JEE Main

The temperature of an ideal gas in 3-dimensions is ….

Leave a comment

The temperature of an ideal gas in 3-dimensions is 300 K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is:
[$m_e $ = mass of electron = $9 \times 10^{-31}$ kg, h = Planck constant = $6.6 \times 10^{-34} $Js, $k_b $ = Boltzmann constant = $1.38 \times 10^{-23} JK^{-1} $]

(A) 3.25 nm
(B) 6.26 nm
(C) 2.26 nm
(D) 8.46 nm

Solution

de-Broglie wavelength $\lambda = \frac {h}{p} = \frac {h}{\sqrt {2m_e K}}$

For ideal gas in 3-dimensions, $K = \frac {3}{2} k_b T $

$\therefore \lambda  = \frac{h}{{\sqrt {3{m_e}{k_b}T} }} = \frac{{6.6 \times {{10}^{ – 34}}}}{{\sqrt {3 \times 9 \times {{10}^{ – 31}} \times 1.38 \times {{10}^{ – 23}} \times 300} }}$

$\lambda  \approx \frac{{2.2 \times {{10}^{ – 34}}}}{{3 \times 1.17 \times {{10}^{ – 26}}}} \approx 6.26 \times {10^{ – 9}}m = 6.26nm$

Answer: (B)

JEE Main

Electric field of a plane electromagnetic wave ….

Leave a comment

Electric field of a plane electromagnetic wave propagating through a non-magnetic medium is given by $E = 20\cos (2 \times {10^{10}}t – 200x)V/m$. The dielectric constant of the medium is equal to: (Take $\mu _r = 1 $)

(A) $\frac {1}{3}$
(B) 3
(C) 2
(D) 9

Solution

We have, $E = 20\cos \left[ {200\left( {\frac{{2 \times {{10}^{10}}}}{{200}}t – x} \right)} \right]V/m$

So, $v = \frac{{2 \times {{10}^{10}}}}{{200}} = 1 \times {10^8}m/s$

$\mu  = \sqrt {{\mu _r}{\epsilon_r}}  = \frac{c}{v} = \frac{{3 \times {{10}^8}}}{{1 \times {{10}^8}}} = 3$

$ \Rightarrow \sqrt {1 \times {\epsilon_r}}  = 3$

$ \Rightarrow {\epsilon_r} = 9$ = Dielectric Constant

Answer: (D)

JEE Main

The half life period of radioactive element x is same ….

Leave a comment

The half life period of radioactive element x is same as the mean life time of another radioactive element y. Initially they have the same number of atoms. Then:

(A) x and y have same decay rate initially and later on different decay rate.
(B) x and y decay at the same rate always.
(C) x will decay faster than y.
(D) y will decay faster than x.

Solution

Given, ${({t_{1/2}})_x} = {({t_{mean}})_y}$

$\therefore \frac{{0.693}}{{{\lambda _x}}} = \frac{1}{{{\lambda _y}}}$

$ \Rightarrow {\lambda _x} = 0.693{\lambda _y}$

$ \Rightarrow {\lambda _x} < {\lambda _y}$

If N is same, $A \propto \lambda $

$\therefore A_x < A_y $

Answer: (D)

Math

Solve Differential Equation, $ \frac{{dy}}{{dx}} = \sqrt {x + y} $

Leave a comment

Let, $ \sqrt {x + y} = t$

So, $ \frac{1}{{2\sqrt {x + y} }}\left( {1 + \frac{{dy}}{{dx}}} \right) = \frac{{dt}}{{dx}}$

Or, $\frac{1}{{2t}}\left( {1 + \frac{{dy}}{{dx}}} \right) = \frac{{dt}}{{dx}}$

$\therefore \frac{{dy}}{{dx}} = 2t\frac{{dt}}{{dx}} – 1$

From the original differential equation,

$2t\frac{{dt}}{{dx}} – 1 = t$

$\therefore 2t\frac{{dt}}{{dx}} = 1 + t$

Or, $\frac{{2t}}{{1 + t}}dt = dx$

$ \Rightarrow 2\int {\frac{{1 + t – 1}}{{1 + t}}dt} = \int {dx} $

$ \Rightarrow 2\int {\left( {1 – \frac{1}{{1 + t}}} \right)dt} = \int {dx} $

$ \Rightarrow 2t – 2\ln (1 + t) = x + C$

$ \Rightarrow 2\sqrt {x + y} – 2\ln (1 + \sqrt {x + y} ) = x + C$

Or, $\sqrt {x + y} – \ln (1 + \sqrt {x + y} ) = \frac{x}{2} + C’$

Math

Given $x+y=2$ & $xy=3$, evaluate $x^5+y^5$

Leave a comment

Question

Given x+y=2 & xy=3, evaluate $x^5+y^5$.

Solution

Putting y=2-x in the equation xy=3,

x(2-x)=3

Or, x2-2x+3=0

Or, $x=\frac {2\pm\sqrt {4-12}}{2}$

Or, $x=1\pm i.\sqrt 2$

Out of these two complex conjugate roots, one should represent x and the other should represent y.

Let $x=1+ i.\sqrt 2$, So $y=1- i.\sqrt 2$

${x^5} + {y^5} = {(1 + i\sqrt 2 )^5} + {(1 – i\sqrt 2 )^5}$

$ = [{}^5{C_0} + {}^5{C_1}i\sqrt 2 + {}^5{C_2}{(i\sqrt 2 )^2} + {}^5{C_3}{(i\sqrt 2 )^3} + {}^5{C_4}{(i\sqrt 2 )^4} + {}^5{C_5}{(i\sqrt 2 )^5}]$$+$$[{}^5{C_0} – {}^5{C_1}i\sqrt 2 + {}^5{C_2}{(i\sqrt 2 )^2} – {}^5{C_3}{(i\sqrt 2 )^3} + {}^5{C_4}{(i\sqrt 2 )^4} – {}^5{C_5}{(i\sqrt 2 )^5}]$

$= 2[{}^5{C_0} + {}^5{C_2}{(i\sqrt 2 )^2} + {}^5{C_4}{(i\sqrt 2 )^4}]$

$ = 2\left[ {1 + \frac{{5!}}{{2!3!}}( – 1)(2) + 5(4)} \right]$

=2(1-20+20)

=2

 

JEE Main

Circular Coil Magnetic Flux

Leave a comment

Consider a circular coil of wire carrying constant current I, forming a magnetic dipole. The magnetic flux through an infinite plane that contains the circular coil and excluding the circular coil area is given by \phi_i. The magnetic flux through the area of the circular coil area is given by \phi_0. Which of the following option is correct?

(A) \phi_i < \phi_0 (B) \phi_i > \phi_0
(C) \phi_i = \phi_0 (D) \phi_i = -\phi_0

[Based on JEE Main 2020]

Continue reading Circular Coil Magnetic Flux →

Physics

L-Shaped Object Equilibrium

Leave a comment

An L-shaped object, made of thin rod of uniform mass density, is suspended with a string as shown in figure. If AB = BC, and the angle made by AB with downward vertical is $\theta$, then:

(A) $tan\theta = \frac {1}{2}$
(B) $tan \theta = \frac {1}{3}$
(C) $tan \theta = \frac {2}{\sqrt 3}$
(D) $tan \theta = \frac {1}{2\sqrt 3}$

[Based on JEE Main 2019]