JEE 2014 :: Advanced 2014 :: Physics 2014

• Advanced 2014 - Physics 2014
• Advanced 2014 - Chemistry 2014
• Advanced 2014 - Mathematics 2014

A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from the person.

 In the following state of the lift's motion is given in List I and the distance where the water jet hits the floor of the lift is given in List II. Match the statements from List I  with those in List II  and select the correct answers using the code given below the lists

Answer:

Explanation:

$d=2\sqrt{h_{1}h_{2}}=\sqrt{4h_{1}h_{2}}$

 This is independent of the value of g

 (P) $g_{eff}> g\Rightarrow d=\sqrt{4h_{1}h_{2}}=1.2 m$

(Q)    $g_{eff}< g\Rightarrow d=\sqrt{4h_{1}h_{2}}=1.2 m$

  (R)    $g_{eff}= g\Rightarrow d=\sqrt{4h_{1}h_{2}}=1.2 m$

 (S)     g_eff   = 0

 No water leaks out of jar. As there will be no pressure difference between top of the container and any other point 

          p₁  =p₂  =p₃  =p₀

A block of mass m₁=1 kg another mass m₂= 2 kg are placed together (see figure) on an inclined plane with the angle of inclination θ. Various values of θ are given in List I.

 The coefficient of friction between the block m₁ and the plane is always zero. The coefficient of static and dynamic friction between the block m₂ and the plane are equal to   $\mu$ =0.3

 In List II expressions for the frictions on the block, m₂ are given. Match the correct expressions of the friction in list II with the angles given in List I, and choose the correct options.

 The acceleration due to gravity  is denoted by g (Useful information tan (5.5°)=0.1:

   tan (11.5°)=0.2:  tan (16.5°)=0.3]

Answer:

Explanation:

Block will not slip if

 $(m_{1}+m_{2})g\sin\theta\leq\mu m_{2}g\cos\theta$

  $\Rightarrow $   $3\sin\theta\leq\left(\frac{3}{10}\right)(2)\cos\theta$

$\tan\theta\leq\frac{1}{5}\Rightarrow\theta\leq 11.5^{0}$ 

(P)     $\theta=5^{0}$   friction is static

               $f=(m_{1}+m_{2})g\sin\theta$

(Q)     $\theta=10^{0}$   friction is static

                      $f=(m_{1}+m_{2})g\sin\theta$

(R)    $\theta=15^{0}$   friction is kinetic

       $f=\mu m_{2}g\cos\theta$

(S)     $\theta=20^{0}$   friction is kinetic 

               $f=\mu m_{2}g\cos\theta$

Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is 1.5. Match lens combinations in List I with their focal length in List-II and select the correct answer using the code given below the lists

Answer:

Explanation:

(P)  $  \frac{1}{r}=\left(\frac{3}{2}-1\right)\left(\frac{1}{r}+\frac{1}{r}\right)=\frac{1}{r}$

          $\Rightarrow f=r$

           $\Rightarrow$           $\frac{1}{f_{eq}}=\frac{1}{f}+\frac{1}{f}=\frac{2}{r}$

            $\Rightarrow$    f_eq=\frac{r}{2}$

(Q)   $  \frac{1}{r}=\left(\frac{3}{2}-1\right)\left(\frac{1}{r}\right)\Rightarrow f=2r$

                $\Rightarrow$     $\frac{1}{f}+\frac{1}{f}=\frac{2}{f}= \frac{1}{r}$

                  $\Rightarrow$    f_eq=r

(R)     $  \frac{1}{f}=\left(\frac{3}{2}-1\right)\left(-\frac{1}{r}\right)\Rightarrow -\frac{1}{2r}$

                      $\Rightarrow$    f=-2r

$\Rightarrow$           $\frac{1}{f_{eq}}=\frac{1}{f}+\frac{1}{f}=\frac{2}{2r}$

$\Rightarrow$    f_eq=-r

(S)          

           $\frac{1}{f_{eq}}=\frac{1}{r}+\frac{1}{-2r}=\frac{1}{2r}$

       $\Rightarrow$    f_eq=2r

Four charges Q₁, Q₂, Q₃ andQ₄ of same magnitude are fixed along the x-axis at x=-2a, -a, +a and +2a respectively. A positive

charge q is placed on the positive y-axis at a distance b>0. Four options of the signs of these charges are given in List I . The direction of the forces on the charge q is given in List II. Match List I with List II and select the correct answer using the code given below. 

Answer:

Explanation:

(p) Component of forces along x-axis  will vanish .Net force along positive Y-axis

(Q)   Component  pf forces along y-axis will vanish .Net force along the positive x-axis

  (R) 

 Component of forces along x-axis will vanish .Net force along negative y-axis

(S)  Component of force along y-axis will vanish. Net force along the negative x-axis

The figure shows a circular loop of radius   $\alpha$  with two long parallel wires (numbered 1 and 2)  all in the plane of the paper. The distance of each wire from the centre of the loop is d, The loop and the wires are carrying the same current I. The  current in the loop is in the counter-clockwise direction if seen from above

Consider d>>a, and the loop is rotated about its diameter parallel to the wire by 30° from the position shown in the figure. If the current in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires  is constant over the loop)

Answer:

Explanation:

 Magnetic field  at mid point of two wires = 2(  magnetic field due to  one wire )

                  $=2\left[\frac{\mu_{0}}{2\pi}\frac{I}{d}\right]=\frac{\mu_{0}I}{\pi d}\otimes$

  Magnetic  moment  of loop 

 M=IA=  $I \pi a^{2}$

   Torque on loop[  = MB sin 150°

                =   $\frac{\mu_{0} I^{2} a^{2}}{2d}$

• Advanced 2014 - Physics 2014
• Advanced 2014 - Chemistry 2014
• Advanced 2014 - Mathematics 2014