QBManager

PHXI14:OSCILLATIONS

364433 A simple second pendulum is mounted in a rocket. Its time period will decrease when the rocket is

1 moving up with uniform velocity

2 moving up with uniform acceleration

3 moving down with uniform acceleration

4 moving around the earth in geostationary orbit

Explanation:

The time period of simple pendulum is
$T = 2 π \sqrt{\frac{l}{g}} \Rightarrow T \propto \frac{1}{\sqrt{g}}$
$T$ will decrease, when $g$ increases and $g$ will increase when rocket moves up with a uniform acceleration.

PHXI14:OSCILLATIONS

364434 Assertion :
The value of acceleration due to gravity is low at the mountain top than at the plane.
Reason :
If a pendulum clock is taken to mountain top, it will gain time.

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.

PHXI14:OSCILLATIONS

364435 The time period of a simple pendulum in a stationary van is $T$ . The time period of a mass attached to a spring is also $T$ . The van accelerates at the rate $5 m s^{- 2} .$ If the new time periods of the pendulum and spring be $T_{p}$ and $T_{s}$ respectively, then

1

T_{p} = T_{s}

2

T_{p} > T_{s}

3

T_{p} < T_{s}

4 Cannot be predicted

Explanation:

Time period of simple pendulum placed in a van accelerating at the rate of a $m s^{- 2}$ is given by
supporting img
$T = 2 π {[\frac{l}{\sqrt{g^{2} + a^{2}}}]}^{\frac{1}{2}}$
However, the time period of mass attached to the spring is
$T = 2 π \sqrt{(\frac{m}{k})}$
It is independent of $g$ as well as $a$ . Hence, when the van accelerates, the time period of the simple pendulum decreases and that of spring remains unchanged.
Hence, $T_{p} < T$ and $T_{s} = T$
i.e. $T_{p} < T_{s}$

PHXI14:OSCILLATIONS

364436 A pendulum of length $1 m$ is released from $θ = 60^{\circ}$ . The rate of change of speed of the bob at $θ = 30^{\circ}$ is (Take, $g = 10 m s^{- 2}$ )

1

10 m s^{- 2}

2

7.5 m s^{- 1}

3

5 m s^{- 2}

4

5 \sqrt{3} m s^{- 2}

Explanation:

For a simple pendulum, time period,
$\begin{aligned} T = 2 π \sqrt{\frac{l}{g}} \Rightarrow \frac{2 π}{T} = \sqrt{\frac{g}{l}} \\ (1) & ∴ ω = \sqrt{\frac{g}{l}}, ω^{2} = \frac{g}{l} \end{aligned}$
Amplitude when angular displacement is $60^{\circ}$
$= \frac{2 π l}{360^{\circ}} \times 60^{\circ} = \frac{2 π l}{6}$
Therefore, displacement when angular displacement is $30^{\circ}$
$\begin{matrix} (2) & = \frac{1}{2} (\frac{2 π l}{6}), y = \frac{π l}{6} \end{matrix}$
Acceleration, $a = - ω^{2} y$
Using Eqs. (1) and (2), we get
$a = - \frac{g}{l} \times \frac{π l}{6} = - \frac{10 \times 3.14}{6}$
$= - 5.2 m s^{- 2} ≃ 5 m s^{- 2}$

PHXI14:OSCILLATIONS

364433 A simple second pendulum is mounted in a rocket. Its time period will decrease when the rocket is

1 moving up with uniform velocity

2 moving up with uniform acceleration

3 moving down with uniform acceleration

4 moving around the earth in geostationary orbit

Explanation:

The time period of simple pendulum is
$T = 2 π \sqrt{\frac{l}{g}} \Rightarrow T \propto \frac{1}{\sqrt{g}}$
$T$ will decrease, when $g$ increases and $g$ will increase when rocket moves up with a uniform acceleration.

PHXI14:OSCILLATIONS

364434 Assertion :
The value of acceleration due to gravity is low at the mountain top than at the plane.
Reason :
If a pendulum clock is taken to mountain top, it will gain time.

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.

PHXI14:OSCILLATIONS

364435 The time period of a simple pendulum in a stationary van is $T$ . The time period of a mass attached to a spring is also $T$ . The van accelerates at the rate $5 m s^{- 2} .$ If the new time periods of the pendulum and spring be $T_{p}$ and $T_{s}$ respectively, then

1

T_{p} = T_{s}

2

T_{p} > T_{s}

3

T_{p} < T_{s}

4 Cannot be predicted

Explanation:

Time period of simple pendulum placed in a van accelerating at the rate of a $m s^{- 2}$ is given by
supporting img
$T = 2 π {[\frac{l}{\sqrt{g^{2} + a^{2}}}]}^{\frac{1}{2}}$
However, the time period of mass attached to the spring is
$T = 2 π \sqrt{(\frac{m}{k})}$
It is independent of $g$ as well as $a$ . Hence, when the van accelerates, the time period of the simple pendulum decreases and that of spring remains unchanged.
Hence, $T_{p} < T$ and $T_{s} = T$
i.e. $T_{p} < T_{s}$

PHXI14:OSCILLATIONS

364436 A pendulum of length $1 m$ is released from $θ = 60^{\circ}$ . The rate of change of speed of the bob at $θ = 30^{\circ}$ is (Take, $g = 10 m s^{- 2}$ )

1

10 m s^{- 2}

2

7.5 m s^{- 1}

3

5 m s^{- 2}

4

5 \sqrt{3} m s^{- 2}

Explanation:

For a simple pendulum, time period,
$\begin{aligned} T = 2 π \sqrt{\frac{l}{g}} \Rightarrow \frac{2 π}{T} = \sqrt{\frac{g}{l}} \\ (1) & ∴ ω = \sqrt{\frac{g}{l}}, ω^{2} = \frac{g}{l} \end{aligned}$
Amplitude when angular displacement is $60^{\circ}$
$= \frac{2 π l}{360^{\circ}} \times 60^{\circ} = \frac{2 π l}{6}$
Therefore, displacement when angular displacement is $30^{\circ}$
$\begin{matrix} (2) & = \frac{1}{2} (\frac{2 π l}{6}), y = \frac{π l}{6} \end{matrix}$
Acceleration, $a = - ω^{2} y$
Using Eqs. (1) and (2), we get
$a = - \frac{g}{l} \times \frac{π l}{6} = - \frac{10 \times 3.14}{6}$
$= - 5.2 m s^{- 2} ≃ 5 m s^{- 2}$

PHXI14:OSCILLATIONS

364433 A simple second pendulum is mounted in a rocket. Its time period will decrease when the rocket is

1 moving up with uniform velocity

2 moving up with uniform acceleration

3 moving down with uniform acceleration

4 moving around the earth in geostationary orbit

Explanation:

The time period of simple pendulum is
$T = 2 π \sqrt{\frac{l}{g}} \Rightarrow T \propto \frac{1}{\sqrt{g}}$
$T$ will decrease, when $g$ increases and $g$ will increase when rocket moves up with a uniform acceleration.

PHXI14:OSCILLATIONS

364434 Assertion :
The value of acceleration due to gravity is low at the mountain top than at the plane.
Reason :
If a pendulum clock is taken to mountain top, it will gain time.

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.

PHXI14:OSCILLATIONS

364435 The time period of a simple pendulum in a stationary van is $T$ . The time period of a mass attached to a spring is also $T$ . The van accelerates at the rate $5 m s^{- 2} .$ If the new time periods of the pendulum and spring be $T_{p}$ and $T_{s}$ respectively, then

1

T_{p} = T_{s}

2

T_{p} > T_{s}

3

T_{p} < T_{s}

4 Cannot be predicted

Explanation:

Time period of simple pendulum placed in a van accelerating at the rate of a $m s^{- 2}$ is given by
supporting img
$T = 2 π {[\frac{l}{\sqrt{g^{2} + a^{2}}}]}^{\frac{1}{2}}$
However, the time period of mass attached to the spring is
$T = 2 π \sqrt{(\frac{m}{k})}$
It is independent of $g$ as well as $a$ . Hence, when the van accelerates, the time period of the simple pendulum decreases and that of spring remains unchanged.
Hence, $T_{p} < T$ and $T_{s} = T$
i.e. $T_{p} < T_{s}$

PHXI14:OSCILLATIONS

364436 A pendulum of length $1 m$ is released from $θ = 60^{\circ}$ . The rate of change of speed of the bob at $θ = 30^{\circ}$ is (Take, $g = 10 m s^{- 2}$ )

1

10 m s^{- 2}

2

7.5 m s^{- 1}

3

5 m s^{- 2}

4

5 \sqrt{3} m s^{- 2}

Explanation:

For a simple pendulum, time period,
$\begin{aligned} T = 2 π \sqrt{\frac{l}{g}} \Rightarrow \frac{2 π}{T} = \sqrt{\frac{g}{l}} \\ (1) & ∴ ω = \sqrt{\frac{g}{l}}, ω^{2} = \frac{g}{l} \end{aligned}$
Amplitude when angular displacement is $60^{\circ}$
$= \frac{2 π l}{360^{\circ}} \times 60^{\circ} = \frac{2 π l}{6}$
Therefore, displacement when angular displacement is $30^{\circ}$
$\begin{matrix} (2) & = \frac{1}{2} (\frac{2 π l}{6}), y = \frac{π l}{6} \end{matrix}$
Acceleration, $a = - ω^{2} y$
Using Eqs. (1) and (2), we get
$a = - \frac{g}{l} \times \frac{π l}{6} = - \frac{10 \times 3.14}{6}$
$= - 5.2 m s^{- 2} ≃ 5 m s^{- 2}$

PHXI14:OSCILLATIONS

364433 A simple second pendulum is mounted in a rocket. Its time period will decrease when the rocket is

1 moving up with uniform velocity

2 moving up with uniform acceleration

3 moving down with uniform acceleration

4 moving around the earth in geostationary orbit

Explanation:

The time period of simple pendulum is
$T = 2 π \sqrt{\frac{l}{g}} \Rightarrow T \propto \frac{1}{\sqrt{g}}$
$T$ will decrease, when $g$ increases and $g$ will increase when rocket moves up with a uniform acceleration.

PHXI14:OSCILLATIONS

364434 Assertion :
The value of acceleration due to gravity is low at the mountain top than at the plane.
Reason :
If a pendulum clock is taken to mountain top, it will gain time.

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.

PHXI14:OSCILLATIONS

364435 The time period of a simple pendulum in a stationary van is $T$ . The time period of a mass attached to a spring is also $T$ . The van accelerates at the rate $5 m s^{- 2} .$ If the new time periods of the pendulum and spring be $T_{p}$ and $T_{s}$ respectively, then

1

T_{p} = T_{s}

2

T_{p} > T_{s}

3

T_{p} < T_{s}

4 Cannot be predicted

Explanation:

Time period of simple pendulum placed in a van accelerating at the rate of a $m s^{- 2}$ is given by
supporting img
$T = 2 π {[\frac{l}{\sqrt{g^{2} + a^{2}}}]}^{\frac{1}{2}}$
However, the time period of mass attached to the spring is
$T = 2 π \sqrt{(\frac{m}{k})}$
It is independent of $g$ as well as $a$ . Hence, when the van accelerates, the time period of the simple pendulum decreases and that of spring remains unchanged.
Hence, $T_{p} < T$ and $T_{s} = T$
i.e. $T_{p} < T_{s}$

PHXI14:OSCILLATIONS

364436 A pendulum of length $1 m$ is released from $θ = 60^{\circ}$ . The rate of change of speed of the bob at $θ = 30^{\circ}$ is (Take, $g = 10 m s^{- 2}$ )

1

10 m s^{- 2}

2

7.5 m s^{- 1}

3

5 m s^{- 2}

4

5 \sqrt{3} m s^{- 2}

Explanation:

For a simple pendulum, time period,
$\begin{aligned} T = 2 π \sqrt{\frac{l}{g}} \Rightarrow \frac{2 π}{T} = \sqrt{\frac{g}{l}} \\ (1) & ∴ ω = \sqrt{\frac{g}{l}}, ω^{2} = \frac{g}{l} \end{aligned}$
Amplitude when angular displacement is $60^{\circ}$
$= \frac{2 π l}{360^{\circ}} \times 60^{\circ} = \frac{2 π l}{6}$
Therefore, displacement when angular displacement is $30^{\circ}$
$\begin{matrix} (2) & = \frac{1}{2} (\frac{2 π l}{6}), y = \frac{π l}{6} \end{matrix}$
Acceleration, $a = - ω^{2} y$
Using Eqs. (1) and (2), we get
$a = - \frac{g}{l} \times \frac{π l}{6} = - \frac{10 \times 3.14}{6}$
$= - 5.2 m s^{- 2} ≃ 5 m s^{- 2}$

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: