Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A null point is found at in potentiometer when cell in secondary circuit is shunted b

A null point is found at 200 \mathrm{~cm} in potentiometer when cell in secondary circuit is shunted by 5 \Omega. When a resistance of 15 \Omega is used for shunting, null point moves to 300 \mathrm{~cm}. The internal resistance of the cell is __________ \Omega.

Option: 1

5


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Potential Gradient  =\frac{\Delta \mathrm{V}}{\mathrm{L}}

\begin{aligned} & E-I r=\left(\frac{\Delta v}{L}\right) x \\ & \frac{E R}{R+r}=\left(\frac{\Delta V}{L}\right) x \\ & \frac{E \times 5}{5+r}=\frac{\Delta V}{L} \times 200 \, \, \, \, \, \, ........(1)\\ & \frac{E \times 15}{15+r}=\frac{\Delta V}{L} \times 300 \, \, \, \, \, \, ..........(2)\\ & =r=5 \Omega \end{aligned}

Posted by

Anam Khan

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: BTech ECE cut-offs for female candidates go up at top NITs
anam-khan

JEE Main 2025 Session 2 City Slip (OUT) Live: Admit card likely on March 29 at jeemain.nta.nic.in; cut-offs
anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads

Latest Question