JEE Main 2019 8th April Morning Shift Question Paper with Solutions

${\overrightarrow a _A} = - a\widehat i$;    ${\overrightarrow a _B} = - a\widehat j$

${\overrightarrow a _C} = - a\widehat i$;    ${\overrightarrow a _D} = - a\widehat j$

${\overrightarrow a _{cm}} = {{{m_a}{{\overrightarrow a }_a} + {m_b}{{\overrightarrow a }_b} + {m_c}{{\overrightarrow a }_c} + {m_d}{{\overrightarrow a }_d}} \over {{m_a} + {m_b} + {m_c} + {m_d}}}$

${\overrightarrow a _{cm}} = {{ - ma\widehat i + 2m\widehat j + 3ma\widehat i - 4ma\widehat j} \over {10m}}$

= ${{2ma\widehat i - 2ma\widehat j} \over {10m}} = {a \over 5}\widehat i - {a \over 5}\widehat j = {a \over 5}\left( {\widehat i - \widehat j} \right)$

P_R = i² × R

P_B = V × i

$\therefore$ P_L = Vi – i²R

$\Rightarrow$ Vi – i²R = i²R

$\Rightarrow$ $i = {V \over {2R}}\,\,and\,\,i = {V \over R}\left( {1 - {e^{ - t/\tau }}} \right)$

$\therefore$ ${V \over {2R}} = {V \over R}\left( {1 - {e^{ - t/\tau }}} \right)$

$\Rightarrow$ t = $\tau$ln(2) = ${{20} \over {10}}ln(2) = 2ln(2)$

$\sqrt {{{{ \in _0}} \over {{\mu _0}}}}$ = ${{{ \in _0}} \over {\sqrt {{\mu _0}{ \in _0}} }}$ = c $\times$ ${{ \in _0}}$

$\therefore$ $\left[ {\sqrt {{{{ \in _0}} \over {{\mu _0}}}} } \right]$ = $\left[ {L{T^{ - 1}}} \right] \times \left[ {{ \in _0}} \right]$

We know, F = ${1 \over {4\pi { \in _0}}}{{{q^2}} \over {{r^2}}}$

$\therefore$ ${ \in _0} = {{{q^2}} \over {4\pi {r^2}F}}$

$\Rightarrow$ $\left[ {{ \in _0}} \right] = {{{{\left[ {AT} \right]}^2}} \over {\left[ {ML{T^{ - 2}}} \right] \times \left[ {{L^2}} \right]}}$ = $\left[ {{A^2}{M^{ - 1}}{L^{ - 3}}{T^4}} \right]$

$\therefore$ $\left[ {\sqrt {{{{ \in _0}} \over {{\mu _0}}}} } \right]$ = $\left[ {L{T^{ - 1}}} \right]$ $\times$ $\left[ {{A^2}{M^{ - 1}}{L^{ - 3}}{T^4}} \right]$

                 = [A²T³M^–1L^–2]

A = 10^–4 m²

E_max = 10⁶ V/m

C = 15 $\mu$F

$C = {{k{\varepsilon _0}A} \over d};{{Cd} \over {{\varepsilon _0}A}} = k$

$k = {{15 \times {{10}^{ - 12}} \times 500 \times {{10}^{ - 6}}} \over {8.86 \times {{10}^{ - 12}} \times {{10}^4}}}$

= ${{15 \times 5} \over {8.86}} = 8.465$

k $\approx$ 8.5

According to the question, the situation can be drawn as

Let the current I is flowing in anti-clockwise direction, then the magnetic moment of the coil is

m = NIA

where, N = number of turns in coil

and A = area of each coil = $\pi$r².

Its direction is perpendicular to the area of coil and is along Y-axis.

Then, torque on the current coil is

$\tau = m \times B = mB\sin 90^\circ = NIAB$

$= NI\pi {r^2}B(N - m)$

Let x grams of water is evaporated.

According to the principle of calorimetry,

Heat lost by freezing water (that turns into ice) = Heat gained by evaporated water

Given, mass of water = 150 g

$\Rightarrow (150 - x) \times {10^{ - 3}} \times 3.36 \times {10^5} = x \times {10^{ - 3}} \times 2.10 \times {10^6}$

$\Rightarrow (150 - x) \times 3.36 = 21x$

$\Rightarrow x = {{150} \over {7.25}} = 20.6$

$\therefore$ $x \approx 20g$

Tensile stress in wire will be

= ${{Tensile{\rm{ }}force} \over {Cross{\rm{ }}section{\rm{ }}Area}}$

= ${{mg} \over {\pi {R^2}}} = {{4 \times 3.1\pi } \over {\pi \times 4 \times {{10}^{ - 6}}}}N{m^{ - 2}}$

= 3.1 × 10⁶ Nm^–2

${{{I_{\max }}} \over {{I_{\min }}}} = {{{{\left( {{a_1} + {a_2}} \right)}^2}} \over {{{\left( {{a_1} - {a_2}} \right)}^2}}} = {{{{\left( {1 + 3} \right)}^2}} \over {{{\left( {1 - 3} \right)}^2}}} = {{16} \over 4}$ = 4

Wavelength of carrier waves in modern optical fiber communication is most widely used near about 1500 nm.

Flow rate of water (Q) = 100 lit/min

= ${{100 \times {{10}^{ - 3}}} \over {60}} = {5 \over 3} \times {10^{ - 3}}{m^3}$

$\therefore$ Velocity of flow (v)

= ${Q \over A} = {{5 \times {{10}^{ - 3}}} \over {3 \times \pi \times {{(5 \times {{10}^{ - 2}})}^2}}}$

$= {{10} \over {15\pi }} = {2 \over {3\pi }}\,m/s$

= 0.2 m/s

$\therefore$ Reynold number (R_e) = ${{Dv\rho } \over \eta }$

$= {{\left( {10 \times {{10}^{ - 2}}} \right) \times {2 \over {3\pi }} \times 1000} \over 1} = 2 \times {10^4}$

$\therefore$ Order of R_e = 10⁴

When rubber cord is stretched, then it stores potential energy and when released, this potential energy is given to the stone as kinetic energy.

So, potential energy of stretched cord = kinetic energy of stone

$\Rightarrow {1 \over 2}Y{\left( {{{\Delta L} \over L}} \right)^2}A\,.\,L = {1 \over 2}m{v^2}$

Here, $\Delta$L = 20 cm = 0.2 m, L = 42 cm = 0.42 m, v = 20 ms^$-$1, m = 0.02 kg, d = 6 mm = 6 $\times$ 10^$-$3 m

$\therefore$ $A = \pi {r^2} = \pi {\left( {{d \over 2}} \right)^2} = \pi {\left( {{{6 \times {{10}^{ - 3}}} \over 2}} \right)^2}$

$= \pi {(3 \times {10^{ - 3}})^2} = 9\pi \times {10^{ - 6}}{m^2}$

On substituting values, we get

$Y = {{m{v^2}L} \over {A{{(\Delta L)}^2}}} = {{0.02 \times {{(20)}^2} \times 0.42} \over {9\pi \times {{10}^{ - 6}} \times {{(0.2)}^2}}}$

$\approx 3.0 \times {10^6}N{m^{ - 2}}$

So, the closest value of Young's modulus is 10⁶ Nm^$-$2.

In an AC resistive circuit, current and voltage are in phase.

So, $I = {V \over R} \Rightarrow I = {{220} \over {50}}\sin (100\pi t)$ ..... (i)

$\therefore$ Time period of one complete cycle of current is

$T = {{2\pi } \over \omega } = {{2\pi } \over {100\pi }} = {1 \over {50}}s$

So, current reaches its maximum value at

${t_1} = {T \over 4} = {1 \over {200}}s$

When current is half of its maximum value, then from Eq.(i), we have

$I = {{{I_{\max }}} \over 2} = {I_{\max }}\sin (100\pi {t_2})$

$\Rightarrow \sin (100\pi {t_2}) = {1 \over 2} \Rightarrow 100\pi {t_2} = {{5\pi } \over 6}$

So, instantaneous time at which current is half of maximum value is ${t_2} = {1 \over {120}}s$

Hence, time duration in which current reaches half of its maximum value after reaching maximum value is

$\Delta t = {t_2} - {t_1} = {1 \over {120}} - {1 \over {200}} = {1 \over {300}}s = 3.3$ ms

Initially when uncharged shell encloses charge Q, charge distribution due to induction will be as shown,

The potential on surface of inner shell is

${V_A} = {{kQ} \over a} + {{k( - Q)} \over b} + {{kQ} \over b}$ ..... (i)

where, k = proportionality constant.

Potential on surface of outer shell is

${V_B} = {{kQ} \over b} + {{k( - Q)} \over b} + {{kQ} \over b}$ ..... (ii)

Then, potential difference is

$\Delta {V_{AB}} = {V_A} - {V_B} = kQ\left( {{1 \over a} - {1 \over b}} \right)$

Given, $\Delta {V_{AB}} = V$

So, $kQ\left( {{1 \over a} - {1 \over b}} \right) = V$ ....... (iii)

Finally after giving charge $-$ 4Q to outer shell, potential difference will be

$\Delta {V_{AB}} = {V_A} - {V_B}$

$= \left( {{{kQ} \over a} + {{k( - 4Q)} \over b}} \right) - \left( {{{kQ} \over b} + {{k( - 4Q)} \over b}} \right)$

$= kQ\left( {{1 \over a} - {1 \over b}} \right) = V$ [from Eq. (iii)]

Hence, we obtain that potential difference does not depend on the charge of outer sphere, hence potential difference remains same.

Net force on mass M at position B towards centre of circle is

${F_{BO\,net}} = {F_{BD}} + {F_{BA}}\sin 45^\circ + {F_{BC}}\sin 45^\circ$

$= {{G{M^2}} \over {{{(\sqrt 2 a)}^2}}} + {{G{M^2}} \over {{a^2}}}\left( {{1 \over {\sqrt 2 }}} \right) + {{G{M^2}} \over {{a^2}}}\left( {{1 \over {\sqrt 2 }}} \right)$

[where, diagonal length BD is $\sqrt2$a]

$= {{G{M^2}} \over {2{a^2}}} + {{G{M^2}} \over {{a^2}}}\left( {{2 \over {\sqrt 2 }}} \right) = {{G{M^2}} \over {{a^2}}}\left( {{1 \over 2} + \sqrt 2 } \right)$

This force will act as centripetal force.

Distance of particle from centre of circle is ${a \over {\sqrt 2 }}$.

Here, ${F_{centripetal}} = {{M{v^2}} \over r} = {{M{v^2}} \over {{a \over {\sqrt 2 }}}} = {{\sqrt 2 M{v^2}} \over a}$ ($\because$ $r = {a \over {\sqrt 2 }}$)

So, for rotation about the centre,

${F_{centripetal}} = {F_{BO(net)}}$

$\Rightarrow \sqrt 2 {{M{v^2}} \over a} = {{G{M^2}} \over {{a^2}}}\left( {{1 \over 2} + \sqrt 2 } \right)$

$\Rightarrow {v^2} = {{GM} \over a}\left( {1 + {1 \over {2\sqrt 2 }}} \right) = {{GM} \over a}(1.35)$

$\Rightarrow v = 1.16\sqrt {{{GM} \over a}}$

Tcos$\theta$ = mg

Tsin$\theta$ = qE

tan$\theta$ = ${{qE} \over {mg}}$

tan$\theta$ = ${{5 \times {{10}^{ - 16}} \times 2000} \over {2 \times {{10}^{ - 3}} \times 10}} = {1 \over 2}$

$\Rightarrow$ ${\tan ^{ - 1}}\left( {{1 \over 2}} \right) = {\tan ^{ - 1}}\left( {0.5} \right)$

The direction of propagation of an EM wave is direction of $\overrightarrow E \times \overrightarrow B$

$\widehat i = \widehat j \times \widehat B$

$\Rightarrow \widehat B = \widehat k$

$C = {E \over B} \Rightarrow B = {E \over C} = {6 \over {3 \times {{10}^8}}}$

B = 2 × 10^–8 T along z direction.

We know, m = V . $\rho$

where, V = volume and $\rho$ = density.

So, we have

${{dQ} \over {dt}} = {h \over {ms}}(T - {T_0}) = {{h(T - {T_0})} \over {V\,.\,\rho s}}$

Since, h, (T $-$ T₀) and V are constant for both beaker.

$\therefore$ ${{dQ} \over {dt}} \propto {1 \over {\rho s}}$

We have given that $\rho$_A = 8 $\times$ 10² kgm^$-$3, $\rho$_B = 10³ kgm^$-$3, s_A = 2000 J kg^$-$1 K^$-$1 and s_B = 4000 J kg^$-$1 K^$-$1, $\rho$_As_A = 16 $\times$ 10⁵ and $\rho$_Bs_B = 4 $\times$ 10⁶

$\rho$_A < $\rho$_B, s_A < s_B and $\rho$_As_A < $\rho$_Bs_B

$\Rightarrow {1 \over {{\rho _A}{s_A}}} > {1 \over {{\rho _B}{s_B}}} \Rightarrow {{d{Q_A}} \over {dt}} > {{d{Q_B}} \over {dt}}$

So, for container B, rate of cooling is smaller than the container A. Hence, graph of B lies above the graph of A and it is not a straight line (slope of A is greater than B).

$M = \int\limits_0^R {{\rho _0}r \times 2\pi rdr = {{2\pi {\rho _0}{R^3}} \over 3}}$

${I_C} = \int\limits_0^R {{\rho _0}r \times 2\pi rdr \times {r^2} = {{2\pi {\rho _0}{R^5}} \over 3}}$

$\therefore$ $I = {I_C} + M{R^2} = 2\pi {\rho _0}{R^5}\left( {{1 \over 3} + {1 \over 5}} \right)$

$\Rightarrow$${{16\pi {\rho _0}{R^5}} \over {15}} = {8 \over 5}\left[ {{2 \over 3}\pi {\rho _0}{R^3}} \right]{R^2} = {8 \over 5}M{R^2}$

There are two forces on slider.

Spring force = kx

where, k = spring constant.

As the slider is kept in a uniform magnetic field B = 0.1 T, hence it will experience a force, i.e.

Magnetic force = Bil

where, l = length of the strip.

Now, using

F_net = ma

We have,

$( - kx) + ( - Bil) = ma$

$\Rightarrow - kx - Bil - ma = 0$

$\Rightarrow - kx - {{{B^2}{l^2}} \over R}.\,v - ma = 0$ [$\because$ $i = {{Blv} \over R}$]

and acceleration, $a = {{{d^2}x} \over {d{t^2}}}$

Hence, the modified equation becomes

$\Rightarrow {{m{d^2}x} \over {d{t^2}}} + {{{B^2}{l^2}} \over R}\left( {{{dx} \over {dt}}} \right) + kx = 0$

This is the equation of damped simple harmonic motion.

So, amplitude of oscillation varies with time as

$A = {A_e}^{ - {{{B^2}{l^2}} \over {2Rm}}.\,t}$

Now, when amplitude is ${{{A_0}} \over e}$, then

${{{A_0}} \over e} = {{{A_0}} \over {{e^{{{{B^2}{l^2}} \over {2Rm}}.\,t}}}}$ (as given)

$\Rightarrow \left( {{{{B^2}{l^2}} \over {2Rm}}} \right)t = 1$ or $t = {{2Rm} \over {{B^2}{l^2}}}$

According to the question, magnetic field B = 0.1 T, mass of strip m = 50 $\times$ 10^$-$3 kg, resistance R = 10 $\Omega$, l = 10 cm = 10 $\times$ 10^$-$2 m

$\therefore$ $t = {{2Rm} \over {{B^2}{l^2}}} = {{2 \times 10 \times 50 \times {{10}^{ - 3}}} \over {{{(0.1)}^2} \times {{(10 \times {{10}^{ - 2}})}^2}}}$

$= {1 \over {{{10}^{ - 4}}}} = 10000$ s

Given, spring constant, k = 0.5 Nm^$-$1

Also, time period of oscillation is

$T = 2\pi \sqrt {{m \over k}} = 2\pi \sqrt {{{50 \times {{10}^{ - 3}}} \over {0.5}}} = {{2\pi } \over {\sqrt {10} }} \approx 2s$

So, number of oscillations is

$N = {t \over T} = {{10000} \over 2} = 5000$

Considering the initial position of ship A as origin, so the velocity and position of ship will be

${\overrightarrow v _A} = (30\widehat i + 50\widehat j)$ and ${\overrightarrow r _A} = (0\widehat i + 0\widehat j)$

Now, as given in the question, velocity and position of ship B will be, ${\overrightarrow v _B} = - 10\widehat i$ and ${\overrightarrow r _B} = (80\widehat i + 150\widehat j)$

Time after which the distance is minimum between A and B can be calculated as

$t = {{|{{\overrightarrow r }_{BA}}.\,{{\overrightarrow v }_{BA}}|} \over {|{{\overrightarrow v }_{BA}}{|^2}}}$

where, ${\overrightarrow r _{BA}} = {\overrightarrow r _B} - {\overrightarrow r _A} = 80\widehat i + 150\widehat j$

and ${\overrightarrow v _{BA}} = - 10\widehat i - (30\widehat i + 50\widehat j)$

$= - 40\widehat i - 50\widehat j$

$\Rightarrow t = {{|(80\widehat i + 150\widehat j)\,.\,( - 40\widehat i - 50\widehat j)|} \over {| - 40\widehat i - 50\widehat j{|^2}}}$

$= {{3200 + 7500} \over {4100}} = {{10700} \over {4100}} = 2.6$ h

According to work energy theorem.

Work done by force on the particle = Change in KE

Work done = Area under F-x graph = $\int {F.dx}$

= $2 \times 2 + {{(2 + 3) \times 1} \over 2}$

W = KE_final – KE_initial = 6.5

KE_initial = 0

KE_final = 6.5 J

200 $\Omega$ = Red + Black + Brown

Green = 5

So, Green + Black + Brown = 500 $\Omega$

9 = Vz + V_R1

V_Z = 5.6 V

V_R1 = 9 - 5.6

V_R1 = 3.4

${I_{{R_1}}} = {{{V_{{R_1}}}} \over R} = {{3.4} \over {200}};\,{I_{{R_1}}} = 17\,mA$

V_z = V_R2 = I_R2 (R₂ )

${{5.6} \over {700}} = {I_{{R_2}}}$;     ${I_{{R_2}}} = 7\,mA$

I_z = (17 – 7) mA = 10 mA

Energy released for tension n = 2 to n = 1 of hydrogen atom

$E = 13.6{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$

Z = 1, n₁ = 1, n₂ = 2

$E = 13.6 \times 1 \times \left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$

$E = 13.6 \times {3 \over 4}eV$ = 10.2 eV

For He⁺ ion z = 2

(A) n = 1 to n = 4
   $E = 13.6 \times {2^2} \times \left( {{1 \over {{1^2}}} - {1 \over {{4^2}}}} \right) = 13.6 \times {{15} \over 4}eV$

(B) n = 2 to n = 4
   $E = 13.6 \times {2^2} \times \left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = 13.6 \times {{3} \over 4}eV$

(C) n = 2 to n = 5
   $E = 13.6 \times {2^2} \times \left( {{1 \over {{2^2}}} - {1 \over {{5^2}}}} \right) = 13.6 \times {{21} \over 25}eV$

(D) n = 2 to n = 3
   $E = 13.6 \times {2^2} \times \left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right) = 13.6 \times {{5} \over 9}eV$

So, possible transition is n = 2 $\to$ n = 4

If we approximate the angle $\theta _2$ as 30° initially then answer will be closer to 57000. but if we solve thoroughly, answer will be close to 55000.
So both the answers must be awarded. Detailed solution as following.

EXACT SOLUTION
By Snell’s law 1.sin 40° = (1.31) sin $\theta _2$

$\sin {\theta _2} = {{0.64} \over {1.31}} = {{64} \over {131}} \approx .49$

Now tan$\theta _2$
= ${{64} \over {\sqrt {{{\left( {131} \right)}^2} - {{\left( {64} \right)}^2}} }} = {{64} \over {\sqrt {13065} }} \approx {{64} \over {114.3}} = {d \over x}$

Now number of reflections

$= {{2 \times 64} \over {114.3 \times 20 \times {{10}^{ - 6}}}} = {{64 \times {{10}^5}} \over {114.3}}$

$\approx 55991 \approx 55000$

APPROXIMATE SOLUTION
By Snell’s law 1.sin 40° = (1.31)sin $\theta _2$

$\sin {\theta _2} = {{0.64} \over {1.31}} = {{64} \over {131}} \approx .49$

If assume $\Rightarrow$ $\theta _2$ $\approx$ 30°

tan 30° = ${d \over x} \Rightarrow x = {\sqrt3} d$

Now number of reflections
$= {t \over {\sqrt 3 d}} = {2 \over {\sqrt 3 \times 20 \times {{10}^{ - 6}}}} = {{{{10}^5}} \over {\sqrt 3 }}$

$\approx 57735 \approx 57000$

Let mass per unit length of wires are $\mu$₁ and $\mu$₂ respectively

$\because$ Materials are same, so density $\rho$ will be same.

$\therefore$ ${\mu _1} = {{\rho \pi {r^2}L} \over L} = \mu$ and ${\mu _2} = {{\rho 4\pi {r^2}L} \over L} = 4\mu$

Tension in both are same = T, let speed of wave in wires are V₁ and V₂

${V_1} = {{{V_1}} \over {2L}} = {V \over {2L}}\,\,\& \,{V_2} = {{{V_2}} \over {2L}} = {V \over {4L}}$

Frequency at which both resonate is L.C.M. of both frequencies (i.e : ${V \over {2L}}$ )

Hence number of loops in wires are 1 and 2 respectively

So, ratio of number of antinodes is 1 : 2.

In given system of lens and mirror, position of object O in front of lens is at a distance 2f. i.e. u = 2f = 40 cm

So, image (I₁) formed is real, inverted and at a distance, v = 2f = 2 $\times$ 20 = 40 cm, (behind lens) magnification, ${m_1} = {v \over u} = {{40} \over {40}} = 1$

Thus, size of image is same as that of that of object. This image (I₁) acts like a real object for mirror.

As object distance for mirror is

u = C = 2f = $-$ 20 cm

where, C = centre of curvature.

So, image (I₂) formed by mirror is at 2f.

$\therefore$ For mirror v = 2f = 2($-$ 10) = $-$ 20 cm

Magnification, ${m_2} = - {v \over u} = - {{( - 20)} \over {( - 20)}} = - 1$

Thus, image size is same as that of object.

The image I₂ formed by the mirror will act like an object for lens.

As the object is at 2f distance from lens, so image (I₃) will be formed at a distance 2f or 40 cm. Thus, magnification,

${m_3} = {v \over u} = {{40} \over {40}} = 1$

So, final magnification, $m = {m_1} \times {m_2} \times {m_3} = - 1$

Hence, final image (I₃) is real, inverted of same size as that of object and coinciding with object.

Let the two particles be moving along x-direction and y-direction.

So, the net momentum initially is $\sqrt {{{{h^2}} \over {\lambda _1^2}} + {{{h^2}} \over {\lambda _2^2}}}$

and final momentum will be ${h \over \lambda }$.

Applying momentum conservation,

${h \over \lambda } = \sqrt {{{{h^2}} \over {\lambda _1^2}} + {{{h^2}} \over {\lambda _2^2}}}$

$\Rightarrow$ ${1 \over {{\lambda ^2}}} = {1 \over {\lambda _1^2}} + {1 \over {\lambda _2^2}}$

Momentum imparted to the surface in one collision,

$\Delta p = ({p_i} - {p_f}) = mv - ( - mv) = 2mv$ ..... (i)

Force on the surface due to n collision per second, $F = {n \over t}(\Delta p) = n\Delta p$ ($\because$ $t = 1s$)

= 2 mnv [from Eq. (i)]

So, pressure on the surface,

$p = {F \over A} = {{2mnv} \over A}$

Here, m = 10^$-$26 kg, n = 10²² s^$-$1, v = 10⁴ ms^$-$1, A = 1 m²

$\therefore$ Pressure, $p = {{2 \times {{10}^{ - 26}} \times {{10}^{22}} \times {{10}^4}} \over 1} = 2$ N/m²

So, pressure exerted is of order of 10⁰.

Given circuit is

Above circuit can be viewed as

This is a parallel combination of three cells or in other words, a parallel grouping of three cells with internal resistances.

So, ${V_{ab}} = {E_{eq}} = {{{I_{eq}}} \over {{r_{eq}}}} = {{{{{E_1}} \over {{r_1}}} + {{{E_2}} \over {{r_2}}} + {{{E_3}} \over {{r_3}}}} \over {{1 \over {{r_1}}} + {1 \over {{r_2}}} + {1 \over {{r_3}}}}}$

$= {{{2 \over 2} + {4 \over 2} + {4 \over 2}} \over {{1 \over 2} + {1 \over 2} + {1 \over 2}}} = {{10} \over 3}V \approx 3.3\,V$

(1) The organic compound does not react with natural FeCl₃ solution. It means phenolic group is absent in the organic compound.

(2) The organic compound does not give Fehling solution test. It means aldehyde group (- CHO) is absent in the organic compound.

From 1^st law of thermodynamics we know,

$\Delta$U = q + W

Option A :

In adiabatic process exchage of heat = 0

$\therefore$ q = 0

$\therefore$ From 1st law of thermodynaics, $\Delta$U = W

So option A is wrong.

Option B :

U is a state function. In cyclic process, initial state and final state both are same. So change in all the state function in cyclic process will be zero.

$\therefore$ $\Delta$U = 0

$\therefore$ From 1st law of thermodynaics,

q + W = 0 $\Rightarrow$ q = -W

So option B is correct..

Option C :

In isochoric process volume (V) is constant. So dV = 0.

We know, W = $- \int {{P_{ex}}} dV$

$\therefore$ W = 0

$\therefore$ From 1st law of thermodynaics,

$\Delta$U = q

So option C is correct.

Option D :

In isothermal process temerature (T) is constant. So dT = 0.

We know, $\Delta$U = nC_vdT

$\therefore$ $\Delta$U = 0

$\therefore$ From 1st law of thermodynaics,

q + W = 0 $\Rightarrow$ q = -W

So option D is correct.

Which electrode have higher value of standard reduction potential (SRP), that electrode will be strongest oxidizing agent.

Tendency to gain electrone is called standard reduction potential. When tendency to gain electron is more then that electrode will have more oxidizing power.

Here given SRP's are :

${E^\Theta }_{{O_2}/{H_2}O} = 1.23\,V$ ;

${E^\Theta }_{{S_2}O_8^{2 - }/SO_4^{2 - }} = 2.05\,V$

${E^\Theta }_{B{r_2}/B{r^ - }} = 1.09\,V$

${E^\Theta }_{A{u^{3 + }}/Au} = 1.4\,V$

You can see ${E^\Theta }_{{S_2}O_8^{2 - }/SO_4^{2 - }} = 2.05$,

have highest SRP so ${S_2}O_8^{2 - }$ is the strongest oxidizing agent.

Maltose on treatment with dil. HCl gives D-glucose. Hydrolysis of maltose yields two moles of $\alpha$- D-glucose. Thus, it is composed of two $\alpha$-D-glucose units in which C-1 of one glucose unit (I) is linked to C-4 of another glucose unit (II). The free aldehyde group can be produced at C-1 of second glucose in solution and it shows reducing properties.
So, it is a reducing sugar.

$\therefore$ Equivalent of KMnO₄ = Eq of FeC₂O₄ + Eq of Fe₂(C₂O₄)₃ + FeSO₄

$\Rightarrow$ moles of KMnO₄ $\times$ 5 = 3 $\times$ 1 + 6 $\times$ 1 + 1 $\times$ 1

$\Rightarrow$ moles of KMnO₄ = ${{10} \over 5}$ = 2