In a meter bridge,the null point is found at | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In a meter bridge,the balance condition is given by $\frac{R}{S} = \frac{\ell}{100-\ell}$,where $\ell$ is the distance of the null point from end

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) In a meter bridge,the balance condition is given by $\frac{R}{S} = \frac{\ell}{100-\ell}$,where $\ell$ is the distance of the null point from end $A$.Case $1$: Null point is at $\ell = 25\, cm$.$\frac{R}{S} = \frac{25}{100-25} = \frac{25}{75} = \frac{1}{3}$$S = 3R$  .......$(1)$Case $2$: $A$ resistance of $10\,\Omega$ is connected in parallel with $S$. The new resistance $S'$ is given by $\frac{1}{S'} = \frac{1}{S} + \frac{1}{10}$,so $S' = \frac{10S}{S+10}$.The null point is now at the midpoint,i.e.,$\ell = 50\, cm$.$\frac{R}{S'} = \frac{50}{100-50} = 1$$R = S' = \frac{10S}{S+10}$ .......$(2)$Substitute $S = 3R$ from $(1)$ into $(2)$:$R = \frac{10(3R)}{3R+10}$$R(3R+10) = 30R$Since $R 
eq 0$,divide by $R$:$3R + 10 = 30$$3R = 20$$R = \frac{20}{3} \approx 6.67\,\Omega$.

In a meter bridge,the null point is found at a distance of $25\, cm$ from $A$. If now a resistance of $10\,\Omega$ is connected in parallel with $S$,the null point occurs at the midpoint of $AB$. The value of $R$ is .............. $\Omega$. (in $.67$)

Similar Questions

The number of lone pairs of electrons present on $Xe$ in $XeF_4$ is

Which of the following is known as the pitcher plant?

The $pH$ of a buffer solution made by mixing $25 \ mL$ of $0.02 \ M \ NH_4OH$ and $25 \ mL$ of $0.2 \ M \ NH_4Cl$ at $25^{\circ}C$ is ($pK_b$ of $NH_4OH = 4.8$).

If $\Delta$ is the area of the triangle formed by the positive $x$-axis and the normal and tangent to the circle $x^2+y^2=4$ at $(1, \sqrt{3})$,then $\Delta$ is equal to

The correct output for the given logic circuit and inputs is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module