Category Archives: JEE Main

A capacitor is connected to a 20 V battery …

A capacitor is connected to a 20 V battery through a resistance of $10 \Omega$. It is found that the potential difference across the capacitor rises to 2 V in $1 \mu s $. The capacitance of the capacitor is _ _ _ _ _ $\mu F$. [Given $\ln (\frac {10}{9})=0.105 $]

(A) 1.85
(B) 0.95
(C) 0.105
(D) 9.52

Solution

In RC charging circuit, $V=E (1-e^{-\frac {t}{\tau}}) $

$\Rightarrow 2=20 (1-e^{-\frac {1}{\tau})}$ where $\tau $ is in $\mu s $

$\Rightarrow 0.9=e^{-\frac {1}{\tau}} $

$\Rightarrow ln (\frac {10}{9})=\frac {1}{\tau} =0.105=\frac {1}{RC}$ where C is in $\mu F $

$\Rightarrow 10C=\frac {1}{0.105}$

$\Rightarrow C=\frac {1}{1.05} \mu F=0.95 \mu F$

Answer: (B)

The ranges and heights for two projectiles projected ….

The ranges and heights for two projectiles projected with the same initial velocity at angles $42^\circ $ and $48^\circ $ with the horizontal are $R_1\,,R_2$ and $H_1\,,H_2$ respectively. Choose the correct option:

(A) $R_1 < R_2 $ and $H_1 < H_2 $
(B) $R_1 > R_2 $ and $H_1 = H_2 $
(C) $R_1 = R_2 $ and $H_1 = H_2 $
(D) $R_1 = R_2 $ and $H_1 < H_2 $

Solution

$R=\frac {u^2 sin 2\theta }{g} $, $H=\frac {u^2 sin^2 \theta }{2g}$

$\frac {R_1}{R_2} = \frac {sin 2 \times 42^\circ}{sin 2\times 48^\circ} = \frac {cos 6^\circ}{cos 6^\circ} = 1 $

$\frac {H_1}{H_2}= \frac {sin^2 42^\circ }{sin^2 48^\circ } = tan^2 42^\circ < 1 $

Answer: (D)