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 $\mathrm{T}_{\mathrm{M}}$. If the Young's modulus of the material of the wire is Y then $\frac{1}{\mathrm{Y}}$ is equal to: ( $g=$ gravitational acceleration)

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

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

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

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

JEE Main · 2015 · 04 Apr 2015 · Offline · Elasticity & SHM

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)

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

Answer: (D)

Solution

Hanging the extra mass

M

stretches the wire by

Δ ℓ = \frac{M g ℓ}{A Y}

, lengthening the pendulum. Writing the two periods as

T = 2 π ℓ / g

and

T_{M} = 2 π (ℓ + Δ ℓ) / g

and dividing,

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

. Isolating the compliance,

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

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option

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JEE Main · 2015 · 04 Apr 2015 · Offline · Elasticity & SHM

Solution

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