Current Electricity Quiz-20
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. In the Wheatstone’s network given, P=10 Ω,Q=20Ω, R=15 Ω, S=30 Ω, the current passing through the battery (of negligible internal resistance) is?
Solution
The balanced condition for Wheatstone bridge is P/Q=R/S as is obvious from the given values. No, current flows through galvanometer is zero. Now, P and R are in series, so Resistance R1=P+R =10+15=25Ω Similarly, Q and S are in series, so Resistance R2=R+S=20+30=50Ω Net resistance of the network as R1 and R2 are in parallel. 1/R=1/R1 +1/R2 ∴R=(25×50)/(25+50)=50/3 Ω Hence, I= V/R=6/(50/3)=0.36A
The balanced condition for Wheatstone bridge is P/Q=R/S as is obvious from the given values. No, current flows through galvanometer is zero. Now, P and R are in series, so Resistance R1=P+R =10+15=25Ω Similarly, Q and S are in series, so Resistance R2=R+S=20+30=50Ω Net resistance of the network as R1 and R2 are in parallel. 1/R=1/R1 +1/R2 ∴R=(25×50)/(25+50)=50/3 Ω Hence, I= V/R=6/(50/3)=0.36A
Q2. A circuit consists of five identical conductors as shown in figure. The two similar conductors are added as indicated by the dotted lines. The ratio of resistances before and after addition will be?
Solution
Before adding, resistance is 5 ohms. After the addition, the central one is a Wheatstone’s network. ∴ Total resistance is 1+(2 & 2 in parallel)+1=3Ω ∴ The ratio of resistances =5/3
Before adding, resistance is 5 ohms. After the addition, the central one is a Wheatstone’s network. ∴ Total resistance is 1+(2 & 2 in parallel)+1=3Ω ∴ The ratio of resistances =5/3
Q3. For what value of R the net resistance of the circuit will be 18 ohms?
Solution
∴(R×16)/(R+16)+10=18, on solving we get, R=16Ω
∴(R×16)/(R+16)+10=18, on solving we get, R=16Ω
Q4. The cold junction of a thermocouple is maintained at 10℃. No thermo e.m.f. is developed when the hot junction is maintained at 530℃. The neutral temperature is?
Solution
Tn=(Ti+Tc)/2=(10+530)/2=270℃
Tn=(Ti+Tc)/2=(10+530)/2=270℃
Q5. Two electric bulbs rated P1 watt V volts and P2 watt V volts are connected in parallel and V volts are applied to it. The total power will be?
Solution
If resistances of bulbs are R1 and R2 respectively then in parallel 1/RP =1/R1 +1/R2 ⇒1/((V2/Pp ) )=1/((V2/P1 ) )+1/((V2/P2 ) ) ⇒PP=P1+P2 .
If resistances of bulbs are R1 and R2 respectively then in parallel 1/RP =1/R1 +1/R2 ⇒1/((V2/Pp ) )=1/((V2/P1 ) )+1/((V2/P2 ) ) ⇒PP=P1+P2 .
Q6. A tap supplies water at 22℃. A man takes 1 L of water per min at 37℃ from the geyser. The power of the geyser is?
Solution
Pt=mSθ P=(1×4200×15)/60W=1050 W
Pt=mSθ P=(1×4200×15)/60W=1050 W
Q7. Kirchoff’s second law for the analysis of circuit is based on?
Solution
Kirchhoff’s second law is ∑V=0 It states that the algebric sum of the potential differences in any loop including those associated emf’s and those of resistive elements, must equal zero. This law represents ‘conservation of energy’.
Kirchhoff’s second law is ∑V=0 It states that the algebric sum of the potential differences in any loop including those associated emf’s and those of resistive elements, must equal zero. This law represents ‘conservation of energy’.
Q8. On passing the current in water voltmeter, hydrogen?
Solution
Because H has positive charge
Because H has positive charge
Q9. 62.5×1018 electrons per second are flowing through a wire of area of cross-section 0.1 m2, the value of current flowing will be?
Solution
i= ne/t=(62.5×1018×1.6×10-19)/1=10 ampere
i= ne/t=(62.5×1018×1.6×10-19)/1=10 ampere
Q10. A certain current passing through a galvanometer produces a deflection of 100 divisions. When a shunt of one ohm is connected, the deflection reduces to 1 division. The galvanometer resistance is?
Solution
Shunt is connected to the galvanometer ig=iS/(S+G) 1= (100×1)/((1+G) ) G=99Ω
Shunt is connected to the galvanometer ig=iS/(S+G) 1= (100×1)/((1+G) ) G=99Ω
