1
JEE Main 2023 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A conducting circular loop of radius $$\frac{10}{\sqrt\pi}$$ cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is :

A
emf = 10 mV
B
emf = 5 mV
C
emf = 100 mV
D
emf = 1 mV
2
JEE Main 2023 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language

Consider the following radioactive decay process

$$_{84}^{218}A\buildrel \alpha \over \longrightarrow {A_1}\buildrel {{\beta ^ - }} \over \longrightarrow {A_2}\buildrel \gamma \over \longrightarrow {A_3}\buildrel \alpha \over \longrightarrow {A_4}\buildrel {{\beta ^ + }} \over \longrightarrow {A_5}\buildrel \gamma \over \longrightarrow {A_6}$$

The mass number and the atomic number of A$$_6$$ are given by :

A
210 and 84
B
210 and 80
C
211 and 80
D
210 and 82
3
JEE Main 2023 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

1 g of a liquid is converted to vapour at 3 $$\times$$ 10$$^5$$ Pa pressure. If 10% of the heat supplied is used for increasing the volume by 1600 cm$$^3$$ during this phase change, then the increase in internal energy in the process will be :

A
4800 J
B
4320 J
C
432000 J
D
4.32 $$\times$$ 10$$^8$$ J
4
JEE Main 2023 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

As shown in the figure, a network of resistors is connected to a battery of 24V with an internal resistance of 3 $$\Omega$$. The currents through the resistors R$$_4$$ and R$$_5$$ are I$$_4$$ and I$$_5$$ respectively. The values of I$$_4$$ and I$$_5$$ are :

JEE Main 2023 (Online) 24th January Morning Shift Physics - Current Electricity Question 74 English

A
$$\mathrm{I_4=\frac{8}{5}A}$$ and $$\mathrm{I_5=\frac{2}{5}A}$$
B
$$\mathrm{I_4=\frac{6}{5}A}$$ and $$\mathrm{I_5=\frac{24}{5}A}$$
C
$$\mathrm{I_4=\frac{2}{5}A}$$ and $$\mathrm{I_5=\frac{8}{5}A}$$
D
$$\mathrm{I_4=\frac{24}{5}A}$$ and $$\mathrm{I_5=\frac{6}{5}A}$$
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