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PHXI14:OSCILLATIONS

364415 A spring of mass $m$ is connected to a rigid support as shown in the figure. If the free end of the spring is given velocity $v_{o}$ then kinetic energy of the spring is
supporting img

1

\frac{m v_{o}^{2}}{4}

2

\frac{m v_{o}^{2}}{3}

3

\frac{1}{2} (\frac{m}{3}) v_{o}^{2}

4

\frac{1}{2} m v_{o}^{2}

Explanation:

The velocity of the spring increases linearly from left to right. The velocity of the spring at a distance $x$ from the left end is
$v = \frac{v_{o}}{l} x$
consider a small element of length $d x$ as shown in the figure.
supporting img
The kinetic energy of the element is
$\begin{aligned} d x = \frac{1}{2} (\frac{m}{l}) d x v^{2} = \frac{1}{2} \frac{m}{l} \frac{v_{o}^{2}}{l^{2}} x^{2} d x \\ k = \frac{m v_{o}^{2}}{2 l^{3}} \int_{x^{2}}^{l} x^{2} d x = \frac{m v_{o}^{2}}{2 l^{3}} [\frac{l^{3}}{3}] = \frac{1}{2} (\frac{m}{3}) v_{o}^{2} \end{aligned}$

PHXI14:OSCILLATIONS

364416 Two bodies $M$ and $N$ of equal masses are suspended from two separate springs of constants $k_{1}$ and $k_{2}$ respectively. If $M$ and $N$ oscillate vertically so that their maximum velocities are equal, the ratio of amplitude of vibration of $M$ to that of $N$ is

1

\frac{k_{2}}{k_{1}}

2

\frac{k_{1}}{k_{2}}

3

\sqrt{\frac{k_{2}}{k_{1}}}

4

\sqrt{\frac{k_{1}}{k_{2}}}

Explanation:

For first case : $v_{max_{1}} = A_{1} ω_{1} = A_{1} \sqrt{\frac{k_{1}}{m}}$
For second case : $v_{max_{2}} = A_{2} ω_{2} = A_{2} \sqrt{\frac{k_{2}}{m}}$
$∵ v_{max_{1}} = v_{max_{2}}$
$A_{1} \sqrt{\frac{k_{1}}{m}} = A_{2} \sqrt{\frac{k_{2}}{m}}$
$\Rightarrow \frac{A_{1}}{A_{2}} = \sqrt{\frac{k_{2}}{k_{1}}}$ .
So correct option is (3)

PHXI14:OSCILLATIONS

364417 One-fourth length of a spring of force constant $k$ is cut away: The force constant of the remaining spring will be

1

\frac{3}{4} k

2

\frac{4}{3} k

3

k

4

4 k

Explanation:

By using, $k \propto \frac{1}{l}$
Since, one-fourth length is cut away,
so remaining length is $\frac{3}{4}$ th, hence, $k$
becomes $\frac{4}{3}$ times, i.e. $k^{'} = \frac{4}{3} k$ .

PHXI14:OSCILLATIONS

364418 A uniform disc of mass $m = 2 k g$ and radius $R = 5 c m$ is pivoted smoothly at its centre of mass. A light spring of stiffness $k$ is attached with the disc tangentially (as shown in figure). Find the angular frequency of torsional oscillation of the disc is
(Take $k = 5 N / s, \sqrt{5} = 2.23)$
supporting img

1

1.67 r a d / s

2

4.25 r a d / s

3

2.23 r a d / s

4

7.35 r a d / s

Explanation:

supporting img
If disc is twisted clockwise for a small angle $θ$ , deformation in spring $(x) = R θ$
$\Rightarrow F_{s} = k x = k (R θ)$
$τ_{r e s} = - F_{s} R$
$= - k R^{2} θ (1)$
Also $τ = I_{c} α (2)$
From (1) & (2)
$\Rightarrow α = \frac{k^{2} θ}{I_{c}}$
$= - \frac{2 k}{m} θ [∴ I_{c} = \frac{m R^{2}}{2}]$
Comparing with standard equation,
$ω^{2} θ = \frac{- 2 k}{m} θ$
$∴ ω = \sqrt{\frac{2 k}{m}}$
$= \sqrt{\frac{2 \times 5}{2}}$
$∴ ω = 2.23 r a d / s$

PHXI14:OSCILLATIONS

364419 A block of mass $m$ attached to one end of the vertical spring produces extension $x$ . If the block is pulled and released, the periodic time of oscillation is

1

2 π \sqrt{\frac{x}{4 g}}

2

2 π \sqrt{\frac{2 x}{g}}

3

2 π \sqrt{\frac{x}{2 g}}

4

2 π \sqrt{\frac{x}{g}}

Explanation:

For vertical spring, the periodic time of oscillation is
$T = 2 π \sqrt{\frac{m}{k}}$
As, restoring force, $| F | = k x$
$\begin{aligned} \Rightarrow k = \frac{F}{x} = \frac{m g}{x} \\ ∴ T = 2 π \sqrt{\frac{\frac{m}{m g}}{x}} = 2 π \sqrt{\frac{x}{g}} \end{aligned}$

MHTCET - 2020

PHXI14:OSCILLATIONS

364415 A spring of mass $m$ is connected to a rigid support as shown in the figure. If the free end of the spring is given velocity $v_{o}$ then kinetic energy of the spring is
supporting img

1

\frac{m v_{o}^{2}}{4}

2

\frac{m v_{o}^{2}}{3}

3

\frac{1}{2} (\frac{m}{3}) v_{o}^{2}

4

\frac{1}{2} m v_{o}^{2}

Explanation:

The velocity of the spring increases linearly from left to right. The velocity of the spring at a distance $x$ from the left end is
$v = \frac{v_{o}}{l} x$
consider a small element of length $d x$ as shown in the figure.
supporting img
The kinetic energy of the element is
$\begin{aligned} d x = \frac{1}{2} (\frac{m}{l}) d x v^{2} = \frac{1}{2} \frac{m}{l} \frac{v_{o}^{2}}{l^{2}} x^{2} d x \\ k = \frac{m v_{o}^{2}}{2 l^{3}} \int_{x^{2}}^{l} x^{2} d x = \frac{m v_{o}^{2}}{2 l^{3}} [\frac{l^{3}}{3}] = \frac{1}{2} (\frac{m}{3}) v_{o}^{2} \end{aligned}$

PHXI14:OSCILLATIONS

364416 Two bodies $M$ and $N$ of equal masses are suspended from two separate springs of constants $k_{1}$ and $k_{2}$ respectively. If $M$ and $N$ oscillate vertically so that their maximum velocities are equal, the ratio of amplitude of vibration of $M$ to that of $N$ is

1

\frac{k_{2}}{k_{1}}

2

\frac{k_{1}}{k_{2}}

3

\sqrt{\frac{k_{2}}{k_{1}}}

4

\sqrt{\frac{k_{1}}{k_{2}}}

Explanation:

For first case : $v_{max_{1}} = A_{1} ω_{1} = A_{1} \sqrt{\frac{k_{1}}{m}}$
For second case : $v_{max_{2}} = A_{2} ω_{2} = A_{2} \sqrt{\frac{k_{2}}{m}}$
$∵ v_{max_{1}} = v_{max_{2}}$
$A_{1} \sqrt{\frac{k_{1}}{m}} = A_{2} \sqrt{\frac{k_{2}}{m}}$
$\Rightarrow \frac{A_{1}}{A_{2}} = \sqrt{\frac{k_{2}}{k_{1}}}$ .
So correct option is (3)

PHXI14:OSCILLATIONS

364417 One-fourth length of a spring of force constant $k$ is cut away: The force constant of the remaining spring will be

1

\frac{3}{4} k

2

\frac{4}{3} k

3

k

4

4 k

Explanation:

By using, $k \propto \frac{1}{l}$
Since, one-fourth length is cut away,
so remaining length is $\frac{3}{4}$ th, hence, $k$
becomes $\frac{4}{3}$ times, i.e. $k^{'} = \frac{4}{3} k$ .

PHXI14:OSCILLATIONS

364418 A uniform disc of mass $m = 2 k g$ and radius $R = 5 c m$ is pivoted smoothly at its centre of mass. A light spring of stiffness $k$ is attached with the disc tangentially (as shown in figure). Find the angular frequency of torsional oscillation of the disc is
(Take $k = 5 N / s, \sqrt{5} = 2.23)$
supporting img

1

1.67 r a d / s

2

4.25 r a d / s

3

2.23 r a d / s

4

7.35 r a d / s

Explanation:

supporting img
If disc is twisted clockwise for a small angle $θ$ , deformation in spring $(x) = R θ$
$\Rightarrow F_{s} = k x = k (R θ)$
$τ_{r e s} = - F_{s} R$
$= - k R^{2} θ (1)$
Also $τ = I_{c} α (2)$
From (1) & (2)
$\Rightarrow α = \frac{k^{2} θ}{I_{c}}$
$= - \frac{2 k}{m} θ [∴ I_{c} = \frac{m R^{2}}{2}]$
Comparing with standard equation,
$ω^{2} θ = \frac{- 2 k}{m} θ$
$∴ ω = \sqrt{\frac{2 k}{m}}$
$= \sqrt{\frac{2 \times 5}{2}}$
$∴ ω = 2.23 r a d / s$

PHXI14:OSCILLATIONS

364419 A block of mass $m$ attached to one end of the vertical spring produces extension $x$ . If the block is pulled and released, the periodic time of oscillation is

1

2 π \sqrt{\frac{x}{4 g}}

2

2 π \sqrt{\frac{2 x}{g}}

3

2 π \sqrt{\frac{x}{2 g}}

4

2 π \sqrt{\frac{x}{g}}

Explanation:

For vertical spring, the periodic time of oscillation is
$T = 2 π \sqrt{\frac{m}{k}}$
As, restoring force, $| F | = k x$
$\begin{aligned} \Rightarrow k = \frac{F}{x} = \frac{m g}{x} \\ ∴ T = 2 π \sqrt{\frac{\frac{m}{m g}}{x}} = 2 π \sqrt{\frac{x}{g}} \end{aligned}$

MHTCET - 2020

PHXI14:OSCILLATIONS

364415 A spring of mass $m$ is connected to a rigid support as shown in the figure. If the free end of the spring is given velocity $v_{o}$ then kinetic energy of the spring is
supporting img

1

\frac{m v_{o}^{2}}{4}

2

\frac{m v_{o}^{2}}{3}

3

\frac{1}{2} (\frac{m}{3}) v_{o}^{2}

4

\frac{1}{2} m v_{o}^{2}

Explanation:

The velocity of the spring increases linearly from left to right. The velocity of the spring at a distance $x$ from the left end is
$v = \frac{v_{o}}{l} x$
consider a small element of length $d x$ as shown in the figure.
supporting img
The kinetic energy of the element is
$\begin{aligned} d x = \frac{1}{2} (\frac{m}{l}) d x v^{2} = \frac{1}{2} \frac{m}{l} \frac{v_{o}^{2}}{l^{2}} x^{2} d x \\ k = \frac{m v_{o}^{2}}{2 l^{3}} \int_{x^{2}}^{l} x^{2} d x = \frac{m v_{o}^{2}}{2 l^{3}} [\frac{l^{3}}{3}] = \frac{1}{2} (\frac{m}{3}) v_{o}^{2} \end{aligned}$

PHXI14:OSCILLATIONS

364416 Two bodies $M$ and $N$ of equal masses are suspended from two separate springs of constants $k_{1}$ and $k_{2}$ respectively. If $M$ and $N$ oscillate vertically so that their maximum velocities are equal, the ratio of amplitude of vibration of $M$ to that of $N$ is

1

\frac{k_{2}}{k_{1}}

2

\frac{k_{1}}{k_{2}}

3

\sqrt{\frac{k_{2}}{k_{1}}}

4

\sqrt{\frac{k_{1}}{k_{2}}}

Explanation:

For first case : $v_{max_{1}} = A_{1} ω_{1} = A_{1} \sqrt{\frac{k_{1}}{m}}$
For second case : $v_{max_{2}} = A_{2} ω_{2} = A_{2} \sqrt{\frac{k_{2}}{m}}$
$∵ v_{max_{1}} = v_{max_{2}}$
$A_{1} \sqrt{\frac{k_{1}}{m}} = A_{2} \sqrt{\frac{k_{2}}{m}}$
$\Rightarrow \frac{A_{1}}{A_{2}} = \sqrt{\frac{k_{2}}{k_{1}}}$ .
So correct option is (3)

PHXI14:OSCILLATIONS

364417 One-fourth length of a spring of force constant $k$ is cut away: The force constant of the remaining spring will be

1

\frac{3}{4} k

2

\frac{4}{3} k

3

k

4

4 k

Explanation:

By using, $k \propto \frac{1}{l}$
Since, one-fourth length is cut away,
so remaining length is $\frac{3}{4}$ th, hence, $k$
becomes $\frac{4}{3}$ times, i.e. $k^{'} = \frac{4}{3} k$ .

PHXI14:OSCILLATIONS

364418 A uniform disc of mass $m = 2 k g$ and radius $R = 5 c m$ is pivoted smoothly at its centre of mass. A light spring of stiffness $k$ is attached with the disc tangentially (as shown in figure). Find the angular frequency of torsional oscillation of the disc is
(Take $k = 5 N / s, \sqrt{5} = 2.23)$
supporting img

1

1.67 r a d / s

2

4.25 r a d / s

3

2.23 r a d / s

4

7.35 r a d / s

Explanation:

supporting img
If disc is twisted clockwise for a small angle $θ$ , deformation in spring $(x) = R θ$
$\Rightarrow F_{s} = k x = k (R θ)$
$τ_{r e s} = - F_{s} R$
$= - k R^{2} θ (1)$
Also $τ = I_{c} α (2)$
From (1) & (2)
$\Rightarrow α = \frac{k^{2} θ}{I_{c}}$
$= - \frac{2 k}{m} θ [∴ I_{c} = \frac{m R^{2}}{2}]$
Comparing with standard equation,
$ω^{2} θ = \frac{- 2 k}{m} θ$
$∴ ω = \sqrt{\frac{2 k}{m}}$
$= \sqrt{\frac{2 \times 5}{2}}$
$∴ ω = 2.23 r a d / s$

PHXI14:OSCILLATIONS

364419 A block of mass $m$ attached to one end of the vertical spring produces extension $x$ . If the block is pulled and released, the periodic time of oscillation is

1

2 π \sqrt{\frac{x}{4 g}}

2

2 π \sqrt{\frac{2 x}{g}}

3

2 π \sqrt{\frac{x}{2 g}}

4

2 π \sqrt{\frac{x}{g}}

Explanation:

For vertical spring, the periodic time of oscillation is
$T = 2 π \sqrt{\frac{m}{k}}$
As, restoring force, $| F | = k x$
$\begin{aligned} \Rightarrow k = \frac{F}{x} = \frac{m g}{x} \\ ∴ T = 2 π \sqrt{\frac{\frac{m}{m g}}{x}} = 2 π \sqrt{\frac{x}{g}} \end{aligned}$

MHTCET - 2020

PHXI14:OSCILLATIONS

364415 A spring of mass $m$ is connected to a rigid support as shown in the figure. If the free end of the spring is given velocity $v_{o}$ then kinetic energy of the spring is
supporting img

1

\frac{m v_{o}^{2}}{4}

2

\frac{m v_{o}^{2}}{3}

3

\frac{1}{2} (\frac{m}{3}) v_{o}^{2}

4

\frac{1}{2} m v_{o}^{2}

Explanation:

The velocity of the spring increases linearly from left to right. The velocity of the spring at a distance $x$ from the left end is
$v = \frac{v_{o}}{l} x$
consider a small element of length $d x$ as shown in the figure.
supporting img
The kinetic energy of the element is
$\begin{aligned} d x = \frac{1}{2} (\frac{m}{l}) d x v^{2} = \frac{1}{2} \frac{m}{l} \frac{v_{o}^{2}}{l^{2}} x^{2} d x \\ k = \frac{m v_{o}^{2}}{2 l^{3}} \int_{x^{2}}^{l} x^{2} d x = \frac{m v_{o}^{2}}{2 l^{3}} [\frac{l^{3}}{3}] = \frac{1}{2} (\frac{m}{3}) v_{o}^{2} \end{aligned}$

PHXI14:OSCILLATIONS

364416 Two bodies $M$ and $N$ of equal masses are suspended from two separate springs of constants $k_{1}$ and $k_{2}$ respectively. If $M$ and $N$ oscillate vertically so that their maximum velocities are equal, the ratio of amplitude of vibration of $M$ to that of $N$ is

1

\frac{k_{2}}{k_{1}}

2

\frac{k_{1}}{k_{2}}

3

\sqrt{\frac{k_{2}}{k_{1}}}

4

\sqrt{\frac{k_{1}}{k_{2}}}

Explanation:

For first case : $v_{max_{1}} = A_{1} ω_{1} = A_{1} \sqrt{\frac{k_{1}}{m}}$
For second case : $v_{max_{2}} = A_{2} ω_{2} = A_{2} \sqrt{\frac{k_{2}}{m}}$
$∵ v_{max_{1}} = v_{max_{2}}$
$A_{1} \sqrt{\frac{k_{1}}{m}} = A_{2} \sqrt{\frac{k_{2}}{m}}$
$\Rightarrow \frac{A_{1}}{A_{2}} = \sqrt{\frac{k_{2}}{k_{1}}}$ .
So correct option is (3)

PHXI14:OSCILLATIONS

364417 One-fourth length of a spring of force constant $k$ is cut away: The force constant of the remaining spring will be

1

\frac{3}{4} k

2

\frac{4}{3} k

3

k

4

4 k

Explanation:

By using, $k \propto \frac{1}{l}$
Since, one-fourth length is cut away,
so remaining length is $\frac{3}{4}$ th, hence, $k$
becomes $\frac{4}{3}$ times, i.e. $k^{'} = \frac{4}{3} k$ .

PHXI14:OSCILLATIONS

364418 A uniform disc of mass $m = 2 k g$ and radius $R = 5 c m$ is pivoted smoothly at its centre of mass. A light spring of stiffness $k$ is attached with the disc tangentially (as shown in figure). Find the angular frequency of torsional oscillation of the disc is
(Take $k = 5 N / s, \sqrt{5} = 2.23)$
supporting img

1

1.67 r a d / s

2

4.25 r a d / s

3

2.23 r a d / s

4

7.35 r a d / s

Explanation:

supporting img
If disc is twisted clockwise for a small angle $θ$ , deformation in spring $(x) = R θ$
$\Rightarrow F_{s} = k x = k (R θ)$
$τ_{r e s} = - F_{s} R$
$= - k R^{2} θ (1)$
Also $τ = I_{c} α (2)$
From (1) & (2)
$\Rightarrow α = \frac{k^{2} θ}{I_{c}}$
$= - \frac{2 k}{m} θ [∴ I_{c} = \frac{m R^{2}}{2}]$
Comparing with standard equation,
$ω^{2} θ = \frac{- 2 k}{m} θ$
$∴ ω = \sqrt{\frac{2 k}{m}}$
$= \sqrt{\frac{2 \times 5}{2}}$
$∴ ω = 2.23 r a d / s$

PHXI14:OSCILLATIONS

364419 A block of mass $m$ attached to one end of the vertical spring produces extension $x$ . If the block is pulled and released, the periodic time of oscillation is

1

2 π \sqrt{\frac{x}{4 g}}

2

2 π \sqrt{\frac{2 x}{g}}

3

2 π \sqrt{\frac{x}{2 g}}

4

2 π \sqrt{\frac{x}{g}}

Explanation:

For vertical spring, the periodic time of oscillation is
$T = 2 π \sqrt{\frac{m}{k}}$
As, restoring force, $| F | = k x$
$\begin{aligned} \Rightarrow k = \frac{F}{x} = \frac{m g}{x} \\ ∴ T = 2 π \sqrt{\frac{\frac{m}{m g}}{x}} = 2 π \sqrt{\frac{x}{g}} \end{aligned}$

MHTCET - 2020

PHXI14:OSCILLATIONS

364415 A spring of mass $m$ is connected to a rigid support as shown in the figure. If the free end of the spring is given velocity $v_{o}$ then kinetic energy of the spring is
supporting img

1

\frac{m v_{o}^{2}}{4}

2

\frac{m v_{o}^{2}}{3}

3

\frac{1}{2} (\frac{m}{3}) v_{o}^{2}

4

\frac{1}{2} m v_{o}^{2}

Explanation:

The velocity of the spring increases linearly from left to right. The velocity of the spring at a distance $x$ from the left end is
$v = \frac{v_{o}}{l} x$
consider a small element of length $d x$ as shown in the figure.
supporting img
The kinetic energy of the element is
$\begin{aligned} d x = \frac{1}{2} (\frac{m}{l}) d x v^{2} = \frac{1}{2} \frac{m}{l} \frac{v_{o}^{2}}{l^{2}} x^{2} d x \\ k = \frac{m v_{o}^{2}}{2 l^{3}} \int_{x^{2}}^{l} x^{2} d x = \frac{m v_{o}^{2}}{2 l^{3}} [\frac{l^{3}}{3}] = \frac{1}{2} (\frac{m}{3}) v_{o}^{2} \end{aligned}$

PHXI14:OSCILLATIONS

364416 Two bodies $M$ and $N$ of equal masses are suspended from two separate springs of constants $k_{1}$ and $k_{2}$ respectively. If $M$ and $N$ oscillate vertically so that their maximum velocities are equal, the ratio of amplitude of vibration of $M$ to that of $N$ is

1

\frac{k_{2}}{k_{1}}

2

\frac{k_{1}}{k_{2}}

3

\sqrt{\frac{k_{2}}{k_{1}}}

4

\sqrt{\frac{k_{1}}{k_{2}}}

Explanation:

For first case : $v_{max_{1}} = A_{1} ω_{1} = A_{1} \sqrt{\frac{k_{1}}{m}}$
For second case : $v_{max_{2}} = A_{2} ω_{2} = A_{2} \sqrt{\frac{k_{2}}{m}}$
$∵ v_{max_{1}} = v_{max_{2}}$
$A_{1} \sqrt{\frac{k_{1}}{m}} = A_{2} \sqrt{\frac{k_{2}}{m}}$
$\Rightarrow \frac{A_{1}}{A_{2}} = \sqrt{\frac{k_{2}}{k_{1}}}$ .
So correct option is (3)

PHXI14:OSCILLATIONS

364417 One-fourth length of a spring of force constant $k$ is cut away: The force constant of the remaining spring will be

1

\frac{3}{4} k

2

\frac{4}{3} k

3

k

4

4 k

Explanation:

By using, $k \propto \frac{1}{l}$
Since, one-fourth length is cut away,
so remaining length is $\frac{3}{4}$ th, hence, $k$
becomes $\frac{4}{3}$ times, i.e. $k^{'} = \frac{4}{3} k$ .

PHXI14:OSCILLATIONS

364418 A uniform disc of mass $m = 2 k g$ and radius $R = 5 c m$ is pivoted smoothly at its centre of mass. A light spring of stiffness $k$ is attached with the disc tangentially (as shown in figure). Find the angular frequency of torsional oscillation of the disc is
(Take $k = 5 N / s, \sqrt{5} = 2.23)$
supporting img

1

1.67 r a d / s

2

4.25 r a d / s

3

2.23 r a d / s

4

7.35 r a d / s

Explanation:

supporting img
If disc is twisted clockwise for a small angle $θ$ , deformation in spring $(x) = R θ$
$\Rightarrow F_{s} = k x = k (R θ)$
$τ_{r e s} = - F_{s} R$
$= - k R^{2} θ (1)$
Also $τ = I_{c} α (2)$
From (1) & (2)
$\Rightarrow α = \frac{k^{2} θ}{I_{c}}$
$= - \frac{2 k}{m} θ [∴ I_{c} = \frac{m R^{2}}{2}]$
Comparing with standard equation,
$ω^{2} θ = \frac{- 2 k}{m} θ$
$∴ ω = \sqrt{\frac{2 k}{m}}$
$= \sqrt{\frac{2 \times 5}{2}}$
$∴ ω = 2.23 r a d / s$

PHXI14:OSCILLATIONS

364419 A block of mass $m$ attached to one end of the vertical spring produces extension $x$ . If the block is pulled and released, the periodic time of oscillation is

1

2 π \sqrt{\frac{x}{4 g}}

2

2 π \sqrt{\frac{2 x}{g}}

3

2 π \sqrt{\frac{x}{2 g}}

4

2 π \sqrt{\frac{x}{g}}

Explanation:

For vertical spring, the periodic time of oscillation is
$T = 2 π \sqrt{\frac{m}{k}}$
As, restoring force, $| F | = k x$
$\begin{aligned} \Rightarrow k = \frac{F}{x} = \frac{m g}{x} \\ ∴ T = 2 π \sqrt{\frac{\frac{m}{m g}}{x}} = 2 π \sqrt{\frac{x}{g}} \end{aligned}$

MHTCET - 2020

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