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(1) 6F (2) 9F (3) F (4) 4F [AIEEE-2009]
. $\mathrm{A} \ell_{1}=3 \mathrm{A} \ell_{2}$ $\Rightarrow \ell_{2}=\frac{\ell_{1}}{3}$ For wire $1: Y=\frac{F / A}{\Delta x / \ell_{1}}$ For wire $2: \mathrm{Y}=\frac{F^{\prime} / 3 A}{\Delta x / \ell_{2}}$ $\Rightarrow \mathrm{F}^{\prime}=9 \mathrm{F}$
[AIEEE 2012]
$Y=\frac{F / S}{\Delta L / L}$ $\frac{\Delta L}{L}=\frac{\Delta R}{R}=\alpha \Delta \mathrm{T} \quad[\mathrm{L}=2 \pi \mathrm{R}]$ $\Rightarrow \mathrm{F}=\mathrm{YS} \alpha \Delta \mathrm{T}$ $\therefore$ The ring is pressing the wheel from both sides, thus $\mathrm{F}_{\mathrm{net}}=2 \mathrm{F}=2 \mathrm{YS} \alpha \Delta \mathrm{T}$
(For steel Young’s modulus is $2 \times 10^{11} \mathrm{Nm}^{-2}$ and coefficient of thermal expansion is $1.1 \times 10^{-5} \mathrm{K}^{-1}$ ) (1) $2.2 \times 10^{7} \mathrm{Pa}$ (2) $2.2 \times 10^{6} \mathrm{Pa}$ (3) $2.2 \times 10^{8} \mathrm{Pa}$ (4) $2.2 \times 10^{9} \mathrm{Pa}$ [JEE-Main-2014]
Thermal strain $=\alpha \Delta \mathrm{T}$ (by $\left.\ell=\ell_{0}(1+\alpha \Delta \mathrm{T})\right)$ $\Rightarrow$ Thermal stress in Rod (Pressure due to Thermal strain $)=\mathrm{Y} \alpha \Delta \mathrm{T}$ $=2 \times 10^{11} \times 1.1 \times 10^{-5} \times 100$ $=2.2 \times 10^{8} \mathrm{Pa}$
( 1)$\left[1-\left(\frac{\mathrm{T}_{\mathrm{M}}}{\mathrm{T}}\right)^{2}\right] \frac{\mathrm{A}}{\mathrm{Mg}}$ ( 2)$\left[1-\left(\frac{\mathrm{T}}{\mathrm{T}_{\mathrm{M}}}\right)^{2}\right] \frac{\mathrm{A}}{\mathrm{Mg}}$ ( 3)$\left[\left(\frac{\mathrm{T}_{\mathrm{M}}}{\mathrm{T}}\right)^{2}-1\right] \frac{\mathrm{A}}{\mathrm{Mg}}$ ( 4)$\left[\left(\frac{\mathrm{T}_{\mathrm{M}}}{\mathrm{T}}\right)^{2}-1\right] \frac{\mathrm{Mg}}{\mathrm{A}}$ [JEE-Main-2015]
(1) $55^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-2} /^{\circ} \mathrm{C}$ (2) $25^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-5} /^{\circ} \mathrm{C}$ (3) $60^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-4} /^{\circ} \mathrm{C}$ (4) $30^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-3} /^{\circ} \mathrm{C}$ [JEE-Main-2016]
$\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$ $\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{1}{2} \frac{\Delta \ell}{\ell}$ When clock gain 12 sec $\frac{12}{\mathrm{T}}=\frac{1}{2} \alpha(40-\theta)$ …. (1) When clock lose 4 sec. $\frac{4}{\mathrm{T}}=\frac{1}{2} \alpha(\theta-20) \quad \ldots(2)$ From equation (1) & (2) $3=\frac{40-\theta}{\theta-20}$ $3 \theta-60=40-\theta$ $4 \theta=100$ $\theta=25^{\circ} \mathrm{C}$ from equation (1) $\frac{12}{\mathrm{T}}=\frac{1}{2} \alpha(40-25)$ $\frac{12}{24 \times 3600}=\frac{1}{2} \alpha \times 15$ $\alpha=\frac{24}{24 \times 3600 \times 15}$ $\alpha=1.85 \times 10^{-15} /^{\circ} \mathrm{C}$
(1) $\frac{3 \alpha}{\mathrm{PK}}$ (2) $3 \mathrm{PK} \alpha$ (3) $\frac{\mathrm{P}}{3 \alpha \mathrm{K}}$ (4) $\frac{\mathrm{P}}{\alpha \mathrm{K}}$ [JEE-Main-2017]
Due to thermal expansion $\frac{\Delta \mathrm{v}}{\mathrm{v}}=3 \alpha \Delta \mathrm{T}$ Due to External pressure $\frac{\Delta \mathrm{v}}{\mathrm{v}}=\frac{\mathrm{P}}{\mathrm{K}}$
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