VITEEE 2019 :: Physics :: General questions

• Physics - General questions

In a common base mode of a transistor, the collector current is 5.488 mA for an emitter current of 5.60 mA. The value of the base current amplification factor (β) will be

Answer:

Explanation:

I_c = 5.488mA I_e = 5.6mA

$\alpha = \frac{I_{c}}{I_{e}}$ = $\frac{5.488}{5.6}$

$\beta = \frac{\alpha}{\left(1 - \alpha\right)}$ = 49

The half-life period and the mean life period of a radioactive element are denoted respectively by T_h and T_m. Then

Answer:

Explanation:

Half-life T_h = $\frac{0.693}{\lambda}, T_{m}= \frac{1}{\lambda}$

Clearly, $T_{h}< T_{m}$

The equivalent capacitance of the combination of the capacitors is

Answer:

Explanation:

Equivalent capacitance of two parallel capacitors 10 μF and 6 μF = (10 + 6)μF = 16 μF . This 16 μF capacitor is in series combination with 4μF capacitor,

.'. Equivalent capacitance of the entire combination 

= $\frac{16\times4}{16+4} = \frac{64}{20}$

= 3.20μF

A point charge q is rotated along a circle in the electric field generated by another point charge Q The work done by the electric field on the rotating charge in one complete revolution is

Answer:

Explanation:

Work done in the displacement of charge,

$W = q\triangle V$

where ΔV : the potential difference between the displacement of charge Hence, work done in complete rotation is zero

Which of the following gates will have an output of I?

Answer:

Explanation:

(A) is a NAND gate so the output is

$\overline{1\times1} =\overline{1} = 0 $

(B) is a NOR gate so output is

$\overline{0+1} =\overline{1} = 0 $

(C) is a NAND gate so output is

 $\overline{0\times1} =\overline{0} = 1$

(D) is a XOR gate so output is

0 ⊕ 0 = 0

• Physics - General questions