JEE 2013 :: Advanced 2013 :: Physics 2013

• Advanced 2013 - Physics 2013
• Advanced 2013 - Chemistry 2013
• Advanced 2013 - Mathematics 2013

Using the expression   $2d \sin\theta=\lambda$, one calculates the values of d by measuring the corresponding angle $\theta$ in the range 0 to 90°.  The wavelength $\lambda$  is exactly known and the error in $\theta$  is constant for all values of $\theta$. As $\theta$  increases from 0°  

Answer:

Explanation:

$2d\sin\theta=\lambda$

 $d=\frac{\lambda}{2\sin\theta}$

 Differentiate    $\partial(d)=\frac{\lambda}{2}\partial( cosec \theta)$

  $\partial(d)=\frac{\lambda}{2}( -cosec \theta\cot\theta)\partial\theta$

 $\partial(d)=\frac{-\lambda\cos\theta}{2\sin^{2}\theta}$

  as $\theta$ = increases,  $\frac{\lambda\cos\theta}{2\sin^{2}\theta}$ decreases

Two vehicles, each moving with speed u on the same horizontal straight road, are approaching each other.The wind blows along the road with velocity w. One of these vehicles blows a whistle of frequency  f_1. An observer in the other vehicle hears the frequency of the whistle to be f₂. The speed of sound in still air is  v. The correct statement(s) is (are)

Answer:

Explanation:

when wind blows from S to O

 $f_{2}=f_{1}\left(\frac{v+w+u}{v+w-u}\right)$

 or f₂ >f₁

 when wind blows from o to S

$f_{2}=f_{1}\left(\frac{v-w+u}{v-w-u}\right)$

 $\therefore$ f₂ >f₁

A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is placed coaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carries a steady current I. Consider a point p  at a distance r from the common axis . The  correct statement (s)  is (are)

Answer:

Explanation:

in the region , 0<r<R

 B_p=0,

 $B_{Q}\neq 0$ , along the axis

 $\therefore$   $B_{net}\neq 0$ 

 in the region R<r<2R

 $B_{p}\neq 0$  tangential to the circle  of radius r, centred on the axis

 $B_{Q}\neq 0$  , along the axis

 $\therefore $  $B_{net}\neq 0$ neither in the directions mentioned in options (b) and (c) 

 In the region  , r>2R

$B_{p}\neq 0$

$B_{Q}\neq 0$

$\therefore$   $B_{net}\neq 0$

A  particle of mass m is attached to one end of the mass-less spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t=0 with an initial velocity $\mu_{0}$.When the speed of the particle is 0.5 $\mu_{0}$, it collides elastically with a rigid wall. After this collision

Answer:

Explanation:

(a) At equilibrium (t=0)  particle has maximum velocity u₀. Therefore  velocity at time t can be written as

$u=u_{max}\cos \omega t+u_{0}\cos \omega t$

  writing, u=0.5 u₀=u₀ cos ω t

$\therefore$   $\omega t=\frac{\pi}{3}$

 $\therefore$    $\frac{2\pi}{T}t=\frac{\pi}{3}$   $\therefore$    $t=\frac{T}{6}$

 (b)     $t=t_{AB}+t_{BA}=\frac{T}{6}+\frac{T}{6}=\frac{T}{3}=\frac{2\pi}{3}\sqrt{\frac{m}{k}}$

  (c)     $t=t_{AB}+t_{BA}+t_{AC}=\frac{T}{6}+\frac{T}{6}+\frac{T}{4}=\frac{7}{12}T=\frac{7\pi}{6}\sqrt{\frac{m}{k}}$

 (d)   $t=t_{AB}+t_{BA}+t_{AC}+t_{CA}=\frac{T}{6}+\frac{T}{6}+\frac{T}{4}+\frac{T}{4}=\frac{5}{6}T=\frac{5\pi}{3}\sqrt{\frac{m}{k}}$

Two bodies, each of mass M, are kept fixed with a separation of 2L. A particle of mass m is projected from the mid-point of the line joining their centres, perpendicular to the line.The  gravitational constant G .The correct statement(s) is (are)

Answer:

Explanation:

Let v is the minimum velocity .From energy conservation

 $U_{c}+K_{c}=U_{\infty}+K_{\infty}$

$\therefore$    $mV_{c}+\frac{1}{2}mv^{2}=0+0$  $  \therefore$   $ v=\sqrt{-2V_{c}}$

 = $\sqrt{(-2)\left(\frac{-2GM}{L}\right)}=2\sqrt{\frac{GM}{L}}$

• Advanced 2013 - Physics 2013
• Advanced 2013 - Chemistry 2013
• Advanced 2013 - Mathematics 2013