$A$ resistance wire connected in the left gap of a meter bridge balances a $10\, \Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio $3: 2$. If the length of the resistance wire is $1.5\, m$,then the length of $1\, \Omega$ of the resistance wire is $....... \times 10^{-2}\, m$.

  • A
    $1.5$
  • B
    $1.0$
  • C
    $10$
  • D
    $15$

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Similar Questions

When the two known resistances $R$ and $S$ are connected in the left and right gaps of a meter bridge respectively,the null point is found at a distance $l_1$ from the zero end of the meter bridge wire. An unknown resistance $X$ is now connected in parallel with $S$,and the null point is found at a distance $l_2$ from the zero end of the meter bridge wire. The unknown resistance $X$ is:

Two resistors $2 \Omega$ and $3 \Omega$ are connected in the gaps of a meter bridge as shown in the figure. The null point is obtained with the contact of the jockey at some point on the wire $XY$. When an unknown resistor $R$ is connected in parallel with the $3 \Omega$ resistor, the null point is shifted by $22.5 \text{ cm}$ toward $Y$. The resistance of the unknown resistor $R$ is . . . . . . $\Omega$.

In the circuit shown, a meter bridge is in its balanced state. The meter bridge wire has a resistance of $0.1 \, \Omega/cm$. The value of the unknown resistance $X$ and the current drawn from the battery of negligible internal resistance are:

$A$ student obtained the following observations in an experiment of a meter bridge to find the unknown resistance of the circuit. The most accurate value of the unknown resistance is ............ $\Omega$.
$S.No.$$R$$l$$100-l$$S = \left( \frac{100-l}{l} \right)R$
$1$$20\,\Omega$$43$$57$$26.51\,\Omega$
$2$$30\,\Omega$$51$$49$$28.82\,\Omega$
$3$$40\,\Omega$$59$$41$$27.80\,\Omega$
$4$$60\,\Omega$$70$$30$$25.71\,\Omega$

$A$ meter bridge is set up as shown,to determine an unknown resistance '$X$' using a standard $10 \ \Omega$ resistor. The galvanometer shows a null point when the tapping key is at the $52 \ cm$ mark. The end corrections are $1 \ cm$ and $2 \ cm$ respectively for the ends $A$ and $B$. The determined value of '$X$' is (in $\Omega$)

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