JAMB Physics Past Questions & Answers - Page 5

 W = mg = 0.5 x 10 = 5 N

 Since it's light, neglect the weight of the metre rule.

 The effective tension T acting in the vertical direction = T sin 30°

 From the second condition of equilibrium, sum of clockwise moments equal sum of anticlockwise moments

 Taking moment at E 

 ⇒ T sin 30° x 30 = 5 x 50

 ⇒ ∴T=250

 T = \(\frac{250}{15}\) = 16.67N 

Ep = 2.2kV = 2200V; Es=110V; Ns=25; Np=?

=\(\frac{E_p}{E_s}\) = \(\frac{N_p}{N_s}\)
 

= \(\frac{2200}{110}\) = \(\frac{N_p}{25}\)

=110 × Np = 2200 × 25

=\(N_p\) = \(\frac{55,000}{110}\)

∴\(N_p\) = 500 turns

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°

     Let the original length L=xm

      ;New length =( x -3 ) m

       \(T_1\) = 5.77s; \(T_2\) = 4.60s,  

       \(T^2\) α  L

       ⇒\(T_2) = kL  where K is constant

       ⇒K = \(\frac{T^2_1}{L_1}\) = \(\frac{T^2_2}{L_2}\)

      ⇒\(\frac{5.77^2}{x}\)  = \(\frac{4.60^2}{x-3}\)

      ⇒ \(\frac{33.29}{x}\)  = \(\frac{4.60^2}{x-3}\)

      ⇒ 33.29(x-3)  = 21.16x
        
      ⇒ 33.29x - 99.87 =21.16x

      ⇒12.13x = 99.87
     
      ;x =\(\frac{99.87}{12.13}\)  = 8.23m
    
      New length of the pendulum 
     
      =x-3 = 8.23-3
   
      =5.23m

     

n=1.458, c=\(3.00 ×10^8 ms^-1\) ,λo = 589nm;  f=?

Speed of light in a medium (v)=\(\frac{c}{n}\)  where n is the refractive index of the medium

⇒λn=\(\frac{589}{1.458}\) = 404nm

  v=fλ
  
⇒f=\(\frac{v}{λ}\) 

=\(\frac{2.06×10^8}{404×10^-9}\) \(1nano=10^{-9}\)

∴f=\(5.09×10^{14}\) Hz

W = 400 N; P = 100 N; θ = 30o; μ = ?

Frictional force (Fr) = μR (where R is the normal reaction)

The forces acting along the horizontal direction are Fr and Px

∴ Pcos 30° - Fr = ma (Pcos 30° is acting in the +ve x-axis while Fr in the -ve x-axis)

⇒ 100cos 30° - μR = ma

Since the box is moving at constant speed, its acceleration is zero

⇒ 100cos 30° - μR = 0

⇒ 100cos 30o = μR ----- (i)

The forces acting in the vertical direction are W, Py and R

∴ R - Psin 30° - W = 0 (R is acting upward (+ve) while Py and W are acting downward (-ve) and they are at equilibrium)

⇒ R - 100sin 30° - 400 = 0

⇒ R = 100sin 30° + 400

⇒ R = 50 + 400 = 450 N

From equation (i)

⇒ 100cos 30° = 450μ

⇒μ=100cos30°

  N = \(\frac{100cos30°}{450}\)

        = μ = 0.19