Tag: measurements and experimentation

Quizzes (17) Questions (185)

Questions Related to measurements and experimentation

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

If R is the radius of the earth and g the acceleration due to gravity on the earth's surface, the mean density of the earth is

  1. 4πG/3gR

  2. 3πR/4gG

  3. 3g/4πRG

  4. πRg/12G
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that

$g=\cfrac{GM}{R^2}$
Also, density $=mass\times volume$
$M=density\times volume\M=P\times\cfrac{4\pi R^3}{3R^2}=P\times\cfrac{4\pi R}{3}$
Put value of m in $g=\cfrac{GM}{R^2}\P=\cfrac{3g}{4\pi RG}$

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

The mass of a bob, suspended in a simple pendulum, is halved from the initial mass, its time period will :

  1. Be less

  2. Be more

  3. Remain unchanged

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The time period of simple pendulum id given by

$T=2\pi \sqrt{\dfrac{l}{g}}$
where, $l=$ length of simple pendulum
$g=$ acceleration due to gravity
$T=$ Time period
The time period of simple pendulum is independent of the mass of bob, the time period remains unchanged,when mass of bob will change.
The correct option is C. 

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

If the length of a seconds pendulum is increased by $2$% then what is loss and gain in a day?

  1. losses $764 \ s$
  2. losses $924 \ s$
  3. gains $236 \ s$
  4. losses $864 \ s$
  5. gains $346 \ s$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$T _0=2\pi\sqrt{\cfrac{l}{g}}\T^1=2\pi\sqrt{\cfrac{l+l\times2/100}{g}}\ \cfrac{T _0}{T^1}=\cfrac{\sqrt{100}}{\sqrt{102}}\ T^1=\cfrac{\sqrt{102}}{\sqrt{100}}T _0\T^1=1.0099T _0\approx  1.01T _0\Loss=(1.01-1)T _0=0.01T _0$

In one second, it looses $0.01sec$
$\Rightarrow$ Total time loose in one day$=(0.01\times24\times3600)seconds\=864seconds$

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

The different equation of simple harmonic motion for a seconds pendulum is:

  1. $\dfrac{d^2 x}{dt^2} + x = 0$
  2. $\dfrac{d^2 x}{dt^2} + \pi x = 0$
  3. $\dfrac{d^2 x}{dt^2} + 4 \pi x = 0$
  4. $\dfrac{d^2 x}{dt^2} + \pi^2 x = 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The differential equation for simple harmonic motion is d^2x/dt^2 + omega^2 x = 0. For a seconds pendulum, the time period T = 2 seconds, so angular frequency omega = 2pi/T = 2pi/2 = pi. Substituting omega^2 into the equation gives d^2x/dt^2 + pi^2 x = 0.

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

A simple pendulum with a bob of mass m swings with an angular amplitude of ${ 60 }^{ 0 }$, when its angular displacement is ${ 30 }^{ 0 }$, the tension of string would be 

  1. $3\sqrt { 3 } mg$
  2. $\frac { 1 }{ 2 } mg(2\sqrt { 3 } -1)$
  3. $\frac { 1 }{ 2 } mg(3\sqrt { 3 } +2)$
  4. $\frac { 1 }{ 2 } mg(3-\sqrt { 2 } )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

The simple pendulum acts as second's pendulum on earth. Its time on a planet, whose mass and diameter are twice that of earth is:

  1. $\sqrt { 2 } s$
  2. $2\sqrt { 2 } s$
  3. $2s$
  4. $\dfrac { 1 }{ \sqrt { 2 } } s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Time period of second's pendulum is two second.
Second's pendulum is that simple pendulum whose time period of vibration is two seconds. The bob of such pendulum while oscillating passes through the mean position after every one second.
Noe,
Time period of simple pendulum is given by
$T=2\pi \sqrt { \left( \dfrac { l }{ g }  \right)  } $
or  $T\propto \dfrac { 1 }{ \sqrt { g }  } $             ......(i)
but  $g=\dfrac { GM }{ { R }^{ 2 } } $      (on earth)
and  ${ g }^{ \prime  }=\dfrac { G\left( 2M \right)  }{ 4{ R }^{ 2 } } $     (on planet)
$=\dfrac { 1 }{ 2 } \dfrac { GM }{ { R }^{ 2 } } =\dfrac { g }{ 2 } $
Equation (i) gives
$\dfrac { { T }^{ \prime  } }{ T } =\dfrac { \sqrt { g }  }{ \sqrt { { g }^{ \prime  } }  } =\sqrt { 2 } $
or  ${ T }^{ \prime  }=\sqrt { 2 } T$
  $=\sqrt { 2 } \times 2              \left( T=2s \right) $
  $=2\sqrt { 2 } s$

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

The length of a second's pendulum at a place where g = 9.8m/s $\displaystyle ^{2}$ is 90.2 cm. State whether true or false.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Time period of pendulum is:

$T =2\pi \sqrt [  ]{ \cfrac { l }{ g }  } $
$l=\cfrac { T^{ 2 }g }{ 4\pi ^{ 2 } } $
$l=\cfrac { 4\times 9.8 }{ 4\times \pi ^{ 2 } } $
$l=0.993m=99.3m$
$l$= length of pendulum 
$g$= $9.8m/s$
$T$ = Time period of seconds pendulum $=2s$
So, our given statement is false.