1
JEE Main 2024 (Online) 4th April Morning Shift
Numerical
+4
-1

Let $$a=1+\frac{{ }^2 \mathrm{C}_2}{3 !}+\frac{{ }^3 \mathrm{C}_2}{4 !}+\frac{{ }^4 \mathrm{C}_2}{5 !}+...., \mathrm{b}=1+\frac{{ }^1 \mathrm{C}_0+{ }^1 \mathrm{C}_1}{1 !}+\frac{{ }^2 \mathrm{C}_0+{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_2}{2 !}+\frac{{ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_2+{ }^3 \mathrm{C}_3}{3 !}+....$$ Then $$\frac{2 b}{a^2}$$ is equal to _________.

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2
JEE Main 2024 (Online) 4th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

An infinitely long positively charged straight thread has a linear charge density $$\lambda \mathrm{~Cm}^{-1}$$. An electron revolves along a circular path having axis along the length of the wire. The graph that correctly represents the variation of the kinetic energy of electron as a function of radius of circular path from the wire is :

A
JEE Main 2024 (Online) 4th April Morning Shift Physics - Electrostatics Question 12 English Option 1
B
JEE Main 2024 (Online) 4th April Morning Shift Physics - Electrostatics Question 12 English Option 2
C
JEE Main 2024 (Online) 4th April Morning Shift Physics - Electrostatics Question 12 English Option 3
D
JEE Main 2024 (Online) 4th April Morning Shift Physics - Electrostatics Question 12 English Option 4
3
JEE Main 2024 (Online) 4th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

An electron is projected with uniform velocity along the axis inside a current carrying long solenoid. Then :

A
the electron will experience a force at $$45^{\circ}$$ to the axis and execute a helical path.
B
the electron will be accelerated along the axis.
C
the electron path will be circular about the axis.
D
the electron will continue to move with uniform velocity along the axis of the solenoid.
4
JEE Main 2024 (Online) 4th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

A metal wire of uniform mass density having length $$L$$ and mass $$M$$ is bent to form a semicircular arc and a particle of mass $$\mathrm{m}$$ is placed at the centre of the arc. The gravitational force on the particle by the wire is :

A
$$\frac{\mathrm{GmM} \pi^2}{\mathrm{~L}^2}$$
B
$$\frac{\mathrm{GMm} \pi}{2 \mathrm{~L}^2}$$
C
0
D
$$\frac{2 \mathrm{GmM} \pi}{\mathrm{L}^2}$$
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