Question

In a meter Bridge experiment resistance x is connected in the right gap. When R₁ and R₂ are connected in the left gap separately the balance points are 40 cm and 50 cm respectively from the left end. Now if both R₁& R₂ are connected in series in the left gap with x in the right gap then the new balance point from the left end will be

• A. 50 cm
• B. 62.5 cm
• C. 75 cm
• D. Data insufficient for calculation

Answer 1

Answer: (B)

62.5 cm

Answer 2

= $\frac { 40 } { 60 }$ = $\frac { 2 } { 3 }$ .... (1), = $\frac { 50 } { 50 }$ = 1 ..... (2)

= $\frac { \ell } { 100 - \ell }$ .... (3)

Add eq (1) & (2)

=$\frac { 2 } { 3 }$ + 1 = $\frac { 5 } { 3 }$... (4)

Placing eq (4) in (3)

5/ 3 =$\frac { \ell } { 100 - \ell }$ ⇒ 500–5l = 3l

⇒ 8l=500 ∴ l = 62.5 cms