IIT-JEE 2011 Paper 2 Offline | JEE Advanced Year Wise Previous Years Questions

IIT-JEE 2011 Paper 2 Offline

MCQ (Single Correct Answer)

-0.75

A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, $${x_1}\left( t \right) = A\sin \omega t$$ and $${x_2}\left( t \right) = A\sin \left( {\omega t + {{2\pi } \over 3}} \right)$$. Adding a third sinusoidal displacement $${x_3}\left( t \right) = B\sin \left( {\omega t + \phi } \right)$$ brings the mass to a complete rest. The values of B and $$\phi $$ are

$$\sqrt 2 A,{{3\pi } \over 4}$$

$$A,{{4\pi } \over 3}$$

$$\sqrt 3 A,{{5\pi } \over 6}$$

$$A,{\pi \over 3}$$

IIT-JEE 2011 Paper 2 Offline

MCQ (Single Correct Answer)

-0.5

Which of the field patterns given below is valid for electric field as well as for magnetic field ?

IIT-JEE 2011 Paper 2 Offline Physics - Electrostatics Question 51 English Option 1

IIT-JEE 2011 Paper 2 Offline Physics - Electrostatics Question 51 English Option 2

IIT-JEE 2011 Paper 2 Offline

MCQ (Single Correct Answer)

-0.5

A wooden block performs $$SHM$$ on a frictionless surface with frequency, $${v_0}.$$ The block carries a charge $$+Q$$ on its surface . If now a uniform electric field $$\overrightarrow E $$ is switched- on as shown, then the $$SHM$$ of the block will be
IIT-JEE 2011 Paper 2 Offline Physics - Simple Harmonic Motion Question 16 English

IIT-JEE 2011 Paper 2 Offline Physics - Simple Harmonic Motion Question 16 English

of the same frequency and with shifted mean position.

of the same frequency and with the same mean position

of changed frequency and with shifted mean position.

of changed frequency and with the same mean position.

IIT-JEE 2011 Paper 2 Offline

MCQ (More than One Correct Answer)

-1

Which of the following statement(s) is/are correct?

If the electric field due to a point charge varies as $${r^{ - 2.5}}$$ instead of $${r^{ - 2}},$$ then the Gauss law will still be valid.

The Gauss law can be used to calculate the field distribution around an electric dipole.

If the electric field between two point charges is zero somewhere, then the sign of the two charges is the same.

The work done by the external force in moving a unit positive charge from point $$A$$ at potential $${V_A}$$ to point $$B$$ at potential $${V_B}$$ is $$\left( {{V_B} - {V_A}} \right).$$

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