A standard cell of emf 1.08 V is balanced by the potential difference across 91 cm of a meter long wire supplied by a cell of emf 2 V through a series resistor of resistance `2Omega`. The internal resistance of the cell is zero. Fnd the reistance per unit length of the potentiometer wire.
1 Answers
Talk Doctor Online in Bissoy App
The standard oxidation potential potential of zinc is 0.76 volt and of silver is - 0.80 volt. Calculate the emf of the cell : `Zn|underset(0.25 M)(Zn(
The cell reaction is `Zn+2Ag^(+) rarr 2Ag+Zn^(2+)` `E_(o x)^(@)` of `Zn =0.76` volt `E_("red")^(@)` of `Ag=0.80` volt `E_(cell)^(@)=E_(o x)^(@)` of `Zn+E_("red")^(@)` of `Ag=0.76+0.80=1.56` volt We know that, `E_(cell)=E_(cell)^(@)-0.0591/n"log" (["Products"])/(["Reactants"])` `=E_(cell)^(@)-0.0591/2"log"...
2 Answers 5 views
If a pole 6 meter high casts a shadow 2√3 meter long on ground, then the angle of elevation of sun is –
Correct answer is (D) 60°
2 Answers 2 views
An accumulator with a steady emf of 2 V is connected across a potentiometer wire at 6.732 m If a resistance of 2.5 `Omega` is put in series with the w
Correct Answer - 6.748 m
2 Answers 1 views
Two, capacitors `A` and `B` are connected in series across a `100 V` supply and it is observed that the potential difference across them are `60 V` an
Correct Answer - `0.16muF, 0.24 muF`
2 Answers 1 views
Two capacitors of capacitances `20 .0pF` and `50.0pF ` are connected in series with a `6.00V bettery . Find (a )the potential difference across each c
Correct Answer - (i). 1.71 V, 4.29V (ii). `184pJ,73.5pJ`
2 Answers 1 views
With a cell of emf 1.5 V, the balance point is obtained at a distance of 60 cm from one end of the potential wire. Calculate the potential difference
Correct Answer - 2.5 V
2 Answers 1 views
A 10 metre long potentiometer wire carries a steady current. A 1.018 volt standard cell is balanced at a length of 850 cm. find (i) the potential grad
Correct Answer - (i). `1.2xx10^(-3)V" "cm^(-1),(ii).1.2V`
2 Answers 5 views
vii) A cell of emf \( 1.1 V \) and internal resistance \( r \) is connected across an external resistor of resistance \( R=10 \). The potential difference across the resistor is a) \( 0.1 V \) b) \( 0.9 V \) c) \( 1.0 V \) d) \( 1.1 V \)
Given E = 1.1 V r = r R = 10 r Potential difference across the resistor = IR = (E/R+r) R = (1.1/10r+r) 10 r \(=\frac{1.1\times10\,r}{11\,r\times10}\) = 11/11 = 1 volt option (c) is correct.
2 Answers 2 views
When a current of 0.5 A is passed through two resistances in series, the potential difference between the ends of the series arrangement is 12.5 V. On
Correct Answer - `20 Omega, 5 Omega`
2 Answers 1 views
The potential difference of a cell in an open circuit is 6 V, which falls to 4 V when a current of 2 A is drawn from the cell. Calculate the emf and t
Correct Answer - `6 V, 1 Omega`
2 Answers 1 views