Advertisements
Advertisements
प्रश्न
In a potentiometer experiment, balancing length is found to be 120 cm for a cell E1 of emf 2V. What will be the balancing length for another cell E2 of emf 1.5V? (No other changes are made in the experiment.)
Advertisements
उत्तर
``E_1/E_2 = l_1/l_2`
`2/1.5 = 120/l_2`
`l_2 = 90 cm`
APPEARS IN
संबंधित प्रश्न
SI unit of potential gradient is _______.
(a) V cm
(b) `V/"cm"`
(c) Vm
(d) `V/m`
In the given circuit, with steady current, calculate the potential drop across the capacitor and the charge stored in it.
When a resistor of 5Ω is connected across the cell, its terminal potential difference is balanced by 150 cm of potentiometer wire and when a resistance of 10 Ω is connected across the cell, the terminal potential difference is balanced by 175 cm same potentiometer wire. Find the balancing length when the cell is in open circuit and the internal resistance of the cell.
In a potentiometer experiment, the balancing length with a resistance of 2Ω is found to be 100 cm, while that of an unknown resistance is 500 cm. Calculate the value of the unknown resistance.
A 10 m long wire of resistance 20 Q is connected in series with a battery of emf 3 V and a resistance of 10 Ω. The potential gradient along the wire in V/m is ________.
The potentiometer is more sensitive, when ______.
A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F. because the method involves ______.
What will a voltmeter of resistance 200 Ω read when connected across a cell of emf 2 V and internal resistance 2 Ω?
A particle carrying 8 electron charges starts from rest and is accelerated through a potential difference of 9000 V. Calculate the KE acquired by it in keV.
A null point is obtained at 200 cm on potentiometer wire when cell in secondary circuit is shunted by 52. When a resistance of 152 is used for shunting, null point moves to 300 cm. The internal resistance of the cell is ______.
