# JEE Advanced 2015

When the bullet reaches maximum height, acceleration due to gravity is $$\dfrac{1}{4}$$th of that at planet's surface. $$r=2R$$

$$\dfrac{GM}{r^2} = \dfrac{1}{4}\dfrac{GM}{R^2}$$

By conservation of mechanical energy,

$$\dfrac{-GMm}{R} + \dfrac{1}{2}mv^2 = \dfrac{-GMm}{r } + 0$$ since velocity is zero at max height.

$$ \Rightarrow \dfrac{1}{2}mv^2 = \dfrac{GMm}{2R}$$

$$ v = \sqrt{\dfrac{GM}{R}} = \sqrt{\dfrac{2GM}{2R}} = \dfrac{1}{\sqrt{2}} \sqrt{\dfrac{2GM}{R}}= \dfrac{1}{\sqrt{2}}v_{esc} $$

$$ \Rightarrow v_{esc} = \sqrt{2}v$$

$$\Rightarrow n=2$$

$$K_{total} = K_{tr} + K_{rot}= \frac{1}{2}mv^2 + \frac{1}{2}I\omega ^2= \frac{1}{2}mv^2 + \frac{1}{2} ( \frac{mR^2}{2})(\frac{v}{R})^2=\frac{3}{4}mv^2$$

Since velocity of both the discs are same at B , using conservation of energy,

$$\frac{3}{4}mv_1^2 + mg(30) = \frac{3}{4}mv_2^2 + mg(27)$$

Putting $$v_1=3$$ and solving

$$v_2 ^2 = 49$$

$$ \Rightarrow v_2 =7$$

Since velocity of both the discs are same at B , using conservation of energy,

$$\frac{3}{4}mv_1^2 + mg(30) = \frac{3}{4}mv_2^2 + mg(27)$$

Putting $$v_1=3$$ and solving

$$v_2 ^2 = 49$$

$$ \Rightarrow v_2 =7$$

From stephan's law:

$$\dfrac{\Delta Q}{\Delta t}= \sigma e A T^4$$

Given $$(\dfrac{\Delta Q}{\Delta t})_A=10^4(\dfrac{\Delta Q}{\Delta t})_B$$

$$ \sigma e A_A T_A^4=10^4\sigma e A_B T_B^4$$

$$(\dfrac{T_A}{T_B})^4=10^4\dfrac{A_B}{A_A}$$

$${\dfrac{T_A}{T_B}}=10(\dfrac{A_B}{A_A})^\dfrac{1}{4}$$

$${\dfrac{T_A}{T_B}}=10(\dfrac{\pi R_B^2}{\pi R_A^2})^\dfrac{1}{4}$$

Given $$\dfrac{R_A}{R_B}=400$$

$$\dfrac{T_A}{T_B}=10(\dfrac{1}{400^2})^\dfrac{1}{4}$$

$$\dfrac{T_A}{T_B}=\dfrac{10}{20}=\dfrac{1}{2}$$

From wein's displcament law

$$\lambda_mT=b$$

$$\lambda \alpha \dfrac{1}{T}$$

$$\dfrac{\lambda_A}{\lambda_B}=\dfrac{T_B}{T_A}=2$$

Hence, correct answer is 2.

Let E and E' be the power produced by the plant and power consumed by the village.

$$ E'= \dfrac{12.5}{100}E= \dfrac{1}{8}E= (\dfrac{1}{2})^3E$$

Thus, the no of half lives is $$3$$ until power requirement is met.

$$\therefore n=3$$

$$ E'= \dfrac{12.5}{100}E= \dfrac{1}{8}E= (\dfrac{1}{2})^3E$$

Thus, the no of half lives is $$3$$ until power requirement is met.

$$\therefore n=3$$

$$X_1 = \sqrt{X^2 + d^2}$$

$$X_2 =\mu \sqrt{X^2 + d^2}= \dfrac{4}{3} \sqrt{X^2 + d^2}$$

Path diff $$\Delta X = ( \dfrac{4}{3} -1) \sqrt{X^2 + d^2}=n\lambda$$

$$ X^2 + d^2 = 9n^2\lambda ^2$$

$$ \Rightarrow p^2 =9$$

$$ \Rightarrow p =3$$

$$X_2 =\mu \sqrt{X^2 + d^2}= \dfrac{4}{3} \sqrt{X^2 + d^2}$$

Path diff $$\Delta X = ( \dfrac{4}{3} -1) \sqrt{X^2 + d^2}=n\lambda$$

$$ X^2 + d^2 = 9n^2\lambda ^2$$

$$ \Rightarrow p^2 =9$$

$$ \Rightarrow p =3$$

The first image from reflection by spherical concave mirror can be found from mirror formula:

$$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$...(i)

For concave surface: u=-15 cm, f=-10 cm

Substituting in equation (i)

$$\frac{1}{v}+\frac{1}{-15}=\frac{1}{-10}=\frac{1}{v}=\frac{10-15}{150}\Rightarrow v=-30 cm$$

the magnification is given as $$m=-\frac{v}{u}=-\frac{-30}{-15}=-2$$

The first image will act as object for lens and it's position can be found using lens formula

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$...(ii)

For concave surface: u=-20 cm, f=10 cm

Substituting in equation (ii)

$$\frac{1}{v}-\frac{1}{-20}=\frac{1}{10} \Rightarrow \frac{1}{v} \frac{1}{10}-\frac{1}{20}=\frac{2-1}{20}\Rightarrow v=20 cm$$

the magnification is given as $$m=\frac{v}{u}=\frac{20}{-20}=-1$$

total magnification$$M_1=-2\times -1=2$$

Now the whole set is immersed in liquid of $$n=\frac{7}{6}$$ the reflection from concave mirror will remain same as their is no change in object position and focal length.

The change in focal length can be found using lens maker formula

$$\frac{1}{f}=(\frac{\mu_2}{\mu_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$$

$$\frac{1}{f_{air}}=(1.5-1)(\frac{1}{R_1}-\frac{1}{R_2})$$...(iii)

$$\frac{1}{f_{liquid}}=(\frac{1.5}{\frac{7}{6}}-1)(\frac{1}{R_1}-\frac{1}{R_2})$$...(iv)

dividing (iii) by (iv)

$$\frac{f_{liquid}}{f_{air}}=\frac{\frac{1}{2}}{\frac{3}{2}\frac{6}{7}-1}$$

$$\frac{f_{liquid}}{f_{air}}=\frac{\frac{1}{2}}{\frac{3}{2}\frac{6}{7}-1}=\frac{7}{4}$$

$$f_{liquid}=f_{air}\times \frac{7}{4} =\frac{70}{4}$$

The first image will act as object for lens and it's position can be found using lens formula

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$...(i)

For concave surface: $$u=-20 cm, f=\frac{70}{4} cm$$

Substituting in equation (i)

$$\frac{1}{v}-\frac{1}{-20}=\frac{4}{70} \Rightarrow \frac{1}{v}\frac{4}{70}-\frac{1}{20}=\frac{80-70}{1400}\Rightarrow v=140 cm$$

the magnification is given as $$m=\frac{v}{u}=\frac{20}{-20}=-7$$

total magnification$$M_2=-2\times -7=14$$

$$\frac{M_2}{M_1}=7$$

$$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$...(i)

For concave surface: u=-15 cm, f=-10 cm

Substituting in equation (i)

$$\frac{1}{v}+\frac{1}{-15}=\frac{1}{-10}=\frac{1}{v}=\frac{10-15}{150}\Rightarrow v=-30 cm$$

the magnification is given as $$m=-\frac{v}{u}=-\frac{-30}{-15}=-2$$

The first image will act as object for lens and it's position can be found using lens formula

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$...(ii)

For concave surface: u=-20 cm, f=10 cm

Substituting in equation (ii)

$$\frac{1}{v}-\frac{1}{-20}=\frac{1}{10} \Rightarrow \frac{1}{v} \frac{1}{10}-\frac{1}{20}=\frac{2-1}{20}\Rightarrow v=20 cm$$

the magnification is given as $$m=\frac{v}{u}=\frac{20}{-20}=-1$$

total magnification$$M_1=-2\times -1=2$$

Now the whole set is immersed in liquid of $$n=\frac{7}{6}$$ the reflection from concave mirror will remain same as their is no change in object position and focal length.

The change in focal length can be found using lens maker formula

$$\frac{1}{f}=(\frac{\mu_2}{\mu_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$$

$$\frac{1}{f_{air}}=(1.5-1)(\frac{1}{R_1}-\frac{1}{R_2})$$...(iii)

$$\frac{1}{f_{liquid}}=(\frac{1.5}{\frac{7}{6}}-1)(\frac{1}{R_1}-\frac{1}{R_2})$$...(iv)

dividing (iii) by (iv)

$$\frac{f_{liquid}}{f_{air}}=\frac{\frac{1}{2}}{\frac{3}{2}\frac{6}{7}-1}$$

$$\frac{f_{liquid}}{f_{air}}=\frac{\frac{1}{2}}{\frac{3}{2}\frac{6}{7}-1}=\frac{7}{4}$$

$$f_{liquid}=f_{air}\times \frac{7}{4} =\frac{70}{4}$$

The first image will act as object for lens and it's position can be found using lens formula

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$...(i)

For concave surface: $$u=-20 cm, f=\frac{70}{4} cm$$

Substituting in equation (i)

$$\frac{1}{v}-\frac{1}{-20}=\frac{4}{70} \Rightarrow \frac{1}{v}\frac{4}{70}-\frac{1}{20}=\frac{80-70}{1400}\Rightarrow v=140 cm$$

the magnification is given as $$m=\frac{v}{u}=\frac{20}{-20}=-7$$

total magnification$$M_2=-2\times -7=14$$

$$\frac{M_2}{M_1}=7$$

Let the perpendicular from O to the line charge intersects the line charge at point Q.

$$OQ = \dfrac{\sqrt{3}a}{2}$$

Angle subtended by AB at Q is $$\angle AQB= 2 \angle AQO$$

$$\tan \angle AQO = \dfrac{AO}{QO} =\dfrac{ \dfrac{a}{2}}{\dfrac{\sqrt{3a}}{2}}=\dfrac{1}{\sqrt{3}} =\tan 30^0$$

$$ \therefore \angle AQB = 60^0$$

Let us consider a cylindrical gaussian surface with the axis along line charge of length L and radius a(since AQ =a ).

$$\phi _{cylinder,curved} = \dfrac{q_{enc}}{\epsilon _0} = \dfrac{\lambda L}{\epsilon _0}$$

Flux through the rectangle is same as the flux through $$\dfrac{1}{6}$$th of the cuboid since it subtends $$60^o$$ at Q and field lines are symmetrically radially outwards.

Hence, $$\phi _{rectangle} = \dfrac{1}{6}\dfrac{\lambda L}{\epsilon _0}$$

$$ \therefore n=6$$

Energy of $$90\ nm$$ wavelength,

$$E = \dfrac{hc}{\lambda} = \dfrac{1242}{90} = 13.8 \: eV$$

$$ 13.8 - \dfrac{13.6}{n^2} = 10.4$$

$$ \Rightarrow n^2 = 4$$

$$ \Rightarrow n=2$$

**Checking option A**

**Average K.E. for 1 mole for mono atomic molecule is given as**

**$$\dfrac{3}{2}RT$$**

**Average K.E. for 1 mole for diatomic molecule is given as**

**$$\dfrac{5}{2}RT$$**

**Taking average K.E.**

**$$K.E._{average}=\dfrac{n_1 K.E._{He}+n_2 K.E._{H_2}}{n_1+n_2}$$**

**Average K.E.$$= \dfrac{\dfrac{3}{2}RT+\dfrac{5}{2}RT}{2}=2RT$$**

**Option A is correct**

**Checking option B**

**The velocity of sound in gas is given as $$V=\sqrt{\dfrac{\gamma RT}{M}}$$ where $$\gamma $$ is ratio of heat capacities, and M is Molecular mass of gas**

**$$\gamma_{average}=\dfrac{n_1\gamma_{He}+n_2\gamma_{H_2}}{n_1+n_2}=\dfrac{1.6+1.4}{2}=1.5$$**

**$$M_{average}=\dfrac{n_1M_{He}+n_2M_{H_2}}{n_1+n_2}=\dfrac{4+2}{2}=3$$**

**Checking option C**

**The R.M.S. speed is given as $$\sqrt{\dfrac{3RT}{M}}$$**

**$$R.M.S. speed \alpha \dfrac{1}{\sqrt{M}}$$**

**$$ \dfrac{V_{R.M.S. \;He}}{V_{R.M.S. \;H_2}}=\sqrt{\dfrac{M_{H_2}}{M_{He}}}=\sqrt{\dfrac{2}{4}}=\dfrac{1}{\sqrt{2}}$$**

**Hence option C is incorrect and option D is correct**

**Correct options are A,B and D.**

$$R_{Al}$$ and $$R_{Fe}$$ are parallel to each other.

$$R =\frac{\rho l}{A}$$

Al:

$$Area = 7^2 -2^2 = 45\: mm^2$$

$$\rho = 2.7 \times 10^{-8} \Omega m$$

$$l =50 \: mm$$

Fe:

$$Area = 2^2 = 4\: mm^2$$

$$\rho = 1.0 \times 10^{-7} \Omega m$$

$$l =50 \: mm$$

Substituting values:

$$R_{Al} = 30 \: \mu \Omega$$

$$R_{Fe} = 1250 \: \mu \Omega$$

$$R_{total} = \frac{R_{Al} R_{Fe}}{R_{Al} + R_{Fe}} = \frac{30 \times 1250}{30 + 1250} = \frac{1875}{64}\: \Omega$$

$$R =\frac{\rho l}{A}$$

Al:

$$Area = 7^2 -2^2 = 45\: mm^2$$

$$\rho = 2.7 \times 10^{-8} \Omega m$$

$$l =50 \: mm$$

Fe:

$$Area = 2^2 = 4\: mm^2$$

$$\rho = 1.0 \times 10^{-7} \Omega m$$

$$l =50 \: mm$$

Substituting values:

$$R_{Al} = 30 \: \mu \Omega$$

$$R_{Fe} = 1250 \: \mu \Omega$$

$$R_{total} = \frac{R_{Al} R_{Fe}}{R_{Al} + R_{Fe}} = \frac{30 \times 1250}{30 + 1250} = \frac{1875}{64}\: \Omega$$

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