JEE Main 2017 Question Paper with Solutions

As we know, for 1 g mole of a gas,

C_p – C_v = R where C_p and C_v are molar specific heat capacities.

So, when n gram moles are given,

C_p – C_v = ${R \over n}$

For hydrogen (n = 2), C_p – C_v = ${R \over 2}$ = $a$

For nitrogen (n = 28), C_p – C_v = ${R \over 28}$ = $b$

$\therefore$ ${a \over b} = 14$

$\Rightarrow$ $a$ = 14 $b$

The potential difference in each loop is zero.

$\therefore$ No current will flow or current in each resistance is Zero.

There is no change in null point, if the cell and the galvanometer are exchanged in a balanced wheatstone bridge.

The volume of the cylinder V = $\pi {R^2}l$

$\therefore$ ${R^2} = {V \over {\pi l}}$

We know, moment of inertia of a uniform cylinder of length $l$ and radius R about its perpendicular bisector is,

$I = {{M{l^2}} \over {12}} + {{M{R^2}} \over 4}$

[ Putting ${R^2} = {V \over {\pi l}}$ in this equation]

$\Rightarrow$ $I = {{M{l^2}} \over {12}} + {{MV} \over {4\pi l}}$

Here $I$ is a function of $l$ as M and V are constant.

$I$ will be maximum or minimum when ${{{dI} \over {dl}}}$ = 0.

$\Rightarrow {{Ml} \over 6} - {{MV} \over {4\pi {l^2}}} = 0$

$\Rightarrow {{Ml} \over 6} = {{MV} \over {4\pi {l^2}}}$

$\Rightarrow {l \over 6} = {{\pi {R^2}l} \over {4\pi {l^2}}}$ [ as ${V = \pi {R^2}l}$ ]

$\Rightarrow {{{R^2}} \over {{l^2}}} = {4 \over 6}$

$\Rightarrow {l \over R} = \sqrt {{3 \over 2}}$

From question, m_A = M; m_B = ${m \over 2}$

u_A = V and u_B = 0

Let after collision velocity of A = V₁ and

velocity of B = V₂

Applying law of conservation of momentum,

mu = mv₁ + $\left( {{m \over 2}} \right){v_2}$

$\Rightarrow$ 24= 2v₁ + v₂ ........(i)

By law of collision

$e = {{{v_2} - {v_1}} \over {u - 0}}$

$\Rightarrow$ u = v₂ - v₁ ..........(ii)

[As collision is elastic, e = 1]

using eqns (i) and (ii)

v₁ = ${4 \over 3}$ and v₂ = ${4 \over 3}v$

We know, de-Broglie wavelength $\lambda$ = ${h \over p}$

$\therefore$ ${{{\lambda _A}} \over {{\lambda _B}}} = {{{P_B}} \over {{P_A}}}$ = ${{{m \over 2} \times {4 \over 3}u} \over {m \times {4 \over 3}}}$ = 2

Torque experienced by the dipole in an electric field,

$T$ = pE sin$\theta$

$\overrightarrow T = \overrightarrow p \times \overrightarrow E$

$\overrightarrow p = p\cos \theta \widehat i + p\sin \theta \widehat j$

$\mathop {{E_1}}\limits^ \to = E\widehat i$

$\overrightarrow {{T _1}} = \overrightarrow P \times {\overrightarrow E _1}$

= ($p\cos \theta \widehat i + p\sin \theta \widehat j$) $\times$ $E\left( {\widehat i} \right)$

= pE sin$\theta$$\left( { - \widehat k} \right)$

$\mathop {{E_2}}\limits^ \to = \sqrt 3 {E_1}\widehat j$

$\overrightarrow {{T _2}} =$($p\cos \theta \widehat i + p\sin \theta \widehat j$) $\times$ $\sqrt 3 {E_1}\widehat j$

= $\sqrt 3 pE\cos \theta \left( {\widehat k} \right)$

Now given, $\overrightarrow {{T _2}}$ = $-\overrightarrow {{T _1}}$

$\Rightarrow$ $\sqrt 3 pE\cos \theta \left( {\widehat k} \right)$ = -pE sin$\theta$$\left( { - \widehat k} \right)$

$\Rightarrow$ $\tan \theta = \sqrt 3$

$\Rightarrow$ $\theta$ = 60^o

As we know, Bulk modulus

K = ${{\Delta P} \over {\left( {{{ - \Delta V} \over V}} \right)}}$

$\Rightarrow$ ${{{\Delta V} \over V} = {P \over K}}$

V = V₀(1 + $\gamma$$\Delta$t)

${{{\Delta V} \over {{V_0}}} = \gamma \Delta t}$

$\therefore$ ${{P \over K} = \gamma \Delta t}$

$\Rightarrow$ ${\Delta t = {P \over {\gamma K}}}$ = ${{P \over {3\alpha K}}}$

To determine how the stress in the leg changes when a man's linear dimensions increase by a factor of 9, we can analyze the relationship between the dimensions and the stress on the legs.

First, let's establish some relationships:

1. Linear dimensions (length, width, height) increase by a factor of 9.


2. Density remains the same.

Stress is defined as force per unit area:

$\text{Stress} = \dfrac{\text{Force}}{\text{Area}}$

Since density remains constant, the volume and thus the mass of the man will change according to the cube of the linear dimensions. Since the linear dimension changes by a factor of 9, the volume (and hence the mass) changes by a factor of:

$9^3 = 729$

Therefore, the weight (force due to gravity) also increases by a factor of 729.

The cross-sectional area of the leg is proportional to the square of the linear dimensions. Thus, if the linear dimension increases by a factor of 9, the cross-sectional area increases by a factor of:

$9^2 = 81$

Now, substituting these factors into the stress formula:

$\text{New Stress} = \dfrac{\text{New Force}}{\text{New Area}} = \dfrac{729 \times \text{Original Force}}{81 \times \text{Original Area}}$

Simplifying, we get:

$\text{New Stress} = 9 \times \dfrac{\text{Original Force}}{\text{Original Area}} = 9 \times \text{Original Stress}$

Thus, the stress in the leg increases by a factor of 9. Therefore, the correct answer is:

Option B: 9

Given : Current through the galvanometer,

i_g = 5 × 10^–3 A

Galvanometer resistance, G = 15 $\Omega$

Let resistance R to be put in series with the galvanometer to convert it into a voltmeter.

V = i_g (R + G)

10 = 5 × 10^–3 (R + 15)

$\therefore$ R = 2000 – 15 = 1985

= 1.985 × 10³ $\Omega$

According to the question, final kinetic energy = ${1 \over 8}mv_0^2$

Let final speed of the body = V_f

So final kinetic energy = ${1 \over 2}mv_f^2$

According to question,

${1 \over 2}mv_f^2$ = ${1 \over 8}mv_0^2$

$\Rightarrow {v_f} = {{{v_0}} \over 2}$ = ${{10} \over 2}$ = 5 m/s

Given that, F = –kv²

$\Rightarrow$ $m\left( {{{dv} \over {dt}}} \right)$$= - k{v^2}$

$\Rightarrow {10^{ - 2}}\left( {{{dv} \over {dt}}} \right) = - k{v^2}$

$\Rightarrow \int\limits_{10}^5 {{{dv} \over {{v^2}}}} = - 100k\int\limits_0^{10} {dt}$

$\Rightarrow {1 \over 5} - {1 \over {10}} = 100k \times 10$

$\Rightarrow k = {10^{ - 4}}kg\,{m^{ - 1}}$

In common emitter configuration for n-p-n transistor input and output signals are 180° out of phase i.e., phase difference between output and input voltage is 180°.

This question is from Doppler's effect of light.

When observer is moving towards the source then the frequency of wave measured by the observer will be

f_observed = f_actual$\sqrt {{{c + v} \over {c - v}}}$

where c = speed of light and v = speed of observer

According to the question, v = ${c \over 2}$

$\therefore$ f_observed = f_actual$\sqrt {{{c + {c \over 2}} \over {c - {c \over 2}}}}$

                     = 10$\times$$\sqrt {{{{{3c} \over 2}} \over {{c \over 2}}}}$

                     = 10$\sqrt 3$ = 17.3 GHz

Modulated carrier wave contains frequency ${\omega _c}$ and ${\omega _c}$ ± ${\omega _m}$.

For a particle executing SHM

At mean position; t = 0, $\omega$t = 0, y = 0, V = V_max = a$\omega$

$\therefore$ K.E = K.E_max = ${1 \over 2}m{\omega ^2}{a^2}$

At extreme position : t = ${T \over 4}$ , $\omega$t = ${\pi \over 2}$ , y = A, V = V_min = 0

$\therefore$ K.E = K.E_min = 0

Hence graph (A) correctly represents kinetic energy - time graph.

Given : Magnetic moment, M = 6.7 × 10^–2 Am²

Magnetic field, B = 0.01 T

Moment of inertia, I = 7.5 × 10^–6 Kgm²

Using, T = $2\pi \sqrt {{I \over {MB}}}$

= $2\pi \sqrt {{{7.5 \times {{10}^{ - 6}}} \over {6.7 \times {{10}^{ - 2}} \times 0.01}}}$

= ${{2\pi } \over {10}} \times 1.06$ s

Time taken for 10 complete oscillations

t = 10T = 2$\pi$ × 1.06

= 6.6568 $\simeq$ 6.65 s

From energy level diagram, using $\Delta E = {{hc} \over \lambda }$

For wavelength ${{\lambda _1}}$, $\Delta E =$ –E – (–2E) = ${{hc} \over {{\lambda _1}}}$

For wavelength ${{\lambda _2}}$, $\Delta E =$ –E – (–${{4E} \over 3}$) = ${{hc} \over {{\lambda _2}}}$

$\therefore$ r = ${{{\lambda _1}} \over {{\lambda _2}}} = {{\left( {{{4E} \over 3} - E} \right)} \over {\left( {2E - E} \right)}}$ = ${1 \over 3}$

Surface tension,

T = ${{rhg} \over 2} \times {10^3}N/m$

Relative error,

${{\Delta T} \over T} = {{\Delta r} \over r} + {{\Delta h} \over h}$

Percentage error,

${{\Delta T} \over T} \times 100 = {{\Delta r} \over r} \times 100 + {{\Delta h} \over h} \times 100$

${{\Delta T} \over T} \times 100 = \left( {{{{{10}^{ - 2}} \times 0.01} \over {1.25 \times {{10}^{ - 2}}}} + {{{{10}^{ - 2}} \times 0.01} \over {1.45 \times {{10}^{ - 2}}}}} \right) \times 100$

= (0.8 + 0.689) = 1.489 % = 1.5 %

In X-ray tube, ${\lambda _{\min }} = {{hc} \over {eV}}$

$\log \left( {{\lambda _{\min }}} \right) = \log \left( {{{hc} \over e}} \right) - \log V$

Clearly, log ($\lambda$_min) versus log V graph slope is negative hence option (b) is correct.

Let n₁^th fringe formed due to first wavelength and n₂^th fringe formed due to second wavelength coincide. So their distance from common central maxima will be same.

y_n1 = y_n2

${{{n_1}{\lambda _1}D} \over d} = {{{n_2}{\lambda _2}D} \over d}$

$\Rightarrow$ ${{{n_1}} \over {{n_2}}} = {{{\lambda _2}} \over {{\lambda _1}}}$ = ${{520 \times {{10}^{ - 9}}} \over {650 \times {{10}^{ - 9}}}} = {4 \over 5}$

Hence, distance of the point of coincidence from the central maxima is

y = ${{{n_1}{\lambda _1}D} \over d} = {{{n_2}{\lambda _2}D} \over d}$ = ${{4 \times 450 \times {{10}^{ - 9}} \times 15} \over {0.5 \times {{10}^{ - 3}}}}$ = 7.8 mm

Given that, F = 6t

We know, F = ma = $m{{dv} \over {dt}}$

$\therefore$ $m{{dv} \over {dt}} = 6t$

$\Rightarrow$ $1.{{dv} \over {dt}} = 6t$ [as m = 1]

$\Rightarrow$ $\int\limits_0^v {dv} = \int {6t} dt$

$\Rightarrow$ $v = 6\left[ {{{{t^2}} \over 2}} \right]_0^1$

$\Rightarrow$ $v = {6 \over 2} = 3$ m/s [ as given t = 1 sec ]

Work done by the body during the first 1 form work-energy theorem,

W = $\Delta$K.E = ${1 \over 2}m\left( {{V^2} - {v^2}} \right)$

= ${1 \over 2}.1.\left( {{3^2} - {0^2}} \right)$ = 4.5 J

To get a capacitance of 2 μF arrangement of capacitors of capacitance 1μF as shown in figure 8 capacitors of 1μF in parallel with four such branches in series i.e., 32 such capacitors are required.

${1 \over {{C_{eq}}}} = {1 \over 8} + {1 \over 8} + {1 \over 8} + {1 \over 8}$

$\Rightarrow$ ${1 \over {{C_{eq}}}} = {1 \over 2}$

$\Rightarrow$ ${{C_{eq}} = 2}$

Let initially there are total N₀ number of nuclei.

At time t ${{{N_B}} \over {{N_A}}}$ = 0.3 (given)

$\Rightarrow$ ${{N_B}}$ = 0.3${{N_A}}$

N₀ = N_A + N_B = N_A + 0.3N_A

$\therefore$ N_A = ${{{N_0}} \over {1.3}}$

As we know N_t = N₀ e^{– $\lambda$t}

$\Rightarrow$ ${{{N_0}} \over {1.3}}$ = N₀ e^{– $\lambda$t}

$\Rightarrow$ ${1 \over {1.3}}$ = e^{– $\lambda$t}

$\Rightarrow$ ln(1.3) = $\lambda$t

$\Rightarrow$ t = ${{\ln \left( {1.3} \right)} \over \lambda }$

If T is half-life, then $\lambda$ = ${{{{\ln} } 2} \over T}$

$\Rightarrow$ t = ${{\ln \left( {1.3} \right)} \over {{{\ln 2} \over T}}}$ = ${{\ln \left( {1.3} \right)} \over {\ln 2}}T$

According to principle of calorimetry,

Heat lost = Heat gain

100 × 0.1( – 75) = 100 × 0.1 × 45 + 170 × 1 × 45

10 – 750 = 450 + 7650

10 = 1200 + 7650 = 8850

T = 885°C

Forces acting on the rod are shown below. Torque about pivot point O due to force N_x and N_y are zero.

Mgcos$\theta$ is passing through O, so torque will be zero due to this force.

So torque about the point O is

$\tau = Mg\sin \theta \times {l \over 2}$

We know, $\tau = I\alpha$

$\therefore$ $I\alpha = Mg\sin \theta \times {l \over 2}$

Moment of inertia of rod about point O, $I$ = ${{M{l^2}} \over 3}$

$\therefore$ ${{M{l^2}} \over 3} \times \alpha = Mg\sin \theta \times {l \over 2}$

$\Rightarrow \alpha = {3 \over 2}{g \over l}\sin \theta$

According to Faraday's law of electromagnetic induction,

$\varepsilon = {{d\phi } \over {dt}}$

Also, $\varepsilon$ = iR

$\therefore$ ${{d\phi } \over {dt}}$ = iR

$\Rightarrow$ $\int {d\phi } = R\int {idt}$

Magnitude of change in flux (d$\phi$) = R × area under current vs time graph

$\Rightarrow$ d$\phi$ = $100 \times {1 \over 2} \times {1 \over 2} \times 10$ = 250 Wb

When d < R means distance of a point is d from the center of the circle where d is inside the earth surface. In this case the value of acceleration, $g = -{{GMd} \over {{R^3}}}$

$\therefore$ $g \propto d$

So inside of the earth surface g - d graph is straight line.

When d > R means distance of a point is d from the center of the circle where d is outside of the earth's surface. In this case the value of acceleration, $g = - {{GM} \over {{d^2}}}$

$\therefore$ $g \propto {1 \over {{d^2}}}$

So outside of the earth surface g - d graph is hyperbolic.

Motion of the particle is shown below



Initially speed of the particle is maximum(v₀) and at heigth h velocity will become zero, then particle will move downward and velocity will increase gradually and become maximum when it reaches the ground. And acceleration in this entire motion will be -g, so slope of v - t will be same and negative.

So only Option (D) will be correct.

As parallel beam incident on diverging lens so the image will be formed at the focus of diverging lens.

$\therefore$ v = –25 cm
The image formed by diverging lens is used as an object for converging lens.

So for converging lens

u = –25 – 15 = –40 cm,

f = 20 cm

So Final image formed by converging lens

${1 \over {20}} = {1 \over V} - {1 \over { - 40}}$

$\Rightarrow$ V = 40 cm

V is positive so image will be real and will form at right side of converging lens at 40 cm.

In steady state, flow of current through capacitor will be zero.

Current through the circuit,

i = ${E \over {r + {r_2}}}$

Potential difference through capacitor

V_c = ${Q \over C}$ = E - ir = E - $\left( {{E \over {r + {r_2}}}} \right)r$

$\therefore$ Q = $CE{{{r_2}} \over {(r + {r_2})}}$

Given: Initial temperature T_i = 17 + 273 = 290 K

Final temperature T_f = 27 + 273 = 300 K

Atmospheric pressure, P₀ = 1 × 10⁵ Pa

Volume of room, V₀ = 30 m³

Difference in number of molecules, N_f – N_i = ?

We know PV = nRT = ${N \over {{N_A}}}$RT

The number of molecules

N = ${{PV{N_A}} \over {RT}}$

$\therefore$ N_f – N_i = ${{{P_0}{V_0}{N_A}} \over R}\left( {{1 \over {{T_f}}} - {1 \over {{T_i}}}} \right)$

= ${{1 \times {{10}^5} \times 30 \times 6.023 \times {{10}^{23}}} \over {8.314}}\left( {{1 \over {300}} - {1 \over {290}}} \right)$

= – 2.5 $\times$ 10²⁵

Permissible limit for SO₄ ^2- = 500 ppm

Permissible limit for NO₃ ^- = 50 ppm

Permissible limit for F^- = 1 ppm

Zinc oxide (ZnO) when react with Na₂O it act as acid while with CO₂ it act as base. Therefore, it is an amphoteric oxide.

(a) ZnO + Na₂O $\to$ Na₂ZnO₂
     Acid      Base          Salt

(b) ZnO + CO₂ $\to$ ZnCO₃
     Base      Acid       Salt

Tyndall effect is observed only when

(i) The diameter of the dispersed particle is not much smaller than the wavelength of the light used.

(ii) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

So, (b) and (d) are correct.

From 1st law of thermodynamics

ΔU = q + w

For adiabatic process :

q = 0

$\therefore$ ΔU = w

So change in internal energy (ΔU) is equal to adiabatic work.

Mg can form basic carbonate like

3MgCO₃ . Mg(OH)₂ . 3H₂O

While Li can not.

$\Delta$T_f = i $\times$ K_f $\times$ m

$\Rightarrow$ 0.45 = i $\times$ 5.12 $\times$ ${{0.2 \times 1000} \over {60 \times 20}}$

$\Rightarrow$ i = 0.527

2CH₃COOH ⇌ (CH₃COOH)₂
      1 - $\alpha$                   ${\alpha \over 2}$

i = 1 - $\alpha$ + ${\alpha \over 2}$

$\Rightarrow$ $\alpha$ = 0.946

$\therefore$ % dissociation is 94.6%.

We know, from arrhenius equation,

k = A.${e^{{{ - {E_a}} \over {RT}}}}$

$\therefore$ k₁ = A.${e^{{{ - {E_{{a_1}}}} \over {RT}}}}$ ......(1)

k₂ = A.${e^{{{ - {E_{{a_2}}}} \over {RT}}}}$ ......(2)

On dividing equation (2) by (1), we get

${{{k_2}} \over {{k_1}}} = {e^{{{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)} \over {RT}}}}$

$\Rightarrow$ $\ln \left( {{{{k_2}} \over {{k_1}}}} \right) = {{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)} \over {RT}}$ = ${{10,000} \over {8.314 \times 300}}$ = 4

A compound having two chiral centres can exist in 4 stereoisomeric forms (2ⁿ).

The salt (AB) given is a salt is of weak acid and weak base. Hence its pH can be calculated by the formula

$\therefore$ pH = 7 + ${1 \over 2}p{K_a} - {1 \over 2}p{K_b}$

= 7 + ${1 \over 2}\left( {3.2} \right) - {1 \over 2}\left( {3.4} \right)$

= 6.9

Those species which have unpaired electrons are called paramagnetic species.

(a)    CO has 14 electrons.

Moleculer orbital configuration of CO is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

Here is no unpaired electron so it is diamagnetic.

(b)   $O_2$ has 16 electrons.

Moleculer orbital configuration of $O_2$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ *$

Here 2 unpaired electron present, so it is paramagnetic.

(c)   B₂ has 10 electrons.

Molecular orbital configuration of B₂ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^1}} = {\pi _{2p_y^1}}$

Here two unpaired electrons present. So it is paramagnetic.

(d)   NO has 15 electrons.

Moleculer orbital configuration of NO is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ *$

Here is 1 unpaired electron, So it is Paramagnetic.

Moles of complex = ${{Molarity \times Volume\left( {mL} \right)} \over {1000}}$

= ${{100 \times 0.1} \over {1000}}$ = 0.01 mole

Moles of ions precipitated with excess of AgNO₃

= ${{1.2 \times {{10}^{22}}} \over {6.023 \times {{10}^{23}}}}$ = 0.02 mole

So 0.01 × n = 0.02

$\Rightarrow$ n = 2

It means 2Cl^– ions present in ionization sphere.

$\therefore$ The formula of complex is [Co(H₂O)₅Cl]Cl₂.H₂O

Since S_N1 reactions involve the formation of carbocation as intermediate in the rate determining step, more is the stability of carbocation higher will be the reactivity of alkyl halides towards S_N1 reaction.


Hence correct order is II < I < III.

Nitration takes place in presence of concentrated HNO₃ + concentrated H₂SO₄.

In strongly acidic nitration medium, the amine is converted into anilinium ion (– NH₃⁺ ); substitution is thus controlled not by – NH₂ group but by – NH₃⁺ group which, because of its positive charge, directs the entering group to the meta-position instead of ortho and para.

$E_{C{l_2}/C{l^ - }}^o$ = 1.36 V, $E_{C{r^{3 + }}/Cr}^o$ = - 0.74 V
$E_{C{r_2}{O_7}^{2 - }/C{r^{3 + }}}^o$ = 1.33 V, $E_{Mn{O_4}^ - /Mn ^{2+}}^o$ = 1.51 V

More negative the E° value of the species, more stronger is the reducing agent. Since Cr³⁺ is having least reducing potential, so Cr would be strongest reducing agent.