quad\left[\because \mu=\frac{\mathrm{M}}{\mathrm{L}}\right]$ Where, $\mu=$ mass per unit length $\mathrm{v}=\sqrt{\frac{\mathrm{TL}}{\mathrm{M}}} \quad[\because \mathrm{M}=\rho \times \text { volume }=\rho \mathrm{AL}]$ $\mathrm{v}=\sqrt{\frac{\mathrm{TL}}{\rho \mathrm{AL}}}$ On squaring both sides $\mathrm{v}^{2}=\frac{\mathrm{T}}{\mathrm{A} \rho}$ Cross-section, $A=\frac{T}{v^{2} \rho}$
MHT-CET 2020
WAVES
172410
A piano wire with a diameter of $0.90 \mathrm{~mm}$ is replaced by another wire of diameter $0.93 \mathrm{~mm}$ of the same material. If tension of wire is kept the same, then the percentage change in frequency of fundamental tone is
172519
Standing waves are produced in $10 \mathrm{~m}$ long stretched string. If the string vibrates in 5 segments and wave velocity of $20 \mathrm{~m} / \mathrm{s}$, then its frequency will be :
1 $5 \mathrm{~Hz}$
2 $2 \mathrm{~Hz}$
3 $10 \mathrm{~Hz}$
4 $2 \mathrm{~Hz}$
Explanation:
A Given that, $l=10 \mathrm{~m}$ No. of segments $=5$ $\therefore \quad$ Length of one loop $=\frac{10}{5}=2 \mathrm{~m}$ $\mathrm{v}=20 \mathrm{~m} / \mathrm{s} \text {. }$ Since, one complete wave in a standing wave pattern consists of two loops. $\therefore \quad \lambda=2 \times 2 \mathrm{~m}= 4 \mathrm{~m}$ $\text { Now, frequency, } \mathrm{f} =\frac{\mathrm{v}}{\lambda}$ $=\frac{20 \mathrm{~m} / \mathrm{s}}{4 \mathrm{~m}}$ $\mathrm{f} =5 \mathrm{~Hz}$
AIIMS-1998
WAVES
172477
A string has a length of $5 \mathrm{~m}$ between fixed points and has fundamental frequency of $\mathbf{2 0}$ Hz. What is the frequency of the second overtone?
1 $40 \mathrm{~Hz}$
2 $50 \mathrm{~Hz}$
3 $60 \mathrm{~Hz}$
4 $30 \mathrm{~Hz}$
Explanation:
C We know that, Length of string between fixed ends $=5 \mathrm{~m}$ Fundamental frequency $(\mathrm{f})=20 \mathrm{~Hz}$ For second overtone $(\mathrm{n})=3$ $\therefore \quad \mathrm{f}^{\prime} =3 \mathrm{f}$ $\mathrm{f}^{\prime} =3 \times 20$ $\mathrm{f}^{\prime} =60 \mathrm{~Hz}$
172409
A uniform metal wire has length ' $L$ ', mass ' $M$ ' density ' $\rho$ '. It is under tension ' $T$ ' and ' $v$ ' is the speed of transverse wave along the wire. The area of cross-section $A$ of the wire is
1 $T^{2} \rho v$
2 $\frac{v^{2} \rho}{T}$
3 $\frac{T}{v^{2} \rho}$
4 $T v^{2} \rho$
Explanation:
C Given, length of wire $=\mathrm{L}$ Mass of wire $=M$ Density of wire $=\rho$, Tension of wire $=\mathrm{T}$ Velocity, $v=\sqrt{\frac{\mathrm{T}}{\mu}} \quad\left[\because \mu=\frac{\mathrm{M}}{\mathrm{L}}\right]$ Where, $\mu=$ mass per unit length $\mathrm{v}=\sqrt{\frac{\mathrm{TL}}{\mathrm{M}}} \quad[\because \mathrm{M}=\rho \times \text { volume }=\rho \mathrm{AL}]$ $\mathrm{v}=\sqrt{\frac{\mathrm{TL}}{\rho \mathrm{AL}}}$ On squaring both sides $\mathrm{v}^{2}=\frac{\mathrm{T}}{\mathrm{A} \rho}$ Cross-section, $A=\frac{T}{v^{2} \rho}$
MHT-CET 2020
WAVES
172410
A piano wire with a diameter of $0.90 \mathrm{~mm}$ is replaced by another wire of diameter $0.93 \mathrm{~mm}$ of the same material. If tension of wire is kept the same, then the percentage change in frequency of fundamental tone is
172519
Standing waves are produced in $10 \mathrm{~m}$ long stretched string. If the string vibrates in 5 segments and wave velocity of $20 \mathrm{~m} / \mathrm{s}$, then its frequency will be :
1 $5 \mathrm{~Hz}$
2 $2 \mathrm{~Hz}$
3 $10 \mathrm{~Hz}$
4 $2 \mathrm{~Hz}$
Explanation:
A Given that, $l=10 \mathrm{~m}$ No. of segments $=5$ $\therefore \quad$ Length of one loop $=\frac{10}{5}=2 \mathrm{~m}$ $\mathrm{v}=20 \mathrm{~m} / \mathrm{s} \text {. }$ Since, one complete wave in a standing wave pattern consists of two loops. $\therefore \quad \lambda=2 \times 2 \mathrm{~m}= 4 \mathrm{~m}$ $\text { Now, frequency, } \mathrm{f} =\frac{\mathrm{v}}{\lambda}$ $=\frac{20 \mathrm{~m} / \mathrm{s}}{4 \mathrm{~m}}$ $\mathrm{f} =5 \mathrm{~Hz}$
AIIMS-1998
WAVES
172477
A string has a length of $5 \mathrm{~m}$ between fixed points and has fundamental frequency of $\mathbf{2 0}$ Hz. What is the frequency of the second overtone?
1 $40 \mathrm{~Hz}$
2 $50 \mathrm{~Hz}$
3 $60 \mathrm{~Hz}$
4 $30 \mathrm{~Hz}$
Explanation:
C We know that, Length of string between fixed ends $=5 \mathrm{~m}$ Fundamental frequency $(\mathrm{f})=20 \mathrm{~Hz}$ For second overtone $(\mathrm{n})=3$ $\therefore \quad \mathrm{f}^{\prime} =3 \mathrm{f}$ $\mathrm{f}^{\prime} =3 \times 20$ $\mathrm{f}^{\prime} =60 \mathrm{~Hz}$
