JEE 2012 :: General Questions :: Physics

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 A person blows into the open-end of a long pipe. As a result, a high-pressure pulse of air travels down the pipe. When this pulse reaches the other end of the pipe. 

Answer:

Explanation:

At the open-end phase of pressure wave changes by $180^{0}$. So, compression returns as rarefaction. At closed-end, there is no phase change. So, compression returns as compression and rarefaction as rarefaction.

For the resistance network shown in the figure, choose the correct option(s)

Answer:

Explanation:

Due to symmetry on upper side and lower side, points P and Q are at same potentials. Similarly, points S and T are at same potentials. Therefore, the simple circuit can be drawn as shown

   $I_{2}= \frac{12}{2+2+2}=2A$

 $I_{3}=\frac{12}{4+4+4}=1A$

 $\therefore$    $I_{1}=I_{2}+I_{3}=3A$

 $I_{PQ}=0$ because $V_{p}=V_{Q}$ Potential drop (from left to right) across each resistance is 

 $\frac{12}{3}=4V$

 $\therefore$    $V_{MS}=2 \times 4=8V$

     $V_{NQ}=1 \times 4=4V$

 or $V_{S} < V_{Q}$

Three very large plates of the same area are kept parallel and close to each other. They are considered ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at temperatures 2T and 3T respectively. The temperature of the middle (i.e., second) plate under steady-state conditions is

Answer:

Explanation:

Let temperature of middle plate in steady state is $T_{0}$

 $Q_{1}=Q_{2}$

 Q= net rate of heat flow

$\therefore$   $\sigma A(3T)^{4}-\sigma AT_{0}^{4}=\sigma AT_{0}^{4}-\sigma A(2T)^{4}$

 Solving this equation , we get

 $T_{0}=\left(\frac{97}{2}\right)^{\frac{1}{4}}T$

A small block is connected to one end of a massless spring of unstretched length 4.9 m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t = 0. It then executes simple harmonic motion with angular frequency $\omega=\frac{\pi}{3}rad/s$. simultaneously at t=0, a small pebble is projected with speed v from point P  at an angle of $45^{0}$ as shown in the figure.  Point P is at a horizontal distance of 10 m from 0. If the pebble hits the block at t = 1 s, the value of v is (take $g=10 m/s^{2})$

Answer:

Explanation:

t= tiime of flight of projectile

       $=\frac{2 v \sin \theta}{g}$             ($\theta=45^{0})$

 $\therefore$    $v= \frac{gt}{2 \sin \theta}=\frac{10 \times 1}{2 \times 1/\sqrt{2}}=$  $\sqrt{50}$ m/s

In the determination of Young's  modulus $\left(Y=\frac{4MLg}{\pi ld^{2}}\right)$ Searle's method, a wire of length L = 2m and diameter d = 0.5 nm is used. For a load M = 2.5 kg, an extension l= 0.25 mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometre, respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100. T1le contributions to the maximum probable error of the Y measurement is

Answer:

Explanation:

$\triangle d=\triangle l=\frac{0.5}{100}mm=0.005mm$

   $Y=\frac{4MLg}{\pi ld^{2}}$

 $\therefore$  $\left(\frac{\triangle Y}{Y}\right)_{max}=\left(\frac{\triangle l}{l}\right)+2\left(\frac{\triangle d}{d}\right)$

  $\left(\frac{\triangle l}{l}\right)=\frac{0.5/100}{0.25}=0.02$

and       $\frac{2 \triangle d}{d}=\frac{(2)(0.5/100)}{0.25}=0.02$

 or   $\frac{\triangle l}{l}=2 \frac{\triangle d}{d}$

• General Questions - Physics
• General Questions - Chemistry
• General Questions - Mathematics