When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10^{–4} ms^{–1}. If the electron density in the wire is 8 × 10^{28} m^{–3}, the resistivity of the material is close to:
1.6 x 10^{-8}Ωm
1.6 x 10^{-7}Ωm
1.6 x 10^{-6}Ωm
1.6 x 10^{-5}Ωm
D.
1.6 x 10^{-5}Ωm
The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is :
3 V/m
6 V/m
9 V/m
12 V/m
B.
6 V/m
Ε = cB
Where E = electric field,
B = magnetic field
c = speed of EM waves
= 3 × 10^{8} × 20 × 10^{–9} = 6 V/m
In the circuit shown, the current in the 1Ω resistor is:
1.3 A from P to Q
0 A
0.13 A, from Q to P
0.13 A, from P to Q
C.
0.13 A, from Q to P
Connect Point Q to ground and by applying Kirchhoff's laws
consider the grounded circuit as shown below,
Applying Kirchhoff's law at point Q,
Incoming current at Q = outgoing current from Q
Thus, current in the 1Ω resistance is 0.13 A, from Q to P
This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement- I: Higher the range, greater is the resistance of ammeter.
Statement- II: To increase the range of ammeter, additional shunt needs to be used across it.
Statement – I is true, Statement – II is true, Statement – II is the correct explanation of statement- I.
Statement – I is true, Statement – II is true, Statement – II is not the correct explanation of Statement–I.
Statement – I is true, statement – II is false.
Statement – I is false, Statement – II is true
D.
Statement – I is false, Statement – II is true
For Ammeter, S =
So for I to increase, S should decrease, so additional S can be connected across it.
In the circuit shown here, the point C is kept connected to point A till the current flowing through the circuit becomes constant. Afterwards, suddenly point C is disconnected from point A and connected to point B at time t= 0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to
1
-1
C.
-1
The supply voltage to a room is 120 V. The resistance of the lead wires is 6 Ω. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?
zero
2.9V
13.3 V
10.04 V
D.
10.04 V
Resistance of bulb =(120x 120)/60 = 240Ω
Resistance of Heater =(120x 120)/240 = 60Ω
Voltage across bulb before heater is switched on,
Voltage across bulb after heater is switched on,
A decrease in the voltage is V_{1} − V_{2} = 10.04 (approximately)
The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 x 10^{3}Am^{-1}.The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is
30 mA
60 mA
3 A
6 A
C.
3 A
For solenoid, the magnetic field needed to be magnetised the magnet,
B =μ_{o}nl
Where, n = 100
l = 10 cm = 10/100 m = 0.1m
The current voltage relation of the diode is given by I = e(1000V/T-1) mA, where the applied voltage V is in volt and the temperature T is in kelvin. If a student makes an error measuring ± 0.01V while measuring the current of 5mA at 300K, what will be the error in the value of current in mA?^{}
0.2 mA
0.02 mA
05 mA
0.05 mA
A.
0.2 mA
Given, I =(e^{1000V/T}-1)mA
dv =±0.01 V
T = 300 K
I = 5 mA
I = e1^{000V/T}-1
I +1 = e^{1000V/T}
Taking log on both sides, we get
log(l+T) = 1000V/T
On differentiating,
So, error in the value of current is 0.2 mA
An inductor (L = 0.03 H) and a resistor (R = 0.15 kΩ) are connected in series to a battery of 15V EMF in a circuit shown. The key K_{1} has been kept closed for a long time. Then at t = 0, K_{1} is opened and key K_{2} is closed simultaneously. At t = 1 ms, the current in the circuit will be:
()
100 mA
67 mA
6.7 mA
0.67 mA
D.
0.67 mA
After long time inductor behaves as short-circuit.
At t = 0, the inductor behaves as short circuit.The current,
As K_{2} is closed, current through the indicator starts decay which is given at any time t as
Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle ‘θ’ with the vertical. If wires have mass λ per unit length then the value of I is: (g = gravitational acceleration)
B.
consider free body diagram of the wire
As the wires are in equilibrium. they must carry current in opposite direction.
Here, , where l is the length of each wire are d is a separation between wires.
From the diagram, d = 2L sin θ
T = cos θ = mg = λlg
(in vertical direction) ..... (i)
(In horizontal direction) ..... (ii)
From eqs. (i) and (ii)