NEET physics

Sponsor Area

Question
CBSEENPH11020812

One end of a string of length l is connected to a particle of mass ‘m’ and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed ‘v’, the net force on the particle (directed towards centre) will be (T represents the tension in the string)

  • T

  • straight T space plus space mv squared over straight I
  • straight T space space minus space mv squared over straight I
  • zero

Solution

A.

T

Centripetal force mv2/l is provided by tension so the net force will be equal to tension i.e., T.

Sponsor Area

Question
CBSEENPH11020813

The x and y coordinates of the particle at any time are x = 5t – 2t2 and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2 s is

  • 0

  • 5 m/s2

  • -4 m/s2

  • –8 m/s2

Solution

C.

-4 m/s2

x = 5t – 2t2    y = 10t
dx over dt space equals space 5 minus 4 straight t space space space space space space space dy over dt space equals space 10
straight v subscript straight x space equals space 5 minus 4 straight t space space space space space space space space space straight V subscript straight y space equals space 10
dv over dt straight x space equals space minus 4 space space space space space space space space dv over dt straight y space equals space 10
straight a subscript straight x space equals space minus space 4 space space space space space space space space space space space straight a subscript straight y space equals space 0
Acceleration of particle at t = 2 s is = –4 m/s2

Question
CBSEENPH11020814

Suppose the charge of a proton and an electron differ slightly. One of them is –e, the other is (e + Δe). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of [Given mass of hydrogen
mh = 1.67 × 10–27 kg]

  • 10–20 C

  • 10–23 C

  • 10–37 C

  • 10–47 C

Solution

C.

10–37 C

Fe = Fg
fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator increment straight e squared over denominator straight d squared end fraction space equals space Gm squared over straight d squared
9 space straight x space 10 to the power of 9 space left parenthesis increment straight e squared right parenthesis space equals space 6.67 space straight x space 10 to the power of negative 11 end exponent space straight x space 1.67
straight x space 10 to the power of negative 27 end exponent space space straight x space 1.67 space straight x space 10 to the power of negative 27 end exponent
increment straight e squared space equals space fraction numerator 6.67 space straight x space 1.67 space straight x space 1.67 over denominator 9 end fraction space straight x space 10 to the power of negative 74 end exponent
increment straight e almost equal to space 10 to the power of negative 37 end exponent

Question
CBSEENPH11020816

The diagrams below show regions of equipotentials



A positive charge is moved from A to B in each diagram.

  • Maximum work is required to move q in figure (c).

  • In all the four cases the work done is the same.

  • Minimum work is required to move q in figure (a).

  • Maximum work is required to move q in figure (b).

Solution

B.

In all the four cases the work done is the same.

Work was done w = qΔV
ΔV is same in all the cases so work is done will be same in all the cases.

Question
CBSEENPH11020819

The physical quantity of the dimensions of length that can be formed out of c, G and e/4πε0 is [c is the velocity of light, G is the universal constant of gravitation and e is charge]

  • 1 over straight c squared space open square brackets straight G fraction numerator straight e squared over denominator 4 πε subscript 0 end fraction close square brackets to the power of 1 half end exponent
  • straight c squared space open square brackets straight G fraction numerator straight e squared over denominator 4 πε subscript 0 end fraction close square brackets to the power of 1 half end exponent
  • straight c squared space open square brackets fraction numerator straight e squared over denominator straight G 4 πε subscript 0 end fraction close square brackets to the power of 1 half end exponent
  • straight c squared space open square brackets fraction numerator straight e squared over denominator straight G 4 πε subscript 0 end fraction close square brackets

Solution

A.

1 over straight c squared space open square brackets straight G fraction numerator straight e squared over denominator 4 πε subscript 0 end fraction close square brackets to the power of 1 half end exponent