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CBSE

Subject

Physics

Class

JEE Class 12

JEE Physics 2012 Exam Questions

Multiple Choice Questions

1.

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


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

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2.

A liquid in a beaker has temperature θ(t) at time t and θ0 is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge (θ – θ0) and t is


A.

According to Newton's law of cooling.

dθ over dt space proportional to left parenthesis straight theta minus straight theta subscript straight o right parenthesis
rightwards double arrow fraction numerator begin display style dθ end style over denominator begin display style dt end style end fraction space equals space minus straight k left parenthesis straight theta minus straight theta subscript straight o right parenthesis
integral fraction numerator begin display style dθ end style over denominator begin display style straight theta minus straight theta subscript straight o end style end fraction space equals space integral negative kdt space
rightwards double arrow space In space left parenthesis straight theta minus straight theta subscript straight o right parenthesis space equals space minus kt plus straight c

Hence the plot of ln(θ – θ0) vs t should be a straight line with negative slope.

473 Views

3.

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


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

369 Views

4.

A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F (t) = F0e–bt in the x direction. Its speed v(t) is depicted by which of the following curves?


C.

straight F equals space straight F subscript 0 straight e to the power of negative bt end exponent
rightwards double arrow space straight a space equals space straight F over straight m space equals space straight F subscript 0 over straight m straight e to the power of negative bt end exponent
rightwards double arrow space dv over dt space equals straight F subscript 0 over straight m straight e to the power of negative bt end exponent
integral dv space equals space integral subscript 0 superscript straight t straight F over straight m straight e to the power of negative bt end exponent dt
rightwards double arrow space straight v space equals straight F over straight m open square brackets table row cell negative 1 end cell row straight b end table close square brackets open square brackets table row cell straight e to the power of negative bt end exponent end cell end table close square brackets subscript 0 superscript straight t
rightwards double arrow space straight v equals space straight F over mb left square bracket straight e to the power of negative bt end exponent right square bracket
straight v equals space 0 space at space straight t equals space 0
and space straight v rightwards arrow space straight F over mb space as space straight t rightwards arrow infinity

So, velocity increases continuously and attains a maximum value of v= F/mb as t ∞
614 Views

5.

The mass of a spaceship is 1000 kg. It is to be launched from the earth's surface out into free space. The value of 'g'and 'R'(radius of earth) are 10 m/s2 and 6400km respectively. The required energy for this work will be;

  • 6.4 x1011 J

  • 6.4 x108 J

  • 6.4 x109 J

  • 6.4 x109 J


D.

6.4 x109 J

Potential energy on the earth's surface is -mgR while in free space, it is zero. So, to free the spaceship minimum required energy is
E =mgR
 = 103 x 10 x 6400 x 103 J
 = 6.4 x 1010 J

621 Views

6.

A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500 K It is desired to have an engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be

  • the efficiency of Carnot engine cannot be made larger than 50%

  • 1200 K

  • 750 K 

  • 600 K


C.

750 K 

Efficiency
straight eta space equals space 1 minus straight T subscript sink over straight T subscript source
Now comma space 0.4 space equals 1 minus fraction numerator straight T subscript sink over denominator 500 straight K end fraction
rightwards double arrow space straight T subscript sink space equals space 0.6 space straight x space 500 space straight K
space equals space 300 space straight K
Thus comma space 0.6 space equals space 1 minus fraction numerator 300 space straight K over denominator straight T apostrophe subscript source end fraction
rightwards double arrow space straight T to the power of apostrophe subscript source space equals space fraction numerator 300 space straight K over denominator 0.4 end fraction
space equals 750 space straight K

690 Views

7.

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.


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

698 Views

8.

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


C.

58.65 degree

1035 Views

9.

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

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

864 Views

10.

A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be

  • 20 square root of 2m
  • 10 m

  • 10 square root of 2 m
  • 20 m


D.

20 m

Maximum speed with which the boy can throw stone is
straight u space equals space square root of 2 gh end root
space equals space square root of 2 space straight x 10 straight x 10 end root
space equals 10 square root of 2 space straight m divided by straight s
Range is maximum when projectile is thrown at angle of 45o, Thus,
straight R subscript max space equals space fraction numerator straight u squared over denominator straight g space end fraction space equals space fraction numerator left parenthesis 10 square root of 2 right parenthesis squared over denominator 10 end fraction space equals space 20 space straight m

254 Views

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