JEE 2016 :: Main 2016 :: Physics 2016

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A point particle of mass m moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals µ. The particle is released from rest, from point P and it comes to rest at a point R. The energies lost by the ball, over the parts PQ and QR of the track, are equal to each other and no energy is lost when particle changes direction from PQ to QR. the values of the coefficient of friction µ and the distance x  (=QR), are respectively close to 

Answer:

Explanation:

As energy loss is the same, thus

     $\mu mg \cos\theta.(PQ)=\mu mg. (QR)$

 $\therefore$    $QR= (PQ) \cos\theta$

$\Rightarrow   $   $QR=4\times\frac{\sqrt{3}}{2}$

                          $=2\sqrt{3}=3.5m$

Furthur, decreases in potential energy = loss due to friction

$\therefore$    $mgh= (\mu mg\cos\theta)d_{1}+(\mu mg)d_{2}$

 $m\times 10\times2= \mu \times m\times10\times \frac{\sqrt{3}}{2}\times 4 +\mu \times m\times 10\times 2\sqrt{3}$

$\Rightarrow$   $4\sqrt{3} \mu =2$

$\Rightarrow$   $\mu= \frac{1}{2\sqrt{3}}$

                  =0.288=.029

A particle of mass m is moving along the side of a square of side a, with a uniform speed v in the XY-plane as shown in the figure

Which of the following statements is false for the angular momentum L about the origin?

Answer:

Explanation:

We can apply L=m(r × v) for different parts

   For example

if part (a) , cooredinates of  A are  $(\frac{R}{\sqrt{2}},\frac{R}{\sqrt{2}})$

there fore ,r = $\frac{R}{\sqrt{2}}\widehat{i}+\frac{R}{\sqrt{2}}\widehat{j}$   and  $v=v\widehat{i}$

So, subsuituting in L=m(r× v) we get,

   $L= -\frac{mvR}{\sqrt{2}}\widehat{R}$

Hence, option (a) is correct, similarly , we can check other options also

A student measures the time period of 100 oscillatons of a simple pendulum four times. The data set is the 90s, 91s, 92s and 95s. If the minimum division in the measuring clock is 1s, then the reported mean times should be

Answer:

Explanation:

True value= $\frac{90+91+95+92}{4}=92$

Mean abosulte error

      = $\frac{\mid92-90\mid+\mid92-91\mid+\mid92-95\mid+\mid92-92\mid}{4}$

  =  $\frac{2+1+3+0}{4}=1.5$

   value $(92\pm1.5)$

Since. least count is 1s

Value  = $(92\pm2s)$

• Main 2016 - Physics 2016
• Main 2016 - Chemistry 2016
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