Physics Questions and Answers

The pinhole camera works on the principle of the rectilinear propagation of light. This principle states that light travels in straight lines. When light passes through the tiny hole in a pinhole camera, it forms an inverted image on the opposite side of the camera. The size of the image depends on the distance between the object and the pinhole

The S.I unit of heat capacity is JK-1 and it is an extensive property as it depends on the amount of substance present.

Therefore, (i) and (iv) only are not true

The patient cannot see clearly an object closer than 50 cm

Therefore, the patient needs a lens that would enable him see clearly, objects placed 25 cm from the lens

So, we take the object to a distance of 25 cm from the lens so that the image forms at 50 cm in front of the lens

u=25cm ;v=-50cm (virtual image); p=

$\frac{1}{f}$=$\frac{1}{u}$ + $\frac{1}{v}$

⇒$\frac{1}{f}$ =$\frac{1}{50}$

⇒$\frac{1}{f}$ = $\frac{1}{25}$-$\frac{1}{50}$

⇒$\frac{1}{f}$ = f = 50cm =0.5m

⇒$\frac{1}{f}$ = $\frac{2-1}{50}$

   p = $\frac{1}{f}$

   p =  $\frac{1}{0.5}$

   p =   2 diopters 
 
  ; The patient needs a converging lens with a power of 2 diopters

A=0.8m; f=16Hz; v=20 $ms^-1$; k=?

v = fλ

λ =1.25m

= k= $\frac{2π}{λ}$ =$\frac{2 × 3.142}{1.25}$
 
 k=  5m ( to 1 s.f)

 

Fx= 12cos32°-16cos26°    =    -4.20N    (right is taken +ve and left is taken -ve)

Fy=12sin32°+16sin26°-21 = -7.63N  (up is taken +ve and down is taken -ve)

$R^2$= 4.202+7.632

⇒$R^2$ = 75.86

⇒ R=√75.86

∴R = 8.71 N

tan θ=$\frac{opp}{adj}$

⇒tan θ=$\frac{7.63}{4.20}$

⇒tan θ =1.8095

⇒ θ = $tan^-1$ (1.8095)

   θ = 61°

∴The angle of the resultant force with the x-axis is 61°