Category Archives: JEE Main

$R_1 = (4\pm 0.8) \Omega $
$R_2 = (4\pm 0.4) \Omega $
$R_{eq} =? $

May 30, 2022 Manish Verma Leave a comment

Two resistors $R_1 = (4\pm 0.8) \Omega $ and $R_2 = (4\pm 0.4) \Omega $ are connected in parallel. The equivalent resistance of their parallel combination will be:

(A) $(4 \pm 0.4 ) \Omega $
(B) $(2 \pm 0.3 ) \Omega $
(C) $(4 \pm 0.3 ) \Omega $
(D) $(2 \pm 0.4 ) \Omega $

Solution

We have, $R=\frac {R_1 R_2}{R_1 + R_2 } = \frac {4 \times 4}{4+4} = 2 \Omega $

Further, $\frac {1}{R} = \frac {1}{R_1} + \frac {1}{R_2} $

$\therefore \frac {|\Delta R |}{R^2 }=\frac {|\Delta R_1 |}{R_1 ^2 } + \frac {|\Delta R_2 | }{R_2 ^2 }$

$\therefore \frac {|\Delta R |}{2^2 }=\frac {0.8}{4^2 } + \frac {0.4}{4^2 } = \frac {1.2}{16}$

$\therefore |\Delta R| = 0.3 \Omega $

$R_{eq} = R \pm |\Delta R | = 2 \pm 0.3 \Omega $

Answer: (B)

JEE Main

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

May 29, 2022 Manish Verma 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 ….

May 28, 2022 Manish Verma 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 ….

May 27, 2022 Manish Verma 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)

JEE Main

Circular Coil Magnetic Flux

February 5, 2020 Manish Verma 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?