Pendulum (Simple Pendulum and Compound Pendulum)
Oscillations

140409 A simple pendulum having bob of mass $m$ and length of string $L$ has time period of $T$. If the mass of the bob in doubled and the length of the string in halved, then the time period of this pendulum will be

1 $\mathrm{T}$
2 $\mathrm{T} / \sqrt{2}$
3 $2 \mathrm{~T}$
4 $\sqrt{2} \mathrm{~T}$
Oscillations

140411 Assume there are two identical simple pendulum Clocks-1 is placed on the earth and Clock-2 is placed on a space station located at a height $h$ above the earth surface. Clock-1 and Clock-2 operate at time periods $4 s$ and $6 s$ respectively. Then the value of $h$ is -
(Consider radius of earth $R_{E}=6400 \mathrm{~km}$ and $g$ on earth $10 \mathrm{~m} / \mathrm{s}^{2}$ )

1 $1200 \mathrm{~km}$
2 $1600 \mathrm{~km}$
3 $3200 \mathrm{~km}$
4 $4800 \mathrm{~km}$
Oscillations

140413 A simple pendulum of length $1 \mathrm{~m}$ and having a bob of mass $100 \mathrm{~g}$ is suspended in a car. moving on a circular track of radius $100 \mathrm{~m}$ with uniform speed $10 \mathrm{~m} / \mathrm{s}$ If the pendulum makes small oscillation in a radial direction about its equilibrium position, then its time period can be given by $T=2 \pi / \alpha^{1 / 4}$ The value of $\alpha$ is
[Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ ]

1 11
2 110
3 101
4 1100
Oscillations

140414 The time periods of a given simple pendulum on the surface of earth $T_{1}$. In international space station $T_{2}$ and on the surface of the moon $T_{3}$ are related as

1 $\mathrm{T}_{2} \lt \mathrm{T}_{1} \lt \mathrm{T}_{3}$
2 $T_{2}>T_{1}>T_{3}$
3 $\mathrm{T}_{1}>\mathrm{T}_{2}>\mathrm{T}_{3}$
4 $\mathrm{T}_{1} \lt \mathrm{T}_{2} \lt \mathrm{T}_{3}$
Oscillations

140415 The time period of a pendulum in a stationary lift is ' $T$ '. If the lift moves upwards with acceleration $\frac{g}{6}$, the time period of the pendulum becomes

1 $\left(\sqrt{\frac{6}{7}}\right) \mathrm{T}$
2 $\left(\frac{\sqrt{5}}{2}\right) \mathrm{T}$
3 $\left(\frac{\sqrt{6}}{7}\right) \mathrm{T}$
4 $\left(\frac{2}{\sqrt{5}}\right) \mathrm{T}$
Oscillations

140409 A simple pendulum having bob of mass $m$ and length of string $L$ has time period of $T$. If the mass of the bob in doubled and the length of the string in halved, then the time period of this pendulum will be

1 $\mathrm{T}$
2 $\mathrm{T} / \sqrt{2}$
3 $2 \mathrm{~T}$
4 $\sqrt{2} \mathrm{~T}$
Oscillations

140411 Assume there are two identical simple pendulum Clocks-1 is placed on the earth and Clock-2 is placed on a space station located at a height $h$ above the earth surface. Clock-1 and Clock-2 operate at time periods $4 s$ and $6 s$ respectively. Then the value of $h$ is -
(Consider radius of earth $R_{E}=6400 \mathrm{~km}$ and $g$ on earth $10 \mathrm{~m} / \mathrm{s}^{2}$ )

1 $1200 \mathrm{~km}$
2 $1600 \mathrm{~km}$
3 $3200 \mathrm{~km}$
4 $4800 \mathrm{~km}$
Oscillations

140413 A simple pendulum of length $1 \mathrm{~m}$ and having a bob of mass $100 \mathrm{~g}$ is suspended in a car. moving on a circular track of radius $100 \mathrm{~m}$ with uniform speed $10 \mathrm{~m} / \mathrm{s}$ If the pendulum makes small oscillation in a radial direction about its equilibrium position, then its time period can be given by $T=2 \pi / \alpha^{1 / 4}$ The value of $\alpha$ is
[Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ ]

1 11
2 110
3 101
4 1100
Oscillations

140414 The time periods of a given simple pendulum on the surface of earth $T_{1}$. In international space station $T_{2}$ and on the surface of the moon $T_{3}$ are related as

1 $\mathrm{T}_{2} \lt \mathrm{T}_{1} \lt \mathrm{T}_{3}$
2 $T_{2}>T_{1}>T_{3}$
3 $\mathrm{T}_{1}>\mathrm{T}_{2}>\mathrm{T}_{3}$
4 $\mathrm{T}_{1} \lt \mathrm{T}_{2} \lt \mathrm{T}_{3}$
Oscillations

140415 The time period of a pendulum in a stationary lift is ' $T$ '. If the lift moves upwards with acceleration $\frac{g}{6}$, the time period of the pendulum becomes

1 $\left(\sqrt{\frac{6}{7}}\right) \mathrm{T}$
2 $\left(\frac{\sqrt{5}}{2}\right) \mathrm{T}$
3 $\left(\frac{\sqrt{6}}{7}\right) \mathrm{T}$
4 $\left(\frac{2}{\sqrt{5}}\right) \mathrm{T}$
Oscillations

140409 A simple pendulum having bob of mass $m$ and length of string $L$ has time period of $T$. If the mass of the bob in doubled and the length of the string in halved, then the time period of this pendulum will be

1 $\mathrm{T}$
2 $\mathrm{T} / \sqrt{2}$
3 $2 \mathrm{~T}$
4 $\sqrt{2} \mathrm{~T}$
Oscillations

140411 Assume there are two identical simple pendulum Clocks-1 is placed on the earth and Clock-2 is placed on a space station located at a height $h$ above the earth surface. Clock-1 and Clock-2 operate at time periods $4 s$ and $6 s$ respectively. Then the value of $h$ is -
(Consider radius of earth $R_{E}=6400 \mathrm{~km}$ and $g$ on earth $10 \mathrm{~m} / \mathrm{s}^{2}$ )

1 $1200 \mathrm{~km}$
2 $1600 \mathrm{~km}$
3 $3200 \mathrm{~km}$
4 $4800 \mathrm{~km}$
Oscillations

140413 A simple pendulum of length $1 \mathrm{~m}$ and having a bob of mass $100 \mathrm{~g}$ is suspended in a car. moving on a circular track of radius $100 \mathrm{~m}$ with uniform speed $10 \mathrm{~m} / \mathrm{s}$ If the pendulum makes small oscillation in a radial direction about its equilibrium position, then its time period can be given by $T=2 \pi / \alpha^{1 / 4}$ The value of $\alpha$ is
[Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ ]

1 11
2 110
3 101
4 1100
Oscillations

140414 The time periods of a given simple pendulum on the surface of earth $T_{1}$. In international space station $T_{2}$ and on the surface of the moon $T_{3}$ are related as

1 $\mathrm{T}_{2} \lt \mathrm{T}_{1} \lt \mathrm{T}_{3}$
2 $T_{2}>T_{1}>T_{3}$
3 $\mathrm{T}_{1}>\mathrm{T}_{2}>\mathrm{T}_{3}$
4 $\mathrm{T}_{1} \lt \mathrm{T}_{2} \lt \mathrm{T}_{3}$
Oscillations

140415 The time period of a pendulum in a stationary lift is ' $T$ '. If the lift moves upwards with acceleration $\frac{g}{6}$, the time period of the pendulum becomes

1 $\left(\sqrt{\frac{6}{7}}\right) \mathrm{T}$
2 $\left(\frac{\sqrt{5}}{2}\right) \mathrm{T}$
3 $\left(\frac{\sqrt{6}}{7}\right) \mathrm{T}$
4 $\left(\frac{2}{\sqrt{5}}\right) \mathrm{T}$
Oscillations

140409 A simple pendulum having bob of mass $m$ and length of string $L$ has time period of $T$. If the mass of the bob in doubled and the length of the string in halved, then the time period of this pendulum will be

1 $\mathrm{T}$
2 $\mathrm{T} / \sqrt{2}$
3 $2 \mathrm{~T}$
4 $\sqrt{2} \mathrm{~T}$
Oscillations

140411 Assume there are two identical simple pendulum Clocks-1 is placed on the earth and Clock-2 is placed on a space station located at a height $h$ above the earth surface. Clock-1 and Clock-2 operate at time periods $4 s$ and $6 s$ respectively. Then the value of $h$ is -
(Consider radius of earth $R_{E}=6400 \mathrm{~km}$ and $g$ on earth $10 \mathrm{~m} / \mathrm{s}^{2}$ )

1 $1200 \mathrm{~km}$
2 $1600 \mathrm{~km}$
3 $3200 \mathrm{~km}$
4 $4800 \mathrm{~km}$
Oscillations

140413 A simple pendulum of length $1 \mathrm{~m}$ and having a bob of mass $100 \mathrm{~g}$ is suspended in a car. moving on a circular track of radius $100 \mathrm{~m}$ with uniform speed $10 \mathrm{~m} / \mathrm{s}$ If the pendulum makes small oscillation in a radial direction about its equilibrium position, then its time period can be given by $T=2 \pi / \alpha^{1 / 4}$ The value of $\alpha$ is
[Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ ]

1 11
2 110
3 101
4 1100
Oscillations

140414 The time periods of a given simple pendulum on the surface of earth $T_{1}$. In international space station $T_{2}$ and on the surface of the moon $T_{3}$ are related as

1 $\mathrm{T}_{2} \lt \mathrm{T}_{1} \lt \mathrm{T}_{3}$
2 $T_{2}>T_{1}>T_{3}$
3 $\mathrm{T}_{1}>\mathrm{T}_{2}>\mathrm{T}_{3}$
4 $\mathrm{T}_{1} \lt \mathrm{T}_{2} \lt \mathrm{T}_{3}$
Oscillations

140415 The time period of a pendulum in a stationary lift is ' $T$ '. If the lift moves upwards with acceleration $\frac{g}{6}$, the time period of the pendulum becomes

1 $\left(\sqrt{\frac{6}{7}}\right) \mathrm{T}$
2 $\left(\frac{\sqrt{5}}{2}\right) \mathrm{T}$
3 $\left(\frac{\sqrt{6}}{7}\right) \mathrm{T}$
4 $\left(\frac{2}{\sqrt{5}}\right) \mathrm{T}$
Oscillations

140409 A simple pendulum having bob of mass $m$ and length of string $L$ has time period of $T$. If the mass of the bob in doubled and the length of the string in halved, then the time period of this pendulum will be

1 $\mathrm{T}$
2 $\mathrm{T} / \sqrt{2}$
3 $2 \mathrm{~T}$
4 $\sqrt{2} \mathrm{~T}$
Oscillations

140411 Assume there are two identical simple pendulum Clocks-1 is placed on the earth and Clock-2 is placed on a space station located at a height $h$ above the earth surface. Clock-1 and Clock-2 operate at time periods $4 s$ and $6 s$ respectively. Then the value of $h$ is -
(Consider radius of earth $R_{E}=6400 \mathrm{~km}$ and $g$ on earth $10 \mathrm{~m} / \mathrm{s}^{2}$ )

1 $1200 \mathrm{~km}$
2 $1600 \mathrm{~km}$
3 $3200 \mathrm{~km}$
4 $4800 \mathrm{~km}$
Oscillations

140413 A simple pendulum of length $1 \mathrm{~m}$ and having a bob of mass $100 \mathrm{~g}$ is suspended in a car. moving on a circular track of radius $100 \mathrm{~m}$ with uniform speed $10 \mathrm{~m} / \mathrm{s}$ If the pendulum makes small oscillation in a radial direction about its equilibrium position, then its time period can be given by $T=2 \pi / \alpha^{1 / 4}$ The value of $\alpha$ is
[Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ ]

1 11
2 110
3 101
4 1100
Oscillations

140414 The time periods of a given simple pendulum on the surface of earth $T_{1}$. In international space station $T_{2}$ and on the surface of the moon $T_{3}$ are related as

1 $\mathrm{T}_{2} \lt \mathrm{T}_{1} \lt \mathrm{T}_{3}$
2 $T_{2}>T_{1}>T_{3}$
3 $\mathrm{T}_{1}>\mathrm{T}_{2}>\mathrm{T}_{3}$
4 $\mathrm{T}_{1} \lt \mathrm{T}_{2} \lt \mathrm{T}_{3}$
Oscillations

140415 The time period of a pendulum in a stationary lift is ' $T$ '. If the lift moves upwards with acceleration $\frac{g}{6}$, the time period of the pendulum becomes

1 $\left(\sqrt{\frac{6}{7}}\right) \mathrm{T}$
2 $\left(\frac{\sqrt{5}}{2}\right) \mathrm{T}$
3 $\left(\frac{\sqrt{6}}{7}\right) \mathrm{T}$
4 $\left(\frac{2}{\sqrt{5}}\right) \mathrm{T}$