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PHXI14:OSCILLATIONS

364338 A rectangular block of mass ' $M$ ' and cross sectional area ' $A$ ' floats on a liquid of density ' $ρ$ '. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency ' $n$ ' then

1

n α A

2

n α A^{2}

3

n α \sqrt{A}

4

n α A^{3}

Explanation:

Time period of SHM is given by
$T = 2 π \sqrt{\frac{l}{g}}$
(where $l$ is length of rectangular block) According to law of floatation,
Weight of block = weight of the displaced liquid
$\Rightarrow M g = A l ρ g \Rightarrow l = \frac{M}{A ρ}$
using value of ' $l$ ' in equation (1), we get
$T = 2 π \sqrt{\frac{M}{A ρ g}}$
Frequency $n = \frac{1}{T}$
$\begin{aligned} n = \frac{1}{2 π} \sqrt{\frac{A ρ g}{M}} \\ \Rightarrow n \propto \sqrt{A} . \end{aligned}$

MHTCET - 2022

PHXI14:OSCILLATIONS

364339 The period of oscillations of mass $m = 200 g$ poured into a bent tube (see figure), whose right arm makes an angle $30^{\circ}$ with the vertical and whose left arm is vertical, is $T$ seconds. The cross-sectional area of the tube is $A = 0.5 {c m}^{2}$ . The value of $10 T$ is Neglect viscosity.
supporting img

1 8

2 11

3 15

4 20

Explanation:

$T = \sqrt{\frac{m}{A ρ g (\sin θ_{1} + \sin θ_{2})}}$
$= \sqrt{\frac{0.2}{0.5 \times 10^{- 4} 9.8 (\sin 90^{\circ} + \sin 60^{\circ}) \times 13.6 \times 10^{3}}}$
$T = \sqrt{\frac{0.2}{0.5 \times 9.8 (1 + \frac{\sqrt{3}}{2}) \times 13.6 \times 10^{- 1}}}$
$= 0.8 s$
$∴ 10 T = 8$

PHXI14:OSCILLATIONS

364340 Assertion :
The height of a liquid column in a U-tube is $0.3 m$ . If the liquid in one of the limbs is depressed and then released the time period of a liquid column will be $1.1 \sec$ .
Reason :
Value of density of liquid is needed to calculate the period.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$T = 2 π \sqrt{\frac{h}{g}} = 2 π \sqrt{\frac{0.3}{9.8}} = 1.1 s$
From the expression we can see that time period does depend on density of the liquid. Assertion is correct and reason is wrong. So correct option is (3).
supporting img

PHXI14:OSCILLATIONS

364341 Motion of an oscillating liquid column in a U tube is

1 Non-periodic

2 Simple harmonic motion with period

3 Simple harmonic and time-period is directlyproportional to the density of the liquid

4 Simple harmonic and time-period is independent of the density of the liquid

Explanation:

Motion of an oscillating liquid column in a $U$ tube is SHM with period, $T = 2 π \sqrt{\frac{l}{g}}$ , where $l$ is the height of liquid column in one arm of $U$ tube in equilibrium position of liquid. Therefore, $T$ is independent of density of liquid.

NCERT Exemplar

PHXI14:OSCILLATIONS

364342 A block of mass $M$ is performing SHM with amplitude A on a smooth horizontal surface. At the extreme position a small block of mass $m$ falls vertically and sticks to $M$ . New amplitude of oscillation will be

1

A

2

A \frac{M + m}{m}

3

A \sqrt{\frac{M}{M + m}}

4

A \frac{M}{M + m}

Explanation:

At extreme position velocity of $M$ will be zero. So amplitude remains same.

PHXI14:OSCILLATIONS

364338 A rectangular block of mass ' $M$ ' and cross sectional area ' $A$ ' floats on a liquid of density ' $ρ$ '. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency ' $n$ ' then

1

n α A

2

n α A^{2}

3

n α \sqrt{A}

4

n α A^{3}

Explanation:

Time period of SHM is given by
$T = 2 π \sqrt{\frac{l}{g}}$
(where $l$ is length of rectangular block) According to law of floatation,
Weight of block = weight of the displaced liquid
$\Rightarrow M g = A l ρ g \Rightarrow l = \frac{M}{A ρ}$
using value of ' $l$ ' in equation (1), we get
$T = 2 π \sqrt{\frac{M}{A ρ g}}$
Frequency $n = \frac{1}{T}$
$\begin{aligned} n = \frac{1}{2 π} \sqrt{\frac{A ρ g}{M}} \\ \Rightarrow n \propto \sqrt{A} . \end{aligned}$

MHTCET - 2022

PHXI14:OSCILLATIONS

364339 The period of oscillations of mass $m = 200 g$ poured into a bent tube (see figure), whose right arm makes an angle $30^{\circ}$ with the vertical and whose left arm is vertical, is $T$ seconds. The cross-sectional area of the tube is $A = 0.5 {c m}^{2}$ . The value of $10 T$ is Neglect viscosity.
supporting img

1 8

2 11

3 15

4 20

Explanation:

$T = \sqrt{\frac{m}{A ρ g (\sin θ_{1} + \sin θ_{2})}}$
$= \sqrt{\frac{0.2}{0.5 \times 10^{- 4} 9.8 (\sin 90^{\circ} + \sin 60^{\circ}) \times 13.6 \times 10^{3}}}$
$T = \sqrt{\frac{0.2}{0.5 \times 9.8 (1 + \frac{\sqrt{3}}{2}) \times 13.6 \times 10^{- 1}}}$
$= 0.8 s$
$∴ 10 T = 8$

PHXI14:OSCILLATIONS

364340 Assertion :
The height of a liquid column in a U-tube is $0.3 m$ . If the liquid in one of the limbs is depressed and then released the time period of a liquid column will be $1.1 \sec$ .
Reason :
Value of density of liquid is needed to calculate the period.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$T = 2 π \sqrt{\frac{h}{g}} = 2 π \sqrt{\frac{0.3}{9.8}} = 1.1 s$
From the expression we can see that time period does depend on density of the liquid. Assertion is correct and reason is wrong. So correct option is (3).
supporting img

PHXI14:OSCILLATIONS

364341 Motion of an oscillating liquid column in a U tube is

1 Non-periodic

2 Simple harmonic motion with period

3 Simple harmonic and time-period is directlyproportional to the density of the liquid

4 Simple harmonic and time-period is independent of the density of the liquid

Explanation:

Motion of an oscillating liquid column in a $U$ tube is SHM with period, $T = 2 π \sqrt{\frac{l}{g}}$ , where $l$ is the height of liquid column in one arm of $U$ tube in equilibrium position of liquid. Therefore, $T$ is independent of density of liquid.

NCERT Exemplar

PHXI14:OSCILLATIONS

364342 A block of mass $M$ is performing SHM with amplitude A on a smooth horizontal surface. At the extreme position a small block of mass $m$ falls vertically and sticks to $M$ . New amplitude of oscillation will be

1

A

2

A \frac{M + m}{m}

3

A \sqrt{\frac{M}{M + m}}

4

A \frac{M}{M + m}

Explanation:

At extreme position velocity of $M$ will be zero. So amplitude remains same.

PHXI14:OSCILLATIONS

364338 A rectangular block of mass ' $M$ ' and cross sectional area ' $A$ ' floats on a liquid of density ' $ρ$ '. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency ' $n$ ' then

1

n α A

2

n α A^{2}

3

n α \sqrt{A}

4

n α A^{3}

Explanation:

Time period of SHM is given by
$T = 2 π \sqrt{\frac{l}{g}}$
(where $l$ is length of rectangular block) According to law of floatation,
Weight of block = weight of the displaced liquid
$\Rightarrow M g = A l ρ g \Rightarrow l = \frac{M}{A ρ}$
using value of ' $l$ ' in equation (1), we get
$T = 2 π \sqrt{\frac{M}{A ρ g}}$
Frequency $n = \frac{1}{T}$
$\begin{aligned} n = \frac{1}{2 π} \sqrt{\frac{A ρ g}{M}} \\ \Rightarrow n \propto \sqrt{A} . \end{aligned}$

MHTCET - 2022

PHXI14:OSCILLATIONS

364339 The period of oscillations of mass $m = 200 g$ poured into a bent tube (see figure), whose right arm makes an angle $30^{\circ}$ with the vertical and whose left arm is vertical, is $T$ seconds. The cross-sectional area of the tube is $A = 0.5 {c m}^{2}$ . The value of $10 T$ is Neglect viscosity.
supporting img

1 8

2 11

3 15

4 20

Explanation:

$T = \sqrt{\frac{m}{A ρ g (\sin θ_{1} + \sin θ_{2})}}$
$= \sqrt{\frac{0.2}{0.5 \times 10^{- 4} 9.8 (\sin 90^{\circ} + \sin 60^{\circ}) \times 13.6 \times 10^{3}}}$
$T = \sqrt{\frac{0.2}{0.5 \times 9.8 (1 + \frac{\sqrt{3}}{2}) \times 13.6 \times 10^{- 1}}}$
$= 0.8 s$
$∴ 10 T = 8$

PHXI14:OSCILLATIONS

364340 Assertion :
The height of a liquid column in a U-tube is $0.3 m$ . If the liquid in one of the limbs is depressed and then released the time period of a liquid column will be $1.1 \sec$ .
Reason :
Value of density of liquid is needed to calculate the period.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$T = 2 π \sqrt{\frac{h}{g}} = 2 π \sqrt{\frac{0.3}{9.8}} = 1.1 s$
From the expression we can see that time period does depend on density of the liquid. Assertion is correct and reason is wrong. So correct option is (3).
supporting img

PHXI14:OSCILLATIONS

364341 Motion of an oscillating liquid column in a U tube is

1 Non-periodic

2 Simple harmonic motion with period

3 Simple harmonic and time-period is directlyproportional to the density of the liquid

4 Simple harmonic and time-period is independent of the density of the liquid

Explanation:

Motion of an oscillating liquid column in a $U$ tube is SHM with period, $T = 2 π \sqrt{\frac{l}{g}}$ , where $l$ is the height of liquid column in one arm of $U$ tube in equilibrium position of liquid. Therefore, $T$ is independent of density of liquid.

NCERT Exemplar

PHXI14:OSCILLATIONS

364342 A block of mass $M$ is performing SHM with amplitude A on a smooth horizontal surface. At the extreme position a small block of mass $m$ falls vertically and sticks to $M$ . New amplitude of oscillation will be

1

A

2

A \frac{M + m}{m}

3

A \sqrt{\frac{M}{M + m}}

4

A \frac{M}{M + m}

Explanation:

At extreme position velocity of $M$ will be zero. So amplitude remains same.

PHXI14:OSCILLATIONS

364338 A rectangular block of mass ' $M$ ' and cross sectional area ' $A$ ' floats on a liquid of density ' $ρ$ '. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency ' $n$ ' then

1

n α A

2

n α A^{2}

3

n α \sqrt{A}

4

n α A^{3}

Explanation:

Time period of SHM is given by
$T = 2 π \sqrt{\frac{l}{g}}$
(where $l$ is length of rectangular block) According to law of floatation,
Weight of block = weight of the displaced liquid
$\Rightarrow M g = A l ρ g \Rightarrow l = \frac{M}{A ρ}$
using value of ' $l$ ' in equation (1), we get
$T = 2 π \sqrt{\frac{M}{A ρ g}}$
Frequency $n = \frac{1}{T}$
$\begin{aligned} n = \frac{1}{2 π} \sqrt{\frac{A ρ g}{M}} \\ \Rightarrow n \propto \sqrt{A} . \end{aligned}$

MHTCET - 2022

PHXI14:OSCILLATIONS

364339 The period of oscillations of mass $m = 200 g$ poured into a bent tube (see figure), whose right arm makes an angle $30^{\circ}$ with the vertical and whose left arm is vertical, is $T$ seconds. The cross-sectional area of the tube is $A = 0.5 {c m}^{2}$ . The value of $10 T$ is Neglect viscosity.
supporting img

1 8

2 11

3 15

4 20

Explanation:

$T = \sqrt{\frac{m}{A ρ g (\sin θ_{1} + \sin θ_{2})}}$
$= \sqrt{\frac{0.2}{0.5 \times 10^{- 4} 9.8 (\sin 90^{\circ} + \sin 60^{\circ}) \times 13.6 \times 10^{3}}}$
$T = \sqrt{\frac{0.2}{0.5 \times 9.8 (1 + \frac{\sqrt{3}}{2}) \times 13.6 \times 10^{- 1}}}$
$= 0.8 s$
$∴ 10 T = 8$

PHXI14:OSCILLATIONS

364340 Assertion :
The height of a liquid column in a U-tube is $0.3 m$ . If the liquid in one of the limbs is depressed and then released the time period of a liquid column will be $1.1 \sec$ .
Reason :
Value of density of liquid is needed to calculate the period.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$T = 2 π \sqrt{\frac{h}{g}} = 2 π \sqrt{\frac{0.3}{9.8}} = 1.1 s$
From the expression we can see that time period does depend on density of the liquid. Assertion is correct and reason is wrong. So correct option is (3).
supporting img

PHXI14:OSCILLATIONS

364341 Motion of an oscillating liquid column in a U tube is

1 Non-periodic

2 Simple harmonic motion with period

3 Simple harmonic and time-period is directlyproportional to the density of the liquid

4 Simple harmonic and time-period is independent of the density of the liquid

Explanation:

Motion of an oscillating liquid column in a $U$ tube is SHM with period, $T = 2 π \sqrt{\frac{l}{g}}$ , where $l$ is the height of liquid column in one arm of $U$ tube in equilibrium position of liquid. Therefore, $T$ is independent of density of liquid.

NCERT Exemplar

PHXI14:OSCILLATIONS

364342 A block of mass $M$ is performing SHM with amplitude A on a smooth horizontal surface. At the extreme position a small block of mass $m$ falls vertically and sticks to $M$ . New amplitude of oscillation will be

1

A

2

A \frac{M + m}{m}

3

A \sqrt{\frac{M}{M + m}}

4

A \frac{M}{M + m}

Explanation:

At extreme position velocity of $M$ will be zero. So amplitude remains same.

PHXI14:OSCILLATIONS

364338 A rectangular block of mass ' $M$ ' and cross sectional area ' $A$ ' floats on a liquid of density ' $ρ$ '. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency ' $n$ ' then

1

n α A

2

n α A^{2}

3

n α \sqrt{A}

4

n α A^{3}

Explanation:

Time period of SHM is given by
$T = 2 π \sqrt{\frac{l}{g}}$
(where $l$ is length of rectangular block) According to law of floatation,
Weight of block = weight of the displaced liquid
$\Rightarrow M g = A l ρ g \Rightarrow l = \frac{M}{A ρ}$
using value of ' $l$ ' in equation (1), we get
$T = 2 π \sqrt{\frac{M}{A ρ g}}$
Frequency $n = \frac{1}{T}$
$\begin{aligned} n = \frac{1}{2 π} \sqrt{\frac{A ρ g}{M}} \\ \Rightarrow n \propto \sqrt{A} . \end{aligned}$

MHTCET - 2022

PHXI14:OSCILLATIONS

364339 The period of oscillations of mass $m = 200 g$ poured into a bent tube (see figure), whose right arm makes an angle $30^{\circ}$ with the vertical and whose left arm is vertical, is $T$ seconds. The cross-sectional area of the tube is $A = 0.5 {c m}^{2}$ . The value of $10 T$ is Neglect viscosity.
supporting img

1 8

2 11

3 15

4 20

Explanation:

$T = \sqrt{\frac{m}{A ρ g (\sin θ_{1} + \sin θ_{2})}}$
$= \sqrt{\frac{0.2}{0.5 \times 10^{- 4} 9.8 (\sin 90^{\circ} + \sin 60^{\circ}) \times 13.6 \times 10^{3}}}$
$T = \sqrt{\frac{0.2}{0.5 \times 9.8 (1 + \frac{\sqrt{3}}{2}) \times 13.6 \times 10^{- 1}}}$
$= 0.8 s$
$∴ 10 T = 8$

PHXI14:OSCILLATIONS

364340 Assertion :
The height of a liquid column in a U-tube is $0.3 m$ . If the liquid in one of the limbs is depressed and then released the time period of a liquid column will be $1.1 \sec$ .
Reason :
Value of density of liquid is needed to calculate the period.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$T = 2 π \sqrt{\frac{h}{g}} = 2 π \sqrt{\frac{0.3}{9.8}} = 1.1 s$
From the expression we can see that time period does depend on density of the liquid. Assertion is correct and reason is wrong. So correct option is (3).
supporting img

PHXI14:OSCILLATIONS

364341 Motion of an oscillating liquid column in a U tube is

1 Non-periodic

2 Simple harmonic motion with period

3 Simple harmonic and time-period is directlyproportional to the density of the liquid

4 Simple harmonic and time-period is independent of the density of the liquid

Explanation:

Motion of an oscillating liquid column in a $U$ tube is SHM with period, $T = 2 π \sqrt{\frac{l}{g}}$ , where $l$ is the height of liquid column in one arm of $U$ tube in equilibrium position of liquid. Therefore, $T$ is independent of density of liquid.

NCERT Exemplar

PHXI14:OSCILLATIONS

364342 A block of mass $M$ is performing SHM with amplitude A on a smooth horizontal surface. At the extreme position a small block of mass $m$ falls vertically and sticks to $M$ . New amplitude of oscillation will be

1

A

2

A \frac{M + m}{m}

3

A \sqrt{\frac{M}{M + m}}

4

A \frac{M}{M + m}

Explanation:

At extreme position velocity of $M$ will be zero. So amplitude remains same.

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