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A pendulum made of a uniform wire of cross-sectional area A ha

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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 $T_{M}$ . If the Young's modulus of the material of the wire is $Y$ , then $\frac{1}{Y}$ is equal to $(g =$ gravitational acceleration) (a) $[(\frac{T _{M}}{T})^{2} - 1] \frac{A}{M g}$ (b) $[(\frac{T _{M}}{T})^{2} - 1] \frac{M g}{A}$ (c) $[1 - (\frac{T _{M}}{T})^{2}] \frac{A}{M g}$ (d) $[1 - (\frac{T}{T _{M}})^{2}] \frac{A}{M g}$

Name: Video solution 1: 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 TM. If the Young's modulus of the material of the wire is Y, then Y1 is equal to (g= gravitational acceleration) (a) [(TTM)2−1]MgA (b) [(TTM)2−1]AMg (c) [1−(TTM)2]MgA (d) [1−(TMT)2]MgA
Uploaded: 2023-01-17T11:58:38.000Z
Duration: 1 min
Description: Video solution 1: 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 TM. If the Young's modulus of the material of the wire is Y, then Y1 is equal to (g= gravitational acceleration) (a) [(TTM)2−1]MgA (b) [(TTM)2−1]AMg (c) [1−(TTM)2]MgA (d) [1−(TMT)2]MgA

Views: 5,726 students

Updated on: Jan 17, 2023

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2015

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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

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) (a)

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

(b)

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

(c)

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

(d)

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

6 mins ago

Discuss this question LIVE

6 mins ago

Text solutionVerified

Exp. (a)
We know that time period,

T = 2 π \frac{L}{g}

When additional mass

M

is added to its bob

T_{M} = 2 π \frac{L + Δ L}{g}

where,

Δ L

is increase in length.
We know that

Y = \frac{M g / A}{Δ L / L} = \frac{M gL}{A Δ L} \Rightarrow Δ L = \frac{M gL}{A Y}

\Rightarrow ∴ T_{M} = 2 π \frac{L + \frac{M gL}{A Y}}{g}

\Rightarrow (\frac{T _{M}}{T})^{2} = 1 + \frac{M g}{A Y}

\frac{M g}{A Y} = (\frac{T _{M}}{T})^{2} - 1

\frac{1}{Y} = \frac{A}{M g} [(\frac{T _{M}}{T})^{2} - 1]

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Question Text	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 $T_{M}$ . If the Young's modulus of the material of the wire is $Y$ , then $\frac{1}{Y}$ is equal to $(g =$ gravitational acceleration) (a) $[(\frac{T _{M}}{T})^{2} - 1] \frac{A}{M g}$ (b) $[(\frac{T _{M}}{T})^{2} - 1] \frac{M g}{A}$ (c) $[1 - (\frac{T _{M}}{T})^{2}] \frac{A}{M g}$ (d) $[1 - (\frac{T}{T _{M}})^{2}] \frac{A}{M g}$
Updated On	Jan 17, 2023
Topic	Mechanical Properties of Solids
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 1
Upvotes	139
Avg. Video Duration	1 min

Text solutionVerified

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