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PHXI15:WAVES

354854 A heavy ball of mass $M$ is suspended from the ceiling of a car by a light string of mass $m (m <<$ $M)$ . When the car is at rest, the speed of transverse waves in the string is $60 m s^{- 1}$ . When the car has acceleration $a$ , the wave speed increases to $60.5 m s^{- 1}$ . The value of $a$ , in terms of gravitational acceleration $g$ , is closest to

1

\frac{g}{5}

2

\frac{g}{20}

3

\frac{g}{30}

4

\frac{g}{10}

Explanation:

When the car is at rest $v_{1} = \sqrt{\frac{T_{1}}{μ}} = \sqrt{\frac{M g}{μ}}$
when the car is in acceleration
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{g}{{(g^{2} + a^{2})}^{1 / 2}}}$
$(\frac{v_{2}}{v_{1}}) = {(\frac{g^{2} + a^{2}}{g^{2}})}^{1 / 4} \approx 1 + \frac{a^{2}}{4 g^{2}}$
$\frac{60.5}{60} = 1 + \frac{1}{4} \frac{a^{2}}{g^{2}} a^{2}$
$\frac{a^{2}}{g^{2}} = \frac{1}{30} \Rightarrow a = \frac{1}{\sqrt{30}} \approx \frac{g}{5}$

JEE - 2019

PHXI15:WAVES

354855 A uniform rope of length $12 m$ and mass $6 k g$ hangs vertically from a rigid support. A block of mass $2 k g$ is attached to the free end of the rope. A transverse pulse of wavelength $0.06 m$ is produced at the lower end of the rope. What is the wavelength of the pulse (in $m$ ) when it reaches the top of the rope?
supporting img

1 0.06

2 0.25

3 0.12

4 0.1

Explanation:

As the rope is heavy, its tension will be different at different points. The tension at the free end will be $2 g$ and that at the upper end it will be $8 g$ .
We have $v = n λ$
or $\sqrt{\frac{T}{μ}} = n λ$
$\sqrt{\frac{T}{λ}} = n \sqrt{μ} (1)$
The frequency of the wave pulse will be the same everywhere on the rope as it depends only on the frequency of the source. The mass per unit length is also the same throughout the rope as it is uniform. Thus, we have
As $\sqrt{\frac{T}{λ}}$ is constant, we have $\frac{\sqrt{2 g}}{0.06 m} = \frac{\sqrt{8 g}}{λ_{1}}$
Solving we get $λ_{1} = 0.12 m$
Where $λ_{1}$ is the wavelength at the top of the rope

PHXI15:WAVES

354856 A string is hanging from a rigid support. A transverse wave pulse is generated at its free end. The speed of wave pulse at distance $x$ from its free and is proportional to

1

x^{2}

2

x

3

\sqrt{x}

4

\frac{1}{x}

Explanation:

supporting img

At $x$ distance from end
$\begin{aligned} v = \sqrt{\frac{T}{μ}} \\ v = \sqrt{\frac{μ x g}{μ}} \\ v = \sqrt{x g} \Rightarrow v \propto \sqrt{x} \end{aligned}$

PHXI15:WAVES

354857 Two strings $A$ and $B$ of same material are stretched by same tension. The radius of the string $A$ is double the radius of string $B$ . Transverse wave travels on string $A$ with speed $v_{A}$ and on string $B$ with speed $v_{B}$ . The ratio $\frac{v_{A}}{v_{B}}$ is

1

\frac{1}{4}

2

\frac{1}{2}

3 2

4 4

Explanation:

According to question, the velocity of wave travelling on string is given by
$\begin{aligned} v = n λ = \frac{λ}{2 L} \sqrt{\frac{T}{μ}} (∵ n = \frac{1}{2 L} \sqrt{\frac{T}{μ}}) \\ ∴ v = \sqrt{\frac{T}{m / l}} \Rightarrow v = \sqrt{\frac{T l}{m}} \end{aligned}$
and Young's modulus, $Y = \frac{T \times l}{A Δ l}$
$∴ T \times l = Y A Δ L$
$∴ v \propto \sqrt{A} (A = area)$
For string $A$ , radius is $2 r$ and for string $B$ radius is $r$ .
So, $\frac{v_{A}}{v_{B}} = \sqrt{\frac{4 r^{2}}{r^{2}}} = \sqrt{4} = 2$
$[∵ Y$ for both the string is same]

MHTCET - 2016

PHXI15:WAVES

354854 A heavy ball of mass $M$ is suspended from the ceiling of a car by a light string of mass $m (m <<$ $M)$ . When the car is at rest, the speed of transverse waves in the string is $60 m s^{- 1}$ . When the car has acceleration $a$ , the wave speed increases to $60.5 m s^{- 1}$ . The value of $a$ , in terms of gravitational acceleration $g$ , is closest to

1

\frac{g}{5}

2

\frac{g}{20}

3

\frac{g}{30}

4

\frac{g}{10}

Explanation:

When the car is at rest $v_{1} = \sqrt{\frac{T_{1}}{μ}} = \sqrt{\frac{M g}{μ}}$
when the car is in acceleration
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{g}{{(g^{2} + a^{2})}^{1 / 2}}}$
$(\frac{v_{2}}{v_{1}}) = {(\frac{g^{2} + a^{2}}{g^{2}})}^{1 / 4} \approx 1 + \frac{a^{2}}{4 g^{2}}$
$\frac{60.5}{60} = 1 + \frac{1}{4} \frac{a^{2}}{g^{2}} a^{2}$
$\frac{a^{2}}{g^{2}} = \frac{1}{30} \Rightarrow a = \frac{1}{\sqrt{30}} \approx \frac{g}{5}$

JEE - 2019

PHXI15:WAVES

354855 A uniform rope of length $12 m$ and mass $6 k g$ hangs vertically from a rigid support. A block of mass $2 k g$ is attached to the free end of the rope. A transverse pulse of wavelength $0.06 m$ is produced at the lower end of the rope. What is the wavelength of the pulse (in $m$ ) when it reaches the top of the rope?
supporting img

1 0.06

2 0.25

3 0.12

4 0.1

Explanation:

As the rope is heavy, its tension will be different at different points. The tension at the free end will be $2 g$ and that at the upper end it will be $8 g$ .
We have $v = n λ$
or $\sqrt{\frac{T}{μ}} = n λ$
$\sqrt{\frac{T}{λ}} = n \sqrt{μ} (1)$
The frequency of the wave pulse will be the same everywhere on the rope as it depends only on the frequency of the source. The mass per unit length is also the same throughout the rope as it is uniform. Thus, we have
As $\sqrt{\frac{T}{λ}}$ is constant, we have $\frac{\sqrt{2 g}}{0.06 m} = \frac{\sqrt{8 g}}{λ_{1}}$
Solving we get $λ_{1} = 0.12 m$
Where $λ_{1}$ is the wavelength at the top of the rope

PHXI15:WAVES

354856 A string is hanging from a rigid support. A transverse wave pulse is generated at its free end. The speed of wave pulse at distance $x$ from its free and is proportional to

1

x^{2}

2

x

3

\sqrt{x}

4

\frac{1}{x}

Explanation:

supporting img

At $x$ distance from end
$\begin{aligned} v = \sqrt{\frac{T}{μ}} \\ v = \sqrt{\frac{μ x g}{μ}} \\ v = \sqrt{x g} \Rightarrow v \propto \sqrt{x} \end{aligned}$

PHXI15:WAVES

354857 Two strings $A$ and $B$ of same material are stretched by same tension. The radius of the string $A$ is double the radius of string $B$ . Transverse wave travels on string $A$ with speed $v_{A}$ and on string $B$ with speed $v_{B}$ . The ratio $\frac{v_{A}}{v_{B}}$ is

1

\frac{1}{4}

2

\frac{1}{2}

3 2

4 4

Explanation:

According to question, the velocity of wave travelling on string is given by
$\begin{aligned} v = n λ = \frac{λ}{2 L} \sqrt{\frac{T}{μ}} (∵ n = \frac{1}{2 L} \sqrt{\frac{T}{μ}}) \\ ∴ v = \sqrt{\frac{T}{m / l}} \Rightarrow v = \sqrt{\frac{T l}{m}} \end{aligned}$
and Young's modulus, $Y = \frac{T \times l}{A Δ l}$
$∴ T \times l = Y A Δ L$
$∴ v \propto \sqrt{A} (A = area)$
For string $A$ , radius is $2 r$ and for string $B$ radius is $r$ .
So, $\frac{v_{A}}{v_{B}} = \sqrt{\frac{4 r^{2}}{r^{2}}} = \sqrt{4} = 2$
$[∵ Y$ for both the string is same]

MHTCET - 2016

PHXI15:WAVES

354854 A heavy ball of mass $M$ is suspended from the ceiling of a car by a light string of mass $m (m <<$ $M)$ . When the car is at rest, the speed of transverse waves in the string is $60 m s^{- 1}$ . When the car has acceleration $a$ , the wave speed increases to $60.5 m s^{- 1}$ . The value of $a$ , in terms of gravitational acceleration $g$ , is closest to

1

\frac{g}{5}

2

\frac{g}{20}

3

\frac{g}{30}

4

\frac{g}{10}

Explanation:

When the car is at rest $v_{1} = \sqrt{\frac{T_{1}}{μ}} = \sqrt{\frac{M g}{μ}}$
when the car is in acceleration
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{g}{{(g^{2} + a^{2})}^{1 / 2}}}$
$(\frac{v_{2}}{v_{1}}) = {(\frac{g^{2} + a^{2}}{g^{2}})}^{1 / 4} \approx 1 + \frac{a^{2}}{4 g^{2}}$
$\frac{60.5}{60} = 1 + \frac{1}{4} \frac{a^{2}}{g^{2}} a^{2}$
$\frac{a^{2}}{g^{2}} = \frac{1}{30} \Rightarrow a = \frac{1}{\sqrt{30}} \approx \frac{g}{5}$

JEE - 2019

PHXI15:WAVES

354855 A uniform rope of length $12 m$ and mass $6 k g$ hangs vertically from a rigid support. A block of mass $2 k g$ is attached to the free end of the rope. A transverse pulse of wavelength $0.06 m$ is produced at the lower end of the rope. What is the wavelength of the pulse (in $m$ ) when it reaches the top of the rope?
supporting img

1 0.06

2 0.25

3 0.12

4 0.1

Explanation:

As the rope is heavy, its tension will be different at different points. The tension at the free end will be $2 g$ and that at the upper end it will be $8 g$ .
We have $v = n λ$
or $\sqrt{\frac{T}{μ}} = n λ$
$\sqrt{\frac{T}{λ}} = n \sqrt{μ} (1)$
The frequency of the wave pulse will be the same everywhere on the rope as it depends only on the frequency of the source. The mass per unit length is also the same throughout the rope as it is uniform. Thus, we have
As $\sqrt{\frac{T}{λ}}$ is constant, we have $\frac{\sqrt{2 g}}{0.06 m} = \frac{\sqrt{8 g}}{λ_{1}}$
Solving we get $λ_{1} = 0.12 m$
Where $λ_{1}$ is the wavelength at the top of the rope

PHXI15:WAVES

354856 A string is hanging from a rigid support. A transverse wave pulse is generated at its free end. The speed of wave pulse at distance $x$ from its free and is proportional to

1

x^{2}

2

x

3

\sqrt{x}

4

\frac{1}{x}

Explanation:

supporting img

At $x$ distance from end
$\begin{aligned} v = \sqrt{\frac{T}{μ}} \\ v = \sqrt{\frac{μ x g}{μ}} \\ v = \sqrt{x g} \Rightarrow v \propto \sqrt{x} \end{aligned}$

PHXI15:WAVES

354857 Two strings $A$ and $B$ of same material are stretched by same tension. The radius of the string $A$ is double the radius of string $B$ . Transverse wave travels on string $A$ with speed $v_{A}$ and on string $B$ with speed $v_{B}$ . The ratio $\frac{v_{A}}{v_{B}}$ is

1

\frac{1}{4}

2

\frac{1}{2}

3 2

4 4

Explanation:

According to question, the velocity of wave travelling on string is given by
$\begin{aligned} v = n λ = \frac{λ}{2 L} \sqrt{\frac{T}{μ}} (∵ n = \frac{1}{2 L} \sqrt{\frac{T}{μ}}) \\ ∴ v = \sqrt{\frac{T}{m / l}} \Rightarrow v = \sqrt{\frac{T l}{m}} \end{aligned}$
and Young's modulus, $Y = \frac{T \times l}{A Δ l}$
$∴ T \times l = Y A Δ L$
$∴ v \propto \sqrt{A} (A = area)$
For string $A$ , radius is $2 r$ and for string $B$ radius is $r$ .
So, $\frac{v_{A}}{v_{B}} = \sqrt{\frac{4 r^{2}}{r^{2}}} = \sqrt{4} = 2$
$[∵ Y$ for both the string is same]

MHTCET - 2016

PHXI15:WAVES

354854 A heavy ball of mass $M$ is suspended from the ceiling of a car by a light string of mass $m (m <<$ $M)$ . When the car is at rest, the speed of transverse waves in the string is $60 m s^{- 1}$ . When the car has acceleration $a$ , the wave speed increases to $60.5 m s^{- 1}$ . The value of $a$ , in terms of gravitational acceleration $g$ , is closest to

1

\frac{g}{5}

2

\frac{g}{20}

3

\frac{g}{30}

4

\frac{g}{10}

Explanation:

When the car is at rest $v_{1} = \sqrt{\frac{T_{1}}{μ}} = \sqrt{\frac{M g}{μ}}$
when the car is in acceleration
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{g}{{(g^{2} + a^{2})}^{1 / 2}}}$
$(\frac{v_{2}}{v_{1}}) = {(\frac{g^{2} + a^{2}}{g^{2}})}^{1 / 4} \approx 1 + \frac{a^{2}}{4 g^{2}}$
$\frac{60.5}{60} = 1 + \frac{1}{4} \frac{a^{2}}{g^{2}} a^{2}$
$\frac{a^{2}}{g^{2}} = \frac{1}{30} \Rightarrow a = \frac{1}{\sqrt{30}} \approx \frac{g}{5}$

JEE - 2019

PHXI15:WAVES

354855 A uniform rope of length $12 m$ and mass $6 k g$ hangs vertically from a rigid support. A block of mass $2 k g$ is attached to the free end of the rope. A transverse pulse of wavelength $0.06 m$ is produced at the lower end of the rope. What is the wavelength of the pulse (in $m$ ) when it reaches the top of the rope?
supporting img

1 0.06

2 0.25

3 0.12

4 0.1

Explanation:

As the rope is heavy, its tension will be different at different points. The tension at the free end will be $2 g$ and that at the upper end it will be $8 g$ .
We have $v = n λ$
or $\sqrt{\frac{T}{μ}} = n λ$
$\sqrt{\frac{T}{λ}} = n \sqrt{μ} (1)$
The frequency of the wave pulse will be the same everywhere on the rope as it depends only on the frequency of the source. The mass per unit length is also the same throughout the rope as it is uniform. Thus, we have
As $\sqrt{\frac{T}{λ}}$ is constant, we have $\frac{\sqrt{2 g}}{0.06 m} = \frac{\sqrt{8 g}}{λ_{1}}$
Solving we get $λ_{1} = 0.12 m$
Where $λ_{1}$ is the wavelength at the top of the rope

PHXI15:WAVES

354856 A string is hanging from a rigid support. A transverse wave pulse is generated at its free end. The speed of wave pulse at distance $x$ from its free and is proportional to

1

x^{2}

2

x

3

\sqrt{x}

4

\frac{1}{x}

Explanation:

supporting img

At $x$ distance from end
$\begin{aligned} v = \sqrt{\frac{T}{μ}} \\ v = \sqrt{\frac{μ x g}{μ}} \\ v = \sqrt{x g} \Rightarrow v \propto \sqrt{x} \end{aligned}$

PHXI15:WAVES

354857 Two strings $A$ and $B$ of same material are stretched by same tension. The radius of the string $A$ is double the radius of string $B$ . Transverse wave travels on string $A$ with speed $v_{A}$ and on string $B$ with speed $v_{B}$ . The ratio $\frac{v_{A}}{v_{B}}$ is

1

\frac{1}{4}

2

\frac{1}{2}

3 2

4 4

Explanation:

According to question, the velocity of wave travelling on string is given by
$\begin{aligned} v = n λ = \frac{λ}{2 L} \sqrt{\frac{T}{μ}} (∵ n = \frac{1}{2 L} \sqrt{\frac{T}{μ}}) \\ ∴ v = \sqrt{\frac{T}{m / l}} \Rightarrow v = \sqrt{\frac{T l}{m}} \end{aligned}$
and Young's modulus, $Y = \frac{T \times l}{A Δ l}$
$∴ T \times l = Y A Δ L$
$∴ v \propto \sqrt{A} (A = area)$
For string $A$ , radius is $2 r$ and for string $B$ radius is $r$ .
So, $\frac{v_{A}}{v_{B}} = \sqrt{\frac{4 r^{2}}{r^{2}}} = \sqrt{4} = 2$
$[∵ Y$ for both the string is same]

MHTCET - 2016