JEE physics

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Question
CBSEENPH11020296

A spectrometer gives the following reading when used to measure the angle of a prism.
Main scale reading: 58.5 degree
Vernier scale reading: 09 divisions
Given that 1 division on the main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism from the above data

  • 58.59 degree

  • 58.77 degree

  • 58.65 degree

  • 59 degree

Solution

C.

58.65 degree

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Question
CBSEENPH11020297

A diatomic molecule is made of two masses m1 and m2 which are separated by a distance r. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by (n is an integer)

  • fraction numerator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis squared straight n squared straight h squared over denominator 2 straight m subscript 1 superscript 2 straight m subscript 2 superscript 2 straight r squared end fraction
  • fraction numerator straight n squared straight h squared over denominator 2 left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight r squared end fraction
  • fraction numerator 2 straight n squared straight h squared over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight r squared end fraction
  • fraction numerator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight n squared straight h squared over denominator 2 straight m subscript 1 straight m subscript 2 straight r squared end fraction

Solution

D.

fraction numerator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight n squared straight h squared over denominator 2 straight m subscript 1 straight m subscript 2 straight r squared end fraction

Rotational kinetic energy of the two body system rotating about their centre of mass is
RKE space equals space 1 half μω squared straight r squared
where comma space straight mu space equals space fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction equals space reduced space mass
and space angular space momentum comma space straight L space equals space μωr squared space equals space fraction numerator nh over denominator 2 straight pi end fraction
straight omega space equals fraction numerator nh over denominator 2 πμr squared end fraction
therefore space RKE space equals space 1 half μω squared straight r squared space equals space 1 half space straight mu. open parentheses fraction numerator nh over denominator 2 πμr squared end fraction close parentheses squared straight r squared
space equals fraction numerator straight n squared straight h squared over denominator 8 straight pi squared μr squared end fraction space equals space fraction numerator straight n squared straight ħ squared over denominator 2 μr squared end fraction space open parentheses where comma straight ħ space equals space fraction numerator straight h over denominator 2 straight pi end fraction close parentheses
fraction numerator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight n squared straight ħ squared over denominator 2 straight m subscript 1 straight m subscript 2 straight r squared end fraction

Question
CBSEENPH11020298

A cylindrical tube, open at both ends, has a fundamental frequency, f, in the air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now

  • m1r1:m2r2

  • m1 :m2

  • r1 :r2

  • 1:1

Solution

C.

r1 :r2

As their period of revolution is same, so its angular speed is also same. Centripetal acceleration is circular path,
a= ω2r
Thus, 
straight a subscript 1 over straight a subscript 2 space equals space fraction numerator straight omega squared straight r subscript 1 over denominator straight omega squared straight r subscript 2 end fraction space equals space straight r subscript 1 over straight r subscript 2

Question
CBSEENPH11020299

This question has statement 1 and statement 2. Of the four choices given after the statements, choose the one that best describes the two statements.
If two springs S1 and S2 of force constants k1 and k2, respectively, are stretched by the same force, it is found that more work is done on spring S1 than on spring S2.

Statement 1: If stretched by the same amount, work done on S1, will be more than that on S2
Statement 2 : k1 < k2

  • Statement 1 is false, Statement 2 is true

  • Statement 1 is true, Statement 2 is false

  • Statement 1 is true, Statement 2 is the correct explanation for statement 1

  • Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for statement 1.

Solution

A.

Statement 1 is false, Statement 2 is true

As no relation between k1 and k2 is given in the question, that is why nothing can be predicted about the statement I. But as in Statment II. k1<k2
Then, for same force
straight W space equals space straight F. straight x space equals space straight F. straight F over straight k space equals space straight F squared over straight k
straight W space proportional to space 1 over straight k space straight i. straight e space for space constant space straight F
straight W subscript 1 over straight W subscript 2 space equals straight K subscript 2 over straight K subscript 1
But space for space same space displacement
straight W space equals space straight F. straight x space equals 1 half space kx. straight x space equals space 1 half kx squared
straight W space proportional to space straight K comma space straight W subscript 1 over straight W subscript 2 space equals space straight K subscript 1 over straight K subscript 2
straight W subscript 1 space less than straight W subscript 2

Question
CBSEENPH11020301

A thin liquid film formed between a U-shaped wire and a light slider supports a weight of 1.5 x10–2N (see figure). The length of the slider is 30 cm and its weight negligible. The surface tension of the liquid film is

  • 0.0125 Nm-1

  • 0.1 Nm-1

  • 0.05 Nm-1

  • 0.025 Nm-1

Solution

D.

0.025 Nm-1


The force of surface tension acting on the slider balances the force due to the weight.

⇒F = 2Tl = w
⇒2T(0.3) = 1.5 x 10–2
⇒T = 2.5 x 10–2 N/m