The resistance per centimeter of a meter bridge wire is $\mathrm{r}$, with $\mathrm{X}\ \Omega$ resistance in left gap. Balancing length from left end is at $40 \mathrm{~cm}$ with $25\ \Omega$ resistance in right gap. Now the wire is replaced by another wire of $2 \mathrm{r}$ resistance per centimeter. The new balancing length for same settings will be at
JEE MAIN 2024, Diffcult
Download our app for free and get started
$\frac{25}{\mathrm{r} \ell_1}=\frac{\mathrm{x}}{\mathrm{r} \ell_2}$ $.....(i)$
$\frac{25}{2 \mathrm{r} \ell_1^{\prime}}=\frac{\mathrm{X}}{2 \mathrm{r} \ell_2^{\prime}}$ $...(ii)$
From $(i)$ and $(ii)$
$\ell_2^{\prime}=\ell_2=40 \mathrm{~cm}$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are $8 \Omega$ and $10 \Omega$ respectively. After inserting in a hot bath of temperature $400^{\circ} \mathrm{C}$, the resistance of platinum wire is:View Solution
- 2View SolutionElectroplating does not help in
- 3Find the equivalent resistance across the terminals of source of $e.m.f$. $24\, V$ for the circuit shown in figure .............. $\Omega$View Solution
- 4A $3\,^oC$ rise in temperature is observed in a conductor by passing a certain current. When the current is doubled, the rise in temperature will be ............. $^oC$View Solution
- 5If resistance of voltmeter is $10000 \,\Omega$ and resistance of ammeter is $2 \,\Omega$ then find $R$ when voltmeter reads $12\,V$ and ammeter reads $0.1\, A$ ............... $\Omega$View Solution
- 6View SolutionWhich of the following is not equal to watt
- 7An electric lamp is marked $60\, W$, $ 230\, V$. The cost of $1\, kilowatt$ hour of power is Rs. $1.25$. The cost of using this lamp for $8$ hours is Rs. ................View Solution
- 8A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of $2.0\,V$ and a negligible internal resistance. The potentiometer wire itself is $4\,m$ long. When the resistance $R,$ connected across the given cell, has values of $(i)$ infinity $(ii)$ $9.5\,\Omega$ the balancing lengths on the potentiometer wire are found to be $3\,m$ and $2.85\,m,$ respectively. The value of internal resistance of the cell is ............... $\Omega$View Solution
- 9A ring is made of a wire having a resistance $R_0 = 12 \,\,\Omega$. Find the points $A$ and $B,$ as shown in the figure, at which a current carrying conductor should be connected so that the resistance $R$ of the sub circuit between these points is equal to $\frac{8}{3}\,\Omega$.View Solution
- 10Under what conditions current passing through a resistance $R$ can be increased by short circuiting the battery of emf $E_2$. The internal resistances of the two batteries are $r_1$ and $r_2$ respectivelyView Solution
