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 Multiple Choice QuestionsMultiple Choice Questions

1.

The period of oscillation of a simple pendulum is straight T space equals space 2 straight pi space square root of straight L over straight g end root . The measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:

  • 2%

  • 3%

  • 1%

  • 5%


B.

3%

Time period,

straight T space equals space 2 straight pi space square root of straight L divided by straight g end root
thus, changes can be expressed as

plus-or-minus fraction numerator 2 increment straight T over denominator straight T end fraction space equals space plus-or-minus fraction numerator increment straight L over denominator straight L end fraction space plus-or-minus fraction numerator increment straight g over denominator straight g end fraction
According to the question, we can write

fraction numerator increment straight L over denominator straight L end fraction space equals space fraction numerator 0.1 over denominator 20.0 end fraction cm space equals space 1 over 200
Again time period

straight T space equals space 90 over 100 straight s
and space increment straight T space equals space 1 over 100 straight s
rightwards double arrow space fraction numerator increment straight T over denominator straight T end fraction space equals space 1 over 90
Now comma space
because space straight T space equals space 2 straight pi space square root of straight L over straight g end root
straight g space equals 4 straight pi squared straight L over straight T squared
fraction numerator increment straight g over denominator straight g end fraction space equals space fraction numerator increment straight L over denominator straight L end fraction space plus space fraction numerator 2 increment straight T over denominator straight T end fraction
or space fraction numerator increment straight g over denominator straight g end fraction straight x space 100 percent sign space equals space open parentheses fraction numerator increment straight L over denominator straight L end fraction close parentheses straight x 100 percent sign space plus space open parentheses fraction numerator 2 increment straight T over denominator straight T end fraction close parentheses straight x space 100 percent sign
space equals space open parentheses 1 over 200 straight x 100 close parentheses percent sign space plus space 2 space straight x space 1 over 90 space straight x space 100 percent sign
space equals space 2.72 percent sign space almost equal to space 3

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2.

A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density σ at the equilibrium position. The extension x0 of the spring when it is in equilibrium is

  • Mg over straight k
  • Mg over straight k open parentheses 1 minus LAσ over straight M close parentheses
  • Mg over straight k open parentheses 1 minus fraction numerator LAσ over denominator 2 straight M end fraction close parentheses
  • Mg over straight k open parentheses 1 plus LAσ over straight M close parentheses

C.

Mg over straight k open parentheses 1 minus fraction numerator LAσ over denominator 2 straight M end fraction close parentheses

At equilibrium ΣF = 0


kx subscript straight o space plus space open parentheses AL over 2 σg close parentheses minus Mg space equals 0
straight x subscript straight o space equals space Mg open square brackets 1 minus fraction numerator LAσ over denominator 2 straight M end fraction close square brackets


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3.

A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be:

  • left parenthesis 92 space plus-or-minus 2 right parenthesis straight s
  • left parenthesis 92 space plus-or-minus 5 right parenthesis straight s
  • left parenthesis 92 space plus-or-minus space 1.8 right parenthesis straight s
  • left parenthesis 92 space plus-or-minus 3 right parenthesis straight s

A.

left parenthesis 92 space plus-or-minus 2 right parenthesis straight s

Arithmetic mean time of an oscillating simple pendulum

fraction numerator straight capital sigma space straight x subscript straight i over denominator straight N end fraction space equals space fraction numerator 90 space plus 91 space plus 92 space plus 95 over denominator 4 end fraction space equals space 92 space straight s
Mean deviation of a simple pendulum

equals fraction numerator straight capital sigma space vertical line space begin display style straight x with minus on top end style space minus space straight x subscript straight i vertical line over denominator straight N end fraction space equals space fraction numerator 2 space plus space 1 space plus 3 space plus space 3 space plus space 0 over denominator 4 end fraction space equals space 1.5
Given, minimum division in the measuring clock, i.e. simple pendulum = 1s. Thus, the reported mean time of an oscillating simple pendulum = left parenthesis space 92 space plus-or-minus space 2 right parenthesis straight s

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4.

The amplitude of a damped oscillator decreases to 0.9 times its original magnitude is 5s. In another 10s it will decrease to α times its original magnitude, where α equals

  • 0.7

  • 0.81

  • 0.729

  • 0.6


C.

0.729

Amplitude of damped oscillator
straight A space equals space straight A subscript 0 straight e to the power of negative fraction numerator bt over denominator 2 straight m end fraction end exponent
After space 5 straight s comma space space space 0.9 space straight A subscript straight o space equals space straight A subscript straight o straight e to the power of negative fraction numerator straight b left parenthesis 5 right parenthesis over denominator 2 straight m end fraction end exponent space space.. space left parenthesis straight i right parenthesis space
After space 10 space more space second comma
straight A space equals space straight A subscript straight o straight e to the power of negative straight b fraction numerator 15 over denominator 2 straight m end fraction end exponent
straight A space equals space straight A subscript straight o open parentheses straight e to the power of negative straight b fraction numerator 15 over denominator 2 straight m end fraction end exponent close parentheses cubed space space... space left parenthesis ii right parenthesis
From space Eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma
straight A space equals space 0.729 space straight A subscript straight o space equals space αA subscript 0
Hence comma space straight alpha space equals space 0.729

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5.

A pendulum clock loses 12 s a day if the temperature is 408C and gains 4 s a day if the temperature is 208C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (α) of the metal of the pendulum shaft are respectively:

  • 25 C; α=1.85×10−5/ °C

  • 60 °C; α=1.85×10−4/ °C

  • 30°C; α=1.85×10−3/°C

  • 55°C; α=1.85×10−2/°8C


A.

25 C; α=1.85×10−5/ °C

Time period of pendulum,
straight T space equals space 2 straight pi square root of straight l over straight g end root
where comma space straight l space is space length space of space pendulum space and space straight g space is space acceleration space due space to space gravity
such space as space change space in space space time space period space of space straight a space pendulum comma
fraction numerator increment straight T over denominator straight T end fraction space equals space 1 half fraction numerator increment straight l over denominator straight l end fraction
When space clock space gains space 12 space straight s comma we space get
12 over straight T space equals space 1 half straight alpha space left parenthesis 40 minus straight theta right parenthesis
when space clock space loses space 4 straight s comma space we space get
4 over straight T space equals space 1 half straight alpha space left parenthesis straight theta minus 20 right parenthesis
Comparing space Eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space get
3 space equals space fraction numerator 40 minus straight theta over denominator straight theta minus 20 end fraction
rightwards double arrow space 30 minus 60 space equals space 40 minus 0
rightwards double arrow space 40 space equals space 100 space rightwards double arrow straight theta space equals space 25 to the power of straight o space straight C
substituting space the space value space of space straight theta space in space eq space left parenthesis straight i right parenthesis space we space have
12 over straight T space equals space 1 half straight alpha space left parenthesis 40 minus 25 right parenthesis
rightwards double arrow space substituting space the space value space of space straight theta space in space eq space left parenthesis straight i right parenthesis space we space have
12 over straight T space equals space 1 half straight alpha space left parenthesis 40 minus 25 right parenthesis
rightwards double arrow space fraction numerator 12 over denominator 24 space straight x space 3600 space end fraction space equals space 1 half straight alpha space left parenthesis 15 right parenthesis
straight alpha space equals space fraction numerator 24 over denominator 24 space straight x space 3600 space straight x space 15 end fraction
straight alpha space equals space 1.85 space straight x space 10 to the power of negative 5 end exponent divided by to the power of straight o straight C


Thus, the coefficient of linear expansion in pendulum clock = 1.85 x 10-5/C

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6.

A particle performs simple harmonic motion with amplitude A. It's speed is trebled at the instant that it is at a distance 2A/3 from the equilibrium position. The new amplitude of the motion is:

  • straight A over 3 square root of 41
  • 3A

  • straight A square root of 3
  • 7 over 3 straight A

D.

7 over 3 straight A

The velocity of a particle executing SHM at any instant is defined as the time rate of change of its displacement at that instant.

straight v space equals space straight omega square root of straight A squared minus straight x squared end root
Where ω is angular frequency, A is amplitude and x are displacements of a particle.

Suppose that the new amplitude of the modules be A'.

The initial velocity of a particle performs SHM. 
straight v squared space equals space straight omega squared space open square brackets straight A squared minus open parentheses fraction numerator 2 straight A over denominator 3 end fraction close parentheses squared close square brackets
Where A, is initial amplitude and ω is angular frequency.
Final velocity,
left parenthesis 3 straight v squared right parenthesis space equals space straight omega squared open square brackets straight A squared minus open parentheses fraction numerator 2 straight A over denominator 3 end fraction close parentheses squared close square brackets
From space Eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space get
1 over 9 space equals space fraction numerator straight A squared minus begin display style fraction numerator 4 straight A squared over denominator 9 end fraction end style over denominator straight A apostrophe squared space minus begin display style fraction numerator 4 straight A squared over denominator 9 end fraction end style end fraction
straight A apostrophe space equals space fraction numerator 7 straight A over denominator 3 end fraction




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7.

If a simple pendulum has the significant amplitude (up to a factor of 1/e of original) only in the period between t = Os to t = τs, then τ may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with 'b' as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds:

  • 0.693/b

  • b

  • 1/b

  • 2/b


D.

2/b

For damped harmonic motion,
ma = - kx - mbv
or
ma + mbv +kx = 0
Solution to above equation is
 straight x space equals space straight A subscript straight o straight e to the power of negative bt over 2 end exponent space sin space ωt semicolon space space with space straight omega squared space equals space straight k over straight m minus straight b squared over 4
Where amplitude drops exponentially with time
straight A subscript straight tau space equals space straight A subscript straight o straight e to the power of negative bτ over 2 end exponent
Average time τ is that duration when amplitude drops by 63%. i.e., becomes Ao/e
Thus, 

straight A subscript straight tau space equals space straight A subscript straight o over straight e space equals space straight A subscript straight o straight e to the power of negative bt over 2 end exponent
or space bτ over 2 space equals 1 space or space straight tau space equals 2 over straight b

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8.

For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly?


A.

During oscillation, the motion of a simple pendulum KE is maximum of mean position where PE is minimum.

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9.

A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of the air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.

  • 12

  • 8

  • 6

  • 4


C.

6


i.e, 
straight lambda over 4 space equals space 0.85
rightwards double arrow straight lambda space 4 space straight x space 0.85
As space we space know comma space straight v space equals space straight c over straight lambda
rightwards double arrow space fraction numerator 340 over denominator 4 space straight x space 0.85 end fraction space equals space 100 space Hz
therefore, possible frequencies = 100 Hz, 300 Hz, 500 Hz, 700Hz, 900Hz, 1100Hz, below 1250 Hz.
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10.

A particle moves with simple harmonic motion in a straight line. In first τ s, after starting from rest it travels a distance a, and in next τ s it travels 2a, in the same direction, then

  • amplitude of motion is 3a

  • time period of oscillations is 8τ

  • amplitude of motion is 4a

  • time period of oscillations is 6τ


D.

time period of oscillations is 6τ

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