In an experiment to determine the temperature coefficient of resistance of a conductor,a coil of wire $X$ is immersed in a liquid. It is heated by an external agent. $A$ meter bridge setup is used to determine the resistance of the coil $X$ at different temperatures. The balancing points measured at temperatures $t_1=0^{\circ} C$ and $t_2=100^{\circ} C$ are $50 \ cm$ and $60 \ cm$ respectively. If the standard resistance taken out is $S=4 \ \Omega$ in both trials,the temperature coefficient of the coil is
- A$0.05^{\circ} C^{-1}$
- B$0.02^{\circ} C^{-1}$
- C$0.005^{\circ} C^{-1}$
- D$2.0^{\circ} C^{-1}$
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Similar Questions
In the given Wheatstone bridge (meter bridge) circuit,the galvanometer shows zero deflection. The value of the unknown resistance $R$ is ................. $\Omega$.
Difficult
View SolutionThe resistances in the two arms of a meter bridge are $5\,\Omega$ and $R\,\Omega$,respectively. When the resistance $R$ is shunted with an equal resistance,the new balance point is at $1.6\,l_1$. The resistance $R$ is ................. $\Omega$.
Difficult
View Solution$STATEMENT-1$: In a Meter Bridge experiment,a null point for an unknown resistance is measured. Now,the unknown resistance is placed inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance.
$STATEMENT-2$: The resistance of a metal increases with an increase in temperature.
$STATEMENT-2$: The resistance of a metal increases with an increase in temperature.
In the meter bridge experiment,for given values of $R_1$ and $R_2$,the balance point is $20 \ cm$ from point $A$. When $R_2$ is shunted by a $20 \ \Omega$ resistance,the balancing point shifts to $50 \ cm$ from point $A$. Find the value of $R_2$ in $\Omega$.
Medium
View Solution$A$ 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 ............ $\Omega$.
$S.No.$ $R$ $l$ $100-l$ $S = \left( \frac{100-l}{l} \right)R$ $1$ $20\,\Omega$ $43$ $57$ $26.51\,\Omega$ $2$ $30\,\Omega$ $51$ $49$ $28.82\,\Omega$ $3$ $40\,\Omega$ $59$ $41$ $27.80\,\Omega$ $4$ $60\,\Omega$ $70$ $30$ $25.71\,\Omega$
| $S.No.$ | $R$ | $l$ | $100-l$ | $S = \left( \frac{100-l}{l} \right)R$ |
| $1$ | $20\,\Omega$ | $43$ | $57$ | $26.51\,\Omega$ |
| $2$ | $30\,\Omega$ | $51$ | $49$ | $28.82\,\Omega$ |
| $3$ | $40\,\Omega$ | $59$ | $41$ | $27.80\,\Omega$ |
| $4$ | $60\,\Omega$ | $70$ | $30$ | $25.71\,\Omega$ |
Medium
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