JEE Advanced 2018 Paper 1 Question Paper with Solutions

Given, intensity of sunlight, I = 1.3 kW m^-2 and focal length of convex lens, f = 20 cm

Now, we know, Intensity = ${{Energy} \over {Area}}$

So, intensity on screen is given as

${I_{screen}} = {{Incident\,energy} \over {{A_2}}}$

But, Incident energy = Intensity of sunlight $\times$ Area of lens

= 1.3 kW m^$-$2 $\times$ A₁

So, intensity on screen is given as

${I_{screen}} = 1.3\,kW\,{m^{ - 2}} \times {{{A_1}} \over {{A_2}}}$

Now from figure, and using similar triangles theorem, we have

${{{A_1}} \over {{A_2}}} = {\left( {{{20} \over 2}} \right)^2} = 100$

Therefore, intensity on screen = 1.3 kW m^$-$2 $\times$ 100 = 130 kW m^$-$2

Given, h = height of the top of inclined plane = ?, $\theta$ = 60$^\circ$, g = 10 m s^$-$2 and time difference between ring and disc reaching ground = ${{2 - \sqrt 3 } \over {\sqrt {10} }}$ s

We know that, $a = {{g\sin \theta } \over {1 + {I \over {M{R^2}}}}}$

For ring, $I = M{R^2}$ and for disc $I = {3 \over 2}M{R^2}$.

So, ${a_{ring}} = {{g\sin \theta } \over 2}$ and ${a_{disk}} = {{2g\sin \theta } \over 3}$

Now using $s = {1 \over 2}a{t^2}$

${s_{ring}} = {1 \over 2}\left( {{{g\sin \theta } \over 2}} \right)t_1^2 \Rightarrow {h \over {\sin \theta }} = {1 \over 2}\left( {{{g\sin \theta } \over 2}} \right)t_1^2$ ...... (1)

${s_{disk}} = {1 \over 2}\left( {{{2g\sin \theta } \over 3}} \right)t_2^2 \Rightarrow {h \over {\sin \theta }} = {1 \over 2}\left( {{{2g\sin \theta } \over 3}} \right)t_2^2$ ...... (2)

Given, ${t_1} - {t_2} = {{2 - \sqrt 3 } \over {\sqrt {10} }}$

From Eq. (1), we have ${t_1} = \sqrt {{{4h} \over {g{{\sin }^2}\theta }}}$

and from Eq. (2), we have ${t_2} = \sqrt {{{3h} \over {g{{\sin }^2}\theta }}}$

So, $\sqrt {{{4h} \over {g{{\sin }^2}\theta }}} - \sqrt {{{3h} \over {g{{\sin }^2}\theta }}} = {{2 - \sqrt 3 } \over {\sqrt {10} }}$

$\Rightarrow \sqrt {{{4h} \over {10{{\sin }^2}60^\circ }}} - \sqrt {{{3h} \over {10{{\sin }^2}60^\circ }}} = {{2 - \sqrt 3 } \over {\sqrt {10} }}$

$\Rightarrow \sqrt {{{16h} \over {3 \times 10}}} - \sqrt {{{4h} \over {10}}} = {{2 - \sqrt 3 } \over {\sqrt {10} }}$

$\Rightarrow {{4\sqrt h } \over {\sqrt 3 }} - 2\sqrt h = 2 - \sqrt 3$

$\Rightarrow \sqrt h \left( {{4 \over {\sqrt 3 }} - 2} \right) = (2 - \sqrt 3 )$

$\Rightarrow \sqrt h {{(4 - 2\sqrt 3 )} \over {\sqrt 3 }} = (2 - \sqrt 3 )$

$\Rightarrow \sqrt h \times {2 \over {\sqrt 3 }}(2 - \sqrt 3 ) = (2 - \sqrt 3 ) \Rightarrow \sqrt h = {{\sqrt 3 } \over 2}$

$\Rightarrow h = {3 \over 4} = 0.75$ m

Given C₁ = C₂ = C₃ = 1.0 $\mu$F and V₀ = 8 V

When S₁ is close and S₂ is open. Charge on C₃ is 8 $\mu$C.

The circuit is

When S₁ is open and S₂ is close. Charge on C₃ is 5 $\mu$C. Therefore, charge on C₁ and C₂ will be 3 $\mu$C.

The potential difference across C₃ is equal to the potential difference across C₁ and C₂ together i.e.,

${{{q_1}} \over {C{'_1}}} + {{{q_2}} \over {{C_2}}} = {{{q_3}} \over {{C_3}}}$,

${{3\mu C} \over {{ \in _r}(1\mu F)}} + {{3\mu C} \over {1\mu F}} = {{5\mu C} \over {1\mu F}}$.

Solve to get ${ \in _r} = 1.5$.

${I_1} = {V \over R}\left( {1 - {e^{ - {{tR} \over L}}}} \right)$

${I_2} = {V \over R}\left( {1 - {e^{ - {{tR} \over {2L}}}}} \right)$

From principle of superposition,

$I = {I_1} - {I_2} \Rightarrow I = {V \over R}{e^{ - {{tR} \over {2L}}}}\left( {1 - {e^{ - {{tR} \over {2L}}}}} \right)$ ...... (i)

I is maximum when ${{dI} \over {dt}} = 0$, which gives

${e^{ - {{tR} \over {2L}}}} = {1 \over 2}$ or $t = {{2L} \over R}\ln 2$

Substituting this time in Eq. (i), we get

${I_{\max }} = {V \over {4R}}$

Given, force applied on body

$\overrightarrow F = (\alpha t\widehat i + \beta \widehat j)$

Here, $\alpha$ = 1.0 N s^$-$1 and $\beta$ = 1.0 N. Therefore,

$\overrightarrow F = t\widehat i + \widehat j$

By Newton's second law of motion, F = ma. Given m = 1.0 kg. Therefore,

$\overrightarrow F = \overrightarrow a \Rightarrow \overrightarrow a = (t\widehat i + \widehat j)$

We know $\overrightarrow a = {{d\overrightarrow v } \over {dt}} \Rightarrow \overrightarrow v = \int {\overrightarrow a dt}$

$\Rightarrow \overrightarrow v = \int {(t\widehat i + \widehat j)dt = {{{t^2}} \over 2}\widehat i + t\widehat j}$ ..... (1)

Also, $\overrightarrow v = {{\overrightarrow {dr} } \over {dt}} \Rightarrow \overrightarrow r = \int {\overrightarrow v dt}$

$\Rightarrow \overrightarrow r = \int {\left( {{{{t^2}} \over 2}\widehat i + t\widehat j} \right)dt = {{{t^3}} \over 6}\widehat i + {{{t^2}} \over 2}\widehat j}$ ..... (2)

Now, \overrightarrow \tau = \overrightarrow r \times \overrightarrow F = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr {{t^3}/6} & {{t^2}/2} & 0 \cr t & 1 & 0 \cr } } \right|

$\overrightarrow \tau = \widehat i(0) - \widehat j(0) + \widehat k\left( {{{{t^3}} \over 6} - {{{t^3}} \over 2}} \right) = {{ - 1} \over 3}{t^3}\widehat k$ N m

At t = 1.0 s, $\overrightarrow \tau = {{ - 1} \over 3}\widehat k$ N m

Hence, option (A) is correct.

Now, from Eq. (1), $\overrightarrow v = {{{t^2}} \over 2}\widehat i + t\widehat j$

At t = 1.0 s, $\overrightarrow v = \left( {{1 \over 2}\widehat i + \widehat j} \right)$ m s^$-$1 = $= {1 \over 2}(\widehat i + 2\widehat j)$ m s^$-$1

Hence, option (C) is correct.

From given data, we can draw the figure below.

Now, frequency of man behind is

${f_1} = f\left( {{v \over {v - {v_s}\cos \theta }}} \right)$

where, x is speed of sound in air, v_s is speed of man and f is frequency of whistle. Therefore,

${f_1} = 1430\left( {{{330} \over {330 - 2 \times {5 \over {13}}}}} \right)$

Frequency of man in front is

${f_2} = f\left( {{v \over {v + {v_s}\cos \theta }}} \right) = 1430\left( {{{330} \over {330 + 1 \times {5 \over {13}}}}} \right)$

Now, frequency of beat is given as

$\Delta f = {f_1} - {f_2}$

$= 1430\left( {{{330} \over {330 - {{10} \over {13}}}}} \right) - 1430\left( {{{330} \over {330 - {5 \over {13}}}}} \right)$

$= 1430 \times 330\left( {\left[ {{1 \over {330 - {{10} \over {13}}}}} \right] - \left[ {{1 \over {330 - {5 \over {13}}}}} \right]} \right)$

$= 1430 \times 330\left[ {{{13} \over {4280}} - {{13} \over {4295}}} \right]$

$= {{1430 \times 330 \times 13 \times 15} \over {4280 \times 4295}}$ = 5.0058 Hz $\approx$ 5.00 Hz

The vector $\overrightarrow A = \alpha \widehat i$ has a magnitude a and its direction is fixed. The vector $\overrightarrow B = a(\cos \omega t\widehat i + \sin \omega t\widehat j)$ rotates with an angular speed $\omega$ = $\pi$ / 6 rad/s in a circle of radius a. The magnitude of the sum and the difference of these vectors are given by

$\left| {\overrightarrow A + \overrightarrow B } \right| = \left| {(a + a\cos \omega t)\widehat i + a\sin \omega t\widehat j} \right|$

$= a\sqrt {{{(1 + \cos \omega t)}^2} + {{\sin }^2}\omega t}$

$= 2a\cos (\omega t/2)$.

$\left| {\overrightarrow A - \overrightarrow B } \right| = \left| {(a - a\cos \omega t)\widehat i - a\sin \omega t\widehat j} \right|$

$= 2a\sin (\omega t/2)$

Hence,

${{\left| {\overrightarrow A - \overrightarrow B } \right|} \over {\left| {\overrightarrow A + \overrightarrow B } \right|}} = \tan {{\omega t} \over 2} = {1 \over {\sqrt 3 }}$,

which gives $t = \tau = (2/\omega )(\pi /6) = 2$ s.

We know $B = {E \over c}$ or $E = Bc$

Now, $[E] = [B][c]$

We know that $[c] = [L{T^{ - 1}}]$

Therefore, $[E] = [B][L{T^{ - 1}}] = [B][L]{[T]^{ - 1}}$

The charged particle enters the magnetic field region at O. Its velocity is perpendicular to the field direction. The magnetic force on the particle, qv₀B₁, provides centripetal acceleration to move on a circular path from O to P. The radius of the circular path is given by

${r_1} = m{v_0}/(q{B_1})$.

The time taken by the particle to travel from O to P is

${T_1} = \pi {r_1}/{v_0} = \pi m/(q{B_1})$

At P, the particle enters the magnetic field B₂ with its velocity perpendicular to the field direction. The magnetic force qv₀B₂ provides the centripetal acceleration to move in a circular path from P to Q. The radius of the circular path is

${r_2} = m{v_0}/(q{B_2})$.

The time taken by the particle to travel from P to Q is

${T_2} = \pi {r_2}/{v_0} = \pi m/(q{B_2})$.

AT Q, the particle crosses x-axis from below for the first time. Thus, T = T₁ + T₂. The average speed of the particle along the x-axis in the time interval T is given by

$v = {{\int_0^T {{v_x}dt} } \over T} = {{\int_0^{{T_1}} {{v_{x1}}dt + \int_{{T_1}}^{{T_1} + {T_2}} {{v_{x2}}dt} } } \over {{T_1} + {T_2}}}$

$= {{2{r_1} + 2{r_2}} \over {{T_1} + {T_2}}} = {{{{2m{v_0}} \over q}\left( {{1 \over {{B_1}}} + {1 \over {{B_2}}}} \right)} \over {{{\pi m} \over q}\left( {{1 \over {{B_1}}} + {1 \over {{B_2}}}} \right)}}$

$= {{2{v_0}} \over \pi } = {{2(\pi )} \over \pi } = 2$ m/s

The magnetic field at the origin O (0, 0, 0) due to the infinitely long straight wire located at x = +R is

${\overrightarrow B _1}(O) = {{{\mu _0}{i_1}} \over {2\pi R}}\widehat k$

and due to the infinitely long straight wire located at x = $-$R is

${\overrightarrow B _2}(O) = {{{\mu _0}{i_2}} \over {2\pi R}}( - \widehat k)$.

The magnetic field at O due to the circular loop is

${\overrightarrow B _3}(O) = {{{\mu _0}i{R^2}} \over {2{{({R^2} + {{(\sqrt 3 R)}^2})}^{3/2}}}}( - \widehat k) = {{{\mu _0}i} \over {16R}}( - \widehat k)$.

The resultant magnetic field at the origin (0, 0, 0) is

${\overrightarrow B _3}(O) = {\overrightarrow B _1}(O) + {\overrightarrow B _2}(O) + {\overrightarrow B _3}(O)$

$= {{{\mu _0}} \over {16\pi R}}(8({i_1} - {i_2}) - \pi i)\widehat k$.

Thus, if i₁ = i₂ then ${\overrightarrow B _1}(O) = - {\overrightarrow B _2}(O)$ and ${\overrightarrow B _1}(O) = {\overrightarrow B _3}(O) \ne \overrightarrow 0$.

If i₁ > 0 and i₂ < 0 then both ${\overrightarrow B _1}(O)$ and ${\overrightarrow B _2}(O)$ are along $\widehat k$ but ${\overrightarrow B _3}(O)$ is along $-$$\widehat k$. Thus, resultant field $\overrightarrow B (O)$ can be zero.

If i₁ < 0 and i₂ > 0 then ${\overrightarrow B _1}(O)$, ${\overrightarrow B _2}(O)$ and ${\overrightarrow B _3}(O)$ are all along $-$$\widehat k$. Thus, resultant field $\overrightarrow B (O)$ cannot be zero. We encourage you to relate these results with the feasible solutions of 8(i₁ $-$ i₂) $-$ $\pi$i = 0 for i > 0.

Now, let us find the magnetic field at the centre of the loop i.e., point C in the figure.

The magnetic field $\overrightarrow B_1$ at the point C due to the infinitely long straight wire located x = +R is perpendicular to the line AC (see figure). It lies in the x-z plane and makes an angle 30$^\circ$ with the x axis. Resolve $\overrightarrow B_1$ along the x and z directions to get

${{\vec B}_1}(C) = {{{\mu _0}{i_1}} \over {2\pi (2R)}}(\cos 30^\circ \widehat i + \sin 30^\circ \widehat k)$.

Similarly, the magnetic field at C due to the infinitely long straight wire located at x = $-$R is

${{\vec B}_2}(C) = {{{\mu _0}{i_2}} \over {2\pi (2R)}}(\cos 30^\circ \widehat i - \sin 30^\circ \widehat k)$.

The magnetic field at C due to the circular loop is

${{\vec B}_3}(C) = {{{\mu _0}i} \over {(2R)}}( - \widehat k)$.

The resultant magnetic field at C is

${{\vec B}_1}(C) = {{\vec B}_1}(C) + {{\vec B}_2}(C) + {{\vec B}_3}(C) = {{{\mu _0}} \over {8\pi R}}\left( {\sqrt 3 ({i_1} + {i_2})\widehat i + ({i_1} - {i_2} - 4\pi i)\widehat k} \right)$.

If i₁ = i₂ then magnitudes of ${{\vec B}_1}(C)$ and ${{\vec B}_2}(C)$ are equal and their z-components are equal in magnitude but opposite in direction. Thus, z-component of the magnetic field at C is due to the circular loop only and its value is $\left( { - {{{\mu _0}i} \over {2R}}} \right)$.

We know that,

$h = {{2\sigma \cos \theta } \over {r\,\rho g}}$..... (1)

where r is inner radius of capillary tube; $\sigma$ is surface tension; $\theta$ is angle between water and wall of capillary tube; h is height of water in capillary tube and g is acceleration due to gravity. From Eq. (1),

$h \propto {1 \over r}$

Therefore, for a given material of the capillary tube, h decreases with increase in r.

Hence, option (A) is correct.

h depends on $\sigma$, therefore option (B) is incorrect.

In a lift going up with a constant acceleration a, the force due to surface tension remains same but weight of the water increases i.e., W_lift = $\pi$r²h$\rho$(g + a). Thus, the height of the water rise in a capillary becomes

$h = {{2\sigma \cos \theta } \over {r\,\rho (g + a)}} = {{2\sigma \cos \theta } \over {r\, \rho{g_{eff}}}}$,

where g_eff is the effective g in the lift. Hence, h decrease if the experiment is performed in a lift going up with a constant acceleration.

From (1), h $\propto$ cos$\theta$ and not $\theta$, therefore, option (D) is incorrect.

In process I, volume is changing. Therefore, it is not isochoric. Therefore 'A' is incorrect.

In process II, q = $\Delta$U + W . $\Delta$U = 0 as temperature is constant. Therefore, q = W. Here W = P(V_f $-$ V_i) is positive therefore q is positive i.e., gas absorbs heat. Therefore 'B' is correct.

For process IV, q = $\Delta$U + W. Here $\Delta$ = 0 and W is negative (volume decreases). Therefore q is negative i.e., gas releases heat. 'C' is correct.

For an isobaric process, V $\propto$ T i.e., we will get a straight inclined line in T-V graph. Therefore I and II are NOT isobaric. 'D' is correct.

Given, ratio $r = {{1 - a} \over {1 + a}}$ ..... (1)

Given, error in measurement of a is $\Delta$a ($\Delta$a/a << 1)

Taking natural log of Eq. (1), we get

$\ln r = \ln \left( {{{1 - a} \over {1 + a}}} \right)$

$\Rightarrow \ln r = \ln (1 - a) - \ln (1 + a)$

${{\Delta r} \over r} = {{\Delta a} \over {1 - a}} + {{\Delta a} \over {1 + a}} = {{(1 + a)\Delta a + (1 - a)\Delta a} \over {(1 - a)(1 + a)}}$

$\Rightarrow {{\Delta r} \over r} = {{2\Delta a} \over {(1 - a)(1 + a)}}$

$\Rightarrow \Delta r = r\,.\,{{2\Delta a} \over {(1 - a)(1 + a)}}$

Substituting value of r, we get

$\Delta r = {{1 - a} \over {1 + a}}{{2\Delta a} \over {(1 - a)(1 + a)}} = {{2\Delta a} \over {{{(1 + a)}^2}}}$

Therefore, error $\Delta$r in determining r is ${{2\Delta a} \over {{{(1 + a)}^2}}}$

After collision the 2.0 kg block will perform simple harmonic oscillation with time period

$T = 2\pi \sqrt {{m \over k}}$

Given m = 2.0 kg and k = 2.0 N m^$-$1, we have

T = 2$\pi$

Thus, the block returns to its original position in time

$t = {T \over 2}$ = $\pi$ s

That is, t = 3.14 s

Now, if v₁ and v₂ are velocities of 1.0 kg block and 2.0 kg block, respectively, before collision; v'₁ and v'₂ are velocities of 1.0 kg block and 2.0 block, respectively, after collision. So, by conservation of momentum

m₁v₁ + m₂v₂ = m₁v'₁ + m₂v'₂

Here, m₁ = 1.0 kg, v₁ is initial speed of 1.0 kg block = 2.0 m s^$-$1, m₂ = 2.0 kg, v₂ is initial speed of 2.0 kg block = 0.0 m s^$-$1, v'₁ is final speed of 1.0 kg block after collision and v'₂ is final speed of 2.0 kg block after collision. Then, 1.0 kg $\times$ 2.0 m/s + 2.0 kg $\times$ 0 m/s = 1.0 v'₁ + 2.0 v'₂

v'₁ + 2v'₂ = 2 ..... (1)

Also, using definition of coefficient of restitution

v'₂ $-$ v'₁ = $\varepsilon$(v₁ $-$ v₂)

Since collision is elastic, So $\varepsilon$ = 1

$\Rightarrow$ v'₂ $-$ v'₁ = v₁ $-$ v₂

$\Rightarrow$ v'₂ $-$ v'₁ = 2 $-$ 0

$\Rightarrow$ v'₂ $-$ v'₁ = 2 ...... (2)

From Eqs. (1) and (2), we get

$v{'_2} = {4 \over 3}$ m s^$-$1 and $v{'_1} = {-2 \over 3}$ m s^$-$1

Therefore, distance between the blocks is given as

$s = v{'_1} \times t = {{ - 2} \over 3} \times 3.14 = 2.09$ m

$c = {1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$

${c^2} = {1 \over {{\mu _0}{\varepsilon _0}}}$

${\mu _0} = \varepsilon _0^{ - 1}\,.\,{c^{ - 2}}$

$[{\mu _0}] = {[{\varepsilon _0}]^{ - 1}}[{L^{ - 2}}{T^2}]$

The intermediate temperature is given by the formula

$T = {{{{{k_1}{A_1}{T_1}} \over {{l_1}}} + {{{k_2}{A_2}{T_2}} \over {{l_2}}}} \over {{{{k_1}{A_1}} \over {{l_1}}} + {{{k_2}{A_2}} \over {{l_2}}}}}$

Here, T = 200 k, T₁ = 300 k, T₂ = 100 k, l₁ = l₂ and A₁ = $\pi$r², A₂ = 4$\pi$r²

$\therefore$ $200 = {{{k_1}\pi {r^2} \times 300 + {k_2}\pi (4{r^2})100} \over {{k_1}\pi {r^2} + {k_2}\pi (4{r^2})}}$

$\therefore$ $200 = {{300{k_1} + 400{k_2}} \over {{k_1} + 4{k_2}}}$

$\therefore$ 200 k₁ + 800 k₂ = 300 k₁ + 400 k₂

$\Rightarrow$ 400 k₂ = 100 k₁

$\therefore$ ${{{k_1}} \over {{k_2}}} = 4.00$

The law of radioactive decay gives number of undecayed nuclei N at time t as $N = {N_0}{e^{ - \lambda t}}$, where N₀ is number of nuclei at t = 0 and $\lambda$ is the decay constant. The number of nuclei decayed till time t is given by ${N_d} = {N_0} - N$. Thus, the decay constant is given by

$\lambda = {1 \over t}\ln {{{N_0}} \over N} = {1 \over t}\ln {{{N_0}} \over {{N_0} - {N_d}}}$.

Since, N₀ = 3000 and t = 1 s are given without any error, we assume that error in $\lambda$ is due to error in N_d only. Hence, we can write above equation as

$\lambda + \Delta \lambda = {1 \over t}\ln {{{N_0}} \over {{N_0} - ({N_d} + \Delta {N_d})}}$

$= {1 \over t}\ln {{{N_0}} \over {({N_0} - {N_d})(1 - \Delta {N_d}/({N_0} - {N_d}))}}$

$= {1 \over t}\ln {{{N_0}} \over {{N_0} - {N_d}}} - {1 \over t}\ln \left( {1 - {{\Delta {N_d}} \over {{N_0} - {N_d}}}} \right)$

$\approx \lambda + {1 \over t}{{\Delta {N_d}} \over {{N_0} - {N_d}}}$,

which gives $\Delta \lambda = {{\Delta {N_d}} \over {t({N_0} - {N_d})}}$. Substitute $\Delta$N_d = 40, t = 1 s, and N_d = 1000 to get $\Delta$$\lambda$ = 0.02.