NEET physics

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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]

We space know space that
surface space tension space left parenthesis straight S right parenthesis space equals space fraction numerator Force space left square bracket straight F right square bracket over denominator Length space left square bracket straight L right square bracket end fraction
so comma space left square bracket straight S right square bracket space equals space fraction numerator MLT to the power of negative 2 end exponent over denominator left square bracket straight L right square bracket end fraction space equals space left square bracket ML to the power of 0 straight T to the power of negative 2 end exponent right square bracket
Energy space left parenthesis straight E right parenthesis space equals space force space straight x space displacement
rightwards double arrow space left square bracket straight E right square bracket space equals space left square bracket ML squared straight T squared right square bracket
velocity space left parenthesis straight v right parenthesis space equals displacement over time
left square bracket straight v right square bracket space equals space left square bracket LT to the power of negative 1 end exponent right square bracket

straight S proportional to space straight E to the power of straight a straight v to the power of straight b straight T to the power of straight c
Where space straight a comma space straight b comma space straight c space are space constants
From space the space principle space of space homogeneity
left square bracket LHS right square bracket space equals left square bracket RHS right square bracket
rightwards double arrow space left square bracket ML to the power of 0 straight T to the power of negative 2 end exponent right square bracket space equals space left square bracket ML squared straight T to the power of negative 2 end exponent right square bracket to the power of straight a space left square bracket LT to the power of negative 1 end exponent right square bracket to the power of straight b left square bracket straight T right square bracket to the power of straight c
rightwards double arrow space left square bracket ML to the power of 0 straight T to the power of negative 2 end exponent right square bracket space equals space left square bracket straight M to the power of straight a straight L to the power of 2 straight a plus straight b end exponent space straight T to the power of negative 2 straight a minus straight b plus straight c end exponent right square bracket
Equating space the space power space on space both space sides comma space we space get
straight a equals space 1 comma space 2 straight a space plus straight b space equals 0 comma space straight b equals negative 2
minus 2 straight a minus straight b plus straight c space equals negative 2

straight c equals space left parenthesis 2 straight a plus straight b right parenthesis minus 2 space equals space 0 minus 2 space equals negative 2
left square bracket straight S right square bracket space equals space left square bracket Ev to the power of negative 2 end exponent straight T squared right square bracket space equals space left square bracket Ev to the power of negative 2 end exponent straight T squared right square bracket

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
straight a equals space dv over dt equals dx over dt. dv over dx
straight a equals straight v dv over dx equals left parenthesis space βx to the power of negative 2 straight n end exponent right parenthesis left parenthesis negative 2 straight n space βx to the power of negative 2 straight n minus 1 end exponent right parenthesis
straight a equals negative 2 nβ squared straight x to the power of negative 4 straight n minus 1 end exponent

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
straight x subscript 1 space equals space straight x subscript 2 space equals straight x
straight W subscript straight P space equals space 1 half straight K subscript straight P straight x squared
and space space straight W subscript straight Q space equals space 1 half straight K subscript straight Q straight x squared
therefore space fraction numerator straight W subscript straight P over denominator straight W subscript straight Q space end fraction space equals space straight K subscript straight P over straight K subscript straight Q greater than 1
rightwards double arrow space straight W subscript straight P greater than straight W subscript straight Q
In space case space left parenthesis straight b right parenthesis space the space spring space force space is space same
straight i. straight e space straight F subscript 1 space equals space straight F subscript 2 equals straight F
straight x subscript 1 equals straight F over straight K subscript straight P comma space straight x subscript 2 space equals space straight F over straight K subscript straight Q
therefore space straight W subscript straight P space equals space 1 half straight K subscript straight P straight x subscript 1 superscript 2 space equals space 1 half straight K subscript straight P space fraction numerator straight F squared over denominator straight K subscript straight P superscript 2 end fraction space equals space 1 half straight F squared over straight K subscript straight P
and space space straight W subscript straight Q space equals space 1 half straight K subscript straight Q straight x subscript 2 superscript 2 space equals space 1 half straight K subscript straight Q space fraction numerator straight F squared over denominator straight K subscript straight Q superscript 2 end fraction space equals space 1 half straight F squared over straight K subscript straight Q
straight W subscript straight P over straight W subscript straight Q equals space straight K subscript straight Q over straight K subscript straight P space less than 1
straight W subscript straight P space less than space straight W subscript straight Q
 

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
rightwards double arrow space straight K subscript straight f end subscript space equals space straight W space plus straight K subscript straight i space end subscript equals space integral subscript straight x subscript 1 end subscript superscript straight x subscript 2 end superscript Fxdx space plus 1 half mv squared
equals integral subscript 20 superscript 30 minus 0.1 space straight x space dx space plus 1 half space straight x space 10 space straight x space space 10 squared
space equals space minus 0.1 open square brackets straight x squared over 2 close square brackets subscript 20 superscript 30 space plus 500
equals negative 0.50 left square bracket 30 squared minus 20 squared right square bracket space plus 500
equals negative 0.05 left square bracket 900 minus 400 right square bracket plus 500
rightwards double arrow space straight K subscript straight f equals negative 25 plus 500 equals 475 space straight J

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)
rightwards double arrow space straight m dv over dt. straight v space equals straight k
rightwards double arrow integral vdv space equals straight k over straight m integral dt
rightwards double arrow straight v squared over 2 equals straight k over straight m straight t space space
rightwards double arrow space straight v equals space square root of fraction numerator 2 straight k over denominator straight m end fraction end root straight t
Now comma space force space on space the space particles space is space given space by
straight F equals space straight m fraction numerator d straight v over denominator d straight t end fraction space equals straight m straight d over dt open parentheses fraction numerator 2 kt over denominator straight m end fraction close parentheses to the power of 1 divided by 2 end exponent
equals space square root of 2 km end root space open parentheses 1 half straight t to the power of negative 1 half end exponent close parentheses
equals space square root of mk over 2 end root. straight t to the power of fraction numerator negative 1 over denominator 2 end fraction end exponent