Physics Test

Result

Physics Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

A transistor is used as an amplifier in CB mode with a load resistance of 5 kW. The current gain of the amplifier is 0.98 and the input resistance is 70 W. The voltage gain and power gain respectively, are

A
70, 68.6

B
80, 75.6

C
60, 66.6

D
90, 96.6

Solution
Question 2
1 / -0

The graph of input characteristics of common emitter amplifier is

A

B

C

D

Solution
Question 3
1 / -0

In a transistor circuit, the collector current is 50 mA and the base current is 1 mA. The current gain α is

A
51/50

B
49/50

C
50/51

D
49/51

Solution
Question 4
1 / -0

A
1.08 mA

B
0.08 mA

C
1 mA

D
1 A

Solution
Question 5
1 / -0

For a transistor, the current amplification factor is 0.8. If the transistor is in common emitter configuration and the base current changes by 6 mA, the change in the collector current is

A
6 mA

B
4.8 mA

C
24 mA

D
8 mA

Solution
Question 6
1 / -0

Which of the following is used for making the filaments of electric bulbs?

A
Copper

B
Nichrome

C
Tungsten

D
None of these

Solution

Tungsten has the highest melting point, that's why it is used for making bulb filaments .
Question 7
1 / -0

A
4 Js^-1

B
1.6 Js^-1

C
2 Js^-1

D
8 Js^-1

Solution
Question 8
1 / -0

The thermo-emf of a thermocouple varies with the temperature θ of the hot junction as E = aθ + bθ² in volts ,where the ratio a/b is 700°C. If the cold junction is kept at 0°C, what is the neutral temperature?

A
700°C

B
350°C

C
1400°C

D
no neutral temperature is possible for this thermocouple

Solution

For the given thermocouple the induced emf increases with the increase in the temperature of the hot junction so no neutral temperature is possible for this thermocouple
Question 9
1 / -0

Three identical bulbs connected in series dissipate a power p. If the bulbs are connected in parallel, the power dissipated will be

A
p/3

B
3p

C
9p

D
p/9

Solution
Question 10
1 / -0

The resistance of bulb filament is 100W at a temperature of 100°C. If its temperature coefficient of resistance is 0.005 per °C, its resistance will become 200 W at a temperature (°C) of _____.

A
200

B
100

C
150

D
300

Solution

100 = R₀(1 + 0.005 × 100) -------- (i)

200 = R₀(1 + 0.005 × t) --------- (ii)

Dividing (i) by (ii) and solving

t = 200 °C