Verified Solution • 2025–26 Session moderate

QJEEM18Ph288a: JEE Main 2018 Physics - Class 12 JEEM18Ph288a Solution: On interchanging the resistances, the balance point of a meter…

On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 k Ω\Omega . How much was the resistance on the left slot before interchanging the resistances?

Multiple Choice Options

  • A 910 Ω\Omega
  • B 990 Ω\Omega
  • C 505 Ω\Omega
  • D 550 Ω\Omega

Step-by-Step Verified Solution

JEE Main 2018 (Offline) Physics - Current Electricity Question 294 English Explanation 1

Xl{X \over l} = 1000−X100−l{{1000 - X} \over {100 - l}} . . . . . (1)

When X and Y are interchanged.

JEE Main 2018 (Offline) Physics - Current Electricity Question 294 English Explanation 2

1000−Xl−10{{1000 - X} \over {l - 10}} = X100−l+10{X \over {100 - l + 10}}

⇒\Rightarrow    \,\,\, 1000−Xl−10{{1000 - X} \over {l - 10}} = X110−l{X \over {110 - l}} . . . . . (2)

From (1) and (2) we get,

l100−l{l \over {100 - l}} = 110−ll−10{{110 - l} \over {l - 10}}

⇒\Rightarrow    \,\,\, ll 2 −- 10 = 11000 −- 100 ll −- 110 ll + ll 2

⇒\Rightarrow    \,\,\, 200 ll = 11000

ll = 55 cm

Putting this value of ll , in equation (1)

X55{X \over {55}} = 1000−X100−55{{1000 - X} \over {100 - 55}}

⇒\Rightarrow    \,\,\, 45X = 55000 −- 55X

⇒\Rightarrow    \,\,\, 100X = 55000

⇒\Rightarrow    \,\,\, X = 550 Ω\Omega

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