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A pendulum made of a uniform wire of cross section area A has time period T. When an additional mass M is added to its bob, the time period changes to TM. If the Young’s modulus of the material of the wire is Y the $\dfrac{1}{Y}$ is equal to (g=gravitational acceleration)
Answer
415.2k+ views
Hint: Young Modulus: Young modulus is the ratio between the stress (ratio of force to area) and strain (change in length to original length). Young modulus is denoted by ‘Y’.
As we know that young’s modulus are:
$ \Rightarrow Y = \dfrac{{Fl}}{{A\Delta l}}$
Here,
Y=young’s modulus
F=force
A=area of cross section
l=original length
$\Delta l = change in length$
Complete step by step solution:
Given,
Time period,
\[ \Rightarrow T = 2\pi \sqrt {\dfrac{l}{g}} \] …(1)
When additional mass is added to its bob then
\[ \Rightarrow {T_m} = 2\pi \sqrt {\dfrac{{l + \Delta l}}{g}} \]
\[ \Rightarrow Y = \dfrac{{mg\Delta l}}{{Al}}\]
\[ \Rightarrow \Delta l = \dfrac{{mgl}}{{AY}}\]
Put this value In equation
\[ \Rightarrow {T_m} = 2\pi \sqrt {\dfrac{{l + \dfrac{{mgl}}{{AY}}}}{g}} \] …(2)
Divide equation 1 from 2:
\[ \Rightarrow {(\dfrac{{{T_m}}}{T})^2} = l + \dfrac{{mgl}}{{AY}}\]
\[ \Rightarrow \dfrac{1}{Y} = \dfrac{A}{{mg}} + \left[ {{{(\dfrac{{{T_m}}}{T})}^2} - 1} \right]\]
Note: Voltage is directly proportional to the current. And current is inversely proportional to the resistance higher the current lower the resistance. Resistance also depends on the length, cross-section and properties of material. There are some applications of current to produce heat. Resistors are also made of various types of material and it counts by colour of the resistor.
As we know that young’s modulus are:
$ \Rightarrow Y = \dfrac{{Fl}}{{A\Delta l}}$
Here,
Y=young’s modulus
F=force
A=area of cross section
l=original length
$\Delta l = change in length$
Complete step by step solution:
Given,
Time period,
\[ \Rightarrow T = 2\pi \sqrt {\dfrac{l}{g}} \] …(1)
When additional mass is added to its bob then
\[ \Rightarrow {T_m} = 2\pi \sqrt {\dfrac{{l + \Delta l}}{g}} \]
\[ \Rightarrow Y = \dfrac{{mg\Delta l}}{{Al}}\]
\[ \Rightarrow \Delta l = \dfrac{{mgl}}{{AY}}\]
Put this value In equation
\[ \Rightarrow {T_m} = 2\pi \sqrt {\dfrac{{l + \dfrac{{mgl}}{{AY}}}}{g}} \] …(2)
Divide equation 1 from 2:
\[ \Rightarrow {(\dfrac{{{T_m}}}{T})^2} = l + \dfrac{{mgl}}{{AY}}\]
\[ \Rightarrow \dfrac{1}{Y} = \dfrac{A}{{mg}} + \left[ {{{(\dfrac{{{T_m}}}{T})}^2} - 1} \right]\]
Note: Voltage is directly proportional to the current. And current is inversely proportional to the resistance higher the current lower the resistance. Resistance also depends on the length, cross-section and properties of material. There are some applications of current to produce heat. Resistors are also made of various types of material and it counts by colour of the resistor.
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