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1.

A uniform cylinder of length L and mass M having cross - sectional area A is suspended, with its length vertical, from a fixed point by a massless spring such that it is half submerged in a liquid of density σ at equilibrium position. The extension x0 of the spring when it is in equilibrium is :
JEE_IIT_2013_question_paper_with_answer_on_testveda_q1

A 1
B 2
C 3
D 4
2.

A metallic rod of length ‘l’ is tied to a string of length 2l and made to rotate with angular speed ω on a horizontal table with one end of the string fixed. If there is a
verticalmagnetic field ‘B’ in the region, the e.m.f. induced across the ends of the rod is:
JEE_IIT_2013_question_paper_with_answer_on_testveda_q2
JEE_IIT_2013_question_paper_with_answer_on_testveda_q2.

A 1
B 2
C 3
D 4
3.

This question has statement I and Statement II. Of the four choices given after the Statements, choose the one that best describes the two Statements.
Statement - I :Apoint particle of mass m moving with speed ∪ collides with stationary point particle of mass M. If the maximum energy loss possible is given as
JEE_IIT_2013_question_paper_with_answer_on_testveda_q3
Statement - II: Maximum energy loss occurs when the particles get stuck together as a result of the collision.

A Statement -I is true, Statment -II is true, Statement -II is the correct explanation of Statement -I.
B Statement -I is true, Statment -II is true, Statement -II is not the correct explanation of Statement -I.
C Statement -I is true, Statment -II is false.
D Statement -I is false, Statment -II is true.
5.

A projectile is given an initial velocity of JEE_IIT_2013_question_paper_with_answer_on_testveda_q5 m/s, where iˆ is along the ground and jˆ∧ is along the vertical.
If g = 10 m/s2, the equation of its trajectory is :

A [∈0] = [M-1 L-3 T2 A]
B [∈0] = [M-1 L-3 T4 A2]
C [∈0] = [M-1 L-2 T-1 A-2]
D [∈0] = [M-1 L-2 T-1 A]
4. Let [∈0] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then :
A y = x – 5x2
B y = 2x – 5x2
C 4y = 2x – 5x2
D 4y = 2x – 25x2

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