NEET physics

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Question
CBSEENPH11020567

When a string is divided into three segments of length l1, l2, and l3 the fundamental frequencies of these three segments are v1, v2, and v3 respectively. The original fundamental frequency (v) of the string is

  • square root of straight v space equals space square root of v subscript 1 end root plus square root of v subscript 2 end root plus square root of v subscript 3 end root
  • v = v1 +v2 +v3

  • 1 over straight v space equals space 1 over v subscript 1 space plus 1 over v subscript 2 plus 1 over v subscript 3
  • fraction numerator 1 over denominator square root of straight v end fraction space equals space fraction numerator 1 over denominator square root of v subscript 1 end root end fraction plus fraction numerator 1 over denominator square root of v subscript 2 end root end fraction plus 1 over square root of v subscript 3

Solution

C.

1 over straight v space equals space 1 over v subscript 1 space plus 1 over v subscript 2 plus 1 over v subscript 3

The fundamental frequency of string
straight v equals space fraction numerator 1 over denominator 2 straight l end fraction square root of straight T over straight m end root
straight v subscript 1 straight l subscript 1 space equals straight v subscript 2 straight l subscript 2 space equals space straight v subscript 3 straight l subscript 3 space equals space straight k
From space eq space left parenthesis straight i right parenthesis
straight l subscript 1 space equals space straight k over straight v subscript 1 comma space straight l subscript 2 space equals space straight k over straight v subscript 2 comma straight l subscript 3 space equals space straight k over straight v subscript 3
original space length
straight l equals straight k over straight v
Here comma space straight l equals straight l subscript 1 plus straight l subscript 2 plus straight l subscript 3
straight k over straight v space equals straight k subscript 1 over straight v subscript 1 space plus straight k subscript 2 over straight v subscript 2 plus straight k subscript 3 over straight v subscript 3
1 over straight v space equals 1 over straight v subscript 1 space plus 1 over straight v subscript 2 plus 1 over straight v subscript 3

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Question
CBSEENPH11020570

A 200 W sodium street lamp emits yellow light of wavelength 0.6 um. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is

  • 1.5 x 1020

  • 6 x 1018

  • 62 x 1020

  • 3 x1019

Solution

A.

1.5 x 1020

 Efficient power P= 
P equals space straight N over straight t space equals space fraction numerator h c over denominator lambda end fraction space equals space 200 space x 0.25
N over t space equals space 50 space x fraction numerator lambda over denominator h c end fraction space equals space 1.5 space x space 10 to the power of 20
space equals fraction numerator 50 space x 0.6 space x space 10 to the power of negative 6 end exponent over denominator 6.6 space x space 10 to the power of negative 34 end exponent space x space 3 x space 10 to the power of 8 end fraction
space equals space 1.5 space x space 10 to the power of 20

Question
CBSEENPH11020571

When a mass is rotating in a plane about a fixed point, its angular momentum is directed along

  • a line perpendicular to the plane of rotation

  • the line making an angle of 45o to the plane of rotation

  • the radius

  • the tangent to the orbit

Solution

A.

a line perpendicular to the plane of rotation

We knows  L = m (r x v)
So, here, angular momentum is directed along a line perpendicular to the plane of rotation.

Question
CBSEENPH12039772

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

  • straight theta space equals space tan to the power of negative 1 end exponent open parentheses 1 fourth close parentheses
  • straight theta space equals tan to the power of negative 1 end exponent space left parenthesis 4 right parenthesis
  • straight theta space equals space tan to the power of negative 1 end exponent space left parenthesis 2 right parenthesis
  • straight theta space equals space 45 to the power of straight o

Solution

B.

straight theta space equals tan to the power of negative 1 end exponent space left parenthesis 4 right parenthesis

Given, R = H
Range space straight R space equals space fraction numerator straight u squared left parenthesis 2 sinθ space cosθ right parenthesis over denominator straight g end fraction
Here space straight H space equals space fraction numerator straight u squared sin squared straight theta over denominator 2 straight g end fraction
Hence comma space fraction numerator straight u squared left parenthesis 2 space sin space straight theta space cos space straight theta right parenthesis over denominator straight g end fraction space equals space fraction numerator straight u squared space sin squared straight a over denominator 2 straight g end fraction
2 space cos space straight theta space equals space fraction numerator sin space straight theta over denominator 2 end fraction
tan space straight theta space equals space 4
straight theta space equals space tan to the power of negative 1 end exponent space left parenthesis 4 right parenthesis

Question
CBSEENPH11020576

Two persons of masses 55 kg and 65 kg respectively. are at the opposite ends of the boat. The length of the boat is 3.0 m and weigh 100 kg. The 55 kg man walks to the 65 kg man and sits with him. If the boat is in still water the centre of mass of the system shifts by

  • 3.0 m

  • 2.3 m

  • zero

  • 0.75 m

Solution

C.

zero

Here on the entire system net external force on the system is zero hence centre of mass remains unchanged.