Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Current Electricity question

2011 · Shift 1 · Q59
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Physics
  4. /Current Electricity
  5. /2011 · Shift 1 · Q59

Current Electricity question

2011 · Shift 1 · Q59

JEE AdvancedPhysicsCurrent ElectricityMCQ+3 / −1
A meter bridge is set up as shown, to determine an unknown resistance X using a standard 10 Ω\OmegaΩ resistor. The galvanometer shows null point when tapping-key is at 52 cm mark. The end-corrections are 1 cm and 2 cm, respectively, for the ends A and B. The determined value of X is IIT-JEE 2011 Paper 1 Offline Physics - Current Electricity Question 14 English
  1. A
    10.2 Ω\OmegaΩ
  2. B
    10.6 Ω\OmegaΩ
  3. C
    10.8 Ω\OmegaΩ
  4. D
    11.1 Ω\OmegaΩ
View written solutionFree

Correct answer: B

  1. Principle of meter bridge

At balance point, the ratio of resistances in the two gaps equals the ratio of resistances of the wire segments:

X10=RAJRJB\frac{X}{10} = \frac{R_{AJ}}{R_{JB}}10X​=RJB​RAJ​​

where JJJ is the null point.

  1. Include end corrections

Given:

  • Null point reading = 52 cm52\,\text{cm}52cm from end AAA
  • End correction at A=1 cmA = 1\,\text{cm}A=1cm
  • End correction at B=2 cmB = 2\,\text{cm}B=2cm

So the effective lengths are:

L1=52+1=53 cmL_1 = 52 + 1 = 53\,\text{cm}L1​=52+1=53cm L2=(100−52)+2=48+2=50 cmL_2 = (100 - 52) + 2 = 48 + 2 = 50\,\text{cm}L2​=(100−52)+2=48+2=50cm
  1. Use balance condition

From the figure/setup, the unknown resistance XXX is in the left gap and the standard 10 Ω10\,\Omega10Ω is in the right gap, so

X10=5350\frac{X}{10} = \frac{53}{50}10X​=5053​

Hence,

X=10×5350=10.6 ΩX = 10 \times \frac{53}{50} = 10.6\,\OmegaX=10×5053​=10.6Ω
  1. Match with options
X=10.6 ΩX = 10.6\,\OmegaX=10.6Ω

So the correct option is B.

PreviousNext

More from Current Electricity

  • Two batteries of different emfs and different internal resistance are connected as shown. The voltage across AB in volts is ​. Includes diagram2011 · Numerical
  • Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100, 60 and 40 W bulbs have filament resistances R100, R60 and R40 respectively,…2010 · MCQ
  • To verify Ohm's law, a student is provided with a test resitor RT, a high resistance R1, a small resistance R2, two identical galvanometers G1 and G2, and a variable voltage source V. The correct circuit to carry out the experiment is2010 · MCQ
  • Consider a thin square sheet of side L and thickness, made of a material of resistivity ρ. The resistance between two opposite faces, shown by the shaded areas in the figure is Includes diagram2010 · MCQ
  • When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25…2010 · Numerical
  • For the circuit shown in the figure Includes diagram2009 · Multiple correct
  • Figure shows three resistor configurations R1, R2 and R3 connected to 3 V battery. If the power dissipated by the configuration R1, R2 and R3 is P1, P2 and P3, respectively, then Includes diagram2008 · MCQ
  • STATEMENT - 1 In a Meter Bridge experiment, null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as…2008 · MCQ