JEE physics

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Question
CBSEENPH11020326

STATEMENT – 1
Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
STATEMENT – 2
Principle of conservation of momentum holds true for all kinds of collisions.

  • The statement I is True, Statement II is False.

  • The statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.

  • The statement I is True, Statement II is True; Statement II is not the correct explanation for Statement I.

  • Statement I is False, Statement II is

Solution

B.

The statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.

Sponsor Area

Question
CBSEENPH12039451

A particle is moving with velocity,straight v space equals space straight k left parenthesis straight y space bold i with hat on top space plus space straight x space bold j with hat on top right parenthesis where K is a constant. The general equation for its path is

  • y = x2 + constant

  • y2 =  x + constant

  • xy = constant

  • y2 = x2 + constant

Solution

D.

y2 = x2 + constant

The velocity of the particle,straight v space equals space straight k left parenthesis straight y space bold i with hat on top space plus space straight x space bold j with hat on top right parenthesis 
WiredFaculty

Question
CBSEENPH11020328

The figure shows the position –time (x – t) graph of one–dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is

WiredFaculty

  • 0.4 (Ns)

  • 0.8 Ns

  • 1.6 Ns

  • 0.2 Ns

Solution

B.

0.8 Ns

From the graph it is a straight line, so uniform motion. Because of impulse direction of velocity changes as can be seen from the slope of the graph.
(∴ impulse (l) = f x t
⇒ I  = mat
 space equals space fraction numerator straight m space left parenthesis straight v subscript 2 minus straight v subscript 1 right parenthesis over denominator straight t end fraction space straight t space equals space mv subscript 2 space minus space mv subscript 1
Initial velocity, v1 = 2/2 = 1 ms-1
Final velocity v2 = 2/2 = - 1 ms-1
pi  = mv1 = 0.4 N-s
pf  = mv2 = -0.4 N-s
J = pf-pi = -0.4-0.4
 = - 0.8 N-s
|J| = 0.8 N-s

Question
CBSEENPH11020330

The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 × 10–3 are

  • 5,1,2

  • 5,1,5

  • 5,5,2

  • 4,4,2

Solution

A.

5,1,2

Question
CBSEENPH11020333

A point p moves in counter -clockwise direction on a circular path as shown in the figure. The movement of P is such that it sweeps out the length s = t3 + 5, where s is metre and t is in second. The radius of the path is 20 m. The acceleration of P when t =2 s is nearly

WiredFaculty

  • 13 ms-2

  • 12 ms-2

  • 7.2 ms-2

  • 14 ms-2

Solution

D.

14 ms-2

Given that, s  =t3 +5
therefore speed v, = ds/st = 3t2
and rate of change of speed, at = dv/dt = 6t
∴ Tangential acceleration at t =2s
at = 6 x 2  = 12 ms-1
and at t = 2s, v = 3 (2)2 = 12 ms-1
∴ Centripetal acceleration, ac = v2/R = 144/20 ms-2
∴ Net acceleration  = square root of straight a subscript straight t superscript 2 space plus straight a subscript straight c superscript 2 end root space almost equal to space 14 space ms to the power of negative 2 end exponent