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.

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

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

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

$\Rightarrow n=2$

1
#351386
Solution
A bullet is fired vertically upwards with velocity $\upsilon$ from the surface of a spherical planet. When it reaches its maximum heights, its acceleration due to the planet's gravity is $1/4$th of its value at the surface of the planet. If the escape velocity from the planet is $\upsilon_{esc} = \upsilon \sqrt N$, then the value of N is (ignore energy loss due to atmosphere)
$K_{total} = K_{tr} + K_{rot}= \frac{1}{2}mv^2 + \frac{1}{2}I\omega ^2= \frac{1}{2}mv^2 + \frac{1}{2} ( \frac{mR^2}{2})(\frac{v}{R})^2=\frac{3}{4}mv^2$
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
$v_2 ^2 = 49$
$\Rightarrow v_2 =7$
2
#351415
Solution
Two identical uniform discs roll without slipping on two different surfaces AB and CD starting at A and C with linear speed $\upsilon_1$ and $\upsilon_2$, respectively, and always remain in contact with the surfaces. If they reach B and D with the same linear speed and $\upsilon_1 = 3 m/s$ then $\upsilon_2$ in m/s is $(g = 10 m/s^2)$
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
$\lambda_mT=b$
$\lambda \alpha \dfrac{1}{T}$
$\dfrac{\lambda_A}{\lambda_B}=\dfrac{T_B}{T_A}=2$

3
#351418
Solution
Two spherical stars $A$ and $B$ emit blackbody radiation. The radius of $A$ is $400$ times that of $B$ and $A$ emits 10$^4$ times the power emitted from $B$. The ratio $\displaystyle \left ( \frac{\lambda_A}{\lambda_B} \right )$ of their wavelengths $\lambda_A$ and $\lambda_B$ at which the peaks occur in their respective radiation curves is
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.
$\therefore n=3$
4
#351420
Solution
A nuclear power plant supplying electrical power to a village uses a radioactive material of half life T years as the fuel. The amount of fuel at the beginning is such that the total power requirement of the village is 12.5% of the electrical power available from the plant at that time. If the plant is able to meet the total power needs of the village for a maximum period of nT years, then the value of n is
$X_1 = \sqrt{X^2 + d^2}$
$X_2 =\mu \sqrt{X^2 + d^2}= \dfrac{4}{3} \sqrt{X^2 + d^2}$
Path diff $\Delta X = ( \dfrac{4}{3} -1) \sqrt{X^2 + d^2}=n\lambda$
$X^2 + d^2 = 9n^2\lambda ^2$
$\Rightarrow p^2 =9$
$\Rightarrow p =3$
5
#351424
Solution
A Young's double slit interference arrangement with slits S$_1$ and S$_2$ is immersed in water (refractive index = 4/3) as shown in the figure. The positions of maxima on the surface of water are given by $x^2 = p^2 m^2 \lambda^2 - d^2$, where $\lambda$ is the wavelength of light in air (refractive index = 1), 2d is the separation between the slits and m is an integer. The value of p is
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$M_2=-2\times -7=14$
$\frac{M_2}{M_1}=7$
6
#351433
Solution
Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification M$_1$. When the set-up is kept in a medium of refractive index 7/6, the magnification becomes M$_2$. The magnitude $\displaystyle \left | \frac{M_2}{M_1} \right |$ is
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, $\phi _{rectangle} = \dfrac{1}{6}\dfrac{\lambda L}{\epsilon _0}$
$\therefore n=6$

7
#351449
Solution
An infinitely long uniform line charge distribution of charge per unit length $\lambda$ lies parallel to the y-axis in the y-z plane at $\displaystyle z = \frac{\sqrt 3}{2} a$. If the magnitude of the flux of the electric field through the rectangular surface ABCD lying in the x-y plane with its centre at the origin is $\displaystyle \frac{\lambda L}{n \varepsilon_0}$ ($\varepsilon_0=$ permittivity of free space), then the value of n is
Energy of $90\ nm$ wavelength,
$E = \dfrac{hc}{\lambda} = \dfrac{1242}{90} = 13.8 \: eV$

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

$\Rightarrow n^2 = 4$

$\Rightarrow n=2$
8
#351455
Solution
Consider a hydrogen atom with its electron in the nth orbital. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of n is (hc = 1242 eV nm)
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.
9
#351460
Solution
A container of fixed volume has a mixture of one mole of hydrogen and one mole of helium in equilibrium at temperature T. Assuming the gases are ideal, the correct statement(s) is (are)
This question has multiple correct options
A
the average energy per mole of the gas mixture is 2RT.
B
the ratio of speed of sound in the gas mixture to that in helium gas is $\sqrt{6/5}$.
C
the ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/2.
D
the ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/$\sqrt 2$.
$R_{Al}$ 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:
$R_{Al} = 30 \: \mu \Omega$
$R_{Fe} = 1250 \: \mu \Omega$
$R_{total} = \frac{R_{Al} R_{Fe}}{R_{Al} + R_{Fe}} = \frac{30 \times 1250}{30 + 1250} = \frac{1875}{64}\: \Omega$
10
#351467
Solution
In an aluminum (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are $2.7 \times 10^{-8} \Omega$ m and $1.0 \times 10^{-7} \Omega$ m, respectively. The electrical resistance between the two faces P and Q of the composite bar is
$\displaystyle \frac{2475}{64} \Omega$
$\displaystyle \frac{1875}{64} \Omega$
$\displaystyle \frac{1875}{49} \Omega$
$\displaystyle \frac{2475}{132} \Omega$