JAMB Physics Past Questions & Answers - Page 10

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