In the given figure, each diode has a forward bias …
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)
$\mathop {\lim }\limits_{x \to \infty} \frac {x^3}{\sqrt {x^2-a^2}}-\frac {x^3}{\sqrt {x^2+a^2}}=?$
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)
Sum of the coefficients in the expansion of $(x+y)^n$ ….
If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ .
Solution
$C_0 + C_1 + C_2 + C_3 + ……………………. + C_n =4096 $
$\therefore 2^n = 4096 =2^{12} $
$\Rightarrow n = 12 $
Greatest coefficient = ${}^{12}{C_6} = 924$
The distance of line $3y-2z-1=0=3x-z+4$ from the point $(2,-1,6)=?$
The distance of line $3y-2z-1=0=3x-z+4$ from the point $(2,-1,6)$ is:
(A) $2\sqrt 6$
(B) $\sqrt {26}$
(C) $2\sqrt 5 $
(D) $4\sqrt 2 $
Solution
We have, $3x-z+4=0$ or $z=3x+4$
& $3y-2z-1=0$ or $3y-2(3x+4)-1=0$ or $y=2x+3$
Any point $(x,y,z)$ on the line $ \equiv (t,2t + 3,3t + 4)$
Let $d$ be the distance between $(2,-1,6)$ & $ (t,2t + 3,3t + 4)$
Then, $d^2=(t-2)^2+(2t+3+1)^2+(3t+4-6)^2=14t^2+24$
Minimum $d$ = Required answer = $\sqrt {24}=2\sqrt 6$ when t = 0.
Answer: (A)