JEE 2013 :: Advanced 2013 :: Physics 2013

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A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of a radius of 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity . The work done in overcoming the friction up to the point Q , as shown in the figure, is 150 J  .(take the acceleration due to gravity , g=10 ms^-2)

The magnitude of the normal reaction that acts on the block at the point Q is 

Answer:

Explanation:

$N= mg \cos 60^{0}=\frac{mv^{2}}{R}$

$\therefore$      $N=\frac{mv^{2}}{R}+\frac{mg}{2}$

  $N=\frac{(1)(10)^{2}}{40}+\frac{(1)(10)}{2}$

  = 7.5 N

A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of a radius of 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure, is 150 J  . (take the acceleration due to gravity, g=10 ms^-2)

The speed of the block when it reaches the point Q is 

Answer:

Explanation:

Height fallen upto Q is   

 $R \sin 30^{0}=\frac{R}{2}=20m$

 E_i- E_f = work done aganist friction

 $\therefore$    $0-[-mgh+\frac{1}{2}mv^{2}]=150$

 $\therefore$    $(1) (10) (20)-\frac{1}{2}\times1\times v^{2}=150$

  or v=  $10ms^{-1}$

 The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a₀ where a₀ is the Bohr radius. Its orbital angular moment is $\frac{3h}{2\pi}$. It is given that h is Planck constant and R is Rydberg constant. The possible  wavelength (s) when the atom de-excites,  is (are)

Answer:

Explanation:

$L=3(\frac{h}{2\pi})$

 $\therefore$   n=3 as  $L=3(\frac{h}{2\pi})$

    $r_{n}\propto\frac{n^{2}}{z}$

 r₃= 4.5 a₀  $\therefore$  z=2

$\frac{1}{\lambda_{1}}=Rz^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=4R\left(\frac{1}{4}-\frac{1}{9}\right)$

$\therefore$   $\lambda_{1}=\frac{9}{5R}$

$\frac{1}{\lambda_{2}}=Rz^{2}\left(\frac{1}{1^{2}}-\frac{1}{3^{2}}\right)=4R\left(1-\frac{1}{9}\right)$

  $\Rightarrow$   $\lambda_{2}=\frac{9}{32R}$

$\frac{1}{\lambda_{3}}=Rz^{2}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=4R\left(1-\frac{1}{4}\right)$

$\Rightarrow$   $\lambda_{3}=\frac{1}{3R}$

The figure below shows the variation of specific heat capacity (C)  of a solid as a function of temperature (T). The temperature is increased continuously from 0 to 500 K  at a constant rate. Ignoring any volume change, the following statement (s)  is (are)  correct to a reasonable approximation.

Answer:

Explanation:

Q=mCT

$\frac{dQ}{dt}=mc\frac{dT}{dt}$

R= rate of absortion of heat =  $\frac{dQ}{dt}\propto C$

 (i) in 0-100K

 C increase, so R increase but not linearly

 (ii) $\triangle Q -mC\triangle T$ as C  is more in (400 K-500K) then

(0-100K) so heat is increasing

(iii) C remains constant so there no change in R from (400 K-500K)

 (iv) C increases so R increases in range (200 K-300K) 

Two non- conducting spheres of radii R₁ and R₂ are carrying uniform volume charge densities +$\rho$ and -$\rho$  respectively, are placed, such that they partially overlap, as shown in the figure. At all points in the overlapping region

Answer:

Explanation:

For electrostatic field,

 E_p= E₁+E₂

  =   $\frac{\rho}{3\epsilon_{0}}C_{1}P+\frac{(-\rho)}{3\epsilon_{0}}C_{1}P$

  =  $\frac{\rho}{3\epsilon_{0}}(C_{1}P+PC_{2})$

  $E_{p}=\frac{\rho}{3\epsilon_{0}}C_{1}C_{2}$

 For  electrostatic potential: since the electric field is non zero so it is not equipotential

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