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1
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
A current of 5 A passes through a copper conductor (resistivity = 1.7 × 10–8 $$\Omega $$m) of radius of cross-section 5 mm. Find the mobility of the charges if their drift velocity is 1.1 × 10–3 m/s.
A
1.3 m2/Vs
B
1.0 m2/Vs
C
1.8 m2/Vs
D
1.5 m2/Vs
2
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
In a meter bridge experiment, the circuit diagram and the corresponding observation table are shown in figure JEE Main 2019 (Online) 10th April Morning Slot Physics - Current Electricity Question 162 English
SI. No. R($$\Omega $$) l(cm)
1. 1000 60
2. 100 13
3. 10 1.5
4. 1 1.0
Which of the readings is inconsistent?
A
4
B
3
C
2
D
1
3
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Two wires A & B are carrying currents I1 & I2 as shown in the figure. The separation between them is d. A third wire C carrying a current I is to be kept parallel to them at a distance x from A such that the net force acting on it is zero. The possible values of x are : JEE Main 2019 (Online) 10th April Morning Slot Physics - Current Electricity Question 161 English
A
$$x = \left( {{{{I_1}} \over {{I_1} + {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} - {I_2}}}} \right)d$$
B
$$x = \left( {{{{I_2}} \over {{I_1} + {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} - {I_2}}}} \right)d$$
C
$$x = \left( {{{{I_1}} \over {{I_1} - {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} + {I_2}}}} \right)d$$
D
$$x = \pm {{{I_1}d} \over {{I_1} - {I_2}}}$$
4
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
In an experiment, the resistance of a material is plotted as a function of temperature (in some range). As shown in the figure, it is a straight line. One may conclude that : JEE Main 2019 (Online) 10th April Morning Slot Physics - Current Electricity Question 160 English
A
$$R(T) = {R_0}{e^{ - {T^2}/T_0^2}}$$
B
$$R(T) = {{{R_0}} \over {{T^2}}}$$
C
$$R(T) = {R_0}{e^{ {T^2}/T_0^2}}$$
D
$$R(T) = {R_0}{e^{ - T_0^2/{T^2}}}$$
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