JEE 2011 :: General Questions :: Physics

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A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, travelling with a velocity v m/s in a horizontal direction hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The initial velocity v of the bullet is

Answer:

Explanation:

Time taken by the bullet and ball to strike the ground is

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2 \times 5}{10}}=1s$

Let $v_{1}$and $v_{2}$ are the velocities of ball and bullet after collision.

Then applying

 x=vt 

 we have   $20=v_{1} \times 1$

 or $ v_{1}= 20ms^{-1}$

 $100=v_{2} \times 1$ or $v_{2}=100m/s^{-1}$

 Now, from conservation of linear momentum before and after collision

we have. 

$0.001 v= (0.2 \times 20) +(0.01 \times 100)$

 On solving, we get 

 $v= 500ms^{-1}$

 $\therefore$  Correct answer is (d).

The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%,the relative percentage error in the density is

Answer:

Explanation:

Least count of screw gauage= $\frac{0.5}{50}$

 =0.01 mm=$\triangle r$

 Diameter , r=2.5mm+20$\times \frac{0.5}{50}$

=270mm

$\frac{\triangle r}{r}=\frac{0.01}{2.70}$

or   $\frac{\triangle r}{r} \times 100=\frac{1}{2.7}$

 now, density   $d= \frac{m}{V}=\frac{m}{\frac{4}{3}\pi \left(\frac{r}{2}\right)^{3}}$

Here r is the diameter

$\therefore$     $\frac{\triangle d}{d} \times 100=\left\{\frac{\triangle m}{m}+3\left(\frac{\triangle r}{r}\right)\right\} \times 100$

    $=\frac{\triangle m}{m} \times 100+3\times\left(\frac{\triangle r}{r}\right)\times100$

   $=2$%$ +3 \times \frac{1}{27}$=3.11%

A wooden block performs SHM on a frictionless surface with frequency $V_{0}$. The block carries a charge + Q on its surface. If now a uniform electric field E is switched on as shown, then the SHM of the block will be

Answer:

Explanation:

FrequencY or time Period of SHM depends on variable forces. It does not depend on constant external force. The constant external force can only change the mean position. For example, in the given question mean Position is at the natural length of spring in the absence of an electric field. Whereas in the presence of electric field mean position will be obtained after compression of $x_{0}$.
Where $x_{0}$ is given by 

    $kx_{0}=QE$

 or   $x_{0}=\frac{QE}{k}$ 

$\therefore$ correct answer is (a)

The activity of a freshly prepared radioactive sample is $10^{10}$  disintegrations per second whose mean life is $10^{-9}$ s . The mass of an atom of this radioisotope is $10^{-25}$kg. The mass (in mg)  of the radioactive sample is 

Answer:

Explanation:

Activity $\left(-\frac{dN}{dt}\right)=\lambda N=\left(\frac{1}{t_{mean}}\right)\times N$

$\therefore$  $ N=\left(-\frac{dN}{dt}\right)\times t_{mean}$

 =Total number of atoms 

 Mass of one atom is  $10^{-25} kg=m(say)$

$\therefore$   Total mass of radioactive substance = (number of atoms) x (mass of onbe atom)

  $ =\left(-\frac{dN}{dt}\right) (t_{mean})(m)$

 Substituting the values , we get

 Total mass of radioactive substance =1 mg

 $\therefore $  Answer is 1

A boy is pushing a ring of mass 2 kg and a radius of 0.5 m with a stick as shown in the figure. The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of 0.3 m/ s2. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is $\frac{P}{10}$. The value of P is 

Answer:

Explanation:

There is no slipping between ring and ground. Hence, $f_{2}$ is not maximum. But there is slipping between ring and stick. Therefore, $f_{1}$  is maximum. Now, let us write the equations.

 $l=mR^{2}=(2)(0.5)^{2}$

 =$\frac{1}{2} kgm^{-2}$

 $N_{1}-F_{2}=ma$

 or $N_{1}=F_{2}=(2)(0.3)=0.6 N$....(i)

 $a= R \alpha =\frac{R \tau}{I}$

 =$ \frac{ R( f_{2}-f_{1})R}{I}=\frac{R^{2}(f_{2}-f_{1})}{I}$

 $\therefore$    $0.3=\frac{ (0.5)^{2}(f_{2}-f_{1})}{(1/2)}$

 or     $f_{2}-f_{1}=0.6N$.....(ii)

 $N_{1}^{2}+f_{1}^{2}=(2)^{2}=4$....(iii)

Further $F_{1}=\mu N_{1}= \left(\frac{P}{10}\right)N_{1}$  ...(iv)

 Solving above four equation , we get  P$\approx$ 3.6

 Therefore , the correct answer should be 4

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