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(1) 6F (2) 9F (3) F (4) 4F

**[AIEEE-2009]**

**Sol.**(2)

. $\mathrm{A} \ell_{1}=3 \mathrm{A} \ell_{2}$

$\Rightarrow \ell_{2}=\frac{\ell_{1}}{3}$

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]**

**Sol.**(1)

$Y=\frac{F / S}{\Delta L / L}$

$\frac{\Delta L}{L}=\frac{\Delta R}{R}=\alpha \Delta \mathrm{T} \quad[\mathrm{L}=2 \pi \mathrm{R}]$

$\Rightarrow \mathrm{F}=\mathrm{YS} \alpha \Delta \mathrm{T}$

$\therefore$ 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]**

**Sol.**(3)

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]**

**Sol.**(3)

(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]**

**Sol.**(2)

$\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$

$\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{1}{2} \frac{\Delta \ell}{\ell}$

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]**

**Sol.**(3)

Due to thermal expansion

$\frac{\Delta \mathrm{v}}{\mathrm{v}}=3 \alpha \Delta \mathrm{T}$

Due to External pressure

$\frac{\Delta \mathrm{v}}{\mathrm{v}}=\frac{\mathrm{P}}{\mathrm{K}}$

this is fucking bullshit!!!!!!!