NEET physics

Sponsor Area

Question
CBSEENPH11020525

If energy (E), velocity (v) and time (T) are chosen  as the fundamental quantities, the dimensional formula of surface tension will be

  • [Ev-2T-1]

  • [Ev-1T-2]

  • [Ev-2T-2]

  • [E-2v-1T-3]

Solution

C.

[Ev-2T-2]

WiredFaculty

Sponsor Area

Question
CBSEENPH11020526

A particle of unit mass undergoes one-dimensional motion such that its velocity according to
V(x) =  βx-2n where β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x is given by 

  • -2nβ2 x-2n-1

  • -2nβx-4n-1

  • -2β x-2n+1

  • -2nβ2e-4n+1

Solution

B.

-2nβx-4n-1

Given, v = βx-2n
WiredFaculty

Question
CBSEENPH11020527

Two similar springs P and Q have spring constants KP and KQ, such that KP>KQ. They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs WP and WQ are related as, in  case (a) and case (b), respectively

  • WP =WQ ;WP> WQ

  • WP =WQ ;WP= WQ

  • WP > WQ ;WQ> WP

  • WP <WQ ; WP< WQ

Solution

C.

WP > WQ ;WQ> WP

Given KP>KQ
In case (a) the elongation is same
WiredFaculty 

Question
CBSEENPH11020528

A block of mass 10 kg, moving in the x-direction with a constant speed of 10 ms-1 , is subjected to a retarding force F= 0.1x J/m during its travel from x = 20 m to 30 m. Its final KE will be

  • 475 J

  • 450 J

  • 275 J

  • 250 J 

Solution

A.

475 J

From work energy theorem
work done = change in KE
⇒ W = Kf -Ki
WiredFaculty

Question
CBSEENPH11020529

A particle of mass m is driven by a machine that delivers a constant power K watts. If the particle starts from rest, the force on the particle at time t is

  • square root of mk over 2 end root t to the power of negative 1 divided by 2 end exponent
  • square root of mk space t to the power of negative 1 divided by 2 end exponent
  • square root of 2 mk end root space t to the power of negative 1 divided by 2 end exponent
  • 1 half square root of m t end root to the power of negative 1 divided by 2 end exponent

Solution

A.

square root of mk over 2 end root t to the power of negative 1 divided by 2 end exponent

As the machine delivers a constant power
So F, v =constant = k (watts)
WiredFaculty