Two resistances are connected in two gaps of a meter bridge. The balance point is $20 \ cm$ from the zero end. $A$ resistance of $15 \ \Omega$ is connected in series with the smaller of the two. The null point shifts to $40 \ cm$. The value of the smaller resistance in $\Omega$ is

  • A
    $3$
  • B
    $6$
  • C
    $9$
  • D
    $12$

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Similar Questions

Assertion $(A):$ In a meter bridge experiment,the unknown resistance is placed inside an enclosure maintained at a higher temperature. The null point can be obtained at the same position as before by decreasing the value of the standard resistance.
Reason $(R):$ The resistance of a metal increases with an increase in temperature.

An unknown resistance $R_1$ is connected in series with a resistance of $10 \,\Omega$. This combination is connected to one gap of a meter bridge while a resistance $R_2$ is connected in the other gap. The balance point is at $50 \, cm$. Now,when the $10 \,\Omega$ resistance is removed,the balance point shifts to $40 \, cm$. The value of $R_1$ is (in $\Omega$):

In a meter bridge,as shown in the figure,it is given that resistance $Y = 12.5 \, \Omega$ and the balance point is obtained at a distance $l_1 = 39.5 \, cm$ from end $A$ (by jockey $J$). After interchanging the resistances $X$ and $Y$,a new balance point is found at a distance $l_2$ from end $A$. What are the values of $X$ and $l_2$?

In a meter bridge,the balancing length from the left end is found to be $25 \ cm$. The value of the unknown resistance is (assume,standard resistance of $1 \ \Omega$ is in the right gap). (in $\Omega$)

Two wires $A$ and $B$ of equal lengths are connected in left and right gap respectively of a metre bridge, null point is obtained at $40 \text{ cm}$ from left end. Diameters of the wires $A$ and $B$ are in the ratio $3:1$ respectively, the ratio of specific resistance of $A$ to that of $B$ is (in $: 1$)

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