JEE 2011 :: General Questions :: Physics

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Four solid spheres each of diameter,$\sqrt{5}$ cm and mass 0.5 kg are placed with their centres at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is $N \times 10^{-4}$ kg-m² then N is

Answer:

Explanation:

 $r= \frac{d}{2}=\frac{\sqrt{5}}{2}$ cm

 =$\frac{\sqrt{5}}{2} \times 10^{-2}$ cm

 $m=0.5 kg$

 a=4cm

 =$4 \times 10^{-2}$ cm

 $I_{XX}=I_{1}+I_{2}+I_{3}+I_{4}$

$=\left[\frac{2}{5}mr^{2}+m\left(\frac{a}{\sqrt{2}}\right)^{2}\right]+\frac{2}{5}mr^{2}$

                   +$\left[\frac{2}{5}mr^{2}+m\left(\frac{a}{\sqrt{2}}\right)^{2}\right]+\frac{2}{5}mr^{2}$

 Substituting the values , we get

 $I_{XX}=9 \times 10^{-4} Kgm^{-2}$

 $\therefore$  N=9

 answer is 9

Steel wire of length L at $40^{0}$ C is suspended from the ceiling and then a mass m is hung from its free end. The wire is cooled down from $40^{0}$C to $30^{0}$C to regain its original length L. The coefficient of linear thermal expansion of the steel is$10^{-5}/^{0}C$,  Young's modulus of steel is $10^{11}$ N/m² and the radius of the wire is 1 mm.
Assume that L >> the diameter of the wire. Then, the value of m in kg is nearly

Answer:

Explanation:

$\triangle l_{1}=\frac{FL}{AY}=\frac{mgL}{\pi r^{2}Y}$ =increase in length

 $\triangle l_{2}=L \alpha \triangle \theta$ = Decrease in length 

 To regain is original length

 $\triangle l_{1}= \triangle l_{2}$

 $\therefore$  $\frac{mgL}{\pi r^{2} Y}=L \propto \triangle \theta$

 $\therefore$    $m=\left(\frac{r^{2} Y \alpha \triangle \theta}{g}\right)$

Substituting  the values , we get

 m= 3kg

 $\therefore$ Answer is 3.

A long circular tube of length 10 m and radius 0.3 m carries a current l along its curved surface as shown. A wire loop of resistance 0.005 $\Omega$  and of radius 0.1 m is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as $l=l_{0}\cos 300 t$ , where  $l_{0}$  is constant . If the magnetic moment of the loop  is $N \mu _{0}l_{0} \sin 300t$, then N is 

Answer:

Explanation:

Take the circular tube as a long solenoid. The wires are closely wound.
Magnetic field inside the solenoid is

  $B= \mu_{0} ni$

 Here, n= number of turns per unit length

 $\therefore$ ni= current per unit length$

 In the given problem,

 $ni= \frac{I}{L}$

 $\therefore$  $B= \frac{\mu_{0} I}{L}$

 Flux passing through the circular coil is 

 $\phi =BS=\left(\frac{\mu_{0}l}{L}\right)(\pi r^{2})$

 Induced emf, $e= -\frac{d \phi}{dt}=-\left(\frac{\mu_{0} \pi r^{2}}{LR}\right). \frac{dl}{dt}$

 magnetic moment   $iA=i \pi r^{2}$

 or    $M=-\left(\frac{\mu_{0} \pi^{2} r^{4}}{LR}\right). \frac{dl}{dt}$....(i)

 Given,  $l=l_{0} \cos 300 t$

$\therefore$  $\frac{dl}{dt}=-300 l_{0} \sin (300 t)$

 Substituting in Eq.(i), we get

 $M=\left(\frac{300 \pi^{2} r^{4}}{LR}\right).\mu_{0}l_{0} \sin 300t$

 $\therefore$  $N= \frac{ 300 \pi^{2} r^{4}}{LR}$

 Substituting the values , we get

 $N= \frac{300(22/7)^{2}(0.1)^{4}}{(10)(0.005)}$=5.926

   or N=6

A block is moving on an inclined plane making an angle $45^{0}$ with the horizontal and the coefficient of friction $\mu$. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define $N= 10 \mu$ , then N is 

Answer:

Explanation:

$F_{1}=mg \sin \theta+\mu mg \cos \theta$

 $F_{2}= mg \sin \theta-\mu mg \cos \theta$

 Given that , $F_{1}=3F_{2}$

 or  $(\sin 45^{0}+\mu \cos 45^{0})$

 =$3(\sin  45^{0}-\mu \cos 45^{0})$

 On solving  we get , $\mu=0.5$

 $\therefore$  N= $10 \mu =5$

 $\therefore$ Answer is 5

Four-point charges, each of +q are rigidly fixed at the four corners of a square planar soap film of side a. The surface tension, of the soap film is $\gamma$ . The system of charges and planar film are in equilibrium and, $a =k\left[\frac{q^{2}}{\gamma}\right]^{1/N}$. where k is a constant. Then N is 

Answer:

Explanation:

$F_{1}=$ Net electrostatic force on anyone charge due to rest of three charges

   =$\frac{1}{4 \pi \epsilon_{0}}\frac{q^{2}}{a^{2}}(\sqrt{2}+\frac{1}{2})$

 $F_{2}$= surface tension force =$\gamma a$

 If we see the equilibrium of line BC, 

then $2F_{1} \cos 45^{0}=F_{2}$

 or   $ \sqrt{2} F_{1}=F_{2}$

 or  $\frac{1}{4 \pi \epsilon_{0}}\frac{q^{2}}{a^{2}}(2+\frac{1}{\sqrt{2}})=\gamma a$

 $\therefore$ $a^{3}= \frac{1}{4 \pi \epsilon_{0}}(2+\frac{1}{\sqrt{2}})\frac{q^{2}}{\gamma} $

 or    $a= [\frac{1}{4 \pi \epsilon_{0}}(2+\frac{1}{\sqrt{2}})]^\frac{1}{3}[\frac{q^{2}}{\gamma}]^\frac{1}{3}$

   $=k\left[\frac{q^{2}}{\gamma}\right]^{1/3}$

 where  , $k=\left\{\frac{1}{4 \pi \epsilon_{0}}(2+\frac{1}{\sqrt{2}})\right\}^{1/3}$

 Therefore    ,N=3

 Answer is 3

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