If the balance point is obtained at the $35^{th} cm$ in a meter bridge,the resistances in the left and right gaps are in the ratio of:

Explain the construction of a meter bridge used in the laboratory.

When two resistances $R_1$ and $R_2$ are connected in series and introduced into the left gap of a meter bridge and a resistance of $10 \ \Omega$ is introduced into the right gap,a null point is found at $60 \ cm$ from the left side. When $R_1$ and $R_2$ are connected in parallel and introduced into the left gap,a resistance of $3 \ \Omega$ is introduced into the right gap to get a null point at $40 \ cm$ from the left end. The product $R_1 R_2$ is $............. \ \Omega$.

In a meter bridge experiment to determine the value of an unknown resistance,first the resistances $2 \Omega$ and $3 \Omega$ are connected in the left and right gaps of the bridge and the null point is obtained at a distance $l \ cm$ from the left. Now when an unknown resistance $x \ \Omega$ is connected in parallel to $3 \ \Omega$ resistance,the null point is shifted by $10 \ cm$ to the right of the wire. The value of unknown resistance $x$ is . . . . . . $\Omega$.

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 experiment,a null point is obtained at $20 \ cm$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$,then where will be the new position of the null point from the same end,if one decides to balance a resistance of $4X$ against $Y$ (in $cm$)?