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QJEEM16Ph2c8c: JEE Main 2016 Physics - Class 11 JEEM16Ph2c8c Solution: A simple pendulum made of a bob of mass m and a metallic wire…

A simple pendulum made of a bob of mass m and a metallic wire of negligible mass has time period 2 s at T=0oC. If the temperature of the wire is increased and the corresponding change in its time period is plotted against its temperature, the resulting graph is a line of slope S. If the coefficient of linear expansion of metal is α\alpha then the value of S is :

Multiple Choice Options

  • A α\alpha
  • B α2{\alpha \over 2}
  • C 2 α\alpha
  • D 1α{1 \over \alpha }

Step-by-Step Verified Solution

Change of length of wire with temperature,

Δ\Delta ℓ\ell   =    αℓΔθ\alpha \ell \Delta \theta

Time period of pendulum at temperature θ\theta ,

T θ\theta = 2 π\pi ℓ+Δℓg\sqrt {{{\ell + \Delta \ell } \over g}}

= 2 π\pi ℓ(1+αΔθ)g\sqrt {{{\ell \left( {1 + \alpha \Delta \theta } \right)} \over g}}

= 2πℓg(1+αΔθ)122\pi \sqrt {{\ell \over g}} {\left( {1 + \alpha \Delta \theta } \right)^{{1 \over 2}}}

= 2πℓg(1+Δℓℓ)122\pi \sqrt {{\ell \over g}} {\left( {1 + {{\Delta \ell } \over \ell }} \right)^{{1 \over 2}}}

≃\simeq   T0 (1+Δℓ2ℓ)\left( {1 + {{\Delta \ell } \over {2\ell }}} \right)

Here T0 = time period at temperature 0oC.

∴\therefore    Change in time period,

Δ\Delta T = T θ\theta −- T0

= T0Δℓ2ℓ{{{T_0}\Delta \ell } \over {2\ell }}

= T0(αℓΔθ)2ℓ{{{T_0}\left( {\alpha \ell \Delta \theta } \right)} \over {2\ell }}

∴\therefore     ΔTΔθ{{\Delta T} \over {\Delta \theta }} = T0α2{{{T_0}\alpha } \over 2}

Given that T0 = 2,

∴\therefore     ΔTΔθ{{\Delta T} \over {\Delta \theta }} = 2α2{{2\alpha } \over 2} = α\alpha

ΔTΔθ{{\Delta T} \over {\Delta \theta }} is the shape of Δ\Delta T and Δθ{\Delta \theta }

curve = S (given)

∴\therefore    S = α\alpha

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