1
JEE Main 2020 (Online) 8th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA. For a cell of emf 5V and internal resistance of 20$$\Omega $$, the null point on it is found to be a 1000 cm. The resistance of whole wire is :
A
80$$\Omega $$
B
60$$\Omega $$
C
120$$\Omega $$
D
100$$\Omega $$
2
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In a building there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. The voltage of electric main is 220 V. The minimum fuse capacity (rated value) of the building will be :
A
15 A
B
20 A
C
25 A
D
10 A
3
JEE Main 2020 (Online) 7th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The current I1 (in A) flowing through 1 $$\Omega $$ resistor in the following circuit is : JEE Main 2020 (Online) 7th January Morning Slot Physics - Current Electricity Question 219 English
A
0.4
B
0.25
C
0.2
D
0.5
4
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A moving coil galvanometer, having a resistance G, produces full scale deflection when a current Ig flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to I0(I0 > Ig) by connecting a shunt resistance RA to it and (ii) into a voltmeter of range 0 to V (V = GI0) by connecting a series resistance RV to it. Then,
A
$${R_A}{R_V} = {G^2}$$ and $${{{R_A}} \over {{R_V}}} = {{{I_g}} \over {\left( {{I_0} - {I_g}} \right)}}$$
B
$${R_A}{R_V} = {G^2}\left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)$$ and $${{{R_A}} \over {{R_V}}} = {\left( {{{{I_0} - {I_g}} \over {{I_g}}}} \right)^2}$$
C
$${R_A}{R_V} = {G^2}\left( {{{{I_0} - {I_g}} \over {{I_g}}}} \right)$$ and $${{{R_A}} \over {{R_V}}} = {\left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)^2}$$
D
$${R_A}{R_V} = {G^2}$$ and $${{{R_A}} \over {{R_V}}} = {\left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)^2}$$
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