A pendulum made of a uniform… - JEE Main 2015 Physics PYQ

JEE Main 2015 — Physics Mechanics

2015mcqhard

A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to ${T}_{M}$ . If the Young's modulus of the material of the wire is $Y$ , then $\frac{1}{Y}$ is equal to:

( $g=$ gravitational acceleration)

Official previous-year question

Held on 4 Apr 2015 · Verified 6 Jul 2026.

Options

A
$[1-{(\frac{T}{{T}_{M}})}^{2}]\frac{A}{Mg}$
B
$[ { ( \frac{ {\text{T}}_{\text{M}} }{ \text{T} } ) }^{2} - 1 ] \frac{ A }{ \mathrm{Mg} }$
C
$[ { ( \frac{ {\text{T}}_{\text{M}} }{ \text{T} } ) }^{2} - 1 ] \frac{ \text{Mg} }{ A }$
D
$[1-{(\frac{{T}_{M}}{T})}^{2}]\frac{A}{Mg}$

Did you get this right?

Solution

JEE Main 2015 Physics, Mechanical Properties of Solids — solution figure

Initial length $=l$ . Time period $\text{T}=2\pi \sqrt{\frac{l}{\text{g}}}...(i)$ . After suspending mass $M,$ . Youngs modulus $\text{Y}=\frac{\text{Stress}}{\text{Strain}}$ $=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}=\frac{Fl}{\Delta l A}$

Change in length of wire $\Delta l=\frac{\text{F}l}{\text{AY}}$

Now time period ${\text{T}}_{\text{M}}=2\pi \sqrt{\frac{l+\Delta l}{\text{g}}}...(\mathrm{ii})$

$\frac{\text{T}}{{\text{T}}_{\text{M}}}=\frac{2\pi \sqrt{\frac{l}{\text{g}}}}{2\pi \sqrt{\frac{(l+\Delta l)}{\text{g}}}}$ $[ \because \frac{ \text{i} }{ \text{ii} } ]$

$\frac{{\text{T}}^{2}}{{\text{T}}_{\text{M}}^{2}}=\frac{l}{l+\Delta l}$

$\Rightarrow \frac{{T}_{m}^{2}}{{T}^{2}}=1+\frac{\Delta l}{l}$

$\begin{matrix}1+\frac{F}{AY}={(\frac{{T}_{m}}{T})}^{2} & [putting value of \Delta l]\end{matrix}$

$\frac{1}{Y}=[{(\frac{{T}_{m}}{T})}^{2}-1]\frac{A}{F}$

$F=Mg$

$\frac{1}{Y}=[{(\frac{{T}_{m}}{T})}^{2}-1]\frac{A}{Mg}$

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Options

$[1-{(\frac{T}{{T}_{M}})}^{2}]\frac{A}{Mg}$

$[ { ( \frac{ {\text{T}}_{\text{M}} }{ \text{T} } ) }^{2} - 1 ] \frac{ A }{ \mathrm{Mg} }$

$[ { ( \frac{ {\text{T}}_{\text{M}} }{ \text{T} } ) }^{2} - 1 ] \frac{ \text{Mg} }{ A }$

$[1-{(\frac{{T}_{M}}{T})}^{2}]\frac{A}{Mg}$

Solution

JEE Main 2015 Physics, Mechanical Properties of Solids — solution figure

Change in length of wire $\Delta l=\frac{\text{F}l}{\text{AY}}$

Now time period ${\text{T}}_{\text{M}}=2\pi \sqrt{\frac{l+\Delta l}{\text{g}}}...(\mathrm{ii})$

$\frac{\text{T}}{{\text{T}}_{\text{M}}}=\frac{2\pi \sqrt{\frac{l}{\text{g}}}}{2\pi \sqrt{\frac{(l+\Delta l)}{\text{g}}}}$ $[ \because \frac{ \text{i} }{ \text{ii} } ]$

$\frac{{\text{T}}^{2}}{{\text{T}}_{\text{M}}^{2}}=\frac{l}{l+\Delta l}$

$\Rightarrow \frac{{T}_{m}^{2}}{{T}^{2}}=1+\frac{\Delta l}{l}$

$\begin{matrix}1+\frac{F}{AY}={(\frac{{T}_{m}}{T})}^{2} & [putting value of \Delta l]\end{matrix}$

$\frac{1}{Y}=[{(\frac{{T}_{m}}{T})}^{2}-1]\frac{A}{F}$

$F=Mg$

$\frac{1}{Y}=[{(\frac{{T}_{m}}{T})}^{2}-1]\frac{A}{Mg}$

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Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

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Thermodynamics

Waves & Oscillations

Keep practicing Mechanics