JEE Advanced 2014 Paper 2 Question Paper with Solutions

The velocity of the efflux depends on the hydrostatic pressure difference and its given by

$v = \sqrt {2{g_{eff}}H}$.

After comin gout the jar, water follows the projectile motion. Time taken in traversing a vertical distance h is

$t = \sqrt {2h/{g_{eff}}}$.

The horizontal distance covered in this time is

$d = vt = \sqrt {2{g_{eff}}H} \sqrt {2h/{g_{eff}}} = 2\sqrt {Hh}$,

which is independent of g_eff. In the case of free fall, hydrostatic pressure difference is zero (g_eff = 0) and hence the water does not come out.

The wavelength of the K_$\alpha$ X-ray line is related to the atomic number of the element by Moseley's law, which can be expressed as:

$\sqrt{\nu} = A(Z - B)$

Here, $\nu$ is the frequency of the emitted X-ray, $Z$ is the atomic number, and $A$ and $B$ are constants. Given that the wavelength $\lambda$ is inversely proportional to the frequency $\nu$ (since $\nu = \frac{c}{\lambda}$, where $c$ is the speed of light), we can rewrite this equation in terms of wavelength:

$\lambda \propto \frac{1}{(Z - B)^2}$

Now, let's consider the ratio of the wavelengths for copper (Cu) and molybdenum (Mo):

$\frac{\lambda_{Cu}}{\lambda_{Mo}} = \left(\frac{Z_{Mo} - B}{Z_{Cu} - B}\right)^2$

For the K_$\alpha$ X-ray line, the constant $B$ can be approximated as 1. Using the atomic numbers $Z_{Cu} = 29$ and $Z_{Mo} = 42$, we can plug these values into the ratio:

$\frac{\lambda_{Cu}}{\lambda_{Mo}} = \left(\frac{42 - 1}{29 - 1}\right)^2 = \left(\frac{41}{28}\right)^2$

Calculating the above expression:

$\frac{41}{28} \approx 1.464$

$\left(1.464\right)^2 \approx 2.14$

Therefore, the ratio $\frac{\lambda_{Cu}}{\lambda_{Mo}}$ is close to 2.14. The correct option is:

Option B 2.14

B_R = B due to ring

B₁ = B due to wire-1

B₂ = B due to wire-2

In magnitudes

${B_1} = {B_2} = {{{\mu _0}I} \over {2\pi r}}$

Resultant of B₁ and B₂

$= 2{B_1}\cos \theta = 2\left( {{{{\mu _0}I} \over {2\pi r}}} \right)\left( {{h \over r}} \right)$

$= {{{\mu _0}Ih} \over {\pi {r^2}}}$

${B_R} = {{{\mu _0}I{R^2}} \over {2{{({R^2} + {x^2})}^{3/2}}}}$

$= {{2{\mu _0}I\pi {a^2}} \over {4\pi {r^3}}}$

As, R = a, x = h and a² + h² = r²

For zero magnetic field at P,

${{{\mu _0}Ih} \over {\pi {r^2}}} = {{2{\mu _0}I\pi {a^2}} \over {4\pi {r^3}}}$

$\Rightarrow \pi {a^2} = 2rh \Rightarrow h \approx 1.2a$

Magnetic field at mid-point of two wires

= 2 (magnetic field due to one wire)

$= 2\left[ {{{{\mu _0}} \over {2\pi }}{1 \over d}} \right] = {{{\mu _0}I} \over {\pi d}} \otimes$

Magnetic moment of loop

M = IA = I $\pi$a²

Torque on loop = MB sin 150$^\circ$ = ${{{\mu _0}{I^2}{a^2}} \over {2d}}$

As tennis ball is dropped, so its initial velocity, u = 0

Suppose its velocity at any time t = v

Acceleration due to gravity = g

$\therefore$ v = u + gt

$\Rightarrow$ v = gt ..... (i)

We know, kinetic energy,

$K = {1 \over 2}m{v^2} = {1 \over 2}m({g^2}{t^2})$ [Using eqn. (i)]

$\Rightarrow K = {1 \over 2}m{g^2}{t^2}$

$\Rightarrow K \propto {t^2}$ (As m and g are constant)

This shows that relation between kinetic energy of tennis ball K and time t is parabolic.

During the collision, velocity of the ball falls sharply to zero as it is compressed and regains maximum velocity in the same short time interval.

This relation is best illustrated by the option (b).

Heat lost by monoatomic gas = Heat gained by diatomic gas

$2 \times {5 \over 2}R \times (700 - T) = 2 \times {5 \over 2}R(T - 400)$

$3500 - 5T = 7T - 2800$

$12T = 6300$ or, $T = 525K$

Work done by monoatomic gas:

$= {n_1}R\Delta T$ ($\because$ PV = nRT)

$= 2 \times R \times (700 - 525)$ ($\Delta$T is negative)

$= - 350R$

Work done by diatomic gas $= {n_2}R\Delta T$

$= 2R(525 - 400) = 250R$. (There is gain in temperature)

The net work done = $-$350 + 250 = $-$100R and $\Delta$T is positive.

(P) Component of forces along x-axis will vanish. Net force along positive y-axis.

(Q) Component of forces along y-axis will vanish. Net force along positive x-axis

(R) Component of forces along x-axis will vanish. Net force along negative y-axis.

(S) Component of forces along y-axis will vanish. Net force along negative x-axis

Electric field due to uniformly charged dielectric solid sphere of radius R at a point P is given by

E = \left\{ {\matrix{ {k\,.\,{Q \over {{R^3}}}\,.\,r,} & {r \le R} \cr {k\,.\,{Q \over {{r^2}}},} & {r > R} \cr } } \right.

In first two cases, P is outside the sphere. Therefore,

${E_1} = {1 \over {4\pi {E_0}}}.\,{Q \over {{R^2}}}$ ..... (1)

${E_2} = {1 \over {4\pi {E_0}}}.\,{{2Q} \over {{R^2}}}$ .... (2)

In Sphere 3, point P located is inside. Therefore,

${E_3} = \left( {{\rho \over {3{E_0}}}} \right)r$

$\rho = {{4Q} \over {(4/3)\pi {{(2R)}^3}}}$

$= {{12Q} \over {4\pi \,.\,8{R^3}}} = {{3Q} \over {8\pi }}$

${E_3} = \left( {{{3Q} \over {\pi {R^3}}}} \right)R$

${E_3} = {1 \over {4\pi {E_0}}}\,.\,{Q \over {2{R^2}}}$ ..... (3)

${E_2} > {E_1} > {E_3}$

From Einstein's photoelectric equation,

kinetic energy of photoelectrons,

$K = {1 \over 2}m{v^2} = {{hc} \over \lambda } - \phi$ ..... (i)

For $\lambda$ = 248 nm, v = u₁

$\therefore$ ${K_1} = {1 \over 2}mu_1^2 = {{hc} \over {248}} - \phi$ ...... (ii)

For $\lambda$ = 310 nm, v = u₂

$\therefore$ ${K_2} = {1 \over 2}mu_2^2 = {{hc} \over {310}} - \phi$ ....... (iii)

Divide eqn. (ii) by eqn. (iii),

${{u_1^2} \over {u_2^2}} = {{{{hc} \over {248}} - \phi } \over {{{hc} \over {310}} - \phi }} = {{{{1240} \over {248}} - \phi } \over {{{1240} \over {310}} - \phi }}$

$\Rightarrow {\left( {{2 \over 1}} \right)^2} = {{5 - \phi } \over {4 - \phi }}$

$\Rightarrow 16 - 4\phi = 5 - \phi \Rightarrow 3\phi = 11$

or, $\phi$ = 3.67 eV $\approx$ 3.7 eV

T is final temperature, when equilibrium is reached

Heat gained = Heat lost

$n{C_p}\Delta T = n{C_v}\Delta T$

$2 \times {{7R} \over 2} \times (T - 400) = 2 \times {{3R} \over 2} \times (700 - T)$

$7T - 2800 = 2100 - 3T$

$10T = 4900$

$T = 490K$

when a metre bridge is balanced, then

${{{R_1}} \over x} = {{{R_2}} \over {(100 - x)}}$ ..... (i)

From figure,

R₁ = R, R₂ = 90 $\Omega$, x = 40 cm

Then, ${R \over {40}} = {{90} \over {(100 - 40)}} = {{90} \over {60}}$

$\Rightarrow$ R = 60 $\Omega$

Now, for $\Delta$R taking natural log on both sides of eqn. (i),

$\ln {R_1} = \ln {R_2} + \ln x - \ln (100 - x)$

or, $\ln R = \ln x - \ln (100 - x) + \ln (90)$

On differentiating,

${{\Delta R} \over R} = {{\Delta x} \over x} - {{\Delta (100 - x)} \over {(100 - x)}}$

$\Rightarrow {{\Delta R} \over R} = {{\Delta x} \over x} + {{\Delta x} \over {(100 - x)}}$

$\Rightarrow \Delta R = \left( {{{0.1} \over {40}} + {{0.1} \over {60}}} \right) \times 60 = {{0.5} \over {120}} \times 60 = 0.25\Omega$

$\therefore$ Required value of R = (60 $\pm$ 0.25) $\Omega$

Let v be speed of the bead when it makes an angle $\theta$ with the vertical direction.

The forces acting on the bead are its weight mg and normal reaction N. Since all the forces acting on the bead are conservative, the mechanical energy of the bead is conserved i.e.,

$mgr(1 - \cos \theta ) = {1 \over 2}m{v^2}$ ..... (1)

The radially inward components of force provide centripetal acceleration to the bead i.e.,

$mg\cos \theta - N = m{v^2}/r$ ...... (2)

Eliminate v² from equations (1) and (2) to get

$N = mg(3\cos \theta - 2)$ ...... (3)

When 3cos$\theta$ > 2, direction of N will be same as shown in figure, i.e., radially outward.

When 3cos$\theta$ < 2, direction of N will be radially inward. Now, normal on the bead is opposite to the force applied on the wire by the bead, following Newton's third law. Therefore, the force applied on the bead by the wire is radially inwards initially and radially outwards later, as it moves from A to B.

The forces on the block of mass m₁ are its weight m₁g, normal reaction from the inclined plane N₁, and the reaction from the second block R.

Similarly, forces on the block of mass m₂ are m₂g, N₂, R, and the frictional force f. If $\theta$ is slowly increased, f starts increasing and attains its maximum value f = $\mu$N₂ at $\theta$ = $\theta$_r (angle of repose). The blocks are stationary if $\theta$ $\le$ $\theta$_r otherwise they are moving. Consider the limiting case, $\theta$ = $\theta$_r, when the blocks are at rest and

$f = \mu {N_2}$ ...... (1)

Apply Newton's second law to m₁,

$R = {m_1}g\sin \theta$ ....... (2)

${N_1} = R = {m_2}g\cos \theta$ ....... (3)

and to m₂,

$f = R + {m_2}g\sin \theta$ ...... (4)

${N_2} = {m_1}g\cos \theta$ ...... (5)

Eliminate R, N₂ and f from equations (1)-(5) to get

${\theta _r} = {\tan ^{ - 1}}\left( {{{\mu {m_2}} \over {{m_1} + {m_2}}}} \right) = {\tan ^{ - 1}}\left( {{{0.3 \times 2} \over {1 + 2}}} \right)$

$= {\tan ^{ - 1}}(0.2) = 11.5^\circ$

Thus, for $\theta$ = 5$^\circ$ and $\theta$ = 10$^\circ$, blocks are at rest with frictional force

$f = R + {m_2}g\sin \theta = ({m_1} + {m_2})g\sin \theta$

For $\theta$ = 15$^\circ$ and $\theta$ = 20$^\circ$, blocks are moving with frictional force

$f = \mu {N_2} = \mu {m_2}g\cos \theta$, (limiting value).

Effective area of incident light = $\pi$r²

Total energy per second = 912 $\times$ $\pi$r²

From Stefan's law, we have

E = Ae $\sigma$ (T⁴ $-$ T$_0^4$)

For black body emissivity, e = 1.

A = 4$\pi$r²

912 $\times$ $\pi$r² = 4$\pi$r² $\times$ 5.7 $\times$ 10^$-$8 [T⁴ $-$ 300⁴]

$({T^4} - {300^4}) = {{912} \over {4 \times 5.7 \times {{10}^{ - 8}}}} = 40 \times {10^8}$

${T^4} = 40 \times {10^8} + {3^4} \times {10^8}$

${T^4} = (90 + 81) \times {10^8} \Rightarrow 121 \times {10^8}$

$T = {(121)^{1/4}} \times 100 = \sqrt {11} \times 100 \Rightarrow 3.3 \times 100$

T = 330 K

The lens maker's formula gives focal length of the bi-convex lens as

${1 \over {{f_1}}} = (\mu - 1)\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right]$

$= (1.5 - 1)\left[ {{1 \over r} - {1 \over { - r}}} \right] = {1 \over r}$ ..... (1)

The focal length of a plano-convex (also convex-plane) lens is given by

${1 \over {{f_2}}} = (1.5 - 1)\left[ {{1 \over \infty } - {1 \over { - r}}} \right] = {1 \over {2r}}$ ...... (2)

and that of a plano-concave (also concave-plano) lens is given by

${1 \over {{f_3}}} = (1.5 - 1)\left[ {{1 \over \infty } - {1 \over r}} \right] = - {1 \over {2r}}$ ..... (3)

Case P:

Therefore, ${1 \over {{f_{eff}}}} = {1 \over r} + {1 \over r}$

${f_{eff}} = {r \over 2}$

(P) $\to$ (2)

Case Q :

${1 \over {{f_{eff}}}} = {1 \over {2r}} + {1 \over {2r}}$

${f_{eff}} = r$

(Q) $\to$ (4)

Case R :

${1 \over {{f_{eff}}}} = - {1 \over {2r}} - {1 \over {2r}}$

${f_{eff}} = - r$

(R) $\to$ (3)

Case S : ${1 \over {{f_{eff}}}} = {1 \over r} - {1 \over {2r}} = {1 \over {2r}}$

${f_{eff}} = 2r$

(S) $\to$ (1)

Given R_e = 10R = 6 $\times$ 10⁶ m and g_e = 10 m/s². Consider a wire element of length dr placed at a distance r from the centre O.

Let $\rho$ be the common mass density of the planet and the earth and m be the mass of the planet inside the sphere of radius r i.e., $m = {4 \over 3}\pi {r^3}\rho$. The gravitational force on the wire element by the planet is equal to the force by a point mass of magnitude m placed at the centre O. Thus, force on the wire element is

$dF = {{Gm\lambda dr} \over {{r^2}}} = {4 \over 3}G\pi \rho \lambda rdr$.

Integrate dF from $r = {4 \over 5}R$ to r = R to get

$F = {4 \over 3}G\pi \rho \lambda \int_{{4 \over 5}R}^R {r\,dr = {4 \over 3}G\pi \rho \lambda \left( {{{9{R^2}} \over {50}}} \right)}$

$= {{{4 \over 3}\pi \rho R_e^3G} \over {R_e^2}}\left( {{{9\lambda {R_e}} \over {5000}}} \right)$

$= {g_e}{{9\lambda {R_e}} \over {5000}} = {{(10)(9)({{10}^{ - 3}})(6 \times {{10}^6})} \over {5000}} = 108$ N.

Using equation of continuity,

a₁v₁ = a₂v₂

Here, r₁ = 20 mm = 20 $\times$ 10^$-$3 m

v₁ = 5 mm s^$-$1 = 5 $\times$ 10^$-$3 m s^$-$1

r₂ = 1 mm = 1 $\times$ 10^$-$3 m, v₂ = ?

${v_2} = {{{a_1}{v_1}} \over {{a_2}}} = {{\pi r_1^2{v_1}} \over {\pi r_2^2}}$

$\Rightarrow {v_2} = {{{{(20 \times {{10}^{ - 3}})}^2}(5 \times {{10}^{ - 3}})} \over {{{(1 \times {{10}^{ - 3}})}^2}}}$

$= 400 \times 5 \times {10^{ - 3}} = 2$ m s^$-$1

Let R be the radius of the meniscus formed with a contact angle $\theta$.

By geometry, this radius makes an angle $\theta + {\alpha \over 2}$ with the horizontal where

$\cos \left( {\theta + {\alpha \over 2}} \right) = b/R$. ...... (1)

Let P₀ be the atmospheric pressure and P₁ be the pressure just below the meniscus. Excess pressure on the concave side of meniscus of radius R is given by

${P_0} - {P_1} = 2S/R$ ..... (2)

The hydrostatic pressure gives

${P_0} - {P_1} = h\rho g$ ..... (3)

From (2) and (3) we get,

${{2S} \over R} = h\rho g$

or, $h\rho g = {{2s} \over b}\cos \left( {\theta + {\alpha \over 2}} \right)$ (By putting value of R)

$h = {{2S} \over {b\rho g}}\cos \left( {\theta + {\alpha \over 2}} \right)$

For horizontal flow, Bernoulli's theorem becomes

${P \over \rho } + gh + {1 \over 2}{v^2} = k'$

$P + {1 \over 2}\rho {v^2} = k$ (h = 0)

Between points 1 and 2 :

${P_1} - {P_2} = {1 \over 2}{\rho _a}v_a^2$

Between points 2 and 3 :

${P_3} - {P_2} = {1 \over 2}{\rho _l}v_l^2$

Since, P₁ = P₃, we have

${1 \over 2}{\rho _l}v_l^2 = {1 \over 2}{\rho _a}v_a^2$

${{{v_l}} \over {{v_a}}} =\sqrt {{{{\rho _a}} \over {{\rho _l}}}}$

In $\Delta$SAB

$\tan \theta = {{AB} \over {AS}} = {{11.54/2} \over {10}}$

tan$\theta$ = 0.577

$\theta$ = 30$^\circ$

Using ${{\sin {\theta _l}} \over {\sin 90}} = {{{n_l}} \over {{n_B}}}$, we get

$\sin {\theta _l} = {{{n_l}} \over {{n_B}}}$ or ${n_l} = {n_B} \times \sin 30 = 2.72 \times {1 \over 2} = 1.36$

The reaction involved is

Zn²⁺ + OH^$-$ $\to$ Zn(OH)₂ $\buildrel {O{H^ - }} \over \longrightarrow$ [Zn(OH)₄]^2$-$ (soluble)

For [Cr(NH₃)₄Cl₂]Cl,

For [Ti(H₂O)₅Cl](NO₃)₂,

The complex is paramagnetic due to presence of unpaired electron.

Since the groups between the complex ion and ions outside can be exchanged, it shows ionisation isomerism.

For [Pt(en)(NH₃)Cl]NO₃,

The complex is square planar and all electrons are paired. So it is diamagnetic. Since the groups between the complex ion and ions outside can be exchanged, it shows ionisation isomerism.

For [Co(NH₃)₄(NO₃)₂]NO₃, NH₃ is a strong ligand and causes pairing

The reaction involved is

Collision frequency of X is greater than Y.

The reactions involved are

The reactions involved are as follows:

Write the molecular orbital electronic configuration keeping in mind that there is no $2 s-2 p$ mixing, then if highest occupied molecular orbital contain unpaired electron then molecule is paramagnetic otherwise diamagnetic.

Note: Since there is no mixing of $2 s$ and $2 p$ orbitals of each of the atom in a diatomic molecule, the energy of bonding molecular orbital $2 p z$, i.e., $2 p$ is less than that of bonding molecular orbital of $2 p x$ and $2 p y$, i.e., $\left(\pi_{2 p x}^*\right.$ and $\left.\pi_{2 p y}^*\right)$.

$\therefore$ The order of molecular orbital energy levels, i.e., bonding and anti-bonding energy levels is as follows:

$$\begin{aligned} \left(\sigma_{1 s}\right)\left(\sigma_{1 s}^*\right)\left(\sigma_{2 s}\right)\left(\sigma_{2 s}^*\right)\left(\sigma_{2 p z}\right)\left(\pi_{2 p x}\right. & \left.\equiv \pi_{2 p y}\right) \left(\pi_{2 p x}^*\right. \left.\equiv \pi_{2 p y}^*\right)\left(\sigma_{2 p z}^*\right) \end{aligned}$$

(A) Total number (no.) of electrons in $\mathrm{Be}_2=4+4=$ 8. The electronic configuration of $\mathrm{Be}_2$ is:

$\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2$

All the electrons in $\mathrm{Be}_2$ are paired up, $\mathrm{Be}_2$ is diamagnetic.

(B) Total number (no.) of electrons in $\mathrm{B}_2=5+5=$ 10. The electronic configuration of $B_2$ is:

$\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p z}\right)^2$

All the electrons in $B_2$ are paired up, $B_2$ is diamagnetic.

(C) Total no. of electrons in $\mathrm{C}_2=6+6=12$ The electronic configuration of $\mathrm{C}_2$ is:

$\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p z}\right)^2\left(\pi_{2 p x}^{* 1} \equiv \pi_{2 p y}^{* 1}\right)$

There are two unpaired electrons in doubly degenerate.

$\pi_{2 p x}$ and $\pi_{2 p y} \mathrm{C}_2$ is paramagnetic.

(D) Total no. of electrons in $\mathrm{N}_2=7+7=14$

The electronic configuration of $\mathrm{N}_2$ is:

$\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p z}\right)^2\left(\pi_{2 p x}^{* 2} \equiv \pi_{2 p y}^{* 2}\right)$

All the electrons in $\mathrm{N}_2$ are paired up; $\mathrm{N}_2$ is diamagnetic.

The reaction involved is

P₄ + 8SOCl₂ $\to$ 4PCl₃ + 4SO₂ + 2S₂Cl₂

Using Graham's law of diffusion, we have that for identical conditions, rate of diffusion varies inversely with square root of mass of the gases. So,

${x \over {24 - x}} = {\left( {{{40} \over {10}}} \right)^{1/2}} \Rightarrow x = 16$

The reaction involved is

To analyze the options provided, let's discuss the changes in entropy for the system and the surroundings separately during the phase change from liquid water (H₂O(l)) to water vapor (H₂O(g)) at 100°C and 1 atmosphere pressure. This is a typical boiling process.

First, let's consider the change in entropy, $\Delta S_{\text{system}}$, of the water itself (the system). When water boils and changes from a liquid to a gas, the molecules spread out significantly, occupying a larger volume and showing greater randomness in their motion. This transition from a more ordered phase (liquid) to a less ordered phase (gas) results in an increase in entropy. Thus, $\Delta S_{\text{system}}>0$.

Next, let's consider the change in entropy of the surroundings, $\Delta S_{\text{surroundings}}$. When the water boils at 100°C, it must absorb heat to do so. This heat is typically taken from the surroundings. As the surroundings lose heat, according to the formula for entropy change $\Delta S = \frac{q}{T}$, where $q$ is the heat transferred and $T$ is the temperature, the surroundings lose entropy if the process is happening at a constant temperature as typically assumed in such cases. Therefore, here $\Delta S_{\text{surroundings}}<0$ as heat loss by surroundings (negative $q$) results in a decrease in entropy.

Combining these insights, we find:

• The entropy of the system increases ($\Delta S_{\text{system}}>0$).
• The entropy of the surroundings decreases ($\Delta S_{\text{surroundings}}<0$).

Therefore, the correct answer is: Option B: $\Delta S_{\text{system}}>0$ and $\Delta S_{\text{surroundings}}<0$.

The sequence of reactions for formation of Y are

As it has a CH₃CO$-$group, it gives positive iodoform test and it is a functional isomer of X as only the functional groups are different.

The reaction is

The rate law for an elementary reaction where the concentration of reactant M impacts the rate of the reaction is expressed as:

$\text{Rate} = k[M]^n$

where k is the rate constant, M is the concentration of the reactant, and n is the order of the reaction with respect to M.

In the scenario described, when the concentration of M is doubled, the rate of disappearance of M increases by a factor of 8. Let’s denote the initial concentration of M by [M] and the doubled concentration by 2[M]. According to the rate law, the initial rate would be:

$\text{Rate}_1 = k[M]^n$

Upon doubling M:

$\text{Rate}_2 = k(2[M])^n = k2^n[M]^n$

It is given that this rate is 8 times the initial rate:

$\text{Rate}_2 = 8 \times \text{Rate}_1$

$k2^n[M]^n = 8k[M]^n$