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WAVES

173131 A solid sphere of lithium is rotating with angular frequency ' $ω$ ' about an axis passing through its diameter. If its temperature is raised by $50^{\circ} C$ then its new angular frequency is $(α_{lithium} = 60 \times 10^{- 5}^{\circ} C^{- 1})$

1

0.99 ω

2

0.73 ω

3

0.83 ω

4

0.94 ω

Explanation:

D Let, $M$ be the mass of the sphere and $R_{o}$ be its radius before increasing the temperature.
After increasing the temperature
$R = R_{o} (1 + α_{B} Δ θ)$
Squaring both side
$R^{2} = R_{o}^{2} [(1 + 2 α_{B} (Δ θ) + α_{B}^{2} (Δ θ)^{2})]$
Neglecting higher term.
$∴ R^{2} = R_{o}^{2} (1 + 2 α_{B} Δ θ)$
$R^{2} = R_{o}^{2} (1 + 2 \times (60 \times 10^{- 5}) \times 50)$
$= R_{o}^{2} (1 + 120 \times 10^{- 5} \times 50)$
$= R_{o}^{2} (1 + 0.06)$
$= R_{o}^{2} (1.06)$
From conservation of angular momentum-
$I. ω = I_{o} ω_{0}$
$[Where I = \frac{2}{5} {MR}^{2}]$
$\frac{2}{5} {MR}^{2} ω = \frac{2}{5} {MR}_{o}^{2} ω_{o}$
$R^{2} ω = R_{o}^{2} ω_{o}$
$ω = \frac{R_{o}^{2} ω_{o}}{R^{2}} = \frac{R_{o}^{2} ω_{o}}{R_{o}^{2} (1.06)}$
$ω = \frac{ω_{o}}{1.06} = 0.94 ω$

AP EAMCET-24.04.2018

WAVES

173133 Amplitude of a wave is represented by
$A = \frac{c}{a + b - c}$
Then, resonance will occur when

1

b = - c / 2

2

b = 0

and

a = c

3

b = - a / 2

4 None of these

Explanation:

B Resonance is a phenomenon when we force an oscillating system to vibrate with natural frequency by applying some external force periodically.
Wave is represented as.
$A = \frac{c}{a + b - c}$
For the resonance to occur, taking the amplitude as infinity
$A \to \infty$
$\frac{c}{a + b - c} \to \infty$
When we put $b = 0$ and $a = c$
$\frac{c}{a + b - c} = \frac{a}{a + 0 - a} = \frac{a}{0} = \infty$
For this value, the amplitude will be infinite so, the resonance will occur only when $b = 0$ and $a = c$ .

UPSEE - 2012

WAVES

173134 A piston fitted pipe is pulled as shown in the figure. A tuning fork is sounded at open end and loudest sound is heard at open length 13 $cm, 41 cm$ and $69 cm$ , the frequency of tuning fork if velocity of sound is $350 m / s$ , is:

1

1250 Hz

.

2

625 Hz

3

417 Hz

.

4

715 Hz

.

Explanation:

B In a closed organ pipe in which length of aircolumn can be increased or decreased, the first resonance occurs at $λ / 4$ and second resonance occurs at $3 λ / 4$ .

Thus, at first resonance
$\frac{λ}{4} = 13$
and at second resonance
$\frac{3 λ}{4} = 41$
Subtraction Eq. (i) from Eq. (ii), we have
$\frac{3 λ}{4} - \frac{λ}{4} = 41 - 13$
$\Rightarrow \frac{λ}{2} = 28$
$∴ λ = 56 cm$
Hence, frequency of tuning fork
$n = \frac{v}{λ} = \frac{350}{56 \times 10^{- 2}} = 625 Hz .$

UPSEE - 2006

WAVES

173135 Two tuning forks $P$ an $Q$ sounded together and 6 beats per second are heard. $P$ is in unison with a $30 cm$ air column open at both ends and $Q$ is in resonance when length of air column is increased by $2 cm$ . The frequencies of forks $P$ and $Q$ are

1

90 Hz

and

84 Hz

2

100 Hz

and

106 Hz

3

96 Hz

and

90 Hz

4

206 Hz

and

200 Hz

Explanation:

C We know that frequency of a tuning fork $f = \frac{V}{2 l}$ where $V$ is the velocity of sound- wave
$∴$ Frequencies of tuning forks $P$ and $Q$ are-
$f_{P} = \frac{v}{2 \times 0.30} Hz$
$f_{Q} = \frac{v}{2 \times 0.32} Hz$
At resonance, beats per second $= 6$
$∴ f_{P} - f_{Q} = 6$
$\Rightarrow \frac{v}{2 \times 0.30} - \frac{v}{2 \times 0.32} = 6$
$\Rightarrow v = \frac{2 \times 0.30 \times 0.32 \times 6}{0.32 - 0.30} = 57.6 m / s$
Thus, frequencies of tuning forks P \& Q are -
$f_{P} = \frac{57.6}{2 \times 0.30} = 96 Hz$
$f_{Q} = \frac{57.6}{2 \times 0.32} = 90 Hz$

JCECE-2015

WAVES

173131 A solid sphere of lithium is rotating with angular frequency ' $ω$ ' about an axis passing through its diameter. If its temperature is raised by $50^{\circ} C$ then its new angular frequency is $(α_{lithium} = 60 \times 10^{- 5}^{\circ} C^{- 1})$

1

0.99 ω

2

0.73 ω

3

0.83 ω

4

0.94 ω

Explanation:

D Let, $M$ be the mass of the sphere and $R_{o}$ be its radius before increasing the temperature.
After increasing the temperature
$R = R_{o} (1 + α_{B} Δ θ)$
Squaring both side
$R^{2} = R_{o}^{2} [(1 + 2 α_{B} (Δ θ) + α_{B}^{2} (Δ θ)^{2})]$
Neglecting higher term.
$∴ R^{2} = R_{o}^{2} (1 + 2 α_{B} Δ θ)$
$R^{2} = R_{o}^{2} (1 + 2 \times (60 \times 10^{- 5}) \times 50)$
$= R_{o}^{2} (1 + 120 \times 10^{- 5} \times 50)$
$= R_{o}^{2} (1 + 0.06)$
$= R_{o}^{2} (1.06)$
From conservation of angular momentum-
$I. ω = I_{o} ω_{0}$
$[Where I = \frac{2}{5} {MR}^{2}]$
$\frac{2}{5} {MR}^{2} ω = \frac{2}{5} {MR}_{o}^{2} ω_{o}$
$R^{2} ω = R_{o}^{2} ω_{o}$
$ω = \frac{R_{o}^{2} ω_{o}}{R^{2}} = \frac{R_{o}^{2} ω_{o}}{R_{o}^{2} (1.06)}$
$ω = \frac{ω_{o}}{1.06} = 0.94 ω$

AP EAMCET-24.04.2018

WAVES

173133 Amplitude of a wave is represented by
$A = \frac{c}{a + b - c}$
Then, resonance will occur when

1

b = - c / 2

2

b = 0

and

a = c

3

b = - a / 2

4 None of these

Explanation:

B Resonance is a phenomenon when we force an oscillating system to vibrate with natural frequency by applying some external force periodically.
Wave is represented as.
$A = \frac{c}{a + b - c}$
For the resonance to occur, taking the amplitude as infinity
$A \to \infty$
$\frac{c}{a + b - c} \to \infty$
When we put $b = 0$ and $a = c$
$\frac{c}{a + b - c} = \frac{a}{a + 0 - a} = \frac{a}{0} = \infty$
For this value, the amplitude will be infinite so, the resonance will occur only when $b = 0$ and $a = c$ .

UPSEE - 2012

WAVES

173134 A piston fitted pipe is pulled as shown in the figure. A tuning fork is sounded at open end and loudest sound is heard at open length 13 $cm, 41 cm$ and $69 cm$ , the frequency of tuning fork if velocity of sound is $350 m / s$ , is:

1

1250 Hz

.

2

625 Hz

3

417 Hz

.

4

715 Hz

.

Explanation:

B In a closed organ pipe in which length of aircolumn can be increased or decreased, the first resonance occurs at $λ / 4$ and second resonance occurs at $3 λ / 4$ .

Thus, at first resonance
$\frac{λ}{4} = 13$
and at second resonance
$\frac{3 λ}{4} = 41$
Subtraction Eq. (i) from Eq. (ii), we have
$\frac{3 λ}{4} - \frac{λ}{4} = 41 - 13$
$\Rightarrow \frac{λ}{2} = 28$
$∴ λ = 56 cm$
Hence, frequency of tuning fork
$n = \frac{v}{λ} = \frac{350}{56 \times 10^{- 2}} = 625 Hz .$

UPSEE - 2006

WAVES

173135 Two tuning forks $P$ an $Q$ sounded together and 6 beats per second are heard. $P$ is in unison with a $30 cm$ air column open at both ends and $Q$ is in resonance when length of air column is increased by $2 cm$ . The frequencies of forks $P$ and $Q$ are

1

90 Hz

and

84 Hz

2

100 Hz

and

106 Hz

3

96 Hz

and

90 Hz

4

206 Hz

and

200 Hz

Explanation:

C We know that frequency of a tuning fork $f = \frac{V}{2 l}$ where $V$ is the velocity of sound- wave
$∴$ Frequencies of tuning forks $P$ and $Q$ are-
$f_{P} = \frac{v}{2 \times 0.30} Hz$
$f_{Q} = \frac{v}{2 \times 0.32} Hz$
At resonance, beats per second $= 6$
$∴ f_{P} - f_{Q} = 6$
$\Rightarrow \frac{v}{2 \times 0.30} - \frac{v}{2 \times 0.32} = 6$
$\Rightarrow v = \frac{2 \times 0.30 \times 0.32 \times 6}{0.32 - 0.30} = 57.6 m / s$
Thus, frequencies of tuning forks P \& Q are -
$f_{P} = \frac{57.6}{2 \times 0.30} = 96 Hz$
$f_{Q} = \frac{57.6}{2 \times 0.32} = 90 Hz$

JCECE-2015

WAVES

173131 A solid sphere of lithium is rotating with angular frequency ' $ω$ ' about an axis passing through its diameter. If its temperature is raised by $50^{\circ} C$ then its new angular frequency is $(α_{lithium} = 60 \times 10^{- 5}^{\circ} C^{- 1})$

1

0.99 ω

2

0.73 ω

3

0.83 ω

4

0.94 ω

Explanation:

D Let, $M$ be the mass of the sphere and $R_{o}$ be its radius before increasing the temperature.
After increasing the temperature
$R = R_{o} (1 + α_{B} Δ θ)$
Squaring both side
$R^{2} = R_{o}^{2} [(1 + 2 α_{B} (Δ θ) + α_{B}^{2} (Δ θ)^{2})]$
Neglecting higher term.
$∴ R^{2} = R_{o}^{2} (1 + 2 α_{B} Δ θ)$
$R^{2} = R_{o}^{2} (1 + 2 \times (60 \times 10^{- 5}) \times 50)$
$= R_{o}^{2} (1 + 120 \times 10^{- 5} \times 50)$
$= R_{o}^{2} (1 + 0.06)$
$= R_{o}^{2} (1.06)$
From conservation of angular momentum-
$I. ω = I_{o} ω_{0}$
$[Where I = \frac{2}{5} {MR}^{2}]$
$\frac{2}{5} {MR}^{2} ω = \frac{2}{5} {MR}_{o}^{2} ω_{o}$
$R^{2} ω = R_{o}^{2} ω_{o}$
$ω = \frac{R_{o}^{2} ω_{o}}{R^{2}} = \frac{R_{o}^{2} ω_{o}}{R_{o}^{2} (1.06)}$
$ω = \frac{ω_{o}}{1.06} = 0.94 ω$

AP EAMCET-24.04.2018

WAVES

173133 Amplitude of a wave is represented by
$A = \frac{c}{a + b - c}$
Then, resonance will occur when

1

b = - c / 2

2

b = 0

and

a = c

3

b = - a / 2

4 None of these

Explanation:

B Resonance is a phenomenon when we force an oscillating system to vibrate with natural frequency by applying some external force periodically.
Wave is represented as.
$A = \frac{c}{a + b - c}$
For the resonance to occur, taking the amplitude as infinity
$A \to \infty$
$\frac{c}{a + b - c} \to \infty$
When we put $b = 0$ and $a = c$
$\frac{c}{a + b - c} = \frac{a}{a + 0 - a} = \frac{a}{0} = \infty$
For this value, the amplitude will be infinite so, the resonance will occur only when $b = 0$ and $a = c$ .

UPSEE - 2012

WAVES

173134 A piston fitted pipe is pulled as shown in the figure. A tuning fork is sounded at open end and loudest sound is heard at open length 13 $cm, 41 cm$ and $69 cm$ , the frequency of tuning fork if velocity of sound is $350 m / s$ , is:

1

1250 Hz

.

2

625 Hz

3

417 Hz

.

4

715 Hz

.

Explanation:

B In a closed organ pipe in which length of aircolumn can be increased or decreased, the first resonance occurs at $λ / 4$ and second resonance occurs at $3 λ / 4$ .

Thus, at first resonance
$\frac{λ}{4} = 13$
and at second resonance
$\frac{3 λ}{4} = 41$
Subtraction Eq. (i) from Eq. (ii), we have
$\frac{3 λ}{4} - \frac{λ}{4} = 41 - 13$
$\Rightarrow \frac{λ}{2} = 28$
$∴ λ = 56 cm$
Hence, frequency of tuning fork
$n = \frac{v}{λ} = \frac{350}{56 \times 10^{- 2}} = 625 Hz .$

UPSEE - 2006

WAVES

173135 Two tuning forks $P$ an $Q$ sounded together and 6 beats per second are heard. $P$ is in unison with a $30 cm$ air column open at both ends and $Q$ is in resonance when length of air column is increased by $2 cm$ . The frequencies of forks $P$ and $Q$ are

1

90 Hz

and

84 Hz

2

100 Hz

and

106 Hz

3

96 Hz

and

90 Hz

4

206 Hz

and

200 Hz

Explanation:

C We know that frequency of a tuning fork $f = \frac{V}{2 l}$ where $V$ is the velocity of sound- wave
$∴$ Frequencies of tuning forks $P$ and $Q$ are-
$f_{P} = \frac{v}{2 \times 0.30} Hz$
$f_{Q} = \frac{v}{2 \times 0.32} Hz$
At resonance, beats per second $= 6$
$∴ f_{P} - f_{Q} = 6$
$\Rightarrow \frac{v}{2 \times 0.30} - \frac{v}{2 \times 0.32} = 6$
$\Rightarrow v = \frac{2 \times 0.30 \times 0.32 \times 6}{0.32 - 0.30} = 57.6 m / s$
Thus, frequencies of tuning forks P \& Q are -
$f_{P} = \frac{57.6}{2 \times 0.30} = 96 Hz$
$f_{Q} = \frac{57.6}{2 \times 0.32} = 90 Hz$

JCECE-2015

WAVES

173131 A solid sphere of lithium is rotating with angular frequency ' $ω$ ' about an axis passing through its diameter. If its temperature is raised by $50^{\circ} C$ then its new angular frequency is $(α_{lithium} = 60 \times 10^{- 5}^{\circ} C^{- 1})$

1

0.99 ω

2

0.73 ω

3

0.83 ω

4

0.94 ω

Explanation:

D Let, $M$ be the mass of the sphere and $R_{o}$ be its radius before increasing the temperature.
After increasing the temperature
$R = R_{o} (1 + α_{B} Δ θ)$
Squaring both side
$R^{2} = R_{o}^{2} [(1 + 2 α_{B} (Δ θ) + α_{B}^{2} (Δ θ)^{2})]$
Neglecting higher term.
$∴ R^{2} = R_{o}^{2} (1 + 2 α_{B} Δ θ)$
$R^{2} = R_{o}^{2} (1 + 2 \times (60 \times 10^{- 5}) \times 50)$
$= R_{o}^{2} (1 + 120 \times 10^{- 5} \times 50)$
$= R_{o}^{2} (1 + 0.06)$
$= R_{o}^{2} (1.06)$
From conservation of angular momentum-
$I. ω = I_{o} ω_{0}$
$[Where I = \frac{2}{5} {MR}^{2}]$
$\frac{2}{5} {MR}^{2} ω = \frac{2}{5} {MR}_{o}^{2} ω_{o}$
$R^{2} ω = R_{o}^{2} ω_{o}$
$ω = \frac{R_{o}^{2} ω_{o}}{R^{2}} = \frac{R_{o}^{2} ω_{o}}{R_{o}^{2} (1.06)}$
$ω = \frac{ω_{o}}{1.06} = 0.94 ω$

AP EAMCET-24.04.2018

WAVES

173133 Amplitude of a wave is represented by
$A = \frac{c}{a + b - c}$
Then, resonance will occur when

1

b = - c / 2

2

b = 0

and

a = c

3

b = - a / 2

4 None of these

Explanation:

B Resonance is a phenomenon when we force an oscillating system to vibrate with natural frequency by applying some external force periodically.
Wave is represented as.
$A = \frac{c}{a + b - c}$
For the resonance to occur, taking the amplitude as infinity
$A \to \infty$
$\frac{c}{a + b - c} \to \infty$
When we put $b = 0$ and $a = c$
$\frac{c}{a + b - c} = \frac{a}{a + 0 - a} = \frac{a}{0} = \infty$
For this value, the amplitude will be infinite so, the resonance will occur only when $b = 0$ and $a = c$ .

UPSEE - 2012

WAVES

173134 A piston fitted pipe is pulled as shown in the figure. A tuning fork is sounded at open end and loudest sound is heard at open length 13 $cm, 41 cm$ and $69 cm$ , the frequency of tuning fork if velocity of sound is $350 m / s$ , is:

1

1250 Hz

.

2

625 Hz

3

417 Hz

.

4

715 Hz

.

Explanation:

B In a closed organ pipe in which length of aircolumn can be increased or decreased, the first resonance occurs at $λ / 4$ and second resonance occurs at $3 λ / 4$ .

Thus, at first resonance
$\frac{λ}{4} = 13$
and at second resonance
$\frac{3 λ}{4} = 41$
Subtraction Eq. (i) from Eq. (ii), we have
$\frac{3 λ}{4} - \frac{λ}{4} = 41 - 13$
$\Rightarrow \frac{λ}{2} = 28$
$∴ λ = 56 cm$
Hence, frequency of tuning fork
$n = \frac{v}{λ} = \frac{350}{56 \times 10^{- 2}} = 625 Hz .$

UPSEE - 2006

WAVES

173135 Two tuning forks $P$ an $Q$ sounded together and 6 beats per second are heard. $P$ is in unison with a $30 cm$ air column open at both ends and $Q$ is in resonance when length of air column is increased by $2 cm$ . The frequencies of forks $P$ and $Q$ are

1

90 Hz

and

84 Hz

2

100 Hz

and

106 Hz

3

96 Hz

and

90 Hz

4

206 Hz

and

200 Hz

Explanation:

C We know that frequency of a tuning fork $f = \frac{V}{2 l}$ where $V$ is the velocity of sound- wave
$∴$ Frequencies of tuning forks $P$ and $Q$ are-
$f_{P} = \frac{v}{2 \times 0.30} Hz$
$f_{Q} = \frac{v}{2 \times 0.32} Hz$
At resonance, beats per second $= 6$
$∴ f_{P} - f_{Q} = 6$
$\Rightarrow \frac{v}{2 \times 0.30} - \frac{v}{2 \times 0.32} = 6$
$\Rightarrow v = \frac{2 \times 0.30 \times 0.32 \times 6}{0.32 - 0.30} = 57.6 m / s$
Thus, frequencies of tuning forks P \& Q are -
$f_{P} = \frac{57.6}{2 \times 0.30} = 96 Hz$
$f_{Q} = \frac{57.6}{2 \times 0.32} = 90 Hz$

JCECE-2015

Question Bank Series

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