Current Electricitry Quiz-11
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 metallic block has no potential difference applied across it, then the mean velocity of free electrons at absolute temperature T is?
Solution
In the absence of electric field (or potential difference applied) across the conductor, the average thermal velocity of electron is zero
In the absence of electric field (or potential difference applied) across the conductor, the average thermal velocity of electron is zero
Q2. Two bulbs of 500 W and 200 W are manufactured to operate on 220 V line. The ratio of heat produced in 500 W and 200 W, in two cases, when firstly they are connected in parallel and secondly in series will be ?
Solution
In parallel, H1/H2 =(P1 t)/(P2 t)=P1/P2 =500/200=5/2 In series, H1/H2 =(I2 R1 t)/(I2 R2 t)=R1/R2 =(V2/P1)/(V2/P2 ) =P2/P1 =200/500=2/5
In parallel, H1/H2 =(P1 t)/(P2 t)=P1/P2 =500/200=5/2 In series, H1/H2 =(I2 R1 t)/(I2 R2 t)=R1/R2 =(V2/P1)/(V2/P2 ) =P2/P1 =200/500=2/5
Q3. The arrangement as shown in figure is called as ?
Solution
Potential divider
Potential divider
Q4. A fuse wire with radius 1 mm blows at 1.5 A. The radius of the fuse wire of the same material to blow at 3 A will be?
Solution
The temperature of the wire increase to such a value at which the heat produced per second equals heat lost per second due to radiation. i2 (ρl/(πr2 ))=H×2πrl Where H is heat lost per second per unit area due to radiation. Hence, i2∝r3 So, (i12)/(i22 ) = (r13)/(r23 ) Or r2=r1 (i2/i1 )2/3 [Here∶r1=1 mm,i1=1.5 A,i2=3 A] r2=1×(3/1.5)2/3=41/3 mm
The temperature of the wire increase to such a value at which the heat produced per second equals heat lost per second due to radiation. i2 (ρl/(πr2 ))=H×2πrl Where H is heat lost per second per unit area due to radiation. Hence, i2∝r3 So, (i12)/(i22 ) = (r13)/(r23 ) Or r2=r1 (i2/i1 )2/3 [Here∶r1=1 mm,i1=1.5 A,i2=3 A] r2=1×(3/1.5)2/3=41/3 mm
Q5. Two cells of same emf E but of different internal resistances r1and r2 are connected in series with an external resistance R. The potential drop across the first cell is found to be zero. The external resistance R is ?
Solution
As both cells are in series, the circuit current i= (E+E)/(r1+r2+R)=2E/(r1+r2=R) As terminal potential drop across 1st cell is zero, hence V1=E-ir1=E-2E/((r1+r2+R)) r1=0
⇒ E=(2Er1)/((r1+r2+R)) orr1+r2+R=2r1=R=(r1-r2)
As both cells are in series, the circuit current i= (E+E)/(r1+r2+R)=2E/(r1+r2=R) As terminal potential drop across 1st cell is zero, hence V1=E-ir1=E-2E/((r1+r2+R)) r1=0
⇒ E=(2Er1)/((r1+r2+R)) orr1+r2+R=2r1=R=(r1-r2)
Q6. In a conductor if 3000 coulomb of charge enters and 3000 coulomb of charge exits in time 10 minutes, then the current is?
Solution
Current, I=q/t=3000C/(10×60s)=5A
Current, I=q/t=3000C/(10×60s)=5A
Q7. The resistivity of alloys =Ralloy; the resistivity of constituent metals Rmetal. Then, usually?
Solution
Ralloy>Rmetal
Ralloy>Rmetal
Q8. If the temperature of cold junction of thermocouple is lowered, then the neutral temperature?
Solution
Neutral temperature is independent of temperature of cold junction.
Neutral temperature is independent of temperature of cold junction.
Q9. For obtaining chlorine by electrolysis a current of 100 kW and 125 V is used. (Electro chemical equivalent of chlorine is 0.367 ×kgC-1). The amount of chlorine obtained in one minute will be?
Solution
The amount of chlorine m=zlt=z(P/V) t [∵P=VI] =0.367×10-6×((100×1000)/125)×60 =0.017616 kg=17.616 g
The amount of chlorine m=zlt=z(P/V) t [∵P=VI] =0.367×10-6×((100×1000)/125)×60 =0.017616 kg=17.616 g
Q10. The current in a conductor varies with time t as I=2t+3t2 where I is in ampere and t in seconds. Electric charge flowing through a section of the conductor during t=2 sec to t=3 sec is?:
Solution
