A letter $A$ is constructed of a uniform wire with resistance $1.0\,\Omega \,per\,cm$. The sides of the letter are $20\, cm$ and the cross piece in the middle is $10\, cm$ long. The apex angle is $60^o$. The resistance between the ends is .............. $\Omega$
Diffcult
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
- 1A null point is found at $200\,cm$ in potentiometer when cell in secondary circuit is shunted by $5\,\Omega$. When a resistance of $15\,\Omega$ is used for shunting null point moves to $300\,cm$. The internal resistance of the cell is $..............\,\Omega$.View Solution
- 2Equivalent resistance between $A$ and $B$ will be ............ $ohm$View Solution
- 3The resistance of a wire is $R$. It is bent at the middle by $180^o$ and both the ends are twisted together to make a shorter wire. The resistance of the new wire isView Solution
- 4In given figure, the potentiometer wire $AB$ has a resistance of $5$ $\Omega$ and length $10\, m$. The balancing length $AM$ for the $emf$ of $0.4\, V$ is ............... $m$View Solution
- 5Masses of three wires of copper are in the ratio of $1 : 3 : 5$ and their lengths are in the ratio of $5:3:1.$ The ratio of their electrical resistances areView Solution
- 6View SolutionA potentiometer is an ideal device of measuring potential difference because
- 7${\sigma _1}$ and ${\sigma _2}$ are the electrical conductivities of $Ge $ and $Na$ respectively. If these substances are heated, thenView Solution
- 8In the adjoining circuit diagram each resistance is of $10$ $\Omega$. The current in the arm $AD$ will beView Solution
- 9A current of $3\,amp$ flows through the $2\,\Omega $ resistor shown in the circuit. The power dissipated in the $5\,\Omega $ resistor is ................. $watt$View Solution
- 10In the given circuit of potentiometer, the potential difference $E$ across $AB$ ( $10\, m$ length) is larger than $E _{1}$ and $E _{2}$ as well. For key $K _{1}$ (closed), the jockey is adjusted to touch the wire at point $J_{1}$ so that there is no deflection in the galvanometer. Now the first battery $\left( E _{1}\right)$ is replaced by second battery $\left( E _{2}\right)$ for working by making $K _{1}$ open and $K _{2}$ closed. The galvanometer gives then null deflection at $J _{2}$. The value of $\frac{ E _{1}}{ E _{2}}$ is $\frac{ a }{ b },$ where $a =$ ...............View Solution
