If a wire is stretched to make it $0.1 \%$ longer, its resistance will
Aincrease by $0.05 \% $
Bincrease by $0.2 \%$
Cdecrease by $0.2 \%$
Ddecrease by $0.05 \% $
AIEEE 2011, Easy
Download our app for free and get started
Bincrease by $0.2 \%$
b Resistance of wire
$R=\frac{\rho l}{A}=\frac{\rho l^{2}}{C}(\text { where } A l=C)$
$\therefore $ Fractional change in resistance
$\frac{\Delta R}{R}=2 \frac{\Delta l}{l}$
$\therefore $ Resistance will increase by $0.2 \%$
Download our app
and 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.*
Resistances of $6\, ohm$ each are connected in the manner shown in adjoining figure. With the current $0.5\,ampere$ as shown in figure, the potential difference ${V_P} - {V_Q}$ is .............. $V$
A generator has armature resistance of $0 .1\,\Omega $ and develops an induced emf of $120 \,V$ when driven at its rated speed. Its terminal voltage when a current of $50\,A$ is being drawn is ................. $V$
We have a galvanometer of resistance $25\,\Omega $. It is shunted by a $2.5\,\Omega $ wire. The part of total current that flows through the galvanometer is given as
$ABCD$ is a square where each side is a uniform wire of resistance $1\,\Omega$ . $A$ point $E$ lies on $CD$ such that if a uniform wire of resistance $1\,\Omega$ is connected across $AE$ and constant potential difference is applied across $A$ and $C$ then $B$ and $E$ are equipotential.
In an electric circuit, a cell of certain emf provides a potential difference of $1.25\, {V}$ across a load resistance of $5\, \Omega .$ However, it provides a potential difference of $1\, {V}$ across a load resistance of $2\, \Omega$. The $emf$ of the cell is given by $\frac{x}{10} v$. Then the value of $x$ is ..... .