JEE Main 2019 Question Paper (09-Jan-2019) (Shift 2)

Question 1

A particle having the same charge as that of an electron moves in a circular path of radius $0.5\ 	ext{cm}$ under the influence of a magnetic field of $0.5\ 	ext{T}$. If an electric field of $100\ 	ext{V/m}$ makes it move in a straight path, then the mass of the particle is (Given charge of electron $= 1.6 	imes 10^{-19}\ 	ext{C}$).

Accepted Answer

$2.0 	imes 10^{-24} \quad 	ext{kg}$

Answer

$9.1 	imes 10^{-31} \quad 	ext{kg}$

Answer

$1.6 	imes 10^{-27} \quad 	ext{kg}$

Answer

$1.6 	imes 10^{-19} \quad 	ext{kg}$

Question 2

Two plane mirrors are inclined to each other such that a ray of light incident on the first mirror $ M_1 $ and parallel to the second mirror $ M_2 $ is finally reflected from the second mirror $ M_2 $ and parallel to the first mirror $ M_1 $. The angle between the two mirrors will be:

Accepted Answer

$60^{\circ}$

Answer

$45^{\circ}$

Answer

$90^{\circ}$

Answer

$75^{\circ}$

Question 3

Charge is distributed within a sphere of radius $R$ with a volume charge density $ \rho(r) = \frac{A}{r^{2}} e^{-\frac{2r}{a}}, $ where $A$ and $a$ are constants. If $Q$ is the total charge of this charge distribution, the radius $R$ is _____.

Accepted Answer

$ \frac{a}{2} \log \left( \frac{1}{1 - \frac{Q}{2 \pi a A}} \right) $

Answer

$a \log\left(\frac{1}{1 - \frac{Q}{2 \pi a A}}\right)$

Answer

$a \log\left(1 - \frac{Q}{2 \pi a A}\right)$

Answer

$ \frac{a}{2} \log \left( 1 - \frac{Q}{2 \pi a A} \right) $

Question 4

A rod of length $50\quad 	ext{cm}$ is pivoted at one end. It is raised such that it makes an angle of $30^\circ$ from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in $	ext{rad}\quad 	ext{s}^{-1}$) will be $\left(g=10\quad 	ext{m}\,	ext{s}^{-2}ight)$.

Accepted Answer

$\sqrt{30}$

Answer

$\frac{\sqrt{20}}{3}$

Answer

$ \sqrt{\frac{30}{2}} $

Answer

$\frac{\sqrt{30}}{2}$

Question 5

In a car race on a straight road, car $A$ takes a time $t$ less than car $B$ at the finish and passes the finishing point with a speed $v$ more than that of car $B$. Both cars start from rest and travel with constant accelerations $a_1$ and $a_2$ respectively. Then $v$ is equal to:

Accepted Answer

$ \sqrt{a_{1} a_{2}} \, t$

Answer

$\frac{2 a_1 a_2}{a_1 + a_2} t$

Answer

$\frac{a_1 + a_2}{2} t$

Answer

$\sqrt{2 a_1 a_2} t$