JEE Main 2014 Question Paper with Solutions

Time taken to reach highest point is $t = {u \over g}$

Time taken by the particle to reach the ground = $nt = {nu \over g}$

Speed on reaching ground $v = \sqrt {{u^2} + 2gH}$

Now, $v = u + at$

$\Rightarrow \sqrt {{u^2} + 2gH} = - u + gt$

$\Rightarrow t = {{u + \sqrt {{u^2} + 2gH} } \over g} = {{nu} \over g}$ (from question)

$\Rightarrow 2gH = n\left( {n - 2} \right){u^2}$

By Lens maker's formula for convex lens

${1 \over f} = \left( {{\mu \over {{\mu _L}}} - 1} \right)\left( {{2 \over R}} \right)$

for, $\mu {L_1} = {4 \over 3},{f_1} = 4R$

for $\mu {L_2} = {5 \over 3},{f_2} = - 5R$

$\Rightarrow {f_2} = \left( - \right)ve$

Torque working on the bob of mass $m$ is, $\tau = mg \times \ell \,\sin \,\theta .$ (Direction of torque by the weight is parallel to plane of rotation of the particle)

As $\tau$ is perpendicular to the angular momentum $\overrightarrow L$ of the bob, so the direction of $L$ changes but magnitude remains same.

Length of pipe $=85$ $cm$ $=0.85m$

Pipe is closed from one end so it behaves as a closed organ pipe

Frequency of oscillations of air column in closed organ pipe is given by,

$f = {{\left( {2n - 1} \right)\upsilon } \over {4L}}$

$f = {{\left( {2n - 1} \right)\upsilon } \over {4L}} \le 1250$

$\Rightarrow {{\left( {2n - 1} \right) \times 340} \over {0.85 \times 4}} \le 1250$

$\Rightarrow 2n - 1 \le 12.5 \approx 6$

Possible value of n = 1, 2, 3, 4, 5, 6

So, number of possible natural frequencies lie below 1250 Hz is 6.

$\left( 1 \right)\,\,\,\,\,\,$ Infrared rays are used to treat muscular strain because these are heat rays.

$\left( 2 \right)\,\,\,\,\,\,$ Radio waves are used for broadcasting because these waves have very long wavelength runging from few centimeters to few hundred kilometers

$\left( 3 \right)\,\,\,\,\,\,$ $X$-rays are used to detect fracture of bones because they have high penetrating power but they can't penetrate through denser medium like bones.

$\left( 4 \right)\,\,\,\,\,\,$ Ultraviolet rays are absorbed by ozone of the atmosphere.

Length of the air column above mercury in the tube is,
$P + x = {P_0}$
$\Rightarrow P = \left( {76 - x} \right)$
$\Rightarrow 8 \times A \times 76 = \left( {76 - x} \right) \times A \times \left( {54 - x} \right)$
$\therefore$ $x=38$
Thus, length of air column $=54-38=16cm.$

When the bubble gets detached, Buoyant force $=$ force due to surface tension

Force due to excess pressure $=$ upthrust
Access pressure in air bubble $= {{2T} \over R}$
${{2T} \over R}\left( {\pi {r^2}} \right) = {{4\pi {R^3}} \over {3T}}{\rho _w}g$
$\Rightarrow {r^2} = {{2{R^4}{\rho _w}g} \over {3T}}$
$\Rightarrow r = {R^2}\sqrt {{{2{\rho _w}g} \over {3T}}}$

Electric field in presence of dielectric between the two plates of a parallel plate capacitor is given by,

$E = {\sigma \over {K{\varepsilon _0}}}$

Then, charge density

$\sigma = K{\varepsilon _0}E$

$= 2.2 \times 8.85 \times {10^{ - 12}} \times 3 \times {10^4} \approx 6 \times {10^{ - 7}}\,\,C/{m^2}$

Applying kirchhoffs law of voltage in closed loop

$- {V_R} - {V_C} = 0 \Rightarrow {{{V_R}} \over {{V_C}}} = - 1$

Given Restoring force, F = ax + bx²

Work done in stretching the rubber-band by a distance $dx$ is

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dW = F\,dx = \left( {ax + b{x^2}} \right)dx$

Intergrating both sides,

$W = \int\limits_0^L {axdx + \int\limits_0^L {b{x^2}dx}}$

= $\left[ {a{{{x^2}} \over 2} + b{{{x^3}} \over 3}} \right]_0^L$

= ${{a{L^2}} \over 2} + {{b{L^3}} \over 3}$

Young's modulus $Y = {{stress} \over {strain}}$
$stress = Y \times strain$
$Stress$ in steel wire $=$ Applied $pressure$
$Pressure$ $=$ $stress$ $=$ $Y \times \,strain$
$Strain = {{\Delta L} \over L} = \alpha \Delta T$ (As length is constant)
$= 2 \times {10^{11}} \times 1.1 \times {10^{ - 5}} \times 100$
$= 2.2 \times {10^8}Pa$

Magnetic field in solenoid $B = {\mu _0}ni$

$\Rightarrow {B \over {{\mu _0}}} = ni$

(Where $n=$ number of turns per unit length)

$\Rightarrow {B \over {{\mu _0}}} = {{Ni} \over L}$

$\Rightarrow 3 \times {10^3} = {{100i} \over {10 \times {{10}^{ - 2}}}}$

$\Rightarrow i = 3A$

At limiting equilibrium, $\mu = \tan \theta$

Equation of the surface,

$y = {{{x^3}} \over 6}$

Slope, $\tan \theta = \mu = {{dy} \over {dx}} = {{{x^2}} \over 2}$


Given that, Coefficient of friction $\mu = 0.5$

$\therefore$ $\,\,\,\,0.5 = {{{x^2}} \over 2}$

$\Rightarrow \,\,\,x = \pm \,1$

Now, $y = {{{x^3}} \over 6} = {1 \over 6}m$

Pressure at interface A must be same from both the sides to be in equilibrium.

$\therefore$ $\left( {R\cos \alpha + R\sin \alpha } \right){d_2}g = \left( {R\cos \alpha - R\sin \alpha } \right){d_1}g$
$\Rightarrow {{{d_1}} \over {{d_2}}} = {{\cos \alpha + \sin \alpha } \over {\cos \alpha - \sin \alpha }} = {{1 + \tan \alpha } \over {1 - \tan \alpha }}$

For critical angle ${\theta _c},$

$\sin {\theta _c} = {1 \over \mu }$

For greater wavelength or lesser frequency $\mu$ is less.



So, critical angle would be more, So, they will not suffer reflection and come out at angles less then ${90^ \circ }.$

In simple harmonic motion, starting from rest,

At $t=0,$ $x=A$

$x = A\cos \omega t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

When $t = \tau ,\,\,x = A - a$

When $t = 2\,\tau ,\,x = A - 3a$

From equation $(i)$

$A - a = A\cos \omega \,\tau \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

$A - 3a = A\cos 2\omega \,\tau \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$

As $\cos 2\omega \,\tau = 2{\cos ^2}\omega \tau - 1...\left( {iv} \right)$

From equation $(ii),$ $(iii)$ and $(iv)$

${{A - 3A} \over A} = 2{\left( {{{A - a} \over A}} \right)^2} - 1$

$\Rightarrow {{A - 3a} \over A} = {{2{A^2} + 2{a^2} - 4Aa - {A^2}} \over {{A^2}}}$

$\Rightarrow {A^2} - 3aA = {A^2} + 2{a^2} - 4Aa$

$\Rightarrow 2{a^2} = aA \Rightarrow \,\,\,\,\,\,\,A = 2a$

$\Rightarrow {a \over A} = {1 \over 2}$

Now, $A-a=A$ $\cos \omega \tau$

$\Rightarrow \cos \omega \tau = {{A - a} \over A} \Rightarrow \,\,\cos \omega \tau = {1 \over 2}$

or, ${{2\pi } \over T}\tau = {\pi \over 3} \Rightarrow \,\,\,T - 6\,\tau$

According to malus law, intensity of emerging beam is given by,

$I = {I_0}{\cos ^2}\theta$

Now, ${I_{A'}} = {I_A}{\cos ^2}{30^ \circ }$

${I_{B'}} = {I_B}{\cos ^2}{60^ \circ }$

As ${I_{A'}} = {I_{B'}}$

$\Rightarrow {I_A} \times {3 \over 4} = {I_B} \times {1 \over 4}$

$\therefore$ ${{{I_A}} \over {{I_B}}} = {1 \over 3}$

Total power consumed by electrical appliances in the building, ${P_{total}} = 2500W$

Watt $=$ Volt $\times$ ampere

$\Rightarrow 2500 = V \times {\rm I}$

$\Rightarrow 2500 = 220$ ${\rm I}$

$\Rightarrow I = {{2500} \over {220}} = 11.36 \approx 12A$

(Minimum capacity of main fuse)

Rate of heat flow is given by,

$Q = {{KA\left( {{\theta _1} - {\theta _2}} \right)} \over l}$

Where, $K=$ coefficient of thermal conductivity $l=$ length of rod and $A=$ Area of cross-section of rod

If the junction temperature is $T,$ then

${Q_{Copper}}\,\, = \,\,{Q_{Brass}}\,\, + \,\,{Q_{Steel}}$

${{0.92 \times 4\left( {100 - T} \right)} \over {46}} = {{0.26 \times 4 \times \left( {T - 0} \right)} \over {13}} + {{0.12 \times 4 \times \left( {T - 0} \right)} \over {12}}$

$\Rightarrow 200 - 2T = 2T + T$

$\Rightarrow T = {40^ \circ }C$

$\therefore$ ${Q_{Copper}} = {{0.92 \times 4 \times 60} \over {46}} = 4.8\,cal/s$

Here string is not slipping over pulley.

From figure,

Acceleration $a = R\alpha \,\,\,\,...(i)$

Applying Newton's second law on the block we get,

$mg-T=ma$ $...(ii)$

Torque on cylinder due to tension in string about the fixed point,

$T \times R = I\alpha$

$T \times R = m{R^2}\alpha = m{R^2}\left( {{a \over R}} \right)$

[ as $I$ = MR² for hollow cylinder]

or $T=ma$

$\Rightarrow \,\,mg - ma = ma$

$\Rightarrow \,\,\,a = {g \over 2}$

For forward bias, $p$-side must be at higher potential than $n$-side. $\Delta V = \left( + \right)Ve$

All those particles are moving due to their mutual gravitational attraction.

The force between each masses are repulsive force.

On mass M at C, due to mass at D the repulsive force is F in the vertical direction.

On mass M at C, due to mass at B the repulsive force is F in the horizontal direction.

On mass M at C, due to mass at A the repulsive force is F'

Net force acting on particle at C,

= $2F\,\cos \,{45^ \circ } + F'$

Where $F = {{G{M^2}} \over {{{\left( {\sqrt 2 R} \right)}^2}}}$ and $F' = {{G{M^2}} \over {4{R^2}}}$

$\Rightarrow {{2 \times G{M^2}} \over {\sqrt 2 {{\left( {R\sqrt 2 } \right)}^2}}} + {{G{M^2}} \over {4{R^2}}}$

$\Rightarrow {{G{M^2}} \over R^2}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right]$

This net force will balance by the centripetal force F_cp = ${{M{v^2}} \over R}$

$\therefore$ ${{M{v^2}} \over R} =$ ${{G{M^2}} \over R^2}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right]$

$\Rightarrow$ $v = \sqrt {{{Gm} \over R}\left( {{{\sqrt 2 + 4} \over {4\sqrt 2 }}} \right)}$

$= {1 \over 2}\sqrt {{{Gm} \over R}\left( {1 + 2\sqrt 2 } \right)}$

${E_0} = C{B_0}$ and $C = {1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$

Electric energy density $= {1 \over 2}{\varepsilon _0}{E_0}^2 = {\mu _E}$

Magnetic energy density $= {1 \over 2}{{{B_0}^2} \over {{\mu _0}}} = {\mu _B}$

Thus, ${\mu _E} = {\mu _B}$

Energy is equally divided between electric and magnetic field

Radius of circular path followed by electron is given by,

$r = {{m\upsilon } \over {qB}} = {{\sqrt {2meV} } \over {eB}} = {1 \over B}\sqrt {{{2m} \over e}V}$

$\Rightarrow V = {{{B^2}{r^2}e} \over {2m}} = 0.8V$

For transition between $3$ to $2.$

$E = 13.6\left( {{1 \over 4} - {1 \over 9}} \right)$

$= {{13.6 \times 5} \over {36}} = 1.88eV$

Work function $= 1.88eV - 0.8eV$

$= 1.08eV \approx 1.1eV$

Wave number ${1 \over \lambda } = R{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {{n^2}}}} \right]$

$\Rightarrow \lambda \propto {1 \over {{Z^2}}}$

By question $n=1$ and ${n_1} = 2$

Then, ${\lambda _1} = {\lambda _2} = 4{\lambda _3} = 9{\lambda _4}$

In cyclic process, change in total internal energy is zero.
$\Delta {U_{cyclic}} = 0$
$\Delta {U_{BC}} = n{C_v}\Delta T = 1 \times {{5R} \over 2}\Delta T$

where, ${C_v} =$ molar specific heat at constant volume.
For $BC,$ $\Delta T = - 200\,K$
$\therefore$ $\Delta {U_{BC}} = - 500R$

Measured length of rod = 3.50 cm

That means least count of the measuring instrument should be 0.01 cm = 0.1 mm

For vernier scale 1 main scale division = 1 mm

And 9 MSD = 10 VSD

Least count = 1 MSD - 1 VSD

= 1 - 0.9 = 0.1 mm

The current voltage relation of diode is

$I = \left( {{e^{1000\,V/T}} - 1} \right)\,\,mA$ (given)

When, $I = 5mA,{e^{1000\,\,V/T}}\, = 6mA$

Also, $dl = \left( {{e^{1000\,\,V/T}}} \right) \times {{1000} \over T}$

(By exponential function)

$= \left( {6\,mA} \right) \times {{1000} \over {300}} \times \left( {0.01} \right)$

$=0.2$ $mA$

Potential difference between any two points in an electric field is given by,

$dV = - \overrightarrow E .\overrightarrow {dx}$

$\int\limits_{{V_O}}^{{V_A}} {dV = - \int\limits_0^2 {30{x^2}} } dx$

${V_A} - {V_O} = \left[ {10{x^3}} \right]_{0}^2 = - 80J/C$