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6 months ago

Overview Class 12th

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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Payal Gupta

Contributor-Level 10

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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6 months ago

Overview Class 12th

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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Payal Gupta

Contributor-Level 10

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

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

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6 months ago

Overview Class 12th

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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Vishal Baghel

Contributor-Level 10

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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New answer posted

6 months ago

Overview Class 12th

The current I flowing through the given circuit will be_________A.

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Vishal Baghel

Contributor-Level 10

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

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6 months ago

Overview Class 12th

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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Vishal Baghel

Contributor-Level 10

$R_{A B} = 20 Ω$

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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New answer posted

6 months ago

Overview Class 12th

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

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Payal Gupta

Contributor-Level 10

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New answer posted

6 months ago

Overview Class 12th

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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Payal Gupta

Contributor-Level 10

$H = I^{2} R t = 2^{2} * R * 15$

$\Rightarrow 300 = 60 R$

$R = 5 Ω$

Now for 3A, time = 10 sec

$H' = l^{2} R t = 3^{2} * 5 * 10 = 450 J$

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New answer posted

6 months ago

Overview Class 12th

A potentiometer wire of length 10m and resistance 20 $Ω$ is connected in series with a 25 V battery and an external resistance $30 Ω$ . A cell of emf E in secondary circuit is balanced by 250cm long potentiometer wire. The value of E (in volt) is $\frac{x}{10} .$ The value of x is ______.

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Payal Gupta

Contributor-Level 10

AB = 10 m

$R_{A B} = 20 Ω$

$L_{A C} = 250 c m$

= 2.5 m

$l = \frac{25}{20 + 30}$

$E = \frac{x}{10} = \frac{25}{10}$

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New answer posted

7 months ago

Overview Class 12th

What will be the most suitable combination of three resistors A = $2 Ω$ , B = $4 Ω$ , C = $6 Ω$ so that $(\frac{22}{3}) Ω$ is equivalent resistance of combination

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Payal Gupta

Contributor-Level 10

Given

$\begin{array}{l} A = 2 Ω \\ B = 4 Ω \\ C = 6 Ω \\ R_{e q} = \frac{2 * 4}{2 + 4} + 6 \end{array}$

$R_{e q} = \frac{22}{3} Ω$

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New answer posted

7 months ago

Overview Class 12th

Two resistors are connected in series across a battery as shown in figure. If a voltmeter of resistance 2000 $Ω$ is used to measure the potential difference across 500 $Ω$ resister, the reading of the voltmeter will be_________ V.