TS EAMCET 2018 :: Physics :: General questions

• Physics - General questions

Calculate the minimum thickness  of a soap  film (n=1.33) that results in constructive  interference in reflected light, If the film is illuminated  with light whose wavelength  in free space is 532 nm

Answer:

Explanation:

 For constructive interference  through a thin film  in reflected ligh, thickness  of film  must be

 $t=\left(m+\frac{1}{2}\right)\frac{\lambda}{2n_{2}}$

 where, m=0,1,2,.....

 and $n_{2}$ = refractive index  of medium (soap film) with given values, we get

 $t= \frac{\lambda}{4n_{2}} $                   ( $t_{min}$ occurs with m=0)

 = $\frac{532 \times 10^{-9}}{4 \times 1.33}=100 n m$

 Two objects P and Q are placed at 10 cm and 30 cm infornt of a convex lens of focal length 20 cm . The correct option for the image of p and Q is 

Answer:

Explanation:

 Focal length of convex lens =20 cm

 Here, object P is  at 10 cm

 and object Q is at 30 cm

 As object P is between focus and pole. Its image is virtual and erect

Object Q is beyond focus, So its image is real and inverted 

 A block of mass ($10 \alpha$) g, where $\alpha$ is a constant is moving with velocity 3m/s to the right collides inelastically with the block on the right with mass 10 g and sticks to it. The right block is connected to three springs as shown in the figure. The spring constant of each spring is 2 N/m. if the amplitude  of the resulting simple harmonic motion is $\frac{1}{2\sqrt{2}}$ m, then the value of $\alpha$ is 

Answer:

Explanation:

 Equivalent spring constant  of  system of springs =$\frac{3k}{2}=3 N/m$

 Momentum is conserved in collision. Let blocks moves with velocity v just after collision  then

 $m_{1}v_{1}=(m_{1}+m_{2})v$

 $ 10 \alpha  \times 10^{-3} \times 3=(10 \alpha +10)\times 10^{-3} \times v$

$\Rightarrow$    $v= \frac{30 \alpha}{10(\alpha+1)}= \frac{3 \alpha}{\alpha+1} ms^{-1}$

 Displacement of blocks is  $\frac{1}{2\sqrt{2}}$m.

 so , energy conservation  gives, 

$\frac{1}{2}(m_{1}+m_{2})v^{2}=\frac{1}{2}k_{eq} x^{2}$ [ friction is absent]

 $\Rightarrow$   $\frac{(10\alpha+10)\times 10^{-3}\times 9\alpha^{2}}{(\alpha+1)^{2}}=3\times \frac{1}{8}$

$\Rightarrow$  $\frac{ 9\alpha^{2}}{100(\alpha+1)^{}}=\frac{3}{8}$

 $24 \alpha^{2}=100 \alpha +100 \Rightarrow \alpha =5$

 A mechanical system consists two springs of stiffness coefficients $k_{1}$ and $k_{2}$  are connected in series . The minimum work to be  performed on the system to stretch it by $\triangle l$ is 

Answer:

Explanation:

 $K_{eq}$(in setries) = $\frac{k_{1}k_{2}}{k_{1}+k_{2}}$

 work done  to stretch  springs with elongation $\triangle l$ is 

$W= \frac{1}{2}K_{eq}.\triangle l^{2}=\frac{1}{2}\left(\frac{k_{1}k_{2}}{k_{1}+k_{2}}\right).\triangle l^{2}$

 A cricket player can throw a ball with an initial speed of 30 m/s. What is the maximum range of the player  can throw the ball?  Neglect air resistance .[Take , g= 10 m/s²]

Answer:

Explanation:

Maximum range occurs , when angle of throw $\theta =45^{0}$

Also , range  maximum= $\frac{ u^{2} \sin 2 \theta}{g}$$|_{\theta=45^{0}}$

 $\frac{u^{2}}{g}=\frac{100}{10}=90 m$

• Physics - General questions