364450
Maximum time period of any simple pendulum on the earth is
1 \(180.5 \mathrm{~min}\)
2 \(100 \mathrm{~min}\)
3 \(90.5 \mathrm{~min}\)
4 \(1.5 \mathrm{~min}\)
Explanation:
The maximum time period of any pendulum on the earth is \(1.5 \mathrm{~min}\).
PHXI14:OSCILLATIONS
364451
The period of a simple pendulum, whose bob is a hollow metallic sphere, is \(T_{1}\). The period is \(T_{2}\) when the bob is filled with sand, \(T_{3}\) when it is filled with mercury and \(T_{4}\) when it is half filled with mercury. Which of the following is true?
1 \(T_{1}=T_{2}=T_{3}>T_{4}\)
2 \(T_{1}=T_{2}=T_{3}>T_{4}\)
3 \(T_{1}=T_{2}=T_{3} < T_{4}\)
4 \(T_{1}>T_{2}>T_{3}=T_{4}\)
Explanation:
Time period of pendulum doesn't depends upon mass but it depends upon length (distance between point of suspension and centre of mass). In first three cases length are same so \(T_{1}=T_{2}=T_{3}\) but in last case centre of mass lowers which in turn increases the length. So option (3) is correct
PHXI14:OSCILLATIONS
364452
A simple pendulum of length \(l\) and having a bob of mass \(M\) is suspended in a car. The car is moving on a circular track of radius \(R\) with a uniform speed \(v\). If the pendulum makes small oscillations about its equilibrium position, what will be its time period?
The bob is subjected to two simultaneous, accelerations perpendicular to each other viz acceleration due to gravity \(g\) and radial acceleration \(a_{R}=\dfrac{v^{2}}{R}\) towards the centre of the circular path. \(\therefore\) Effective acceleration \(a_{e f f}=\sqrt{g^{2}+\left(\dfrac{v^{2}}{R}\right)^{2}}\) \(\therefore\) Time period of the simple pendulum \( = 2\pi \sqrt {\frac{l}{{{a_{{\text{eff }}}}}}} \) \( = 2\pi \sqrt {\frac{l}{{\sqrt {{g^2} + {{\left( {\frac{{{v^2}}}{R}} \right)}^2}} }}} = 2\pi \sqrt {\frac{l}{{\sqrt {{g^2} + \frac{{{v^4}}}{{{R^2}}}} }}} \)
PHXI14:OSCILLATIONS
364453
The acceleration due to gravity on a planet is \(3{\rm{/}}2\) times that on the earth. If length of a seconds pendulum on earth is \(1\;m\), length of seconds pendulum on the planet is
1 \(0.7\;m\)
2 \(1\;m\)
3 \(1.7\;m\)
4 \(1.5\;m\)
Explanation:
\(\sqrt{\dfrac{l}{g}}=\) constant \(\Rightarrow \dfrac{l_{1}}{g_{1}}=\dfrac{l_{2}}{g_{2}}\) \(\Rightarrow \dfrac{1}{g}=\dfrac{l}{3 g / 2}\) \( \Rightarrow l = 1.5\;m\)
364450
Maximum time period of any simple pendulum on the earth is
1 \(180.5 \mathrm{~min}\)
2 \(100 \mathrm{~min}\)
3 \(90.5 \mathrm{~min}\)
4 \(1.5 \mathrm{~min}\)
Explanation:
The maximum time period of any pendulum on the earth is \(1.5 \mathrm{~min}\).
PHXI14:OSCILLATIONS
364451
The period of a simple pendulum, whose bob is a hollow metallic sphere, is \(T_{1}\). The period is \(T_{2}\) when the bob is filled with sand, \(T_{3}\) when it is filled with mercury and \(T_{4}\) when it is half filled with mercury. Which of the following is true?
1 \(T_{1}=T_{2}=T_{3}>T_{4}\)
2 \(T_{1}=T_{2}=T_{3}>T_{4}\)
3 \(T_{1}=T_{2}=T_{3} < T_{4}\)
4 \(T_{1}>T_{2}>T_{3}=T_{4}\)
Explanation:
Time period of pendulum doesn't depends upon mass but it depends upon length (distance between point of suspension and centre of mass). In first three cases length are same so \(T_{1}=T_{2}=T_{3}\) but in last case centre of mass lowers which in turn increases the length. So option (3) is correct
PHXI14:OSCILLATIONS
364452
A simple pendulum of length \(l\) and having a bob of mass \(M\) is suspended in a car. The car is moving on a circular track of radius \(R\) with a uniform speed \(v\). If the pendulum makes small oscillations about its equilibrium position, what will be its time period?
The bob is subjected to two simultaneous, accelerations perpendicular to each other viz acceleration due to gravity \(g\) and radial acceleration \(a_{R}=\dfrac{v^{2}}{R}\) towards the centre of the circular path. \(\therefore\) Effective acceleration \(a_{e f f}=\sqrt{g^{2}+\left(\dfrac{v^{2}}{R}\right)^{2}}\) \(\therefore\) Time period of the simple pendulum \( = 2\pi \sqrt {\frac{l}{{{a_{{\text{eff }}}}}}} \) \( = 2\pi \sqrt {\frac{l}{{\sqrt {{g^2} + {{\left( {\frac{{{v^2}}}{R}} \right)}^2}} }}} = 2\pi \sqrt {\frac{l}{{\sqrt {{g^2} + \frac{{{v^4}}}{{{R^2}}}} }}} \)
PHXI14:OSCILLATIONS
364453
The acceleration due to gravity on a planet is \(3{\rm{/}}2\) times that on the earth. If length of a seconds pendulum on earth is \(1\;m\), length of seconds pendulum on the planet is
1 \(0.7\;m\)
2 \(1\;m\)
3 \(1.7\;m\)
4 \(1.5\;m\)
Explanation:
\(\sqrt{\dfrac{l}{g}}=\) constant \(\Rightarrow \dfrac{l_{1}}{g_{1}}=\dfrac{l_{2}}{g_{2}}\) \(\Rightarrow \dfrac{1}{g}=\dfrac{l}{3 g / 2}\) \( \Rightarrow l = 1.5\;m\)
364450
Maximum time period of any simple pendulum on the earth is
1 \(180.5 \mathrm{~min}\)
2 \(100 \mathrm{~min}\)
3 \(90.5 \mathrm{~min}\)
4 \(1.5 \mathrm{~min}\)
Explanation:
The maximum time period of any pendulum on the earth is \(1.5 \mathrm{~min}\).
PHXI14:OSCILLATIONS
364451
The period of a simple pendulum, whose bob is a hollow metallic sphere, is \(T_{1}\). The period is \(T_{2}\) when the bob is filled with sand, \(T_{3}\) when it is filled with mercury and \(T_{4}\) when it is half filled with mercury. Which of the following is true?
1 \(T_{1}=T_{2}=T_{3}>T_{4}\)
2 \(T_{1}=T_{2}=T_{3}>T_{4}\)
3 \(T_{1}=T_{2}=T_{3} < T_{4}\)
4 \(T_{1}>T_{2}>T_{3}=T_{4}\)
Explanation:
Time period of pendulum doesn't depends upon mass but it depends upon length (distance between point of suspension and centre of mass). In first three cases length are same so \(T_{1}=T_{2}=T_{3}\) but in last case centre of mass lowers which in turn increases the length. So option (3) is correct
PHXI14:OSCILLATIONS
364452
A simple pendulum of length \(l\) and having a bob of mass \(M\) is suspended in a car. The car is moving on a circular track of radius \(R\) with a uniform speed \(v\). If the pendulum makes small oscillations about its equilibrium position, what will be its time period?
The bob is subjected to two simultaneous, accelerations perpendicular to each other viz acceleration due to gravity \(g\) and radial acceleration \(a_{R}=\dfrac{v^{2}}{R}\) towards the centre of the circular path. \(\therefore\) Effective acceleration \(a_{e f f}=\sqrt{g^{2}+\left(\dfrac{v^{2}}{R}\right)^{2}}\) \(\therefore\) Time period of the simple pendulum \( = 2\pi \sqrt {\frac{l}{{{a_{{\text{eff }}}}}}} \) \( = 2\pi \sqrt {\frac{l}{{\sqrt {{g^2} + {{\left( {\frac{{{v^2}}}{R}} \right)}^2}} }}} = 2\pi \sqrt {\frac{l}{{\sqrt {{g^2} + \frac{{{v^4}}}{{{R^2}}}} }}} \)
PHXI14:OSCILLATIONS
364453
The acceleration due to gravity on a planet is \(3{\rm{/}}2\) times that on the earth. If length of a seconds pendulum on earth is \(1\;m\), length of seconds pendulum on the planet is
1 \(0.7\;m\)
2 \(1\;m\)
3 \(1.7\;m\)
4 \(1.5\;m\)
Explanation:
\(\sqrt{\dfrac{l}{g}}=\) constant \(\Rightarrow \dfrac{l_{1}}{g_{1}}=\dfrac{l_{2}}{g_{2}}\) \(\Rightarrow \dfrac{1}{g}=\dfrac{l}{3 g / 2}\) \( \Rightarrow l = 1.5\;m\)
364450
Maximum time period of any simple pendulum on the earth is
1 \(180.5 \mathrm{~min}\)
2 \(100 \mathrm{~min}\)
3 \(90.5 \mathrm{~min}\)
4 \(1.5 \mathrm{~min}\)
Explanation:
The maximum time period of any pendulum on the earth is \(1.5 \mathrm{~min}\).
PHXI14:OSCILLATIONS
364451
The period of a simple pendulum, whose bob is a hollow metallic sphere, is \(T_{1}\). The period is \(T_{2}\) when the bob is filled with sand, \(T_{3}\) when it is filled with mercury and \(T_{4}\) when it is half filled with mercury. Which of the following is true?
1 \(T_{1}=T_{2}=T_{3}>T_{4}\)
2 \(T_{1}=T_{2}=T_{3}>T_{4}\)
3 \(T_{1}=T_{2}=T_{3} < T_{4}\)
4 \(T_{1}>T_{2}>T_{3}=T_{4}\)
Explanation:
Time period of pendulum doesn't depends upon mass but it depends upon length (distance between point of suspension and centre of mass). In first three cases length are same so \(T_{1}=T_{2}=T_{3}\) but in last case centre of mass lowers which in turn increases the length. So option (3) is correct
PHXI14:OSCILLATIONS
364452
A simple pendulum of length \(l\) and having a bob of mass \(M\) is suspended in a car. The car is moving on a circular track of radius \(R\) with a uniform speed \(v\). If the pendulum makes small oscillations about its equilibrium position, what will be its time period?
The bob is subjected to two simultaneous, accelerations perpendicular to each other viz acceleration due to gravity \(g\) and radial acceleration \(a_{R}=\dfrac{v^{2}}{R}\) towards the centre of the circular path. \(\therefore\) Effective acceleration \(a_{e f f}=\sqrt{g^{2}+\left(\dfrac{v^{2}}{R}\right)^{2}}\) \(\therefore\) Time period of the simple pendulum \( = 2\pi \sqrt {\frac{l}{{{a_{{\text{eff }}}}}}} \) \( = 2\pi \sqrt {\frac{l}{{\sqrt {{g^2} + {{\left( {\frac{{{v^2}}}{R}} \right)}^2}} }}} = 2\pi \sqrt {\frac{l}{{\sqrt {{g^2} + \frac{{{v^4}}}{{{R^2}}}} }}} \)
PHXI14:OSCILLATIONS
364453
The acceleration due to gravity on a planet is \(3{\rm{/}}2\) times that on the earth. If length of a seconds pendulum on earth is \(1\;m\), length of seconds pendulum on the planet is
1 \(0.7\;m\)
2 \(1\;m\)
3 \(1.7\;m\)
4 \(1.5\;m\)
Explanation:
\(\sqrt{\dfrac{l}{g}}=\) constant \(\Rightarrow \dfrac{l_{1}}{g_{1}}=\dfrac{l_{2}}{g_{2}}\) \(\Rightarrow \dfrac{1}{g}=\dfrac{l}{3 g / 2}\) \( \Rightarrow l = 1.5\;m\)
