Question

Force acting on a perticle is 5 N. If units of length and time are double and unit of mass is holved then find the numerical value of force in the new system of unit .

Answer 1

The numerical value of force in the new system of units is 1.25

->   Given: Force (F) = 5 N = 5 kgm/s^2

->   Dimensions of force are = [ML$T^{-2}$]

->   When length and time are doubled and mass is halved.

     $1/2*M* 2*L*2^{-2}*T^{-2}$ = (1/4)*[ML$T^{-2}$]

->   New Force = 5/4 = 1.25 Units.

Answer 2

Given :

The magnitude of force acting on particle = F = 5 N

The unit length = L' = 2 L  meters

The unit mass = M' = $\dfrac{M}{2}$   Kg

The time  = T' = 2 T sec

To Find :

The value of force in new system

Solution :

∵  A force is defined as product of magnitude of mass and acceleration

i.e       Force = mass × acceleration

And  Acceleration = $\dfrac{velocity}{time}$

and    velocity = $\dfrac{distance}{time}$

So,       Force = mass × $\dfrac{distance}{time^{2} }$

now, At old value

               F = M × $\dfrac{L}{T^{2} }$

i.e            F = $\dfrac{ML}{T^{2} }$

At New value              

             F' = M' × $\dfrac{L'}{T'^{2} }$

i.e          F' = $\dfrac{M}{2}$  ×  $\dfrac{2L}{(2T)^{2} }$                ( ∵ M' = $\dfrac{M}{2}$  and  L' = 2 L   and T' = 2 T)

Or,         F' = $\dfrac{M}{2}$  × $\dfrac{2L}{4T^{2} }$

Or,         F' = $\dfrac{ML}{4T^{2} }$

∵           F' = $\dfrac{F}{4}$                              ( ∵ F = $\dfrac{ML}{T^{2} }$ )

i.e     New Force value = one-fourth of old force force

So, The numerical value of force in the new system of unit = one-fourth of old value of force

Hence, The numerical value of force in the new system of unit is one-fourth of old value of force  Answer