JEE 2011 :: General Questions :: Physics

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A train is moving along a straight line with a constant acceleration a. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of$60^{0}$ to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train in m/s², is

Answer:

Explanation:

 $t= T= \frac{2 u \sin \theta}{g}$

   =$\frac{2 \times 10 \times \sin 60^{0}}{10}= \sqrt{3} s$

Displacement of train in time  t = $\frac{1}{2}at^{2}$

Displacement of boy with respect to train = 1.15 m

$\therefore$ Displacement of boy with respect to ground = $(1.15+\frac{1}{2}at^{2})$

Displacement of ball with respect to  ground = $(u \cos 60^{0} ) t$

 To catch the ball back at initial height, 

 $1.15+\frac{1}{2} at^{2}= (u \cos 60^{0}) t$

 $\therefore$    $1.15+\frac{1}{2} a ( \sqrt{3})^{2}=10 \times \frac{1}{2} \times \sqrt{3}$

 Solving this equation , we get

 a=5 $ms^{-2}$

 $\therefore$ answer is 5

A series R-C combination is connected to an AC voltage of angular frequency $\omega$= 500 rad/s. If the impedance of the R-C circuit is $R\sqrt{1.25}$, the time constant (in millisecond) of the circuit is

Answer:

Explanation:

$Z=\sqrt{R^{2}+X_{C}^{2}}=R\sqrt{1.25}$

 $\therefore$  $R^{2}+X_{C}^{2}=1.25R^{2}$

 or $X_{C}=\frac{R}{2}$

 or   $ \frac{ 1}{ \omega  C}=\frac{R}{2}$

 $\therefore$  Time constant  = CR= $\frac{2}{\omega}$

       $\frac{2}{500} s=4ms$

 $\therefore$  Answer is 4

Two batteries of different emfs and different internal resistances are connected as shown. The voltage across AB in volt is

Answer:

Explanation:

$V_{AB}$= Equivalent emf of two batteries in parallel.

=  $ \frac{ E_{1}/r_{1}+E_{1}/r_{2}}{1/r_{1}+1/r_{2}}$

 $= \frac{(6/1)+(3/2)}{(1/1)+(1/20}=5 V$

 $\therefore$ Answer is 5

A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with  velocity 1 m/s. A small ball of mass 0.1 kg, moving with velocity 20 mls in the opposite directions, hits the ring at a height of 0.75 m and goes vertically up with velocity 10m/s imrnediately after the collision

Answer:

Explanation:

The data is incomplete. Let us assume that friction from ground on ring is not
impulsive during impact.
From linear momentum conservation in horizontal direction, we have

$(2 \times 1)+(0.1 \times 20)= (0.1 \times 0)+(2 \times v) \leftarrow^{-v}\rightarrow^{+ve}$

 Here, v is the ve;locity of CM of ring after impact 

Solving the above equation, we have
                            u=0
Thus, CM becomes stationary.

$\therefore$ correct answer is (a)

Linear impulse during impact
(i) In horizontal direction

 $J_{1}=\triangle P=0.1 \times v20=2Ns$

 (ii) In vertical direction

 $J_{2}=\triangle P = 0.1 \times 10 =1 Ns$

 Writing the equation (about CM) 

Angular impulse= change in angular momentum

    $1\times\left(\frac{\sqrt{3}}{2}\times\frac{1}{2}\right)-2\times0.5\times\frac{1}{2}$

  $=2\times(0.5)^{2}\left[ \omega-\frac{1}{0.5}\right]$

Solving this equation ω comes out to be positive or ω anti-clockwise. So just after collision rightwards slipping is taking place.
Hence, friction is leftwards.
Therefore, option (c) is also correct

Correct options are (a) and (c).

A series R-C circuit is connected to AC voltage source. Consider two cases (A) when C is without a dielectric medium and (B) when C is filled with a dielectric of constant 4. The current I_R through the resistor and voltage V_C across the capacitor are compared in the two cases. Which of the following is/are true?

Answer:

Explanation:

$Z= \sqrt{R^{2}+X_{C}^{2}}=\sqrt{R^{2}+\left(\frac{1}{\omega C}\right)^{2}}$

In case (b) capacitance C will be more. Therefore, impedance Z will be less.
Hence, current will be more.

$\therefore$ Option(b) is correct 

 Further   $V_{C}=\sqrt{V^{2}-V_{R}^{2}}$

                         $\sqrt{V^{2}-(IR)^{2}}$

 In case (b), since current l is more

 Therefore  , $V_{C}$ will be less

 $\therefore$  Option (c) is correct

$\therefore$  Correct options are (b) and (c)

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