Current Electricity Quiz 19

Current Electricity Quiz-19

Dear Readers,

As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .

Q1. Corresponding to the resistance 4.7×10⁶ Ω±5%, which is order of colour coding on carbon resistors?

•  Yellow, violet, blue, gold
•  Yellow, violet, green, gold
•  Orange, blue, green, gold
•  Orange, blue, violet, gold

Solution
Yellow, Violet and Gold

Q2. A 25 W,220 V bulb and a 100 W,220 V bulb are connected in parallel across a 440 V line?

•  Only 100 watt bulb will fuse
•  Only 25 watt bulb will fuse
•  Both bulbs will fuse
•  None of the bulbs will fuse

Solution
Because given voltage is very high

Q3. A battery of emf 2V and internal resistance 0.1 Ω is being charged by a current of 5A. the potential difference between the terminals of the battery is?

•  2.5 Ω
•  1.5 Ω
•  0.5 Ω
•  1 Ω

Solution
When r is internal resistance of the battery and i the charging current, then V=E+ir V=2+5×0.1=2.5V

Q4. Two identical cells send the same current in 3 Ω resistance, whether connected in series or in parallel. The internal resistance on the cell should be?

•  1 Ω
•  3 Ω
•  1/2 Ω
•  3.5 Ω

Solution
From the Ohm’s law I=2E/(2r+R) In parallel combination of two cells, the current through the external resistance r will be I^'=E/(r/2+R)=2E/(r+2R) If I=I’ then 2r+R=r+2R r=R=3Ω

Q5. Resistors of resistance 20Ωand 30Ω are joined in series with a battery of emf 3V. It is desired to measure current and voltage across the 20Ω resistor with the help of an ammeter and voltmeter. Identify the correct arrangement of ammeter (A) and voltmeter (V) out of four possible arrangements shown in figure. Given below?

• 
  
  
•  
  
  
•  
  
  
•  
  
  

Solution
 As ammeter must be connected in series of 20Ω resistor, and the voltmeter in parallel to 20Ω resistor, the correct arrangement is as shown in figure(c).

Q6. For a thermocouple, the inversion temperature is 600℃ and the neutral temperature is 320℃. Find the temperature of the cold junction?

•  40℃
•  20℃
•  80℃
•  60℃

Solution
Neutral temperature T_n=(T_c + T_f)/2 320°= (T_c+ 600°)/2 640°=T_c+600° T_c=40℃

Q7. Two bulbs, one of 50 watt and another of 25 watt are connected in series to the mains. The ratio of the currents through them is?

•  2 :1
•  1:2
•  1:1
•  Without voltage, cannot be calculated

Solution
The bulbs are in series, hence they will have the same current through them

Q8. The total current supplied to the given circuit by the battery is?

•  9 A
•  6 A
•  2 A
•  4 A

Solution
The equivalent circuit of the given circuit is as shown

Resistances 6Ω and 2Ω are in parallel ∴ R^'=(6×2)/(6+2)=3/2 Ω Resistances 3/2 Ω and 1.5Ω are in series ∴ R^''=3/2+1.5=3Ω Resistances 3Ω and 3Ω are in parallel ∴ R=(3×3)/(3+3)=3/2 The current, I=V/R = 9/(3/2)=6A

Q9. A resistance of 2 Ω is connected across one gap of a meter-bridge(the length of the wire is 100cm) and an unknown resistance, greater than 2 Ω is connected across the other gap. When these resistances are interchanged, the unknown resistance is?

•  3 Ω
•  2 Ω
•  4 Ω
•  6 Ω

Solution
R>2Ω ∴ 100- x> x Applying, P/Q=R/S

We have 2/R=x/(100-x) ….(i) R/2=(x+20)/(80-x) ……..(ii) Solving Eqs. (i) and (ii), we get R=3Ω

Q10. A bulb rated at (100W-200V) is used on a 100V line. The current in the bulb is?

•  1/4 amp
•  4 amp
•  1/2 amp
•  2 amp

Solution
P= V²/R⇒100=(200)²/R⇒R=(4×10⁴)/10² =400 Ω Now,i=V/R=100/400=1/4 amp