JEE 2018 :: Physics (2018) :: Solved questions 2018

• Physics (2018) - Solved questions 2018

In a radioactive decay chain, $_{90}^{232}Th$ nucleus  decays to $_{82}^{212}Pb$ nucleus. Let  $N_{\alpha}$ and $N_{\beta}$ be the number of $\alpha$ and $\beta$- particles repectively, emitted in this decay process. Which of the following statements is(are) true?

Answer:

Explanation:

 $_{90}^{232}Th$  is converting into $_{82}^{212}Pb$ .

   Change in mass number (A)=20

$\therefore$     Number of $\alpha$ -particle emitted = $\frac{20}{4}=5$

Due to 5 $\alpha$ -particles, Z will change by 10  units.

  Since, given change is 8, therefore number of $\beta$ -particles emitted is 2

An infinitely long thin non-conducting wire is parallel to the Z-axis and carries a uniform line charge density λ . It pierces a thin non-conducting spherical shell of radius R in such a way that the arc PQ  subtends an angle 120° at the centre O  of  the spherical shell, as shown in the figure. The permittivity of free space is ε₀ . Which of the following statements is (are) true?

Answer:

Explanation:

PQ= (2) R sin 60°

             = $(2R)\frac{\sqrt{3}}{2}=(\sqrt{3}R)$

             $q_{enclosed}=\lambda(\sqrt{3}R)$

 We have,        $\phi =\frac{q_{enclosed}}{\epsilon_{0}}$

$\Rightarrow$               $\phi =(\frac{\sqrt{3}\lambda R_{}}{\epsilon_{0}})$

Also,   electric field is perpendicular to wire, so Z- component will be zero.

A  particle of mass m is initially at rest at the origin . It is subjected to a force and starts moving along the X-axis . Its kinectic energy K changes with time as $\frac{\text{d}K}{\text{d}t}=\gamma t$  , where $\gamma $ is a positive constant of appropriate dimensions. Which of the following statements is (are) true?

Answer:

Explanation:

$K=\frac{1}{2}mv^{2}\Rightarrow\frac{\text{d}K}{\text{d}t}\Rightarrow mv \frac{\text{d}v}{\text{d}t}$

 Given, $\frac{\text{d}K}{\text{d}t}=\gamma t\Rightarrow mv\frac{\text{d}v}{\text{d}t}=\gamma t$

   $\Rightarrow$   $\int_{0}^{v} v dv=\int_{0}^{t} \frac{\gamma}{m}t dt\Rightarrow\frac{v^{2}}{2}\Rightarrow\frac{\gamma}{m}\frac{t^{2}}{2}$

  $\Rightarrow$    $v=\sqrt{\frac{\gamma}{m}}t \Rightarrow a\Rightarrow\frac{\text{d}v}{\text{d}t}\Rightarrow\sqrt{\frac{\gamma}{m}}$

 $\therefore$    $F=ma=\sqrt{\gamma m}= constant$

 $\therefore$    $V=\frac{\text{d}s}{\text{d}t}=\sqrt{\frac{\gamma}{m}}t\Rightarrow s\Rightarrow\sqrt{\frac{\gamma}{m}}\frac{t^{2}}{2}$

 Note : Force is constant. In the website of IIT, option (d) is given correct. In the opinion of author all constant forces are not necessarily conservative. For example : viscous force at terminal velocity is a constant force but it is not conservative.

The relation between $[\epsilon_{0}]$ and $[\mu_{0}]$

Answer:

Explanation:

 $c= \frac{1}{\sqrt{\mu_{0}\epsilon_{0}}}$

       $c^{2}= \frac{1}{{\mu_{0}\epsilon_{0}}}$

    $\mu_{0}=\epsilon_{0}^{-1} c^{-2}$

$[\mu_{0}]= [\epsilon_{0}^{}]^{-1}[L]^{-2}[T]^{2}$

The realtion between [E] and [B] is 

Answer:

Explanation:

In terms of dimension. F_e = F_m

$\Rightarrow$      qE  = qvB    or E=vB

[E]=[B] [L][T^-1]

• Physics (2018) - Solved questions 2018