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**Previous Years AIEEE/JEE Mains Questions**

(1) $6 \times 10^{4} \mathrm{J}$

(2) $7 \times 10^{4} \mathrm{J}$

(3) $3 \times 10^{4} \mathrm{J}$

(4) $5 \times 10^{4} \mathrm{J}$

**Directions : **Question number 11, 12 and 13 are based on the following paragraph.

Two moles of helium gas are taken over the cycle ABCDA, as shown in the P–T diagram.

**[AIEEE-2009]**

**Sol.**(3 )

$\frac{P M}{R T}=\rho$

$\frac{P}{\rho}=\frac{R T}{M}$

$\mathbf{U}=\frac{3}{2} n R T=\frac{3}{2} m\left(\frac{R T}{M}\right)$

(1) $\frac{\gamma \mathrm{Mv}^{2}}{2 \mathrm{R}}$

(2) $\frac{(\gamma-1)}{2 \mathrm{R}} \mathrm{Mv}^{2}$

(3) $\frac{(\gamma-1)}{2(\gamma+1) \mathrm{R}} \mathrm{Mv}^{2}$

(4) $\frac{(\gamma-1)}{2 \gamma \mathrm{R}} \mathrm{Mv}^{2}$

**[AIEEE-2011]**

**Sol.**(2)

$\frac{1}{2} m v^{2}=\frac{n R \Delta T}{\gamma-1}$

$\Delta \mathrm{T}=\frac{1}{2} \frac{m}{n} \frac{(\gamma-1) v^{2}}{R}$

(1) $\frac{n_{1} \mathrm{T}_{1}^{2}+\mathrm{n}_{2} \mathrm{T}_{2}^{2}+\mathrm{n}_{3} \mathrm{T}_{3}^{2}}{\mathrm{n}_{1} \mathrm{T}_{1}+\mathrm{n}_{2} \mathrm{T}_{2}+\mathrm{n}_{3} \mathrm{T}_{3}}$

(2) $\frac{n_{1}^{2} T_{1}^{2}+n_{2}^{2} T_{2}^{2}+n_{3}^{2} T_{3}^{2}}{n_{1} T_{1}+n_{2} T_{2}+n_{3} T_{3}}$

(3) $\frac{\mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3}}{3}$

(4) $\frac{\mathrm{n}_{1} \mathrm{T}_{1}+\mathrm{n}_{2} \mathrm{T}_{2}+\mathrm{n}_{3} \mathrm{T}_{3}}{\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}}$

**[AIEEE-2011]**

**Sol.**(4)

$\mathrm{n}_{1} \mathrm{C}_{\mathrm{V}} \mathrm{T}_{1}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{V}} \mathrm{T}_{2}+\mathrm{n}_{3} \mathrm{C}_{\mathrm{V}} \mathrm{T}_{3}=\left(\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}\right) \mathrm{C}_{\mathrm{v}} \mathrm{T}$

(Atmospheric pressure = 76 cm of Hg)

(1) 38 cm (2) 6 cm (3) 16 cm (4) 22 cm

**[jEE-Mains-2014]**

**Sol.**(3)

(1) $\frac{\gamma+1}{2}$

( 2)$\frac{\gamma-1}{2}$

(3) $\frac{3 y+5}{6}$

(4) $\frac{3 \gamma-5}{6}$

**[jEE-Mains-2015]**

**Sol.**(1)

(1) $2.5 \times 10^{25}$

(2) $-2.5 \times 10^{25}$

(3) $-1.61 \times 10^{23}$

(4) $1.38 \times 10^{23}$

**[jEE-Mains-2017]**

**Sol.**(2)

(1) $4.70 \times 10^{3} \mathrm{N} / \mathrm{m}^{2}$

(2) $2.35 \times 10^{2} \mathrm{N} / \mathrm{m}^{2}$

(3) $4.70 \times 10^{2} \mathrm{N} / \mathrm{m}^{2}$

(4) $2.35 \times 10^{3} \mathrm{N} / \mathrm{m}^{2}$

**[jEE-Mains-2018]**

**Sol.**(4)