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As we learnt in Mole Fraction - It is the ratio of moles of solute or moles of solvent to moles of the solution. - wherein If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are  and  respectively; then the mole fractions of A and B are given as Mole fraction of A = (number of moles of A)/(number of moles of solution ) = /(+ )    At   ,                                      ...

a lone pair refers to a pair of valence electrons that are not shared with another atom and is sometimes called an unshared pair or non-bonding pair. Lone pairs are found in the outermost electron shell of atoms. They can be identified by using a Lewis structure.
Engineering
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. An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of 8 : 27. The ratio of the radii of the nuclei (assumed to be spherical) is :

• Option 1)

8 : 27

• Option 2)

4 : 9

• Option 3)

3 : 2

• Option 4)

2 : 3

$m_{1}v_{1}=m_{2}v_{2}$

$\therefore \frac{m_{1}}{m_{2}}=\frac{v_{2}}{v_{1}}= \frac{27}{8}$

$\therefore \frac{A_{1}}{A_{2}}= \frac{27}{8}$

$\therefore \frac{r_{1}}{r_{2}}= \frac{r_{0}A_{1}^{1/3}}{r_{0}A_{2}^{1/3}}=\left ( \frac{A_{1}}{A_{2}} \right )^{1/3}=\frac{3}{2}$

Option 1)

8 : 27

Option 2)

4 : 9

Option 3)

3 : 2

Option 4)

2 : 3

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Engineering
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Which of the following arrangements shows the schematic alignment of magnetic moments of antiferromagnetic substance?

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we learned

The substances which are expected to passes para-magnetism or ferromagnetism on the basis of unpaired electrons but actually, they possess zero net magnetic moments are called anti ferromagnetic substance.

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
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A man on the top of a vertical tower observes a car moving at a uniform speed
towards the tower on a horizontal road. If it takes 18 min. for the angle of
depression of the car to change from $30 \degree$ to $45 \degree$
; then after this, the time taken
(in min.) by the car to reach the foot of the
tower, is :

• Option 1)

$9(1+\sqrt3)$

• Option 2)

$18(1+\sqrt3)$

• Option 3)

$18(\sqrt3-1)$

• Option 4)

$9/2(\sqrt3-1)$

As we have learned

Angle of Elevation -

If an object is above the horizontal line from the eye, we have to raise our head to view the object.

- wherein

$\tan 45\degree= h/x$

$\tan 30\degree= h/x+18$

$\sqrt 3 x = x+18$

$x= 9(\sqrt3+1)$

Option 1)

$9(1+\sqrt3)$

This is correct

Option 2)

$18(1+\sqrt3)$

This is incorrect

Option 3)

$18(\sqrt3-1)$

This is incorrect

Option 4)

$9/2(\sqrt3-1)$

This is incorrect

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Engineering
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If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm2) of this cone is :

• Option 1)

$6\sqrt{2}\: \pi$

• Option 2)

$6\sqrt{3}\: \pi$

• Option 3)

$8\sqrt{2}\: \pi$

• Option 4)

$8\sqrt{3}\: \pi$

As we learned

Method for maxima or minima -

First and second derivative method :

$Step\:1.\:\:find\:values\:of\:x\:for\:\frac{dy}{dx}=0$

$Step\:2.\:\:x=x_{\circ }\:\:is\:a\:point\:of\:local\:maximum\:if\:f'(x)>0$  $and\:local\:minimum\:if\;f'(x)<0.$

$Step\:\:3.\:\:\:x=x_{\circ }\:\:is\:a\:point\:of\:local\:miximum\:if$  $f''(x)<0\:\:and\:local\:minimum\:if\:f''(x)>0$

- wherein

$Where\:\:y=f(x)$

$\frac{dy}{dx}=f'(x)$

$Sphere's\: radius =3cm$

r,h be the radius and height of sphere

$Volume \: of\: cone =\frac{1}{3}\pi b^{2}h$

in $\bigtriangleup ABC$  ,using pythagorus theorem

$\left ( h-r \right )^{2}+b^{2}=r^{2}$

$b^{2}=r^{2}-\left ( h-r \right )^{2}-2hr-r^{2}$

$volume=\frac{1}{3}\left [ 2h^{2}r-h^{3} \right ]$

$\frac{\mathrm{dv} }{\mathrm{d} h}=h\left ( 4r-3h \right )=0\: \: \frac{d^{2}v}{dh^{2}}=\frac{1}{3}\left [ 4r-6h \right ]$

So at $h=\frac{4r}{3}$   ; we get max  r=3   h = 4

thus  $b=2\sqrt{2}$

$CSA = \pi bl=\pi 2\sqrt{2}\sqrt{4^{2}+\left ( 2\sqrt{2} \right )^{2}}=\pi 2\sqrt{2}\sqrt{24}=\pi 2\sqrt{2}\cdot 2\sqrt{3}\sqrt{2}$

$=8\sqrt{3}\pi$

Option 1)

$6\sqrt{2}\: \pi$

Option 2)

$6\sqrt{3}\: \pi$

Option 3)

$8\sqrt{2}\: \pi$

Option 4)

$8\sqrt{3}\: \pi$

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Engineering
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An ideal capacitor of capacitance 0.2 μF is charged to a potential difference of 10 V. The charging battery is then
disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5 mH. The current at a time
when the potential difference across the capacitor is 5 V, is :

• Option 1)

0.34 A

• Option 2)

0.25 A

• Option 3)

0.17 A

• Option 4)

0.15 A

As we learned @3935   From energy conservation   Option 1) 0.34 A Option 2) 0.25 A Option 3) 0.17 A Option 4) 0.15 A
Engineering
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A force of 40 N acts on a point B at the
end of an L-shaped object, as shown in
the figure. The angle θ that will produce
maximum moment of the force about
point A is given by :

• Option 1)

$tan\theta =\frac{1}{2}$

• Option 2)

$tan\theta =2$

• Option 3)

$tan\theta =4$

• Option 4)

$tan\theta =\frac{1}{4}$

As we learned

Torque -

$\underset{\tau }{\rightarrow}= \underset{r}{\rightarrow}\times \underset{F}{\rightarrow}$

- wherein

This can be calculated by using either  $\tau=r_{1}F\; or\; \tau=r\cdot F_{1}$

$r_{1}$ = perpendicular distance from origin to the line of force.

$F_{1}$ = component of force perpendicular to line joining force.

torque about A= $(F\cos \theta )\cdot 4+(F\sin \theta )\cdot 2$

for maximum torque

$\frac{dI}{do}=0$$or 4f(-\sin \theta ) + 2f\cos \theta =0$

$2\sin \theta =\cos \theta or \tan \theta =\frac{1}{2}$

Option 1)

$tan\theta =\frac{1}{2}$

Option 2)

$tan\theta =2$

Option 3)

$tan\theta =4$

Option 4)

$tan\theta =\frac{1}{4}$

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Engineering
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If $x^{2}+y^{2}+\sin y=4$,  then the value of $\frac{d^{2}y}{dx^{2}}$ at the point (-2,0) is :

• Option 1)

$-34$

• Option 2)

$-32$

• Option 3)

$4$

• Option 4)

$-2$

As we learned

Derivative at a point -

The value of  f'(x) obtained by putting  x = a is called the derivative of  f(x) at  x = a  and it is denoted by  f'(a)  or

$\frac{dy}{dx}$     at  x = a.

-

$x^{2}+y^{2}+\sin y=4$

On Differentiating , we get

$\frac{\mathrm{dy} }{\mathrm{d} x}=-\left ( \frac{2x}{2y+\cos y} \right )$

at (-2,0)

$\frac{\mathrm{dy} }{\mathrm{d} x}=4$

$\frac{\mathrm{d^{2}y} }{\mathrm{d} x^{2}}=-\left [ \frac{\left ( 2y+\cos y \right )\left ( 2x \right )}{\left ( 2y+\cos y \right )^{2}}-\left ( 2y+\cos y \right ){}'\left ( 2x \right ) \right ]$

at (-2,0)

$\frac{\mathrm{d^{2}y} }{\mathrm{d} x^{2}}=-34$

Option 1)

$-34$

Option 2)

$-32$

Option 3)

$4$

Option 4)

$-2$

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Engineering
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It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :

• Option 1)

(0, 1)

• Option 2)

(⋅89, ⋅28)

• Option 3)

(⋅28, ⋅89)

• Option 4)

(0, 0)

Perfectly Elastic Collision -

Law of conservation of momentum and that of Kinetic Energy hold good.

- wherein

$\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}= \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$m_{1},m_{2}:masses$

$u_{1},v_{1}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{1}$

$u_{2},v_{2}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{2}$

Before Collision

for elastic collision :

$\\*mv=2mv_{2}-mv_{1}\\* or v=2v_{2}-v_{1}\: \: ---(1)\\* Velocity \: o\! f \: sepration=Velocity \: o\! f \:approach\\* v_{2}+v_{1}=v---(2)\\* From (1)\: and\: (2)\Rightarrow 3v_{2}= 2v\: \: or\: \: v_{2}= \frac{2v}{3}\\* and \: v_{1}= \frac{v}{3}$

$P_{d}= \frac{\frac{1}{2}mv^{2}-\frac{1}{2}mv_{1}^{2}}{\frac{1}{2}mv^{2}}= \frac{1-\frac{1}{9}}{1}= \frac{8}{9}= 0.89$

For collision with carbon

$\\*2v_{2}-v_{1}= v\\*and\: v_{2}+v_{1}= v$       $\Rightarrow$      $\Rightarrow \: \: \: \: v_{2}= \frac{2v}{13}\: and\: v_{1}= \frac{11v}{13}$

Fraction Loss =$p_{c}= \frac{\frac{1}{2}mv^{2}-\frac{1}{2}mv_{1}^{2}}{\frac{1}{2}mv^{2}}=$

$= \frac{48}{169}= 0.28$

Option 1)

(0, 1)

Option 2)

(⋅89, ⋅28)

Option 3)

(⋅28, ⋅89)

Option 4)

(0, 0)

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Engineering
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If $f\left ( x \right )=\sin ^{-1}\left ( \frac{2\times 3^{x}}{1+9^{x}} \right )$, then $f'\left ( -\frac{1}{2} \right )$ equals :

• Option 1)

$-\sqrt{3}\log_{e}\sqrt{3}$

• Option 2)

$\sqrt{3}\log_{e}\sqrt{3}$

• Option 3)

$-\sqrt{3}\log_{e}3$

• Option 4)

$\sqrt{3}\log_{e}3$

As we learned,

$f\left ( x \right )=\sin ^{-1}\left ( \frac{2\times 3^{x}}{1+9^{x}} \right )$

$=2\tan ^{-1}3^{x}$

$f'\left ( x \right )=2\times \frac{1}{1+\left ( 3x \right )^{2}}\times 3^{n}\times \ln 3$

$f'\left ( \frac{-1}{2} \right )=2\times \frac{1}{1+\left ( 3^{-1} \right )}\times 3^{\frac{-1}{2}}\times \ln 3$

$=2\times \frac{3}{4}\times \frac{1}{\sqrt{3}}\times \ln 3$

$=\sqrt{3}\times \frac{1}{2}\ln 3$

$=\sqrt{3}\ln\sqrt{3}$

Option 1)

$-\sqrt{3}\log_{e}\sqrt{3}$

Option 2)

$\sqrt{3}\log_{e}\sqrt{3}$

Option 3)

$-\sqrt{3}\log_{e}3$

Option 4)

$\sqrt{3}\log_{e}3$

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Engineering
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Let $A_{n}=\left ( \frac{3}{4} \right )-\left ( \frac{3}{4} \right )^{2}+\left ( \frac{3}{4} \right )^{3}- \cdots \cdots + \left ( -1 \right )^{n-1}\left ( \frac{3}{4} \right )^{n}$ and $B_{n}=1-A_{n}.$ Then,
the least odd natural number p, so that $B_{n}> A_{n}$, for all $n\geq p$, is:

• Option 1)

9

• Option 2)

7

• Option 3)

11

• Option 4)

5

As we learned,     If  i.e.   Thus least value of p=7 Option 1) 9 Option 2) 7 Option 3) 11 Option 4) 5
Engineering
101 Views   |

Let $f:A\rightarrow B$ be a function defined as $f\left ( x \right )=\frac{x-1}{x-2}$ where $A=R-\left ( 2 \right )$ and $B=R-\left ( 1 \right )$. Then f is :

• Option 1)

invertible and $f^{-1}\left ( y \right )=\frac{3y-1}{y-1}$

• Option 2)

invertible and $f^{-1}\left ( y \right )=\frac{2y-1}{y-1}$

• Option 3)

invertible and $f^{-1}\left ( y \right )=\frac{2y+1}{y-1}$

• Option 4)

not invertible

As we learned,

$f\left ( x \right )=\frac{x-1}{x-2}=y$

$\Rightarrow \: xy-2y=x-1$

$\Rightarrow \: x\left ( y-1 \right )=2y-1\: \Rightarrow \: x=\frac{2y-1}{y-1}$

The function is invertible and $f^{-1}\cdot \left ( y \right )=\frac{2y-1}{y-1}$

Option 1)

invertible and $f^{-1}\left ( y \right )=\frac{3y-1}{y-1}$

Option 2)

invertible and $f^{-1}\left ( y \right )=\frac{2y-1}{y-1}$

Option 3)

invertible and $f^{-1}\left ( y \right )=\frac{2y+1}{y-1}$

Option 4)

not invertible

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Engineering
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The characteristic distance at which quantum gravitational effects are significant, the Planck length, can be determined from a suitable combination of the fundamental physical constants G, h and c. Which of the following correctly gives the Planck length ?

• Option 1)

$Gh^{^{2}}c^{3}$

• Option 2)

$G^{2}hc$

• Option 3)

$G^{1/2}h^{2}c$

• Option 4)

$\left ( \frac{Gh}{c^{3}} \right )^{1/2}$

As we have learnt @1035

let planck length is

$L = K.G^{^{x}}h^{y}c^{z}$

$\left [M^{0}L^{1}T^{0} \right ] = \left [M^{-1}L^{3}T^{-2} \right ]^{x}\left [ML^{2}T^{-1} \right ]^{y}\left [LT^{-1} \right ]^{z}$

$= \left [ M^{-x+y}.L^{3x+2y+z}.T^{-2x-y-z} \right ]$

$\therefore -x+y=0 \: or\: x=y$

$\therefore -2x-y-z=0\: or\: 2x+y+z=0\: or \: z=-3x$

$\therefore 3x+2y+z=1 or 2x=1 or x=1/2$

$\therefore y=1/2$

$\therefore z=-3/2$

Planck Length = $\left [\frac{Gh}{c^{3}} \right ]^{1/2}$

Option 1)

$Gh^{^{2}}c^{3}$

Option 2)

$G^{2}hc$

Option 3)

$G^{1/2}h^{2}c$

Option 4)

$\left ( \frac{Gh}{c^{3}} \right )^{1/2}$

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Engineering
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In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale
divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is :

• Option 1)

0.4300 cm

• Option 2)

0.2150 cm

• Option 3)

0.3150 cm

• Option 4)

0.0430 cm

As we learned

To measure the diameter of small spherical cylindrical body using Vernier Callipers -

Vernier Constant

= 1 Main scale division - 1 V.S. Division

V.C= 1 M.S.D - 1 V.S.D

- wherein

Total observed reading = $N+n \times V.C$

N= Nth division

Observations:

1.    Vernier constant (least count) of the Vernier Callipers:

1 M.S.D. = 1 mm

10 vernier scale divisions = 9 main scale divisions

i.e.     10 V.S.D. = 9 M.S.D.

$\therefore$        1 V.S.D. = $=\frac{9}{10}$ M.S.D.

Vernier Constant (L.C.) = 1 M.S.D. - 1 V.S.D. = 1 M.S.D. $-\frac{9}{10}$ M.S.D.

$=\left(1-\frac{9}{10} \right )M.S.D.=\frac{1}{10}\times1M.S.D.$

$=\frac{1}{10}\times1mm=0.1mm=0.01cm$

2.    Zero error: (i) ........ cm (ii) ............cm (iii) ..........cm

Mean Zero Error (e) = ............ cm

Mean Zero Correction (c) = - (Mean Zero Error)

= .......... cm

$least count =\frac{0.25}{5\times 100}cm = 5\times 10^{-4}cm$

$Reading = 4\times 0.05 cm +30\times 5\times 10^{-4}cm$

$=(0.2+0.0150)cm$

= 0.2150cm

Option 1)

0.4300 cm

Option 2)

0.2150 cm

Option 3)

0.3150 cm

Option 4)

0.0430 cm

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Engineering
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A uniform rod AB is suspended from a point X, at a variable distance x from A, as shown. To make the rod horizontal, a mass m is suspended from its end A. A set of (m, x) values is recorded. The appropriate variables that give a straight line, when plotted, are :

• Option 1)

m, x

• Option 2)

$m,\frac{1}{x}$

• Option 3)

$m,\frac{1}{x^{2}}$

• Option 4)

$m,{x^{2}}$

As we learned

Torque -

$\underset{\tau }{\rightarrow}= \underset{r}{\rightarrow}\times \underset{F}{\rightarrow}$

- wherein

This can be calculated by using either  $\tau=r_{1}F\; or\; \tau=r\cdot F_{1}$

$r_{1}$ = perpendicular distance from origin to the line of force.

$F_{1}$ = component of force perpendicular to line joining force.

Balancing Torque w.r.t point of suspension

$mgx = Mg\left ( \frac{l}{2}-x \right )$

$\Rightarrow mx = M\frac{l}{2}-Mx \Rightarrow m = \frac{Ml}{2}\cdot \frac{1}{x}-M$

This represents a straight line

Option 1)

m, x

Option 2)

$m,\frac{1}{x}$

Option 3)

$m,\frac{1}{x^{2}}$

Option 4)

$m,{x^{2}}$

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Engineering
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The number of amplitude modulated broadcast stations that can be accomodated in a 300 kHz band width for the highest modulating frequency 15 kHz will be :

• Option 1)

20

• Option 2)

15

• Option 3)

10

• Option 4)

8

As we learned

Band width -

It is frequency range over which an equipment operate or the portion of the spectrum occupied by the signal.

-

Frequency required for one station

$2 \nu _{m}= 30 KHZ$

Number of channel = 10

Option 1)

20

Option 2)

15

Option 3)

10

Option 4)

8

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Engineering
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In a common emitter configuration with suitable bias, it is given that $R_{L}$ is the load resistance and $R_{BE}$ is small signal dynamic
resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by : β is current gain, $I_{B}$ , $I_{C}$ and $I_{E}$ arerespectively base, collector and emitter currents.

• Option 1)

$\beta \frac{R_{L}}{R_{BE}}, \frac{\Delta I_{c}}{\Delta I_{B}}, \beta ^{2 }\frac{R_{L}}{R_{BE}}$

• Option 2)

$\beta \frac{R_{L}}{R_{BE}}, \frac{\Delta I_{E}}{\Delta I_{B}}, \beta ^{2 }\frac{R_{L}}{R_{BE}}$

• Option 3)

$\beta^{2} \frac{R_{L}}{R_{BC}}, \frac{\Delta I_{C}}{\Delta I_{E}}, \beta ^{2 }\frac{R_{L}}{R_{BE}}$

• Option 4)

$\beta^{2} \frac{R_{L}}{R_{BC}}, \frac{\Delta I_{C}}{\Delta I_{B}}, \beta ^{2 }\frac{R_{L}}{R_{BE}}$

As we learned

Relation between α and β -

$\beta = \frac{\alpha }{1-\alpha }$

- wherein

$\alpha = \frac{I_{C}}{I_{E}}$

$\beta = \frac{I_{C}}{I_{B}}$ (current gain )

Current Gain $\beta =\frac{\Delta I_{c}}{\Delta I_{B}}$

Vol + age Gain = $\frac{v_{o}}{v_{1}} = \frac{R_{L}-\Delta I_{c}}{R_{BE}-\Delta I_{B}} = \beta \frac{R_{L}}{R_{BE}}$

power gain = (voltage gain) (Current gain)

$\beta ^{2}\cdot \frac{R_{L}}{R_{BE}}$

Option 1)

$\beta \frac{R_{L}}{R_{BE}}, \frac{\Delta I_{c}}{\Delta I_{B}}, \beta ^{2 }\frac{R_{L}}{R_{BE}}$

Option 2)

$\beta \frac{R_{L}}{R_{BE}}, \frac{\Delta I_{E}}{\Delta I_{B}}, \beta ^{2 }\frac{R_{L}}{R_{BE}}$

Option 3)

$\beta^{2} \frac{R_{L}}{R_{BC}}, \frac{\Delta I_{C}}{\Delta I_{E}}, \beta ^{2 }\frac{R_{L}}{R_{BE}}$

Option 4)

$\beta^{2} \frac{R_{L}}{R_{BC}}, \frac{\Delta I_{C}}{\Delta I_{B}}, \beta ^{2 }\frac{R_{L}}{R_{BE}}$

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