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A wire of resistance R₁ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:

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$R_{2} = ρ \frac{2 l}{A / 2} = \frac{4 ρ l}{A} = 4 R_{1}$
$\frac{R_{2}}{R_{1}} = \frac{4}{1} \to 4 : 1$

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2 months ago

As shown in the figure, in steady state, the charge stored in the capacitor is ____________* 10^-⁶C.

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$q = C V_{100 Ω}$

$= (1.1 * 1 0^{- 6}) (\frac{10}{R + r} R)$

$= 1.1 * 1 0^{- 6} (\frac{10}{110} * 100)$

$= 10 μ C$

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2 months ago

In the given figure of meter bridge experiment, the balancing length AC corresponding to null deflection of the galvanometer is 40 cm. The balancing length, if the radius of the wire AB is doubled, will be__________cm.

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Independent of are a is case of uniform wire

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2 months ago

A 1 m long copper wire carries a current of 1 A. If the cross section of the wire is 2.0mm² and the resistivity of copper is 1.7 * 10^-8 $Ω m,$ the force experienced by moving electron in the wire is _______* 10^-23N.

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i = 1 A $j = σ E$

A = 2mm²

$ρ = 1.7 * 1 0^{- 8} Ω m$ $F = e E = \frac{e j}{σ} = \frac{e i}{A σ} = \frac{e i ρ}{A}$

= $\frac{1.6 * 1 0^{- 19} * 1.7 * 1 0^{- 8}}{2 * 1 0^{- 6}}$

= 136 * 10^-23 N

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2 months ago

In a potentiometer arrangement, a cell of emf 1.20 V gives a balance point at 36cm length of wire. This cell is now replaced by another cell of emf 1.80 V. The difference in balancing length of potentiometer wire in above conditions will be ________cm.

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$V = K l \Rightarrow \frac{V_{2}}{V_{1}} = \frac{l_{2}}{l_{1}} \Rightarrow \frac{l_{2}}{36} = \frac{1.80}{1.20} = \frac{3}{2} \Rightarrow l_{2} = 54 c m$

$Δ l = l_{2} - l_{1} = 54 - 36 = 18 c m$

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2 months ago

Eight copper wire of length $l$ and diameter d are joined in parallel to from a single composite conductor of resistance R. If a single copper wire of length 2 $l$ have the same resistance (R) then its diameter will be ________d.

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Each wile has resistance = $\frac{ρ 4 l}{π d^{2}} = r$ $\frac{}{^{}}$

Eight wire in parallel, then equivalent resistance is

$\frac{r}{8} = \frac{ρ l}{2 π d^{2}}$

Single copper wire of length $2 l$ has resistance

$R = ρ \frac{2 l * 4}{π d_{1}^{2}} = \frac{ρ l}{2 π d^{2}}$

d1 = 4d

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2 months ago

The current I flowing through the given circuit will be_________A.

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$i = \frac{v}{R_{N e t}} = \frac{6}{3} = 2 A$

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2 months ago

As shown in the figure, a potentiometer wire of resistance $20 Ω$ and length 300cm is connected with resistance box (R.B) and a standard cell of emf 4 V. For a resistance 'R' of resistance box introduced into the circuit, the null point for a cell of 20 mV is found to be 60 cm. The value of 'R' is __________ $Ω .$

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$R_{A B} = 20 Ω$

L = 300 cm

Null point is at 60 cm from A, so

$20 m V = \frac{4}{R + 20} * 20 * (\frac{60}{300}) * 1 0^{3}$

$\Rightarrow 20 = \frac{16}{R + 20} * 1 0^{3}$

$R = \frac{15600}{20} = 780 Ω$

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2 months ago

The total current supplied to the circuit is shown in figure by the 5 V battery is…………….A.

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A resistor develops 300 J of thermal energy in 15s, when a current of 2 A is passed through it. If the current increases to 3 A, the energy developed in 10s is…………….J.

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