9394949438

[LATEST]$type=sticky$show=home$rm=0$va=0$count=4$va=0






Current Electricity Quiz-14

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. 4 cells each of emf 2 V and internal resistance of 1Ω are connected in parallel to a load resistor of 2Ω. Then the current through the load resistor is?
  •  2 A
  •  1.5 A
  •  1 A
  •  0.888 A
Solution
I= E/(R+r/4)=2/(2+1/4)=2/2.25=0.888A

Q2. For ensuring dissipation of same energy in all three resistors (R1,R2,R3)connected as shown in figure, their values be related as?

  •  R1=R2=R3
  •  R2=R3 and R1=4 R2
  •  R2=R3 and R1 =R2/4
  •  R1= R2+R3
Solution
As voltage across the resistors R2 and R3 is same and they show same dissipation of energy, so using the relation for energy, H=V2/R t,we have R2=R3. Thus, the current in each resistor R2 and R3 will be I/2 ie,I1=I/2 and I2=I/2 

Since the energy dissipation is same in all the three resistors, so I2 R1 t=I12 R2 t or I2 R1 t=(I/2)2 R2 t or R1=R2/4


Q3. n identical cells, each of emf E and internal resistance r, are connected in series a cell A is joined with reverse polarity. The potential difference across each cell, except A is?
  •  2nE/(n-2)
  •  (n-2)E/n
  •  (n-1)E/n
  •  2E/n
Solution
When one call is wrongly connected in series, the emf of cells decrease by 2 E, but internal resistance of cells remains the same for all the cells. Current in the circuit is i=(n-2)E/nr×r Potential difference across each cell is V=E-Ir=E-(n-2)E/nr×r=2E/n

Q4. A current of A enters one corner one corner P of an equilateral triangle PQR having 3 wires of resistance 2 Ω each and leaves by the corner R. then the current I1 and I2 are?

  •  2 A, 4 A
  •  4 A, 2 A
  •  1 A, 2 A
  •  2 A, 3 A
Solution
From Kirchhoff’s first law at junction P, I1+I2=6 …(i) From Kirchhoff’s second law to the closed circuit PQRP, -2I1-2I1+2I2=0 -4I1+2I2=0 2I1-I2=0 ……(ii) Adding Eqs. (i) and (ii), we get 3I1=6 I1=2A From Eq. (i), I2=6-2=4A

Q5. 160W-60V lamp is connected at 60 V DC supply. The number of electrons passing through the lamp in 1 min is (the charge of electron e=1.6×10-19 C)?
  •  4.5 ohm1019
  •  1021
  •  1.6×1019
  •  1.4×1020
Solution
 Power, P=V2/R R=V2/P=(60)2/160=22.5Ω Now, according to Ohm’s law V=IR ∴ I=60/22.5 I=2.6A Here, t=60s As I=ne/t n=(I×t)/e = (26×60)/(1.6×10-19 )≈1021

Q6. If a 30 V,90 W bulb is to be worked on a 120 V line, a resistance of how many ohms should be connected in series with the bulb?
  •  10 ohm
  •  20 ohm
  •  30 ohm
  •  40 ohm
Solution
Suppose resistance R is corrected in series with bulb Current through the bulb i=90/30=3 A

Hence for resistance V=iR⇒90=3×R⇒R=30Ω


Q7. In a thermo-couple, one junction which is at 0℃ and the othe at t℃ the emf is given by E=at2-bt2. The neutral temperature is given by?
  •  a/b
  •  2 a/3b
  •  3a/2b
  •  b/2a
Solution
dT/dt=d/dt (at2-bt3 )=2at-3bt2 When t=tn (ie, neutral temperature), dE/dt=0 ∴0=2atn-3btn2 ortn=2a/3b.

Q8. 5 cells, each of emf 0.2 V and internal resistance 1 Ω are connected to an external circuit of resistance of 10 Ω. Find the current through external circuit?
  •  1/2.5 A
  •  1/10 A
  •  1/15 A
  •  1/2 A
Solution


Here, emf of each cell, ε=0.2V Internal resistance of each cell, r=1Ω External resistance, R=10Ω The total emf of 5 cells =5ε=5(0.2)V=1V Total internal resistance of 5 cells =5r=5(1)Ω=5Ω Total resistance of the circuit =R+5r=10+5=15Ω The current in the external circuit, I=5ε/(R+5r)=1V/15Ω=1/15 A


Q9. Two heater wires of equal length are first connected in series and then in parallel. The ratio of heat produced in the two cases is?
  •  1 : 4
  •  4 : 1
  •  1 : 2
  •  2 : 1
Solution
Let the resistance of each heater wire is R. When two wires are connected in series, the heat developed is H1=(V2 t)/2R …(i) When two heater wires are connected in parallel, the heat developed is H2=(V2 t)/(R/2)=(2V2 t)/R …(ii) Dividing Eq. (i) by Eq. (ii), we get H1/H2 =1/4 or H1 ∶H2=1∶4 .

Q10. A combination of two resistance of 2 W and 2/3 W connected in parallel is joined across a battery of emf of 3 V and of negligible internal resistance. The energy given out per sec will be?
  •  1/2×3×3 J
  •  1/2×1/3×3×3
  •  2×3 J
  •  3×3×2 J
Solution
Total power spend across two resistors connected in parallel to battery =V2/R1 +V2/R2 =(3×3)/2+(3×3)/(2/3)=36/2=18 =3×3×2 J

Want to know more

Want to Know More
Please fill in the details below:

INNER POST ADS

Latest IITJEE Articles$type=three$c=3$author=hide$comment=hide$rm=hide$date=hide$snippet=hide

Latest NEET Articles$type=three$c=3$author=hide$comment=hide$rm=hide$date=hide$snippet=hide

Name

Admissions,1,Alternating Current,60,AP EAMCET 2020,1,Basic Maths,2,BCECE 2020,1,best books for iit jee,2,best coaching institute for iit,1,best coaching institute for iit jee preparation,1,best iit jee coaching delhi,1,best iit jee coaching in delhi,2,best study material for iit jee,4,BITSAT Registration 2020,1,Blog,62,books for jee preparation,1,books recommended by iit toppers,3,Capacitance,3,CBSE,1,CBSE accounts exam,1,CBSE boards,1,CBSE NEET,9,cbse neet 2019,3,CBSE NEET 2020,1,cbse neet nic,1,Centre of Mass,2,Chemistry,58,Class 12 Physics,15,coaching for jee advanced,1,coaching institute for iit jee,2,Collision,2,COMEDK UGET 2020 Application Form,1,COMEDK UGET 2020 Exam Form,1,COMEDK UGET news,1,CUCET 2020,2,Current Electricity,4,CVR college,1,Electromagnetic Induction,3,Electronics,1,Electrostatics,3,Energy,1,Fluid Mechanics,4,Gravitation,2,GUJCET 2020 Application Form,1,Heat,4,iit admission,1,iit advanced,1,iit coaching centre,3,iit coaching centre in delhi,2,iit coaching classes,2,iit coaching in delhi,1,iit coaching institute in delhi,1,iit entrance exam,1,iit entrance exam syllabus,2,iit exam pattern,2,iit jee,5,iit jee 2019,3,iit jee advanced,2,iit jee books,3,iit jee coaching,2,iit jee exam,3,iit jee exam 2019,1,iit jee exam pattern,3,iit jee institute,1,iit jee main 2019,2,iit jee mains,3,iit jee mains syllabus,2,iit jee material,1,iit jee online test,3,iit jee practice test,3,iit jee preparation,6,iit jee preparation in delhi,2,iit jee preparation time,1,iit jee preparation tips by toppers,2,iit jee question paper,1,iit jee study material,3,iit jee study materials,2,iit jee syllabus,2,iit jee syllabus 2019,2,iit jee test,3,iit preparation,2,iit preparation books,5,iit preparation time table,2,iit preparation tips,2,iit syllabus,2,iit test series,3,IITJEE,100,IPU CET,1,JEE Advanced,83,jee advanced exam,2,jee advanced exam pattern,1,jee advanced paper,1,JEE Books,1,JEE Coaching Delhi,3,jee exam,3,jee exam 2019,6,JEE Exam Pattern,2,jee exam pattern 2019,1,jee exam preparation,1,JEE Main,85,jee main 2019,4,JEE Main 2020,1,JEE Main 2020 Application Form,2,JEE Main 2020 news,2,JEE Main 2020 Official Answer Key,1,JEE Main 2020 Registration,1,JEE Main 2020 Score,1,JEE Main application form,1,jee main coaching,1,JEE Main eligibility criteria,3,jee main exam,1,jee main exam 2019,3,jee main online question paper,1,jee main online test,3,JEE Main Paper-2 Result,1,jee main registration,2,jee main syllabus,2,JEE mains 2020,1,jee mains question bank,1,jee mains test papers,3,JEE Mock Test,2,jee notes,1,jee past papers,1,JEE Preparation,2,jee preparation in delhi,1,jee preparation material,4,JEE Study Material,1,jee syllabus,6,JEE Syllabus Chemistry,1,JEE Syllabus Maths,1,JEE Syllabus Physics,1,jee test series,3,KCET - 2020,1,Kinematics,1,Latest article,5,Latest Articles,61,Latest News,34,latest news about neet exam,1,Laws of Motion,2,Magnetic Effect of Current,3,Magnetism,3,MHT CET 2020,2,MHT CET 2020 exam schedule,1,Modern Physics,1,NCERT Solutions,15,neet,3,neet 2019,1,neet 2019 eligibility criteria,1,neet 2019 exam date,2,neet 2019 test series,2,NEET 2020,2,NEET 2020 Application Form,1,NEET 2020 Eligibility Criteria,1,NEET 2020 Registration,1,neet application form,1,neet application form 2019 last date,1,Neet Biology Syllabus,1,Neet Books,3,neet eligibility criteria,3,neet exam 2019,7,neet exam application,1,neet exam date,1,neet exam details,1,neet exam pattern,6,neet exam pattern 2019,2,neet examination,1,neet mock test 2019,1,Neet Notes,3,Neet Online Application Form,3,neet online test,2,neet past papers,1,neet physics syllabus,1,neet practice test,2,NEET preparation books,1,neet qualification marks,1,NEET question paper 2019,1,neet question papers,1,neet registration,1,Neet Study Material,3,neet syllabus,6,neet syllabus 2019,5,NEET Syllabus 2020,1,neet syllabus chemistry,1,neet syllabus for biology,1,neet syllabus for physics,1,neet test series,1,neet ug 2019,2,news,5,online study material for iit jee,1,Optical Instruments,1,Physics,110,physics books for iit jee,1,Power,1,Practical Physics,1,Quiz,5,Ray Optics,1,Rotational Motion,3,SHM,3,Simple Harmonic Motion,3,study materials for iit jee,1,Study Notes,110,study notes for iit jee,1,Thermodynamics,4,TS EAMCET Notification,2,Units and Dimensions,1,UPSEE 2020,1,UPSEE 2020 Application Form,2,UPSEE EXAM,1,Vectors,2,VITEE Application form,1,Wave Motion,3,Wave Optics,1,WBJEE 2020 Admit Card,1,WBJEE 2020 Answer Key,1,Work,1,
ltr
item
BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET, IIT JEE COACHING INSTITUTE: Current Electricity Quiz 14
Current Electricity Quiz 14
https://1.bp.blogspot.com/-wj6ENZ4Xxw8/X60UNOiSwqI/AAAAAAAADoI/rAG7r17sqY4CtrlvdPzFAV_6LS7KeM-1ACLcBGAsYHQ/w467-h245/Quiz%2BImage%2BTemplate%2B%252813%2529%2B%25282%2529.jpg
https://1.bp.blogspot.com/-wj6ENZ4Xxw8/X60UNOiSwqI/AAAAAAAADoI/rAG7r17sqY4CtrlvdPzFAV_6LS7KeM-1ACLcBGAsYHQ/s72-w467-c-h245/Quiz%2BImage%2BTemplate%2B%252813%2529%2B%25282%2529.jpg
BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET, IIT JEE COACHING INSTITUTE
https://www.cleariitmedical.com/2020/11/CurrentElectricityQuiz14.html
https://www.cleariitmedical.com/
https://www.cleariitmedical.com/
https://www.cleariitmedical.com/2020/11/CurrentElectricityQuiz14.html
true
7783647550433378923
UTF-8
Loaded All Posts Not found any posts VIEW ALL Readmore Reply Cancel reply Delete By Home PAGES POSTS View All RECOMMENDED FOR YOU LABEL ARCHIVE SEARCH ALL POSTS Not found any post match with your request Back Home Sunday Monday Tuesday Wednesday Thursday Friday Saturday Sun Mon Tue Wed Thu Fri Sat January February March April May June July August September October November December Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec just now 1 minute ago $$1$$ minutes ago 1 hour ago $$1$$ hours ago Yesterday $$1$$ days ago $$1$$ weeks ago more than 5 weeks ago Followers Follow THIS CONTENT IS PREMIUM Please share to unlock Copy All Code Select All Code All codes were copied to your clipboard Can not copy the codes / texts, please press [CTRL]+[C] (or CMD+C with Mac) to copy