Current Electricity Quiz-12
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.
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Q1. A wire of length 5m and radius 1 mm has a resistance of 1 ohm. What length of the wire of the same material at the same temperature and of radius 2 mm will also have a resistance of 1 ohm?
Solution
R∝ 1/r2 ⇒R1/R2 =l1/l2 ×(r22)/(r12 )⇒1/1=5/l2 ×(2/1)2⇒l2=20m
R∝ 1/r2 ⇒R1/R2 =l1/l2 ×(r22)/(r12 )⇒1/1=5/l2 ×(2/1)2⇒l2=20m
Q2. The measurement of voltmeter in the following circuit is?
Solution
If the voltmeter is ideal then given circuit is an open circuit, so reading of voltmeter is equal to the e.m.f. of cell i.e.,6V
If the voltmeter is ideal then given circuit is an open circuit, so reading of voltmeter is equal to the e.m.f. of cell i.e.,6V
Q3. An electron (charge =1.6×10-19coulomb) is moving in a circle of radius 5.1×10-11 m at a frequency of 6.8×1015 revolutions/sec. The equivalent current is approximately?
Solution
i=ev=1.6×10-19×6.8×1015=1.1×10-3 amp
i=ev=1.6×10-19×6.8×1015=1.1×10-3 amp
Q4. In the following Wheatstone bridge P/Q=R/S. If key K is closed, then the galvanometer will show deflection?
Solution
Pressing the key does not disturb current in all resistances as the bridge is balanced. Therefore, deflection in the galvanometer in whatever direction it was, will stay
Pressing the key does not disturb current in all resistances as the bridge is balanced. Therefore, deflection in the galvanometer in whatever direction it was, will stay
Q5. The amount of heat generated in 500Ω resistance, when the key is thrown over from contact 1 to 2, as shown in figure is?
Solution
When the key is in contact with 1, then energy stored in the condenser =1/2 CE2 But when the key is thrown to contact 2, total heat H=I2 (500+330)=1/2 CE2 H1=I2 (500) H1/H=R1/((R1+R2)) H1=500/830×1/2×5×10-6×(200)2 H1=60×10-3 J
When the key is in contact with 1, then energy stored in the condenser =1/2 CE2 But when the key is thrown to contact 2, total heat H=I2 (500+330)=1/2 CE2 H1=I2 (500) H1/H=R1/((R1+R2)) H1=500/830×1/2×5×10-6×(200)2 H1=60×10-3 J
Q6. In potentiometer a balance point is obtained, when?
Solution
The p.d. of the wire between the +ve end to jockey becomes equal to the e.m.f. of the experimental cell
The p.d. of the wire between the +ve end to jockey becomes equal to the e.m.f. of the experimental cell
Q7. With the rise of temperature the resistivity of a semiconductor?
Solution
For semiconductor the temperature coefficient of resistance (α) is negative. Hence, resistivity will decrease with the temperature rise.
For semiconductor the temperature coefficient of resistance (α) is negative. Hence, resistivity will decrease with the temperature rise.
Q8. Five equal resistances, each of resistance R,are connected as shown in figure below. A battery of V volt is connected between A and B.The current flowing in FC will be?
Solution
I= V/R
∴ Current in FC=1/2=V/2R
I= V/R
∴ Current in FC=1/2=V/2R
Q9. I-Vcharacteristic of a copper wire of length L and area of cross-section A is shown in figure. The slope of the curve becomes?
Solution
Slope of graph =I/V=1/R If experiment is performed at higher temperature then resistance increase and hence slope decrease, choice (a) is wrong. Similarly in choice (b) and (c) resistance increase. But for choice (d) resistance R increases, so slope decreases
Slope of graph =I/V=1/R If experiment is performed at higher temperature then resistance increase and hence slope decrease, choice (a) is wrong. Similarly in choice (b) and (c) resistance increase. But for choice (d) resistance R increases, so slope decreases
Q10. If 2.2 kilowatt power is transmitted through a 10 ohm line at 22000 volt, the power loss in the form of heat will be?:
Solution
For power transmission power loss in line PL=i2 R If power of electricity is P and it is transmitted at voltage V, then P=Vi⇒i=P/V PL=(P/V)2 R=(P2 R)/V2 =(2.2×103×2.2×103×10)/(22000×22000)=0.1W
For power transmission power loss in line PL=i2 R If power of electricity is P and it is transmitted at voltage V, then P=Vi⇒i=P/V PL=(P/V)2 R=(P2 R)/V2 =(2.2×103×2.2×103×10)/(22000×22000)=0.1W
