JEE Main 2019 12th January Evening Shift Question Paper with Solutions

Phase difference between I₂ and V, i.e. C $-$ R₂ circuit is given by

$\tan \phi = {{{X_C}} \over {{R_2}}} \Rightarrow \tan \phi = {1 \over {C\omega {R_2}}}$

Substituting the given values, we get

$\tan \phi = {1 \over {{{\sqrt 3 } \over 2} \times {{10}^{ - 6}} \times 100 \times 20}} = {{{{10}^3}} \over {\sqrt 3 }}$

$\therefore$ $\phi$₁ is nearly 90$^\circ$.

Phase difference between I₁ and V, i.e. in L $-$ R₁ circuit is given by

$\tan {\phi _2} = - {{{X_L}} \over {{R_1}}} = - {{L\omega } \over R}$

Substituting the given values, we get

$\tan {\phi _2} = - {{{{\sqrt 3 } \over {10}} \times 100} \over {10}} = - \sqrt 3$

As, $\tan {\phi _2} = - \sqrt 3$

$\therefore$ ${\phi _2} = 120^\circ$

Now, phase difference between I₁ and I₂ is

$\Delta \phi = {\phi _2} - {\phi _1} = 120^\circ - 90^\circ = 30^\circ$

When soap bubble is being inflated and its temperature remains constant, then it follows Boyle's law, so

pV = constant (k)

$\Rightarrow p = {k \over V}$

Differentiating above equation with time, we get

${{dp} \over {dt}} = k.\,{d \over {dt}}\left( {{1 \over V}} \right)$

$\Rightarrow {{dp} \over {dt}} = k\left( {{{ - 1} \over {{V^2}}}} \right).\,{{dV} \over {dt}}$

It is given that, ${{dV} \over {dt}} = c$ (a constant)

So, ${{dp} \over {dt}} = {{ - kc} \over {{V^2}}}$ ..... (i)

Now, from ${{dV} \over {dt}} = c$, we get

$dV = cdt$

or, $\int {dV = \int {cdt} }$ or, $V = ct$ ..... (ii)

From Eqs. (i) and (ii), we get

${{dp} \over {dt}} = {{ - kc} \over {{c^2}{t^2}}}$

or, ${{dp} \over {dt}} = - {\left( {{k \over c}} \right)^{{t^{ - 2}}}}$

$\Rightarrow dp = - {k \over c}.\,{t^{ - 2}}dt$

Integrating both sides, we get

$\int {dp = - {k \over c}\int {{t^{ - 2}}dt} }$

$p = - {k \over c}.\,\left( {{{{t^{ - 2 + 1}}} \over { - 2 + 1}}} \right)$

$= - {k \over c}.\,{{ - 1} \over t} = {k \over {ct}}$

or, $p \propto {1 \over t}$

Hence, p versus ${1 \over t}$ graph is a straight line, which is correctly represented in option (b).

In first resonance, length of air column $= {\lambda \over 4}$.

So, ${l_1} + e = {\lambda \over 4}$ or $11 \times 4 + 4e = \lambda$

So, speed of sound is

$\Rightarrow v = {f_1}\lambda = 512(44 + 4e)$ ...... (i)

And in second case,

$l{'_1} + e = {{\lambda '} \over 4}$ or $27 \times 4 + 4e = \lambda '$

$\Rightarrow v = {f_2}\lambda ' = 256(108 + 4e)$ ..... (ii)

Dividing both Eqs. (i) and (ii), we get

$1 = {{512(44 + 4e)} \over {256(108 + 4e)}} \Rightarrow e = 5$ cm

Substituting value of e in Eq. (i), we get

Speed of sound $v = 512(44 + 4e)$

$= 512(44 + 4 \times 5)$

$= 512 \times 64$ cm s^$-$1 = 327.68 ms^$-$1 $\approx$ 328 ms^$-$1

The correct answer is Option C. Here's why :

The moment of inertia of a solid sphere about an axis passing through its center is given by :

$I_0 = \frac{2}{5}MR^2$

Where :

• $I_0$ is the moment of inertia about the center
• $M$ is the mass of the sphere
• $R$ is the radius of the sphere

Now, using the parallel axis theorem, we can find the moment of inertia about an axis parallel to the diameter and at a distance of 'x' from it :

$I(x) = I_0 + Mx^2$

Substituting the value of $I_0$ :

$I(x) = \frac{2}{5}MR^2 + Mx^2$

This equation represents a parabola, where :

• The coefficient of $x^2$ is positive (M), indicating an upward-opening parabola.
• The y-intercept (the value of $I(x)$ when $x = 0$) is $\frac{2}{5}MR^2$, which is a positive constant.

Therefore, the graph of $I(x)$ vs. $x$ should be an upward-opening parabola, and Option C accurately represents this.

As we know, Current, $I = {{dq} \over {dt}}$

= Slope of q versus t graph

= Zero at t = 4s; (as graph is a line parallel to time axis at t = 4s)

Range of TV transmitting tower $d = \sqrt {2hR}$ where, h is the height of the transmission tower. When range is doubled,

$\therefore$ $2d = 2\sqrt {2hR} = \sqrt {2(4h)R}$

So, height must be multiplied with 4.

All the given compounds are acid derivatives, thus contain carbonyl group in them. LiAlH₄ reduces these compounds through nucleophilic substitution via addition elimination (S_NAE) reaction. The rate of reaction depends upon the following factors:

(i) Size of alkyl group.

(ii) Steric hinderance around the > C == O group.

(iii) (+) ve charge on the C-atom of > C == O group.

The alkyl groups are more or less same in the given compounds. Thus, the reactivity order of given compounds depends upon 2nd and 3rd factor written above. The cumulative effect of these two factors results to leaving group ability (LGA) of the substituents in the following order :

This leaving group ability (weak conjugate base) corresponds directly to the reactivity order.

Hence, the correct reactivity order is: