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

Units and Measurements question

2023 · 6 Apr · Shift 1 · Q56
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 Main
  3. /Physics
  4. /Units and Measurements
  5. /2023 · 6 Apr · Shift 1 · Q56

Units and Measurements question

2023 · 6 Apr · Shift 1 · Q56

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
Two resistances are given as R1=(10±0.5)Ω\mathrm{R}_{1}=(10 \pm 0.5) \OmegaR1​=(10±0.5)Ω and R2=(15±0.5)Ω\mathrm{R}_{2}=(15 \pm 0.5) \OmegaR2​=(15±0.5)Ω. The percentage error in the measurement of equivalent resistance when they are connected in parallel is -
  1. A
    2.33
  2. B
    5.33
  3. C
    4.33
  4. D
    6.33
View written solutionFree

Correct answer: C

  1. Equivalent resistance in parallel

For two resistors in parallel,

R=R1R2R1+R2R = \frac{R_1R_2}{R_1+R_2}R=R1​+R2​R1​R2​​

Given:

R1=(10±0.5) Ω,R2=(15±0.5) ΩR_1 = (10 \pm 0.5)\,\Omega, \qquad R_2 = (15 \pm 0.5)\,\OmegaR1​=(10±0.5)Ω,R2​=(15±0.5)Ω
  1. First compute the equivalent resistance
R=10×1510+15=15025=6 ΩR = \frac{10\times 15}{10+15} = \frac{150}{25} = 6\,\OmegaR=10+1510×15​=25150​=6Ω
  1. Find fractional error using error propagation

Write

R=R1R2R1+R2R = \frac{R_1R_2}{R_1+R_2}R=R1​+R2​R1​R2​​

Taking logarithm is not convenient directly because of the sum term, so use differential form:

ΔRR=Δ(R1R2)R1R2+Δ(1R1+R2)1R1+R2\frac{\Delta R}{R} = \frac{\Delta(R_1R_2)}{R_1R_2} + \frac{\Delta\left(\frac{1}{R_1+R_2}\right)}{\frac{1}{R_1+R_2}}RΔR​=R1​R2​Δ(R1​R2​)​+R1​+R2​1​Δ(R1​+R2​1​)​

More simply,

R=R1R2R1+R2R = \frac{R_1R_2}{R_1+R_2}R=R1​+R2​R1​R2​​

so maximum fractional error is

ΔRR=ΔR1R1+ΔR2R2+Δ(R1+R2)R1+R2\frac{\Delta R}{R} = \frac{\Delta R_1}{R_1} + \frac{\Delta R_2}{R_2} + \frac{\Delta(R_1+R_2)}{R_1+R_2}RΔR​=R1​ΔR1​​+R2​ΔR2​​+R1​+R2​Δ(R1​+R2​)​

Since for a sum, absolute errors add:

Δ(R1+R2)=ΔR1+ΔR2=0.5+0.5=1\Delta(R_1+R_2)=\Delta R_1+\Delta R_2 = 0.5+0.5=1Δ(R1​+R2​)=ΔR1​+ΔR2​=0.5+0.5=1

Therefore,

ΔRR=0.510+0.515+125\frac{\Delta R}{R} = \frac{0.5}{10} + \frac{0.5}{15} + \frac{1}{25}RΔR​=100.5​+150.5​+251​

Now calculate:

0.510=0.05\frac{0.5}{10} = 0.05100.5​=0.05 0.515≈0.0333\frac{0.5}{15} \approx 0.0333150.5​≈0.0333 125=0.04\frac{1}{25} = 0.04251​=0.04

So,

ΔRR=0.05+0.0333+0.04=0.1233\frac{\Delta R}{R} = 0.05 + 0.0333 + 0.04 = 0.1233RΔR​=0.05+0.0333+0.04=0.1233

Hence percentage error:

0.1233×100=12.33%0.1233 \times 100 = 12.33\%0.1233×100=12.33%

This does not match any option, which suggests the above direct rule has been overapplied to the denominator term.

  1. Use the standard shortcut for parallel combination

For two resistors in parallel,

R=R1R2R1+R2R = \frac{R_1R_2}{R_1+R_2}R=R1​+R2​R1​R2​​

Using differential method correctly:

dR=R22 dR1+R12 dR2(R1+R2)2dR = \frac{R_2^2\,dR_1 + R_1^2\,dR_2}{(R_1+R_2)^2}dR=(R1​+R2​)2R22​dR1​+R12​dR2​​

Thus maximum error,

ΔR=R22 ΔR1+R12 ΔR2(R1+R2)2\Delta R = \frac{R_2^2\,\Delta R_1 + R_1^2\,\Delta R_2}{(R_1+R_2)^2}ΔR=(R1​+R2​)2R22​ΔR1​+R12​ΔR2​​

Substitute values:

ΔR=152(0.5)+102(0.5)(25)2\Delta R = \frac{15^2(0.5) + 10^2(0.5)}{(25)^2}ΔR=(25)2152(0.5)+102(0.5)​ =225(0.5)+100(0.5)625= \frac{225(0.5) + 100(0.5)}{625}=625225(0.5)+100(0.5)​ =112.5+50625=162.5625=0.26 Ω= \frac{112.5 + 50}{625} = \frac{162.5}{625} = 0.26\,\Omega=625112.5+50​=625162.5​=0.26Ω

Now percentage error in equivalent resistance:

ΔRR×100=0.266×100\frac{\Delta R}{R}\times 100 = \frac{0.26}{6}\times 100RΔR​×100=60.26​×100 =4.33%= 4.33\%=4.33%
  1. Option check
  • A: 2.332.332.33 ❌
  • B: 5.335.335.33 ❌
  • C: 4.334.334.33 ✅
  • D: 6.336.336.33 ❌

Therefore, the correct answer is Option C.

PreviousNext

More from Units and Measurements

  • Dimension of μ0​∈0​1​ should be equal to2023 · MCQ
  • A cylindrical wire of mass (0.4±0.01)g has length (8±0.04)cm and radius (6±0.03)mm. The maximum error in its density will be:2023 · MCQ
  • Match List I with List II Choose the correct answer from the options given below: Includes table2023 · MCQ
  • A physical quantity P is given as P=cd​a2b3​ The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4% respectively. The percentage error in the measurement of quantity P will be2023 · MCQ
  • In an experiment with vernier callipers of least count 0.1 mm, when two jaws are joined together the zero of vernier scale lies right to the zero of the main scale and 6th division of vernier scale coincides with the main scale…2023 · MCQ
  • Given below are two statements : Statements I : Astronomical unit (Au), Parsec (Pc) and Light year (ly) are units for measuring astronomical distances. Statements II : $$\mathrm{Au} In the light of the above statements, choose the most…2023 · MCQ
  • If force (F), velocity (V) and time (T) are considered as fundamental physical quantity, then dimensional formula of density will be :2023 · MCQ
  • Match List I with List II Choose the correct answer from the options given below: Includes table2023 · MCQ