Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is 
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is 
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A 
   = (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
   = 4.5 cm
Answer: 4.5 cm
        
                    
             
        
        
        
According to Newton, an object will only accelerate if there is a net or unbalanced forceacting upon it. The presence of an unbalanced force will accelerate an object - changing its speed, its direction, or both its speed and direction.
        
                    
             
        
        
        
Answer:

Explanation:
As we know that the combination is maintained at rest position
So we will take net torque on the system to be ZERO
so we know that

here we will have

so we have

so we have


 
        
             
        
        
        
Answer:
0.12m/s
Explanation:
v=λf
Given that, λ = 12cm = 0.12m
T = 1second
(A period T is the time required for one complete cycle of vibration to pass a given point)
frequency 'f' is unknown but we can get frequency from f = 1/T = 1/1 = 1Hz
therefore, v= 0.12 × 1 = 0.12m/s